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]

-1/4*arcsin(cos(b*x+a)-sin(b*x+a))/b+1/4*ln(cos(b*x+a)+sin(b*x+a)+sin(2*b*x+2*a)^(1/2))/b-1/2*cos(b*x+a)*sin(2
*b*x+2*a)^(1/2)/b

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Rubi [A]  time = 0.04, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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*}

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Mathematica [A]  time = 0.08, 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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fricas [B]  time = 0.78, size = 266, normalized size = 3.17 \[ -\frac {8 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} \cos \left (b x + a\right ) - 2 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) + 2 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) + \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{16 \, b} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\sin \left (2 \, b x + 2 \, a\right )} \sin \left (b x + a\right )\,{d x} \]

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)

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maple [B]  time = 2.85, size = 6214674, normalized size = 73984.21 \[ \text {output too large to display} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\sin \left (2 \, b x + 2 \, a\right )} \sin \left (b x + a\right )\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sin \left (a+b\,x\right )\,\sqrt {\sin \left (2\,a+2\,b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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