3.1.11 \(\int \frac {\sin ^5(x)}{a+b \cos ^2(x)} \, dx\) [11]

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

[Out]

(a+2*b)*cos(x)/b^2-1/3*cos(x)^3/b-(a+b)^2*arctan(cos(x)*b^(1/2)/a^(1/2))/b^(5/2)/a^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 54, 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 = {3269, 398, 211} \begin {gather*} -\frac {(a+b)^2 \text {ArcTan}\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 {gather*}

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 211

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

Rule 398

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 3269

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]

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(116\) vs. \(2(54)=108\).
time = 0.19, size = 116, normalized size = 2.15 \begin {gather*} \frac {-\frac {12 (a+b)^2 \text {ArcTan}\left (\frac {\sqrt {b}-\sqrt {a+b} \tan \left (\frac {x}{2}\right )}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {12 (a+b)^2 \text {ArcTan}\left (\frac {\sqrt {b}+\sqrt {a+b} \tan \left (\frac {x}{2}\right )}{\sqrt {a}}\right )}{\sqrt {a}}+3 \sqrt {b} (4 a+7 b) \cos (x)-b^{3/2} \cos (3 x)}{12 b^{5/2}} \end {gather*}

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 = 0.16, size = 57, normalized size = 1.06

method	result	size
default	\(\frac {-\frac {b \left (\cos ^{3}\left (x \right )\right )}{3}+a \cos \left (x \right )+2 b \cos \left (x \right )}{b^{2}}+\frac {\left (-a^{2}-2 a b -b^{2}\right ) \arctan \left (\frac {b \cos \left (x \right )}{\sqrt {a b}}\right )}{b^{2} \sqrt {a b}}\)	\(57\)
risch	\(\frac {a \,{\mathrm e}^{i x}}{2 b^{2}}+\frac {7 \,{\mathrm e}^{i x}}{8 b}+\frac {{\mathrm e}^{-i x} a}{2 b^{2}}+\frac {7 \,{\mathrm e}^{-i x}}{8 b}-\frac {i \ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {a b}}+1\right ) a^{2}}{2 \sqrt {a b}\, b^{2}}-\frac {i \ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {a b}}+1\right ) a}{\sqrt {a b}\, b}-\frac {i \ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {a b}}+1\right )}{2 \sqrt {a b}}+\frac {i \ln \left ({\mathrm e}^{2 i x}-\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {a b}}+1\right ) a^{2}}{2 \sqrt {a b}\, b^{2}}+\frac {i \ln \left ({\mathrm e}^{2 i x}-\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {a b}}+1\right ) a}{\sqrt {a b}\, b}+\frac {i \ln \left ({\mathrm e}^{2 i x}-\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {a b}}+1\right )}{2 \sqrt {a b}}-\frac {\cos \left (3 x \right )}{12 b}\)	\(253\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 53, normalized size = 0.98 \begin {gather*} -\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 \cos \left (x\right )^{3} - 3 \, {\left (a + 2 \, b\right )} \cos \left (x\right )}{3 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.43, size = 152, normalized size = 2.81 \begin {gather*} \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 {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.41, size = 59, normalized size = 1.09 \begin {gather*} -\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 {gather*}

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

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Mupad [B]
time = 0.10, size = 65, normalized size = 1.20 \begin {gather*} \cos \left (x\right )\,\left (\frac {a}{b^2}+\frac {2}{b}\right )-\frac {{\cos \left (x\right )}^3}{3\,b}-\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\cos \left (x\right )\,{\left (a+b\right )}^2}{\sqrt {a}\,\left (a^2+2\,a\,b+b^2\right )}\right )\,{\left (a+b\right )}^2}{\sqrt {a}\,b^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^5/(a + b*cos(x)^2),x)

[Out]

cos(x)*(a/b^2 + 2/b) - cos(x)^3/(3*b) - (atan((b^(1/2)*cos(x)*(a + b)^2)/(a^(1/2)*(2*a*b + a^2 + b^2)))*(a + b
)^2)/(a^(1/2)*b^(5/2))

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