Integrand size = 22, antiderivative size = 69 \[ \int \sin ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\frac {3 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{10 b}-\frac {\cos (2 a+2 b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{10 b}-\frac {\sin ^{\frac {7}{2}}(2 a+2 b x)}{14 b} \] Output:
-3/10*EllipticE(cos(a+1/4*Pi+b*x),2^(1/2))/b-1/10*cos(2*b*x+2*a)*sin(2*b*x +2*a)^(3/2)/b-1/14*sin(2*b*x+2*a)^(7/2)/b
Time = 0.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.96 \[ \int \sin ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\frac {84 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )+\sqrt {\sin (2 (a+b x))} (-15 \sin (2 (a+b x))-14 \sin (4 (a+b x))+5 \sin (6 (a+b x)))}{280 b} \] Input:
Integrate[Sin[a + b*x]^2*Sin[2*a + 2*b*x]^(5/2),x]
Output:
(84*EllipticE[a - Pi/4 + b*x, 2] + Sqrt[Sin[2*(a + b*x)]]*(-15*Sin[2*(a + b*x)] - 14*Sin[4*(a + b*x)] + 5*Sin[6*(a + b*x)]))/(280*b)
Time = 0.33 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 4786, 3042, 3115, 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.
\(\displaystyle \int \sin ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (a+b x)^2 \sin (2 a+2 b x)^{5/2}dx\) |
\(\Big \downarrow \) 4786 |
\(\displaystyle \frac {1}{2} \int \sin ^{\frac {5}{2}}(2 a+2 b x)dx-\frac {\sin ^{\frac {7}{2}}(2 a+2 b x)}{14 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \int \sin (2 a+2 b x)^{5/2}dx-\frac {\sin ^{\frac {7}{2}}(2 a+2 b x)}{14 b}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{5} \int \sqrt {\sin (2 a+2 b x)}dx-\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{5 b}\right )-\frac {\sin ^{\frac {7}{2}}(2 a+2 b x)}{14 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{5} \int \sqrt {\sin (2 a+2 b x)}dx-\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{5 b}\right )-\frac {\sin ^{\frac {7}{2}}(2 a+2 b x)}{14 b}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {1}{2} \left (\frac {3 E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{5 b}-\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{5 b}\right )-\frac {\sin ^{\frac {7}{2}}(2 a+2 b x)}{14 b}\) |
Input:
Int[Sin[a + b*x]^2*Sin[2*a + 2*b*x]^(5/2),x]
Output:
-1/14*Sin[2*a + 2*b*x]^(7/2)/b + ((3*EllipticE[a - Pi/4 + b*x, 2])/(5*b) - (Cos[2*a + 2*b*x]*Sin[2*a + 2*b*x]^(3/2))/(5*b))/2
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 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[((e_.)*sin[(a_.) + (b_.)*(x_)])^(m_)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p _), x_Symbol] :> Simp[(-e^2)*(e*Sin[a + b*x])^(m - 2)*((g*Sin[c + d*x])^(p + 1)/(2*b*g*(m + 2*p))), x] + Simp[e^2*((m + p - 1)/(m + 2*p)) Int[(e*Sin [a + b*x])^(m - 2)*(g*Sin[c + d*x])^p, x], x] /; FreeQ[{a, b, c, d, e, g, p }, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && !IntegerQ[p] && GtQ[m, 1] && NeQ[m + 2*p, 0] && IntegersQ[2*m, 2*p]
Timed out.
hanged
Input:
int(sin(b*x+a)^2*sin(2*b*x+2*a)^(5/2),x)
Output:
int(sin(b*x+a)^2*sin(2*b*x+2*a)^(5/2),x)
\[ \int \sin ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}} \sin \left (b x + a\right )^{2} \,d x } \] Input:
integrate(sin(b*x+a)^2*sin(2*b*x+2*a)^(5/2),x, algorithm="fricas")
Output:
integral(((cos(b*x + a)^2 - 1)*cos(2*b*x + 2*a)^2 - cos(b*x + a)^2 + 1)*sq rt(sin(2*b*x + 2*a)), x)
Timed out. \[ \int \sin ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\text {Timed out} \] Input:
integrate(sin(b*x+a)**2*sin(2*b*x+2*a)**(5/2),x)
Output:
Timed out
\[ \int \sin ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}} \sin \left (b x + a\right )^{2} \,d x } \] Input:
integrate(sin(b*x+a)^2*sin(2*b*x+2*a)^(5/2),x, algorithm="maxima")
Output:
integrate(sin(2*b*x + 2*a)^(5/2)*sin(b*x + a)^2, x)
\[ \int \sin ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}} \sin \left (b x + a\right )^{2} \,d x } \] Input:
integrate(sin(b*x+a)^2*sin(2*b*x+2*a)^(5/2),x, algorithm="giac")
Output:
integrate(sin(2*b*x + 2*a)^(5/2)*sin(b*x + a)^2, x)
Timed out. \[ \int \sin ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int {\sin \left (a+b\,x\right )}^2\,{\sin \left (2\,a+2\,b\,x\right )}^{5/2} \,d x \] Input:
int(sin(a + b*x)^2*sin(2*a + 2*b*x)^(5/2),x)
Output:
int(sin(a + b*x)^2*sin(2*a + 2*b*x)^(5/2), x)
\[ \int \sin ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int \sqrt {\sin \left (2 b x +2 a \right )}\, \sin \left (2 b x +2 a \right )^{2} \sin \left (b x +a \right )^{2}d x \] Input:
int(sin(b*x+a)^2*sin(2*b*x+2*a)^(5/2),x)
Output:
int(sqrt(sin(2*a + 2*b*x))*sin(2*a + 2*b*x)**2*sin(a + b*x)**2,x)