Integrand size = 33, antiderivative size = 199 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {35 i \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{64 \sqrt {2} a^2 c^{3/2} f}-\frac {35 i}{96 a^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {7 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {35 i}{64 a^2 c f \sqrt {c-i c \tan (e+f x)}} \] Output:
35/128*I*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)/a^2 /c^(3/2)/f-35/96*I/a^2/f/(c-I*c*tan(f*x+e))^(3/2)+1/4*I/a^2/f/(1+I*tan(f*x +e))^2/(c-I*c*tan(f*x+e))^(3/2)+7/16*I/a^2/f/(1+I*tan(f*x+e))/(c-I*c*tan(f *x+e))^(3/2)-35/64*I/a^2/c/f/(c-I*c*tan(f*x+e))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.85 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.27 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {i \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},3,-\frac {1}{2},-\frac {1}{2} i (i+\tan (e+f x))\right )}{12 a^2 f (c-i c \tan (e+f x))^{3/2}} \] Input:
Integrate[1/((a + I*a*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(3/2)),x]
Output:
((-1/12*I)*Hypergeometric2F1[-3/2, 3, -1/2, (-1/2*I)*(I + Tan[e + f*x])])/ (a^2*f*(c - I*c*Tan[e + f*x])^(3/2))
Time = 0.44 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.97, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3042, 4005, 3042, 3968, 52, 52, 61, 61, 73, 219}
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.
| \(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 4005 |
| \(\displaystyle \frac {\int \cos ^4(e+f x) \sqrt {c-i c \tan (e+f x)}dx}{a^2 c^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {c-i c \tan (e+f x)}}{\sec (e+f x)^4}dx}{a^2 c^2}\) |
| \(\Big \downarrow \) 3968 |
| \(\displaystyle \frac {i c^3 \int \frac {1}{(c-i c \tan (e+f x))^{5/2} (i \tan (e+f x) c+c)^3}d(-i c \tan (e+f x))}{a^2 f}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {i c^3 \left (\frac {7 \int \frac {1}{(c-i c \tan (e+f x))^{5/2} (i \tan (e+f x) c+c)^2}d(-i c \tan (e+f x))}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))^2}\right )}{a^2 f}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {i c^3 \left (\frac {7 \left (\frac {5 \int \frac {1}{(c-i c \tan (e+f x))^{5/2} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))^2}\right )}{a^2 f}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {i c^3 \left (\frac {7 \left (\frac {5 \left (\frac {\int \frac {1}{(c-i c \tan (e+f x))^{3/2} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{2 c}-\frac {1}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))^2}\right )}{a^2 f}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {i c^3 \left (\frac {7 \left (\frac {5 \left (\frac {\frac {\int \frac {1}{\sqrt {c-i c \tan (e+f x)} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{2 c}-\frac {1}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {1}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))^2}\right )}{a^2 f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {i c^3 \left (\frac {7 \left (\frac {5 \left (\frac {\frac {\int \frac {1}{c^2 \tan ^2(e+f x)+2 c}d\sqrt {c-i c \tan (e+f x)}}{c}-\frac {1}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {1}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))^2}\right )}{a^2 f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {i c^3 \left (\frac {7 \left (\frac {5 \left (\frac {-\frac {i \arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2}}\right )}{\sqrt {2} c^{3/2}}-\frac {1}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {1}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))^2}\right )}{a^2 f}\) |
Input:
Int[1/((a + I*a*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(3/2)),x]
Output:
(I*c^3*(1/(4*c*(c - I*c*Tan[e + f*x])^(3/2)*(c + I*c*Tan[e + f*x])^2) + (7 *(1/(2*c*(c - I*c*Tan[e + f*x])^(3/2)*(c + I*c*Tan[e + f*x])) + (5*(-1/3*1 /(c*(c - I*c*Tan[e + f*x])^(3/2)) + (((-I)*ArcTan[(Sqrt[c]*Tan[e + f*x])/S qrt[2]])/(Sqrt[2]*c^(3/2)) - 1/(c*Sqrt[c - I*c*Tan[e + f*x]]))/(2*c)))/(4* c)))/(8*c)))/(a^2*f)
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
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]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[1/(a^(m - 2)*b*f) Subst[Int[(a - x)^(m/2 - 1)*(a + x )^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] && !(IGtQ[n, 0] && (LtQ[ m, 0] || GtQ[m, n]))
Time = 0.33 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.70
| method | result | size |
| derivativedivides | \(\frac {2 i c^{3} \left (-\frac {3}{16 c^{4} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{24 c^{3} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {\frac {-\frac {11 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{8}+\frac {13 \sqrt {c -i c \tan \left (f x +e \right )}\, c}{4}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {35 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{16 \sqrt {c}}}{16 c^{4}}\right )}{f \,a^{2}}\) | \(139\) |
| default | \(\frac {2 i c^{3} \left (-\frac {3}{16 c^{4} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{24 c^{3} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {\frac {-\frac {11 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{8}+\frac {13 \sqrt {c -i c \tan \left (f x +e \right )}\, c}{4}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {35 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{16 \sqrt {c}}}{16 c^{4}}\right )}{f \,a^{2}}\) | \(139\) |
Input:
int(1/(a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(3/2),x,method=_RETURNVERBOS E)
Output:
2*I/f/a^2*c^3*(-3/16/c^4/(c-I*c*tan(f*x+e))^(1/2)-1/24/c^3/(c-I*c*tan(f*x+ e))^(3/2)+1/16/c^4*(4*(-11/32*(c-I*c*tan(f*x+e))^(3/2)+13/16*(c-I*c*tan(f* x+e))^(1/2)*c)/(c+I*c*tan(f*x+e))^2+35/16*2^(1/2)/c^(1/2)*arctanh(1/2*(c-I *c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (148) = 296\).
Time = 0.09 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.61 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {{\left (-105 i \, \sqrt {\frac {1}{2}} a^{2} c^{2} f \sqrt {\frac {1}{a^{4} c^{3} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {35 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{2} c f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a^{2} c f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{4} c^{3} f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{32 \, a^{2} c f}\right ) + 105 i \, \sqrt {\frac {1}{2}} a^{2} c^{2} f \sqrt {\frac {1}{a^{4} c^{3} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {35 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{2} c f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{2} c f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{4} c^{3} f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{32 \, a^{2} c f}\right ) + \sqrt {2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-8 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 88 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 41 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 45 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 6 i\right )}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{384 \, a^{2} c^{2} f} \] Input:
integrate(1/(a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fr icas")
Output:
1/384*(-105*I*sqrt(1/2)*a^2*c^2*f*sqrt(1/(a^4*c^3*f^2))*e^(4*I*f*x + 4*I*e )*log(-35/32*(sqrt(2)*sqrt(1/2)*(I*a^2*c*f*e^(2*I*f*x + 2*I*e) + I*a^2*c*f )*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^4*c^3*f^2)) - I)*e^(-I*f*x - I*e)/(a^2*c*f)) + 105*I*sqrt(1/2)*a^2*c^2*f*sqrt(1/(a^4*c^3*f^2))*e^(4*I* f*x + 4*I*e)*log(-35/32*(sqrt(2)*sqrt(1/2)*(-I*a^2*c*f*e^(2*I*f*x + 2*I*e) - I*a^2*c*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^4*c^3*f^2)) - I) *e^(-I*f*x - I*e)/(a^2*c*f)) + sqrt(2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*( -8*I*e^(8*I*f*x + 8*I*e) - 88*I*e^(6*I*f*x + 6*I*e) - 41*I*e^(4*I*f*x + 4* I*e) + 45*I*e^(2*I*f*x + 2*I*e) + 6*I))*e^(-4*I*f*x - 4*I*e)/(a^2*c^2*f)
\[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}} \, dx=- \frac {\int \frac {1}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} - c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx}{a^{2}} \] Input:
integrate(1/(a+I*a*tan(f*x+e))**2/(c-I*c*tan(f*x+e))**(3/2),x)
Output:
-Integral(1/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 - c*sqrt(-I* c*tan(e + f*x) + c)*tan(e + f*x)**2 - I*c*sqrt(-I*c*tan(e + f*x) + c)*tan( e + f*x) - c*sqrt(-I*c*tan(e + f*x) + c)), x)/a**2
Time = 0.15 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.92 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {i \, {\left (\frac {4 \, {\left (105 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} - 350 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} c + 224 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} c^{2} + 64 \, c^{3}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{2} - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{2} c + 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{2} c^{2}} + \frac {105 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{2} \sqrt {c}}\right )}}{768 \, c f} \] Input:
integrate(1/(a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="ma xima")
Output:
-1/768*I*(4*(105*(-I*c*tan(f*x + e) + c)^3 - 350*(-I*c*tan(f*x + e) + c)^2 *c + 224*(-I*c*tan(f*x + e) + c)*c^2 + 64*c^3)/((-I*c*tan(f*x + e) + c)^(7 /2)*a^2 - 4*(-I*c*tan(f*x + e) + c)^(5/2)*a^2*c + 4*(-I*c*tan(f*x + e) + c )^(3/2)*a^2*c^2) + 105*sqrt(2)*log(-(sqrt(2)*sqrt(c) - sqrt(-I*c*tan(f*x + e) + c))/(sqrt(2)*sqrt(c) + sqrt(-I*c*tan(f*x + e) + c)))/(a^2*sqrt(c)))/ (c*f)
Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/(a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="gi ac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad Argument TypeDone
Time = 2.28 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.91 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {-\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,175{}\mathrm {i}}{96\,a^2\,f}+\frac {c^2\,1{}\mathrm {i}}{3\,a^2\,f}+\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,35{}\mathrm {i}}{64\,a^2\,c\,f}+\frac {c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,7{}\mathrm {i}}{6\,a^2\,f}}{-4\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}+{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}+4\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,35{}\mathrm {i}}{128\,a^2\,{\left (-c\right )}^{3/2}\,f} \] Input:
int(1/((a + a*tan(e + f*x)*1i)^2*(c - c*tan(e + f*x)*1i)^(3/2)),x)
Output:
(2^(1/2)*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*35i) /(128*a^2*(-c)^(3/2)*f) - ((c^2*1i)/(3*a^2*f) - ((c - c*tan(e + f*x)*1i)^2 *175i)/(96*a^2*f) + ((c - c*tan(e + f*x)*1i)^3*35i)/(64*a^2*c*f) + (c*(c - c*tan(e + f*x)*1i)*7i)/(6*a^2*f))/((c - c*tan(e + f*x)*1i)^(7/2) - 4*c*(c - c*tan(e + f*x)*1i)^(5/2) + 4*c^2*(c - c*tan(e + f*x)*1i)^(3/2))
\[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {\sqrt {c}\, \left (-\sqrt {-\tan \left (f x +e \right ) i +1}\, i -7 \left (\int -\frac {\sqrt {-\tan \left (f x +e \right ) i +1}}{2 \tan \left (f x +e \right )^{5} i +2 \tan \left (f x +e \right )^{4}+4 \tan \left (f x +e \right )^{3} i +4 \tan \left (f x +e \right )^{2}+2 \tan \left (f x +e \right ) i +2}d x \right ) \tan \left (f x +e \right )^{4} f -14 \left (\int -\frac {\sqrt {-\tan \left (f x +e \right ) i +1}}{2 \tan \left (f x +e \right )^{5} i +2 \tan \left (f x +e \right )^{4}+4 \tan \left (f x +e \right )^{3} i +4 \tan \left (f x +e \right )^{2}+2 \tan \left (f x +e \right ) i +2}d x \right ) \tan \left (f x +e \right )^{2} f -7 \left (\int -\frac {\sqrt {-\tan \left (f x +e \right ) i +1}}{2 \tan \left (f x +e \right )^{5} i +2 \tan \left (f x +e \right )^{4}+4 \tan \left (f x +e \right )^{3} i +4 \tan \left (f x +e \right )^{2}+2 \tan \left (f x +e \right ) i +2}d x \right ) f -7 \left (\int -\frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}}{2 \tan \left (f x +e \right )^{5} i +2 \tan \left (f x +e \right )^{4}+4 \tan \left (f x +e \right )^{3} i +4 \tan \left (f x +e \right )^{2}+2 \tan \left (f x +e \right ) i +2}d x \right ) \tan \left (f x +e \right )^{4} f -14 \left (\int -\frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}}{2 \tan \left (f x +e \right )^{5} i +2 \tan \left (f x +e \right )^{4}+4 \tan \left (f x +e \right )^{3} i +4 \tan \left (f x +e \right )^{2}+2 \tan \left (f x +e \right ) i +2}d x \right ) \tan \left (f x +e \right )^{2} f -7 \left (\int -\frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}}{2 \tan \left (f x +e \right )^{5} i +2 \tan \left (f x +e \right )^{4}+4 \tan \left (f x +e \right )^{3} i +4 \tan \left (f x +e \right )^{2}+2 \tan \left (f x +e \right ) i +2}d x \right ) f \right )}{3 a^{2} c^{2} f \left (\tan \left (f x +e \right )^{4}+2 \tan \left (f x +e \right )^{2}+1\right )} \] Input:
int(1/(a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(3/2),x)
Output:
(sqrt(c)*( - sqrt( - tan(e + f*x)*i + 1)*i - 7*int(( - sqrt( - tan(e + f*x )*i + 1))/(2*tan(e + f*x)**5*i + 2*tan(e + f*x)**4 + 4*tan(e + f*x)**3*i + 4*tan(e + f*x)**2 + 2*tan(e + f*x)*i + 2),x)*tan(e + f*x)**4*f - 14*int(( - sqrt( - tan(e + f*x)*i + 1))/(2*tan(e + f*x)**5*i + 2*tan(e + f*x)**4 + 4*tan(e + f*x)**3*i + 4*tan(e + f*x)**2 + 2*tan(e + f*x)*i + 2),x)*tan(e + f*x)**2*f - 7*int(( - sqrt( - tan(e + f*x)*i + 1))/(2*tan(e + f*x)**5*i + 2*tan(e + f*x)**4 + 4*tan(e + f*x)**3*i + 4*tan(e + f*x)**2 + 2*tan(e + f*x)*i + 2),x)*f - 7*int(( - sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2)/ (2*tan(e + f*x)**5*i + 2*tan(e + f*x)**4 + 4*tan(e + f*x)**3*i + 4*tan(e + f*x)**2 + 2*tan(e + f*x)*i + 2),x)*tan(e + f*x)**4*f - 14*int(( - sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2)/(2*tan(e + f*x)**5*i + 2*tan(e + f*x )**4 + 4*tan(e + f*x)**3*i + 4*tan(e + f*x)**2 + 2*tan(e + f*x)*i + 2),x)* tan(e + f*x)**2*f - 7*int(( - sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2) /(2*tan(e + f*x)**5*i + 2*tan(e + f*x)**4 + 4*tan(e + f*x)**3*i + 4*tan(e + f*x)**2 + 2*tan(e + f*x)*i + 2),x)*f))/(3*a**2*c**2*f*(tan(e + f*x)**4 + 2*tan(e + f*x)**2 + 1))