\(\int \frac {1}{(a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}} \, dx\) [1123]

Optimal result

Integrand size = 30, antiderivative size = 221 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{4 a^2 \sqrt {c-i d} f}+\frac {\left (2 i c^2-6 c d-7 i d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{8 a^2 (c+i d)^{5/2} f}+\frac {(2 i c-5 d) \sqrt {c+d \tan (e+f x)}}{8 a^2 (c+i d)^2 f (1+i \tan (e+f x))}-\frac {\sqrt {c+d \tan (e+f x)}}{4 (i c-d) f (a+i a \tan (e+f x))^2} \] Output:

-1/4*I*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/a^2/(c-I*d)^(1/2)/f+1 
/8*(2*I*c^2-6*c*d-7*I*d^2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/a 
^2/(c+I*d)^(5/2)/f+1/8*(2*I*c-5*d)*(c+d*tan(f*x+e))^(1/2)/a^2/(c+I*d)^2/f/ 
(1+I*tan(f*x+e))-1/4*(c+d*tan(f*x+e))^(1/2)/(I*c-d)/f/(a+I*a*tan(f*x+e))^2
 

Mathematica [A] (verified)

Time = 1.12 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.92 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}} \, dx=-\frac {\frac {2 i (c+i d)^2 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d}}+\frac {\left (-2 i c^2+6 c d+7 i d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d}}+\frac {2 i (c+i d) \sqrt {c+d \tan (e+f x)}}{(-i+\tan (e+f x))^2}-\frac {(2 c+5 i d) \sqrt {c+d \tan (e+f x)}}{-i+\tan (e+f x)}}{8 a^2 (c+i d)^2 f} \] Input:

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

Output:

-1/8*(((2*I)*(c + I*d)^2*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/ 
Sqrt[c - I*d] + (((-2*I)*c^2 + 6*c*d + (7*I)*d^2)*ArcTanh[Sqrt[c + d*Tan[e 
 + f*x]]/Sqrt[c + I*d]])/Sqrt[c + I*d] + ((2*I)*(c + I*d)*Sqrt[c + d*Tan[e 
 + f*x]])/(-I + Tan[e + f*x])^2 - ((2*c + (5*I)*d)*Sqrt[c + d*Tan[e + f*x] 
])/(-I + Tan[e + f*x]))/(a^2*(c + I*d)^2*f)
 

Rubi [A] (warning: unable to verify)

Time = 1.19 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4042

\(\displaystyle -\frac {\int -\frac {a (4 i c-7 d)+3 i a d \tan (e+f x)}{2 (i \tan (e+f x) a+a) \sqrt {c+d \tan (e+f x)}}dx}{4 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{4 f (-d+i c) (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4079

\(\displaystyle \frac {-\frac {\int \frac {\left (4 c^2+10 i d c-9 d^2\right ) a^2+(2 c+5 i d) d \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}-\frac {(2 c+5 i d) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}}{8 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{4 f (-d+i c) (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\int \frac {\left (4 c^2+10 i d c-9 d^2\right ) a^2+(2 c+5 i d) d \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}-\frac {(2 c+5 i d) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}}{8 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{4 f (-d+i c) (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {-\frac {a^2 \left (2 c^2+6 i c d-7 d^2\right ) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+2 a^2 (c+i d)^2 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}-\frac {(2 c+5 i d) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}}{8 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{4 f (-d+i c) (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {a^2 \left (2 c^2+6 i c d-7 d^2\right ) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+2 a^2 (c+i d)^2 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}-\frac {(2 c+5 i d) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}}{8 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{4 f (-d+i c) (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {-\frac {\frac {2 i a^2 (c+i d)^2 \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{f}-\frac {i a^2 \left (2 c^2+6 i c d-7 d^2\right ) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{f}}{2 a^2 (-d+i c)}-\frac {(2 c+5 i d) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}}{8 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{4 f (-d+i c) (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-\frac {\frac {i a^2 \left (2 c^2+6 i c d-7 d^2\right ) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{f}-\frac {2 i a^2 (c+i d)^2 \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{f}}{2 a^2 (-d+i c)}-\frac {(2 c+5 i d) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}}{8 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{4 f (-d+i c) (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {-\frac {\frac {2 a^2 \left (2 c^2+6 i c d-7 d^2\right ) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {4 a^2 (c+i d)^2 \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}}{2 a^2 (-d+i c)}-\frac {(2 c+5 i d) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}}{8 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{4 f (-d+i c) (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {-\frac {\frac {2 a^2 \left (2 c^2+6 i c d-7 d^2\right ) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}+\frac {4 a^2 (c+i d)^2 \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}}{2 a^2 (-d+i c)}-\frac {(2 c+5 i d) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}}{8 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{4 f (-d+i c) (a+i a \tan (e+f x))^2}\)

Input:

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

Output:

-1/4*Sqrt[c + d*Tan[e + f*x]]/((I*c - d)*f*(a + I*a*Tan[e + f*x])^2) + (-1 
/2*((4*a^2*(c + I*d)^2*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/(Sqrt[c - I*d]* 
f) + (2*a^2*(2*c^2 + (6*I)*c*d - 7*d^2)*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]] 
)/(Sqrt[c + I*d]*f))/(a^2*(I*c - d)) - ((2*c + (5*I)*d)*Sqrt[c + d*Tan[e + 
 f*x]])/((c + I*d)*f*(1 + I*Tan[e + f*x])))/(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 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 = 0.48 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(\frac {2 d^{3} \left (\frac {i \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{8 d^{3} \sqrt {i d -c}}+\frac {\frac {\frac {d \left (5 i d +2 c \right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{4 i c d +2 c^{2}-2 d^{2}}-\frac {d \left (9 i c d +2 c^{2}-7 d^{2}\right ) \sqrt {c +d \tan \left (f x +e \right )}}{2 \left (2 i c d +c^{2}-d^{2}\right )}}{\left (-d \tan \left (f x +e \right )+i d \right )^{2}}-\frac {\left (2 i c^{2}-7 i d^{2}-6 c d \right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{2 \left (2 i c d +c^{2}-d^{2}\right ) \sqrt {-i d -c}}}{8 d^{3}}\right )}{f \,a^{2}}\)	\(232\)
default	\(\frac {2 d^{3} \left (\frac {i \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{8 d^{3} \sqrt {i d -c}}+\frac {\frac {\frac {d \left (5 i d +2 c \right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{4 i c d +2 c^{2}-2 d^{2}}-\frac {d \left (9 i c d +2 c^{2}-7 d^{2}\right ) \sqrt {c +d \tan \left (f x +e \right )}}{2 \left (2 i c d +c^{2}-d^{2}\right )}}{\left (-d \tan \left (f x +e \right )+i d \right )^{2}}-\frac {\left (2 i c^{2}-7 i d^{2}-6 c d \right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{2 \left (2 i c d +c^{2}-d^{2}\right ) \sqrt {-i d -c}}}{8 d^{3}}\right )}{f \,a^{2}}\)	\(232\)

Input:

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

Output:

2/f/a^2*d^3*(1/8*I/d^3/(I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(I*d-c) 
^(1/2))+1/8/d^3*((1/2*d*(5*I*d+2*c)/(2*I*c*d+c^2-d^2)*(c+d*tan(f*x+e))^(3/ 
2)-1/2*d*(9*I*c*d+2*c^2-7*d^2)/(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*(2*I*c^2-6*c*d-7*I*d^2)/(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))))
 

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. 1400 vs. \(2 (173) = 346\).

Time = 0.26 (sec) , antiderivative size = 1400, normalized size of antiderivative = 6.33 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/32*(8*(a^2*c^2 + 2*I*a^2*c*d - a^2*d^2)*f*sqrt(1/16*I/((-I*a^4*c - a^4* 
d)*f^2))*e^(4*I*f*x + 4*I*e)*log(-2*(4*((I*a^2*c + a^2*d)*f*e^(2*I*f*x + 2 
*I*e) + (I*a^2*c + a^2*d)*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 - a^4*d)*f^2)) - (c - 
I*d)*e^(2*I*f*x + 2*I*e) - c)*e^(-2*I*f*x - 2*I*e)) - 8*(a^2*c^2 + 2*I*a^2 
*c*d - a^2*d^2)*f*sqrt(1/16*I/((-I*a^4*c - a^4*d)*f^2))*e^(4*I*f*x + 4*I*e 
)*log(-2*(4*((-I*a^2*c - a^2*d)*f*e^(2*I*f*x + 2*I*e) + (-I*a^2*c - a^2*d) 
*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 - a^4*d)*f^2)) - (c - I*d)*e^(2*I*f*x + 2*I*e) 
- c)*e^(-2*I*f*x - 2*I*e)) + (a^2*c^2 + 2*I*a^2*c*d - a^2*d^2)*f*sqrt(-(4* 
I*c^4 - 24*c^3*d - 64*I*c^2*d^2 + 84*c*d^3 + 49*I*d^4)/((I*a^4*c^5 - 5*a^4 
*c^4*d - 10*I*a^4*c^3*d^2 + 10*a^4*c^2*d^3 + 5*I*a^4*c*d^4 - a^4*d^5)*f^2) 
)*e^(4*I*f*x + 4*I*e)*log(1/8*(2*c^3 + 8*I*c^2*d - 13*c*d^2 - 7*I*d^3 + (( 
I*a^2*c^3 - 3*a^2*c^2*d - 3*I*a^2*c*d^2 + a^2*d^3)*f*e^(2*I*f*x + 2*I*e) + 
 (I*a^2*c^3 - 3*a^2*c^2*d - 3*I*a^2*c*d^2 + a^2*d^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(-(4*I*c^4 - 2 
4*c^3*d - 64*I*c^2*d^2 + 84*c*d^3 + 49*I*d^4)/((I*a^4*c^5 - 5*a^4*c^4*d - 
10*I*a^4*c^3*d^2 + 10*a^4*c^2*d^3 + 5*I*a^4*c*d^4 - a^4*d^5)*f^2)) + (2*c^ 
3 + 6*I*c^2*d - 7*c*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/((-I*a^ 
2*c^3 + 3*a^2*c^2*d + 3*I*a^2*c*d^2 - a^2*d^3)*f)) - (a^2*c^2 + 2*I*a^2...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(1/(a+I*a*tan(f*x+e))^2/(c+d*tan(f*x+e))^(1/2),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. 471 vs. \(2 (173) = 346\).

Time = 0.59 (sec) , antiderivative size = 471, normalized size of antiderivative = 2.13 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}} \, dx=-\frac {\frac {8 \, \sqrt {2} {\left (2 \, c^{2} + 6 i \, c d - 7 \, d^{2}\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} + i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{-8 \, {\left (i \, c^{2} - 2 \, c d - i \, d^{2}\right )} \sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {2 i \, \sqrt {2} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} - i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{\sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {2 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c d - 2 \, \sqrt {d \tan \left (f x + e\right ) + c} c^{2} d + 5 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} d^{2} - 9 i \, \sqrt {d \tan \left (f x + e\right ) + c} c d^{2} + 7 \, \sqrt {d \tan \left (f x + e\right ) + c} d^{3}}{{\left (c^{2} + 2 i \, c d - d^{2}\right )} {\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2}}}{8 \, a^{2} f} \] Input:

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

Output:

-1/8*(8*sqrt(2)*(2*c^2 + 6*I*c*d - 7*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^2 + 16*c*d + 8*I*d^2)*sqrt(-c 
+ sqrt(c^2 + d^2))*(I*d/(c - sqrt(c^2 + d^2)) + 1)) - 2*I*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))))/(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 + 5*I*(d*tan(f*x + e) + c)^(3/ 
2)*d^2 - 9*I*sqrt(d*tan(f*x + e) + c)*c*d^2 + 7*sqrt(d*tan(f*x + e) + c)*d 
^3)/((c^2 + 2*I*c*d - d^2)*(d*tan(f*x + e) - I*d)^2))/(a^2*f)
 

Mupad [B] (verification not implemented)

Time = 6.17 (sec) , antiderivative size = 32178, normalized size of antiderivative = 145.60 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

log(a^2*d^10*f*35i - ((-(45*d^9 - c*d^8*15i + 60*c^2*d^7 - c^3*d^6*80i - 4 
0*c^4*d^5 + c^5*d^4*8i + a^4*c^4*f^2*(((165*c*d^12 + 70*c^3*d^10 + 73*c^5* 
d^8 + 32*c^7*d^6 + 8*c^9*d^4)/(a^4*c^8*f^2 + a^4*d^8*f^2 + 4*a^4*c^2*d^6*f 
^2 + 6*a^4*c^4*d^4*f^2 + 4*a^4*c^6*d^2*f^2) + ((150*c^2*d^11 - 45*d^13 + 9 
5*c^4*d^9 + 52*c^6*d^7 + 8*c^8*d^5)*1i)/(a^4*c^8*f^2 + a^4*d^8*f^2 + 4*a^4 
*c^2*d^6*f^2 + 6*a^4*c^4*d^4*f^2 + 4*a^4*c^6*d^2*f^2))^2 - 4*(256*d^6 + 25 
6*c^2*d^4)*((((7*c*d^11)/4 + (19*c^3*d^9)/16 + (11*c^5*d^7)/16 + (c^7*d^5) 
/8)*1i)/(a^8*c^8*f^4 + a^8*d^8*f^4 + 4*a^8*c^2*d^6*f^4 + 6*a^8*c^4*d^4*f^4 
 + 4*a^8*c^6*d^2*f^4) + ((49*d^12)/64 - (11*c^2*d^10)/32 + (5*c^4*d^8)/64 
+ (c^6*d^6)/8 + (c^8*d^4)/16)/(a^8*c^8*f^4 + a^8*d^8*f^4 + 4*a^8*c^2*d^6*f 
^4 + 6*a^8*c^4*d^4*f^4 + 4*a^8*c^6*d^2*f^4)))^(1/2)*1i + a^4*d^4*f^2*(((16 
5*c*d^12 + 70*c^3*d^10 + 73*c^5*d^8 + 32*c^7*d^6 + 8*c^9*d^4)/(a^4*c^8*f^2 
 + a^4*d^8*f^2 + 4*a^4*c^2*d^6*f^2 + 6*a^4*c^4*d^4*f^2 + 4*a^4*c^6*d^2*f^2 
) + ((150*c^2*d^11 - 45*d^13 + 95*c^4*d^9 + 52*c^6*d^7 + 8*c^8*d^5)*1i)/(a 
^4*c^8*f^2 + a^4*d^8*f^2 + 4*a^4*c^2*d^6*f^2 + 6*a^4*c^4*d^4*f^2 + 4*a^4*c 
^6*d^2*f^2))^2 - 4*(256*d^6 + 256*c^2*d^4)*((((7*c*d^11)/4 + (19*c^3*d^9)/ 
16 + (11*c^5*d^7)/16 + (c^7*d^5)/8)*1i)/(a^8*c^8*f^4 + a^8*d^8*f^4 + 4*a^8 
*c^2*d^6*f^4 + 6*a^8*c^4*d^4*f^4 + 4*a^8*c^6*d^2*f^4) + ((49*d^12)/64 - (1 
1*c^2*d^10)/32 + (5*c^4*d^8)/64 + (c^6*d^6)/8 + (c^8*d^4)/16)/(a^8*c^8*f^4 
 + a^8*d^8*f^4 + 4*a^8*c^2*d^6*f^4 + 6*a^8*c^4*d^4*f^4 + 4*a^8*c^6*d^2*...
 

Reduce [F]

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

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

Output:

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