3.28 \(\int \frac{\cos ^7(x)}{a+b \cos ^2(x)} \, dx\)

Optimal. Leaf size=78 \[ \frac{\left (a^2-a b+b^2\right ) \sin (x)}{b^3}-\frac{a^3 \tanh ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a+b}}\right )}{b^{7/2} \sqrt{a+b}}+\frac{(a-2 b) \sin ^3(x)}{3 b^2}+\frac{\sin ^5(x)}{5 b} \]

[Out]

-((a^3*ArcTanh[(Sqrt[b]*Sin[x])/Sqrt[a + b]])/(b^(7/2)*Sqrt[a + b])) + ((a^2 - a*b + b^2)*Sin[x])/b^3 + ((a -
2*b)*Sin[x]^3)/(3*b^2) + Sin[x]^5/(5*b)

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Rubi [A]  time = 0.0840385, antiderivative size = 78, 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} \[ \frac{\left (a^2-a b+b^2\right ) \sin (x)}{b^3}-\frac{a^3 \tanh ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a+b}}\right )}{b^{7/2} \sqrt{a+b}}+\frac{(a-2 b) \sin ^3(x)}{3 b^2}+\frac{\sin ^5(x)}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]^7/(a + b*Cos[x]^2),x]

[Out]

-((a^3*ArcTanh[(Sqrt[b]*Sin[x])/Sqrt[a + b]])/(b^(7/2)*Sqrt[a + b])) + ((a^2 - a*b + b^2)*Sin[x])/b^3 + ((a -
2*b)*Sin[x]^3)/(3*b^2) + Sin[x]^5/(5*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 ^7(x)}{a+b \cos ^2(x)} \, dx &=\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^3}{a+b-b x^2} \, dx,x,\sin (x)\right )\\ &=\operatorname{Subst}\left (\int \left (\frac{a^2-a b+b^2}{b^3}+\frac{(a-2 b) x^2}{b^2}+\frac{x^4}{b}-\frac{a^3}{b^3 \left (a+b-b x^2\right )}\right ) \, dx,x,\sin (x)\right )\\ &=\frac{\left (a^2-a b+b^2\right ) \sin (x)}{b^3}+\frac{(a-2 b) \sin ^3(x)}{3 b^2}+\frac{\sin ^5(x)}{5 b}-\frac{a^3 \operatorname{Subst}\left (\int \frac{1}{a+b-b x^2} \, dx,x,\sin (x)\right )}{b^3}\\ &=-\frac{a^3 \tanh ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a+b}}\right )}{b^{7/2} \sqrt{a+b}}+\frac{\left (a^2-a b+b^2\right ) \sin (x)}{b^3}+\frac{(a-2 b) \sin ^3(x)}{3 b^2}+\frac{\sin ^5(x)}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.385071, size = 111, normalized size = 1.42 \[ \frac{\left (8 a^2-6 a b+5 b^2\right ) \sin (x)}{8 b^3}+\frac{a^3 \left (\log \left (\sqrt{a+b}-\sqrt{b} \sin (x)\right )-\log \left (\sqrt{a+b}+\sqrt{b} \sin (x)\right )\right )}{2 b^{7/2} \sqrt{a+b}}+\frac{(5 b-4 a) \sin (3 x)}{48 b^2}+\frac{\sin (5 x)}{80 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]^7/(a + b*Cos[x]^2),x]

[Out]

(a^3*(Log[Sqrt[a + b] - Sqrt[b]*Sin[x]] - Log[Sqrt[a + b] + Sqrt[b]*Sin[x]]))/(2*b^(7/2)*Sqrt[a + b]) + ((8*a^
2 - 6*a*b + 5*b^2)*Sin[x])/(8*b^3) + ((-4*a + 5*b)*Sin[3*x])/(48*b^2) + Sin[5*x]/(80*b)

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Maple [A]  time = 0.02, size = 78, normalized size = 1. \begin{align*}{\frac{1}{{b}^{3}} \left ({\frac{ \left ( \sin \left ( x \right ) \right ) ^{5}{b}^{2}}{5}}+{\frac{ \left ( \sin \left ( x \right ) \right ) ^{3}ba}{3}}-{\frac{2\, \left ( \sin \left ( x \right ) \right ) ^{3}{b}^{2}}{3}}+{a}^{2}\sin \left ( x \right ) -ab\sin \left ( x \right ) +{b}^{2}\sin \left ( x \right ) \right ) }-{\frac{{a}^{3}}{{b}^{3}}{\it Artanh} \left ({b\sin \left ( x \right ){\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)^7/(a+b*cos(x)^2),x)

[Out]

1/b^3*(1/5*sin(x)^5*b^2+1/3*sin(x)^3*b*a-2/3*sin(x)^3*b^2+a^2*sin(x)-a*b*sin(x)+b^2*sin(x))-a^3/b^3/((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)^7/(a+b*cos(x)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.91061, size = 597, normalized size = 7.65 \begin{align*} \left [\frac{15 \, \sqrt{a b + b^{2}} a^{3} \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 \,{\left (a b^{3} + b^{4}\right )} \cos \left (x\right )^{4} + 15 \, a^{3} b + 5 \, a^{2} b^{2} - 2 \, a b^{3} + 8 \, b^{4} -{\left (5 \, a^{2} b^{2} + a b^{3} - 4 \, b^{4}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{30 \,{\left (a b^{4} + b^{5}\right )}}, \frac{15 \, \sqrt{-a b - b^{2}} a^{3} \arctan \left (\frac{\sqrt{-a b - b^{2}} \sin \left (x\right )}{a + b}\right ) +{\left (3 \,{\left (a b^{3} + b^{4}\right )} \cos \left (x\right )^{4} + 15 \, a^{3} b + 5 \, a^{2} b^{2} - 2 \, a b^{3} + 8 \, b^{4} -{\left (5 \, a^{2} b^{2} + a b^{3} - 4 \, b^{4}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{15 \,{\left (a b^{4} + b^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^7/(a+b*cos(x)^2),x, algorithm="fricas")

[Out]

[1/30*(15*sqrt(a*b + b^2)*a^3*log(-(b*cos(x)^2 + 2*sqrt(a*b + b^2)*sin(x) - a - 2*b)/(b*cos(x)^2 + a)) + 2*(3*
(a*b^3 + b^4)*cos(x)^4 + 15*a^3*b + 5*a^2*b^2 - 2*a*b^3 + 8*b^4 - (5*a^2*b^2 + a*b^3 - 4*b^4)*cos(x)^2)*sin(x)
)/(a*b^4 + b^5), 1/15*(15*sqrt(-a*b - b^2)*a^3*arctan(sqrt(-a*b - b^2)*sin(x)/(a + b)) + (3*(a*b^3 + b^4)*cos(
x)^4 + 15*a^3*b + 5*a^2*b^2 - 2*a*b^3 + 8*b^4 - (5*a^2*b^2 + a*b^3 - 4*b^4)*cos(x)^2)*sin(x))/(a*b^4 + b^5)]

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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)**7/(a+b*cos(x)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.16356, size = 130, normalized size = 1.67 \begin{align*} \frac{a^{3} \arctan \left (\frac{b \sin \left (x\right )}{\sqrt{-a b - b^{2}}}\right )}{\sqrt{-a b - b^{2}} b^{3}} + \frac{3 \, b^{4} \sin \left (x\right )^{5} + 5 \, a b^{3} \sin \left (x\right )^{3} - 10 \, b^{4} \sin \left (x\right )^{3} + 15 \, a^{2} b^{2} \sin \left (x\right ) - 15 \, a b^{3} \sin \left (x\right ) + 15 \, b^{4} \sin \left (x\right )}{15 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^7/(a+b*cos(x)^2),x, algorithm="giac")

[Out]

a^3*arctan(b*sin(x)/sqrt(-a*b - b^2))/(sqrt(-a*b - b^2)*b^3) + 1/15*(3*b^4*sin(x)^5 + 5*a*b^3*sin(x)^3 - 10*b^
4*sin(x)^3 + 15*a^2*b^2*sin(x) - 15*a*b^3*sin(x) + 15*b^4*sin(x))/b^5