Integrand size = 28, antiderivative size = 164 \[ \int \frac {1}{(e \cos (c+d x))^{11/2} (a+i a \tan (c+d x))^2} \, dx=-\frac {14 \cos ^{\frac {11}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d (e \cos (c+d x))^{11/2}}+\frac {14 \cos ^3(c+d x) \sin (c+d x)}{15 a^2 d (e \cos (c+d x))^{11/2}}+\frac {14 \cos ^5(c+d x) \sin (c+d x)}{5 a^2 d (e \cos (c+d x))^{11/2}}-\frac {4 i \cos ^2(c+d x)}{3 d (e \cos (c+d x))^{11/2} \left (a^2+i a^2 \tan (c+d x)\right )} \] Output:
-14/5*cos(d*x+c)^(11/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d/(e*cos (d*x+c))^(11/2)+14/15*cos(d*x+c)^3*sin(d*x+c)/a^2/d/(e*cos(d*x+c))^(11/2)+ 14/5*cos(d*x+c)^5*sin(d*x+c)/a^2/d/(e*cos(d*x+c))^(11/2)-4/3*I*cos(d*x+c)^ 2/d/(e*cos(d*x+c))^(11/2)/(a^2+I*a^2*tan(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 3.59 (sec) , antiderivative size = 414, normalized size of antiderivative = 2.52 \[ \int \frac {1}{(e \cos (c+d x))^{11/2} (a+i a \tan (c+d x))^2} \, dx=\frac {\cos ^3(c+d x) (\cos (d x)+i \sin (d x))^2 \left (7 \cos (c) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c)))-\frac {7}{2} (3 \cos (c-d x-\arctan (\tan (c)))+\cos (c+d x+\arctan (\tan (c)))) \cot (c) \sqrt {\sin ^2(d x+\arctan (\tan (c)))}+\frac {1}{6} \csc (c) \sqrt {\sec ^2(c)} \sec ^2(c+d x) (\cos (2 c)+i \sin (2 c)) (36 \cos (d x)+27 \cos (2 c+d x)+21 \cos (2 c+3 d x)+20 i \sin (d x)-20 i \sin (2 c+d x)) \sqrt {\sin ^2(d x+\arctan (\tan (c)))}+7 i \left (2 \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (c) \sin (d x+\arctan (\tan (c)))-(3 \cos (c-d x-\arctan (\tan (c)))+\cos (c+d x+\arctan (\tan (c)))) \sqrt {\sin ^2(d x+\arctan (\tan (c)))}\right )+\frac {7}{2} \left (-2 \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (c) \sin (d x+\arctan (\tan (c)))+(3 \cos (c-d x-\arctan (\tan (c)))+\cos (c+d x+\arctan (\tan (c)))) \sqrt {\sin ^2(d x+\arctan (\tan (c)))}\right ) \tan (c)\right )}{5 d (e \cos (c+d x))^{11/2} \sqrt {\sec ^2(c)} \sqrt {\sin ^2(d x+\arctan (\tan (c)))} (a+i a \tan (c+d x))^2} \] Input:
Integrate[1/((e*Cos[c + d*x])^(11/2)*(a + I*a*Tan[c + d*x])^2),x]
Output:
(Cos[c + d*x]^3*(Cos[d*x] + I*Sin[d*x])^2*(7*Cos[c]*HypergeometricPFQ[{-1/ 2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]] - (7*(3*Cos[c - d*x - ArcTan[Tan[c]]] + Cos[c + d*x + ArcTan[Tan[c]]])*Cot[c ]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2])/2 + (Csc[c]*Sqrt[Sec[c]^2]*Sec[c + d* x]^2*(Cos[2*c] + I*Sin[2*c])*(36*Cos[d*x] + 27*Cos[2*c + d*x] + 21*Cos[2*c + 3*d*x] + (20*I)*Sin[d*x] - (20*I)*Sin[2*c + d*x])*Sqrt[Sin[d*x + ArcTan [Tan[c]]]^2])/6 + (7*I)*(2*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[c]*Sin[d*x + ArcTan[Tan[c]]] - (3*Cos[c - d*x - A rcTan[Tan[c]]] + Cos[c + d*x + ArcTan[Tan[c]]])*Sqrt[Sin[d*x + ArcTan[Tan[ c]]]^2]) + (7*(-2*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[ Tan[c]]]^2]*Sin[c]*Sin[d*x + ArcTan[Tan[c]]] + (3*Cos[c - d*x - ArcTan[Tan [c]]] + Cos[c + d*x + ArcTan[Tan[c]]])*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2])* Tan[c])/2))/(5*d*(e*Cos[c + d*x])^(11/2)*Sqrt[Sec[c]^2]*Sqrt[Sin[d*x + Arc Tan[Tan[c]]]^2]*(a + I*a*Tan[c + d*x])^2)
Time = 0.90 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.09, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3998, 3042, 3981, 3042, 4255, 3042, 4255, 3042, 4258, 3042, 3119}
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.
| \(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^2 (e \cos (c+d x))^{11/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^2 (e \cos (c+d x))^{11/2}}dx\) |
| \(\Big \downarrow \) 3998 |
| \(\displaystyle \frac {\int \frac {(e \sec (c+d x))^{11/2}}{(i \tan (c+d x) a+a)^2}dx}{(e \cos (c+d x))^{11/2} (e \sec (c+d x))^{11/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(e \sec (c+d x))^{11/2}}{(i \tan (c+d x) a+a)^2}dx}{(e \cos (c+d x))^{11/2} (e \sec (c+d x))^{11/2}}\) |
| \(\Big \downarrow \) 3981 |
| \(\displaystyle \frac {\frac {7 e^2 \int (e \sec (c+d x))^{7/2}dx}{3 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}}{(e \cos (c+d x))^{11/2} (e \sec (c+d x))^{11/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {7 e^2 \int \left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}dx}{3 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}}{(e \cos (c+d x))^{11/2} (e \sec (c+d x))^{11/2}}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\frac {7 e^2 \left (\frac {3}{5} e^2 \int (e \sec (c+d x))^{3/2}dx+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )}{3 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}}{(e \cos (c+d x))^{11/2} (e \sec (c+d x))^{11/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {7 e^2 \left (\frac {3}{5} e^2 \int \left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )}{3 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}}{(e \cos (c+d x))^{11/2} (e \sec (c+d x))^{11/2}}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\frac {7 e^2 \left (\frac {3}{5} e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-e^2 \int \frac {1}{\sqrt {e \sec (c+d x)}}dx\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )}{3 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}}{(e \cos (c+d x))^{11/2} (e \sec (c+d x))^{11/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {7 e^2 \left (\frac {3}{5} e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-e^2 \int \frac {1}{\sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )}{3 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}}{(e \cos (c+d x))^{11/2} (e \sec (c+d x))^{11/2}}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {\frac {7 e^2 \left (\frac {3}{5} e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {e^2 \int \sqrt {\cos (c+d x)}dx}{\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )}{3 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}}{(e \cos (c+d x))^{11/2} (e \sec (c+d x))^{11/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {7 e^2 \left (\frac {3}{5} e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {e^2 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )}{3 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}}{(e \cos (c+d x))^{11/2} (e \sec (c+d x))^{11/2}}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\frac {7 e^2 \left (\frac {3}{5} e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )}{3 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}}{(e \cos (c+d x))^{11/2} (e \sec (c+d x))^{11/2}}\) |
Input:
Int[1/((e*Cos[c + d*x])^(11/2)*(a + I*a*Tan[c + d*x])^2),x]
Output:
((7*e^2*((2*e*(e*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(5*d) + (3*e^2*((-2*e^2 *EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (2*e*Sqrt[e*Sec[c + d*x]]*Sin[c + d*x])/d))/5))/(3*a^2) - (((4*I)/3)*e^2*( e*Sec[c + d*x])^(7/2))/(d*(a^2 + I*a^2*Tan[c + d*x])))/((e*Cos[c + d*x])^( 11/2)*(e*Sec[c + d*x])^(11/2))
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x _)])^(n_), x_Symbol] :> Simp[2*d^2*(d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + f*x])^(n + 1)/(b*f*(m + 2*n))), x] - Simp[d^2*((m - 2)/(b^2*(m + 2*n))) Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[ {a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (IntegersQ[n, m + 1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x _)])^(n_.), x_Symbol] :> Simp[(d*Cos[e + f*x])^m*(d*Sec[e + f*x])^m Int[( a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m , n}, x] && !IntegerQ[m]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (146 ) = 292\).
Time = 5.31 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.96
| method | result | size |
| default | \(\frac {\frac {112 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}-\frac {56 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{5}-\frac {112 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}+\frac {56 \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{5}+\frac {8 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+\frac {24 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}-\frac {14 \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}}{5}-\frac {4 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )}{3}}{\left (4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} e +e}\, e^{5} d}\) | \(321\) |
Input:
int(1/(e*cos(d*x+c))^(11/2)/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
2/15/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)/a^2/sin(1/2*d*x+1/2 *c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/e^5*(168*sin(1/2*d*x+1/2*c)^6*cos( 1/2*d*x+1/2*c)-84*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c ),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-168*sin(1 /2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+84*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(si n(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d* x+1/2*c)^2+20*I*sin(1/2*d*x+1/2*c)^3+36*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1 /2*c)-21*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^ (1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)-10*I*sin(1/2*d*x+1/2*c))/d
Time = 0.08 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.14 \[ \int \frac {1}{(e \cos (c+d x))^{11/2} (a+i a \tan (c+d x))^2} \, dx=-\frac {4 \, {\left (\sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} {\left (21 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 56 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 47 i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} + 21 \, \sqrt {\frac {1}{2}} \sqrt {e} {\left (i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 3 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )}}{15 \, {\left (a^{2} d e^{6} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a^{2} d e^{6} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a^{2} d e^{6} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d e^{6}\right )}} \] Input:
integrate(1/(e*cos(d*x+c))^(11/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="frica s")
Output:
-4/15*(sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*(21*I*e^(6*I*d*x + 6*I*c) + 56*I*e^(4*I*d*x + 4*I*c) + 47*I*e^(2*I*d*x + 2*I*c))*e^(-1/2*I*d*x - 1/ 2*I*c) + 21*sqrt(1/2)*sqrt(e)*(I*e^(6*I*d*x + 6*I*c) + 3*I*e^(4*I*d*x + 4* I*c) + 3*I*e^(2*I*d*x + 2*I*c) + I)*weierstrassZeta(-4, 0, weierstrassPInv erse(-4, 0, e^(I*d*x + I*c))))/(a^2*d*e^6*e^(6*I*d*x + 6*I*c) + 3*a^2*d*e^ 6*e^(4*I*d*x + 4*I*c) + 3*a^2*d*e^6*e^(2*I*d*x + 2*I*c) + a^2*d*e^6)
Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{11/2} (a+i a \tan (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate(1/(e*cos(d*x+c))**(11/2)/(a+I*a*tan(d*x+c))**2,x)
Output:
Timed out
Exception generated. \[ \int \frac {1}{(e \cos (c+d x))^{11/2} (a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(1/(e*cos(d*x+c))^(11/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="maxim a")
Output:
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un defined.
\[ \int \frac {1}{(e \cos (c+d x))^{11/2} (a+i a \tan (c+d x))^2} \, dx=\int { \frac {1}{\left (e \cos \left (d x + c\right )\right )^{\frac {11}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:
integrate(1/(e*cos(d*x+c))^(11/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac" )
Output:
integrate(1/((e*cos(d*x + c))^(11/2)*(I*a*tan(d*x + c) + a)^2), x)
Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{11/2} (a+i a \tan (c+d x))^2} \, dx=\int \frac {1}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{11/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \] Input:
int(1/((e*cos(c + d*x))^(11/2)*(a + a*tan(c + d*x)*1i)^2),x)
Output:
int(1/((e*cos(c + d*x))^(11/2)*(a + a*tan(c + d*x)*1i)^2), x)
\[ \int \frac {1}{(e \cos (c+d x))^{11/2} (a+i a \tan (c+d x))^2} \, dx=-\frac {\int \frac {1}{\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{5} \tan \left (d x +c \right )^{2}-2 \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{5} \tan \left (d x +c \right ) i -\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{5}}d x}{\sqrt {e}\, a^{2} e^{5}} \] Input:
int(1/(e*cos(d*x+c))^(11/2)/(a+I*a*tan(d*x+c))^2,x)
Output:
( - int(1/(sqrt(cos(c + d*x))*cos(c + d*x)**5*tan(c + d*x)**2 - 2*sqrt(cos (c + d*x))*cos(c + d*x)**5*tan(c + d*x)*i - sqrt(cos(c + d*x))*cos(c + d*x )**5),x))/(sqrt(e)*a**2*e**5)