\(\int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [127]

Optimal result

Integrand size = 26, antiderivative size = 176 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}-\frac {\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {17 i \tan ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {151 i}{60 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {83 i \sqrt {a+i a \tan (c+d x)}}{30 a^3 d} \] Output:

-1/8*I*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/a^(5/ 
2)/d-1/5*tan(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(5/2)+17/30*I*tan(d*x+c)^2/a/d/ 
(a+I*a*tan(d*x+c))^(3/2)+151/60*I/a^2/d/(a+I*a*tan(d*x+c))^(1/2)+83/30*I*( 
a+I*a*tan(d*x+c))^(1/2)/a^3/d
 

Mathematica [A] (verified)

Time = 2.27 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.85 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {e^{-6 i (c+d x)} \left (-\frac {60 i e^{7 i (c+d x)} \text {arcsinh}\left (e^{i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}+i \left (1+e^{2 i (c+d x)}\right ) \left (3-26 e^{2 i (c+d x)}+194 e^{4 i (c+d x)}+463 e^{6 i (c+d x)}\right ) \sec ^2(c+d x)\right )}{240 a^2 d \sqrt {a+i a \tan (c+d x)}} \] Input:

Integrate[Tan[c + d*x]^4/(a + I*a*Tan[c + d*x])^(5/2),x]
 

Output:

(((-60*I)*E^((7*I)*(c + d*x))*ArcSinh[E^(I*(c + d*x))])/Sqrt[1 + E^((2*I)* 
(c + d*x))] + I*(1 + E^((2*I)*(c + d*x)))*(3 - 26*E^((2*I)*(c + d*x)) + 19 
4*E^((4*I)*(c + d*x)) + 463*E^((6*I)*(c + d*x)))*Sec[c + d*x]^2)/(240*a^2* 
d*E^((6*I)*(c + d*x))*Sqrt[a + I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.03, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4041, 27, 3042, 4078, 27, 3042, 4075, 3042, 4009, 3042, 3961, 219}

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.

\(\displaystyle \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^4}{(a+i a \tan (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 4041

\(\displaystyle -\frac {\int -\frac {\tan ^2(c+d x) (6 a-11 i a \tan (c+d x))}{2 (i \tan (c+d x) a+a)^{3/2}}dx}{5 a^2}-\frac {\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\tan ^2(c+d x) (6 a-11 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}-\frac {\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\tan (c+d x)^2 (6 a-11 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}-\frac {\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4078

\(\displaystyle \frac {\frac {17 i a \tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {\tan (c+d x) \left (83 \tan (c+d x) a^2+68 i a^2\right )}{2 \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}}{10 a^2}-\frac {\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {17 i a \tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {\tan (c+d x) \left (83 \tan (c+d x) a^2+68 i a^2\right )}{\sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}}{10 a^2}-\frac {\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {17 i a \tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {\tan (c+d x) \left (83 \tan (c+d x) a^2+68 i a^2\right )}{\sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}}{10 a^2}-\frac {\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4075

\(\displaystyle \frac {\frac {17 i a \tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {68 i a^2 \tan (c+d x)-83 a^2}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {166 i a \sqrt {a+i a \tan (c+d x)}}{d}}{6 a^2}}{10 a^2}-\frac {\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {17 i a \tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {68 i a^2 \tan (c+d x)-83 a^2}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {166 i a \sqrt {a+i a \tan (c+d x)}}{d}}{6 a^2}}{10 a^2}-\frac {\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4009

\(\displaystyle \frac {\frac {17 i a \tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {-\frac {15}{2} a \int \sqrt {i \tan (c+d x) a+a}dx-\frac {151 i a^2}{d \sqrt {a+i a \tan (c+d x)}}-\frac {166 i a \sqrt {a+i a \tan (c+d x)}}{d}}{6 a^2}}{10 a^2}-\frac {\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {17 i a \tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {-\frac {15}{2} a \int \sqrt {i \tan (c+d x) a+a}dx-\frac {151 i a^2}{d \sqrt {a+i a \tan (c+d x)}}-\frac {166 i a \sqrt {a+i a \tan (c+d x)}}{d}}{6 a^2}}{10 a^2}-\frac {\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3961

\(\displaystyle \frac {\frac {17 i a \tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {15 i a^2 \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}-\frac {151 i a^2}{d \sqrt {a+i a \tan (c+d x)}}-\frac {166 i a \sqrt {a+i a \tan (c+d x)}}{d}}{6 a^2}}{10 a^2}-\frac {\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {17 i a \tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {15 i a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} d}-\frac {151 i a^2}{d \sqrt {a+i a \tan (c+d x)}}-\frac {166 i a \sqrt {a+i a \tan (c+d x)}}{d}}{6 a^2}}{10 a^2}-\frac {\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

Input:

Int[Tan[c + d*x]^4/(a + I*a*Tan[c + d*x])^(5/2),x]
 

Output:

-1/5*Tan[c + d*x]^3/(d*(a + I*a*Tan[c + d*x])^(5/2)) + ((((17*I)/3)*a*Tan[ 
c + d*x]^2)/(d*(a + I*a*Tan[c + d*x])^(3/2)) - (((15*I)*a^(3/2)*ArcTanh[Sq 
rt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*d) - ((151*I)*a^2)/( 
d*Sqrt[a + I*a*Tan[c + d*x]]) - ((166*I)*a*Sqrt[a + I*a*Tan[c + d*x]])/d)/ 
(6*a^2))/(10*a^2)
 

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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 3961

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

rule 4009

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

rule 4041

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

rule 4075

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

rule 4078

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

Maple [A] (verified)

Time = 1.38 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.63

method	result	size
derivativedivides	\(\frac {2 i \left (\sqrt {a +i a \tan \left (d x +c \right )}-\frac {\sqrt {a}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16}+\frac {17 a}{8 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {7 a^{2}}{12 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {a^{3}}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\right )}{d \,a^{3}}\)	\(111\)
default	\(\frac {2 i \left (\sqrt {a +i a \tan \left (d x +c \right )}-\frac {\sqrt {a}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16}+\frac {17 a}{8 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {7 a^{2}}{12 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {a^{3}}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\right )}{d \,a^{3}}\)	\(111\)

Input:

int(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

2*I/d/a^3*((a+I*a*tan(d*x+c))^(1/2)-1/16*a^(1/2)*2^(1/2)*arctanh(1/2*(a+I* 
a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))+17/8*a/(a+I*a*tan(d*x+c))^(1/2)-7/12* 
a^2/(a+I*a*tan(d*x+c))^(3/2)+1/10*a^3/(a+I*a*tan(d*x+c))^(5/2))
 

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. 283 vs. \(2 (131) = 262\).

Time = 0.11 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.61 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {{\left (-15 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 15 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (463 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 194 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 26 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \] Input:

integrate(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

1/120*(-15*I*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(4*( 
sqrt(2)*sqrt(1/2)*(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 
 2*I*c) + 1))*sqrt(1/(a^5*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + 1 
5*I*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-4*(sqrt(2)* 
sqrt(1/2)*(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) 
+ 1))*sqrt(1/(a^5*d^2)) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + sqrt(2)*s 
qrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(463*I*e^(6*I*d*x + 6*I*c) + 194*I*e^(4*I 
*d*x + 4*I*c) - 26*I*e^(2*I*d*x + 2*I*c) + 3*I))*e^(-5*I*d*x - 5*I*c)/(a^3 
*d)
 

Sympy [F]

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

integrate(tan(d*x+c)**4/(a+I*a*tan(d*x+c))**(5/2),x)
 

Output:

Integral(tan(c + d*x)**4/(I*a*(tan(c + d*x) - I))**(5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.79 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {i \, {\left (15 \, \sqrt {2} a^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 480 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{2} + \frac {4 \, {\left (255 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3} - 70 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} + 12 \, a^{5}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\right )}}{240 \, a^{5} d} \] Input:

integrate(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

1/240*I*(15*sqrt(2)*a^(5/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) 
+ a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a))) + 480*sqrt(I*a*tan(d 
*x + c) + a)*a^2 + 4*(255*(I*a*tan(d*x + c) + a)^2*a^3 - 70*(I*a*tan(d*x + 
 c) + a)*a^4 + 12*a^5)/(I*a*tan(d*x + c) + a)^(5/2))/(a^5*d)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(tan(d*x+c)^4/(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 [B] (verification not implemented)

Time = 1.05 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.73 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\frac {1{}\mathrm {i}}{5\,d}+\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2\,17{}\mathrm {i}}{4\,a^2\,d}-\frac {\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,7{}\mathrm {i}}{6\,a\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{a^3\,d}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{8\,{\left (-a\right )}^{5/2}\,d} \] Input:

int(tan(c + d*x)^4/(a + a*tan(c + d*x)*1i)^(5/2),x)
 

Output:

(1i/(5*d) + ((a + a*tan(c + d*x)*1i)^2*17i)/(4*a^2*d) - ((a + a*tan(c + d* 
x)*1i)*7i)/(6*a*d))/(a + a*tan(c + d*x)*1i)^(5/2) + ((a + a*tan(c + d*x)*1 
i)^(1/2)*2i)/(a^3*d) + (2^(1/2)*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2 
))/(2*(-a)^(1/2)))*1i)/(8*(-a)^(5/2)*d)
 

Reduce [F]

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

int(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^(5/2),x)
 

Output:

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