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

Optimal. Leaf size=56 \[ \frac{a^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a+b}}\right )}{b^{5/2} \sqrt{a+b}}-\frac{(a-b) \sin (x)}{b^2}-\frac{\sin ^3(x)}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.071169, antiderivative size = 56, 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 = {3186, 390, 208} \[ \frac{a^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a+b}}\right )}{b^{5/2} \sqrt{a+b}}-\frac{(a-b) \sin (x)}{b^2}-\frac{\sin ^3(x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]^5/(a + b*Cos[x]^2),x]

[Out]

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

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 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 208

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

Rubi steps

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

Mathematica [A]  time = 0.184982, size = 86, normalized size = 1.54 \[ \frac{\frac{6 a^2 \left (\log \left (\sqrt{a+b}+\sqrt{b} \sin (x)\right )-\log \left (\sqrt{a+b}-\sqrt{b} \sin (x)\right )\right )}{\sqrt{a+b}}+3 \sqrt{b} (3 b-4 a) \sin (x)+b^{3/2} \sin (3 x)}{12 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*a^2*(-Log[Sqrt[a + b] - Sqrt[b]*Sin[x]] + Log[Sqrt[a + b] + Sqrt[b]*Sin[x]]))/Sqrt[a + b] + 3*Sqrt[b]*(-4*
a + 3*b)*Sin[x] + b^(3/2)*Sin[3*x])/(12*b^(5/2))

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Maple [A]  time = 0.015, size = 50, normalized size = 0.9 \begin{align*} -{\frac{1}{{b}^{2}} \left ({\frac{ \left ( \sin \left ( x \right ) \right ) ^{3}b}{3}}+\sin \left ( x \right ) a-\sin \left ( x \right ) b \right ) }+{\frac{{a}^{2}}{{b}^{2}}{\it Artanh} \left ({\sin \left ( x \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.87134, size = 441, normalized size = 7.88 \begin{align*} \left [\frac{3 \, \sqrt{a b + b^{2}} a^{2} \log \left (-\frac{b \cos \left (x\right )^{2} - 2 \, \sqrt{a b + b^{2}} \sin \left (x\right ) - a - 2 \, b}{b \cos \left (x\right )^{2} + a}\right ) - 2 \,{\left (3 \, a^{2} b + a b^{2} - 2 \, b^{3} -{\left (a b^{2} + b^{3}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{6 \,{\left (a b^{3} + b^{4}\right )}}, -\frac{3 \, \sqrt{-a b - b^{2}} a^{2} \arctan \left (\frac{\sqrt{-a b - b^{2}} \sin \left (x\right )}{a + b}\right ) +{\left (3 \, a^{2} b + a b^{2} - 2 \, b^{3} -{\left (a b^{2} + b^{3}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{3 \,{\left (a b^{3} + b^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*sqrt(a*b + b^2)*a^2*log(-(b*cos(x)^2 - 2*sqrt(a*b + b^2)*sin(x) - a - 2*b)/(b*cos(x)^2 + a)) - 2*(3*a^
2*b + a*b^2 - 2*b^3 - (a*b^2 + b^3)*cos(x)^2)*sin(x))/(a*b^3 + b^4), -1/3*(3*sqrt(-a*b - b^2)*a^2*arctan(sqrt(
-a*b - b^2)*sin(x)/(a + b)) + (3*a^2*b + a*b^2 - 2*b^3 - (a*b^2 + b^3)*cos(x)^2)*sin(x))/(a*b^3 + 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*cos(x)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.16901, size = 88, normalized size = 1.57 \begin{align*} -\frac{a^{2} \arctan \left (\frac{b \sin \left (x\right )}{\sqrt{-a b - b^{2}}}\right )}{\sqrt{-a b - b^{2}} b^{2}} - \frac{b^{2} \sin \left (x\right )^{3} + 3 \, a b \sin \left (x\right ) - 3 \, b^{2} \sin \left (x\right )}{3 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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