\(\int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx\) [398]

Optimal result

Integrand size = 33, antiderivative size = 264 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {(a (A-B)+b (A+B)) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}-\frac {(a (A-B)+b (A+B)) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}-\frac {2 a^{5/2} (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{5/2} \left (a^2+b^2\right ) d}-\frac {(b (A-B)-a (A+B)) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{b^2 d}+\frac {2 B \tan ^{\frac {3}{2}}(c+d x)}{3 b d} \] Output:

-1/2*(a*(A-B)+b*(A+B))*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)/(a^2+b^ 
2)/d-1/2*(a*(A-B)+b*(A+B))*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)/(a^2 
+b^2)/d-2*a^(5/2)*(A*b-B*a)*arctan(b^(1/2)*tan(d*x+c)^(1/2)/a^(1/2))/b^(5/ 
2)/(a^2+b^2)/d-1/2*(b*(A-B)-a*(A+B))*arctanh(2^(1/2)*tan(d*x+c)^(1/2)/(1+t 
an(d*x+c)))*2^(1/2)/(a^2+b^2)/d+2*(A*b-B*a)*tan(d*x+c)^(1/2)/b^2/d+2/3*B*t 
an(d*x+c)^(3/2)/b/d
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.78 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.71 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {-3 \sqrt [4]{-1} (a+i b) b^{5/2} (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+6 a^{5/2} (-A b+a B) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )+3 \sqrt [4]{-1} b^{5/2} (i a+b) (A+i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+2 \sqrt {b} \left (a^2+b^2\right ) \sqrt {\tan (c+d x)} (3 A b-3 a B+b B \tan (c+d x))}{3 b^{5/2} \left (a^2+b^2\right ) d} \] Input:

Integrate[(Tan[c + d*x]^(5/2)*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x]),x 
]
 

Output:

(-3*(-1)^(1/4)*(a + I*b)*b^(5/2)*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + 
d*x]]] + 6*a^(5/2)*(-(A*b) + a*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt 
[a]] + 3*(-1)^(1/4)*b^(5/2)*(I*a + b)*(A + I*B)*ArcTanh[(-1)^(3/4)*Sqrt[Ta 
n[c + d*x]]] + 2*Sqrt[b]*(a^2 + b^2)*Sqrt[Tan[c + d*x]]*(3*A*b - 3*a*B + b 
*B*Tan[c + d*x]))/(3*b^(5/2)*(a^2 + b^2)*d)
 

Rubi [A] (verified)

Time = 1.44 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.09, number of steps used = 23, number of rules used = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 4090, 27, 3042, 4130, 27, 3042, 4136, 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.

\(\displaystyle \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^{5/2} (A+B \tan (c+d x))}{a+b \tan (c+d x)}dx\)

\(\Big \downarrow \) 4090

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4130

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4136

\(\displaystyle \frac {2 B \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {\frac {\int \frac {(A b-a B) b^2+(a A+b B) \tan (c+d x) b^2}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}+\frac {a^3 (A b-a B) \int \frac {\tan ^2(c+d x)+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{b}-\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{b d}}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 B \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {\frac {\int \frac {(A b-a B) b^2+(a A+b B) \tan (c+d x) b^2}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}+\frac {a^3 (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{b}-\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{b d}}{b}\)

\(\Big \downarrow \) 4017

\(\displaystyle \frac {2 B \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {\frac {2 \int \frac {b^2 (A b-a B+(a A+b B) \tan (c+d x))}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}+\frac {a^3 (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{b}-\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{b d}}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 B \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {\frac {2 b^2 \int \frac {A b-a B+(a A+b B) \tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}+\frac {a^3 (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{b}-\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{b d}}{b}\)

\(\Big \downarrow \) 1482

\(\displaystyle \frac {2 B \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {\frac {2 b^2 \left (\frac {1}{2} (b (A-B)-a (A+B)) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} (a (A-B)+b (A+B)) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}+\frac {a^3 (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{b}-\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{b d}}{b}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {2 B \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {\frac {2 b^2 \left (\frac {1}{2} (b (A-B)-a (A+B)) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d \left (a^2+b^2\right )}+\frac {a^3 (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{b}-\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{b d}}{b}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {2 B \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {\frac {2 b^2 \left (\frac {1}{2} (b (A-B)-a (A+B)) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {a^3 (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{b}-\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{b d}}{b}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {2 B \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {\frac {2 b^2 \left (\frac {1}{2} (b (A-B)-a (A+B)) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {a^3 (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{b}-\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{b d}}{b}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {2 B \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {\frac {2 b^2 \left (\frac {1}{2} (b (A-B)-a (A+B)) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {a^3 (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{b}-\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{b d}}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 B \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {\frac {2 b^2 \left (\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {a^3 (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{b}-\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{b d}}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 B \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {\frac {2 b^2 \left (\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )+\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {a^3 (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{b}-\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{b d}}{b}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {2 B \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {\frac {a^3 (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}+\frac {2 b^2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{b}-\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{b d}}{b}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {2 B \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {\frac {a^3 (A b-a B) \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}d\tan (c+d x)}{d \left (a^2+b^2\right )}+\frac {2 b^2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{b}-\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{b d}}{b}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 B \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {\frac {2 a^3 (A b-a B) \int \frac {1}{a+b \tan (c+d x)}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}+\frac {2 b^2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{b}-\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{b d}}{b}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {2 B \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {\frac {2 b^2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {2 a^{5/2} (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {b} d \left (a^2+b^2\right )}}{b}-\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{b d}}{b}\)

Input:

Int[(Tan[c + d*x]^(5/2)*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x]),x]
 

Output:

-((((2*a^(5/2)*(A*b - a*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/( 
Sqrt[b]*(a^2 + b^2)*d) + (2*b^2*(((a*(A - B) + b*(A + B))*(-(ArcTan[1 - Sq 
rt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] 
]/Sqrt[2]))/2 + ((b*(A - B) - a*(A + B))*(-1/2*Log[1 - Sqrt[2]*Sqrt[Tan[c 
+ d*x]] + Tan[c + d*x]]/Sqrt[2] + Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan 
[c + d*x]]/(2*Sqrt[2])))/2))/((a^2 + b^2)*d))/b - (2*(A*b - a*B)*Sqrt[Tan[ 
c + d*x]])/(b*d))/b) + (2*B*Tan[c + d*x]^(3/2))/(3*b*d)
 

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 4090

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[b*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + 
 n))), x] + Simp[1/(d*(m + n))   Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Ta 
n[e + f*x])^n*Simp[a^2*A*d*(m + n) - b*B*(b*c*(m - 1) + a*d*(n + 1)) + d*(m 
 + n)*(2*a*A*b + B*(a^2 - b^2))*Tan[e + f*x] - (b*B*(b*c - a*d)*(m - 1) - b 
*(A*b + a*B)*d*(m + n))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, 
 f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2 
, 0] && GtQ[m, 1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 1] 
&& ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
 

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 4130

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[C*(a + b*Tan[e + f*x])^m*((c + d*Tan[ 
e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a 
+ b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C 
*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f*x] - (C* 
m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[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] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !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 = 0.12 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.11

method	result	size
derivativedivides	\(\frac {\frac {\frac {2 B b \tan \left (d x +c \right )^{\frac {3}{2}}}{3}+2 A b \sqrt {\tan \left (d x +c \right )}-2 B a \sqrt {\tan \left (d x +c \right )}}{b^{2}}-\frac {2 a^{3} \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right )}{b^{2} \left (a^{2}+b^{2}\right ) \sqrt {a b}}+\frac {\frac {\left (-A b +B a \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (-a A -B b \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{a^{2}+b^{2}}}{d}\)	\(292\)
default	\(\frac {\frac {\frac {2 B b \tan \left (d x +c \right )^{\frac {3}{2}}}{3}+2 A b \sqrt {\tan \left (d x +c \right )}-2 B a \sqrt {\tan \left (d x +c \right )}}{b^{2}}-\frac {2 a^{3} \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right )}{b^{2} \left (a^{2}+b^{2}\right ) \sqrt {a b}}+\frac {\frac {\left (-A b +B a \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (-a A -B b \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{a^{2}+b^{2}}}{d}\)	\(292\)

Input:

int(tan(d*x+c)^(5/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x,method=_RETURNVER 
BOSE)
 

Output:

1/d*(2/b^2*(1/3*B*b*tan(d*x+c)^(3/2)+A*b*tan(d*x+c)^(1/2)-B*a*tan(d*x+c)^( 
1/2))-2/b^2*a^3*(A*b-B*a)/(a^2+b^2)/(a*b)^(1/2)*arctan(b*tan(d*x+c)^(1/2)/ 
(a*b)^(1/2))+2/(a^2+b^2)*(1/8*(-A*b+B*a)*2^(1/2)*(ln((tan(d*x+c)+2^(1/2)*t 
an(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*a-B* 
b)*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)))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1324 vs. \(2 (230) = 460\).

Time = 20.06 (sec) , antiderivative size = 2674, normalized size of antiderivative = 10.13 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(5/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm= 
"fricas")
 

Output:

[-1/6*(6*sqrt(1/2)*(a^2*b^2 + b^4)*d*sqrt(((A^2 - 2*A*B + B^2)*a^2 + 2*(A^ 
2 - B^2)*a*b + (A^2 + 2*A*B + B^2)*b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*arc 
tan(((a^4 + 2*a^2*b^2 + b^4)*d^2*sqrt(((A^2 + 2*A*B + B^2)*a^2 - 2*(A^2 - 
B^2)*a*b + (A^2 - 2*A*B + B^2)*b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*sqrt((( 
A^2 - 2*A*B + B^2)*a^2 + 2*(A^2 - B^2)*a*b + (A^2 + 2*A*B + B^2)*b^2)/((a^ 
4 + 2*a^2*b^2 + b^4)*d^2)) + 2*sqrt(1/2)*((A + B)*a^3 - (A - B)*a^2*b + (A 
 + B)*a*b^2 - (A - B)*b^3)*d*sqrt(((A^2 - 2*A*B + B^2)*a^2 + 2*(A^2 - B^2) 
*a*b + (A^2 + 2*A*B + B^2)*b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*sqrt(tan(d* 
x + c)))/(4*A*B*a*b + (A^2 - B^2)*a^2 - (A^2 - B^2)*b^2)) + 6*sqrt(1/2)*(a 
^2*b^2 + b^4)*d*sqrt(((A^2 - 2*A*B + B^2)*a^2 + 2*(A^2 - B^2)*a*b + (A^2 + 
 2*A*B + B^2)*b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*arctan(-((a^4 + 2*a^2*b^ 
2 + b^4)*d^2*sqrt(((A^2 + 2*A*B + B^2)*a^2 - 2*(A^2 - B^2)*a*b + (A^2 - 2* 
A*B + B^2)*b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*sqrt(((A^2 - 2*A*B + B^2)*a 
^2 + 2*(A^2 - B^2)*a*b + (A^2 + 2*A*B + B^2)*b^2)/((a^4 + 2*a^2*b^2 + b^4) 
*d^2)) - 2*sqrt(1/2)*((A + B)*a^3 - (A - B)*a^2*b + (A + B)*a*b^2 - (A - B 
)*b^3)*d*sqrt(((A^2 - 2*A*B + B^2)*a^2 + 2*(A^2 - B^2)*a*b + (A^2 + 2*A*B 
+ B^2)*b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*sqrt(tan(d*x + c)))/(4*A*B*a*b 
+ (A^2 - B^2)*a^2 - (A^2 - B^2)*b^2)) - 3*sqrt(1/2)*(a^2*b^2 + b^4)*d*sqrt 
(((A^2 + 2*A*B + B^2)*a^2 - 2*(A^2 - B^2)*a*b + (A^2 - 2*A*B + B^2)*b^2)/( 
(a^4 + 2*a^2*b^2 + b^4)*d^2))*log(2*sqrt(1/2)*(a^2 + b^2)*d*sqrt(((A^2 ...
 

Sympy [F]

\[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{\frac {5}{2}}{\left (c + d x \right )}}{a + b \tan {\left (c + d x \right )}}\, dx \] Input:

integrate(tan(d*x+c)**(5/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x)
 

Output:

Integral((A + B*tan(c + d*x))*tan(c + d*x)**(5/2)/(a + b*tan(c + d*x)), x)
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 258, normalized size of antiderivative = 0.98 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {\frac {24 \, {\left (B a^{4} - A a^{3} b\right )} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{2} b^{2} + b^{4}\right )} \sqrt {a b}} - \frac {3 \, {\left (2 \, \sqrt {2} {\left ({\left (A - B\right )} a + {\left (A + B\right )} b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left ({\left (A - B\right )} a + {\left (A + B\right )} b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left ({\left (A + B\right )} a - {\left (A - B\right )} b\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left ({\left (A + B\right )} a - {\left (A - B\right )} b\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )}}{a^{2} + b^{2}} + \frac {8 \, {\left (B b \tan \left (d x + c\right )^{\frac {3}{2}} - 3 \, {\left (B a - A b\right )} \sqrt {\tan \left (d x + c\right )}\right )}}{b^{2}}}{12 \, d} \] Input:

integrate(tan(d*x+c)^(5/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm= 
"maxima")
 

Output:

1/12*(24*(B*a^4 - A*a^3*b)*arctan(b*sqrt(tan(d*x + c))/sqrt(a*b))/((a^2*b^ 
2 + b^4)*sqrt(a*b)) - 3*(2*sqrt(2)*((A - B)*a + (A + B)*b)*arctan(1/2*sqrt 
(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*((A - B)*a + (A + B)*b)* 
arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) - sqrt(2)*((A + B)*a 
 - (A - B)*b)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + sqrt(2) 
*((A + B)*a - (A - B)*b)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 
1))/(a^2 + b^2) + 8*(B*b*tan(d*x + c)^(3/2) - 3*(B*a - A*b)*sqrt(tan(d*x + 
 c)))/b^2)/d
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(tan(d*x+c)^(5/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm= 
"giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [B] (verification not implemented)

Time = 10.64 (sec) , antiderivative size = 16441, normalized size of antiderivative = 62.28 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\text {Too large to display} \] Input:

int((tan(c + d*x)^(5/2)*(A + B*tan(c + d*x)))/(a + b*tan(c + d*x)),x)
 

Output:

atan(((((((32*(4*A*a*b^8*d^4 + 8*A*a^3*b^6*d^4 + 4*A*a^5*b^4*d^4))/(b*d^5) 
 - (32*tan(c + d*x)^(1/2)*(((64*A^4*a^2*b^2*d^4 - A^4*(16*a^4*d^4 + 16*b^4 
*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*A^2*a*b*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2* 
a^2*b^2*d^4)))^(1/2)*(16*b^10*d^4 + 16*a^2*b^8*d^4 - 16*a^4*b^6*d^4 - 16*a 
^6*b^4*d^4))/(b*d^4))*(((64*A^4*a^2*b^2*d^4 - A^4*(16*a^4*d^4 + 16*b^4*d^4 
 + 32*a^2*b^2*d^4))^(1/2) - 8*A^2*a*b*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2* 
b^2*d^4)))^(1/2) + (32*tan(c + d*x)^(1/2)*(4*A^2*a^3*b^5*d^2 + 2*A^2*a^5*b 
^3*d^2 - 14*A^2*a*b^7*d^2 + 16*A^2*a^7*b*d^2))/(b*d^4))*(((64*A^4*a^2*b^2* 
d^4 - A^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*A^2*a*b*d^ 
2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2) - (32*(A^3*a^2*b^5*d^2 
- 15*A^3*a^4*b^3*d^2 + 12*A^3*a^6*b*d^2))/(b*d^5))*(((64*A^4*a^2*b^2*d^4 - 
 A^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*A^2*a*b*d^2)/(1 
6*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2) + (32*tan(c + d*x)^(1/2)*(2* 
A^4*a^6 - A^4*b^6))/(b*d^4))*(((64*A^4*a^2*b^2*d^4 - A^4*(16*a^4*d^4 + 16* 
b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*A^2*a*b*d^2)/(16*(a^4*d^4 + b^4*d^4 + 
 2*a^2*b^2*d^4)))^(1/2)*1i - (((((32*(4*A*a*b^8*d^4 + 8*A*a^3*b^6*d^4 + 4* 
A*a^5*b^4*d^4))/(b*d^5) + (32*tan(c + d*x)^(1/2)*(((64*A^4*a^2*b^2*d^4 - A 
^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*A^2*a*b*d^2)/(16* 
(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2)*(16*b^10*d^4 + 16*a^2*b^8*d^4 
- 16*a^4*b^6*d^4 - 16*a^6*b^4*d^4))/(b*d^4))*(((64*A^4*a^2*b^2*d^4 - A^...
 

Reduce [F]

\[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\int \sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{2}d x \] Input:

int(tan(d*x+c)^(5/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x)
 

Output:

int(sqrt(tan(c + d*x))*tan(c + d*x)**2,x)