Integrand size = 28, antiderivative size = 70 \[ \int \frac {(e \sec (c+d x))^{3/2}}{a+i a \tan (c+d x)} \, dx=\frac {2 i e^2}{a d \sqrt {e \sec (c+d x)}}+\frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}} \] Output:
2*I*e^2/a/d/(e*sec(d*x+c))^(1/2)+2*e^2*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2 ))/a/d/cos(d*x+c)^(1/2)/(e*sec(d*x+c))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 1.24 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.06 \[ \int \frac {(e \sec (c+d x))^{3/2}}{a+i a \tan (c+d x)} \, dx=\frac {2 i e e^{-i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right ) \sqrt {e \sec (c+d x)}}{a d} \] Input:
Integrate[(e*Sec[c + d*x])^(3/2)/(a + I*a*Tan[c + d*x]),x]
Output:
((2*I)*e*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, - E^((2*I)*(c + d*x))]*Sqrt[e*Sec[c + d*x]])/(a*d*E^(I*(c + d*x)))
Time = 0.39 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3042, 3982, 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.
| \(\displaystyle \int \frac {(e \sec (c+d x))^{3/2}}{a+i a \tan (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(e \sec (c+d x))^{3/2}}{a+i a \tan (c+d x)}dx\) |
| \(\Big \downarrow \) 3982 |
| \(\displaystyle \frac {e^2 \int \frac {1}{\sqrt {e \sec (c+d x)}}dx}{a}+\frac {2 i e^2}{a d \sqrt {e \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^2 \int \frac {1}{\sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}+\frac {2 i e^2}{a d \sqrt {e \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {e^2 \int \sqrt {\cos (c+d x)}dx}{a \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 i e^2}{a d \sqrt {e \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^2 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 i e^2}{a d \sqrt {e \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 i e^2}{a d \sqrt {e \sec (c+d x)}}\) |
Input:
Int[(e*Sec[c + d*x])^(3/2)/(a + I*a*Tan[c + d*x]),x]
Output:
((2*I)*e^2)/(a*d*Sqrt[e*Sec[c + d*x]]) + (2*e^2*EllipticE[(c + d*x)/2, 2]) /(a*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]])
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[d^2*(d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + f*x])^(n + 1)/(b*f*(m + n - 1))), x] + Simp[d^2*((m - 2)/(a*(m + n - 1))) Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ [{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && LtQ[n, 0] && GtQ[m, 1] && !IL tQ[m + n, 0] && NeQ[m + n - 1, 0] && IntegersQ[2*m, 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. 476 vs. \(2 (65 ) = 130\).
Time = 1.97 (sec) , antiderivative size = 477, normalized size of antiderivative = 6.81
| method | result | size |
| default | \(-\frac {e \left (i \left (4 \cos \left (d x +c \right )^{2}+8 \cos \left (d x +c \right )+4\right ) \operatorname {EllipticE}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-i \ln \left (\frac {2 \cos \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}-\cos \left (d x +c \right )+1}{\cos \left (d x +c \right )+1}\right ) \cos \left (d x +c \right )+i \left (-4 \cos \left (d x +c \right )^{2}-8 \cos \left (d x +c \right )-4\right ) \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+i \ln \left (\frac {4 \cos \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}+4 \sqrt {-\frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}-2 \cos \left (d x +c \right )+2}{\cos \left (d x +c \right )+1}\right ) \cos \left (d x +c \right )+i \cos \left (d x +c \right ) \left (-4 \cos \left (d x +c \right )-4\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}-4 \sqrt {-\frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) \sqrt {e \sec \left (d x +c \right )}}{2 a d \sqrt {-\frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, \left (\cos \left (d x +c \right )+1\right )}\) | \(477\) |
Input:
int((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
-1/2*e/a/d*(I*(4*cos(d*x+c)^2+8*cos(d*x+c)+4)*EllipticE(I*(cot(d*x+c)-csc( d*x+c)),I)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*( cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-I*ln((2*cos(d*x+c)*(-cos(d*x+c)/(cos(d*x+ c)+1)^2)^(1/2)+2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)-cos(d*x+c)+1)/(cos(d *x+c)+1))*cos(d*x+c)+I*(-4*cos(d*x+c)^2-8*cos(d*x+c)-4)*EllipticF(I*(cot(d *x+c)-csc(d*x+c)),I)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(1/(cos(d*x+c)+1 ))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+I*ln(2*(2*cos(d*x+c)*(-cos(d*x+ c)/(cos(d*x+c)+1)^2)^(1/2)+2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)-cos(d*x+ c)+1)/(cos(d*x+c)+1))*cos(d*x+c)+I*cos(d*x+c)*(-4*cos(d*x+c)-4)*(-cos(d*x+ c)/(cos(d*x+c)+1)^2)^(1/2)-4*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+ c)*sin(d*x+c))*(e*sec(d*x+c))^(1/2)/(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)/( cos(d*x+c)+1)
Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.37 \[ \int \frac {(e \sec (c+d x))^{3/2}}{a+i a \tan (c+d x)} \, dx=-\frac {2 \, {\left (-i \, \sqrt {2} e^{\frac {3}{2}} e^{\left (i \, d x + i \, c\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \sqrt {2} {\left (-i \, e e^{\left (2 i \, d x + 2 i \, c\right )} - i \, e\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{a d} \] Input:
integrate((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c)),x, algorithm="fricas")
Output:
-2*(-I*sqrt(2)*e^(3/2)*e^(I*d*x + I*c)*weierstrassZeta(-4, 0, weierstrassP Inverse(-4, 0, e^(I*d*x + I*c))) + sqrt(2)*(-I*e*e^(2*I*d*x + 2*I*c) - I*e )*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))*e^(-I*d*x - I *c)/(a*d)
\[ \int \frac {(e \sec (c+d x))^{3/2}}{a+i a \tan (c+d x)} \, dx=- \frac {i \int \frac {\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\tan {\left (c + d x \right )} - i}\, dx}{a} \] Input:
integrate((e*sec(d*x+c))**(3/2)/(a+I*a*tan(d*x+c)),x)
Output:
-I*Integral((e*sec(c + d*x))**(3/2)/(tan(c + d*x) - I), x)/a
Exception generated. \[ \int \frac {(e \sec (c+d x))^{3/2}}{a+i a \tan (c+d x)} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c)),x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un defined.
\[ \int \frac {(e \sec (c+d x))^{3/2}}{a+i a \tan (c+d x)} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {3}{2}}}{i \, a \tan \left (d x + c\right ) + a} \,d x } \] Input:
integrate((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c)),x, algorithm="giac")
Output:
integrate((e*sec(d*x + c))^(3/2)/(I*a*tan(d*x + c) + a), x)
Timed out. \[ \int \frac {(e \sec (c+d x))^{3/2}}{a+i a \tan (c+d x)} \, dx=\int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \] Input:
int((e/cos(c + d*x))^(3/2)/(a + a*tan(c + d*x)*1i),x)
Output:
int((e/cos(c + d*x))^(3/2)/(a + a*tan(c + d*x)*1i), x)
\[ \int \frac {(e \sec (c+d x))^{3/2}}{a+i a \tan (c+d x)} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )}{\tan \left (d x +c \right ) i +1}d x \right ) e}{a} \] Input:
int((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c)),x)
Output:
(sqrt(e)*int((sqrt(sec(c + d*x))*sec(c + d*x))/(tan(c + d*x)*i + 1),x)*e)/ a