3.364 \(\int \frac{\cos ^3(e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=73 \[ \frac{2 \sin (e+f x)}{3 a^2 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\sin (e+f x) \cos ^2(e+f x)}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

[Out]

(Cos[e + f*x]^2*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 = 0.0922958, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {3190, 378, 191} \[ \frac{2 \sin (e+f x)}{3 a^2 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\sin (e+f x) \cos ^2(e+f x)}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^3/(a + b*Sin[e + f*x]^2)^(5/2),x]

[Out]

(Cos[e + f*x]^2*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 378

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^q)/(a*n*(p + 1)), x] - Dist[(c*q)/(a*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 0] && GtQ[q, 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 ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1-x^2}{\left (a+b x^2\right )^{5/2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\cos ^2(e+f x) \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{\cos ^2(e+f x) \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 = 0.106822, size = 51, normalized size = 0.7 \[ \frac{3 a \sin (e+f x)-(a-2 b) \sin ^3(e+f x)}{3 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^3/(a + b*Sin[e + f*x]^2)^(5/2),x]

[Out]

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

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Maple [A]  time = 3.362, size = 120, normalized size = 1.6 \begin{align*}{\frac{\sin \left ( fx+e \right ) \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2\,b \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\,a+2\,b \right ) }{3\,{a}^{2} \left ({b}^{2} \left ( \cos \left ( fx+e \right ) \right ) ^{4}-2\,ab \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2\,{b}^{2} \left ( \cos \left ( fx+e \right ) \right ) ^{2}+{a}^{2}+2\,ab+{b}^{2} \right ) f}\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}+{\frac{a{b}^{2}+{b}^{3}}{{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^3/(a+b*sin(f*x+e)^2)^(5/2),x)

[Out]

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

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Maxima [A]  time = 0.97476, size = 144, normalized size = 1.97 \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} + \frac{\sin \left (f x + e\right )}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} b} - \frac{\sin \left (f x + e\right )}{\sqrt{b \sin \left (f x + e\right )^{2} + a} a b}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]  time = 4.28294, size = 250, normalized size = 3.42 \begin{align*} \frac{{\left ({\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{2} + 2 \, 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)^3/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

1/3*((a - 2*b)*cos(f*x + e)^2 + 2*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)**3/(a+b*sin(f*x+e)**2)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.39039, size = 78, normalized size = 1.07 \begin{align*} -\frac{{\left (\frac{{\left (a b - 2 \, b^{2}\right )} \sin \left (f x + e\right )^{2}}{a^{2} b} - \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)^3/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

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