3.90 \(\int \sin ^3(a+b x) \sqrt{\sin (2 a+2 b x)} \, dx\)

Optimal. Leaf size=110 \[ -\frac{\sin (a+b x) \sin ^{\frac{3}{2}}(2 a+2 b x)}{8 b}-\frac{5 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{32 b}-\frac{5 \sqrt{\sin (2 a+2 b x)} \cos (a+b x)}{16 b}+\frac{5 \log \left (\sin (a+b x)+\sqrt{\sin (2 a+2 b x)}+\cos (a+b x)\right )}{32 b} \]

[Out]

(-5*ArcSin[Cos[a + b*x] - Sin[a + b*x]])/(32*b) + (5*Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]]
)/(32*b) - (5*Cos[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(16*b) - (Sin[a + b*x]*Sin[2*a + 2*b*x]^(3/2))/(8*b)

________________________________________________________________________________________

Rubi [A]  time = 0.0782088, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {4298, 4302, 4305} \[ -\frac{\sin (a+b x) \sin ^{\frac{3}{2}}(2 a+2 b x)}{8 b}-\frac{5 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{32 b}-\frac{5 \sqrt{\sin (2 a+2 b x)} \cos (a+b x)}{16 b}+\frac{5 \log \left (\sin (a+b x)+\sqrt{\sin (2 a+2 b x)}+\cos (a+b x)\right )}{32 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^3*Sqrt[Sin[2*a + 2*b*x]],x]

[Out]

(-5*ArcSin[Cos[a + b*x] - Sin[a + b*x]])/(32*b) + (5*Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]]
)/(32*b) - (5*Cos[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(16*b) - (Sin[a + b*x]*Sin[2*a + 2*b*x]^(3/2))/(8*b)

Rule 4298

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] + Dist[(e^2*(m + p - 1))/(m + 2*p), Int[(e*S
in[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]

Rule 4302

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] + Dist[(2*p*g)/(2*p + 1), Int[Cos[a + b*x]*(g*Sin[c + d*x])^(p - 1), x], x] /; Fr
eeQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p] && GtQ[p, 0] && IntegerQ[2*p]

Rule 4305

Int[cos[(a_.) + (b_.)*(x_)]/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> -Simp[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]

Rubi steps

\begin{align*} \int \sin ^3(a+b x) \sqrt{\sin (2 a+2 b x)} \, dx &=-\frac{\sin (a+b x) \sin ^{\frac{3}{2}}(2 a+2 b x)}{8 b}+\frac{5}{8} \int \sin (a+b x) \sqrt{\sin (2 a+2 b x)} \, dx\\ &=-\frac{5 \cos (a+b x) \sqrt{\sin (2 a+2 b x)}}{16 b}-\frac{\sin (a+b x) \sin ^{\frac{3}{2}}(2 a+2 b x)}{8 b}+\frac{5}{16} \int \frac{\cos (a+b x)}{\sqrt{\sin (2 a+2 b x)}} \, dx\\ &=-\frac{5 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{32 b}+\frac{5 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt{\sin (2 a+2 b x)}\right )}{32 b}-\frac{5 \cos (a+b x) \sqrt{\sin (2 a+2 b x)}}{16 b}-\frac{\sin (a+b x) \sin ^{\frac{3}{2}}(2 a+2 b x)}{8 b}\\ \end{align*}

Mathematica [A]  time = 0.219184, size = 86, normalized size = 0.78 \[ \frac{2 \sqrt{\sin (2 (a+b x))} (\cos (3 (a+b x))-6 \cos (a+b x))+5 \left (\log \left (\sin (a+b x)+\sqrt{\sin (2 (a+b x))}+\cos (a+b x)\right )-\sin ^{-1}(\cos (a+b x)-\sin (a+b x))\right )}{32 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^3*Sqrt[Sin[2*a + 2*b*x]],x]

[Out]

(5*(-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*(-6*
Cos[a + b*x] + Cos[3*(a + b*x)])*Sqrt[Sin[2*(a + b*x)]])/(32*b)

________________________________________________________________________________________

Maple [B]  time = 15.251, size = 47430975, normalized size = 431190.7 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^3*sin(2*b*x+2*a)^(1/2),x)

[Out]

result too large to display

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\sin \left (2 \, b x + 2 \, a\right )} \sin \left (b x + a\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3*sin(2*b*x+2*a)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(sin(2*b*x + 2*a))*sin(b*x + a)^3, x)

________________________________________________________________________________________

Fricas [B]  time = 0.550574, size = 774, normalized size = 7.04 \begin{align*} \frac{8 \, \sqrt{2}{\left (4 \, \cos \left (b x + a\right )^{3} - 9 \, \cos \left (b x + a\right )\right )} \sqrt{\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 10 \, \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 ) - 10 \, \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 ) - 5 \, \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 )}{128 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3*sin(2*b*x+2*a)^(1/2),x, algorithm="fricas")

[Out]

1/128*(8*sqrt(2)*(4*cos(b*x + a)^3 - 9*cos(b*x + a))*sqrt(cos(b*x + a)*sin(b*x + a)) + 10*arctan(-(sqrt(2)*sqr
t(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*co
s(b*x + a)*sin(b*x + a) - 1)) - 10*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))) - 5*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*cos(b*x + a)^2 + 16*cos(b*x + a)
*sin(b*x + a) + 1))/b

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)**3*sin(2*b*x+2*a)**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3*sin(2*b*x+2*a)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError