\(\int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {11}{2}}(c+d x)} \, dx\) [209]

Optimal result

Integrand size = 28, antiderivative size = 239 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {11}{2}}(c+d x)} \, dx=\frac {(4+4 i) a^{5/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {38 i a^2 \sqrt {a+i a \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {92 a^2 \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {472 i a^2 \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {1576 a^2 \sqrt {a+i a \tan (c+d x)}}{315 d \sqrt {\tan (c+d x)}} \] Output:

(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-2/9*a^2*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(9/2)-38/63*I*a^2*( 
a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(7/2)+92/105*a^2*(a+I*a*tan(d*x+c))^( 
1/2)/d/tan(d*x+c)^(5/2)+472/315*I*a^2*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c 
)^(3/2)-1576/315*a^2*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 4.83 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.87 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {11}{2}}(c+d x)} \, dx=\frac {a^2 \left (\frac {10080 e^{-i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \text {arctanh}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )}{\sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}-\csc ^5(c+d x) (1650 \cos (c+d x)-2051 \cos (3 (c+d x))+961 \cos (5 (c+d x))-282 i \sin (c+d x)+49 i \sin (3 (c+d x))+331 i \sin (5 (c+d x))) \sqrt {\tan (c+d x)}\right ) \sqrt {a+i a \tan (c+d x)}}{2520 d} \] Input:

Integrate[(a + I*a*Tan[c + d*x])^(5/2)/Tan[c + d*x]^(11/2),x]
 

Output:

(a^2*((10080*Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[E^(I*(c + d*x))/Sqrt[- 
1 + E^((2*I)*(c + d*x))]])/(E^(I*(c + d*x))*Sqrt[((-I)*(-1 + E^((2*I)*(c + 
 d*x))))/(1 + E^((2*I)*(c + d*x)))]) - Csc[c + d*x]^5*(1650*Cos[c + d*x] - 
 2051*Cos[3*(c + d*x)] + 961*Cos[5*(c + d*x)] - (282*I)*Sin[c + d*x] + (49 
*I)*Sin[3*(c + d*x)] + (331*I)*Sin[5*(c + d*x)])*Sqrt[Tan[c + d*x]])*Sqrt[ 
a + I*a*Tan[c + d*x]])/(2520*d)
 

Rubi [A] (verified)

Time = 1.50 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.11, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.607, Rules used = {3042, 4036, 27, 3042, 4081, 27, 3042, 4081, 3042, 4081, 27, 3042, 4081, 27, 3042, 4027, 218}

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.

\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {11}{2}}(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan (c+d x)^{11/2}}dx\)

\(\Big \downarrow \) 4036

\(\displaystyle -\frac {2}{9} \int -\frac {\sqrt {i \tan (c+d x) a+a} \left (19 i a^2-17 a^2 \tan (c+d x)\right )}{2 \tan ^{\frac {9}{2}}(c+d x)}dx-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{9} \int \frac {\sqrt {i \tan (c+d x) a+a} \left (19 i a^2-17 a^2 \tan (c+d x)\right )}{\tan ^{\frac {9}{2}}(c+d x)}dx-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \int \frac {\sqrt {i \tan (c+d x) a+a} \left (19 i a^2-17 a^2 \tan (c+d x)\right )}{\tan (c+d x)^{9/2}}dx-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 4081

\(\displaystyle \frac {1}{9} \left (\frac {2 \int -\frac {3 \sqrt {i \tan (c+d x) a+a} \left (19 i \tan (c+d x) a^3+23 a^3\right )}{\tan ^{\frac {7}{2}}(c+d x)}dx}{7 a}-\frac {38 i a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{9} \left (-\frac {6 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (19 i \tan (c+d x) a^3+23 a^3\right )}{\tan ^{\frac {7}{2}}(c+d x)}dx}{7 a}-\frac {38 i a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \left (-\frac {6 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (19 i \tan (c+d x) a^3+23 a^3\right )}{\tan (c+d x)^{7/2}}dx}{7 a}-\frac {38 i a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 4081

\(\displaystyle \frac {1}{9} \left (-\frac {6 \left (\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (59 i a^4-46 a^4 \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)}dx}{5 a}-\frac {46 a^3 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{7 a}-\frac {38 i a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \left (-\frac {6 \left (\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (59 i a^4-46 a^4 \tan (c+d x)\right )}{\tan (c+d x)^{5/2}}dx}{5 a}-\frac {46 a^3 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{7 a}-\frac {38 i a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 4081

\(\displaystyle \frac {1}{9} \left (-\frac {6 \left (\frac {2 \left (\frac {2 \int -\frac {\sqrt {i \tan (c+d x) a+a} \left (118 i \tan (c+d x) a^5+197 a^5\right )}{2 \tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {118 i a^4 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{5 a}-\frac {46 a^3 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{7 a}-\frac {38 i a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{9} \left (-\frac {6 \left (\frac {2 \left (-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (118 i \tan (c+d x) a^5+197 a^5\right )}{\tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {118 i a^4 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{5 a}-\frac {46 a^3 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{7 a}-\frac {38 i a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \left (-\frac {6 \left (\frac {2 \left (-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (118 i \tan (c+d x) a^5+197 a^5\right )}{\tan (c+d x)^{3/2}}dx}{3 a}-\frac {118 i a^4 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{5 a}-\frac {46 a^3 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{7 a}-\frac {38 i a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 4081

\(\displaystyle \frac {1}{9} \left (-\frac {6 \left (\frac {2 \left (-\frac {\frac {2 \int \frac {315 i a^6 \sqrt {i \tan (c+d x) a+a}}{2 \sqrt {\tan (c+d x)}}dx}{a}-\frac {394 a^5 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {118 i a^4 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{5 a}-\frac {46 a^3 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{7 a}-\frac {38 i a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{9} \left (-\frac {6 \left (\frac {2 \left (-\frac {315 i a^5 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {394 a^5 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {118 i a^4 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{5 a}-\frac {46 a^3 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{7 a}-\frac {38 i a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \left (-\frac {6 \left (\frac {2 \left (-\frac {315 i a^5 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {394 a^5 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {118 i a^4 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{5 a}-\frac {46 a^3 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{7 a}-\frac {38 i a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 4027

\(\displaystyle \frac {1}{9} \left (-\frac {6 \left (\frac {2 \left (-\frac {\frac {630 a^7 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {394 a^5 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {118 i a^4 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{5 a}-\frac {46 a^3 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{7 a}-\frac {38 i a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {1}{9} \left (-\frac {38 i a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 \left (\frac {2 \left (-\frac {118 i a^4 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {\frac {(315+315 i) a^{11/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {394 a^5 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}\right )}{5 a}-\frac {46 a^3 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{7 a}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\)

Input:

Int[(a + I*a*Tan[c + d*x])^(5/2)/Tan[c + d*x]^(11/2),x]
 

Output:

(-2*a^2*Sqrt[a + I*a*Tan[c + d*x]])/(9*d*Tan[c + d*x]^(9/2)) + ((((-38*I)/ 
7)*a^2*Sqrt[a + I*a*Tan[c + d*x]])/(d*Tan[c + d*x]^(7/2)) - (6*((-46*a^3*S 
qrt[a + I*a*Tan[c + d*x]])/(5*d*Tan[c + d*x]^(5/2)) + (2*((((-118*I)/3)*a^ 
4*Sqrt[a + I*a*Tan[c + d*x]])/(d*Tan[c + d*x]^(3/2)) - (((315 + 315*I)*a^( 
11/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d* 
x]]])/d - (394*a^5*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]))/(3* 
a)))/(5*a)))/(7*a))/9
 

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 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 4036

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-a^2)*(b*c - a*d)*(a + b*Tan[e + f*x] 
)^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x] + Si 
mp[a/(d*(b*c + a*d)*(n + 1))   Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[ 
e + f*x])^(n + 1)*Simp[b*(b*c*(m - 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) 
 + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[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[m, 1] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
 

rule 4081

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 500 vs. \(2 (193 ) = 386\).

Time = 1.61 (sec) , antiderivative size = 501, normalized size of antiderivative = 2.10

method	result	size
derivativedivides	\(-\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (315 i \sqrt {i a}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, a \tan \left (d x +c \right )^{5}+315 \sqrt {i a}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, a \tan \left (d x +c \right )^{5}+1260 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )^{5}-472 i \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{3} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+1576 \tan \left (d x +c \right )^{4} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+190 i \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-276 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )^{2}+70 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{315 d \tan \left (d x +c \right )^{\frac {9}{2}} \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}}\)	\(501\)
default	\(-\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (315 i \sqrt {i a}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, a \tan \left (d x +c \right )^{5}+315 \sqrt {i a}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, a \tan \left (d x +c \right )^{5}+1260 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )^{5}-472 i \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{3} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+1576 \tan \left (d x +c \right )^{4} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+190 i \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-276 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )^{2}+70 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{315 d \tan \left (d x +c \right )^{\frac {9}{2}} \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}}\)	\(501\)

Input:

int((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(11/2),x,method=_RETURNVERBOSE)
 

Output:

-1/315/d*(a*(1+I*tan(d*x+c)))^(1/2)*a^2*(315*I*(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))/(t 
an(d*x+c)+I))*2^(1/2)*a*tan(d*x+c)^5+315*(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))*2^(1/2)*a*tan(d*x+c)^5+1260*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)*a*tan( 
d*x+c)^5-472*I*(I*a)^(1/2)*tan(d*x+c)^3*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1 
/2)*(-I*a)^(1/2)+1576*tan(d*x+c)^4*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*( 
I*a)^(1/2)*(-I*a)^(1/2)+190*I*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^( 
1/2)*(I*a)^(1/2)*(-I*a)^(1/2)-276*(-I*a)^(1/2)*(I*a)^(1/2)*(a*tan(d*x+c)*( 
1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2+70*(-I*a)^(1/2)*(I*a)^(1/2)*(a*tan(d*x 
+c)*(1+I*tan(d*x+c)))^(1/2))/tan(d*x+c)^(9/2)/(I*a)^(1/2)/(a*tan(d*x+c)*(1 
+I*tan(d*x+c)))^(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. 566 vs. \(2 (181) = 362\).

Time = 0.11 (sec) , antiderivative size = 566, normalized size of antiderivative = 2.37 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {11}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:

integrate((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(11/2),x, algorithm="fricas" 
)
 

Output:

-1/630*(8*sqrt(2)*(646*I*a^2*e^(11*I*d*x + 11*I*c) - 1001*I*a^2*e^(9*I*d*x 
 + 9*I*c) + 684*I*a^2*e^(7*I*d*x + 7*I*c) + 966*I*a^2*e^(5*I*d*x + 5*I*c) 
- 1050*I*a^2*e^(3*I*d*x + 3*I*c) + 315*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)) + 315*sqrt(32*I*a^5/d^2)*(d*e^(10*I*d*x + 10*I*c) - 5*d*e^(8*I*d* 
x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) - 10*d*e^(4*I*d*x + 4*I*c) + 5*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)*s 
qrt(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) - 315*sqrt(32*I*a^5/d^2)*(d*e^(10*I*d*x + 10*I*c) - 5*d*e^(8*I*d 
*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) - 10*d*e^(4*I*d*x + 4*I*c) + 5*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))/(d*e^(10*I*d*x + 10*I*c) - 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6 
*I*d*x + 6*I*c) - 10*d*e^(4*I*d*x + 4*I*c) + 5*d*e^(2*I*d*x + 2*I*c) - d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {11}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate((a+I*a*tan(d*x+c))**(5/2)/tan(d*x+c)**(11/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 3508 vs. \(2 (181) = 362\).

Time = 0.92 (sec) , antiderivative size = 3508, normalized size of antiderivative = 14.68 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {11}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(11/2),x, algorithm="maxima" 
)
 

Output:

-1/1260*(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) 
 + 1)*(((5040*I - 5040)*a^2*cos(7*d*x + 7*c) - (16800*I - 16800)*a^2*cos(5 
*d*x + 5*c) + (20496*I - 20496)*a^2*cos(3*d*x + 3*c) - (9071*I - 9071)*a^2 
*cos(d*x + c) - (5040*I + 5040)*a^2*sin(7*d*x + 7*c) + (16800*I + 16800)*a 
^2*sin(5*d*x + 5*c) - (20496*I + 20496)*a^2*sin(3*d*x + 3*c) + (9071*I + 9 
071)*a^2*sin(d*x + c))*cos(7/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) 
 + 1)) + 8*(121*(-(I - 1)*a^2*cos(d*x + c) + (I + 1)*a^2*sin(d*x + c))*cos 
(2*d*x + 2*c)^2 - (121*I - 121)*a^2*cos(d*x + c) + 121*(-(I - 1)*a^2*cos(d 
*x + c) + (I + 1)*a^2*sin(d*x + c))*sin(2*d*x + 2*c)^2 + (121*I + 121)*a^2 
*sin(d*x + c) + 630*((I - 1)*a^2*cos(2*d*x + 2*c)^2 + (I - 1)*a^2*sin(2*d* 
x + 2*c)^2 - (2*I - 2)*a^2*cos(2*d*x + 2*c) + (I - 1)*a^2)*cos(3*d*x + 3*c 
) + 242*((I - 1)*a^2*cos(d*x + c) - (I + 1)*a^2*sin(d*x + c))*cos(2*d*x + 
2*c) + 630*(-(I + 1)*a^2*cos(2*d*x + 2*c)^2 - (I + 1)*a^2*sin(2*d*x + 2*c) 
^2 + (2*I + 2)*a^2*cos(2*d*x + 2*c) - (I + 1)*a^2)*sin(3*d*x + 3*c))*cos(3 
/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + (-(5040*I + 5040)*a 
^2*cos(7*d*x + 7*c) + (16800*I + 16800)*a^2*cos(5*d*x + 5*c) - (20496*I + 
20496)*a^2*cos(3*d*x + 3*c) + (9071*I + 9071)*a^2*cos(d*x + c) - (5040*I - 
 5040)*a^2*sin(7*d*x + 7*c) + (16800*I - 16800)*a^2*sin(5*d*x + 5*c) - (20 
496*I - 20496)*a^2*sin(3*d*x + 3*c) + (9071*I - 9071)*a^2*sin(d*x + c))*si 
n(7/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + 8*(121*((I + ...
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {11}{2}}(c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(11/2),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. \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {11}{2}}(c+d x)} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{11/2}} \,d x \] Input:

int((a + a*tan(c + d*x)*1i)^(5/2)/tan(c + d*x)^(11/2),x)
                                                                                    
                                                                                    
 

Output:

int((a + a*tan(c + d*x)*1i)^(5/2)/tan(c + d*x)^(11/2), x)
 

Reduce [F]

\[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {11}{2}}(c+d x)} \, dx=\frac {2 \sqrt {a}\, a^{2} \left (-368 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{4}-184 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{3} i +138 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}-95 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -35 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}-630 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{\tan \left (d x +c \right )^{3}}d x \right ) \tan \left (d x +c \right )^{5} d i \right )}{315 \tan \left (d x +c \right )^{5} d} \] Input:

int((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(11/2),x)
 

Output:

(2*sqrt(a)*a**2*( - 368*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c 
+ d*x)**4 - 184*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)** 
3*i + 138*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 - 95 
*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)*i - 35*sqrt(tan( 
c + d*x))*sqrt(tan(c + d*x)*i + 1) - 630*int((sqrt(tan(c + d*x))*sqrt(tan( 
c + d*x)*i + 1))/tan(c + d*x)**3,x)*tan(c + d*x)**5*d*i))/(315*tan(c + d*x 
)**5*d)