Integrand size = 28, antiderivative size = 85 \[ \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{3/2}} \, dx=-\frac {2 a^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{3 d (e \sec (c+d x))^{3/2}} \] Output:
-2/3*a^2*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*(e*sec(d* x+c))^(1/2)/d/e^2-4/3*I*(a^2+I*a^2*tan(d*x+c))/d/(e*sec(d*x+c))^(3/2)
Time = 1.48 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.34 \[ \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{3/2}} \, dx=-\frac {2 a^2 \sec ^2(c+d x) \left (2 i \cos (c+d x)+\sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (c+d x)-i \sin (c+d x))\right ) (\cos (c+3 d x)+i \sin (c+3 d x))}{3 d (e \sec (c+d x))^{3/2} (\cos (d x)+i \sin (d x))^2} \] Input:
Integrate[(a + I*a*Tan[c + d*x])^2/(e*Sec[c + d*x])^(3/2),x]
Output:
(-2*a^2*Sec[c + d*x]^2*((2*I)*Cos[c + d*x] + Sqrt[Cos[c + d*x]]*EllipticF[ (c + d*x)/2, 2]*(Cos[c + d*x] - I*Sin[c + d*x]))*(Cos[c + 3*d*x] + I*Sin[c + 3*d*x]))/(3*d*(e*Sec[c + d*x])^(3/2)*(Cos[d*x] + I*Sin[d*x])^2)
Time = 0.39 (sec) , antiderivative size = 85, 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, 3977, 3042, 4258, 3042, 3120}
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.
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3977 |
| \(\displaystyle -\frac {a^2 \int \sqrt {e \sec (c+d x)}dx}{3 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{3 d (e \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a^2 \int \sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{3 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{3 d (e \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle -\frac {a^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{3 d (e \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{3 d (e \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle -\frac {2 a^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{3 d (e \sec (c+d x))^{3/2}}\) |
Input:
Int[(a + I*a*Tan[c + d*x])^2/(e*Sec[c + d*x])^(3/2),x]
Output:
(-2*a^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]]) /(3*d*e^2) - (((4*I)/3)*(a^2 + I*a^2*Tan[c + d*x]))/(d*(e*Sec[c + d*x])^(3 /2))
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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*b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^( n - 1)/(f*m)), x] - Simp[b^2*((m + 2*n - 2)/(d^2*m)) Int[(d*Sec[e + f*x]) ^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILtQ[m, 0] & & LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]
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]
Time = 10.62 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.20
| method | result | size |
| default | \(\frac {a^{2} \left (-\frac {2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\sec \left (d x +c \right )+1\right )}{3}-\frac {4 i \cos \left (d x +c \right )}{3}+\frac {4 \sin \left (d x +c \right )}{3}\right )}{d \sqrt {e \sec \left (d x +c \right )}\, e}\) | \(102\) |
| parts | \(\frac {a^{2} \left (-\frac {2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (\sec \left (d x +c \right )+1\right )}{3}+\frac {2 \sin \left (d x +c \right )}{3}\right )}{d \sqrt {e \sec \left (d x +c \right )}\, e}-\frac {4 i a^{2}}{3 d \left (e \sec \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {a^{2} \left (-\frac {2 i \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (2+2 \sec \left (d x +c \right )\right )}{3}-\frac {2 \sin \left (d x +c \right )}{3}\right )}{d \sqrt {e \sec \left (d x +c \right )}\, e}\) | \(208\) |
| risch | \(-\frac {2 i {\mathrm e}^{i \left (d x +c \right )} a^{2} \sqrt {2}}{3 d e \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}-\frac {2 \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, \operatorname {EllipticF}\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right ) a^{2} \sqrt {e \,{\mathrm e}^{i \left (d x +c \right )} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}}{3 d \sqrt {e \,{\mathrm e}^{3 i \left (d x +c \right )}+e \,{\mathrm e}^{i \left (d x +c \right )}}\, e \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}\) | \(232\) |
Input:
int((a+I*a*tan(d*x+c))^2/(e*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
a^2/d*(-2/3*I*(1/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cot(d*x+c)-csc(d*x+c)) ,I)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(sec(d*x+c)+1)-4/3*I*cos(d*x+c)+4/3* sin(d*x+c))/(e*sec(d*x+c))^(1/2)/e
Time = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.96 \[ \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left (-i \, \sqrt {2} a^{2} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \sqrt {2} {\left (i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{2}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )}}{3 \, d e^{2}} \] Input:
integrate((a+I*a*tan(d*x+c))^2/(e*sec(d*x+c))^(3/2),x, algorithm="fricas")
Output:
-2/3*(-I*sqrt(2)*a^2*sqrt(e)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)) + sqrt(2)*(I*a^2*e^(2*I*d*x + 2*I*c) + I*a^2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))/(d*e^2)
\[ \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{3/2}} \, dx=- a^{2} \left (\int \left (- \frac {1}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\right )\, dx + \int \frac {\tan ^{2}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \left (- \frac {2 i \tan {\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\right )\, dx\right ) \] Input:
integrate((a+I*a*tan(d*x+c))**2/(e*sec(d*x+c))**(3/2),x)
Output:
-a**2*(Integral(-1/(e*sec(c + d*x))**(3/2), x) + Integral(tan(c + d*x)**2/ (e*sec(c + d*x))**(3/2), x) + Integral(-2*I*tan(c + d*x)/(e*sec(c + d*x))* *(3/2), x))
\[ \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}}{\left (e \sec \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((a+I*a*tan(d*x+c))^2/(e*sec(d*x+c))^(3/2),x, algorithm="maxima")
Output:
integrate((I*a*tan(d*x + c) + a)^2/(e*sec(d*x + c))^(3/2), x)
\[ \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}}{\left (e \sec \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((a+I*a*tan(d*x+c))^2/(e*sec(d*x+c))^(3/2),x, algorithm="giac")
Output:
integrate((I*a*tan(d*x + c) + a)^2/(e*sec(d*x + c))^(3/2), x)
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{3/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:
int((a + a*tan(c + d*x)*1i)^2/(e/cos(c + d*x))^(3/2),x)
Output:
int((a + a*tan(c + d*x)*1i)^2/(e/cos(c + d*x))^(3/2), x)
\[ \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{3/2}} \, dx=\frac {\sqrt {e}\, a^{2} \left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{2}}d x -\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{2}}d x \right )+2 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )}{\sec \left (d x +c \right )^{2}}d x \right ) i \right )}{e^{2}} \] Input:
int((a+I*a*tan(d*x+c))^2/(e*sec(d*x+c))^(3/2),x)
Output:
(sqrt(e)*a**2*(int(sqrt(sec(c + d*x))/sec(c + d*x)**2,x) - int((sqrt(sec(c + d*x))*tan(c + d*x)**2)/sec(c + d*x)**2,x) + 2*int((sqrt(sec(c + d*x))*t an(c + d*x))/sec(c + d*x)**2,x)*i))/e**2