\(\int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx\) [181]

Optimal result

Integrand size = 28, antiderivative size = 273 \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {d^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}-\frac {d^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}+\frac {d^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}+\sqrt {d} \tan (e+f x)}\right )}{8 \sqrt {2} a^3 f}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )} \] Output:

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

Mathematica [A] (verified)

Time = 2.06 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.85 \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {d^3 \sec ^4(e+f x) \left (i-i \cos (4 (e+f x))+6 i \cos (3 (e+f x)) \log \left (\cos (e+f x)+\sin (e+f x)+\sqrt {\sin (2 (e+f x))}\right ) \sqrt {\sin (2 (e+f x))}-6 \sin (2 (e+f x))+6 i \arcsin (\cos (e+f x)-\sin (e+f x)) \sqrt {\sin (2 (e+f x))} (\cos (3 (e+f x))+i \sin (3 (e+f x)))-6 \log \left (\cos (e+f x)+\sin (e+f x)+\sqrt {\sin (2 (e+f x))}\right ) \sqrt {\sin (2 (e+f x))} \sin (3 (e+f x))+3 \sin (4 (e+f x))\right )}{96 a^3 f \sqrt {d \tan (e+f x)} (-i+\tan (e+f x))^3} \] Input:

Integrate[(d*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^3,x]
 

Output:

(d^3*Sec[e + f*x]^4*(I - I*Cos[4*(e + f*x)] + (6*I)*Cos[3*(e + f*x)]*Log[C 
os[e + f*x] + Sin[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sqrt[Sin[2*(e + f*x)] 
] - 6*Sin[2*(e + f*x)] + (6*I)*ArcSin[Cos[e + f*x] - Sin[e + f*x]]*Sqrt[Si 
n[2*(e + f*x)]]*(Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)]) - 6*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)] + 3*Sin[4*(e + f*x)]))/(96*a^3*f*Sqrt[d*Tan[e + f*x]]*(-I + Tan[ 
e + f*x])^3)
 

Rubi [A] (warning: unable to verify)

Time = 1.15 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.12, number of steps used = 22, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3042, 4041, 27, 3042, 4078, 27, 3042, 4079, 27, 2030, 3042, 3957, 266, 826, 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.

\(\displaystyle \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4041

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4078

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4079

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3957

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[(d*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^3,x]
 

Output:

-1/6*(d*(d*Tan[e + f*x])^(3/2))/(f*(a + I*a*Tan[e + f*x])^3) + ((I*a*d^2*S 
qrt[d*Tan[e + f*x]])/(f*(a + I*a*Tan[e + f*x])^2) - ((2*a*d^3*((-(ArcTan[1 
 - Sqrt[2]*Sqrt[d]*Tan[e + f*x]]/(Sqrt[2]*Sqrt[d])) + ArcTan[1 + Sqrt[2]*S 
qrt[d]*Tan[e + f*x]]/(Sqrt[2]*Sqrt[d]))/2 + (Log[d - Sqrt[2]*d^(3/2)*Tan[e 
 + f*x] + d^2*Tan[e + f*x]^2]/(2*Sqrt[2]*Sqrt[d]) - Log[d + Sqrt[2]*d^(3/2 
)*Tan[e + f*x] + d^2*Tan[e + f*x]^2]/(2*Sqrt[2]*Sqrt[d]))/2))/f + ((2*I)*a 
^2*d^2*Sqrt[d*Tan[e + f*x]])/(f*(a + I*a*Tan[e + f*x])))/(2*a^2))/(4*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 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ 
a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] 
 && AtomQ[SplitProduct[SumBaseQ, 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 2030

Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m   Int[(b*v) 
^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
 

rule 3042

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

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Subst[Int 
[x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && 
!IntegerQ[n]
 

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]
 

rule 4079

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*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*( 
b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d))   Int[(a + b*Tan[e + f*x])^(m 
 + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m 
- b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Free 
Q[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] 
 && LtQ[m, 0] &&  !GtQ[n, 0]
 

Maple [A] (verified)

Time = 1.46 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.42

method	result	size
derivativedivides	\(\frac {2 d^{4} \left (-\frac {\frac {2 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}-\frac {2 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}}{\left (d \tan \left (f x +e \right )-i d \right )^{3}}+\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{16 d}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 d \sqrt {i d}}\right )}{f \,a^{3}}\)	\(116\)
default	\(\frac {2 d^{4} \left (-\frac {\frac {2 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}-\frac {2 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}}{\left (d \tan \left (f x +e \right )-i d \right )^{3}}+\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{16 d}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 d \sqrt {i d}}\right )}{f \,a^{3}}\)	\(116\)

Input:

int((d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^3,x,method=_RETURNVERBOSE)
 

Output:

2/f/a^3*d^4*(-1/16/d*((2*(d*tan(f*x+e))^(5/2)-2/3*I*d*(d*tan(f*x+e))^(3/2) 
)/(d*tan(f*x+e)-I*d)^3+1/(-I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d)^( 
1/2)))-1/16/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. 587 vs. \(2 (213) = 426\).

Time = 0.09 (sec) , antiderivative size = 587, normalized size of antiderivative = 2.15 \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx =\text {Too large to display} \] Input:

integrate((d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="fricas")
 

Output:

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

\[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {i \int \frac {\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\, dx}{a^{3}} \] Input:

integrate((d*tan(f*x+e))**(5/2)/(a+I*a*tan(f*x+e))**3,x)
 

Output:

I*Integral((d*tan(e + f*x))**(5/2)/(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. \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((d*tan(f*x+e))^(5/2)/(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 = 180, normalized size of antiderivative = 0.66 \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {i \, d^{2} {\left (-\frac {3 i \, \sqrt {2} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{\frac {i \, d}{{\left | d \right |}} + 1} - \frac {3 i \, \sqrt {2} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{-i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{-\frac {i \, d}{{\left | d \right |}} + 1} - \frac {2 \, {\left (3 i \, \sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right )^{2} + \sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right )\right )}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3}}\right )}}{24 \, a^{3} f} \] Input:

integrate((d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="giac")
 

Output:

-1/24*I*d^2*(-3*I*sqrt(2)*sqrt(d)*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) - 3*I*sqrt(2)* 
sqrt(d)*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) - 2*(3*I*sqrt(d*tan(f*x + e))*d^3*tan( 
f*x + e)^2 + sqrt(d*tan(f*x + e))*d^3*tan(f*x + e))/(d*tan(f*x + e) - I*d) 
^3)/(a^3*f)
 

Mupad [B] (verification not implemented)

Time = 1.11 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.58 \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {{\left (-1\right )}^{1/4}\,d^{5/2}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,f}-\frac {{\left (-1\right )}^{1/4}\,d^{5/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,f}-\frac {-\frac {d^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{4\,a^3\,f}+\frac {d^4\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,1{}\mathrm {i}}{12\,a^3\,f}}{-d^3\,{\mathrm {tan}\left (e+f\,x\right )}^3+d^3\,{\mathrm {tan}\left (e+f\,x\right )}^2\,3{}\mathrm {i}+3\,d^3\,\mathrm {tan}\left (e+f\,x\right )-d^3\,1{}\mathrm {i}} \] Input:

int((d*tan(e + f*x))^(5/2)/(a + a*tan(e + f*x)*1i)^3,x)
 

Output:

((-1)^(1/4)*d^(5/2)*atanh(((-1)^(1/4)*(d*tan(e + f*x))^(1/2))/d^(1/2)))/(8 
*a^3*f) - ((-1)^(1/4)*d^(5/2)*atan(((-1)^(1/4)*(d*tan(e + f*x))^(1/2))/d^( 
1/2)))/(8*a^3*f) - ((d^4*(d*tan(e + f*x))^(3/2)*1i)/(12*a^3*f) - (d^3*(d*t 
an(e + f*x))^(5/2))/(4*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]

\[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {\sqrt {d}\, \left (\int \frac {\sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{3} i +3 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i -1}d x \right ) d^{2}}{a^{3}} \] Input:

int((d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^3,x)
 

Output:

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