Integrand size = 21, antiderivative size = 66 \[ \int (b \sec (e+f x))^{3/2} \sin ^2(e+f x) \, dx=-\frac {4 b^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}+\frac {2 b \sqrt {b \sec (e+f x)} \sin (e+f x)}{f} \] Output:
-4*b^2*EllipticE(sin(1/2*f*x+1/2*e),2^(1/2))/f/cos(f*x+e)^(1/2)/(b*sec(f*x +e))^(1/2)+2*b*(b*sec(f*x+e))^(1/2)*sin(f*x+e)/f
Time = 0.35 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.73 \[ \int (b \sec (e+f x))^{3/2} \sin ^2(e+f x) \, dx=\frac {2 b \sqrt {b \sec (e+f x)} \left (-2 \sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+\sin (e+f x)\right )}{f} \] Input:
Integrate[(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^2,x]
Output:
(2*b*Sqrt[b*Sec[e + f*x]]*(-2*Sqrt[Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2] + Sin[e + f*x]))/f
Time = 0.57 (sec) , antiderivative size = 66, 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, 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.
\(\displaystyle \int \sin ^2(e+f x) (b \sec (e+f x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(b \sec (e+f x))^{3/2}}{\csc (e+f x)^2}dx\) |
\(\Big \downarrow \) 3104 |
\(\displaystyle \frac {2 b \sin (e+f x) \sqrt {b \sec (e+f x)}}{f}-2 b^2 \int \frac {1}{\sqrt {b \sec (e+f x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b \sin (e+f x) \sqrt {b \sec (e+f x)}}{f}-2 b^2 \int \frac {1}{\sqrt {b \csc \left (e+f x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {2 b \sin (e+f x) \sqrt {b \sec (e+f x)}}{f}-\frac {2 b^2 \int \sqrt {\cos (e+f x)}dx}{\sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b \sin (e+f x) \sqrt {b \sec (e+f x)}}{f}-\frac {2 b^2 \int \sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}dx}{\sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {2 b \sin (e+f x) \sqrt {b \sec (e+f x)}}{f}-\frac {4 b^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}\) |
Input:
Int[(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^2,x]
Output:
(-4*b^2*EllipticE[(e + f*x)/2, 2])/(f*Sqrt[Cos[e + f*x]]*Sqrt[b*Sec[e + f* x]]) + (2*b*Sqrt[b*Sec[e + f*x]]*Sin[e + f*x])/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[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[(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 = 2.54 (sec) , antiderivative size = 193, normalized size of antiderivative = 2.92
method | result | size |
default | \(-\frac {2 \left (\left (\cos \left (f x +e \right )-1\right ) \sin \left (f x +e \right )+i \left (2 \cos \left (f x +e \right )^{2}+4 \cos \left (f x +e \right )+2\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right )+i \left (-2 \cos \left (f x +e \right )^{2}-4 \cos \left (f x +e \right )-2\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \operatorname {EllipticE}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right )\right ) b \sqrt {b \sec \left (f x +e \right )}}{f \left (\cos \left (f x +e \right )+1\right )}\) | \(193\) |
Input:
int((b*sec(f*x+e))^(3/2)*sin(f*x+e)^2,x,method=_RETURNVERBOSE)
Output:
-2/f*((cos(f*x+e)-1)*sin(f*x+e)+I*(2*cos(f*x+e)^2+4*cos(f*x+e)+2)*(1/(cos( f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cot(f*x+e) -csc(f*x+e)),I)+I*(-2*cos(f*x+e)^2-4*cos(f*x+e)-2)*(cos(f*x+e)/(cos(f*x+e) +1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(cot(f*x+e)-csc(f*x+e)),I) )*b*(b*sec(f*x+e))^(1/2)/(cos(f*x+e)+1)
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.29 \[ \int (b \sec (e+f x))^{3/2} \sin ^2(e+f x) \, dx=-\frac {2 \, {\left (i \, \sqrt {2} b^{\frac {3}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) - i \, \sqrt {2} b^{\frac {3}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) - b \sqrt {\frac {b}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )\right )}}{f} \] Input:
integrate((b*sec(f*x+e))^(3/2)*sin(f*x+e)^2,x, algorithm="fricas")
Output:
-2*(I*sqrt(2)*b^(3/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, co s(f*x + e) + I*sin(f*x + e))) - I*sqrt(2)*b^(3/2)*weierstrassZeta(-4, 0, w eierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))) - b*sqrt(b/cos(f *x + e))*sin(f*x + e))/f
Timed out. \[ \int (b \sec (e+f x))^{3/2} \sin ^2(e+f x) \, dx=\text {Timed out} \] Input:
integrate((b*sec(f*x+e))**(3/2)*sin(f*x+e)**2,x)
Output:
Timed out
\[ \int (b \sec (e+f x))^{3/2} \sin ^2(e+f x) \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}} \sin \left (f x + e\right )^{2} \,d x } \] Input:
integrate((b*sec(f*x+e))^(3/2)*sin(f*x+e)^2,x, algorithm="maxima")
Output:
integrate((b*sec(f*x + e))^(3/2)*sin(f*x + e)^2, x)
\[ \int (b \sec (e+f x))^{3/2} \sin ^2(e+f x) \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}} \sin \left (f x + e\right )^{2} \,d x } \] Input:
integrate((b*sec(f*x+e))^(3/2)*sin(f*x+e)^2,x, algorithm="giac")
Output:
integrate((b*sec(f*x + e))^(3/2)*sin(f*x + e)^2, x)
Timed out. \[ \int (b \sec (e+f x))^{3/2} \sin ^2(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^2\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2} \,d x \] Input:
int(sin(e + f*x)^2*(b/cos(e + f*x))^(3/2),x)
Output:
int(sin(e + f*x)^2*(b/cos(e + f*x))^(3/2), x)
\[ \int (b \sec (e+f x))^{3/2} \sin ^2(e+f x) \, dx=\sqrt {b}\, \left (\int \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right ) \sin \left (f x +e \right )^{2}d x \right ) b \] Input:
int((b*sec(f*x+e))^(3/2)*sin(f*x+e)^2,x)
Output:
sqrt(b)*int(sqrt(sec(e + f*x))*sec(e + f*x)*sin(e + f*x)**2,x)*b