3.75 \(\int \sin (a+b x) \sqrt{\sin (2 a+2 b x)} \, dx\)

Optimal. Leaf size=84 \[ -\frac{\sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{4 b}-\frac{\sqrt{\sin (2 a+2 b x)} \cos (a+b x)}{2 b}+\frac{\log \left (\sin (a+b x)+\sqrt{\sin (2 a+2 b x)}+\cos (a+b x)\right )}{4 b} \]

[Out]

-ArcSin[Cos[a + b*x] - Sin[a + b*x]]/(4*b) + Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]]/(4*b) -
(Cos[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(2*b)

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Rubi [A]  time = 0.0431275, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {4302, 4305} \[ -\frac{\sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{4 b}-\frac{\sqrt{\sin (2 a+2 b x)} \cos (a+b x)}{2 b}+\frac{\log \left (\sin (a+b x)+\sqrt{\sin (2 a+2 b x)}+\cos (a+b x)\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcSin[Cos[a + b*x] - Sin[a + b*x]]/(4*b) + Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]]/(4*b) -
(Cos[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(2*b)

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 (a+b x) \sqrt{\sin (2 a+2 b x)} \, dx &=-\frac{\cos (a+b x) \sqrt{\sin (2 a+2 b x)}}{2 b}+\frac{1}{2} \int \frac{\cos (a+b x)}{\sqrt{\sin (2 a+2 b x)}} \, dx\\ &=-\frac{\sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{4 b}+\frac{\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt{\sin (2 a+2 b x)}\right )}{4 b}-\frac{\cos (a+b x) \sqrt{\sin (2 a+2 b x)}}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.0721811, size = 72, normalized size = 0.86 \[ \frac{-\sin ^{-1}(\cos (a+b x)-\sin (a+b x))-2 \sqrt{\sin (2 (a+b x))} \cos (a+b x)+\log \left (\sin (a+b x)+\sqrt{\sin (2 (a+b x))}+\cos (a+b x)\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-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*Cos[a +
b*x]*Sqrt[Sin[2*(a + b*x)]])/(4*b)

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Maple [B]  time = 1.171, size = 6219390, normalized size = 74040.4 \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)*sin(2*b*x+2*a)^(1/2),x)

[Out]

result too large to display

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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 )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 0.538193, size = 737, normalized size = 8.77 \begin{align*} -\frac{8 \, \sqrt{2} \sqrt{\cos \left (b x + a\right ) \sin \left (b x + a\right )} \cos \left (b x + a\right ) - 2 \, \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 ) + 2 \, \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 ) + \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 )}{16 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(8*sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a))*cos(b*x + a) - 2*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)) + 2*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) - si
n(b*x + a))) + 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

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

[Out]

Timed out

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Giac [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 )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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