Integrand size = 32, antiderivative size = 269 \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} (c-i d)^{3/2} f}-\frac {1}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}+\frac {3 i c-11 d}{6 a (c+i d)^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {(3 c-5 i d) (c+5 i d) d \sqrt {a+i a \tan (e+f x)}}{6 a^2 (c-i d) (c+i d)^3 f \sqrt {c+d \tan (e+f x)}} \]
-1/4*I*arctanh(2^(1/2)*a^(1/2)*(c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2)/(a+I*a *tan(f*x+e))^(1/2))/a^(3/2)/(c-I*d)^(3/2)/f*2^(1/2)+1/6*(3*I*c-11*d)/a/(c+ I*d)^2/f/(a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(1/2)+1/6*(3*c-5*I*d)*( c+5*I*d)*d*(a+I*a*tan(f*x+e))^(1/2)/a^2/(c-I*d)/(c+I*d)^3/f/(c+d*tan(f*x+e ))^(1/2)-1/3/(I*c-d)/f/(c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(3/2)
Time = 5.02 (sec) , antiderivative size = 257, normalized size of antiderivative = 0.96 \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=\frac {\frac {4 i (c+i d)}{(a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}+\frac {6 i c-22 d}{a \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {\frac {3 i \sqrt {2} a (c+i d)^3 \arctan \left (\frac {\sqrt {-a (c-i d)} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} a \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a (c-i d)}}-\frac {2 d \left (3 c^2+10 i c d+25 d^2\right ) \sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}}{a^2 \left (c^2+d^2\right )}}{12 (c+i d)^2 f} \]
(((4*I)*(c + I*d))/((a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c + d*Tan[e + f*x]]) + ((6*I)*c - 22*d)/(a*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]] ) - (((3*I)*Sqrt[2]*a*(c + I*d)^3*ArcTan[(Sqrt[-(a*(c - I*d))]*Sqrt[a + I* a*Tan[e + f*x]])/(Sqrt[2]*a*Sqrt[c + d*Tan[e + f*x]])])/Sqrt[-(a*(c - I*d) )] - (2*d*(3*c^2 + (10*I)*c*d + 25*d^2)*Sqrt[a + I*a*Tan[e + f*x]])/Sqrt[c + d*Tan[e + f*x]])/(a^2*(c^2 + d^2)))/(12*(c + I*d)^2*f)
Time = 1.36 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.15, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 4042, 27, 3042, 4079, 27, 3042, 4081, 27, 3042, 4027, 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.
\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}dx\) |
\(\Big \downarrow \) 4042 |
\(\displaystyle -\frac {\int -\frac {a (3 i c-7 d)+4 i a d \tan (e+f x)}{2 \sqrt {i \tan (e+f x) a+a} (c+d \tan (e+f x))^{3/2}}dx}{3 a^2 (-d+i c)}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a (3 i c-7 d)+4 i a d \tan (e+f x)}{\sqrt {i \tan (e+f x) a+a} (c+d \tan (e+f x))^{3/2}}dx}{6 a^2 (-d+i c)}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {a (3 i c-7 d)+4 i a d \tan (e+f x)}{\sqrt {i \tan (e+f x) a+a} (c+d \tan (e+f x))^{3/2}}dx}{6 a^2 (-d+i c)}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 4079 |
\(\displaystyle \frac {-\frac {\int \frac {\sqrt {i \tan (e+f x) a+a} \left (\left (3 c^2+12 i d c-25 d^2\right ) a^2+2 (3 c+11 i d) d \tan (e+f x) a^2\right )}{2 (c+d \tan (e+f x))^{3/2}}dx}{a^2 (-d+i c)}-\frac {a (3 c+11 i d)}{f (c+i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}}{6 a^2 (-d+i c)}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {-\frac {\int \frac {\sqrt {i \tan (e+f x) a+a} \left (\left (3 c^2+12 i d c-25 d^2\right ) a^2+2 (3 c+11 i d) d \tan (e+f x) a^2\right )}{(c+d \tan (e+f x))^{3/2}}dx}{2 a^2 (-d+i c)}-\frac {a (3 c+11 i d)}{f (c+i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}}{6 a^2 (-d+i c)}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-\frac {\int \frac {\sqrt {i \tan (e+f x) a+a} \left (\left (3 c^2+12 i d c-25 d^2\right ) a^2+2 (3 c+11 i d) d \tan (e+f x) a^2\right )}{(c+d \tan (e+f x))^{3/2}}dx}{2 a^2 (-d+i c)}-\frac {a (3 c+11 i d)}{f (c+i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}}{6 a^2 (-d+i c)}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 4081 |
\(\displaystyle \frac {-\frac {\frac {2 \int \frac {3 a^3 (c+i d)^3 \sqrt {i \tan (e+f x) a+a}}{2 \sqrt {c+d \tan (e+f x)}}dx}{a \left (c^2+d^2\right )}+\frac {2 a^2 d \left (3 c^2+10 i c d+25 d^2\right ) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{2 a^2 (-d+i c)}-\frac {a (3 c+11 i d)}{f (c+i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}}{6 a^2 (-d+i c)}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {-\frac {\frac {3 a^2 (c+i d)^3 \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {2 a^2 d \left (3 c^2+10 i c d+25 d^2\right ) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{2 a^2 (-d+i c)}-\frac {a (3 c+11 i d)}{f (c+i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}}{6 a^2 (-d+i c)}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-\frac {\frac {3 a^2 (c+i d)^3 \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {2 a^2 d \left (3 c^2+10 i c d+25 d^2\right ) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{2 a^2 (-d+i c)}-\frac {a (3 c+11 i d)}{f (c+i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}}{6 a^2 (-d+i c)}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 4027 |
\(\displaystyle \frac {-\frac {\frac {2 a^2 d \left (3 c^2+10 i c d+25 d^2\right ) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {6 i a^4 (c+i d)^3 \int \frac {1}{a (c-i d)-\frac {2 a^2 (c+d \tan (e+f x))}{i \tan (e+f x) a+a}}d\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {i \tan (e+f x) a+a}}}{f \left (c^2+d^2\right )}}{2 a^2 (-d+i c)}-\frac {a (3 c+11 i d)}{f (c+i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}}{6 a^2 (-d+i c)}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {-\frac {\frac {2 a^2 d \left (3 c^2+10 i c d+25 d^2\right ) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {3 i \sqrt {2} a^{5/2} (c+i d)^3 \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f \sqrt {c-i d} \left (c^2+d^2\right )}}{2 a^2 (-d+i c)}-\frac {a (3 c+11 i d)}{f (c+i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}}{6 a^2 (-d+i c)}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}\) |
-1/3*1/((I*c - d)*f*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c + d*Tan[e + f*x]]) + (-((a*(3*c + (11*I)*d))/((c + I*d)*f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])) - (((-3*I)*Sqrt[2]*a^(5/2)*(c + I*d)^3*ArcTanh[(Sqrt[2 ]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f* x]])])/(Sqrt[c - I*d]*(c^2 + d^2)*f) + (2*a^2*d*(3*c^2 + (10*I)*c*d + 25*d ^2)*Sqrt[a + I*a*Tan[e + f*x]])/((c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]]))/ (2*a^2*(I*c - d)))/(6*a^2*(I*c - d))
3.12.65.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*a*(b/f) Subst[Int[1/(a*c - b*d - 2* a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && N eQ[c^2 + d^2, 0]
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])
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]
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*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 + d^2))), x] - Simp[1/(a*(n + 1)*(c^2 + d^2)) Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c* m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Fr eeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 4834 vs. \(2 (219 ) = 438\).
Time = 1.39 (sec) , antiderivative size = 4835, normalized size of antiderivative = 17.97
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(4835\) |
default | \(\text {Expression too large to display}\) | \(4835\) |
1/24/f*(-12*I*c^6*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*tan(f*x+e)^2 -256*I*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*d^6*tan(f*x+e)^2+40*I*( a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*c^4*d^2+68*I*(a*(1+I*tan(f*x+e) )*(c+d*tan(f*x+e)))^(1/2)*c^2*d^4-3*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d *tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+ e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^6*tan(f*x+e)^3-9* ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^( 1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)* (-a*(I*d-c))^(1/2)*d^6*tan(f*x+e)^3+9*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a *d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f* x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^6*tan(f*x+e)+3* ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^( 1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)* (-a*(I*d-c))^(1/2)*d^6*tan(f*x+e)+15*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a* d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x +e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^5*d+20*I*(a*(1+I *tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*c^6+48*I*(a*(1+I*tan(f*x+e))*(c+d*tan (f*x+e)))^(1/2)*d^6-30*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2 *2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/( tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^3*d^3+3*ln((3*a*c+I*a*tan(f...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 982 vs. \(2 (205) = 410\).
Time = 0.30 (sec) , antiderivative size = 982, normalized size of antiderivative = 3.65 \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \]
-1/12*(sqrt(2)*(c^3 + I*c^2*d + c*d^2 + I*d^3 + 2*(2*c^3 + 3*I*c^2*d + 12* c*d^2 - 19*I*d^3)*e^(6*I*f*x + 6*I*e) + (9*c^3 + 19*I*c^2*d + 29*c*d^2 - 2 5*I*d^3)*e^(4*I*f*x + 4*I*e) + 2*(3*c^3 + 7*I*c^2*d + 3*c*d^2 + 7*I*d^3)*e ^(2*I*f*x + 2*I*e))*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(a/(e^(2*I*f*x + 2*I*e) + 1)) - 3*((-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^(5 *I*f*x + 5*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^(3*I*f*x + 3*I*e))*sqrt(-1/2*I/((I*a^3* c^3 + 3*a^3*c^2*d - 3*I*a^3*c*d^2 - a^3*d^3)*f^2))*log(-2*(I*a^2*c^2 + 2*a ^2*c*d - I*a^2*d^2)*f*sqrt(-1/2*I/((I*a^3*c^3 + 3*a^3*c^2*d - 3*I*a^3*c*d^ 2 - a^3*d^3)*f^2))*e^(I*f*x + I*e) + sqrt(2)*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(a/(e^(2*I*f*x + 2*I*e) + 1))*(e^(2*I*f*x + 2*I*e) + 1)) - 3*((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^(5*I*f*x + 5*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^(3*I*f*x + 3*I*e))*sqrt(-1/2*I/((I*a^3*c^3 + 3*a^3*c^2*d - 3*I* a^3*c*d^2 - a^3*d^3)*f^2))*log(-2*(-I*a^2*c^2 - 2*a^2*c*d + I*a^2*d^2)*f*s qrt(-1/2*I/((I*a^3*c^3 + 3*a^3*c^2*d - 3*I*a^3*c*d^2 - a^3*d^3)*f^2))*e^(I *f*x + I*e) + sqrt(2)*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(a/(e^(2*I*f*x + 2*I*e) + 1))*(e^(2*I*f*x + 2...
\[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=\int \frac {1}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: RuntimeError} \]
Timed out. \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=\int \frac {1}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \]