3.24 \(\int \frac{\sin ^4(x)}{a+b \cos (x)} \, dx\)

Optimal. Leaf size=104 \[ -\frac{a x \left (2 a^2-3 b^2\right )}{2 b^4}+\frac{\sin (x) \left (2 \left (a^2-b^2\right )-a b \cos (x)\right )}{2 b^3}+\frac{2 (a-b)^{3/2} (a+b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{b^4}-\frac{\sin ^3(x)}{3 b} \]

[Out]

-(a*(2*a^2 - 3*b^2)*x)/(2*b^4) + (2*(a - b)^(3/2)*(a + b)^(3/2)*ArcTan[(Sqrt[a - b]*Tan[x/2])/Sqrt[a + b]])/b^
4 + ((2*(a^2 - b^2) - a*b*Cos[x])*Sin[x])/(2*b^3) - Sin[x]^3/(3*b)

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Rubi [A]  time = 0.256214, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {2695, 2865, 2735, 2659, 205} \[ -\frac{a x \left (2 a^2-3 b^2\right )}{2 b^4}+\frac{\sin (x) \left (2 \left (a^2-b^2\right )-a b \cos (x)\right )}{2 b^3}+\frac{2 (a-b)^{3/2} (a+b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{b^4}-\frac{\sin ^3(x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^4/(a + b*Cos[x]),x]

[Out]

-(a*(2*a^2 - 3*b^2)*x)/(2*b^4) + (2*(a - b)^(3/2)*(a + b)^(3/2)*ArcTan[(Sqrt[a - b]*Tan[x/2])/Sqrt[a + b]])/b^
4 + ((2*(a^2 - b^2) - a*b*Cos[x])*Sin[x])/(2*b^3) - Sin[x]^3/(3*b)

Rule 2695

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + p)), x] + Dist[(g^2*(p - 1))/(b*(m + p)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*(b + a*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] &&
NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && IntegersQ[2*m, 2*p]

Rule 2865

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c*(m + p + 1) -
a*d*p + b*d*(m + p)*Sin[e + f*x]))/(b^2*f*(m + p)*(m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(m + p)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1
) - d*(a^2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2,
0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*m]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

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 ^4(x)}{a+b \cos (x)} \, dx &=-\frac{\sin ^3(x)}{3 b}-\frac{\int \frac{(-b-a \cos (x)) \sin ^2(x)}{a+b \cos (x)} \, dx}{b}\\ &=\frac{\left (2 \left (a^2-b^2\right )-a b \cos (x)\right ) \sin (x)}{2 b^3}-\frac{\sin ^3(x)}{3 b}-\frac{\int \frac{b \left (a^2-2 b^2\right )+a \left (2 a^2-3 b^2\right ) \cos (x)}{a+b \cos (x)} \, dx}{2 b^3}\\ &=-\frac{a \left (2 a^2-3 b^2\right ) x}{2 b^4}+\frac{\left (2 \left (a^2-b^2\right )-a b \cos (x)\right ) \sin (x)}{2 b^3}-\frac{\sin ^3(x)}{3 b}+\frac{\left (a^2-b^2\right )^2 \int \frac{1}{a+b \cos (x)} \, dx}{b^4}\\ &=-\frac{a \left (2 a^2-3 b^2\right ) x}{2 b^4}+\frac{\left (2 \left (a^2-b^2\right )-a b \cos (x)\right ) \sin (x)}{2 b^3}-\frac{\sin ^3(x)}{3 b}+\frac{\left (2 \left (a^2-b^2\right )^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{b^4}\\ &=-\frac{a \left (2 a^2-3 b^2\right ) x}{2 b^4}+\frac{2 (a-b)^{3/2} (a+b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{b^4}+\frac{\left (2 \left (a^2-b^2\right )-a b \cos (x)\right ) \sin (x)}{2 b^3}-\frac{\sin ^3(x)}{3 b}\\ \end{align*}

Mathematica [A]  time = 0.205923, size = 96, normalized size = 0.92 \[ \frac{3 b \left (4 a^2-5 b^2\right ) \sin (x)-24 \left (b^2-a^2\right )^{3/2} \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )-12 a^3 x+18 a b^2 x-3 a b^2 \sin (2 x)+b^3 \sin (3 x)}{12 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^4/(a + b*Cos[x]),x]

[Out]

(-12*a^3*x + 18*a*b^2*x - 24*(-a^2 + b^2)^(3/2)*ArcTanh[((a - b)*Tan[x/2])/Sqrt[-a^2 + b^2]] + 3*b*(4*a^2 - 5*
b^2)*Sin[x] - 3*a*b^2*Sin[2*x] + b^3*Sin[3*x])/(12*b^4)

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Maple [B]  time = 0.093, size = 315, normalized size = 3. \begin{align*} 2\,{\frac{ \left ( \tan \left ( x/2 \right ) \right ) ^{5}{a}^{2}}{{b}^{3} \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+{\frac{a}{{b}^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{5} \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-3}}-2\,{\frac{ \left ( \tan \left ( x/2 \right ) \right ) ^{5}}{b \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+4\,{\frac{ \left ( \tan \left ( x/2 \right ) \right ) ^{3}{a}^{2}}{{b}^{3} \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}-{\frac{20}{3\,b} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{3} \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-3}}+2\,{\frac{\tan \left ( x/2 \right ){a}^{2}}{{b}^{3} \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}-2\,{\frac{\tan \left ( x/2 \right ) }{b \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}-{\frac{a}{{b}^{2}}\tan \left ({\frac{x}{2}} \right ) \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-3}}-2\,{\frac{\arctan \left ( \tan \left ( x/2 \right ) \right ){a}^{3}}{{b}^{4}}}+3\,{\frac{\arctan \left ( \tan \left ( x/2 \right ) \right ) a}{{b}^{2}}}+2\,{\frac{{a}^{4}}{{b}^{4}\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( x/2 \right ) \left ( a-b \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) }-4\,{\frac{{a}^{2}}{{b}^{2}\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( x/2 \right ) \left ( a-b \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) }+2\,{\frac{1}{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( x/2 \right ) \left ( a-b \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^4/(a+b*cos(x)),x)

[Out]

2/b^3/(tan(1/2*x)^2+1)^3*tan(1/2*x)^5*a^2+1/b^2/(tan(1/2*x)^2+1)^3*tan(1/2*x)^5*a-2/b/(tan(1/2*x)^2+1)^3*tan(1
/2*x)^5+4/b^3/(tan(1/2*x)^2+1)^3*tan(1/2*x)^3*a^2-20/3/b/(tan(1/2*x)^2+1)^3*tan(1/2*x)^3+2/b^3/(tan(1/2*x)^2+1
)^3*tan(1/2*x)*a^2-2/b/(tan(1/2*x)^2+1)^3*tan(1/2*x)-1/b^2/(tan(1/2*x)^2+1)^3*tan(1/2*x)*a-2/b^4*arctan(tan(1/
2*x))*a^3+3/b^2*arctan(tan(1/2*x))*a+2/b^4/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*x)*(a-b)/((a-b)*(a+b))^(1/2))*a^
4-4/b^2/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*x)*(a-b)/((a-b)*(a+b))^(1/2))*a^2+2/((a-b)*(a+b))^(1/2)*arctan(tan(
1/2*x)*(a-b)/((a-b)*(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)^4/(a+b*cos(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.86173, size = 587, normalized size = 5.64 \begin{align*} \left [-\frac{3 \,{\left (a^{2} - b^{2}\right )} \sqrt{-a^{2} + b^{2}} \log \left (\frac{2 \, a b \cos \left (x\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (x\right ) + b\right )} \sin \left (x\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}}\right ) + 3 \,{\left (2 \, a^{3} - 3 \, a b^{2}\right )} x -{\left (2 \, b^{3} \cos \left (x\right )^{2} - 3 \, a b^{2} \cos \left (x\right ) + 6 \, a^{2} b - 8 \, b^{3}\right )} \sin \left (x\right )}{6 \, b^{4}}, \frac{6 \,{\left (a^{2} - b^{2}\right )}^{\frac{3}{2}} \arctan \left (-\frac{a \cos \left (x\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (x\right )}\right ) - 3 \,{\left (2 \, a^{3} - 3 \, a b^{2}\right )} x +{\left (2 \, b^{3} \cos \left (x\right )^{2} - 3 \, a b^{2} \cos \left (x\right ) + 6 \, a^{2} b - 8 \, b^{3}\right )} \sin \left (x\right )}{6 \, b^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^4/(a+b*cos(x)),x, algorithm="fricas")

[Out]

[-1/6*(3*(a^2 - b^2)*sqrt(-a^2 + b^2)*log((2*a*b*cos(x) + (2*a^2 - b^2)*cos(x)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(x
) + b)*sin(x) - a^2 + 2*b^2)/(b^2*cos(x)^2 + 2*a*b*cos(x) + a^2)) + 3*(2*a^3 - 3*a*b^2)*x - (2*b^3*cos(x)^2 -
3*a*b^2*cos(x) + 6*a^2*b - 8*b^3)*sin(x))/b^4, 1/6*(6*(a^2 - b^2)^(3/2)*arctan(-(a*cos(x) + b)/(sqrt(a^2 - b^2
)*sin(x))) - 3*(2*a^3 - 3*a*b^2)*x + (2*b^3*cos(x)^2 - 3*a*b^2*cos(x) + 6*a^2*b - 8*b^3)*sin(x))/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(sin(x)**4/(a+b*cos(x)),x)

[Out]

Timed out

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Giac [B]  time = 1.18432, size = 262, normalized size = 2.52 \begin{align*} -\frac{{\left (2 \, a^{3} - 3 \, a b^{2}\right )} x}{2 \, b^{4}} - \frac{2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, x\right ) - b \tan \left (\frac{1}{2} \, x\right )}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{\sqrt{a^{2} - b^{2}} b^{4}} + \frac{6 \, a^{2} \tan \left (\frac{1}{2} \, x\right )^{5} + 3 \, a b \tan \left (\frac{1}{2} \, x\right )^{5} - 6 \, b^{2} \tan \left (\frac{1}{2} \, x\right )^{5} + 12 \, a^{2} \tan \left (\frac{1}{2} \, x\right )^{3} - 20 \, b^{2} \tan \left (\frac{1}{2} \, x\right )^{3} + 6 \, a^{2} \tan \left (\frac{1}{2} \, x\right ) - 3 \, a b \tan \left (\frac{1}{2} \, x\right ) - 6 \, b^{2} \tan \left (\frac{1}{2} \, x\right )}{3 \,{\left (\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right )}^{3} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^4/(a+b*cos(x)),x, algorithm="giac")

[Out]

-1/2*(2*a^3 - 3*a*b^2)*x/b^4 - 2*(a^4 - 2*a^2*b^2 + b^4)*(pi*floor(1/2*x/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(
a*tan(1/2*x) - b*tan(1/2*x))/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*b^4) + 1/3*(6*a^2*tan(1/2*x)^5 + 3*a*b*tan(1/2
*x)^5 - 6*b^2*tan(1/2*x)^5 + 12*a^2*tan(1/2*x)^3 - 20*b^2*tan(1/2*x)^3 + 6*a^2*tan(1/2*x) - 3*a*b*tan(1/2*x) -
6*b^2*tan(1/2*x))/((tan(1/2*x)^2 + 1)^3*b^3)