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

Optimal. Leaf size=40 \[ \frac{\sin ^{\frac{3}{2}}(2 a+2 b x)}{6 b}+\frac{E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{2 b} \]

[Out]

EllipticE[a - Pi/4 + b*x, 2]/(2*b) + Sin[2*a + 2*b*x]^(3/2)/(6*b)

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Rubi [A]  time = 0.0345386, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {4297, 2639} \[ \frac{\sin ^{\frac{3}{2}}(2 a+2 b x)}{6 b}+\frac{E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

EllipticE[a - Pi/4 + b*x, 2]/(2*b) + Sin[2*a + 2*b*x]^(3/2)/(6*b)

Rule 4297

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(e^2*(e*Cos[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*Co
s[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 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int \cos ^2(a+b x) \sqrt{\sin (2 a+2 b x)} \, dx &=\frac{\sin ^{\frac{3}{2}}(2 a+2 b x)}{6 b}+\frac{1}{2} \int \sqrt{\sin (2 a+2 b x)} \, dx\\ &=\frac{E\left (\left .a-\frac{\pi }{4}+b x\right |2\right )}{2 b}+\frac{\sin ^{\frac{3}{2}}(2 a+2 b x)}{6 b}\\ \end{align*}

Mathematica [A]  time = 0.0577439, size = 34, normalized size = 0.85 \[ \frac{\sin ^{\frac{3}{2}}(2 (a+b x))+3 E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{6 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*EllipticE[a - Pi/4 + b*x, 2] + Sin[2*(a + b*x)]^(3/2))/(6*b)

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Maple [B]  time = 3.975, size = 24847123, normalized size = 621178.1 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^2*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 \cos \left (b x + a\right )^{2} \sqrt{\sin \left (2 \, b x + 2 \, a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\cos \left (b x + a\right )^{2} \sqrt{\sin \left (2 \, b x + 2 \, a\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(cos(b*x + a)^2*sqrt(sin(2*b*x + 2*a)), x)

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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(cos(b*x+a)**2*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 \cos \left (b x + a\right )^{2} \sqrt{\sin \left (2 \, b x + 2 \, a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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