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

Optimal. Leaf size=81 \[ -\frac{2 \cos (e+f x)}{3 f (a+b)^2 \sqrt{a-b \cos ^2(e+f x)+b}}-\frac{\sin ^2(e+f x) \cos (e+f x)}{3 f (a+b) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0951856, antiderivative size = 81, 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 = {3186, 378, 191} \[ -\frac{2 \cos (e+f x)}{3 f (a+b)^2 \sqrt{a-b \cos ^2(e+f x)+b}}-\frac{\sin ^2(e+f x) \cos (e+f x)}{3 f (a+b) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3186

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, -Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos
[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{\sin ^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-b x^2\right )^{5/2}} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{\cos (e+f x) \sin ^2(e+f x)}{3 (a+b) f \left (a+b-b \cos ^2(e+f x)\right )^{3/2}}-\frac{2 \operatorname{Subst}\left (\int \frac{1}{\left (a+b-b x^2\right )^{3/2}} \, dx,x,\cos (e+f x)\right )}{3 (a+b) f}\\ &=-\frac{2 \cos (e+f x)}{3 (a+b)^2 f \sqrt{a+b-b \cos ^2(e+f x)}}-\frac{\cos (e+f x) \sin ^2(e+f x)}{3 (a+b) f \left (a+b-b \cos ^2(e+f x)\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.310118, size = 64, normalized size = 0.79 \[ \frac{\sqrt{2} \cos (e+f x) ((a+3 b) \cos (2 (e+f x))-5 a-3 b)}{3 f (a+b)^2 (2 a-b \cos (2 (e+f x))+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*Cos[e + f*x]*(-5*a - 3*b + (a + 3*b)*Cos[2*(e + f*x)]))/(3*(a + b)^2*f*(2*a + b - b*Cos[2*(e + f*x)])
^(3/2))

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Maple [A]  time = 1.161, size = 56, normalized size = 0.7 \begin{align*} -{\frac{ \left ( a \left ( \sin \left ( fx+e \right ) \right ) ^{2}+3\,b \left ( \sin \left ( fx+e \right ) \right ) ^{2}+2\,a \right ) \cos \left ( fx+e \right ) }{3\, \left ( a+b \right ) ^{2}f} \left ( a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.99422, size = 163, normalized size = 2.01 \begin{align*} -\frac{\frac{2 \, \cos \left (f x + e\right )}{\sqrt{-b \cos \left (f x + e\right )^{2} + a + b}{\left (a + b\right )}^{2}} + \frac{\cos \left (f x + e\right )}{{\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac{3}{2}}{\left (a + b\right )}} - \frac{\cos \left (f x + e\right )}{{\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac{3}{2}} b} + \frac{\cos \left (f x + e\right )}{\sqrt{-b \cos \left (f x + e\right )^{2} + a + b}{\left (a + b\right )} b}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^3/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.67331, size = 323, normalized size = 3.99 \begin{align*} \frac{{\left ({\left (a + 3 \, b\right )} \cos \left (f x + e\right )^{3} - 3 \,{\left (a + b\right )} \cos \left (f x + e\right )\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{3 \,{\left ({\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{4} - 2 \,{\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^3/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

1/3*((a + 3*b)*cos(f*x + e)^3 - 3*(a + b)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b)/((a^2*b^2 + 2*a*b^3 +
b^4)*f*cos(f*x + e)^4 - 2*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*f*cos(f*x + e)^2 + (a^4 + 4*a^3*b + 6*a^2*b^2 +
4*a*b^3 + b^4)*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(sin(f*x+e)**3/(a+b*sin(f*x+e)**2)**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.35355, size = 201, normalized size = 2.48 \begin{align*} -\frac{\sqrt{-{\left (\cos \left (f x + e\right )^{2} - 1\right )} b + a}{\left (\frac{3 \,{\left (a b f^{2} + b^{2} f^{2}\right )}}{a^{2} b f^{2} + 2 \, a b^{2} f^{2} + b^{3} f^{2}} - \frac{{\left (a b f^{4} + 3 \, b^{2} f^{4}\right )} \cos \left (f x + e\right )^{2}}{{\left (a^{2} b f^{2} + 2 \, a b^{2} f^{2} + b^{3} f^{2}\right )} f^{2}}\right )} \cos \left (f x + e\right )}{3 \,{\left ({\left (\cos \left (f x + e\right )^{2} - 1\right )} b - a\right )}^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^3/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

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