3.201 \(\int \frac{\sin ^2(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx\)

Optimal. Leaf size=72 \[ \frac{2 \sin (a+b x)}{3 b d (d \cos (a+b x))^{3/2}}-\frac{4 \sqrt{\cos (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{3 b d^2 \sqrt{d \cos (a+b x)}} \]

[Out]

(-4*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2])/(3*b*d^2*Sqrt[d*Cos[a + b*x]]) + (2*Sin[a + b*x])/(3*b*d*(d*
Cos[a + b*x])^(3/2))

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Rubi [A]  time = 0.0644769, antiderivative size = 72, 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, 2642, 2641} \[ \frac{2 \sin (a+b x)}{3 b d (d \cos (a+b x))^{3/2}}-\frac{4 \sqrt{\cos (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{3 b d^2 \sqrt{d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^2/(d*Cos[a + b*x])^(5/2),x]

[Out]

(-4*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2])/(3*b*d^2*Sqrt[d*Cos[a + b*x]]) + (2*Sin[a + b*x])/(3*b*d*(d*
Cos[a + b*x])^(3/2))

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 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(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))^{5/2}} \, dx &=\frac{2 \sin (a+b x)}{3 b d (d \cos (a+b x))^{3/2}}-\frac{2 \int \frac{1}{\sqrt{d \cos (a+b x)}} \, dx}{3 d^2}\\ &=\frac{2 \sin (a+b x)}{3 b d (d \cos (a+b x))^{3/2}}-\frac{\left (2 \sqrt{\cos (a+b x)}\right ) \int \frac{1}{\sqrt{\cos (a+b x)}} \, dx}{3 d^2 \sqrt{d \cos (a+b x)}}\\ &=-\frac{4 \sqrt{\cos (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{3 b d^2 \sqrt{d \cos (a+b x)}}+\frac{2 \sin (a+b x)}{3 b d (d \cos (a+b x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0809506, size = 60, normalized size = 0.83 \[ \frac{\sin ^3(a+b x) \cos ^2(a+b x)^{3/4} \, _2F_1\left (\frac{3}{2},\frac{7}{4};\frac{5}{2};\sin ^2(a+b x)\right )}{3 b d (d \cos (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^2/(d*Cos[a + b*x])^(5/2),x]

[Out]

((Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[3/2, 7/4, 5/2, Sin[a + b*x]^2]*Sin[a + b*x]^3)/(3*b*d*(d*Cos[a + b*x
])^(3/2))

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Maple [B]  time = 0.063, size = 242, normalized size = 3.4 \begin{align*} -{\frac{4}{3\,{d}^{2}b} \left ( 2\,\sqrt{ \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,bx+a/2 \right ) ,\sqrt{2} \right ) \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-\sqrt{ \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) ,\sqrt{2} \right ) - \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}\cos \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) \sqrt{d \left ( 2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}} \left ( 2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) ^{-1}{\frac{1}{\sqrt{-d \left ( 2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}- \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2} \right ) }}} \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{d \left ( 2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^2/(d*cos(b*x+a))^(5/2),x)

[Out]

-4/3*(2*(sin(1/2*b*x+1/2*a)^2)^(1/2)*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*EllipticF(cos(1/2*b*x+1/2*a),2^(1/2))*si
n(1/2*b*x+1/2*a)^2-(sin(1/2*b*x+1/2*a)^2)^(1/2)*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*EllipticF(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*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2
)/(2*cos(1/2*b*x+1/2*a)^2-1)/(-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{5}{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))^(5/2),x, algorithm="maxima")

[Out]

integrate(sin(b*x + a)^2/(d*cos(b*x + a))^(5/2), 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^{3} \cos \left (b x + a\right )^{3}}, 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))^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(d*cos(b*x + a))*(cos(b*x + a)^2 - 1)/(d^3*cos(b*x + a)^3), 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))**(5/2),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{5}{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))^(5/2),x, algorithm="giac")

[Out]

integrate(sin(b*x + a)^2/(d*cos(b*x + a))^(5/2), x)