\(\int \frac {(d \tan (e+f x))^{7/2}}{a+i a \tan (e+f x)} \, dx\) [164]

Optimal result

Integrand size = 28, antiderivative size = 252 \[ \int \frac {(d \tan (e+f x))^{7/2}}{a+i a \tan (e+f x)} \, dx=\frac {\left (\frac {5}{4}-\frac {7 i}{4}\right ) d^{7/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a f}-\frac {\left (\frac {5}{4}-\frac {7 i}{4}\right ) d^{7/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a f}-\frac {\left (\frac {5}{4}+\frac {7 i}{4}\right ) d^{7/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}+\sqrt {d} \tan (e+f x)}\right )}{\sqrt {2} a f}+\frac {5 d^3 \sqrt {d \tan (e+f x)}}{2 a f}-\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{6 a f}-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))} \] Output:

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

Mathematica [A] (verified)

Time = 1.65 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.54 \[ \int \frac {(d \tan (e+f x))^{7/2}}{a+i a \tan (e+f x)} \, dx=\frac {d^3 \left (-3 \sqrt [4]{-1} \sqrt {d} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )+18 \sqrt [4]{-1} \sqrt {d} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )+\frac {\sqrt {d \tan (e+f x)} \left (-15 i+8 \tan (e+f x)-4 i \tan ^2(e+f x)\right )}{-i+\tan (e+f x)}\right )}{6 a f} \] Input:

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

Output:

(d^3*(-3*(-1)^(1/4)*Sqrt[d]*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[ 
d]] + 18*(-1)^(1/4)*Sqrt[d]*ArcTanh[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt 
[d]] + (Sqrt[d*Tan[e + f*x]]*(-15*I + 8*Tan[e + f*x] - (4*I)*Tan[e + f*x]^ 
2))/(-I + Tan[e + f*x])))/(6*a*f)
 

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.15, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.679, Rules used = {3042, 4033, 27, 3042, 4011, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4033

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

\(\displaystyle \frac {\int \sqrt {d \tan (e+f x)} \left (7 i a d^3+5 a \tan (e+f x) d^3\right )dx-\frac {14 i a d^2 (d \tan (e+f x))^{3/2}}{3 f}}{4 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \sqrt {d \tan (e+f x)} \left (7 i a d^3+5 a \tan (e+f x) d^3\right )dx-\frac {14 i a d^2 (d \tan (e+f x))^{3/2}}{3 f}}{4 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {\int \frac {7 i a d^4 \tan (e+f x)-5 a d^4}{\sqrt {d \tan (e+f x)}}dx+\frac {10 a d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {14 i a d^2 (d \tan (e+f x))^{3/2}}{3 f}}{4 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {7 i a d^4 \tan (e+f x)-5 a d^4}{\sqrt {d \tan (e+f x)}}dx+\frac {10 a d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {14 i a d^2 (d \tan (e+f x))^{3/2}}{3 f}}{4 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))}\)

\(\Big \downarrow \) 4017

\(\displaystyle \frac {\frac {2 \int -\frac {a d^4 (5 d-7 i d \tan (e+f x))}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{f}+\frac {10 a d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {14 i a d^2 (d \tan (e+f x))^{3/2}}{3 f}}{4 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-\frac {2 \int \frac {a d^4 (5 d-7 i d \tan (e+f x))}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{f}+\frac {10 a d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {14 i a d^2 (d \tan (e+f x))^{3/2}}{3 f}}{4 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {2 a d^4 \int \frac {5 d-7 i d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{f}+\frac {10 a d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {14 i a d^2 (d \tan (e+f x))^{3/2}}{3 f}}{4 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))}\)

\(\Big \downarrow \) 1482

\(\displaystyle \frac {-\frac {2 a d^4 \left (\left (\frac {5}{2}+\frac {7 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {5}{2}-\frac {7 i}{2}\right ) \int \frac {\tan (e+f x) d+d}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}+\frac {10 a d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {14 i a d^2 (d \tan (e+f x))^{3/2}}{3 f}}{4 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {-\frac {2 a d^4 \left (\left (\frac {5}{2}+\frac {7 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {5}{2}-\frac {7 i}{2}\right ) \left (\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}+\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}\right )\right )}{f}+\frac {10 a d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {14 i a d^2 (d \tan (e+f x))^{3/2}}{3 f}}{4 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {-\frac {2 a d^4 \left (\left (\frac {5}{2}+\frac {7 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {5}{2}-\frac {7 i}{2}\right ) \left (\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}+\frac {10 a d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {14 i a d^2 (d \tan (e+f x))^{3/2}}{3 f}}{4 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {-\frac {2 a d^4 \left (\left (\frac {5}{2}+\frac {7 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {5}{2}-\frac {7 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}+\frac {10 a d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {14 i a d^2 (d \tan (e+f x))^{3/2}}{3 f}}{4 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {-\frac {2 a d^4 \left (\left (\frac {5}{2}+\frac {7 i}{2}\right ) \left (-\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )+\left (\frac {5}{2}-\frac {7 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}+\frac {10 a d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {14 i a d^2 (d \tan (e+f x))^{3/2}}{3 f}}{4 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-\frac {2 a d^4 \left (\left (\frac {5}{2}+\frac {7 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )+\left (\frac {5}{2}-\frac {7 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}+\frac {10 a d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {14 i a d^2 (d \tan (e+f x))^{3/2}}{3 f}}{4 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {2 a d^4 \left (\left (\frac {5}{2}+\frac {7 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {d}}\right )+\left (\frac {5}{2}-\frac {7 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}+\frac {10 a d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {14 i a d^2 (d \tan (e+f x))^{3/2}}{3 f}}{4 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {-\frac {2 a d^4 \left (\left (\frac {5}{2}-\frac {7 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )+\left (\frac {5}{2}+\frac {7 i}{2}\right ) \left (\frac {\log \left (d \tan (e+f x)+\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (d \tan (e+f x)-\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}+\frac {10 a d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {14 i a d^2 (d \tan (e+f x))^{3/2}}{3 f}}{4 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))}\)

Input:

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

Output:

-1/2*(d*(d*Tan[e + f*x])^(5/2))/(f*(a + I*a*Tan[e + f*x])) + ((-2*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 
]*Sqrt[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]))))/f + (10*a*d^3*Sqr 
t[d*Tan[e + f*x]])/f - (((14*I)/3)*a*d^2*(d*Tan[e + f*x])^(3/2))/f)/(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 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 4033

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

Maple [A] (verified)

Time = 1.35 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.50

method	result	size
derivativedivides	\(\frac {2 d^{2} \left (-\frac {i \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+d \sqrt {d \tan \left (f x +e \right )}+\frac {i d^{2} \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{4 \sqrt {i d}}+\frac {d^{2} \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{i d \tan \left (f x +e \right )+d}+\frac {6 i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}\right )}{4}\right )}{f a}\)	\(126\)
default	\(\frac {2 d^{2} \left (-\frac {i \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+d \sqrt {d \tan \left (f x +e \right )}+\frac {i d^{2} \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{4 \sqrt {i d}}+\frac {d^{2} \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{i d \tan \left (f x +e \right )+d}+\frac {6 i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}\right )}{4}\right )}{f a}\)	\(126\)

Input:

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

Output:

2/f/a*d^2*(-1/3*I*(d*tan(f*x+e))^(3/2)+d*(d*tan(f*x+e))^(1/2)+1/4*I*d^2/(I 
*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(I*d)^(1/2))+1/4*d^2*((d*tan(f*x+e)) 
^(1/2)/(I*d*tan(f*x+e)+d)+6*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. 614 vs. \(2 (186) = 372\).

Time = 0.10 (sec) , antiderivative size = 614, normalized size of antiderivative = 2.44 \[ \int \frac {(d \tan (e+f x))^{7/2}}{a+i a \tan (e+f x)} \, dx =\text {Too large to display} \] Input:

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

Output:

1/12*(3*sqrt(9*I*d^7/(a^2*f^2))*(a*f*e^(4*I*f*x + 4*I*e) + a*f*e^(2*I*f*x 
+ 2*I*e))*log((-3*I*d^4 + sqrt(9*I*d^7/(a^2*f^2))*(a*f*e^(2*I*f*x + 2*I*e) 
 + a*f)*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*f)) - 3*sqrt(9*I*d^7/(a^2*f^2))*(a*f*e^(4*I*f*x + 
4*I*e) + a*f*e^(2*I*f*x + 2*I*e))*log((-3*I*d^4 - sqrt(9*I*d^7/(a^2*f^2))* 
(a*f*e^(2*I*f*x + 2*I*e) + a*f)*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*f)) - 3*sqrt(-1/4*I*d^7/(a 
^2*f^2))*(a*f*e^(4*I*f*x + 4*I*e) + a*f*e^(2*I*f*x + 2*I*e))*log(-2*(I*d^4 
*e^(2*I*f*x + 2*I*e) + 2*sqrt(-1/4*I*d^7/(a^2*f^2))*(a*f*e^(2*I*f*x + 2*I* 
e) + a*f)*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) + 3*sqrt(-1/4*I*d^7/(a^2*f^2))*(a*f*e^(4*I*f*x 
 + 4*I*e) + a*f*e^(2*I*f*x + 2*I*e))*log(-2*(I*d^4*e^(2*I*f*x + 2*I*e) - 2 
*sqrt(-1/4*I*d^7/(a^2*f^2))*(a*f*e^(2*I*f*x + 2*I*e) + a*f)*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) + (19*d^3*e^(4*I*f*x + 4*I*e) + 38*d^3*e^(2*I*f*x + 2*I*e) + 3*d^3)*sq 
rt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/(a*f*e^(4* 
I*f*x + 4*I*e) + a*f*e^(2*I*f*x + 2*I*e))
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.73 \[ \int \frac {(d \tan (e+f x))^{7/2}}{a+i a \tan (e+f x)} \, dx=-\frac {i \, {\left (4 \, \sqrt {d \tan \left (f x + e\right )} d^{4} \tan \left (f x + e\right ) - \frac {3 \, \sqrt {2} d^{\frac {9}{2}} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{\frac {i \, d}{{\left | d \right |}} + 1} - \frac {18 \, \sqrt {2} d^{\frac {9}{2}} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{-i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{-\frac {i \, d}{{\left | d \right |}} + 1} + 12 i \, \sqrt {d \tan \left (f x + e\right )} d^{4} + \frac {3 \, \sqrt {d \tan \left (f x + e\right )} d^{5}}{d \tan \left (f x + e\right ) - i \, d}\right )}}{6 \, a d f} \] Input:

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

Output:

-1/6*I*(4*sqrt(d*tan(f*x + e))*d^4*tan(f*x + e) - 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)*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) + 12* 
I*sqrt(d*tan(f*x + e))*d^4 + 3*sqrt(d*tan(f*x + e))*d^5/(d*tan(f*x + e) - 
I*d))/(a*d*f)
 

Mupad [B] (verification not implemented)

Time = 3.16 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.71 \[ \int \frac {(d \tan (e+f x))^{7/2}}{a+i a \tan (e+f x)} \, dx=\mathrm {atan}\left (\frac {2\,a\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {d^7\,9{}\mathrm {i}}{4\,a^2\,f^2}}}{3\,d^4}\right )\,\sqrt {\frac {d^7\,9{}\mathrm {i}}{4\,a^2\,f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {4\,a\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {d^7\,1{}\mathrm {i}}{16\,a^2\,f^2}}}{d^4}\right )\,\sqrt {-\frac {d^7\,1{}\mathrm {i}}{16\,a^2\,f^2}}\,2{}\mathrm {i}+\frac {2\,d^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{a\,f}-\frac {d^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,2{}\mathrm {i}}{3\,a\,f}+\frac {d^4\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{2\,a\,f\,\left (-d\,\mathrm {tan}\left (e+f\,x\right )+d\,1{}\mathrm {i}\right )} \] Input:

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

Output:

atan((2*a*f*(d*tan(e + f*x))^(1/2)*((d^7*9i)/(4*a^2*f^2))^(1/2))/(3*d^4))* 
((d^7*9i)/(4*a^2*f^2))^(1/2)*2i + atan((4*a*f*(d*tan(e + f*x))^(1/2)*(-(d^ 
7*1i)/(16*a^2*f^2))^(1/2))/d^4)*(-(d^7*1i)/(16*a^2*f^2))^(1/2)*2i + (2*d^3 
*(d*tan(e + f*x))^(1/2))/(a*f) - (d^2*(d*tan(e + f*x))^(3/2)*2i)/(3*a*f) + 
 (d^4*(d*tan(e + f*x))^(1/2)*1i)/(2*a*f*(d*1i - d*tan(e + f*x)))
 

Reduce [F]

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

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

Output:

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