Integrand size = 22, antiderivative size = 84 \[ \int \frac {\sin ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=-\frac {3 \arcsin (\cos (a+b x)-\sin (a+b x))}{8 b}-\frac {3 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{8 b}-\frac {\sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{4 b} \] Output:
-3/8*arcsin(cos(b*x+a)-sin(b*x+a))/b-3/8*ln(cos(b*x+a)+sin(b*x+a)+sin(2*b* x+2*a)^(1/2))/b-1/4*sin(b*x+a)*sin(2*b*x+2*a)^(1/2)/b
Time = 0.18 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.88 \[ \int \frac {\sin ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=-\frac {3 \arcsin (\cos (a+b x)-\sin (a+b x))+3 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )+2 \sin (a+b x) \sqrt {\sin (2 (a+b x))}}{8 b} \] Input:
Integrate[Sin[a + b*x]^3/Sqrt[Sin[2*a + 2*b*x]],x]
Output:
-1/8*(3*ArcSin[Cos[a + b*x] - Sin[a + b*x]] + 3*Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*(a + b*x)]]] + 2*Sin[a + b*x]*Sqrt[Sin[2*(a + b*x)]])/b
Time = 0.31 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3042, 4786, 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.
\(\displaystyle \int \frac {\sin ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (a+b x)^3}{\sqrt {\sin (2 a+2 b x)}}dx\) |
\(\Big \downarrow \) 4786 |
\(\displaystyle \frac {3}{4} \int \frac {\sin (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {\sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{4 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3}{4} \int \frac {\sin (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {\sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{4 b}\) |
\(\Big \downarrow \) 4794 |
\(\displaystyle \frac {3}{4} \left (-\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{2 b}-\frac {\log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{2 b}\right )-\frac {\sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{4 b}\) |
Input:
Int[Sin[a + b*x]^3/Sqrt[Sin[2*a + 2*b*x]],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)))/4 - (Sin[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(4*b)
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]
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]
Timed out.
hanged
Input:
int(sin(b*x+a)^3/sin(2*b*x+2*a)^(1/2),x)
Output:
int(sin(b*x+a)^3/sin(2*b*x+2*a)^(1/2),x)
Leaf count of result is larger than twice the leaf count of optimal. 268 vs. \(2 (74) = 148\).
Time = 0.09 (sec) , antiderivative size = 268, normalized size of antiderivative = 3.19 \[ \int \frac {\sin ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=-\frac {8 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} \sin \left (b x + a\right ) - 6 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) + 6 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) - 3 \, \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{32 \, b} \] Input:
integrate(sin(b*x+a)^3/sin(2*b*x+2*a)^(1/2),x, algorithm="fricas")
Output:
-1/32*(8*sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a))*sin(b*x + a) - 6*arctan(- (sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a))*(cos(b*x + a) - sin(b*x + a)) + c os(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*co s(b*x + a))*sqrt(cos(b*x + a)*sin(b*x + a)) + 32*cos(b*x + a)^2 + 16*cos(b *x + a)*sin(b*x + a) + 1))/b
Timed out. \[ \int \frac {\sin ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=\text {Timed out} \] Input:
integrate(sin(b*x+a)**3/sin(2*b*x+2*a)**(1/2),x)
Output:
Timed out
\[ \int \frac {\sin ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=\int { \frac {\sin \left (b x + a\right )^{3}}{\sqrt {\sin \left (2 \, b x + 2 \, a\right )}} \,d x } \] Input:
integrate(sin(b*x+a)^3/sin(2*b*x+2*a)^(1/2),x, algorithm="maxima")
Output:
integrate(sin(b*x + a)^3/sqrt(sin(2*b*x + 2*a)), x)
Timed out. \[ \int \frac {\sin ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=\text {Timed out} \] Input:
integrate(sin(b*x+a)^3/sin(2*b*x+2*a)^(1/2),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {\sin ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=\int \frac {{\sin \left (a+b\,x\right )}^3}{\sqrt {\sin \left (2\,a+2\,b\,x\right )}} \,d x \] Input:
int(sin(a + b*x)^3/sin(2*a + 2*b*x)^(1/2),x)
Output:
int(sin(a + b*x)^3/sin(2*a + 2*b*x)^(1/2), x)
\[ \int \frac {\sin ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=\int \frac {\sqrt {\sin \left (2 b x +2 a \right )}\, \sin \left (b x +a \right )^{3}}{\sin \left (2 b x +2 a \right )}d x \] Input:
int(sin(b*x+a)^3/sin(2*b*x+2*a)^(1/2),x)
Output:
int((sqrt(sin(2*a + 2*b*x))*sin(a + b*x)**3)/sin(2*a + 2*b*x),x)