Integrand size = 28, antiderivative size = 175 \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx=\frac {26 a^3 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{21 d}+\frac {26 i a^3 (e \sec (c+d x))^{5/2}}{35 d}+\frac {26 a^3 e (e \sec (c+d x))^{3/2} \sin (c+d x)}{21 d}+\frac {2 i a (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2}{9 d}+\frac {26 i (e \sec (c+d x))^{5/2} \left (a^3+i a^3 \tan (c+d x)\right )}{63 d} \] Output:
26/21*a^3*e^2*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*(e*s ec(d*x+c))^(1/2)/d+26/35*I*a^3*(e*sec(d*x+c))^(5/2)/d+26/21*a^3*e*(e*sec(d *x+c))^(3/2)*sin(d*x+c)/d+2/9*I*a*(e*sec(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^ 2/d+26/63*I*(e*sec(d*x+c))^(5/2)*(a^3+I*a^3*tan(d*x+c))/d
Time = 2.59 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.51 \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx=\frac {a^3 \sec ^2(c+d x) (e \sec (c+d x))^{5/2} \left (728 i+1008 i \cos (2 (c+d x))+1560 \cos ^{\frac {9}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-150 \sin (2 (c+d x))+195 \sin (4 (c+d x))\right )}{1260 d} \] Input:
Integrate[(e*Sec[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x])^3,x]
Output:
(a^3*Sec[c + d*x]^2*(e*Sec[c + d*x])^(5/2)*(728*I + (1008*I)*Cos[2*(c + d* x)] + 1560*Cos[c + d*x]^(9/2)*EllipticF[(c + d*x)/2, 2] - 150*Sin[2*(c + d *x)] + 195*Sin[4*(c + d*x)]))/(1260*d)
Time = 0.89 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3979, 3042, 3979, 3042, 3967, 3042, 4255, 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.
| \(\displaystyle \int (a+i a \tan (c+d x))^3 (e \sec (c+d x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a+i a \tan (c+d x))^3 (e \sec (c+d x))^{5/2}dx\) |
| \(\Big \downarrow \) 3979 |
| \(\displaystyle \frac {13}{9} a \int (e \sec (c+d x))^{5/2} (i \tan (c+d x) a+a)^2dx+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{5/2}}{9 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {13}{9} a \int (e \sec (c+d x))^{5/2} (i \tan (c+d x) a+a)^2dx+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{5/2}}{9 d}\) |
| \(\Big \downarrow \) 3979 |
| \(\displaystyle \frac {13}{9} a \left (\frac {9}{7} a \int (e \sec (c+d x))^{5/2} (i \tan (c+d x) a+a)dx+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}{7 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{5/2}}{9 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {13}{9} a \left (\frac {9}{7} a \int (e \sec (c+d x))^{5/2} (i \tan (c+d x) a+a)dx+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}{7 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{5/2}}{9 d}\) |
| \(\Big \downarrow \) 3967 |
| \(\displaystyle \frac {13}{9} a \left (\frac {9}{7} a \left (a \int (e \sec (c+d x))^{5/2}dx+\frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}{7 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{5/2}}{9 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {13}{9} a \left (\frac {9}{7} a \left (a \int \left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}dx+\frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}{7 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{5/2}}{9 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {13}{9} a \left (\frac {9}{7} a \left (a \left (\frac {1}{3} e^2 \int \sqrt {e \sec (c+d x)}dx+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}{7 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{5/2}}{9 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {13}{9} a \left (\frac {9}{7} a \left (a \left (\frac {1}{3} e^2 \int \sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}{7 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{5/2}}{9 d}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {13}{9} a \left (\frac {9}{7} a \left (a \left (\frac {1}{3} e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}{7 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{5/2}}{9 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {13}{9} a \left (\frac {9}{7} a \left (a \left (\frac {1}{3} e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}{7 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{5/2}}{9 d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {13}{9} a \left (\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}{7 d}+\frac {9}{7} a \left (a \left (\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d}+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}\right )\right )+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{5/2}}{9 d}\) |
Input:
Int[(e*Sec[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x])^3,x]
Output:
(((2*I)/9)*a*(e*Sec[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x])^2)/d + (13*a*(( 9*a*((((2*I)/5)*a*(e*Sec[c + d*x])^(5/2))/d + a*((2*e^2*Sqrt[Cos[c + d*x]] *EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(3*d) + (2*e*(e*Sec[c + d *x])^(3/2)*Sin[c + d*x])/(3*d))))/7 + (((2*I)/7)*(e*Sec[c + d*x])^(5/2)*(a ^2 + I*a^2*Tan[c + d*x]))/d))/9
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_)]), x_Symbol] :> Simp[b*((d*Sec[e + f*x])^m/(f*m)), x] + Simp[a Int[(d *Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2*m] || NeQ[a^2 + b^2, 0])
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^(n_), x_Symbol] :> Simp[b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] + Simp[a*((m + 2*n - 2)/(m + n - 1)) Int[(d*Se c[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 0] && NeQ[m + n - 1, 0] && IntegersQ [2*m, 2*n]
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]
Time = 5.25 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.77
\[-\frac {a^{3} \left (-\frac {26 \tan \left (d x +c \right )}{21}+\frac {6 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{7}+i \left (\frac {2 \sec \left (d x +c \right )^{4}}{9}-\frac {8 \sec \left (d x +c \right )^{2}}{5}\right )+\frac {26 i \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )}{21}\right ) e^{2} \sqrt {e \sec \left (d x +c \right )}}{d}\]
Input:
int((e*sec(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^3,x)
Output:
-a^3/d*(-26/21*tan(d*x+c)+6/7*tan(d*x+c)*sec(d*x+c)^2+I*(2/9*sec(d*x+c)^4- 8/5*sec(d*x+c)^2)+26/21*I*(1/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(csc(d*x+c) -cot(d*x+c)),I)*(cos(d*x+c)+1)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*e^2*(e*s ec(d*x+c))^(1/2)
Time = 0.10 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.47 \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (195 i \, a^{3} e^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 1158 i \, a^{3} e^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 1456 i \, a^{3} e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 858 i \, a^{3} e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 195 i \, a^{3} e^{2}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 195 \, \sqrt {2} {\left (i \, a^{3} e^{2} e^{\left (8 i \, d x + 8 i \, c\right )} + 4 i \, a^{3} e^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 i \, a^{3} e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 4 i \, a^{3} e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3} e^{2}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )}}{315 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:
integrate((e*sec(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
Output:
-2/315*(sqrt(2)*(195*I*a^3*e^2*e^(8*I*d*x + 8*I*c) - 1158*I*a^3*e^2*e^(6*I *d*x + 6*I*c) - 1456*I*a^3*e^2*e^(4*I*d*x + 4*I*c) - 858*I*a^3*e^2*e^(2*I* d*x + 2*I*c) - 195*I*a^3*e^2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d *x + 1/2*I*c) + 195*sqrt(2)*(I*a^3*e^2*e^(8*I*d*x + 8*I*c) + 4*I*a^3*e^2*e ^(6*I*d*x + 6*I*c) + 6*I*a^3*e^2*e^(4*I*d*x + 4*I*c) + 4*I*a^3*e^2*e^(2*I* d*x + 2*I*c) + I*a^3*e^2)*sqrt(e)*weierstrassPInverse(-4, 0, e^(I*d*x + I* c)))/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4 *I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)
Timed out. \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx=\text {Timed out} \] Input:
integrate((e*sec(d*x+c))**(5/2)*(a+I*a*tan(d*x+c))**3,x)
Output:
Timed out
\[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \,d x } \] Input:
integrate((e*sec(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
Output:
integrate((e*sec(d*x + c))^(5/2)*(I*a*tan(d*x + c) + a)^3, x)
\[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \,d x } \] Input:
integrate((e*sec(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
Output:
integrate((e*sec(d*x + c))^(5/2)*(I*a*tan(d*x + c) + a)^3, x)
Timed out. \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx=\int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3 \,d x \] Input:
int((e/cos(c + d*x))^(5/2)*(a + a*tan(c + d*x)*1i)^3,x)
Output:
int((e/cos(c + d*x))^(5/2)*(a + a*tan(c + d*x)*1i)^3, x)
\[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx=\frac {\sqrt {e}\, a^{3} e^{2} \left (-10 \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2} \tan \left (d x +c \right )^{2} i +62 \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2} i -135 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2} \tan \left (d x +c \right )^{2}d x \right ) d +45 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) d \right )}{45 d} \] Input:
int((e*sec(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^3,x)
Output:
(sqrt(e)*a**3*e**2*( - 10*sqrt(sec(c + d*x))*sec(c + d*x)**2*tan(c + d*x)* *2*i + 62*sqrt(sec(c + d*x))*sec(c + d*x)**2*i - 135*int(sqrt(sec(c + d*x) )*sec(c + d*x)**2*tan(c + d*x)**2,x)*d + 45*int(sqrt(sec(c + d*x))*sec(c + d*x)**2,x)*d))/(45*d)