\(\int \frac {1}{(a+i a \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}} \, dx\) [1129]

Optimal result

Integrand size = 30, antiderivative size = 281 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{4 a^2 (c-i d)^{3/2} f}+\frac {\left (2 i c^2-10 c d-23 i d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{8 a^2 (c+i d)^{7/2} f}+\frac {d \left (2 c^2+7 i c d+25 d^2\right )}{8 a^2 (c-i d) (c+i d)^3 f \sqrt {c+d \tan (e+f x)}}+\frac {2 i c-7 d}{8 a^2 (c+i d)^2 f (1+i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}-\frac {1}{4 (i c-d) f (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}} \] Output:

-1/4*I*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/a^2/(c-I*d)^(3/2)/f+1 
/8*(2*I*c^2-10*c*d-23*I*d^2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2)) 
/a^2/(c+I*d)^(7/2)/f+1/8*d*(2*c^2+7*I*c*d+25*d^2)/a^2/(c-I*d)/(c+I*d)^3/f/ 
(c+d*tan(f*x+e))^(1/2)+1/8*(2*I*c-7*d)/a^2/(c+I*d)^2/f/(1+I*tan(f*x+e))/(c 
+d*tan(f*x+e))^(1/2)-1/4/(I*c-d)/f/(a+I*a*tan(f*x+e))^2/(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 = 0.89 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.67 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}} \, dx=-\frac {-\frac {2 i (c+i d)^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {c+d \tan (e+f x)}{c-i d}\right )}{c-i d}+\frac {i \left (2 c^2+10 i c d-23 d^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {c+d \tan (e+f x)}{c+i d}\right )}{c+i d}+\frac {2 i (c+i d)}{(-i+\tan (e+f x))^2}+\frac {-2 c-7 i d}{-i+\tan (e+f x)}}{8 a^2 (c+i d)^2 f \sqrt {c+d \tan (e+f x)}} \] Input:

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

Output:

-1/8*(((-2*I)*(c + I*d)^2*Hypergeometric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f 
*x])/(c - I*d)])/(c - I*d) + (I*(2*c^2 + (10*I)*c*d - 23*d^2)*Hypergeometr 
ic2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c + I*d)])/(c + I*d) + ((2*I)*(c 
 + I*d))/(-I + Tan[e + f*x])^2 + (-2*c - (7*I)*d)/(-I + Tan[e + f*x]))/(a^ 
2*(c + I*d)^2*f*Sqrt[c + d*Tan[e + f*x]])
 

Rubi [A] (warning: unable to verify)

Time = 1.55 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.09, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {3042, 4042, 27, 3042, 4079, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4042

\(\displaystyle -\frac {\int -\frac {a (4 i c-9 d)+5 i a d \tan (e+f x)}{2 (i \tan (e+f x) a+a) (c+d \tan (e+f x))^{3/2}}dx}{4 a^2 (-d+i c)}-\frac {1}{4 f (-d+i c) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (4 i c-9 d)+5 i a d \tan (e+f x)}{(i \tan (e+f x) a+a) (c+d \tan (e+f x))^{3/2}}dx}{8 a^2 (-d+i c)}-\frac {1}{4 f (-d+i c) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (4 i c-9 d)+5 i a d \tan (e+f x)}{(i \tan (e+f x) a+a) (c+d \tan (e+f x))^{3/2}}dx}{8 a^2 (-d+i c)}-\frac {1}{4 f (-d+i c) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {-\frac {\int \frac {\left (4 c^2+14 i d c-25 d^2\right ) a^2+3 (2 c+7 i d) d \tan (e+f x) a^2}{(c+d \tan (e+f x))^{3/2}}dx}{2 a^2 (-d+i c)}-\frac {2 c+7 i d}{f (c+i d) (1+i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}}{8 a^2 (-d+i c)}-\frac {1}{4 f (-d+i c) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\int \frac {\left (4 c^2+14 i d c-25 d^2\right ) a^2+3 (2 c+7 i d) d \tan (e+f x) a^2}{(c+d \tan (e+f x))^{3/2}}dx}{2 a^2 (-d+i c)}-\frac {2 c+7 i d}{f (c+i d) (1+i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}}{8 a^2 (-d+i c)}-\frac {1}{4 f (-d+i c) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 4012

\(\displaystyle \frac {-\frac {\frac {\int \frac {\left (4 c^3+14 i d c^2-19 d^2 c+21 i d^3\right ) a^2+d \left (2 c^2+7 i d c+25 d^2\right ) \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {2 a^2 d \left (2 c^2+7 i c d+25 d^2\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{2 a^2 (-d+i c)}-\frac {2 c+7 i d}{f (c+i d) (1+i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}}{8 a^2 (-d+i c)}-\frac {1}{4 f (-d+i c) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {\int \frac {\left (4 c^3+14 i d c^2-19 d^2 c+21 i d^3\right ) a^2+d \left (2 c^2+7 i d c+25 d^2\right ) \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {2 a^2 d \left (2 c^2+7 i c d+25 d^2\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{2 a^2 (-d+i c)}-\frac {2 c+7 i d}{f (c+i d) (1+i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}}{8 a^2 (-d+i c)}-\frac {1}{4 f (-d+i c) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {-\frac {\frac {a^2 (c-i d) \left (2 c^2+10 i c d-23 d^2\right ) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+2 a^2 (c+i d)^3 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {2 a^2 d \left (2 c^2+7 i c d+25 d^2\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{2 a^2 (-d+i c)}-\frac {2 c+7 i d}{f (c+i d) (1+i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}}{8 a^2 (-d+i c)}-\frac {1}{4 f (-d+i c) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {a^2 (c-i d) \left (2 c^2+10 i c d-23 d^2\right ) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+2 a^2 (c+i d)^3 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {2 a^2 d \left (2 c^2+7 i c d+25 d^2\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{2 a^2 (-d+i c)}-\frac {2 c+7 i d}{f (c+i d) (1+i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}}{8 a^2 (-d+i c)}-\frac {1}{4 f (-d+i c) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {-\frac {\frac {\frac {2 i a^2 (c+i d)^3 \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{f}-\frac {i a^2 (c-i d) \left (2 c^2+10 i c d-23 d^2\right ) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{f}}{c^2+d^2}+\frac {2 a^2 d \left (2 c^2+7 i c d+25 d^2\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{2 a^2 (-d+i c)}-\frac {2 c+7 i d}{f (c+i d) (1+i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}}{8 a^2 (-d+i c)}-\frac {1}{4 f (-d+i c) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-\frac {\frac {\frac {i a^2 (c-i d) \left (2 c^2+10 i c d-23 d^2\right ) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{f}-\frac {2 i a^2 (c+i d)^3 \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{f}}{c^2+d^2}+\frac {2 a^2 d \left (2 c^2+7 i c d+25 d^2\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{2 a^2 (-d+i c)}-\frac {2 c+7 i d}{f (c+i d) (1+i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}}{8 a^2 (-d+i c)}-\frac {1}{4 f (-d+i c) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {-\frac {\frac {\frac {2 a^2 (c-i d) \left (2 c^2+10 i c d-23 d^2\right ) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {4 a^2 (c+i d)^3 \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}}{c^2+d^2}+\frac {2 a^2 d \left (2 c^2+7 i c d+25 d^2\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{2 a^2 (-d+i c)}-\frac {2 c+7 i d}{f (c+i d) (1+i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}}{8 a^2 (-d+i c)}-\frac {1}{4 f (-d+i c) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {-\frac {\frac {\frac {2 a^2 (c-i d) \left (2 c^2+10 i c d-23 d^2\right ) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}+\frac {4 a^2 (c+i d)^3 \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}}{c^2+d^2}+\frac {2 a^2 d \left (2 c^2+7 i c d+25 d^2\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{2 a^2 (-d+i c)}-\frac {2 c+7 i d}{f (c+i d) (1+i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}}{8 a^2 (-d+i c)}-\frac {1}{4 f (-d+i c) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}\)

Input:

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

Output:

-1/4*1/((I*c - d)*f*(a + I*a*Tan[e + f*x])^2*Sqrt[c + d*Tan[e + f*x]]) + ( 
-((2*c + (7*I)*d)/((c + I*d)*f*(1 + I*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x 
]])) - (((4*a^2*(c + I*d)^3*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/(Sqrt[c - 
I*d]*f) + (2*a^2*(c - I*d)*(2*c^2 + (10*I)*c*d - 23*d^2)*ArcTan[Tan[e + f* 
x]/Sqrt[c + I*d]])/(Sqrt[c + I*d]*f))/(c^2 + d^2) + (2*a^2*d*(2*c^2 + (7*I 
)*c*d + 25*d^2))/((c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]]))/(2*a^2*(I*c - d 
)))/(8*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 4042

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(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))   In 
t[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m 
 + n + 1) + b*d*(m + n + 1)*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] 
 && LtQ[m, 0] && (IntegerQ[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 = 0.38 (sec) , antiderivative size = 399, normalized size of antiderivative = 1.42

method	result	size
derivativedivides	\(\frac {2 d^{3} \left (-\frac {1}{\left (i c -d \right ) \left (i c +d \right ) \left (i d +c \right )^{2} \sqrt {c +d \tan \left (f x +e \right )}}-\frac {i \left (\frac {-\frac {d \left (2 i c^{4}-7 i c^{2} d^{2}-9 i d^{4}-11 c^{3} d -11 c \,d^{3}\right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2 \left (2 i c d +c^{2}-d^{2}\right )}+\frac {d \left (2 i c^{5}-22 i c^{3} d^{2}-24 i c \,d^{4}-15 c^{4} d -4 c^{2} d^{3}+11 d^{5}\right ) \sqrt {c +d \tan \left (f x +e \right )}}{4 i c d +2 c^{2}-2 d^{2}}}{\left (-d \tan \left (f x +e \right )+i d \right )^{2}}-\frac {\left (12 i c^{4} d -11 i c^{2} d^{3}-23 i d^{5}+2 c^{5}-31 c^{3} d^{2}-33 c \,d^{4}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{2 \left (2 i c d +c^{2}-d^{2}\right ) \sqrt {-i d -c}}\right )}{8 d^{3} \left (i d +c \right )^{3} \left (i d -c \right )}-\frac {i \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{8 \left (i d -c \right )^{\frac {3}{2}} \left (i d +c \right )^{3} d^{3}}\right )}{f \,a^{2}}\)	\(399\)
default	\(\frac {2 d^{3} \left (-\frac {1}{\left (i c -d \right ) \left (i c +d \right ) \left (i d +c \right )^{2} \sqrt {c +d \tan \left (f x +e \right )}}-\frac {i \left (\frac {-\frac {d \left (2 i c^{4}-7 i c^{2} d^{2}-9 i d^{4}-11 c^{3} d -11 c \,d^{3}\right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2 \left (2 i c d +c^{2}-d^{2}\right )}+\frac {d \left (2 i c^{5}-22 i c^{3} d^{2}-24 i c \,d^{4}-15 c^{4} d -4 c^{2} d^{3}+11 d^{5}\right ) \sqrt {c +d \tan \left (f x +e \right )}}{4 i c d +2 c^{2}-2 d^{2}}}{\left (-d \tan \left (f x +e \right )+i d \right )^{2}}-\frac {\left (12 i c^{4} d -11 i c^{2} d^{3}-23 i d^{5}+2 c^{5}-31 c^{3} d^{2}-33 c \,d^{4}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{2 \left (2 i c d +c^{2}-d^{2}\right ) \sqrt {-i d -c}}\right )}{8 d^{3} \left (i d +c \right )^{3} \left (i d -c \right )}-\frac {i \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{8 \left (i d -c \right )^{\frac {3}{2}} \left (i d +c \right )^{3} d^{3}}\right )}{f \,a^{2}}\)	\(399\)

Input:

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

Output:

2/f/a^2*d^3*(-1/(I*c-d)/(I*c+d)/(c+I*d)^2/(c+d*tan(f*x+e))^(1/2)-1/8*I/d^3 
/(c+I*d)^3/(I*d-c)*((-1/2*d*(2*I*c^4-7*I*c^2*d^2-9*I*d^4-11*c^3*d-11*c*d^3 
)/(2*I*c*d+c^2-d^2)*(c+d*tan(f*x+e))^(3/2)+1/2*d*(2*I*c^5-22*I*c^3*d^2-24* 
I*c*d^4-15*c^4*d-4*c^2*d^3+11*d^5)/(2*I*c*d+c^2-d^2)*(c+d*tan(f*x+e))^(1/2 
))/(-d*tan(f*x+e)+I*d)^2-1/2*(-31*c^3*d^2-33*c*d^4+12*I*c^4*d-11*I*c^2*d^3 
-23*I*d^5+2*c^5)/(2*I*c*d+c^2-d^2)/(-c-I*d)^(1/2)*arctan((c+d*tan(f*x+e))^ 
(1/2)/(-c-I*d)^(1/2)))-1/8*I/(I*d-c)^(3/2)/(c+I*d)^3*(3*I*c^2*d-I*d^3+c^3- 
3*c*d^2)/d^3*arctan((c+d*tan(f*x+e))^(1/2)/(I*d-c)^(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. 2263 vs. \(2 (223) = 446\).

Time = 0.96 (sec) , antiderivative size = 2263, normalized size of antiderivative = 8.05 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/32*(8*((-I*a^2*c^5 + a^2*c^4*d - 2*I*a^2*c^3*d^2 + 2*a^2*c^2*d^3 - I*a^2 
*c*d^4 + a^2*d^5)*f*e^(6*I*f*x + 6*I*e) + (-I*a^2*c^5 + 3*a^2*c^4*d + 2*I* 
a^2*c^3*d^2 + 2*a^2*c^2*d^3 + 3*I*a^2*c*d^4 - a^2*d^5)*f*e^(4*I*f*x + 4*I* 
e))*sqrt(-1/16*I/((I*a^4*c^3 + 3*a^4*c^2*d - 3*I*a^4*c*d^2 - a^4*d^3)*f^2) 
)*log(-2*(4*((I*a^2*c^2 + 2*a^2*c*d - I*a^2*d^2)*f*e^(2*I*f*x + 2*I*e) + ( 
I*a^2*c^2 + 2*a^2*c*d - I*a^2*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(-1/16*I/((I*a^4*c^3 + 3*a^4*c^2 
*d - 3*I*a^4*c*d^2 - a^4*d^3)*f^2)) - (c - I*d)*e^(2*I*f*x + 2*I*e) - c)*e 
^(-2*I*f*x - 2*I*e)) + 8*((I*a^2*c^5 - a^2*c^4*d + 2*I*a^2*c^3*d^2 - 2*a^2 
*c^2*d^3 + I*a^2*c*d^4 - a^2*d^5)*f*e^(6*I*f*x + 6*I*e) + (I*a^2*c^5 - 3*a 
^2*c^4*d - 2*I*a^2*c^3*d^2 - 2*a^2*c^2*d^3 - 3*I*a^2*c*d^4 + a^2*d^5)*f*e^ 
(4*I*f*x + 4*I*e))*sqrt(-1/16*I/((I*a^4*c^3 + 3*a^4*c^2*d - 3*I*a^4*c*d^2 
- a^4*d^3)*f^2))*log(-2*(4*((-I*a^2*c^2 - 2*a^2*c*d + I*a^2*d^2)*f*e^(2*I* 
f*x + 2*I*e) + (-I*a^2*c^2 - 2*a^2*c*d + I*a^2*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(-1/16*I/((I*a^ 
4*c^3 + 3*a^4*c^2*d - 3*I*a^4*c*d^2 - a^4*d^3)*f^2)) - (c - I*d)*e^(2*I*f* 
x + 2*I*e) - c)*e^(-2*I*f*x - 2*I*e)) + ((I*a^2*c^5 - a^2*c^4*d + 2*I*a^2* 
c^3*d^2 - 2*a^2*c^2*d^3 + I*a^2*c*d^4 - a^2*d^5)*f*e^(6*I*f*x + 6*I*e) + ( 
I*a^2*c^5 - 3*a^2*c^4*d - 2*I*a^2*c^3*d^2 - 2*a^2*c^2*d^3 - 3*I*a^2*c*d^4 
+ a^2*d^5)*f*e^(4*I*f*x + 4*I*e))*sqrt(-(-4*I*c^4 + 40*c^3*d + 192*I*c^...
 

Sympy [F]

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

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

Output:

-Integral(1/(c*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2 - 2*I*c*sqrt(c + d 
*tan(e + f*x))*tan(e + f*x) - c*sqrt(c + d*tan(e + f*x)) + d*sqrt(c + d*ta 
n(e + f*x))*tan(e + f*x)**3 - 2*I*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)* 
*2 - d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)), x)/a**2
 

Maxima [F(-2)]

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

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

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. 535 vs. \(2 (223) = 446\).

Time = 0.65 (sec) , antiderivative size = 535, normalized size of antiderivative = 1.90 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}} \, dx=\frac {\frac {16 \, d^{3}}{{\left (c^{4} + 2 i \, c^{3} d + 2 i \, c d^{3} - d^{4}\right )} \sqrt {d \tan \left (f x + e\right ) + c}} + \frac {8 \, \sqrt {2} {\left (2 \, c^{2} + 10 i \, c d - 23 \, d^{2}\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} + i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{-8 \, {\left (-i \, c^{3} + 3 \, c^{2} d + 3 i \, c d^{2} - d^{3}\right )} \sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {8 \, \sqrt {2} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} - i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{-4 \, {\left (-i \, c - d\right )} \sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {2 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c d - 2 \, \sqrt {d \tan \left (f x + e\right ) + c} c^{2} d + 9 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} d^{2} - 13 i \, \sqrt {d \tan \left (f x + e\right ) + c} c d^{2} + 11 \, \sqrt {d \tan \left (f x + e\right ) + c} d^{3}}{{\left (c^{3} + 3 i \, c^{2} d - 3 \, c d^{2} - i \, d^{3}\right )} {\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2}}}{8 \, a^{2} f} \] Input:

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

Output:

1/8*(16*d^3/((c^4 + 2*I*c^3*d + 2*I*c*d^3 - d^4)*sqrt(d*tan(f*x + e) + c)) 
 + 8*sqrt(2)*(2*c^2 + 10*I*c*d - 23*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))))/((8*I*c^3 - 24*c^2*d - 24*I*c*d^2 + 8*d^ 
3)*sqrt(-c + sqrt(c^2 + d^2))*(I*d/(c - sqrt(c^2 + d^2)) + 1)) - 8*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))))/((4*I* 
c + 4*d)*sqrt(-c + sqrt(c^2 + d^2))*(-I*d/(c - sqrt(c^2 + d^2)) + 1)) + (2 
*(d*tan(f*x + e) + c)^(3/2)*c*d - 2*sqrt(d*tan(f*x + e) + c)*c^2*d + 9*I*( 
d*tan(f*x + e) + c)^(3/2)*d^2 - 13*I*sqrt(d*tan(f*x + e) + c)*c*d^2 + 11*s 
qrt(d*tan(f*x + e) + c)*d^3)/((c^3 + 3*I*c^2*d - 3*c*d^2 - I*d^3)*(d*tan(f 
*x + e) - I*d)^2))/(a^2*f)
 

Mupad [B] (verification not implemented)

Time = 12.06 (sec) , antiderivative size = 66456, normalized size of antiderivative = 236.50 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

log(575*a^2*d^11*f - ((-(c*d^10*1155i + 525*d^11 - 315*c^2*d^9 + c^3*d^8*1 
75i - 140*c^4*d^7 + c^5*d^6*168i + 56*c^6*d^5 - c^7*d^4*8i - a^4*c^8*f^2*( 
(((525*d^15 - 8085*c^2*d^13 + 6195*c^4*d^11 + 609*c^6*d^9 + 228*c^8*d^7 + 
24*c^10*d^5)*1i)/(a^4*c^12*f^2 + a^4*d^12*f^2 + 6*a^4*c^2*d^10*f^2 + 15*a^ 
4*c^4*d^8*f^2 + 20*a^4*c^6*d^6*f^2 + 15*a^4*c^8*d^4*f^2 + 6*a^4*c^10*d^2*f 
^2) + (10115*c^3*d^12 - 3255*c*d^14 - 973*c^5*d^10 + 57*c^7*d^8 + 8*c^9*d^ 
6 + 8*c^11*d^4)/(a^4*c^12*f^2 + a^4*d^12*f^2 + 6*a^4*c^2*d^10*f^2 + 15*a^4 
*c^4*d^8*f^2 + 20*a^4*c^6*d^6*f^2 + 15*a^4*c^8*d^4*f^2 + 6*a^4*c^10*d^2*f^ 
2))^2 - 4*((((207*c*d^11)/8 - (21*c^3*d^9)/16 + (21*c^5*d^7)/16 + (3*c^7*d 
^5)/8)*1i)/(a^8*c^12*f^4 + a^8*d^12*f^4 + 6*a^8*c^2*d^10*f^4 + 15*a^8*c^4* 
d^8*f^4 + 20*a^8*c^6*d^6*f^4 + 15*a^8*c^8*d^4*f^4 + 6*a^8*c^10*d^2*f^4) - 
((763*c^2*d^10)/32 - (529*d^12)/64 + (315*c^4*d^8)/64 + (7*c^6*d^6)/8 - (c 
^8*d^4)/16)/(a^8*c^12*f^4 + a^8*d^12*f^4 + 6*a^8*c^2*d^10*f^4 + 15*a^8*c^4 
*d^8*f^4 + 20*a^8*c^6*d^6*f^4 + 15*a^8*c^8*d^4*f^4 + 6*a^8*c^10*d^2*f^4))* 
(256*d^6 + 256*c^2*d^4))^(1/2)*1i - a^4*d^8*f^2*((((525*d^15 - 8085*c^2*d^ 
13 + 6195*c^4*d^11 + 609*c^6*d^9 + 228*c^8*d^7 + 24*c^10*d^5)*1i)/(a^4*c^1 
2*f^2 + a^4*d^12*f^2 + 6*a^4*c^2*d^10*f^2 + 15*a^4*c^4*d^8*f^2 + 20*a^4*c^ 
6*d^6*f^2 + 15*a^4*c^8*d^4*f^2 + 6*a^4*c^10*d^2*f^2) + (10115*c^3*d^12 - 3 
255*c*d^14 - 973*c^5*d^10 + 57*c^7*d^8 + 8*c^9*d^6 + 8*c^11*d^4)/(a^4*c^12 
*f^2 + a^4*d^12*f^2 + 6*a^4*c^2*d^10*f^2 + 15*a^4*c^4*d^8*f^2 + 20*a^4*...
 

Reduce [F]

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

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

Output:

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