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

Optimal. Leaf size=72 \[ -\frac{(3 a-b) (a+b) \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{2 a^{3/2} b^{5/2}}+\frac{(a+b)^2 \sin (x)}{2 a b^2 \left (a+b \sin ^2(x)\right )}+\frac{\sin (x)}{b^2} \]

[Out]

-((3*a - b)*(a + b)*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/(2*a^(3/2)*b^(5/2)) + Sin[x]/b^2 + ((a + b)^2*Sin[x])/(2
*a*b^2*(a + b*Sin[x]^2))

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Rubi [A]  time = 0.118633, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {3190, 390, 385, 205} \[ -\frac{(3 a-b) (a+b) \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{2 a^{3/2} b^{5/2}}+\frac{(a+b)^2 \sin (x)}{2 a b^2 \left (a+b \sin ^2(x)\right )}+\frac{\sin (x)}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]^5/(a + b*Sin[x]^2)^2,x]

[Out]

-((3*a - b)*(a + b)*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/(2*a^(3/2)*b^(5/2)) + Sin[x]/b^2 + ((a + b)^2*Sin[x])/(2
*a*b^2*(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 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

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

Mathematica [A]  time = 0.328314, size = 118, normalized size = 1.64 \[ \frac{\frac{\left (-3 a^2-2 a b+b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{a^{3/2}}+\frac{\left (3 a^2+2 a b-b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a} \csc (x)}{\sqrt{b}}\right )}{a^{3/2}}+\frac{4 \sqrt{b} (a+b)^2 \sin (x)}{a (2 a-b \cos (2 x)+b)}+4 \sqrt{b} \sin (x)}{4 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((3*a^2 + 2*a*b - b^2)*ArcTan[(Sqrt[a]*Csc[x])/Sqrt[b]])/a^(3/2) + ((-3*a^2 - 2*a*b + b^2)*ArcTan[(Sqrt[b]*Si
n[x])/Sqrt[a]])/a^(3/2) + 4*Sqrt[b]*Sin[x] + (4*Sqrt[b]*(a + b)^2*Sin[x])/(a*(2*a + b - b*Cos[2*x])))/(4*b^(5/
2))

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Maple [A]  time = 0.05, size = 120, normalized size = 1.7 \begin{align*}{\frac{\sin \left ( x \right ) }{{b}^{2}}}+{\frac{a\sin \left ( x \right ) }{2\,{b}^{2} \left ( a+b \left ( \sin \left ( x \right ) \right ) ^{2} \right ) }}+{\frac{\sin \left ( x \right ) }{b \left ( a+b \left ( \sin \left ( x \right ) \right ) ^{2} \right ) }}+{\frac{\sin \left ( x \right ) }{2\,a \left ( a+b \left ( \sin \left ( x \right ) \right ) ^{2} \right ) }}-{\frac{3\,a}{2\,{b}^{2}}\arctan \left ({\sin \left ( x \right ) b{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{1}{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)^5/(a+b*sin(x)^2)^2,x)

[Out]

sin(x)/b^2+1/2/b^2*a*sin(x)/(a+b*sin(x)^2)+1/b*sin(x)/(a+b*sin(x)^2)+1/2*sin(x)/a/(a+b*sin(x)^2)-3/2/b^2*a/(a*
b)^(1/2)*arctan(sin(x)*b/(a*b)^(1/2))-1/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)^5/(a+b*sin(x)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.12144, size = 641, normalized size = 8.9 \begin{align*} \left [-\frac{{\left (3 \, a^{3} + 5 \, a^{2} b + a b^{2} - b^{3} -{\left (3 \, a^{2} b + 2 \, a b^{2} - b^{3}\right )} \cos \left (x\right )^{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 (2 \, a^{2} b^{2} \cos \left (x\right )^{2} - 3 \, a^{3} b - 4 \, a^{2} b^{2} - a b^{3}\right )} \sin \left (x\right )}{4 \,{\left (a^{2} b^{4} \cos \left (x\right )^{2} - a^{3} b^{3} - a^{2} b^{4}\right )}}, \frac{{\left (3 \, a^{3} + 5 \, a^{2} b + a b^{2} - b^{3} -{\left (3 \, a^{2} b + 2 \, a b^{2} - b^{3}\right )} \cos \left (x\right )^{2}\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} \sin \left (x\right )}{a}\right ) +{\left (2 \, a^{2} b^{2} \cos \left (x\right )^{2} - 3 \, a^{3} b - 4 \, a^{2} b^{2} - a b^{3}\right )} \sin \left (x\right )}{2 \,{\left (a^{2} b^{4} \cos \left (x\right )^{2} - a^{3} b^{3} - a^{2} b^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((3*a^3 + 5*a^2*b + a*b^2 - b^3 - (3*a^2*b + 2*a*b^2 - b^3)*cos(x)^2)*sqrt(-a*b)*log(-(b*cos(x)^2 + 2*sq
rt(-a*b)*sin(x) + a - b)/(b*cos(x)^2 - a - b)) - 2*(2*a^2*b^2*cos(x)^2 - 3*a^3*b - 4*a^2*b^2 - a*b^3)*sin(x))/
(a^2*b^4*cos(x)^2 - a^3*b^3 - a^2*b^4), 1/2*((3*a^3 + 5*a^2*b + a*b^2 - b^3 - (3*a^2*b + 2*a*b^2 - b^3)*cos(x)
^2)*sqrt(a*b)*arctan(sqrt(a*b)*sin(x)/a) + (2*a^2*b^2*cos(x)^2 - 3*a^3*b - 4*a^2*b^2 - a*b^3)*sin(x))/(a^2*b^4
*cos(x)^2 - a^3*b^3 - a^2*b^4)]

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

[Out]

Timed out

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Giac [A]  time = 1.12598, size = 111, normalized size = 1.54 \begin{align*} \frac{\sin \left (x\right )}{b^{2}} - \frac{{\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \arctan \left (\frac{b \sin \left (x\right )}{\sqrt{a b}}\right )}{2 \, \sqrt{a b} a b^{2}} + \frac{a^{2} \sin \left (x\right ) + 2 \, a b \sin \left (x\right ) + b^{2} \sin \left (x\right )}{2 \,{\left (b \sin \left (x\right )^{2} + a\right )} a b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sin(x)/b^2 - 1/2*(3*a^2 + 2*a*b - b^2)*arctan(b*sin(x)/sqrt(a*b))/(sqrt(a*b)*a*b^2) + 1/2*(a^2*sin(x) + 2*a*b*
sin(x) + b^2*sin(x))/((b*sin(x)^2 + a)*a*b^2)