\(\int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx\) [601]

Optimal result

Integrand size = 23, antiderivative size = 332 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {a^{3/2} \left (3 a^4+6 a^2 b^2+35 b^4\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 b^{5/2} \left (a^2+b^2\right )^3 d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {a^2 \left (3 a^2+11 b^2\right ) \sqrt {\tan (c+d x)}}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))} \] Output:

1/2*(a+b)*(a^2-4*a*b+b^2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)/(a^2 
+b^2)^3/d+1/2*(a+b)*(a^2-4*a*b+b^2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^( 
1/2)/(a^2+b^2)^3/d+1/4*a^(3/2)*(3*a^4+6*a^2*b^2+35*b^4)*arctan(b^(1/2)*tan 
(d*x+c)^(1/2)/a^(1/2))/b^(5/2)/(a^2+b^2)^3/d+1/2*(a-b)*(a^2+4*a*b+b^2)*arc 
tanh(2^(1/2)*tan(d*x+c)^(1/2)/(1+tan(d*x+c)))*2^(1/2)/(a^2+b^2)^3/d-1/2*a^ 
2*tan(d*x+c)^(3/2)/b/(a^2+b^2)/d/(a+b*tan(d*x+c))^2-1/4*a^2*(3*a^2+11*b^2) 
*tan(d*x+c)^(1/2)/b^2/(a^2+b^2)^2/d/(a+b*tan(d*x+c))
 

Mathematica [A] (warning: unable to verify)

Time = 9.02 (sec) , antiderivative size = 608, normalized size of antiderivative = 1.83 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {-\frac {a^2 \left (3 a^3+11 a b^2+\left (3 a^3+11 a b^2\right ) \cos (2 (c+d x))+\left (5 a^2 b+13 b^3\right ) \sin (2 (c+d x))\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))^2}+\frac {2 \csc (c+d x) (a+b \tan (c+d x)) \left (\left (a^2+b^2\right )^2 \left (3 a^2+4 b^2\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \cos (2 (c+d x)) \sec (c+d x)-b^2 \left (a^2-b^2\right ) \cos (2 (c+d x)) \left (4 \left (a^2-b^2\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )+\sqrt {2} \sqrt {a} \sqrt {b} \left (2 (a-b) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-2 (a-b) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )+(a+b) \left (\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right )\right )\right ) \sec (c+d x)+a^{3/2} b^{7/2} \csc (c+d x) \left (8 \sqrt {a} \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )-\sqrt {2} \left (-2 (a+b) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )+2 (a+b) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )+(a-b) \left (\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right )\right )\right ) \sin (2 (c+d x)) \left (1-\tan ^2(c+d x)\right )\right )}{\sqrt {a} b^{5/2} \left (a^2+b^2\right )^3 (b+a \cot (c+d x)) \left (1-\tan ^2(c+d x)\right )}}{8 d} \] Input:

Integrate[Tan[c + d*x]^(7/2)/(a + b*Tan[c + d*x])^3,x]
 

Output:

(-((a^2*(3*a^3 + 11*a*b^2 + (3*a^3 + 11*a*b^2)*Cos[2*(c + d*x)] + (5*a^2*b 
 + 13*b^3)*Sin[2*(c + d*x)])*Sqrt[Tan[c + d*x]])/(b^2*(a^2 + b^2)^2*(a*Cos 
[c + d*x] + b*Sin[c + d*x])^2)) + (2*Csc[c + d*x]*(a + b*Tan[c + d*x])*((a 
^2 + b^2)^2*(3*a^2 + 4*b^2)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*C 
os[2*(c + d*x)]*Sec[c + d*x] - b^2*(a^2 - b^2)*Cos[2*(c + d*x)]*(4*(a^2 - 
b^2)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]] + Sqrt[2]*Sqrt[a]*Sqrt[b 
]*(2*(a - b)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - 2*(a - b)*ArcTan[1 + 
 Sqrt[2]*Sqrt[Tan[c + d*x]]] + (a + b)*(Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] 
 + Tan[c + d*x]] - Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])))*S 
ec[c + d*x] + a^(3/2)*b^(7/2)*Csc[c + d*x]*(8*Sqrt[a]*Sqrt[b]*ArcTan[(Sqrt 
[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]] - Sqrt[2]*(-2*(a + b)*ArcTan[1 - Sqrt[2]* 
Sqrt[Tan[c + d*x]]] + 2*(a + b)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] + ( 
a - b)*(Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] - Log[1 + Sqrt[ 
2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])))*Sin[2*(c + d*x)]*(1 - Tan[c + d*x 
]^2)))/(Sqrt[a]*b^(5/2)*(a^2 + b^2)^3*(b + a*Cot[c + d*x])*(1 - Tan[c + d* 
x]^2)))/(8*d)
 

Rubi [A] (verified)

Time = 1.67 (sec) , antiderivative size = 376, normalized size of antiderivative = 1.13, number of steps used = 24, number of rules used = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3042, 4048, 27, 3042, 4128, 27, 3042, 4136, 27, 3042, 4017, 27, 1482, 1476, 1082, 217, 1479, 25, 27, 1103, 4117, 73, 218}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^{7/2}}{(a+b \tan (c+d x))^3}dx\)

\(\Big \downarrow \) 4048

\(\displaystyle \frac {\int \frac {\sqrt {\tan (c+d x)} \left (3 a^2-4 b \tan (c+d x) a+\left (3 a^2+4 b^2\right ) \tan ^2(c+d x)\right )}{2 (a+b \tan (c+d x))^2}dx}{2 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sqrt {\tan (c+d x)} \left (3 a^2-4 b \tan (c+d x) a+\left (3 a^2+4 b^2\right ) \tan ^2(c+d x)\right )}{(a+b \tan (c+d x))^2}dx}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sqrt {\tan (c+d x)} \left (3 a^2-4 b \tan (c+d x) a+\left (3 a^2+4 b^2\right ) \tan (c+d x)^2\right )}{(a+b \tan (c+d x))^2}dx}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 4128

\(\displaystyle \frac {\frac {\int \frac {-16 a \tan (c+d x) b^3+\left (3 a^4+3 b^2 a^2+8 b^4\right ) \tan ^2(c+d x)+a^2 \left (3 a^2+11 b^2\right )}{2 \sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{b \left (a^2+b^2\right )}-\frac {a^2 \left (3 a^2+11 b^2\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {-16 a \tan (c+d x) b^3+\left (3 a^4+3 b^2 a^2+8 b^4\right ) \tan ^2(c+d x)+a^2 \left (3 a^2+11 b^2\right )}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (3 a^2+11 b^2\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {-16 a \tan (c+d x) b^3+\left (3 a^4+3 b^2 a^2+8 b^4\right ) \tan (c+d x)^2+a^2 \left (3 a^2+11 b^2\right )}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (3 a^2+11 b^2\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 4136

\(\displaystyle \frac {\frac {\frac {\int \frac {8 \left (a b^2 \left (a^2-3 b^2\right )-b^3 \left (3 a^2-b^2\right ) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}+\frac {a^2 \left (3 a^4+6 a^2 b^2+35 b^4\right ) \int \frac {\tan ^2(c+d x)+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (3 a^2+11 b^2\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {8 \int \frac {a b^2 \left (a^2-3 b^2\right )-b^3 \left (3 a^2-b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}+\frac {a^2 \left (3 a^4+6 a^2 b^2+35 b^4\right ) \int \frac {\tan ^2(c+d x)+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (3 a^2+11 b^2\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {8 \int \frac {a b^2 \left (a^2-3 b^2\right )-b^3 \left (3 a^2-b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}+\frac {a^2 \left (3 a^4+6 a^2 b^2+35 b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (3 a^2+11 b^2\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 4017

\(\displaystyle \frac {\frac {\frac {16 \int \frac {b^2 \left (a \left (a^2-3 b^2\right )-b \left (3 a^2-b^2\right ) \tan (c+d x)\right )}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}+\frac {a^2 \left (3 a^4+6 a^2 b^2+35 b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (3 a^2+11 b^2\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {16 b^2 \int \frac {a \left (a^2-3 b^2\right )-b \left (3 a^2-b^2\right ) \tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}+\frac {a^2 \left (3 a^4+6 a^2 b^2+35 b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (3 a^2+11 b^2\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 1482

\(\displaystyle \frac {\frac {\frac {16 b^2 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}+\frac {a^2 \left (3 a^4+6 a^2 b^2+35 b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (3 a^2+11 b^2\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {\frac {\frac {16 b^2 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d \left (a^2+b^2\right )}+\frac {a^2 \left (3 a^4+6 a^2 b^2+35 b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (3 a^2+11 b^2\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {\frac {\frac {16 b^2 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {a^2 \left (3 a^4+6 a^2 b^2+35 b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (3 a^2+11 b^2\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {\frac {16 b^2 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {a^2 \left (3 a^4+6 a^2 b^2+35 b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (3 a^2+11 b^2\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {\frac {\frac {16 b^2 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {a^2 \left (3 a^4+6 a^2 b^2+35 b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (3 a^2+11 b^2\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {16 b^2 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {a^2 \left (3 a^4+6 a^2 b^2+35 b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (3 a^2+11 b^2\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {16 b^2 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )+\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {a^2 \left (3 a^4+6 a^2 b^2+35 b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (3 a^2+11 b^2\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\frac {\frac {a^2 \left (3 a^4+6 a^2 b^2+35 b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}+\frac {16 b^2 \left (\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (3 a^2+11 b^2\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {\frac {\frac {a^2 \left (3 a^4+6 a^2 b^2+35 b^4\right ) \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}d\tan (c+d x)}{d \left (a^2+b^2\right )}+\frac {16 b^2 \left (\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (3 a^2+11 b^2\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {\frac {2 a^2 \left (3 a^4+6 a^2 b^2+35 b^4\right ) \int \frac {1}{a+b \tan (c+d x)}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}+\frac {16 b^2 \left (\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (3 a^2+11 b^2\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {16 b^2 \left (\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {2 a^{3/2} \left (3 a^4+6 a^2 b^2+35 b^4\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {b} d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (3 a^2+11 b^2\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

Input:

Int[Tan[c + d*x]^(7/2)/(a + b*Tan[c + d*x])^3,x]
 

Output:

-1/2*(a^2*Tan[c + d*x]^(3/2))/(b*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) + ( 
((2*a^(3/2)*(3*a^4 + 6*a^2*b^2 + 35*b^4)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x] 
])/Sqrt[a]])/(Sqrt[b]*(a^2 + b^2)*d) + (16*b^2*(((a + b)*(a^2 - 4*a*b + b^ 
2)*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2] 
*Sqrt[Tan[c + d*x]]]/Sqrt[2]))/2 + ((a - b)*(a^2 + 4*a*b + b^2)*(-1/2*Log[ 
1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/Sqrt[2] + Log[1 + Sqrt[2]*S 
qrt[Tan[c + d*x]] + Tan[c + d*x]]/(2*Sqrt[2])))/2))/((a^2 + b^2)*d))/(2*b* 
(a^2 + b^2)) - (a^2*(3*a^2 + 11*b^2)*Sqrt[Tan[c + d*x]])/(b*(a^2 + b^2)*d* 
(a + b*Tan[c + d*x])))/(4*b*(a^2 + b^2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 1482

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
a*c, 2]}, Simp[(d*q + a*e)/(2*a*c)   Int[(q + c*x^2)/(a + c*x^4), x], x] + 
Simp[(d*q - a*e)/(2*a*c)   Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ[{a 
, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(- 
a)*c]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4017

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ 
)]], x_Symbol] :> Simp[2/f   Subst[Int[(b*c + d*x^2)/(b^2 + x^4), x], x, Sq 
rt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] & 
& NeQ[c^2 + d^2, 0]
 

rule 4048

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m 
 - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Simp[1 
/(d*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + 
f*x])^(n + 1)*Simp[a^2*d*(b*d*(m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c 
*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3*a*b^2*d) 
*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*( 
n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[ 
b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 2] && LtQ 
[n, -1] && IntegerQ[2*m]
 

rule 4117

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
 Simp[A/f   Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; 
FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
 

rule 4128

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*d^2 + c*(c*C - B*d))*(a + b*Tan[e + 
 f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Sim 
p[1/(d*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e 
 + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c*m + a*d* 
(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b 
*(d*(B*c - A*d)*(m + n + 1) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], 
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ 
[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
 

rule 4136

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) 
+ (f_.)*(x_)]), x_Symbol] :> Simp[1/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^ 
n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Simp[ 
(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^n*((1 + Tan[ 
e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, 
 C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] & 
&  !GtQ[n, 0] &&  !LeQ[n, -1]
 

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (a^{3}-3 a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (-3 a^{2} b +b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{3}}+\frac {2 a^{2} \left (\frac {-\frac {\left (5 a^{4}+18 a^{2} b^{2}+13 b^{4}\right ) \tan \left (d x +c \right )^{\frac {3}{2}}}{8 b}-\frac {a \left (3 a^{4}+14 a^{2} b^{2}+11 b^{4}\right ) \sqrt {\tan \left (d x +c \right )}}{8 b^{2}}}{\left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {\left (3 a^{4}+6 a^{2} b^{2}+35 b^{4}\right ) \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right )}{8 b^{2} \sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right )^{3}}}{d}\)	\(347\)
default	\(\frac {\frac {\frac {\left (a^{3}-3 a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (-3 a^{2} b +b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{3}}+\frac {2 a^{2} \left (\frac {-\frac {\left (5 a^{4}+18 a^{2} b^{2}+13 b^{4}\right ) \tan \left (d x +c \right )^{\frac {3}{2}}}{8 b}-\frac {a \left (3 a^{4}+14 a^{2} b^{2}+11 b^{4}\right ) \sqrt {\tan \left (d x +c \right )}}{8 b^{2}}}{\left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {\left (3 a^{4}+6 a^{2} b^{2}+35 b^{4}\right ) \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right )}{8 b^{2} \sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right )^{3}}}{d}\)	\(347\)

Input:

int(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*(2/(a^2+b^2)^3*(1/8*(a^3-3*a*b^2)*2^(1/2)*(ln((tan(d*x+c)+2^(1/2)*tan( 
d*x+c)^(1/2)+1)/(tan(d*x+c)-2^(1/2)*tan(d*x+c)^(1/2)+1))+2*arctan(1+2^(1/2 
)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))+1/8*(-3*a^2*b+b 
^3)*2^(1/2)*(ln((tan(d*x+c)-2^(1/2)*tan(d*x+c)^(1/2)+1)/(tan(d*x+c)+2^(1/2 
)*tan(d*x+c)^(1/2)+1))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^ 
(1/2)*tan(d*x+c)^(1/2))))+2*a^2/(a^2+b^2)^3*((-1/8*(5*a^4+18*a^2*b^2+13*b^ 
4)/b*tan(d*x+c)^(3/2)-1/8*a*(3*a^4+14*a^2*b^2+11*b^4)/b^2*tan(d*x+c)^(1/2) 
)/(a+b*tan(d*x+c))^2+1/8*(3*a^4+6*a^2*b^2+35*b^4)/b^2/(a*b)^(1/2)*arctan(b 
*tan(d*x+c)^(1/2)/(a*b)^(1/2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2395 vs. \(2 (295) = 590\).

Time = 0.96 (sec) , antiderivative size = 4816, normalized size of antiderivative = 14.51 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^3,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\int \frac {\tan ^{\frac {7}{2}}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{3}}\, dx \] Input:

integrate(tan(d*x+c)**(7/2)/(a+b*tan(d*x+c))**3,x)
 

Output:

Integral(tan(c + d*x)**(7/2)/(a + b*tan(c + d*x))**3, x)
 

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 420, normalized size of antiderivative = 1.27 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\frac {{\left (3 \, a^{6} + 6 \, a^{4} b^{2} + 35 \, a^{2} b^{4}\right )} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} \sqrt {a b}} + \frac {2 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {{\left (5 \, a^{4} b + 13 \, a^{2} b^{3}\right )} \tan \left (d x + c\right )^{\frac {3}{2}} + {\left (3 \, a^{5} + 11 \, a^{3} b^{2}\right )} \sqrt {\tan \left (d x + c\right )}}{a^{6} b^{2} + 2 \, a^{4} b^{4} + a^{2} b^{6} + {\left (a^{4} b^{4} + 2 \, a^{2} b^{6} + b^{8}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b^{3} + 2 \, a^{3} b^{5} + a b^{7}\right )} \tan \left (d x + c\right )}}{4 \, d} \] Input:

integrate(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^3,x, algorithm="maxima")
 

Output:

1/4*((3*a^6 + 6*a^4*b^2 + 35*a^2*b^4)*arctan(b*sqrt(tan(d*x + c))/sqrt(a*b 
))/((a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*sqrt(a*b)) + (2*sqrt(2)*(a^3 - 
 3*a^2*b - 3*a*b^2 + b^3)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c 
)))) + 2*sqrt(2)*(a^3 - 3*a^2*b - 3*a*b^2 + b^3)*arctan(-1/2*sqrt(2)*(sqrt 
(2) - 2*sqrt(tan(d*x + c)))) + sqrt(2)*(a^3 + 3*a^2*b - 3*a*b^2 - b^3)*log 
(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - sqrt(2)*(a^3 + 3*a^2*b - 
 3*a*b^2 - b^3)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))/(a^6 
+ 3*a^4*b^2 + 3*a^2*b^4 + b^6) - ((5*a^4*b + 13*a^2*b^3)*tan(d*x + c)^(3/2 
) + (3*a^5 + 11*a^3*b^2)*sqrt(tan(d*x + c)))/(a^6*b^2 + 2*a^4*b^4 + a^2*b^ 
6 + (a^4*b^4 + 2*a^2*b^6 + b^8)*tan(d*x + c)^2 + 2*(a^5*b^3 + 2*a^3*b^5 + 
a*b^7)*tan(d*x + c)))/d
 

Giac [F]

\[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\int { \frac {\tan \left (d x + c\right )^{\frac {7}{2}}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:

integrate(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^3,x, algorithm="giac")
 

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 7.47 (sec) , antiderivative size = 17912, normalized size of antiderivative = 53.95 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\text {Too large to display} \] Input:

int(tan(c + d*x)^(7/2)/(a + b*tan(c + d*x))^3,x)
 

Output:

atan((((1i/(4*(b^6*d^2 - a^6*d^2 + a*b^5*d^2*6i + a^5*b*d^2*6i - 15*a^2*b^ 
4*d^2 - a^3*b^3*d^2*20i + 15*a^4*b^2*d^2)))^(1/2)*((1i/(4*(b^6*d^2 - a^6*d 
^2 + a*b^5*d^2*6i + a^5*b*d^2*6i - 15*a^2*b^4*d^2 - a^3*b^3*d^2*20i + 15*a 
^4*b^2*d^2)))^(1/2)*((1i/(4*(b^6*d^2 - a^6*d^2 + a*b^5*d^2*6i + a^5*b*d^2* 
6i - 15*a^2*b^4*d^2 - a^3*b^3*d^2*20i + 15*a^4*b^2*d^2)))^(1/2)*((1600*a^2 
*b^23*d^4 + 12864*a^4*b^21*d^4 + 45312*a^6*b^19*d^4 + 91392*a^8*b^17*d^4 + 
 115584*a^10*b^15*d^4 + 94080*a^12*b^13*d^4 + 48384*a^14*b^11*d^4 + 14592* 
a^16*b^9*d^4 + 2112*a^18*b^7*d^4 + 64*a^20*b^5*d^4)/(b^19*d^5 + 8*a^2*b^17 
*d^5 + 28*a^4*b^15*d^5 + 56*a^6*b^13*d^5 + 70*a^8*b^11*d^5 + 56*a^10*b^9*d 
^5 + 28*a^12*b^7*d^5 + 8*a^14*b^5*d^5 + a^16*b^3*d^5) + (tan(c + d*x)^(1/2 
)*(1i/(4*(b^6*d^2 - a^6*d^2 + a*b^5*d^2*6i + a^5*b*d^2*6i - 15*a^2*b^4*d^2 
 - a^3*b^3*d^2*20i + 15*a^4*b^2*d^2)))^(1/2)*(512*b^28*d^4 + 4608*a^2*b^26 
*d^4 + 17920*a^4*b^24*d^4 + 38400*a^6*b^22*d^4 + 46080*a^8*b^20*d^4 + 2150 
4*a^10*b^18*d^4 - 21504*a^12*b^16*d^4 - 46080*a^14*b^14*d^4 - 38400*a^16*b 
^12*d^4 - 17920*a^18*b^10*d^4 - 4608*a^20*b^8*d^4 - 512*a^22*b^6*d^4))/(b^ 
19*d^4 + 8*a^2*b^17*d^4 + 28*a^4*b^15*d^4 + 56*a^6*b^13*d^4 + 70*a^8*b^11* 
d^4 + 56*a^10*b^9*d^4 + 28*a^12*b^7*d^4 + 8*a^14*b^5*d^4 + a^16*b^3*d^4)) 
- (tan(c + d*x)^(1/2)*(1472*a*b^21*d^2 + 72*a^21*b*d^2 + 1024*a^3*b^19*d^2 
 + 1352*a^5*b^17*d^2 + 28224*a^7*b^15*d^2 + 70240*a^9*b^13*d^2 + 72640*a^1 
1*b^11*d^2 + 39088*a^13*b^9*d^2 + 13248*a^15*b^7*d^2 + 3488*a^17*b^5*d^...
 

Reduce [F]

\[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\int \frac {\sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{3}}{\tan \left (d x +c \right )^{3} b^{3}+3 \tan \left (d x +c \right )^{2} a \,b^{2}+3 \tan \left (d x +c \right ) a^{2} b +a^{3}}d x \] Input:

int(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^3,x)
 

Output:

int((sqrt(tan(c + d*x))*tan(c + d*x)**3)/(tan(c + d*x)**3*b**3 + 3*tan(c + 
 d*x)**2*a*b**2 + 3*tan(c + d*x)*a**2*b + a**3),x)