3.365 \(\int \frac{\cos (e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=65 \[ \frac{2 \sin (e+f x)}{3 a^2 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\sin (e+f x)}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

[Out]

Sin[e + f*x]/(3*a*f*(a + b*Sin[e + f*x]^2)^(3/2)) + (2*Sin[e + f*x])/(3*a^2*f*Sqrt[a + b*Sin[e + f*x]^2])

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Rubi [A]  time = 0.0552965, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3190, 192, 191} \[ \frac{2 \sin (e+f x)}{3 a^2 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\sin (e+f x)}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sin[e + f*x]/(3*a*f*(a + b*Sin[e + f*x]^2)^(3/2)) + (2*Sin[e + f*x])/(3*a^2*f*Sqrt[a + b*Sin[e + f*x]^2])

Rule 3190

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e +
f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

\begin{align*} \int \frac{\cos (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right )^{5/2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\sin (e+f x)}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{3 a f}\\ &=\frac{\sin (e+f x)}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{2 \sin (e+f x)}{3 a^2 f \sqrt{a+b \sin ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.0467914, size = 47, normalized size = 0.72 \[ \frac{\sin (e+f x) \left (3 a+2 b \sin ^2(e+f x)\right )}{3 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sin[e + f*x]*(3*a + 2*b*Sin[e + f*x]^2))/(3*a^2*f*(a + b*Sin[e + f*x]^2)^(3/2))

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Maple [A]  time = 0.087, size = 56, normalized size = 0.9 \begin{align*}{\frac{1}{f} \left ({\frac{\sin \left ( fx+e \right ) }{3\,a} \left ( a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,\sin \left ( fx+e \right ) }{3\,{a}^{2}}{\frac{1}{\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(1/3*sin(f*x+e)/a/(a+b*sin(f*x+e)^2)^(3/2)+2/3/a^2*sin(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2))

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Maxima [A]  time = 0.97846, size = 74, normalized size = 1.14 \begin{align*} \frac{\frac{2 \, \sin \left (f x + e\right )}{\sqrt{b \sin \left (f x + e\right )^{2} + a} a^{2}} + \frac{\sin \left (f x + e\right )}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} a}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

1/3*(2*sin(f*x + e)/(sqrt(b*sin(f*x + e)^2 + a)*a^2) + sin(f*x + e)/((b*sin(f*x + e)^2 + a)^(3/2)*a))/f

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Fricas [A]  time = 3.64586, size = 243, normalized size = 3.74 \begin{align*} -\frac{{\left (2 \, b \cos \left (f x + e\right )^{2} - 3 \, a - 2 \, b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{3 \,{\left (a^{2} b^{2} f \cos \left (f x + e\right )^{4} - 2 \,{\left (a^{3} b + a^{2} b^{2}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

-1/3*(2*b*cos(f*x + e)^2 - 3*a - 2*b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x + e)/(a^2*b^2*f*cos(f*x + e)^4 -
2*(a^3*b + a^2*b^2)*f*cos(f*x + e)^2 + (a^4 + 2*a^3*b + a^2*b^2)*f)

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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(f*x+e)/(a+b*sin(f*x+e)**2)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.35162, size = 65, normalized size = 1. \begin{align*} \frac{{\left (\frac{2 \, b \sin \left (f x + e\right )^{2}}{a^{2}} + \frac{3}{a}\right )} \sin \left (f x + e\right )}{3 \,{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

1/3*(2*b*sin(f*x + e)^2/a^2 + 3/a)*sin(f*x + e)/((b*sin(f*x + e)^2 + a)^(3/2)*f)