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

Optimal. Leaf size=45 \[ \frac{\sin ^2(a+b x)}{b \sqrt{\sin (2 a+2 b x)}}-\frac{E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.0373112, antiderivative size = 45, 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 = {4296, 2639} \[ \frac{\sin ^2(a+b x)}{b \sqrt{\sin (2 a+2 b x)}}-\frac{E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^2/Sin[2*a + 2*b*x]^(3/2),x]

[Out]

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

Rule 4296

Int[((e_.)*sin[(a_.) + (b_.)*(x_)])^(m_)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> -Simp[((e*Sin[a +
b*x])^m*(g*Sin[c + d*x])^(p + 1))/(2*b*g*(p + 1)), x] + Dist[(e^2*(m + 2*p + 2))/(4*g^2*(p + 1)), Int[(e*Sin[a
+ b*x])^(m - 2)*(g*Sin[c + d*x])^(p + 2), x], x] /; FreeQ[{a, b, c, d, e, g}, x] && EqQ[b*c - a*d, 0] && EqQ[
d/b, 2] &&  !IntegerQ[p] && GtQ[m, 1] && LtQ[p, -1] && NeQ[m + 2*p + 2, 0] && (LtQ[p, -2] || EqQ[m, 2]) && Int
egersQ[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 \frac{\sin ^2(a+b x)}{\sin ^{\frac{3}{2}}(2 a+2 b x)} \, dx &=\frac{\sin ^2(a+b x)}{b \sqrt{\sin (2 a+2 b x)}}-\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 ^2(a+b x)}{b \sqrt{\sin (2 a+2 b x)}}\\ \end{align*}

Mathematica [A]  time = 0.0955152, size = 41, normalized size = 0.91 \[ \frac{\sqrt{\sin (2 (a+b x))} \tan (a+b x)-E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^2/Sin[2*a + 2*b*x]^(3/2),x]

[Out]

(-EllipticE[a - Pi/4 + b*x, 2] + Sqrt[Sin[2*(a + b*x)]]*Tan[a + b*x])/(2*b)

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Maple [B]  time = 9.329, size = 83117872, normalized size = 1847063.8 \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)^2/sin(2*b*x+2*a)^(3/2),x)

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (\cos \left (b x + a\right )^{2} - 1\right )} \sqrt{\sin \left (2 \, b x + 2 \, a\right )}}{\cos \left (2 \, b x + 2 \, a\right )^{2} - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((cos(b*x + a)^2 - 1)*sqrt(sin(2*b*x + 2*a))/(cos(2*b*x + 2*a)^2 - 1), 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(sin(b*x+a)**2/sin(2*b*x+2*a)**(3/2),x)

[Out]

Timed out

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

[Out]

Timed out