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

Optimal. Leaf size=54 \[ \frac{(a+2 b) \cos (x)}{b^2}-\frac{(a+b)^2 \tan ^{-1}\left (\frac{\sqrt{b} \cos (x)}{\sqrt{a}}\right )}{\sqrt{a} b^{5/2}}-\frac{\cos ^3(x)}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.0726949, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3190, 390, 205} \[ \frac{(a+2 b) \cos (x)}{b^2}-\frac{(a+b)^2 \tan ^{-1}\left (\frac{\sqrt{b} \cos (x)}{\sqrt{a}}\right )}{\sqrt{a} b^{5/2}}-\frac{\cos ^3(x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^5/(a + b*Cos[x]^2),x]

[Out]

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

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

Mathematica [B]  time = 0.16841, size = 116, normalized size = 2.15 \[ \frac{3 \sqrt{b} (4 a+7 b) \cos (x)-\frac{12 (a+b)^2 \tan ^{-1}\left (\frac{\sqrt{b}-\sqrt{a+b} \tan \left (\frac{x}{2}\right )}{\sqrt{a}}\right )}{\sqrt{a}}-\frac{12 (a+b)^2 \tan ^{-1}\left (\frac{\sqrt{a+b} \tan \left (\frac{x}{2}\right )+\sqrt{b}}{\sqrt{a}}\right )}{\sqrt{a}}-b^{3/2} \cos (3 x)}{12 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-12*(a + b)^2*ArcTan[(Sqrt[b] - Sqrt[a + b]*Tan[x/2])/Sqrt[a]])/Sqrt[a] - (12*(a + b)^2*ArcTan[(Sqrt[b] + Sq
rt[a + b]*Tan[x/2])/Sqrt[a]])/Sqrt[a] + 3*Sqrt[b]*(4*a + 7*b)*Cos[x] - b^(3/2)*Cos[3*x])/(12*b^(5/2))

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Maple [A]  time = 0.019, size = 86, normalized size = 1.6 \begin{align*} -{\frac{ \left ( \cos \left ( x \right ) \right ) ^{3}}{3\,b}}+{\frac{a\cos \left ( x \right ) }{{b}^{2}}}+2\,{\frac{\cos \left ( x \right ) }{b}}-{\frac{{a}^{2}}{{b}^{2}}\arctan \left ({b\cos \left ( x \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-2\,{\frac{a}{b\sqrt{ab}}\arctan \left ({\frac{b\cos \left ( x \right ) }{\sqrt{ab}}} \right ) }-{\arctan \left ({b\cos \left ( x \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.77727, size = 383, normalized size = 7.09 \begin{align*} \left [-\frac{2 \, a b^{2} \cos \left (x\right )^{3} + 3 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt{-a b} \log \left (-\frac{b \cos \left (x\right )^{2} + 2 \, \sqrt{-a b} \cos \left (x\right ) - a}{b \cos \left (x\right )^{2} + a}\right ) - 6 \,{\left (a^{2} b + 2 \, a b^{2}\right )} \cos \left (x\right )}{6 \, a b^{3}}, -\frac{a b^{2} \cos \left (x\right )^{3} + 3 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} \cos \left (x\right )}{a}\right ) - 3 \,{\left (a^{2} b + 2 \, a b^{2}\right )} \cos \left (x\right )}{3 \, a b^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(2*a*b^2*cos(x)^3 + 3*(a^2 + 2*a*b + b^2)*sqrt(-a*b)*log(-(b*cos(x)^2 + 2*sqrt(-a*b)*cos(x) - a)/(b*cos(
x)^2 + a)) - 6*(a^2*b + 2*a*b^2)*cos(x))/(a*b^3), -1/3*(a*b^2*cos(x)^3 + 3*(a^2 + 2*a*b + b^2)*sqrt(a*b)*arcta
n(sqrt(a*b)*cos(x)/a) - 3*(a^2*b + 2*a*b^2)*cos(x))/(a*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(sin(x)**5/(a+b*cos(x)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.12684, size = 80, normalized size = 1.48 \begin{align*} -\frac{{\left (a^{2} + 2 \, a b + b^{2}\right )} \arctan \left (\frac{b \cos \left (x\right )}{\sqrt{a b}}\right )}{\sqrt{a b} b^{2}} - \frac{b^{2} \cos \left (x\right )^{3} - 3 \, a b \cos \left (x\right ) - 6 \, b^{2} \cos \left (x\right )}{3 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^2 + 2*a*b + b^2)*arctan(b*cos(x)/sqrt(a*b))/(sqrt(a*b)*b^2) - 1/3*(b^2*cos(x)^3 - 3*a*b*cos(x) - 6*b^2*cos
(x))/b^3