Integrand size = 21, antiderivative size = 70 \[ \int (b \sec (e+f x))^{5/2} \sin ^2(e+f x) \, dx=-\frac {4 b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{3 f}+\frac {2 b (b \sec (e+f x))^{3/2} \sin (e+f x)}{3 f} \] Output:
-4/3*b^2*cos(f*x+e)^(1/2)*InverseJacobiAM(1/2*f*x+1/2*e,2^(1/2))*(b*sec(f* x+e))^(1/2)/f+2/3*b*(b*sec(f*x+e))^(3/2)*sin(f*x+e)/f
Time = 0.42 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.74 \[ \int (b \sec (e+f x))^{5/2} \sin ^2(e+f x) \, dx=\frac {2 b^2 \sqrt {b \sec (e+f x)} \left (-2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )+\tan (e+f x)\right )}{3 f} \] Input:
Integrate[(b*Sec[e + f*x])^(5/2)*Sin[e + f*x]^2,x]
Output:
(2*b^2*Sqrt[b*Sec[e + f*x]]*(-2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2] + Tan[e + f*x]))/(3*f)
Time = 0.35 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3104, 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.
\(\displaystyle \int \sin ^2(e+f x) (b \sec (e+f x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(b \sec (e+f x))^{5/2}}{\csc (e+f x)^2}dx\) |
\(\Big \downarrow \) 3104 |
\(\displaystyle \frac {2 b \sin (e+f x) (b \sec (e+f x))^{3/2}}{3 f}-\frac {2}{3} b^2 \int \sqrt {b \sec (e+f x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b \sin (e+f x) (b \sec (e+f x))^{3/2}}{3 f}-\frac {2}{3} b^2 \int \sqrt {b \csc \left (e+f x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {2 b \sin (e+f x) (b \sec (e+f x))^{3/2}}{3 f}-\frac {2}{3} b^2 \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b \sin (e+f x) (b \sec (e+f x))^{3/2}}{3 f}-\frac {2}{3} b^2 \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {2 b \sin (e+f x) (b \sec (e+f x))^{3/2}}{3 f}-\frac {4 b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{3 f}\) |
Input:
Int[(b*Sec[e + f*x])^(5/2)*Sin[e + f*x]^2,x]
Output:
(-4*b^2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*Sec[e + f*x]]) /(3*f) + (2*b*(b*Sec[e + f*x])^(3/2)*Sin[e + f*x])/(3*f)
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[b*(a*Csc[e + f*x])^(m + 1)*((b*Sec[e + f*x])^(n - 1)/ (f*a*(n - 1))), x] + Simp[b^2*((m + 1)/(a^2*(n - 1))) Int[(a*Csc[e + f*x] )^(m + 2)*(b*Sec[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ [n, 1] && LtQ[m, -1] && IntegersQ[2*m, 2*n]
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[(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]
Result contains complex when optimal does not.
Time = 14.86 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.29
method | result | size |
default | \(\frac {\left (\frac {2 \tan \left (f x +e \right )}{3}-\frac {4 i \left (\cos \left (f x +e \right )+1\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right )}{3}\right ) b^{2} \sqrt {b \sec \left (f x +e \right )}}{f}\) | \(90\) |
Input:
int((b*sec(f*x+e))^(5/2)*sin(f*x+e)^2,x,method=_RETURNVERBOSE)
Output:
1/f*(2/3*tan(f*x+e)-4/3*I*EllipticF(I*(cot(f*x+e)-csc(f*x+e)),I)*(1/(cos(f *x+e)+1))^(1/2)*(cos(f*x+e)+1)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2))*b^2*(b*s ec(f*x+e))^(1/2)
Result contains complex when optimal does not.
Time = 0.12 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.44 \[ \int (b \sec (e+f x))^{5/2} \sin ^2(e+f x) \, dx=-\frac {2 \, {\left (-i \, \sqrt {2} b^{\frac {5}{2}} \cos \left (f x + e\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + i \, \sqrt {2} b^{\frac {5}{2}} \cos \left (f x + e\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - b^{2} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )\right )}}{3 \, f \cos \left (f x + e\right )} \] Input:
integrate((b*sec(f*x+e))^(5/2)*sin(f*x+e)^2,x, algorithm="fricas")
Output:
-2/3*(-I*sqrt(2)*b^(5/2)*cos(f*x + e)*weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e)) + I*sqrt(2)*b^(5/2)*cos(f*x + e)*weierstrassPInverse (-4, 0, cos(f*x + e) - I*sin(f*x + e)) - b^2*sqrt(b/cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e))
Timed out. \[ \int (b \sec (e+f x))^{5/2} \sin ^2(e+f x) \, dx=\text {Timed out} \] Input:
integrate((b*sec(f*x+e))**(5/2)*sin(f*x+e)**2,x)
Output:
Timed out
\[ \int (b \sec (e+f x))^{5/2} \sin ^2(e+f x) \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{\frac {5}{2}} \sin \left (f x + e\right )^{2} \,d x } \] Input:
integrate((b*sec(f*x+e))^(5/2)*sin(f*x+e)^2,x, algorithm="maxima")
Output:
integrate((b*sec(f*x + e))^(5/2)*sin(f*x + e)^2, x)
\[ \int (b \sec (e+f x))^{5/2} \sin ^2(e+f x) \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{\frac {5}{2}} \sin \left (f x + e\right )^{2} \,d x } \] Input:
integrate((b*sec(f*x+e))^(5/2)*sin(f*x+e)^2,x, algorithm="giac")
Output:
integrate((b*sec(f*x + e))^(5/2)*sin(f*x + e)^2, x)
Timed out. \[ \int (b \sec (e+f x))^{5/2} \sin ^2(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^2\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{5/2} \,d x \] Input:
int(sin(e + f*x)^2*(b/cos(e + f*x))^(5/2),x)
Output:
int(sin(e + f*x)^2*(b/cos(e + f*x))^(5/2), x)
\[ \int (b \sec (e+f x))^{5/2} \sin ^2(e+f x) \, dx=\sqrt {b}\, \left (\int \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{2} \sin \left (f x +e \right )^{2}d x \right ) b^{2} \] Input:
int((b*sec(f*x+e))^(5/2)*sin(f*x+e)^2,x)
Output:
sqrt(b)*int(sqrt(sec(e + f*x))*sec(e + f*x)**2*sin(e + f*x)**2,x)*b**2