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

Optimal. Leaf size=59 \[ \frac{(a+b) \sin (x)}{2 a b \left (a+b \sin ^2(x)\right )}-\frac{(a-b) \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{2 a^{3/2} b^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0581525, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3190, 385, 205} \[ \frac{(a+b) \sin (x)}{2 a b \left (a+b \sin ^2(x)\right )}-\frac{(a-b) \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{2 a^{3/2} b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]^3/(a + b*Sin[x]^2)^2,x]

[Out]

-((a - b)*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/(2*a^(3/2)*b^(3/2)) + ((a + b)*Sin[x])/(2*a*b*(a + b*Sin[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 385

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.0652649, size = 59, normalized size = 1. \[ \frac{(a+b) \sin (x)}{2 a b \left (a+b \sin ^2(x)\right )}-\frac{(a-b) \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{2 a^{3/2} b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.05, size = 65, normalized size = 1.1 \begin{align*}{\frac{ \left ( a+b \right ) \sin \left ( x \right ) }{2\,ab \left ( a+b \left ( \sin \left ( x \right ) \right ) ^{2} \right ) }}-{\frac{1}{2\,b}\arctan \left ({\sin \left ( x \right ) b{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{1}{2\,a}\arctan \left ({\sin \left ( x \right ) b{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a+b)*sin(x)/a/b/(a+b*sin(x)^2)-1/2/b/(a*b)^(1/2)*arctan(sin(x)*b/(a*b)^(1/2))+1/2/a/(a*b)^(1/2)*arctan(si
n(x)*b/(a*b)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.1153, size = 455, normalized size = 7.71 \begin{align*} \left [\frac{{\left ({\left (a b - b^{2}\right )} \cos \left (x\right )^{2} - a^{2} + b^{2}\right )} \sqrt{-a b} \log \left (-\frac{b \cos \left (x\right )^{2} + 2 \, \sqrt{-a b} \sin \left (x\right ) + a - b}{b \cos \left (x\right )^{2} - a - b}\right ) - 2 \,{\left (a^{2} b + a b^{2}\right )} \sin \left (x\right )}{4 \,{\left (a^{2} b^{3} \cos \left (x\right )^{2} - a^{3} b^{2} - a^{2} b^{3}\right )}}, -\frac{{\left ({\left (a b - b^{2}\right )} \cos \left (x\right )^{2} - a^{2} + b^{2}\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} \sin \left (x\right )}{a}\right ) +{\left (a^{2} b + a b^{2}\right )} \sin \left (x\right )}{2 \,{\left (a^{2} b^{3} \cos \left (x\right )^{2} - a^{3} b^{2} - a^{2} b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(((a*b - b^2)*cos(x)^2 - a^2 + b^2)*sqrt(-a*b)*log(-(b*cos(x)^2 + 2*sqrt(-a*b)*sin(x) + a - b)/(b*cos(x)^
2 - a - b)) - 2*(a^2*b + a*b^2)*sin(x))/(a^2*b^3*cos(x)^2 - a^3*b^2 - a^2*b^3), -1/2*(((a*b - b^2)*cos(x)^2 -
a^2 + b^2)*sqrt(a*b)*arctan(sqrt(a*b)*sin(x)/a) + (a^2*b + a*b^2)*sin(x))/(a^2*b^3*cos(x)^2 - a^3*b^2 - a^2*b^
3)]

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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(x)**3/(a+b*sin(x)**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.11243, size = 76, normalized size = 1.29 \begin{align*} -\frac{{\left (a - b\right )} \arctan \left (\frac{b \sin \left (x\right )}{\sqrt{a b}}\right )}{2 \, \sqrt{a b} a b} + \frac{a \sin \left (x\right ) + b \sin \left (x\right )}{2 \,{\left (b \sin \left (x\right )^{2} + a\right )} a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a - b)*arctan(b*sin(x)/sqrt(a*b))/(sqrt(a*b)*a*b) + 1/2*(a*sin(x) + b*sin(x))/((b*sin(x)^2 + a)*a*b)