\(\int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx\) [172]

Optimal result

Integrand size = 28, antiderivative size = 266 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {\left (\frac {25}{16}-\frac {21 i}{16}\right ) d^{7/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {25}{16}-\frac {21 i}{16}\right ) d^{7/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {25}{16}+\frac {21 i}{16}\right ) d^{7/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}+\sqrt {d} \tan (e+f x)}\right )}{\sqrt {2} a^2 f}-\frac {25 d^3 \sqrt {d \tan (e+f x)}}{8 a^2 f}+\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2} \] Output:

(-25/32+21/32*I)*d^(7/2)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))*2^ 
(1/2)/a^2/f+(25/32-21/32*I)*d^(7/2)*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/ 
d^(1/2))*2^(1/2)/a^2/f+(25/32+21/32*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^2/f-25/8*d^3*(d*tan(f*x+ 
e))^(1/2)/a^2/f+7/8*I*d^2*(d*tan(f*x+e))^(3/2)/a^2/f/(1+I*tan(f*x+e))-1/4* 
d*(d*tan(f*x+e))^(5/2)/f/(a+I*a*tan(f*x+e))^2
 

Mathematica [A] (verified)

Time = 2.27 (sec) , antiderivative size = 233, normalized size of antiderivative = 0.88 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {i d^4 \sec ^3(e+f x) \left (-43 \cos (e+f x)+43 \cos (3 (e+f x))-23 i \sin (e+f x)+(21-25 i) \cos (2 (e+f x)) \log \left (\cos (e+f x)+\sin (e+f x)+\sqrt {\sin (2 (e+f x))}\right ) \sqrt {\sin (2 (e+f x))}+(21+25 i) \arcsin (\cos (e+f x)-\sin (e+f x)) (\cos (2 (e+f x))+i \sin (2 (e+f x))) \sqrt {\sin (2 (e+f x))}+(25+21 i) \log \left (\cos (e+f x)+\sin (e+f x)+\sqrt {\sin (2 (e+f x))}\right ) \sin ^{\frac {3}{2}}(2 (e+f x))+41 i \sin (3 (e+f x))\right )}{32 a^2 f \sqrt {d \tan (e+f x)} (-i+\tan (e+f x))^2} \] Input:

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

Output:

((-1/32*I)*d^4*Sec[e + f*x]^3*(-43*Cos[e + f*x] + 43*Cos[3*(e + f*x)] - (2 
3*I)*Sin[e + f*x] + (21 - 25*I)*Cos[2*(e + f*x)]*Log[Cos[e + f*x] + Sin[e 
+ f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sqrt[Sin[2*(e + f*x)]] + (21 + 25*I)*ArcS 
in[Cos[e + f*x] - Sin[e + f*x]]*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)])*Sq 
rt[Sin[2*(e + f*x)]] + (25 + 21*I)*Log[Cos[e + f*x] + Sin[e + f*x] + Sqrt[ 
Sin[2*(e + f*x)]]]*Sin[2*(e + f*x)]^(3/2) + (41*I)*Sin[3*(e + f*x)]))/(a^2 
*f*Sqrt[d*Tan[e + f*x]]*(-I + Tan[e + f*x])^2)
 

Rubi [A] (verified)

Time = 1.07 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.18, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.679, Rules used = {3042, 4041, 27, 3042, 4078, 3042, 4011, 3042, 4017, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4041

\(\displaystyle -\frac {\int -\frac {(d \tan (e+f x))^{3/2} \left (5 a d^2-9 i a d^2 \tan (e+f x)\right )}{2 (i \tan (e+f x) a+a)}dx}{4 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(d \tan (e+f x))^{3/2} \left (5 a d^2-9 i a d^2 \tan (e+f x)\right )}{i \tan (e+f x) a+a}dx}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(d \tan (e+f x))^{3/2} \left (5 a d^2-9 i a d^2 \tan (e+f x)\right )}{i \tan (e+f x) a+a}dx}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 4078

\(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\int \sqrt {d \tan (e+f x)} \left (21 i a^2 d^3+25 a^2 \tan (e+f x) d^3\right )dx}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\int \sqrt {d \tan (e+f x)} \left (21 i a^2 d^3+25 a^2 \tan (e+f x) d^3\right )dx}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}+\int \frac {21 i a^2 d^4 \tan (e+f x)-25 a^2 d^4}{\sqrt {d \tan (e+f x)}}dx}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}+\int \frac {21 i a^2 d^4 \tan (e+f x)-25 a^2 d^4}{\sqrt {d \tan (e+f x)}}dx}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 4017

\(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}+\frac {2 \int -\frac {a^2 d^4 (25 d-21 i d \tan (e+f x))}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {2 \int \frac {a^2 d^4 (25 d-21 i d \tan (e+f x))}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {2 a^2 d^4 \int \frac {25 d-21 i d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 1482

\(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {2 a^2 d^4 \left (\left (\frac {25}{2}+\frac {21 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {25}{2}-\frac {21 i}{2}\right ) \int \frac {\tan (e+f x) d+d}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {2 a^2 d^4 \left (\left (\frac {25}{2}+\frac {21 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {25}{2}-\frac {21 i}{2}\right ) \left (\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}+\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}\right )\right )}{f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {2 a^2 d^4 \left (\left (\frac {25}{2}+\frac {21 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {25}{2}-\frac {21 i}{2}\right ) \left (\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {2 a^2 d^4 \left (\left (\frac {25}{2}+\frac {21 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {25}{2}-\frac {21 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {2 a^2 d^4 \left (\left (\frac {25}{2}+\frac {21 i}{2}\right ) \left (-\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )+\left (\frac {25}{2}-\frac {21 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {2 a^2 d^4 \left (\left (\frac {25}{2}+\frac {21 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )+\left (\frac {25}{2}-\frac {21 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {2 a^2 d^4 \left (\left (\frac {25}{2}+\frac {21 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {d}}\right )+\left (\frac {25}{2}-\frac {21 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {2 a^2 d^4 \left (\left (\frac {25}{2}-\frac {21 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )+\left (\frac {25}{2}+\frac {21 i}{2}\right ) \left (\frac {\log \left (d \tan (e+f x)+\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (d \tan (e+f x)-\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\)

Input:

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

Output:

-1/4*(d*(d*Tan[e + f*x])^(5/2))/(f*(a + I*a*Tan[e + f*x])^2) + (((7*I)*d^2 
*(d*Tan[e + f*x])^(3/2))/(f*(1 + I*Tan[e + f*x])) - ((-2*a^2*d^4*((25/2 - 
(21*I)/2)*(-(ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(Sqrt[2]*S 
qrt[d])) + ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(Sqrt[2]*Sqr 
t[d])) + (25/2 + (21*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]))))/f + (50*a^2*d^3*Sqrt 
[d*Tan[e + f*x]])/f)/(2*a^2))/(8*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 4011

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

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 = 1.45 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.47

method	result	size
derivativedivides	\(\frac {2 d^{3} \left (-\sqrt {d \tan \left (f x +e \right )}-\frac {d \left (\frac {\frac {11 i \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2}+\frac {9 d \sqrt {d \tan \left (f x +e \right )}}{2}}{\left (i d \tan \left (f x +e \right )+d \right )^{2}}+\frac {23 i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{2 \sqrt {-i d}}\right )}{8}+\frac {i d \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{8 \sqrt {i d}}\right )}{f \,a^{2}}\)	\(126\)
default	\(\frac {2 d^{3} \left (-\sqrt {d \tan \left (f x +e \right )}-\frac {d \left (\frac {\frac {11 i \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2}+\frac {9 d \sqrt {d \tan \left (f x +e \right )}}{2}}{\left (i d \tan \left (f x +e \right )+d \right )^{2}}+\frac {23 i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{2 \sqrt {-i d}}\right )}{8}+\frac {i d \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{8 \sqrt {i d}}\right )}{f \,a^{2}}\)	\(126\)

Input:

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

Output:

2/f/a^2*d^3*(-(d*tan(f*x+e))^(1/2)-1/8*d*((11/2*I*(d*tan(f*x+e))^(3/2)+9/2 
*d*(d*tan(f*x+e))^(1/2))/(I*d*tan(f*x+e)+d)^2+23/2*I/(-I*d)^(1/2)*arctan(( 
d*tan(f*x+e))^(1/2)/(-I*d)^(1/2)))+1/8*I*d/(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. 573 vs. \(2 (198) = 396\).

Time = 0.10 (sec) , antiderivative size = 573, normalized size of antiderivative = 2.15 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/16*(4*a^2*sqrt(-1/16*I*d^7/(a^4*f^2))*f*e^(4*I*f*x + 4*I*e)*log(-2*(I*d 
^4*e^(2*I*f*x + 2*I*e) + 4*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(-1/16* 
I*d^7/(a^4*f^2))*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e 
) + 1)))*e^(-2*I*f*x - 2*I*e)/d^3) - 4*a^2*sqrt(-1/16*I*d^7/(a^4*f^2))*f*e 
^(4*I*f*x + 4*I*e)*log(-2*(I*d^4*e^(2*I*f*x + 2*I*e) - 4*(a^2*f*e^(2*I*f*x 
 + 2*I*e) + a^2*f)*sqrt(-1/16*I*d^7/(a^4*f^2))*sqrt((-I*d*e^(2*I*f*x + 2*I 
*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/d^3) - 4*a^2*s 
qrt(529/64*I*d^7/(a^4*f^2))*f*e^(4*I*f*x + 4*I*e)*log(1/8*(23*I*d^4 + 8*(a 
^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(529/64*I*d^7/(a^4*f^2))*sqrt((-I*d* 
e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e 
)/(a^2*f)) + 4*a^2*sqrt(529/64*I*d^7/(a^4*f^2))*f*e^(4*I*f*x + 4*I*e)*log( 
1/8*(23*I*d^4 - 8*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(529/64*I*d^7/(a 
^4*f^2))*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))) 
*e^(-2*I*f*x - 2*I*e)/(a^2*f)) + (42*d^3*e^(4*I*f*x + 4*I*e) + 9*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^(-4*I*f*x - 4*I*e)/(a^2*f)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

-Integral((d*tan(e + f*x))**(7/2)/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1) 
, x)/a**2
 

Maxima [F(-2)]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.70 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=\frac {\frac {2 i \, \sqrt {2} d^{\frac {9}{2}} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{\frac {i \, d}{{\left | d \right |}} + 1} - \frac {23 i \, \sqrt {2} d^{\frac {9}{2}} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{-i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{-\frac {i \, d}{{\left | d \right |}} + 1} - 16 \, \sqrt {d \tan \left (f x + e\right )} d^{4} + \frac {11 i \, \sqrt {d \tan \left (f x + e\right )} d^{6} \tan \left (f x + e\right ) + 9 \, \sqrt {d \tan \left (f x + e\right )} d^{6}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2}}}{8 \, a^{2} d f} \] Input:

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

Output:

1/8*(2*I*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) - 23*I*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) - 16*sqrt(d*tan(f*x + e))*d^4 + (11*I*sqrt(d*t 
an(f*x + e))*d^6*tan(f*x + e) + 9*sqrt(d*tan(f*x + e))*d^6)/(d*tan(f*x + e 
) - I*d)^2)/(a^2*d*f)
 

Mupad [B] (verification not implemented)

Time = 2.75 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.76 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {\frac {9\,d^5\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{8\,a^2\,f}+\frac {d^4\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,11{}\mathrm {i}}{8\,a^2\,f}}{-d^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+d^2\,\mathrm {tan}\left (e+f\,x\right )\,2{}\mathrm {i}+d^2}-\frac {2\,d^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{a^2\,f}+\mathrm {atan}\left (\frac {8\,a^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {d^7\,1{}\mathrm {i}}{64\,a^4\,f^2}}}{d^4}\right )\,\sqrt {-\frac {d^7\,1{}\mathrm {i}}{64\,a^4\,f^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {16\,a^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {d^7\,529{}\mathrm {i}}{256\,a^4\,f^2}}}{23\,d^4}\right )\,\sqrt {\frac {d^7\,529{}\mathrm {i}}{256\,a^4\,f^2}}\,2{}\mathrm {i} \] Input:

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

Output:

atan((8*a^2*f*(d*tan(e + f*x))^(1/2)*(-(d^7*1i)/(64*a^4*f^2))^(1/2))/d^4)* 
(-(d^7*1i)/(64*a^4*f^2))^(1/2)*2i - ((9*d^5*(d*tan(e + f*x))^(1/2))/(8*a^2 
*f) + (d^4*(d*tan(e + f*x))^(3/2)*11i)/(8*a^2*f))/(d^2*tan(e + f*x)*2i + d 
^2 - d^2*tan(e + f*x)^2) - atan((16*a^2*f*(d*tan(e + f*x))^(1/2)*((d^7*529 
i)/(256*a^4*f^2))^(1/2))/(23*d^4))*((d^7*529i)/(256*a^4*f^2))^(1/2)*2i - ( 
2*d^3*(d*tan(e + f*x))^(1/2))/(a^2*f)
 

Reduce [F]

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

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

Output:

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