3.200 \(\int \frac{\sin ^2(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=68 \[ \frac{2 \sin (a+b x)}{b d \sqrt{d \cos (a+b x)}}-\frac{4 E\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{d \cos (a+b x)}}{b d^2 \sqrt{\cos (a+b x)}} \]

[Out]

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

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Rubi [A]  time = 0.0637746, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2566, 2640, 2639} \[ \frac{2 \sin (a+b x)}{b d \sqrt{d \cos (a+b x)}}-\frac{4 E\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{d \cos (a+b x)}}{b d^2 \sqrt{\cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2566

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(a*Sin[e
+ f*x])^(m - 1)*(b*Cos[e + f*x])^(n + 1))/(b*f*(n + 1)), x] + Dist[(a^2*(m - 1))/(b^2*(n + 1)), Int[(a*Sin[e +
f*x])^(m - 2)*(b*Cos[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && (Integ
ersQ[2*m, 2*n] || EqQ[m + n, 0])

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

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

Mathematica [C]  time = 0.0817393, size = 60, normalized size = 0.88 \[ \frac{\sin ^3(a+b x) \sqrt [4]{\cos ^2(a+b x)} \, _2F_1\left (\frac{5}{4},\frac{3}{2};\frac{5}{2};\sin ^2(a+b x)\right )}{3 b d \sqrt{d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[5/4, 3/2, 5/2, Sin[a + b*x]^2]*Sin[a + b*x]^3)/(3*b*d*Sqrt[d*Cos[a +
b*x]])

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Maple [A]  time = 0.065, size = 168, normalized size = 2.5 \begin{align*} -4\,{\frac{\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}d+ \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d} \left ( \sqrt{2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}}{\it EllipticE} \left ( \cos \left ( 1/2\,bx+a/2 \right ) ,\sqrt{2} \right ) - \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}\cos \left ( 1/2\,bx+a/2 \right ) \right ) }{d\sqrt{-d \left ( 2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}- \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2} \right ) }\sin \left ( 1/2\,bx+a/2 \right ) \sqrt{d \left ( 2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) }b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/d*(-2*sin(1/2*b*x+1/2*a)^4*d+sin(1/2*b*x+1/2*a)^2*d)^(1/2)*((2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*(sin(1/2*b*x+1
/2*a)^2)^(1/2)*EllipticE(cos(1/2*b*x+1/2*a),2^(1/2))-sin(1/2*b*x+1/2*a)^2*cos(1/2*b*x+1/2*a))/(-d*(2*sin(1/2*b
*x+1/2*a)^4-sin(1/2*b*x+1/2*a)^2))^(1/2)/sin(1/2*b*x+1/2*a)/(d*(2*cos(1/2*b*x+1/2*a)^2-1))^(1/2)/b

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (b x + a\right )^{2}}{\left (d \cos \left (b x + a\right )\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/(d*cos(b*x+a))^(3/2),x, algorithm="giac")

[Out]

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