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

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

[Out]

(a^2-a*b+b^2)*sin(x)/b^3+1/3*(a-2*b)*sin(x)^3/b^2+1/5*sin(x)^5/b-a^3*arctanh(sin(x)*b^(1/2)/(a+b)^(1/2))/b^(7/
2)/(a+b)^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.42, size = 111, normalized size = 1.42 \[ \frac {a^3 \left (\log \left (\sqrt {a+b}-\sqrt {b} \sin (x)\right )-\log \left (\sqrt {a+b}+\sqrt {b} \sin (x)\right )\right )}{2 b^{7/2} \sqrt {a+b}}+\frac {\left (8 a^2-6 a b+5 b^2\right ) \sin (x)}{8 b^3}+\frac {(5 b-4 a) \sin (3 x)}{48 b^2}+\frac {\sin (5 x)}{80 b} \]

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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fricas [A]  time = 2.13, size = 259, normalized size = 3.32 \[ \left [\frac {15 \, \sqrt {a b + b^{2}} a^{3} \log \left (-\frac {b \cos \relax (x)^{2} + 2 \, \sqrt {a b + b^{2}} \sin \relax (x) - a - 2 \, b}{b \cos \relax (x)^{2} + a}\right ) + 2 \, {\left (3 \, {\left (a b^{3} + b^{4}\right )} \cos \relax (x)^{4} + 15 \, a^{3} b + 5 \, a^{2} b^{2} - 2 \, a b^{3} + 8 \, b^{4} - {\left (5 \, a^{2} b^{2} + a b^{3} - 4 \, b^{4}\right )} \cos \relax (x)^{2}\right )} \sin \relax (x)}{30 \, {\left (a b^{4} + b^{5}\right )}}, \frac {15 \, \sqrt {-a b - b^{2}} a^{3} \arctan \left (\frac {\sqrt {-a b - b^{2}} \sin \relax (x)}{a + b}\right ) + {\left (3 \, {\left (a b^{3} + b^{4}\right )} \cos \relax (x)^{4} + 15 \, a^{3} b + 5 \, a^{2} b^{2} - 2 \, a b^{3} + 8 \, b^{4} - {\left (5 \, a^{2} b^{2} + a b^{3} - 4 \, b^{4}\right )} \cos \relax (x)^{2}\right )} \sin \relax (x)}{15 \, {\left (a b^{4} + b^{5}\right )}}\right ] \]

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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giac [A]  time = 0.18, size = 96, normalized size = 1.23 \[ \frac {a^{3} \arctan \left (\frac {b \sin \relax (x)}{\sqrt {-a b - b^{2}}}\right )}{\sqrt {-a b - b^{2}} b^{3}} + \frac {3 \, b^{4} \sin \relax (x)^{5} + 5 \, a b^{3} \sin \relax (x)^{3} - 10 \, b^{4} \sin \relax (x)^{3} + 15 \, a^{2} b^{2} \sin \relax (x) - 15 \, a b^{3} \sin \relax (x) + 15 \, b^{4} \sin \relax (x)}{15 \, b^{5}} \]

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

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maple [A]  time = 0.06, size = 78, normalized size = 1.00 \[ \frac {\frac {\left (\sin ^{5}\relax (x )\right ) b^{2}}{5}+\frac {\left (\sin ^{3}\relax (x )\right ) a b}{3}-\frac {2 \left (\sin ^{3}\relax (x )\right ) b^{2}}{3}+a^{2} \sin \relax (x )-a b \sin \relax (x )+b^{2} \sin \relax (x )}{b^{3}}-\frac {a^{3} \arctanh \left (\frac {\sin \relax (x ) b}{\sqrt {\left (a +b \right ) b}}\right )}{b^{3} \sqrt {\left (a +b \right ) b}} \]

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

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maxima [A]  time = 1.08, size = 91, normalized size = 1.17 \[ \frac {a^{3} \log \left (\frac {b \sin \relax (x) - \sqrt {{\left (a + b\right )} b}}{b \sin \relax (x) + \sqrt {{\left (a + b\right )} b}}\right )}{2 \, \sqrt {{\left (a + b\right )} b} b^{3}} + \frac {3 \, b^{2} \sin \relax (x)^{5} + 5 \, {\left (a b - 2 \, b^{2}\right )} \sin \relax (x)^{3} + 15 \, {\left (a^{2} - a b + b^{2}\right )} \sin \relax (x)}{15 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.13, size = 86, normalized size = 1.10 \[ \frac {{\sin \relax (x)}^5}{5\,b}+{\sin \relax (x)}^3\,\left (\frac {a+b}{3\,b^2}-\frac {1}{b}\right )+\sin \relax (x)\,\left (\frac {3}{b}+\frac {\left (a+b\right )\,\left (\frac {a+b}{b^2}-\frac {3}{b}\right )}{b}\right )+\frac {a^3\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sin \relax (x)\,1{}\mathrm {i}}{\sqrt {a+b}}\right )\,1{}\mathrm {i}}{b^{7/2}\,\sqrt {a+b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^7/(a + b*cos(x)^2),x)

[Out]

sin(x)^5/(5*b) + sin(x)^3*((a + b)/(3*b^2) - 1/b) + sin(x)*(3/b + ((a + b)*((a + b)/b^2 - 3/b))/b) + (a^3*atan
((b^(1/2)*sin(x)*1i)/(a + b)^(1/2))*1i)/(b^(7/2)*(a + b)^(1/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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