Integrand size = 28, antiderivative size = 138 \[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2 \, dx=-\frac {14 a^2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {14 i a^2 (e \sec (c+d x))^{3/2}}{15 d}+\frac {14 a^2 e \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 i (e \sec (c+d x))^{3/2} \left (a^2+i a^2 \tan (c+d x)\right )}{5 d} \] Output:
-14/5*a^2*e^2*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/cos(d*x+c)^(1/2)/(e* sec(d*x+c))^(1/2)+14/15*I*a^2*(e*sec(d*x+c))^(3/2)/d+14/5*a^2*e*(e*sec(d*x +c))^(1/2)*sin(d*x+c)/d+2/5*I*(e*sec(d*x+c))^(3/2)*(a^2+I*a^2*tan(d*x+c))/ d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 2.99 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.93 \[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2 \, dx=\frac {(e \sec (c+d x))^{3/2} \left (-\frac {14 i \sqrt {2} \left (3 \sqrt {1+e^{2 i (c+d x)}}-e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )}{\left (-1+e^{2 i c}\right ) \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}}}+\frac {1}{2} \csc (c) \sec ^{\frac {5}{2}}(c+d x) (\cos (2 c)-i \sin (2 c)) (36 \cos (d x)+27 \cos (2 c+d x)+21 \cos (2 c+3 d x)-20 i \sin (d x)+20 i \sin (2 c+d x))\right ) (a+i a \tan (c+d x))^2}{15 d \sec ^{\frac {7}{2}}(c+d x) (\cos (d x)+i \sin (d x))^2} \] Input:
Integrate[(e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^2,x]
Output:
((e*Sec[c + d*x])^(3/2)*(((-14*I)*Sqrt[2]*(3*Sqrt[1 + E^((2*I)*(c + d*x))] - E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^(( 2*I)*(c + d*x))]))/((-1 + E^((2*I)*c))*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)* (c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]) + (Csc[c]*Sec[c + d*x]^(5/2)*( 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]))/2)*(a + I*a*Tan[c + d*x])^ 2)/(15*d*Sec[c + d*x]^(7/2)*(Cos[d*x] + I*Sin[d*x])^2)
Time = 0.66 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.98, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 3979, 3042, 3967, 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.
| \(\displaystyle \int (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{3/2}dx\) |
| \(\Big \downarrow \) 3979 |
| \(\displaystyle \frac {7}{5} a \int (e \sec (c+d x))^{3/2} (i \tan (c+d x) a+a)dx+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7}{5} a \int (e \sec (c+d x))^{3/2} (i \tan (c+d x) a+a)dx+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3967 |
| \(\displaystyle \frac {7}{5} a \left (a \int (e \sec (c+d x))^{3/2}dx+\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7}{5} a \left (a \int \left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx+\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {7}{5} a \left (a \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-e^2 \int \frac {1}{\sqrt {e \sec (c+d x)}}dx\right )+\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7}{5} a \left (a \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-e^2 \int \frac {1}{\sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {7}{5} a \left (a \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {e^2 \int \sqrt {\cos (c+d x)}dx}{\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\right )+\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7}{5} a \left (a \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {e^2 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\right )+\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}+\frac {7}{5} a \left (a \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\right )+\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}\right )\) |
Input:
Int[(e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^2,x]
Output:
(7*a*((((2*I)/3)*a*(e*Sec[c + d*x])^(3/2))/d + a*((-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 + (((2*I)/5)*(e*Sec[c + d*x])^(3/2)*(a^2 + I *a^2*Tan[c + d*x]))/d
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_)]), 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 = 13.82 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.59
| method | result | size |
| default | \(\frac {2 a^{2} \left (21 \sin \left (d x +c \right )-3 \tan \left (d x +c \right )-3 \sec \left (d x +c \right ) \tan \left (d x +c \right )+10 i+10 i \sec \left (d x +c \right )+21 i \left (-\cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+21 i \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\right ) \sqrt {e \sec \left (d x +c \right )}\, e}{15 d \left (\cos \left (d x +c \right )+1\right )}\) | \(220\) |
| parts | \(\frac {2 a^{2} \left (i \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+i \left (-\cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+\sin \left (d x +c \right )\right ) \sqrt {e \sec \left (d x +c \right )}\, e}{d \left (\cos \left (d x +c \right )+1\right )}+\frac {4 i a^{2} \left (e \sec \left (d x +c \right )\right )^{\frac {3}{2}}}{3 d}+\frac {2 a^{2} \sqrt {e \sec \left (d x +c \right )}\, e \left (2 \sin \left (d x +c \right )-\tan \left (d x +c \right )-\sec \left (d x +c \right ) \tan \left (d x +c \right )+i \left (2 \cos \left (d x +c \right )^{2}+4 \cos \left (d x +c \right )+2\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+i \left (-2 \cos \left (d x +c \right )^{2}-4 \cos \left (d x +c \right )-2\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )\right )}{5 d \left (\cos \left (d x +c \right )+1\right )}\) | \(415\) |
Input:
int((e*sec(d*x+c))^(3/2)*(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
2/15*a^2/d*(21*sin(d*x+c)-3*tan(d*x+c)-3*sec(d*x+c)*tan(d*x+c)+10*I+10*I*s ec(d*x+c)+21*I*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*(cos(d*x+c)/(cos(d*x+c)+1))^ (1/2)*EllipticF(I*(cot(d*x+c)-csc(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)+21*I *(cos(d*x+c)^2+2*cos(d*x+c)+1)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE (I*(cot(d*x+c)-csc(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2))*(e*sec(d*x+c))^(1/ 2)/(cos(d*x+c)+1)*e
Time = 0.09 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.20 \[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2 \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (21 i \, a^{2} e e^{\left (5 i \, d x + 5 i \, c\right )} + 16 i \, a^{2} e e^{\left (3 i \, d x + 3 i \, c\right )} + 7 i \, a^{2} e e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 21 \, \sqrt {2} {\left (i \, a^{2} e e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, a^{2} e e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{2} e\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )}}{15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:
integrate((e*sec(d*x+c))^(3/2)*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")
Output:
-2/15*(sqrt(2)*(21*I*a^2*e*e^(5*I*d*x + 5*I*c) + 16*I*a^2*e*e^(3*I*d*x + 3 *I*c) + 7*I*a^2*e*e^(I*d*x + I*c))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/ 2*I*d*x + 1/2*I*c) + 21*sqrt(2)*(I*a^2*e*e^(4*I*d*x + 4*I*c) + 2*I*a^2*e*e ^(2*I*d*x + 2*I*c) + I*a^2*e)*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPI nverse(-4, 0, e^(I*d*x + I*c))))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)
\[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2 \, dx=- a^{2} \left (\int \left (- \left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}\right )\, dx + \int \left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan ^{2}{\left (c + d x \right )}\, dx + \int \left (- 2 i \left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan {\left (c + d x \right )}\right )\, dx\right ) \] Input:
integrate((e*sec(d*x+c))**(3/2)*(a+I*a*tan(d*x+c))**2,x)
Output:
-a**2*(Integral(-(e*sec(c + d*x))**(3/2), x) + Integral((e*sec(c + d*x))** (3/2)*tan(c + d*x)**2, x) + Integral(-2*I*(e*sec(c + d*x))**(3/2)*tan(c + d*x), x))
\[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2 \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{\frac {3}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} \,d x } \] Input:
integrate((e*sec(d*x+c))^(3/2)*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")
Output:
integrate((e*sec(d*x + c))^(3/2)*(I*a*tan(d*x + c) + a)^2, x)
\[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2 \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{\frac {3}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} \,d x } \] Input:
integrate((e*sec(d*x+c))^(3/2)*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")
Output:
integrate((e*sec(d*x + c))^(3/2)*(I*a*tan(d*x + c) + a)^2, x)
Timed out. \[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2 \, dx=\int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2 \,d x \] Input:
int((e/cos(c + d*x))^(3/2)*(a + a*tan(c + d*x)*1i)^2,x)
Output:
int((e/cos(c + d*x))^(3/2)*(a + a*tan(c + d*x)*1i)^2, x)
\[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2 \, dx=\frac {\sqrt {e}\, a^{2} e \left (4 \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right ) i -3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right ) \tan \left (d x +c \right )^{2}d x \right ) d +3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) d \right )}{3 d} \] Input:
int((e*sec(d*x+c))^(3/2)*(a+I*a*tan(d*x+c))^2,x)
Output:
(sqrt(e)*a**2*e*(4*sqrt(sec(c + d*x))*sec(c + d*x)*i - 3*int(sqrt(sec(c + d*x))*sec(c + d*x)*tan(c + d*x)**2,x)*d + 3*int(sqrt(sec(c + d*x))*sec(c + d*x),x)*d))/(3*d)