\(\int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) [201]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [B] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [F(-2)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 28, antiderivative size = 258 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {363 (-1)^{3/4} a^{5/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{64 d}+\frac {(4+4 i) a^{5/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {149 i a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{64 d}+\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{96 d}+\frac {17 i a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 d}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d} \] Output:

363/64*(-1)^(3/4)*a^(5/2)*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I* 
a*tan(d*x+c))^(1/2))/d+(4+4*I)*a^(5/2)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1 
/2)/(a+I*a*tan(d*x+c))^(1/2))/d-149/64*I*a^2*tan(d*x+c)^(1/2)*(a+I*a*tan(d 
*x+c))^(1/2)/d+107/96*a^2*tan(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(1/2)/d+17/2 
4*I*a^2*tan(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^(1/2)/d-1/4*a^2*tan(d*x+c)^(7/ 
2)*(a+I*a*tan(d*x+c))^(1/2)/d
 

Mathematica [A] (verified)

Time = 5.03 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.19 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {a^2 \left (129 \sqrt [4]{-1} a \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) (1+i \tan (c+d x)) \sqrt {\tan (c+d x)}+960 \sqrt {a} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) (1+i \tan (c+d x)) \sqrt {i a \tan (c+d x)}+\sqrt {1+i \tan (c+d x)} \left (-768 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}+a \tan (c+d x) \left (447 i-661 \tan (c+d x)-350 i \tan ^2(c+d x)+184 \tan ^3(c+d x)+48 i \tan ^4(c+d x)\right )\right )\right )}{192 d \sqrt {1+i \tan (c+d x)} \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \] Input:

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

Output:

-1/192*(a^2*(129*(-1)^(1/4)*a*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x]]]*(1 + 
I*Tan[c + d*x])*Sqrt[Tan[c + d*x]] + 960*Sqrt[a]*ArcSinh[Sqrt[I*a*Tan[c + 
d*x]]/Sqrt[a]]*(1 + I*Tan[c + d*x])*Sqrt[I*a*Tan[c + d*x]] + Sqrt[1 + I*Ta 
n[c + d*x]]*(-768*Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a 
+ I*a*Tan[c + d*x]]]*Sqrt[I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]] + a 
*Tan[c + d*x]*(447*I - 661*Tan[c + d*x] - (350*I)*Tan[c + d*x]^2 + 184*Tan 
[c + d*x]^3 + (48*I)*Tan[c + d*x]^4))))/(d*Sqrt[1 + I*Tan[c + d*x]]*Sqrt[T 
an[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 1.75 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.09, number of steps used = 21, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 4039, 27, 3042, 4080, 27, 3042, 4080, 27, 3042, 4080, 27, 3042, 4084, 3042, 4027, 218, 4082, 65, 216}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (c+d x)^{5/2} (a+i a \tan (c+d x))^{5/2}dx\)

\(\Big \downarrow \) 4039

\(\displaystyle \frac {1}{4} a \int \frac {1}{2} \tan ^{\frac {5}{2}}(c+d x) \sqrt {i \tan (c+d x) a+a} (17 i \tan (c+d x) a+15 a)dx-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{8} a \int \tan ^{\frac {5}{2}}(c+d x) \sqrt {i \tan (c+d x) a+a} (17 i \tan (c+d x) a+15 a)dx-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{8} a \int \tan (c+d x)^{5/2} \sqrt {i \tan (c+d x) a+a} (17 i \tan (c+d x) a+15 a)dx-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}\)

\(\Big \downarrow \) 4080

\(\displaystyle \frac {1}{8} a \left (\frac {\int -\frac {1}{2} \tan ^{\frac {3}{2}}(c+d x) \sqrt {i \tan (c+d x) a+a} \left (85 i a^2-107 a^2 \tan (c+d x)\right )dx}{3 a}+\frac {17 i a \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{8} a \left (\frac {17 i a \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {\int \tan ^{\frac {3}{2}}(c+d x) \sqrt {i \tan (c+d x) a+a} \left (85 i a^2-107 a^2 \tan (c+d x)\right )dx}{6 a}\right )-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{8} a \left (\frac {17 i a \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {\int \tan (c+d x)^{3/2} \sqrt {i \tan (c+d x) a+a} \left (85 i a^2-107 a^2 \tan (c+d x)\right )dx}{6 a}\right )-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}\)

\(\Big \downarrow \) 4080

\(\displaystyle \frac {1}{8} a \left (\frac {17 i a \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {\frac {\int \frac {3}{2} \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} \left (149 i \tan (c+d x) a^3+107 a^3\right )dx}{2 a}-\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}\right )-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{8} a \left (\frac {17 i a \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {\frac {3 \int \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} \left (149 i \tan (c+d x) a^3+107 a^3\right )dx}{4 a}-\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}\right )-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{8} a \left (\frac {17 i a \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {\frac {3 \int \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} \left (149 i \tan (c+d x) a^3+107 a^3\right )dx}{4 a}-\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}\right )-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}\)

\(\Big \downarrow \) 4080

\(\displaystyle \frac {1}{8} a \left (\frac {17 i a \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {\frac {3 \left (\frac {\int -\frac {\sqrt {i \tan (c+d x) a+a} \left (149 i a^4-363 a^4 \tan (c+d x)\right )}{2 \sqrt {\tan (c+d x)}}dx}{a}+\frac {149 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}\right )-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{8} a \left (\frac {17 i a \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {\frac {3 \left (\frac {149 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (149 i a^4-363 a^4 \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}}dx}{2 a}\right )}{4 a}-\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}\right )-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{8} a \left (\frac {17 i a \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {\frac {3 \left (\frac {149 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (149 i a^4-363 a^4 \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}}dx}{2 a}\right )}{4 a}-\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}\right )-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}\)

\(\Big \downarrow \) 4084

\(\displaystyle \frac {1}{8} a \left (\frac {17 i a \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {\frac {3 \left (\frac {149 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {512 i a^4 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-363 i a^3 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a}\right )}{4 a}-\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}\right )-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{8} a \left (\frac {17 i a \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {\frac {3 \left (\frac {149 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {512 i a^4 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-363 i a^3 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a}\right )}{4 a}-\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}\right )-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}\)

\(\Big \downarrow \) 4027

\(\displaystyle \frac {1}{8} a \left (\frac {17 i a \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {\frac {3 \left (\frac {149 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\frac {1024 a^6 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-363 i a^3 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a}\right )}{4 a}-\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}\right )-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {1}{8} a \left (\frac {17 i a \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {\frac {3 \left (\frac {149 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\frac {(512+512 i) a^{9/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-363 i a^3 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a}\right )}{4 a}-\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}\right )-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}\)

\(\Big \downarrow \) 4082

\(\displaystyle \frac {1}{8} a \left (\frac {17 i a \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {\frac {3 \left (\frac {149 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\frac {(512+512 i) a^{9/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {363 i a^5 \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}}{2 a}\right )}{4 a}-\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}\right )-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}\)

\(\Big \downarrow \) 65

\(\displaystyle \frac {1}{8} a \left (\frac {17 i a \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {\frac {3 \left (\frac {149 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\frac {(512+512 i) a^{9/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {726 i a^5 \int \frac {1}{1-\frac {i a \tan (c+d x)}{i \tan (c+d x) a+a}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}}{2 a}\right )}{4 a}-\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}\right )-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{8} a \left (\frac {17 i a \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {\frac {3 \left (\frac {149 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\frac {726 (-1)^{3/4} a^{9/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {(512+512 i) a^{9/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{2 a}\right )}{4 a}-\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}\right )-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}\)

Input:

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

Output:

-1/4*(a^2*Tan[c + d*x]^(7/2)*Sqrt[a + I*a*Tan[c + d*x]])/d + (a*((((17*I)/ 
3)*a*Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]])/d - ((-107*a^2*Tan[c + 
 d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(2*d) + (3*(-1/2*((726*(-1)^(3/4)* 
a^(9/2)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c 
+ d*x]]])/d + ((512 + 512*I)*a^(9/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + 
 d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d)/a + ((149*I)*a^3*Sqrt[Tan[c + d*x] 
]*Sqrt[a + I*a*Tan[c + d*x]])/d))/(4*a))/(6*a)))/8
 

Defintions of rubi rules used

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 65
Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2   Sub 
st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d 
}, x] &&  !GtQ[c, 0]
 

rule 216
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 218
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4027
Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) 
 + (f_.)*(x_)]], x_Symbol] :> Simp[-2*a*(b/f)   Subst[Int[1/(a*c - b*d - 2* 
a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x] /; 
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && N 
eQ[c^2 + d^2, 0]
 

rule 4039
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + 
 d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[a/(d*(m + n - 1)) 
Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + 
a*d*(m + 2*n) + (a*c*(m - 2) + b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x 
] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 
 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] 
 && (IntegerQ[m] || IntegersQ[2*m, 2*n])
 

rule 4080
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[B*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(f*(m + n))), x] + Simp[ 
1/(a*(m + n))   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Sim 
p[a*A*c*(m + n) - B*(b*c*m + a*d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*T 
an[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c 
 - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[n, 0]
 

rule 4082
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[b*(B/f)   Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x]], 
x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[ 
a^2 + b^2, 0] && EqQ[A*b + a*B, 0]
 

rule 4084
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)/b   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x], x] 
 - Simp[B/b   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[ 
e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - 
a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]
 
Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 491 vs. \(2 (204 ) = 408\).

Time = 1.60 (sec) , antiderivative size = 492, normalized size of antiderivative = 1.91

method result size
derivativedivides \(-\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (-272 i \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \tan \left (d x +c \right )^{2}+96 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{3}+384 i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a +1089 i \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a +894 i \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+384 \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a -428 \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \tan \left (d x +c \right )+1536 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}\right )}{384 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \sqrt {i a}}\) \(492\)
default \(-\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (-272 i \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \tan \left (d x +c \right )^{2}+96 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{3}+384 i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a +1089 i \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a +894 i \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+384 \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a -428 \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \tan \left (d x +c \right )+1536 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}\right )}{384 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \sqrt {i a}}\) \(492\)

Input:

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

Output:

-1/384/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*a^2*(-272*I*tan(d*x+c 
)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)+96*(a*t 
an(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3+38 
4*I*2^(1/2)*(I*a)^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan( 
d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*a+1089*I*a*ln(1/2*(2*I* 
a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^ 
(1/2))*(-I*a)^(1/2)+894*I*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2 
)*(-I*a)^(1/2)+384*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*t 
an(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*(I*a)^(1/2)*a-428*(- 
I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*tan(d*x+c)+15 
36*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^ 
(1/2)+a)/(I*a)^(1/2))*a*(-I*a)^(1/2))/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2 
)/(-I*a)^(1/2)/(I*a)^(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. 769 vs. \(2 (192) = 384\).

Time = 0.10 (sec) , antiderivative size = 769, normalized size of antiderivative = 2.98 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/192*(sqrt(2)*(-845*I*a^2*e^(7*I*d*x + 7*I*c) - 1275*I*a^2*e^(5*I*d*x + 5 
*I*c) - 1135*I*a^2*e^(3*I*d*x + 3*I*c) - 321*I*a^2*e^(I*d*x + I*c))*sqrt(a 
/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x 
+ 2*I*c) + 1)) + 96*sqrt(131769/4096*I*a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3 
*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*log(1/363*(363*sqrt( 
2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt( 
(-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) + 128*I*sqrt(13176 
9/4096*I*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a^2) - 96*sqrt(13176 
9/4096*I*a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e 
^(2*I*d*x + 2*I*c) + d)*log(1/363*(363*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + 
a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/( 
e^(2*I*d*x + 2*I*c) + 1)) - 128*I*sqrt(131769/4096*I*a^5/d^2)*d*e^(I*d*x + 
 I*c))*e^(-I*d*x - I*c)/a^2) - 96*sqrt(32*I*a^5/d^2)*(d*e^(6*I*d*x + 6*I*c 
) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*log(1/4*(4*sqrt 
(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt 
((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) + I*sqrt(32*I*a^5 
/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a^2) + 96*sqrt(32*I*a^5/d^2)*(d* 
e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + 
d)*log(1/4*(4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 
 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + ...
 

Sympy [F(-1)]

Timed out. \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:

integrate(tan(d*x+c)**(5/2)*(a+I*a*tan(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

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

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

Output:

integrate((I*a*tan(d*x + c) + a)^(5/2)*tan(d*x + c)^(5/2), x)
 

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeDone
 

Mupad [F(-1)]

Timed out. \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\sqrt {a}\, a^{2} \left (-\left (\int \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{4}d x \right )+2 \left (\int \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{3}d x \right ) i +\int \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}d x \right ) \] Input:

int(tan(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^(5/2),x)
 

Output:

sqrt(a)*a**2*( - int(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d 
*x)**4,x) + 2*int(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x) 
**3,x)*i + int(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 
,x))