Integrand size = 20, antiderivative size = 110 \[ \int \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=-\frac {3 \arcsin (\cos (a+b x)-\sin (a+b x))}{16 b}-\frac {3 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{16 b}+\frac {3 \sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{8 b}-\frac {\cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b} \] Output:
-3/16*arcsin(cos(b*x+a)-sin(b*x+a))/b-3/16*ln(cos(b*x+a)+sin(b*x+a)+sin(2* b*x+2*a)^(1/2))/b+3/8*sin(b*x+a)*sin(2*b*x+2*a)^(1/2)/b-1/4*cos(b*x+a)*sin (2*b*x+2*a)^(3/2)/b
Time = 0.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.78 \[ \int \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\frac {-3 \left (\arcsin (\cos (a+b x)-\sin (a+b x))+\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )\right )+2 \sqrt {\sin (2 (a+b x))} (2 \sin (a+b x)-\sin (3 (a+b x)))}{16 b} \] Input:
Integrate[Sin[a + b*x]*Sin[2*a + 2*b*x]^(3/2),x]
Output:
(-3*(ArcSin[Cos[a + b*x] - Sin[a + b*x]] + Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*(a + b*x)]]]) + 2*Sqrt[Sin[2*(a + b*x)]]*(2*Sin[a + b*x] - S in[3*(a + b*x)]))/(16*b)
Time = 0.40 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3042, 4790, 3042, 4789, 3042, 4794}
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.
\(\displaystyle \int \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (a+b x) \sin (2 a+2 b x)^{3/2}dx\) |
\(\Big \downarrow \) 4790 |
\(\displaystyle \frac {3}{4} \int \cos (a+b x) \sqrt {\sin (2 a+2 b x)}dx-\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \cos (a+b x)}{4 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3}{4} \int \cos (a+b x) \sqrt {\sin (2 a+2 b x)}dx-\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \cos (a+b x)}{4 b}\) |
\(\Big \downarrow \) 4789 |
\(\displaystyle \frac {3}{4} \left (\frac {1}{2} \int \frac {\sin (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx+\frac {\sqrt {\sin (2 a+2 b x)} \sin (a+b x)}{2 b}\right )-\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \cos (a+b x)}{4 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3}{4} \left (\frac {1}{2} \int \frac {\sin (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx+\frac {\sqrt {\sin (2 a+2 b x)} \sin (a+b x)}{2 b}\right )-\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \cos (a+b x)}{4 b}\) |
\(\Big \downarrow \) 4794 |
\(\displaystyle \frac {3}{4} \left (\frac {1}{2} \left (-\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{2 b}-\frac {\log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{2 b}\right )+\frac {\sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{2 b}\right )-\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \cos (a+b x)}{4 b}\) |
Input:
Int[Sin[a + b*x]*Sin[2*a + 2*b*x]^(3/2),x]
Output:
(3*((-1/2*ArcSin[Cos[a + b*x] - Sin[a + b*x]]/b - Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]]/(2*b))/2 + (Sin[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(2*b)))/4 - (Cos[a + b*x]*Sin[2*a + 2*b*x]^(3/2))/(4*b)
Int[cos[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[2*Sin[a + b*x]*((g*Sin[c + d*x])^p/(d*(2*p + 1))), x] + Simp[2*p*( g/(2*p + 1)) Int[Sin[a + b*x]*(g*Sin[c + d*x])^(p - 1), x], x] /; FreeQ[{ a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && !IntegerQ[p] && GtQ[p, 0] && IntegerQ[2*p]
Int[sin[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[-2*Cos[a + b*x]*((g*Sin[c + d*x])^p/(d*(2*p + 1))), x] + Simp[2*p* (g/(2*p + 1)) Int[Cos[a + b*x]*(g*Sin[c + d*x])^(p - 1), x], x] /; FreeQ[ {a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && !IntegerQ[p] && GtQ[p, 0] && IntegerQ[2*p]
Int[sin[(a_.) + (b_.)*(x_)]/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Sim p[-ArcSin[Cos[a + b*x] - Sin[a + b*x]]/d, x] - Simp[Log[Cos[a + b*x] + Sin[ a + b*x] + Sqrt[Sin[c + d*x]]]/d, x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 24.99 (sec) , antiderivative size = 94440518, normalized size of antiderivative = 858550.16
Input:
int(sin(b*x+a)*sin(2*b*x+2*a)^(3/2),x,method=_RETURNVERBOSE)
Output:
result too large to display
Leaf count of result is larger than twice the leaf count of optimal. 280 vs. \(2 (96) = 192\).
Time = 0.09 (sec) , antiderivative size = 280, normalized size of antiderivative = 2.55 \[ \int \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=-\frac {8 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{2} - 3\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} \sin \left (b x + a\right ) - 6 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) + 6 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) - 3 \, \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{64 \, b} \] Input:
integrate(sin(b*x+a)*sin(2*b*x+2*a)^(3/2),x, algorithm="fricas")
Output:
-1/64*(8*sqrt(2)*(4*cos(b*x + a)^2 - 3)*sqrt(cos(b*x + a)*sin(b*x + a))*si n(b*x + a) - 6*arctan(-(sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a))*(cos(b*x + a) - sin(b*x + a)) + cos(b*x + a)*sin(b*x + a))/(cos(b*x + a)^2 + 2*cos(b *x + a)*sin(b*x + a) - 1)) + 6*arctan(-(2*sqrt(2)*sqrt(cos(b*x + a)*sin(b* x + a)) - cos(b*x + a) - sin(b*x + a))/(cos(b*x + a) - sin(b*x + a))) - 3* log(-32*cos(b*x + a)^4 + 4*sqrt(2)*(4*cos(b*x + a)^3 - (4*cos(b*x + a)^2 + 1)*sin(b*x + a) - 5*cos(b*x + a))*sqrt(cos(b*x + a)*sin(b*x + a)) + 32*co s(b*x + a)^2 + 16*cos(b*x + a)*sin(b*x + a) + 1))/b
Timed out. \[ \int \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\text {Timed out} \] Input:
integrate(sin(b*x+a)*sin(2*b*x+2*a)**(3/2),x)
Output:
Timed out
\[ \int \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}} \sin \left (b x + a\right ) \,d x } \] Input:
integrate(sin(b*x+a)*sin(2*b*x+2*a)^(3/2),x, algorithm="maxima")
Output:
integrate(sin(2*b*x + 2*a)^(3/2)*sin(b*x + a), x)
\[ \int \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}} \sin \left (b x + a\right ) \,d x } \] Input:
integrate(sin(b*x+a)*sin(2*b*x+2*a)^(3/2),x, algorithm="giac")
Output:
integrate(sin(2*b*x + 2*a)^(3/2)*sin(b*x + a), x)
Timed out. \[ \int \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int \sin \left (a+b\,x\right )\,{\sin \left (2\,a+2\,b\,x\right )}^{3/2} \,d x \] Input:
int(sin(a + b*x)*sin(2*a + 2*b*x)^(3/2),x)
Output:
int(sin(a + b*x)*sin(2*a + 2*b*x)^(3/2), x)
\[ \int \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int \sqrt {\sin \left (2 b x +2 a \right )}\, \sin \left (2 b x +2 a \right ) \sin \left (b x +a \right )d x \] Input:
int(sin(b*x+a)*sin(2*b*x+2*a)^(3/2),x)
Output:
int(sqrt(sin(2*a + 2*b*x))*sin(2*a + 2*b*x)*sin(a + b*x),x)