\(\int \frac {\sqrt {\tan (c+d x)}}{(a+b \tan (c+d x))^3} \, dx\) [604]

Optimal result

Integrand size = 23, antiderivative size = 328 \[ \int \frac {\sqrt {\tan (c+d x)}}{(a+b \tan (c+d x))^3} \, dx=-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {\sqrt {b} \left (15 a^4-18 a^2 b^2-b^4\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 a^{3/2} \left (a^2+b^2\right )^3 d}-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {b \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{4 a \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))} \] 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*b^(1/2)*(15*a^4-18*a^2*b^2-b^4)*arctan(b^(1/2)*tan( 
d*x+c)^(1/2)/a^(1/2))/a^(3/2)/(a^2+b^2)^3/d-1/2*(a+b)*(a^2-4*a*b+b^2)*arct 
anh(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*b*t 
an(d*x+c)^(1/2)/(a^2+b^2)/d/(a+b*tan(d*x+c))^2-1/4*b*(7*a^2-b^2)*tan(d*x+c 
)^(1/2)/a/(a^2+b^2)^2/d/(a+b*tan(d*x+c))
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 3.66 (sec) , antiderivative size = 259, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {\tan (c+d x)}}{(a+b \tan (c+d x))^3} \, dx=-\frac {-\frac {b^{3/2} \left (-15 a^4+18 a^2 b^2+b^4\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} \left (a^2+b^2\right )^2}+\frac {4 \sqrt [4]{-1} a b \left ((i a-b)^3 \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )-(i a+b)^3 \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )\right )}{\left (a^2+b^2\right )^2}-\frac {2 b^3 \tan ^{\frac {3}{2}}(c+d x)}{(a+b \tan (c+d x))^2}+\frac {2 b^2 \sqrt {\tan (c+d x)}}{a+b \tan (c+d x)}+\frac {\left (7 a^2 b^2-b^4\right ) \sqrt {\tan (c+d x)}}{\left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 a b \left (a^2+b^2\right ) d} \] Input:

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

Output:

-1/4*(-((b^(3/2)*(-15*a^4 + 18*a^2*b^2 + b^4)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + 
 d*x]])/Sqrt[a]])/(Sqrt[a]*(a^2 + b^2)^2)) + (4*(-1)^(1/4)*a*b*((I*a - b)^ 
3*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] - (I*a + b)^3*ArcTanh[(-1)^(3/4)*S 
qrt[Tan[c + d*x]]]))/(a^2 + b^2)^2 - (2*b^3*Tan[c + d*x]^(3/2))/(a + b*Tan 
[c + d*x])^2 + (2*b^2*Sqrt[Tan[c + d*x]])/(a + b*Tan[c + d*x]) + ((7*a^2*b 
^2 - b^4)*Sqrt[Tan[c + d*x]])/((a^2 + b^2)*(a + b*Tan[c + d*x])))/(a*b*(a^ 
2 + b^2)*d)
 

Rubi [A] (verified)

Time = 1.60 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.11, number of steps used = 24, number of rules used = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3042, 4051, 27, 3042, 4132, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4051

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4132

\(\displaystyle \frac {\frac {\int \frac {-b \left (7 a^2-b^2\right ) \tan ^2(c+d x)+8 a \left (a^2-b^2\right ) \tan (c+d x)+b \left (9 a^2+b^2\right )}{2 \sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a \left (a^2+b^2\right )}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 \left (a^2+b^2\right )}-\frac {b \sqrt {\tan (c+d x)}}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {-b \left (7 a^2-b^2\right ) \tan ^2(c+d x)+8 a \left (a^2-b^2\right ) \tan (c+d x)+b \left (9 a^2+b^2\right )}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{2 a \left (a^2+b^2\right )}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 \left (a^2+b^2\right )}-\frac {b \sqrt {\tan (c+d x)}}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {-b \left (7 a^2-b^2\right ) \tan (c+d x)^2+8 a \left (a^2-b^2\right ) \tan (c+d x)+b \left (9 a^2+b^2\right )}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{2 a \left (a^2+b^2\right )}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 \left (a^2+b^2\right )}-\frac {b \sqrt {\tan (c+d x)}}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 4136

\(\displaystyle \frac {\frac {\frac {\int \frac {8 \left (\left (a^2-3 b^2\right ) \tan (c+d x) a^2+b \left (3 a^2-b^2\right ) a\right )}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}-\frac {b \left (15 a^4-18 a^2 b^2-b^4\right ) \int \frac {\tan ^2(c+d x)+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 a \left (a^2+b^2\right )}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 \left (a^2+b^2\right )}-\frac {b \sqrt {\tan (c+d x)}}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {8 \int \frac {\left (a^2-3 b^2\right ) \tan (c+d x) a^2+b \left (3 a^2-b^2\right ) a}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}-\frac {b \left (15 a^4-18 a^2 b^2-b^4\right ) \int \frac {\tan ^2(c+d x)+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 a \left (a^2+b^2\right )}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 \left (a^2+b^2\right )}-\frac {b \sqrt {\tan (c+d x)}}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {8 \int \frac {\left (a^2-3 b^2\right ) \tan (c+d x) a^2+b \left (3 a^2-b^2\right ) a}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}-\frac {b \left (15 a^4-18 a^2 b^2-b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 a \left (a^2+b^2\right )}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 \left (a^2+b^2\right )}-\frac {b \sqrt {\tan (c+d x)}}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 4017

\(\displaystyle \frac {\frac {\frac {16 \int \frac {a \left (b \left (3 a^2-b^2\right )+a \left (a^2-3 b^2\right ) \tan (c+d x)\right )}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}-\frac {b \left (15 a^4-18 a^2 b^2-b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 a \left (a^2+b^2\right )}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 \left (a^2+b^2\right )}-\frac {b \sqrt {\tan (c+d x)}}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {16 a \int \frac {b \left (3 a^2-b^2\right )+a \left (a^2-3 b^2\right ) \tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}-\frac {b \left (15 a^4-18 a^2 b^2-b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 a \left (a^2+b^2\right )}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 \left (a^2+b^2\right )}-\frac {b \sqrt {\tan (c+d x)}}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 1482

\(\displaystyle \frac {\frac {\frac {16 a \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}-\frac {b \left (15 a^4-18 a^2 b^2-b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 a \left (a^2+b^2\right )}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 \left (a^2+b^2\right )}-\frac {b \sqrt {\tan (c+d x)}}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {\frac {\frac {16 a \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}-\frac {b \left (15 a^4-18 a^2 b^2-b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 a \left (a^2+b^2\right )}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 \left (a^2+b^2\right )}-\frac {b \sqrt {\tan (c+d x)}}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {\frac {\frac {16 a \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}-\frac {b \left (15 a^4-18 a^2 b^2-b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 a \left (a^2+b^2\right )}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 \left (a^2+b^2\right )}-\frac {b \sqrt {\tan (c+d x)}}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {\frac {16 a \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}-\frac {b \left (15 a^4-18 a^2 b^2-b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 a \left (a^2+b^2\right )}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 \left (a^2+b^2\right )}-\frac {b \sqrt {\tan (c+d x)}}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {\frac {\frac {16 a \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}-\frac {b \left (15 a^4-18 a^2 b^2-b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 a \left (a^2+b^2\right )}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 \left (a^2+b^2\right )}-\frac {b \sqrt {\tan (c+d x)}}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {16 a \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}-\frac {b \left (15 a^4-18 a^2 b^2-b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 a \left (a^2+b^2\right )}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 \left (a^2+b^2\right )}-\frac {b \sqrt {\tan (c+d x)}}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {16 a \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d \left (a^2+b^2\right )}-\frac {b \left (15 a^4-18 a^2 b^2-b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 a \left (a^2+b^2\right )}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 \left (a^2+b^2\right )}-\frac {b \sqrt {\tan (c+d x)}}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\frac {\frac {16 a \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}-\frac {b \left (15 a^4-18 a^2 b^2-b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 a \left (a^2+b^2\right )}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 \left (a^2+b^2\right )}-\frac {b \sqrt {\tan (c+d x)}}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {\frac {\frac {16 a \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}-\frac {b \left (15 a^4-18 a^2 b^2-b^4\right ) \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}d\tan (c+d x)}{d \left (a^2+b^2\right )}}{2 a \left (a^2+b^2\right )}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 \left (a^2+b^2\right )}-\frac {b \sqrt {\tan (c+d x)}}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {\frac {16 a \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}-\frac {2 b \left (15 a^4-18 a^2 b^2-b^4\right ) \int \frac {1}{a+b \tan (c+d x)}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}}{2 a \left (a^2+b^2\right )}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 \left (a^2+b^2\right )}-\frac {b \sqrt {\tan (c+d x)}}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {16 a \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}-\frac {2 \sqrt {b} \left (15 a^4-18 a^2 b^2-b^4\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d \left (a^2+b^2\right )}}{2 a \left (a^2+b^2\right )}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 \left (a^2+b^2\right )}-\frac {b \sqrt {\tan (c+d x)}}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

Input:

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

Output:

-1/2*(b*Sqrt[Tan[c + d*x]])/((a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) + (((-2 
*Sqrt[b]*(15*a^4 - 18*a^2*b^2 - b^4)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/S 
qrt[a]])/(Sqrt[a]*(a^2 + b^2)*d) + (16*a*(((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 - Sq 
rt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/Sqrt[2] + Log[1 + Sqrt[2]*Sqrt[Ta 
n[c + d*x]] + Tan[c + d*x]]/(2*Sqrt[2])))/2))/((a^2 + b^2)*d))/(2*a*(a^2 + 
 b^2)) - (b*(7*a^2 - b^2)*Sqrt[Tan[c + d*x]])/(a*(a^2 + b^2)*d*(a + b*Tan[ 
c + d*x])))/(4*(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 4051

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a + b*Tan[e + f*x])^(m + 1)*((c + 
d*Tan[e + f*x])^n/(f*(m + 1)*(a^2 + b^2))), x] + Simp[1/((m + 1)*(a^2 + b^2 
))   Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*c 
*(m + 1) - b*d*n - (b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(m + 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] && LtQ[m, -1] && GtQ[n, 0] && Int 
egerQ[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 4132

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*b^2 - a*(b*B - a*C))*(a + b*Tan[e + 
 f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + 
b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 + b^2))   Int[(a + b*Tan[e + 
f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1) - b^2*d* 
(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d 
)*(A*b - a*B - b*C)*Tan[e + f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Ta 
n[e + f*x]^2, 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] && LtQ[m, -1] && 
!(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
 

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 = 0.15 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (3 a^{2} b -b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (a^{3}-3 a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{3}}-\frac {2 b \left (\frac {\frac {b \left (7 a^{4}+6 a^{2} b^{2}-b^{4}\right ) \tan \left (d x +c \right )^{\frac {3}{2}}}{8 a}+\left (\frac {9}{8} a^{4}+\frac {5}{4} a^{2} b^{2}+\frac {1}{8} b^{4}\right ) \sqrt {\tan \left (d x +c \right )}}{\left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {\left (15 a^{4}-18 a^{2} b^{2}-b^{4}\right ) \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right )}{8 a \sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right )^{3}}}{d}\)	\(343\)
default	\(\frac {\frac {\frac {\left (3 a^{2} b -b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (a^{3}-3 a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{3}}-\frac {2 b \left (\frac {\frac {b \left (7 a^{4}+6 a^{2} b^{2}-b^{4}\right ) \tan \left (d x +c \right )^{\frac {3}{2}}}{8 a}+\left (\frac {9}{8} a^{4}+\frac {5}{4} a^{2} b^{2}+\frac {1}{8} b^{4}\right ) \sqrt {\tan \left (d x +c \right )}}{\left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {\left (15 a^{4}-18 a^{2} b^{2}-b^{4}\right ) \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right )}{8 a \sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right )^{3}}}{d}\)	\(343\)

Input:

int(tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*(2/(a^2+b^2)^3*(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)))+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))))-2*b/(a^2+b^2)^3*((1/8*b*(7*a^4+6*a^2*b^2-b^4)/a*t 
an(d*x+c)^(3/2)+(9/8*a^4+5/4*a^2*b^2+1/8*b^4)*tan(d*x+c)^(1/2))/(a+b*tan(d 
*x+c))^2+1/8*(15*a^4-18*a^2*b^2-b^4)/a/(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. 2360 vs. \(2 (290) = 580\).

Time = 0.65 (sec) , antiderivative size = 4751, normalized size of antiderivative = 14.48 \[ \int \frac {\sqrt {\tan (c+d x)}}{(a+b \tan (c+d x))^3} \, dx=\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^3,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

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

integrate(tan(d*x+c)**(1/2)/(a+b*tan(d*x+c))**3,x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 411, normalized size of antiderivative = 1.25 \[ \int \frac {\sqrt {\tan (c+d x)}}{(a+b \tan (c+d x))^3} \, dx=-\frac {\frac {{\left (15 \, a^{4} b - 18 \, a^{2} b^{3} - b^{5}\right )} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \sqrt {a b}} - \frac {2 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {{\left (7 \, a^{2} b^{2} - b^{4}\right )} \tan \left (d x + c\right )^{\frac {3}{2}} + {\left (9 \, a^{3} b + a b^{3}\right )} \sqrt {\tan \left (d x + c\right )}}{a^{7} + 2 \, a^{5} b^{2} + a^{3} b^{4} + {\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{6} b + 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} \tan \left (d x + c\right )}}{4 \, d} \] Input:

integrate(tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^3,x, algorithm="maxima")
 

Output:

-1/4*((15*a^4*b - 18*a^2*b^3 - b^5)*arctan(b*sqrt(tan(d*x + c))/sqrt(a*b)) 
/((a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*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(sqr 
t(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) + ((7*a^2*b^2 - b^4)*tan(d*x + c)^(3/2) + (9*a^ 
3*b + a*b^3)*sqrt(tan(d*x + c)))/(a^7 + 2*a^5*b^2 + a^3*b^4 + (a^5*b^2 + 2 
*a^3*b^4 + a*b^6)*tan(d*x + c)^2 + 2*(a^6*b + 2*a^4*b^3 + a^2*b^5)*tan(d*x 
 + c)))/d
 

Giac [F]

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

integrate(tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^3,x, algorithm="giac")
 

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 14.78 (sec) , antiderivative size = 12659, normalized size of antiderivative = 38.59 \[ \int \frac {\sqrt {\tan (c+d x)}}{(a+b \tan (c+d x))^3} \, dx=\text {Too large to display} \] Input:

int(tan(c + d*x)^(1/2)/(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)*((64*a* 
b^23*d^4 + 1472*a^3*b^21*d^4 + 8832*a^5*b^19*d^4 + 25344*a^7*b^17*d^4 + 40 
320*a^9*b^15*d^4 + 34944*a^11*b^13*d^4 + 10752*a^13*b^11*d^4 - 8448*a^15*b 
^9*d^4 - 10176*a^17*b^7*d^4 - 4160*a^19*b^5*d^4 - 640*a^21*b^3*d^4)/(a^18* 
d^5 + a^2*b^16*d^5 + 8*a^4*b^14*d^5 + 28*a^6*b^12*d^5 + 56*a^8*b^10*d^5 + 
70*a^10*b^8*d^5 + 56*a^12*b^6*d^5 + 28*a^14*b^4*d^5 + 8*a^16*b^2*d^5) + (t 
an(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*a^2*b^2 
5*d^4 + 4608*a^4*b^23*d^4 + 17920*a^6*b^21*d^4 + 38400*a^8*b^19*d^4 + 4608 
0*a^10*b^17*d^4 + 21504*a^12*b^15*d^4 - 21504*a^14*b^13*d^4 - 46080*a^16*b 
^11*d^4 - 38400*a^18*b^9*d^4 - 17920*a^20*b^7*d^4 - 4608*a^22*b^5*d^4 - 51 
2*a^24*b^3*d^4))/(a^18*d^4 + a^2*b^16*d^4 + 8*a^4*b^14*d^4 + 28*a^6*b^12*d 
^4 + 56*a^8*b^10*d^4 + 70*a^10*b^8*d^4 + 56*a^12*b^6*d^4 + 28*a^14*b^4*d^4 
 + 8*a^16*b^2*d^4)) - (tan(c + d*x)^(1/2)*(8*a*b^20*d^2 - 1152*a^3*b^18*d^ 
2 + 2528*a^5*b^16*d^2 + 15296*a^7*b^14*d^2 + 14128*a^9*b^12*d^2 - 5056*a^1 
1*b^10*d^2 - 9248*a^13*b^8*d^2 + 64*a^15*b^6*d^2 + 1800*a^17*b^4*d^2 + ...
 

Reduce [F]

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

int(tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^3,x)
 

Output:

int(sqrt(tan(c + d*x))/(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)