\(\int \frac {(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^3} \, dx\) [1112]

Optimal result

Integrand size = 30, antiderivative size = 274 \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {i (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{8 a^3 f}+\frac {i c \left (2 c^2+3 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{16 a^3 (c+i d)^{3/2} f}+\frac {(i c-d) \sqrt {c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac {(3 i c+4 d) \sqrt {c+d \tan (e+f x)}}{24 a f (a+i a \tan (e+f x))^2}-\frac {\left (2 c^2-i c d+2 d^2\right ) \sqrt {c+d \tan (e+f x)}}{16 (i c-d) f \left (a^3+i a^3 \tan (e+f x)\right )} \] Output:

-1/8*I*(c-I*d)^(3/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/a^3/f+1 
/16*I*c*(2*c^2+3*d^2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/a^3/(c 
+I*d)^(3/2)/f+1/6*(I*c-d)*(c+d*tan(f*x+e))^(1/2)/f/(a+I*a*tan(f*x+e))^3+1/ 
24*(3*I*c+4*d)*(c+d*tan(f*x+e))^(1/2)/a/f/(a+I*a*tan(f*x+e))^2-1/16*(2*c^2 
-I*c*d+2*d^2)*(c+d*tan(f*x+e))^(1/2)/(I*c-d)/f/(a^3+I*a^3*tan(f*x+e))
 

Mathematica [A] (verified)

Time = 2.80 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.25 \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {i \left (6 (c-i d)^{3/2} (c+i d) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+\frac {2 (c+i d) (3 c-4 i d) \sqrt {c+d \tan (e+f x)}}{(-i+\tan (e+f x))^2}+\frac {8 i (c+i d) d \sqrt {c+d \tan (e+f x)}}{(-i+\tan (e+f x))^2}+\frac {8 i d (c+d \tan (e+f x))^{3/2}}{(-i+\tan (e+f x))^2}-\frac {8 i (c+d \tan (e+f x))^{5/2}}{(-i+\tan (e+f x))^3}+\frac {3 i \left (c \sqrt {c+i d} \left (2 c^2+3 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right ) (1+i \tan (e+f x))+\left (2 c^3+i c^2 d+3 c d^2+2 i d^3\right ) \sqrt {c+d \tan (e+f x)}\right )}{(c+i d) (-i+\tan (e+f x))}\right )}{48 a^3 (c+i d) f} \] Input:

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

Output:

((-1/48*I)*(6*(c - I*d)^(3/2)*(c + I*d)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/S 
qrt[c - I*d]] + (2*(c + I*d)*(3*c - (4*I)*d)*Sqrt[c + d*Tan[e + f*x]])/(-I 
 + Tan[e + f*x])^2 + ((8*I)*(c + I*d)*d*Sqrt[c + d*Tan[e + f*x]])/(-I + Ta 
n[e + f*x])^2 + ((8*I)*d*(c + d*Tan[e + f*x])^(3/2))/(-I + Tan[e + f*x])^2 
 - ((8*I)*(c + d*Tan[e + f*x])^(5/2))/(-I + Tan[e + f*x])^3 + ((3*I)*(c*Sq 
rt[c + I*d]*(2*c^2 + 3*d^2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d] 
]*(1 + I*Tan[e + f*x]) + (2*c^3 + I*c^2*d + 3*c*d^2 + (2*I)*d^3)*Sqrt[c + 
d*Tan[e + f*x]]))/((c + I*d)*(-I + Tan[e + f*x]))))/(a^3*(c + I*d)*f)
 

Rubi [A] (warning: unable to verify)

Time = 1.73 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.05, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4041, 27, 3042, 4079, 27, 3042, 4079, 3042, 4022, 3042, 4020, 25, 73, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4041

\(\displaystyle \frac {(-d+i c) \sqrt {c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\int -\frac {a \left (6 c^2-7 i d c+d^2\right )+a (5 c-7 i d) d \tan (e+f x)}{2 (i \tan (e+f x) a+a)^2 \sqrt {c+d \tan (e+f x)}}dx}{6 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a \left (6 c^2-7 i d c+d^2\right )+a (5 c-7 i d) d \tan (e+f x)}{(i \tan (e+f x) a+a)^2 \sqrt {c+d \tan (e+f x)}}dx}{12 a^2}+\frac {(-d+i c) \sqrt {c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a \left (6 c^2-7 i d c+d^2\right )+a (5 c-7 i d) d \tan (e+f x)}{(i \tan (e+f x) a+a)^2 \sqrt {c+d \tan (e+f x)}}dx}{12 a^2}+\frac {(-d+i c) \sqrt {c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\frac {a (4 d+3 i c) \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))^2}-\frac {\int -\frac {3 \left (c (i c-d) (4 c-5 i d) a^2+(c+i d) d (3 i c+4 d) \tan (e+f x) a^2\right )}{(i \tan (e+f x) a+a) \sqrt {c+d \tan (e+f x)}}dx}{4 a^2 (-d+i c)}}{12 a^2}+\frac {(-d+i c) \sqrt {c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 \int \frac {c (c+i d) (4 i c+5 d) a^2+(c+i d) d (3 i c+4 d) \tan (e+f x) a^2}{(i \tan (e+f x) a+a) \sqrt {c+d \tan (e+f x)}}dx}{4 a^2 (-d+i c)}+\frac {a (4 d+3 i c) \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))^2}}{12 a^2}+\frac {(-d+i c) \sqrt {c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 \int \frac {c (c+i d) (4 i c+5 d) a^2+(c+i d) d (3 i c+4 d) \tan (e+f x) a^2}{(i \tan (e+f x) a+a) \sqrt {c+d \tan (e+f x)}}dx}{4 a^2 (-d+i c)}+\frac {a (4 d+3 i c) \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))^2}}{12 a^2}+\frac {(-d+i c) \sqrt {c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\frac {3 \left (-\frac {\int \frac {(c+i d) \left (4 c^3-2 i d c^2+5 d^2 c-2 i d^3\right ) a^3+(c+i d) d \left (2 c^2-i d c+2 d^2\right ) \tan (e+f x) a^3}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}-\frac {a^2 \left (2 c^2-i c d+2 d^2\right ) \sqrt {c+d \tan (e+f x)}}{f (a+i a \tan (e+f x))}\right )}{4 a^2 (-d+i c)}+\frac {a (4 d+3 i c) \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))^2}}{12 a^2}+\frac {(-d+i c) \sqrt {c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 \left (-\frac {\int \frac {(c+i d) \left (4 c^3-2 i d c^2+5 d^2 c-2 i d^3\right ) a^3+(c+i d) d \left (2 c^2-i d c+2 d^2\right ) \tan (e+f x) a^3}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}-\frac {a^2 \left (2 c^2-i c d+2 d^2\right ) \sqrt {c+d \tan (e+f x)}}{f (a+i a \tan (e+f x))}\right )}{4 a^2 (-d+i c)}+\frac {a (4 d+3 i c) \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))^2}}{12 a^2}+\frac {(-d+i c) \sqrt {c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {\frac {3 \left (-\frac {a^3 c (c+i d) \left (2 c^2+3 d^2\right ) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+2 a^3 \left (c^2+d^2\right )^2 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}-\frac {a^2 \left (2 c^2-i c d+2 d^2\right ) \sqrt {c+d \tan (e+f x)}}{f (a+i a \tan (e+f x))}\right )}{4 a^2 (-d+i c)}+\frac {a (4 d+3 i c) \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))^2}}{12 a^2}+\frac {(-d+i c) \sqrt {c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 \left (-\frac {a^3 c (c+i d) \left (2 c^2+3 d^2\right ) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+2 a^3 \left (c^2+d^2\right )^2 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}-\frac {a^2 \left (2 c^2-i c d+2 d^2\right ) \sqrt {c+d \tan (e+f x)}}{f (a+i a \tan (e+f x))}\right )}{4 a^2 (-d+i c)}+\frac {a (4 d+3 i c) \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))^2}}{12 a^2}+\frac {(-d+i c) \sqrt {c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {\frac {3 \left (-\frac {\frac {2 i a^3 \left (c^2+d^2\right )^2 \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{f}-\frac {i a^3 c (c+i d) \left (2 c^2+3 d^2\right ) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{f}}{2 a^2 (-d+i c)}-\frac {a^2 \left (2 c^2-i c d+2 d^2\right ) \sqrt {c+d \tan (e+f x)}}{f (a+i a \tan (e+f x))}\right )}{4 a^2 (-d+i c)}+\frac {a (4 d+3 i c) \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))^2}}{12 a^2}+\frac {(-d+i c) \sqrt {c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {3 \left (-\frac {\frac {i a^3 c (c+i d) \left (2 c^2+3 d^2\right ) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{f}-\frac {2 i a^3 \left (c^2+d^2\right )^2 \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{f}}{2 a^2 (-d+i c)}-\frac {a^2 \left (2 c^2-i c d+2 d^2\right ) \sqrt {c+d \tan (e+f x)}}{f (a+i a \tan (e+f x))}\right )}{4 a^2 (-d+i c)}+\frac {a (4 d+3 i c) \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))^2}}{12 a^2}+\frac {(-d+i c) \sqrt {c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {3 \left (-\frac {\frac {2 a^3 c (c+i d) \left (2 c^2+3 d^2\right ) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {4 a^3 \left (c^2+d^2\right )^2 \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}}{2 a^2 (-d+i c)}-\frac {a^2 \left (2 c^2-i c d+2 d^2\right ) \sqrt {c+d \tan (e+f x)}}{f (a+i a \tan (e+f x))}\right )}{4 a^2 (-d+i c)}+\frac {a (4 d+3 i c) \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))^2}}{12 a^2}+\frac {(-d+i c) \sqrt {c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {3 \left (-\frac {a^2 \left (2 c^2-i c d+2 d^2\right ) \sqrt {c+d \tan (e+f x)}}{f (a+i a \tan (e+f x))}-\frac {\frac {4 a^3 \left (c^2+d^2\right )^2 \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}+\frac {2 a^3 c \sqrt {c+i d} \left (2 c^2+3 d^2\right ) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}}{2 a^2 (-d+i c)}\right )}{4 a^2 (-d+i c)}+\frac {a (4 d+3 i c) \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))^2}}{12 a^2}+\frac {(-d+i c) \sqrt {c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}\)

Input:

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

Output:

((I*c - d)*Sqrt[c + d*Tan[e + f*x]])/(6*f*(a + I*a*Tan[e + f*x])^3) + ((a* 
((3*I)*c + 4*d)*Sqrt[c + d*Tan[e + f*x]])/(2*f*(a + I*a*Tan[e + f*x])^2) + 
 (3*(-1/2*((4*a^3*(c^2 + d^2)^2*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/(Sqrt[ 
c - I*d]*f) + (2*a^3*c*Sqrt[c + I*d]*(2*c^2 + 3*d^2)*ArcTan[Tan[e + f*x]/S 
qrt[c + I*d]])/f)/(a^2*(I*c - d)) - (a^2*(2*c^2 - I*c*d + 2*d^2)*Sqrt[c + 
d*Tan[e + f*x]])/(f*(a + I*a*Tan[e + f*x]))))/(4*a^2*(I*c - d)))/(12*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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

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

rule 3042

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

rule 4020

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

rule 4022

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

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 4079

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

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.32

method	result	size
derivativedivides	\(\frac {2 d^{4} \left (\frac {i \left (i d -c \right )^{\frac {3}{2}} \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{16 d^{4}}+\frac {\frac {-\frac {d \left (3 i c^{3} d +5 i c \,d^{3}+2 c^{4}+2 c^{2} d^{2}-2 d^{4}\right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{2 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right )}+\frac {2 \left (3 c^{2}+5 d^{2}\right ) d \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {d \left (9 i c^{5} d -6 i c^{3} d^{3}-7 i c \,d^{5}+2 c^{6}-14 c^{4} d^{2}-6 c^{2} d^{4}+2 d^{6}\right ) \sqrt {c +d \tan \left (f x +e \right )}}{2 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right )}}{\left (-d \tan \left (f x +e \right )+i d \right )^{3}}-\frac {c \left (2 i c^{4}+i c^{2} d^{2}-3 i d^{4}-4 c^{3} d -6 c \,d^{3}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{2 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right ) \sqrt {-i d -c}}}{16 d^{4}}\right )}{f \,a^{3}}\)	\(362\)
default	\(\frac {2 d^{4} \left (\frac {i \left (i d -c \right )^{\frac {3}{2}} \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{16 d^{4}}+\frac {\frac {-\frac {d \left (3 i c^{3} d +5 i c \,d^{3}+2 c^{4}+2 c^{2} d^{2}-2 d^{4}\right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{2 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right )}+\frac {2 \left (3 c^{2}+5 d^{2}\right ) d \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {d \left (9 i c^{5} d -6 i c^{3} d^{3}-7 i c \,d^{5}+2 c^{6}-14 c^{4} d^{2}-6 c^{2} d^{4}+2 d^{6}\right ) \sqrt {c +d \tan \left (f x +e \right )}}{2 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right )}}{\left (-d \tan \left (f x +e \right )+i d \right )^{3}}-\frac {c \left (2 i c^{4}+i c^{2} d^{2}-3 i d^{4}-4 c^{3} d -6 c \,d^{3}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{2 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right ) \sqrt {-i d -c}}}{16 d^{4}}\right )}{f \,a^{3}}\)	\(362\)

Input:

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

Output:

2/f/a^3*d^4*(1/16*I*(I*d-c)^(3/2)/d^4*arctan((c+d*tan(f*x+e))^(1/2)/(I*d-c 
)^(1/2))+1/16/d^4*((-1/2*d*(3*I*c^3*d+5*I*c*d^3+2*c^4+2*c^2*d^2-2*d^4)/(3* 
I*c^2*d-I*d^3+c^3-3*c*d^2)*(c+d*tan(f*x+e))^(5/2)+2/3*(3*c^2+5*d^2)*d*(c+d 
*tan(f*x+e))^(3/2)-1/2*d*(9*I*c^5*d-6*I*c^3*d^3-7*I*c*d^5+2*c^6-14*c^4*d^2 
-6*c^2*d^4+2*d^6)/(3*I*c^2*d-I*d^3+c^3-3*c*d^2)*(c+d*tan(f*x+e))^(1/2))/(- 
d*tan(f*x+e)+I*d)^3-1/2*c*(2*I*c^4+I*c^2*d^2-3*I*d^4-4*c^3*d-6*c*d^3)/(3*I 
*c^2*d-I*d^3+c^3-3*c*d^2)/(-c-I*d)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(-c 
-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. 1238 vs. \(2 (218) = 436\).

Time = 0.36 (sec) , antiderivative size = 1238, normalized size of antiderivative = 4.52 \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

1/192*(6*(I*a^3*c - a^3*d)*f*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a^ 
6*f^2))*e^(6*I*f*x + 6*I*e)*log(-2*(-I*c^2 - c*d + (a^3*f*e^(2*I*f*x + 2*I 
*e) + a^3*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 
2*I*e) + 1))*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a^6*f^2)) + (-I*c^ 
2 - 2*c*d + I*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(I*c + d)) + 
6*(-I*a^3*c + a^3*d)*f*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a^6*f^2) 
)*e^(6*I*f*x + 6*I*e)*log(-2*(-I*c^2 - c*d - (a^3*f*e^(2*I*f*x + 2*I*e) + 
a^3*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) 
 + 1))*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a^6*f^2)) + (-I*c^2 - 2* 
c*d + I*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(I*c + d)) + 3*(-I* 
a^3*c + a^3*d)*f*sqrt(-(-4*I*c^6 - 12*I*c^4*d^2 - 9*I*c^2*d^4)/((-I*a^6*c^ 
3 + 3*a^6*c^2*d + 3*I*a^6*c*d^2 - a^6*d^3)*f^2))*e^(6*I*f*x + 6*I*e)*log(- 
1/16*(-2*I*c^4 + 2*c^3*d - 3*I*c^2*d^2 + 3*c*d^3 + ((a^3*c^2 + 2*I*a^3*c*d 
 - a^3*d^2)*f*e^(2*I*f*x + 2*I*e) + (a^3*c^2 + 2*I*a^3*c*d - a^3*d^2)*f)*s 
qrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*s 
qrt(-(-4*I*c^6 - 12*I*c^4*d^2 - 9*I*c^2*d^4)/((-I*a^6*c^3 + 3*a^6*c^2*d + 
3*I*a^6*c*d^2 - a^6*d^3)*f^2)) + (-2*I*c^4 - 3*I*c^2*d^2)*e^(2*I*f*x + 2*I 
*e))*e^(-2*I*f*x - 2*I*e)/((a^3*c^2 + 2*I*a^3*c*d - a^3*d^2)*f)) + 3*(I*a^ 
3*c - a^3*d)*f*sqrt(-(-4*I*c^6 - 12*I*c^4*d^2 - 9*I*c^2*d^4)/((-I*a^6*c^3 
+ 3*a^6*c^2*d + 3*I*a^6*c*d^2 - a^6*d^3)*f^2))*e^(6*I*f*x + 6*I*e)*log(...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

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

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 600 vs. \(2 (218) = 436\).

Time = 0.67 (sec) , antiderivative size = 600, normalized size of antiderivative = 2.19 \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {\frac {6 \, \sqrt {2} {\left (-i \, c^{2} - 2 \, c d + i \, d^{2}\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} - i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{\sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {24 \, \sqrt {2} {\left (2 \, c^{3} + 3 \, c d^{2}\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} + i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{-8 \, {\left (-i \, c + d\right )} \sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {6 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} c^{2} d - 12 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c^{3} d + 6 \, \sqrt {d \tan \left (f x + e\right ) + c} c^{4} d - 3 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} c d^{2} - 12 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c^{2} d^{2} + 15 i \, \sqrt {d \tan \left (f x + e\right ) + c} c^{3} d^{2} + 6 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} d^{3} - 20 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c d^{3} - 6 \, \sqrt {d \tan \left (f x + e\right ) + c} c^{2} d^{3} - 20 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} d^{4} + 9 i \, \sqrt {d \tan \left (f x + e\right ) + c} c d^{4} - 6 \, \sqrt {d \tan \left (f x + e\right ) + c} d^{5}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3} {\left (c + i \, d\right )}}}{48 \, a^{3} f} \] Input:

integrate((c+d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="giac")
 

Output:

-1/48*(6*sqrt(2)*(-I*c^2 - 2*c*d + I*d^2)*arctan(2*(sqrt(d*tan(f*x + e) + 
c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(sqrt(2)*c*sqrt(-c + sqrt 
(c^2 + d^2)) - I*sqrt(2)*sqrt(-c + sqrt(c^2 + d^2))*d - sqrt(2)*sqrt(c^2 + 
 d^2)*sqrt(-c + sqrt(c^2 + d^2))))/(sqrt(-c + sqrt(c^2 + d^2))*(-I*d/(c - 
sqrt(c^2 + d^2)) + 1)) - 24*sqrt(2)*(2*c^3 + 3*c*d^2)*arctan(2*(sqrt(d*tan 
(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(sqrt(2)*c*sq 
rt(-c + sqrt(c^2 + d^2)) + I*sqrt(2)*sqrt(-c + sqrt(c^2 + d^2))*d - sqrt(2 
)*sqrt(c^2 + d^2)*sqrt(-c + sqrt(c^2 + d^2))))/((8*I*c - 8*d)*sqrt(-c + sq 
rt(c^2 + d^2))*(I*d/(c - sqrt(c^2 + d^2)) + 1)) - (6*(d*tan(f*x + e) + c)^ 
(5/2)*c^2*d - 12*(d*tan(f*x + e) + c)^(3/2)*c^3*d + 6*sqrt(d*tan(f*x + e) 
+ c)*c^4*d - 3*I*(d*tan(f*x + e) + c)^(5/2)*c*d^2 - 12*I*(d*tan(f*x + e) + 
 c)^(3/2)*c^2*d^2 + 15*I*sqrt(d*tan(f*x + e) + c)*c^3*d^2 + 6*(d*tan(f*x + 
 e) + c)^(5/2)*d^3 - 20*(d*tan(f*x + e) + c)^(3/2)*c*d^3 - 6*sqrt(d*tan(f* 
x + e) + c)*c^2*d^3 - 20*I*(d*tan(f*x + e) + c)^(3/2)*d^4 + 9*I*sqrt(d*tan 
(f*x + e) + c)*c*d^4 - 6*sqrt(d*tan(f*x + e) + c)*d^5)/((d*tan(f*x + e) - 
I*d)^3*(c + I*d)))/(a^3*f)
 

Mupad [B] (verification not implemented)

Time = 6.11 (sec) , antiderivative size = 16296, normalized size of antiderivative = 59.47 \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

(((c + d*tan(e + f*x))^(1/2)*(5*c*d^3 + 10*c^3*d + d^4*10i + c^2*d^2*15i)) 
/(80*a^3*f) - (d*(c + d*tan(e + f*x))^(5/2)*(c*d + c^2*2i + d^2*2i)*1i)/(1 
6*(a^3*c*f + a^3*d*f*1i)) + (d*(c^2*6i + d^2*10i)*(c + d*tan(e + f*x))^(3/ 
2)*1i)/(24*a^3*f))/((c + d*tan(e + f*x))*(c*d*6i + 3*c^2 - 3*d^2) + (c + d 
*tan(e + f*x))^3 + 3*c*d^2 - c^2*d*3i - (3*c + d*3i)*(c + d*tan(e + f*x))^ 
2 - c^3 + d^3*1i) - log((((12*c*d^12 - d^13*4i - c^2*d^11*9i + 59*c^3*d^10 
 + c^4*d^9*39i + 51*c^5*d^8 + c^6*d^7*64i + c^8*d^5*24i - 8*c^9*d^4 - 4*a^ 
6*c^4*f^2*((216*c^2*d^24 - 16*d^26 - c*d^25*96i + c^3*d^23*176i + 111*c^4* 
d^22 + c^5*d^21*330i - 209*c^6*d^20 + c^7*d^19*36i - 111*c^8*d^18 - c^9*d^ 
17*54i + 9*c^10*d^16)/(16*a^12*c^8*f^4 + 16*a^12*d^8*f^4 + 64*a^12*c^2*d^6 
*f^4 + 96*a^12*c^4*d^4*f^4 + 64*a^12*c^6*d^2*f^4))^(1/2) - 4*a^6*d^4*f^2*( 
(216*c^2*d^24 - 16*d^26 - c*d^25*96i + c^3*d^23*176i + 111*c^4*d^22 + c^5* 
d^21*330i - 209*c^6*d^20 + c^7*d^19*36i - 111*c^8*d^18 - c^9*d^17*54i + 9* 
c^10*d^16)/(16*a^12*c^8*f^4 + 16*a^12*d^8*f^4 + 64*a^12*c^2*d^6*f^4 + 96*a 
^12*c^4*d^4*f^4 + 64*a^12*c^6*d^2*f^4))^(1/2) - 8*a^6*c^2*d^2*f^2*((216*c^ 
2*d^24 - 16*d^26 - c*d^25*96i + c^3*d^23*176i + 111*c^4*d^22 + c^5*d^21*33 
0i - 209*c^6*d^20 + c^7*d^19*36i - 111*c^8*d^18 - c^9*d^17*54i + 9*c^10*d^ 
16)/(16*a^12*c^8*f^4 + 16*a^12*d^8*f^4 + 64*a^12*c^2*d^6*f^4 + 96*a^12*c^4 
*d^4*f^4 + 64*a^12*c^6*d^2*f^4))^(1/2))/(2048*a^6*d^10*f^2 + 6144*a^6*c^2* 
d^8*f^2 + 6144*a^6*c^4*d^6*f^2 + 2048*a^6*c^6*d^4*f^2))^(1/2)*(2048*a^9...
 

Reduce [F]

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

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

Output:

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