\(\int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [118]

Optimal result

Integrand size = 26, antiderivative size = 174 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {\tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {5 i \tan ^2(c+d x)}{2 a d \sqrt {a+i a \tan (c+d x)}}-\frac {10 i \sqrt {a+i a \tan (c+d x)}}{a^2 d}+\frac {7 i (a+i a \tan (c+d x))^{3/2}}{2 a^3 d} \] Output:

-1/4*I*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/a^(3/ 
2)/d-1/3*tan(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(3/2)+5/2*I*tan(d*x+c)^2/a/d/(a 
+I*a*tan(d*x+c))^(1/2)-10*I*(a+I*a*tan(d*x+c))^(1/2)/a^2/d+7/2*I*(a+I*a*ta 
n(d*x+c))^(3/2)/a^3/d
 

Mathematica [A] (verified)

Time = 2.14 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.95 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {i e^{-4 i (c+d x)} \left (\sqrt {1+e^{2 i (c+d x)}} \left (-1+18 e^{2 i (c+d x)}+87 e^{4 i (c+d x)}+52 e^{6 i (c+d x)}\right )+3 e^{3 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^2 \text {arcsinh}\left (e^{i (c+d x)}\right )\right ) \sec ^2(c+d x)}{24 a d \sqrt {1+e^{2 i (c+d x)}} \sqrt {a+i a \tan (c+d x)}} \] Input:

Integrate[Tan[c + d*x]^4/(a + I*a*Tan[c + d*x])^(3/2),x]
 

Output:

((-1/24*I)*(Sqrt[1 + E^((2*I)*(c + d*x))]*(-1 + 18*E^((2*I)*(c + d*x)) + 8 
7*E^((4*I)*(c + d*x)) + 52*E^((6*I)*(c + d*x))) + 3*E^((3*I)*(c + d*x))*(1 
 + E^((2*I)*(c + d*x)))^2*ArcSinh[E^(I*(c + d*x))])*Sec[c + d*x]^2)/(a*d*E 
^((4*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Sqrt[a + I*a*Tan[c + d*x] 
])
 

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 180, 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, 4010, 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.

\(\displaystyle \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^4}{(a+i a \tan (c+d x))^{3/2}}dx\)

\(\Big \downarrow \) 4041

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4078

\(\displaystyle \frac {\frac {5 i a \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \frac {1}{2} \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (21 \tan (c+d x) a^2+20 i a^2\right )dx}{a^2}}{2 a^2}-\frac {\tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {5 i a \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (21 \tan (c+d x) a^2+20 i a^2\right )dx}{2 a^2}}{2 a^2}-\frac {\tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5 i a \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (21 \tan (c+d x) a^2+20 i a^2\right )dx}{2 a^2}}{2 a^2}-\frac {\tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4075

\(\displaystyle \frac {\frac {5 i a \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \sqrt {i \tan (c+d x) a+a} \left (20 i a^2 \tan (c+d x)-21 a^2\right )dx-\frac {14 i a (a+i a \tan (c+d x))^{3/2}}{d}}{2 a^2}}{2 a^2}-\frac {\tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5 i a \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \sqrt {i \tan (c+d x) a+a} \left (20 i a^2 \tan (c+d x)-21 a^2\right )dx-\frac {14 i a (a+i a \tan (c+d x))^{3/2}}{d}}{2 a^2}}{2 a^2}-\frac {\tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4010

\(\displaystyle \frac {\frac {5 i a \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \left (-\int \sqrt {i \tan (c+d x) a+a}dx\right )+\frac {40 i a^2 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {14 i a (a+i a \tan (c+d x))^{3/2}}{d}}{2 a^2}}{2 a^2}-\frac {\tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5 i a \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \left (-\int \sqrt {i \tan (c+d x) a+a}dx\right )+\frac {40 i a^2 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {14 i a (a+i a \tan (c+d x))^{3/2}}{d}}{2 a^2}}{2 a^2}-\frac {\tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3961

\(\displaystyle \frac {\frac {5 i a \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {2 i a^3 \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}+\frac {40 i a^2 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {14 i a (a+i a \tan (c+d x))^{3/2}}{d}}{2 a^2}}{2 a^2}-\frac {\tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {5 i a \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {40 i a^2 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {14 i a (a+i a \tan (c+d x))^{3/2}}{d}}{2 a^2}}{2 a^2}-\frac {\tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

Input:

Int[Tan[c + d*x]^4/(a + I*a*Tan[c + d*x])^(3/2),x]
 

Output:

-1/3*Tan[c + d*x]^3/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (((5*I)*a*Tan[c + d 
*x]^2)/(d*Sqrt[a + I*a*Tan[c + d*x]]) - ((I*Sqrt[2]*a^(5/2)*ArcTanh[Sqrt[a 
 + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + ((40*I)*a^2*Sqrt[a + I*a*Tan[ 
c + d*x]])/d - ((14*I)*a*(a + I*a*Tan[c + d*x])^(3/2))/d)/(2*a^2))/(2*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 4010

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] + Simp 
[(b*c + a*d)/b   Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e 
, f, m}, 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 = 1.34 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.65

method	result	size
derivativedivides	\(\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 a \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8}-\frac {7 a^{2}}{4 \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {a^{3}}{6 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}\right )}{d \,a^{3}}\)	\(113\)
default	\(\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 a \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8}-\frac {7 a^{2}}{4 \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {a^{3}}{6 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}\right )}{d \,a^{3}}\)	\(113\)

Input:

int(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

2*I/d/a^3*(1/3*(a+I*a*tan(d*x+c))^(3/2)-2*a*(a+I*a*tan(d*x+c))^(1/2)-1/8*a 
^(3/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))-7/4*a 
^2/(a+I*a*tan(d*x+c))^(1/2)+1/6*a^3/(a+I*a*tan(d*x+c))^(3/2))
 

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

Time = 0.09 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.93 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {3 \, \sqrt {\frac {1}{2}} {\left (i \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + i \, a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {\frac {1}{a^{3} d^{2}}} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 3 \, \sqrt {\frac {1}{2}} {\left (-i \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} - i \, a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {\frac {1}{a^{3} d^{2}}} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-52 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 87 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 18 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )}}{12 \, {\left (a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )}} \] Input:

integrate(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

-1/12*(3*sqrt(1/2)*(I*a^2*d*e^(5*I*d*x + 5*I*c) + I*a^2*d*e^(3*I*d*x + 3*I 
*c))*sqrt(1/(a^3*d^2))*log(4*(sqrt(2)*sqrt(1/2)*(a^2*d*e^(2*I*d*x + 2*I*c) 
 + a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^3*d^2)) + a*e^(I*d*x 
 + I*c))*e^(-I*d*x - I*c)) + 3*sqrt(1/2)*(-I*a^2*d*e^(5*I*d*x + 5*I*c) - I 
*a^2*d*e^(3*I*d*x + 3*I*c))*sqrt(1/(a^3*d^2))*log(-4*(sqrt(2)*sqrt(1/2)*(a 
^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1 
/(a^3*d^2)) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - sqrt(2)*sqrt(a/(e^(2* 
I*d*x + 2*I*c) + 1))*(-52*I*e^(6*I*d*x + 6*I*c) - 87*I*e^(4*I*d*x + 4*I*c) 
 - 18*I*e^(2*I*d*x + 2*I*c) + I))/(a^2*d*e^(5*I*d*x + 5*I*c) + a^2*d*e^(3* 
I*d*x + 3*I*c))
 

Sympy [F]

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

integrate(tan(d*x+c)**4/(a+I*a*tan(d*x+c))**(3/2),x)
 

Output:

Integral(tan(c + d*x)**4/(I*a*(tan(c + d*x) - I))**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.80 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {i \, {\left (3 \, \sqrt {2} a^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 16 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{2} - 96 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{3} - \frac {4 \, {\left (21 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} - 2 \, a^{5}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\right )}}{24 \, a^{5} d} \] Input:

integrate(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

1/24*I*(3*sqrt(2)*a^(7/2)*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))) + 16*(I*a*tan(d*x + c) 
 + a)^(3/2)*a^2 - 96*sqrt(I*a*tan(d*x + c) + a)*a^3 - 4*(21*(I*a*tan(d*x + 
 c) + a)*a^4 - 2*a^5)/(I*a*tan(d*x + c) + a)^(3/2))/(a^5*d)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^(3/2),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 TypeError: Bad Argument TypeError: Bad Argument TypeDone
 

Mupad [B] (verification not implemented)

Time = 1.10 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.74 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\frac {1{}\mathrm {i}}{3\,d}-\frac {\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,7{}\mathrm {i}}{2\,a\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,4{}\mathrm {i}}{a^2\,d}+\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,2{}\mathrm {i}}{3\,a^3\,d}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{4\,{\left (-a\right )}^{3/2}\,d} \] Input:

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

Output:

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

Reduce [F]

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

int(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^(3/2),x)
 

Output:

(sqrt(a)*( - int((sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**5)/(tan(c + d*x)* 
*3*i + tan(c + d*x)**2 + tan(c + d*x)*i + 1),x)*i + int((sqrt(tan(c + d*x) 
*i + 1)*tan(c + d*x)**4)/(tan(c + d*x)**3*i + tan(c + d*x)**2 + tan(c + d* 
x)*i + 1),x)))/a**2