\(\int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx\) [180]

Optimal result

Integrand size = 28, antiderivative size = 283 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {\left (\frac {5}{16}-\frac {7 i}{16}\right ) d^{7/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 f}-\frac {\left (\frac {5}{16}-\frac {7 i}{16}\right ) d^{7/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 f}-\frac {\left (\frac {5}{16}+\frac {7 i}{16}\right ) d^{7/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}+\sqrt {d} \tan (e+f x)}\right )}{\sqrt {2} a^3 f}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}+\frac {5 d^3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )} \] Output:

(5/32-7/32*I)*d^(7/2)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))*2^(1/ 
2)/a^3/f+(-5/32+7/32*I)*d^(7/2)*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1 
/2))*2^(1/2)/a^3/f-(5/32+7/32*I)*d^(7/2)*arctanh(2^(1/2)*(d*tan(f*x+e))^(1 
/2)/(d^(1/2)+d^(1/2)*tan(f*x+e)))*2^(1/2)/a^3/f-1/6*d*(d*tan(f*x+e))^(5/2) 
/f/(a+I*a*tan(f*x+e))^3+1/3*I*d^2*(d*tan(f*x+e))^(3/2)/a/f/(a+I*a*tan(f*x+ 
e))^2+5/8*d^3*(d*tan(f*x+e))^(1/2)/f/(a^3+I*a^3*tan(f*x+e))
 

Mathematica [A] (verified)

Time = 2.28 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.83 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {d^4 \sec ^4(e+f x) \left (-19+19 \cos (4 (e+f x))+(21-15 i) \cos (3 (e+f x)) \log \left (\cos (e+f x)+\sin (e+f x)+\sqrt {\sin (2 (e+f x))}\right ) \sqrt {\sin (2 (e+f x))}-12 i \sin (2 (e+f x))+(21+15 i) \arcsin (\cos (e+f x)-\sin (e+f x)) \sqrt {\sin (2 (e+f x))} (\cos (3 (e+f x))+i \sin (3 (e+f x)))+(15+21 i) \log \left (\cos (e+f x)+\sin (e+f x)+\sqrt {\sin (2 (e+f x))}\right ) \sqrt {\sin (2 (e+f x))} \sin (3 (e+f x))+21 i \sin (4 (e+f x))\right )}{96 a^3 f \sqrt {d \tan (e+f x)} (-i+\tan (e+f x))^3} \] Input:

Integrate[(d*Tan[e + f*x])^(7/2)/(a + I*a*Tan[e + f*x])^3,x]
 

Output:

(d^4*Sec[e + f*x]^4*(-19 + 19*Cos[4*(e + f*x)] + (21 - 15*I)*Cos[3*(e + f* 
x)]*Log[Cos[e + f*x] + Sin[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sqrt[Sin[2*( 
e + f*x)]] - (12*I)*Sin[2*(e + f*x)] + (21 + 15*I)*ArcSin[Cos[e + f*x] - S 
in[e + f*x]]*Sqrt[Sin[2*(e + f*x)]]*(Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)] 
) + (15 + 21*I)*Log[Cos[e + f*x] + Sin[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]* 
Sqrt[Sin[2*(e + f*x)]]*Sin[3*(e + f*x)] + (21*I)*Sin[4*(e + f*x)]))/(96*a^ 
3*f*Sqrt[d*Tan[e + f*x]]*(-I + Tan[e + f*x])^3)
 

Rubi [A] (verified)

Time = 1.19 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.17, number of steps used = 21, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 4041, 27, 3042, 4078, 27, 3042, 4078, 27, 3042, 4017, 27, 1482, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

\(\displaystyle \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3}dx\)

\(\Big \downarrow \) 4041

\(\displaystyle -\frac {\int -\frac {(d \tan (e+f x))^{3/2} \left (5 a d^2-11 i a d^2 \tan (e+f x)\right )}{2 (i \tan (e+f x) a+a)^2}dx}{6 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(d \tan (e+f x))^{3/2} \left (5 a d^2-11 i a d^2 \tan (e+f x)\right )}{(i \tan (e+f x) a+a)^2}dx}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(d \tan (e+f x))^{3/2} \left (5 a d^2-11 i a d^2 \tan (e+f x)\right )}{(i \tan (e+f x) a+a)^2}dx}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 4078

\(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {\int \frac {12 \sqrt {d \tan (e+f x)} \left (2 i a^2 d^3+3 a^2 \tan (e+f x) d^3\right )}{i \tan (e+f x) a+a}dx}{4 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \int \frac {\sqrt {d \tan (e+f x)} \left (2 i a^2 d^3+3 a^2 \tan (e+f x) d^3\right )}{i \tan (e+f x) a+a}dx}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \int \frac {\sqrt {d \tan (e+f x)} \left (2 i a^2 d^3+3 a^2 \tan (e+f x) d^3\right )}{i \tan (e+f x) a+a}dx}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 4078

\(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (-\frac {\int -\frac {5 a^3 d^4-7 i a^3 d^4 \tan (e+f x)}{2 \sqrt {d \tan (e+f x)}}dx}{2 a^2}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (\frac {\int \frac {5 a^3 d^4-7 i a^3 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx}{4 a^2}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (\frac {\int \frac {5 a^3 d^4-7 i a^3 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx}{4 a^2}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 4017

\(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (\frac {\int \frac {a^3 d^4 (5 d-7 i d \tan (e+f x))}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{2 a^2 f}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (\frac {a d^4 \int \frac {5 d-7 i d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{2 f}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 1482

\(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (\frac {a d^4 \left (\left (\frac {5}{2}+\frac {7 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {5}{2}-\frac {7 i}{2}\right ) \int \frac {\tan (e+f x) d+d}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{2 f}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (\frac {a d^4 \left (\left (\frac {5}{2}+\frac {7 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {5}{2}-\frac {7 i}{2}\right ) \left (\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}+\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}\right )\right )}{2 f}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (\frac {a d^4 \left (\left (\frac {5}{2}+\frac {7 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {5}{2}-\frac {7 i}{2}\right ) \left (\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{2 f}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (\frac {a d^4 \left (\left (\frac {5}{2}+\frac {7 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {5}{2}-\frac {7 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{2 f}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (\frac {a d^4 \left (\left (\frac {5}{2}+\frac {7 i}{2}\right ) \left (-\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )+\left (\frac {5}{2}-\frac {7 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{2 f}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (\frac {a d^4 \left (\left (\frac {5}{2}+\frac {7 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )+\left (\frac {5}{2}-\frac {7 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{2 f}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (\frac {a d^4 \left (\left (\frac {5}{2}+\frac {7 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {d}}\right )+\left (\frac {5}{2}-\frac {7 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{2 f}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (\frac {a d^4 \left (\left (\frac {5}{2}-\frac {7 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )+\left (\frac {5}{2}+\frac {7 i}{2}\right ) \left (\frac {\log \left (d \tan (e+f x)+\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (d \tan (e+f x)-\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}\right )\right )}{2 f}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\)

Input:

Int[(d*Tan[e + f*x])^(7/2)/(a + I*a*Tan[e + f*x])^3,x]
 

Output:

-1/6*(d*(d*Tan[e + f*x])^(5/2))/(f*(a + I*a*Tan[e + f*x])^3) + (((4*I)*a*d 
^2*(d*Tan[e + f*x])^(3/2))/(f*(a + I*a*Tan[e + f*x])^2) - (3*((a*d^4*((5/2 
 - (7*I)/2)*(-(ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(Sqrt[2] 
*Sqrt[d])) + ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(Sqrt[2]*S 
qrt[d])) + (5/2 + (7*I)/2)*(-1/2*Log[d + d*Tan[e + f*x] - Sqrt[2]*Sqrt[d]* 
Sqrt[d*Tan[e + f*x]]]/(Sqrt[2]*Sqrt[d]) + Log[d + d*Tan[e + f*x] + Sqrt[2] 
*Sqrt[d]*Sqrt[d*Tan[e + f*x]]]/(2*Sqrt[2]*Sqrt[d]))))/(2*f) - (5*a^2*d^3*S 
qrt[d*Tan[e + f*x]])/(2*f*(a + I*a*Tan[e + f*x]))))/a^2)/(12*a^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 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 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 4041

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

rule 4078

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

Maple [A] (verified)

Time = 1.44 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.44

method	result	size
derivativedivides	\(\frac {2 d^{4} \left (-\frac {9 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}-\frac {38 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-5 d^{2} \sqrt {d \tan \left (f x +e \right )}}{16 \left (i d \tan \left (f x +e \right )+d \right )^{3}}+\frac {3 i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{8 \sqrt {-i d}}+\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 \sqrt {i d}}\right )}{f \,a^{3}}\)	\(125\)
default	\(\frac {2 d^{4} \left (-\frac {9 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}-\frac {38 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-5 d^{2} \sqrt {d \tan \left (f x +e \right )}}{16 \left (i d \tan \left (f x +e \right )+d \right )^{3}}+\frac {3 i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{8 \sqrt {-i d}}+\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 \sqrt {i d}}\right )}{f \,a^{3}}\)	\(125\)

Input:

int((d*tan(f*x+e))^(7/2)/(a+I*a*tan(f*x+e))^3,x,method=_RETURNVERBOSE)
 

Output:

2/f/a^3*d^4*(-1/16*(9*(d*tan(f*x+e))^(5/2)-38/3*I*d*(d*tan(f*x+e))^(3/2)-5 
*d^2*(d*tan(f*x+e))^(1/2))/(I*d*tan(f*x+e)+d)^3+3/8*I/(-I*d)^(1/2)*arctan( 
(d*tan(f*x+e))^(1/2)/(-I*d)^(1/2))+1/16*I/(I*d)^(1/2)*arctan((d*tan(f*x+e) 
)^(1/2)/(I*d)^(1/2)))
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 586 vs. \(2 (213) = 426\).

Time = 0.09 (sec) , antiderivative size = 586, normalized size of antiderivative = 2.07 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx =\text {Too large to display} \] Input:

integrate((d*tan(f*x+e))^(7/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="fricas")
 

Output:

-1/48*(12*a^3*f*sqrt(-1/64*I*d^7/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(-2*(I* 
d^4*e^(2*I*f*x + 2*I*e) + 8*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt((-I*d 
*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-1/64*I*d^7/(a 
^6*f^2)))*e^(-2*I*f*x - 2*I*e)/d^3) - 12*a^3*f*sqrt(-1/64*I*d^7/(a^6*f^2)) 
*e^(6*I*f*x + 6*I*e)*log(-2*(I*d^4*e^(2*I*f*x + 2*I*e) - 8*(a^3*f*e^(2*I*f 
*x + 2*I*e) + a^3*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2 
*I*e) + 1))*sqrt(-1/64*I*d^7/(a^6*f^2)))*e^(-2*I*f*x - 2*I*e)/d^3) - 12*a^ 
3*f*sqrt(9/16*I*d^7/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/4*(-3*I*d^4 + 4*( 
a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/( 
e^(2*I*f*x + 2*I*e) + 1))*sqrt(9/16*I*d^7/(a^6*f^2)))*e^(-2*I*f*x - 2*I*e) 
/(a^3*f)) + 12*a^3*f*sqrt(9/16*I*d^7/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/ 
4*(-3*I*d^4 - 4*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt((-I*d*e^(2*I*f*x 
+ 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(9/16*I*d^7/(a^6*f^2)))*e^( 
-2*I*f*x - 2*I*e)/(a^3*f)) - (20*d^3*e^(6*I*f*x + 6*I*e) + 14*d^3*e^(4*I*f 
*x + 4*I*e) - 5*d^3*e^(2*I*f*x + 2*I*e) + d^3)*sqrt((-I*d*e^(2*I*f*x + 2*I 
*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-6*I*f*x - 6*I*e)/(a^3*f)
 

Sympy [F]

\[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {i \int \frac {\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\, dx}{a^{3}} \] Input:

integrate((d*tan(f*x+e))**(7/2)/(a+I*a*tan(f*x+e))**3,x)
 

Output:

I*Integral((d*tan(e + f*x))**(7/2)/(tan(e + f*x)**3 - 3*I*tan(e + f*x)**2 
- 3*tan(e + f*x) + I), x)/a**3
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((d*tan(f*x+e))^(7/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.69 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {i \, {\left (\frac {3 \, \sqrt {2} d^{\frac {9}{2}} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{\frac {i \, d}{{\left | d \right |}} + 1} + \frac {18 \, \sqrt {2} d^{\frac {9}{2}} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{-i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{-\frac {i \, d}{{\left | d \right |}} + 1} - \frac {27 \, \sqrt {d \tan \left (f x + e\right )} d^{7} \tan \left (f x + e\right )^{2} - 38 i \, \sqrt {d \tan \left (f x + e\right )} d^{7} \tan \left (f x + e\right ) - 15 \, \sqrt {d \tan \left (f x + e\right )} d^{7}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3}}\right )}}{24 \, a^{3} d f} \] Input:

integrate((d*tan(f*x+e))^(7/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="giac")
 

Output:

1/24*I*(3*sqrt(2)*d^(9/2)*arctan(2*sqrt(d*tan(f*x + e))*abs(d)/(I*sqrt(2)* 
d^(3/2) + sqrt(2)*sqrt(d)*abs(d)))/(I*d/abs(d) + 1) + 18*sqrt(2)*d^(9/2)*a 
rctan(2*sqrt(d*tan(f*x + e))*abs(d)/(-I*sqrt(2)*d^(3/2) + sqrt(2)*sqrt(d)* 
abs(d)))/(-I*d/abs(d) + 1) - (27*sqrt(d*tan(f*x + e))*d^7*tan(f*x + e)^2 - 
 38*I*sqrt(d*tan(f*x + e))*d^7*tan(f*x + e) - 15*sqrt(d*tan(f*x + e))*d^7) 
/(d*tan(f*x + e) - I*d)^3)/(a^3*d*f)
 

Mupad [B] (verification not implemented)

Time = 2.79 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.77 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx=\mathrm {atan}\left (\frac {8\,a^3\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {d^7\,9{}\mathrm {i}}{64\,a^6\,f^2}}}{3\,d^4}\right )\,\sqrt {\frac {d^7\,9{}\mathrm {i}}{64\,a^6\,f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {16\,a^3\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {d^7\,1{}\mathrm {i}}{256\,a^6\,f^2}}}{d^4}\right )\,\sqrt {-\frac {d^7\,1{}\mathrm {i}}{256\,a^6\,f^2}}\,2{}\mathrm {i}+\frac {\frac {19\,d^5\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{12\,a^3\,f}-\frac {d^6\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,5{}\mathrm {i}}{8\,a^3\,f}+\frac {d^4\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,9{}\mathrm {i}}{8\,a^3\,f}}{-d^3\,{\mathrm {tan}\left (e+f\,x\right )}^3+d^3\,{\mathrm {tan}\left (e+f\,x\right )}^2\,3{}\mathrm {i}+3\,d^3\,\mathrm {tan}\left (e+f\,x\right )-d^3\,1{}\mathrm {i}} \] Input:

int((d*tan(e + f*x))^(7/2)/(a + a*tan(e + f*x)*1i)^3,x)
 

Output:

atan((8*a^3*f*(d*tan(e + f*x))^(1/2)*((d^7*9i)/(64*a^6*f^2))^(1/2))/(3*d^4 
))*((d^7*9i)/(64*a^6*f^2))^(1/2)*2i + atan((16*a^3*f*(d*tan(e + f*x))^(1/2 
)*(-(d^7*1i)/(256*a^6*f^2))^(1/2))/d^4)*(-(d^7*1i)/(256*a^6*f^2))^(1/2)*2i 
 + ((19*d^5*(d*tan(e + f*x))^(3/2))/(12*a^3*f) - (d^6*(d*tan(e + f*x))^(1/ 
2)*5i)/(8*a^3*f) + (d^4*(d*tan(e + f*x))^(5/2)*9i)/(8*a^3*f))/(3*d^3*tan(e 
 + f*x) - d^3*1i + d^3*tan(e + f*x)^2*3i - d^3*tan(e + f*x)^3)
 

Reduce [F]

\[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {\sqrt {d}\, \left (\int \frac {\sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{3} i +3 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i -1}d x \right ) d^{3}}{a^{3}} \] Input:

int((d*tan(f*x+e))^(7/2)/(a+I*a*tan(f*x+e))^3,x)
 

Output:

( - sqrt(d)*int((sqrt(tan(e + f*x))*tan(e + f*x)**3)/(tan(e + f*x)**3*i + 
3*tan(e + f*x)**2 - 3*tan(e + f*x)*i - 1),x)*d**3)/a**3