\(\int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx\) [1128]

Optimal result

Integrand size = 30, antiderivative size = 205 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{2 a (c-i d)^{3/2} f}+\frac {(i c-4 d) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{2 a (c+i d)^{5/2} f}+\frac {(c-5 i d) d}{2 a (c-i d) (c+i d)^2 f \sqrt {c+d \tan (e+f x)}}-\frac {1}{2 (i c-d) f (a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \] Output:

-1/2*I*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/a/(c-I*d)^(3/2)/f+1/2 
*(I*c-4*d)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/a/(c+I*d)^(5/2)/f 
+1/2*(c-5*I*d)*d/a/(c-I*d)/(c+I*d)^2/f/(c+d*tan(f*x+e))^(1/2)-1/2/(I*c-d)/ 
f/(a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.68 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.70 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx=-\frac {i \left (-\frac {(c+i d) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {c+d \tan (e+f x)}{c-i d}\right )}{c-i d}+\frac {(c+4 i d) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {c+d \tan (e+f x)}{c+i d}\right )}{c+i d}+\frac {i}{-i+\tan (e+f x)}\right )}{2 a (c+i d) f \sqrt {c+d \tan (e+f x)}} \] Input:

Integrate[1/((a + I*a*Tan[e + f*x])*(c + d*Tan[e + f*x])^(3/2)),x]
 

Output:

((-1/2*I)*(-(((c + I*d)*Hypergeometric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x 
])/(c - I*d)])/(c - I*d)) + ((c + (4*I)*d)*Hypergeometric2F1[-1/2, 1, 1/2, 
 (c + d*Tan[e + f*x])/(c + I*d)])/(c + I*d) + I/(-I + Tan[e + f*x])))/(a*( 
c + I*d)*f*Sqrt[c + d*Tan[e + f*x]])
 

Rubi [A] (warning: unable to verify)

Time = 1.03 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.04, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4035, 27, 3042, 4012, 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.

\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 4035

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (2 i c-5 d)+3 i a d \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}}dx}{4 a^2 (-d+i c)}-\frac {1}{2 f (-d+i c) (a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (2 i c-5 d)+3 i a d \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}}dx}{4 a^2 (-d+i c)}-\frac {1}{2 f (-d+i c) (a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 4012

\(\displaystyle \frac {\frac {\int \frac {a (c+3 i d) (2 i c+d)+a d (i c+5 d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {2 a d (5 d+i c)}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{4 a^2 (-d+i c)}-\frac {1}{2 f (-d+i c) (a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {a (c+3 i d) (2 i c+d)+a d (i c+5 d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {2 a d (5 d+i c)}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{4 a^2 (-d+i c)}-\frac {1}{2 f (-d+i c) (a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {\frac {i a (c+i d)^2 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+a (c+4 i d) (d+i c) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {2 a d (5 d+i c)}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{4 a^2 (-d+i c)}-\frac {1}{2 f (-d+i c) (a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {i a (c+i d)^2 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+a (c+4 i d) (d+i c) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {2 a d (5 d+i c)}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{4 a^2 (-d+i c)}-\frac {1}{2 f (-d+i c) (a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {\frac {-\frac {a (c+i d)^2 \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{f}-\frac {i a (c+4 i d) (d+i c) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{f}}{c^2+d^2}+\frac {2 a d (5 d+i c)}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{4 a^2 (-d+i c)}-\frac {1}{2 f (-d+i c) (a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {a (c+i d)^2 \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{f}+\frac {i a (c+4 i d) (d+i c) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{f}}{c^2+d^2}+\frac {2 a d (5 d+i c)}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{4 a^2 (-d+i c)}-\frac {1}{2 f (-d+i c) (a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {\frac {2 i a (c+i d)^2 \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {2 a (c+4 i d) (d+i c) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}}{c^2+d^2}+\frac {2 a d (5 d+i c)}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{4 a^2 (-d+i c)}-\frac {1}{2 f (-d+i c) (a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {2 i a (c+i d)^2 \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}+\frac {2 a (c+4 i d) (d+i c) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}}{c^2+d^2}+\frac {2 a d (5 d+i c)}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{4 a^2 (-d+i c)}-\frac {1}{2 f (-d+i c) (a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}\)

Input:

Int[1/((a + I*a*Tan[e + f*x])*(c + d*Tan[e + f*x])^(3/2)),x]
 

Output:

-1/2*1/((I*c - d)*f*(a + I*a*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]]) + ((( 
(2*I)*a*(c + I*d)^2*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f) 
+ (2*a*(c + (4*I)*d)*(I*c + d)*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/(Sqrt[c 
 + I*d]*f))/(c^2 + d^2) + (2*a*d*(I*c + 5*d))/((c^2 + d^2)*f*Sqrt[c + d*Ta 
n[e + f*x]]))/(4*a^2*(I*c - d))
 

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 4012

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

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 4035

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(-a)*((c + d*Tan[e + f*x])^(n + 1)/(2*f*(b* 
c - a*d)*(a + b*Tan[e + f*x]))), x] + Simp[1/(2*a*(b*c - a*d))   Int[(c + d 
*Tan[e + f*x])^n*Simp[b*c + a*d*(n - 1) - b*d*n*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] &&  !GtQ[n, 0]
 

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.25

method	result	size
derivativedivides	\(\frac {2 d^{2} \left (\frac {\left (-i c^{2}+i d^{2}+2 c d \right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{4 \left (i d -c \right )^{\frac {3}{2}} \left (i d +c \right )^{2} d^{2}}+\frac {i}{\left (i d +c \right ) \left (i c -d \right ) \left (i c +d \right ) \sqrt {c +d \tan \left (f x +e \right )}}-\frac {-\frac {\left (c^{2}+d^{2}\right ) d \sqrt {c +d \tan \left (f x +e \right )}}{\left (i d +c \right ) \left (-d \tan \left (f x +e \right )+i d \right )}-\frac {\left (i c^{3}+i c \,d^{2}-4 c^{2} d -4 d^{3}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{\left (i d +c \right ) \sqrt {-i d -c}}}{4 \left (i d -c \right ) \left (i d +c \right )^{2} d^{2}}\right )}{f a}\)	\(257\)
default	\(\frac {2 d^{2} \left (\frac {\left (-i c^{2}+i d^{2}+2 c d \right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{4 \left (i d -c \right )^{\frac {3}{2}} \left (i d +c \right )^{2} d^{2}}+\frac {i}{\left (i d +c \right ) \left (i c -d \right ) \left (i c +d \right ) \sqrt {c +d \tan \left (f x +e \right )}}-\frac {-\frac {\left (c^{2}+d^{2}\right ) d \sqrt {c +d \tan \left (f x +e \right )}}{\left (i d +c \right ) \left (-d \tan \left (f x +e \right )+i d \right )}-\frac {\left (i c^{3}+i c \,d^{2}-4 c^{2} d -4 d^{3}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{\left (i d +c \right ) \sqrt {-i d -c}}}{4 \left (i d -c \right ) \left (i d +c \right )^{2} d^{2}}\right )}{f a}\)	\(257\)

Input:

int(1/(a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

2/f/a*d^2*(1/4*(-I*c^2+I*d^2+2*c*d)/(I*d-c)^(3/2)/(c+I*d)^2/d^2*arctan((c+ 
d*tan(f*x+e))^(1/2)/(I*d-c)^(1/2))+I/(c+I*d)/(I*c-d)/(I*c+d)/(c+d*tan(f*x+ 
e))^(1/2)-1/4/(I*d-c)/(c+I*d)^2/d^2*(-(c^2+d^2)*d/(c+I*d)*(c+d*tan(f*x+e)) 
^(1/2)/(-d*tan(f*x+e)+I*d)-(-4*c^2*d-4*d^3+I*c^3+I*c*d^2)/(c+I*d)/(-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. 1555 vs. \(2 (159) = 318\).

Time = 0.35 (sec) , antiderivative size = 1555, normalized size of antiderivative = 7.59 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))^(3/2),x, algorithm="fricas 
")
 

Output:

-1/8*(2*((a*c^4 + 2*a*c^2*d^2 + a*d^4)*f*e^(4*I*f*x + 4*I*e) + (a*c^4 + 2* 
I*a*c^3*d + 2*I*a*c*d^3 - a*d^4)*f*e^(2*I*f*x + 2*I*e))*sqrt(-1/4*I/((I*a^ 
2*c^3 + 3*a^2*c^2*d - 3*I*a^2*c*d^2 - a^2*d^3)*f^2))*log(-2*(2*((I*a*c^2 + 
 2*a*c*d - I*a*d^2)*f*e^(2*I*f*x + 2*I*e) + (I*a*c^2 + 2*a*c*d - I*a*d^2)* 
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(-1/4*I/((I*a^2*c^3 + 3*a^2*c^2*d - 3*I*a^2*c*d^2 - a^2*d^3)*f^2)) 
- (c - I*d)*e^(2*I*f*x + 2*I*e) - c)*e^(-2*I*f*x - 2*I*e)) - 2*((a*c^4 + 2 
*a*c^2*d^2 + a*d^4)*f*e^(4*I*f*x + 4*I*e) + (a*c^4 + 2*I*a*c^3*d + 2*I*a*c 
*d^3 - a*d^4)*f*e^(2*I*f*x + 2*I*e))*sqrt(-1/4*I/((I*a^2*c^3 + 3*a^2*c^2*d 
 - 3*I*a^2*c*d^2 - a^2*d^3)*f^2))*log(-2*(2*((-I*a*c^2 - 2*a*c*d + I*a*d^2 
)*f*e^(2*I*f*x + 2*I*e) + (-I*a*c^2 - 2*a*c*d + I*a*d^2)*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(-1/4*I/(( 
I*a^2*c^3 + 3*a^2*c^2*d - 3*I*a^2*c*d^2 - a^2*d^3)*f^2)) - (c - I*d)*e^(2* 
I*f*x + 2*I*e) - c)*e^(-2*I*f*x - 2*I*e)) + ((a*c^4 + 2*a*c^2*d^2 + a*d^4) 
*f*e^(4*I*f*x + 4*I*e) + (a*c^4 + 2*I*a*c^3*d + 2*I*a*c*d^3 - a*d^4)*f*e^( 
2*I*f*x + 2*I*e))*sqrt(-(I*c^2 - 8*c*d - 16*I*d^2)/((I*a^2*c^5 - 5*a^2*c^4 
*d - 10*I*a^2*c^3*d^2 + 10*a^2*c^2*d^3 + 5*I*a^2*c*d^4 - a^2*d^5)*f^2))*lo 
g(1/2*(c^2 + 5*I*c*d - 4*d^2 + ((I*a*c^3 - 3*a*c^2*d - 3*I*a*c*d^2 + a*d^3 
)*f*e^(2*I*f*x + 2*I*e) + (I*a*c^3 - 3*a*c^2*d - 3*I*a*c*d^2 + a*d^3)*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)...
 

Sympy [F]

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

integrate(1/(a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))**(3/2),x)
 

Output:

-I*Integral(1/(c*sqrt(c + d*tan(e + f*x))*tan(e + f*x) - I*c*sqrt(c + d*ta 
n(e + f*x)) + d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2 - I*d*sqrt(c + d* 
tan(e + f*x))*tan(e + f*x)), x)/a
                                                                                    
                                                                                    
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(1/(a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))^(3/2),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. 463 vs. \(2 (159) = 318\).

Time = 0.63 (sec) , antiderivative size = 463, normalized size of antiderivative = 2.26 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx=-\frac {i \, {\left (\frac {2 \, \sqrt {2} {\left (i \, c - 4 \, d\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} + i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{-2 \, {\left (-i \, c^{2} + 2 \, c d + i \, d^{2}\right )} \sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {{\left (i \, d \tan \left (f x + e\right ) + i \, c\right )} c d + 5 \, {\left (d \tan \left (f x + e\right ) + c\right )} d^{2} - 4 \, c d^{2} - 4 i \, d^{3}}{{\left (c^{3} + i \, c^{2} d + c d^{2} + i \, d^{3}\right )} {\left ({\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} - \sqrt {d \tan \left (f x + e\right ) + c} c - i \, \sqrt {d \tan \left (f x + e\right ) + c} d\right )}} - \frac {\sqrt {2} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} - i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{{\left (c - i \, d\right )} \sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}}\right )}}{2 \, a f} \] Input:

integrate(1/(a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))^(3/2),x, algorithm="giac")
 

Output:

-1/2*I*(2*sqrt(2)*(I*c - 4*d)*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))))/((2*I*c^2 - 4*c*d - 2*I*d^2)*sqrt(-c + sqrt(c^2 + d 
^2))*(I*d/(c - sqrt(c^2 + d^2)) + 1)) + ((I*d*tan(f*x + e) + I*c)*c*d + 5* 
(d*tan(f*x + e) + c)*d^2 - 4*c*d^2 - 4*I*d^3)/((c^3 + I*c^2*d + c*d^2 + I* 
d^3)*((d*tan(f*x + e) + c)^(3/2) - sqrt(d*tan(f*x + e) + c)*c - I*sqrt(d*t 
an(f*x + e) + c)*d)) - sqrt(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))))/((c - I*d)*sqrt(-c + sqrt(c^2 + d^2))*(-I*d/(c - s 
qrt(c^2 + d^2)) + 1)))/(a*f)
 

Mupad [B] (verification not implemented)

Time = 12.67 (sec) , antiderivative size = 35674, normalized size of antiderivative = 174.02 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

log(10*a*d^7*f - ((-(240*c^2*d^7 - 240*d^9 - c*d^8*720i + c^3*d^6*80i + 16 
0*c^4*d^5 - c^5*d^4*32i - a^2*c^6*f^2*(((1280*c^3*d^8 - 1200*c*d^10 + 208* 
c^5*d^6 + 32*c^7*d^4)/(a^2*c^8*f^2 + a^2*d^8*f^2 + 4*a^2*c^2*d^6*f^2 + 6*a 
^2*c^4*d^4*f^2 + 4*a^2*c^6*d^2*f^2) + ((240*d^11 - 1920*c^2*d^9 + 240*c^4* 
d^7 + 96*c^6*d^5)*1i)/(a^2*c^8*f^2 + a^2*d^8*f^2 + 4*a^2*c^2*d^6*f^2 + 6*a 
^2*c^4*d^4*f^2 + 4*a^2*c^6*d^2*f^2))^2 - 4*(256*d^6 + 256*c^2*d^4)*(((24*c 
*d^7 + 6*c^3*d^5)*1i)/(a^4*c^8*f^4 + a^4*d^8*f^4 + 4*a^4*c^2*d^6*f^4 + 6*a 
^4*c^4*d^4*f^4 + 4*a^4*c^6*d^2*f^4) + (16*d^8 - c^2*d^6 + c^4*d^4)/(a^4*c^ 
8*f^4 + a^4*d^8*f^4 + 4*a^4*c^2*d^6*f^4 + 6*a^4*c^4*d^4*f^4 + 4*a^4*c^6*d^ 
2*f^4)))^(1/2)*1i + a^2*d^6*f^2*(((1280*c^3*d^8 - 1200*c*d^10 + 208*c^5*d^ 
6 + 32*c^7*d^4)/(a^2*c^8*f^2 + a^2*d^8*f^2 + 4*a^2*c^2*d^6*f^2 + 6*a^2*c^4 
*d^4*f^2 + 4*a^2*c^6*d^2*f^2) + ((240*d^11 - 1920*c^2*d^9 + 240*c^4*d^7 + 
96*c^6*d^5)*1i)/(a^2*c^8*f^2 + a^2*d^8*f^2 + 4*a^2*c^2*d^6*f^2 + 6*a^2*c^4 
*d^4*f^2 + 4*a^2*c^6*d^2*f^2))^2 - 4*(256*d^6 + 256*c^2*d^4)*(((24*c*d^7 + 
 6*c^3*d^5)*1i)/(a^4*c^8*f^4 + a^4*d^8*f^4 + 4*a^4*c^2*d^6*f^4 + 6*a^4*c^4 
*d^4*f^4 + 4*a^4*c^6*d^2*f^4) + (16*d^8 - c^2*d^6 + c^4*d^4)/(a^4*c^8*f^4 
+ a^4*d^8*f^4 + 4*a^4*c^2*d^6*f^4 + 6*a^4*c^4*d^4*f^4 + 4*a^4*c^6*d^2*f^4) 
))^(1/2)*1i + 2*a^2*c*d^5*f^2*(((1280*c^3*d^8 - 1200*c*d^10 + 208*c^5*d^6 
+ 32*c^7*d^4)/(a^2*c^8*f^2 + a^2*d^8*f^2 + 4*a^2*c^2*d^6*f^2 + 6*a^2*c^4*d 
^4*f^2 + 4*a^2*c^6*d^2*f^2) + ((240*d^11 - 1920*c^2*d^9 + 240*c^4*d^7 +...
 

Reduce [F]

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

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

Output:

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