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

Optimal. Leaf size=110 \[ -\frac{a x \left (2 a^2+b^2\right )}{2 b^4}-\frac{\left (3 a^2+2 b^2\right ) \cos (x)}{3 b^3}+\frac{2 a^4 \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{b^4 \sqrt{a^2-b^2}}+\frac{a \sin (x) \cos (x)}{2 b^2}-\frac{\sin ^2(x) \cos (x)}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.277666, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.538, Rules used = {2793, 3049, 3023, 2735, 2660, 618, 204} \[ -\frac{a x \left (2 a^2+b^2\right )}{2 b^4}-\frac{\left (3 a^2+2 b^2\right ) \cos (x)}{3 b^3}+\frac{2 a^4 \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{b^4 \sqrt{a^2-b^2}}+\frac{a \sin (x) \cos (x)}{2 b^2}-\frac{\sin ^2(x) \cos (x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^4/(a + b*Sin[x]),x]

[Out]

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

Rule 2793

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d
*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a*d
*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n -
2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] ||
(EqQ[a, 0] && NeQ[c, 0])))

Rule 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

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 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sin ^4(x)}{a+b \sin (x)} \, dx &=-\frac{\cos (x) \sin ^2(x)}{3 b}+\frac{\int \frac{\sin (x) \left (2 a+2 b \sin (x)-3 a \sin ^2(x)\right )}{a+b \sin (x)} \, dx}{3 b}\\ &=\frac{a \cos (x) \sin (x)}{2 b^2}-\frac{\cos (x) \sin ^2(x)}{3 b}+\frac{\int \frac{-3 a^2+a b \sin (x)+2 \left (3 a^2+2 b^2\right ) \sin ^2(x)}{a+b \sin (x)} \, dx}{6 b^2}\\ &=-\frac{\left (3 a^2+2 b^2\right ) \cos (x)}{3 b^3}+\frac{a \cos (x) \sin (x)}{2 b^2}-\frac{\cos (x) \sin ^2(x)}{3 b}+\frac{\int \frac{-3 a^2 b-3 a \left (2 a^2+b^2\right ) \sin (x)}{a+b \sin (x)} \, dx}{6 b^3}\\ &=-\frac{a \left (2 a^2+b^2\right ) x}{2 b^4}-\frac{\left (3 a^2+2 b^2\right ) \cos (x)}{3 b^3}+\frac{a \cos (x) \sin (x)}{2 b^2}-\frac{\cos (x) \sin ^2(x)}{3 b}+\frac{a^4 \int \frac{1}{a+b \sin (x)} \, dx}{b^4}\\ &=-\frac{a \left (2 a^2+b^2\right ) x}{2 b^4}-\frac{\left (3 a^2+2 b^2\right ) \cos (x)}{3 b^3}+\frac{a \cos (x) \sin (x)}{2 b^2}-\frac{\cos (x) \sin ^2(x)}{3 b}+\frac{\left (2 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{b^4}\\ &=-\frac{a \left (2 a^2+b^2\right ) x}{2 b^4}-\frac{\left (3 a^2+2 b^2\right ) \cos (x)}{3 b^3}+\frac{a \cos (x) \sin (x)}{2 b^2}-\frac{\cos (x) \sin ^2(x)}{3 b}-\frac{\left (4 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{x}{2}\right )\right )}{b^4}\\ &=-\frac{a \left (2 a^2+b^2\right ) x}{2 b^4}+\frac{2 a^4 \tan ^{-1}\left (\frac{b+a \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{b^4 \sqrt{a^2-b^2}}-\frac{\left (3 a^2+2 b^2\right ) \cos (x)}{3 b^3}+\frac{a \cos (x) \sin (x)}{2 b^2}-\frac{\cos (x) \sin ^2(x)}{3 b}\\ \end{align*}

Mathematica [A]  time = 0.255606, size = 98, normalized size = 0.89 \[ \frac{-6 a x \left (2 a^2+b^2\right )-3 b \left (4 a^2+3 b^2\right ) \cos (x)+\frac{24 a^4 \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+3 a b^2 \sin (2 x)+b^3 \cos (3 x)}{12 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^4/(a + b*Sin[x]),x]

[Out]

(-6*a*(2*a^2 + b^2)*x + (24*a^4*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] - 3*b*(4*a^2 + 3*b^2
)*Cos[x] + b^3*Cos[3*x] + 3*a*b^2*Sin[2*x])/(12*b^4)

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Maple [B]  time = 0.039, size = 213, normalized size = 1.9 \begin{align*} -{\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{{a}^{2} \left ( \tan \left ( x/2 \right ) \right ) ^{4}}{{b}^{3} \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}-4\,{\frac{ \left ( \tan \left ( x/2 \right ) \right ) ^{2}{a}^{2}}{{b}^{3} \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}-4\,{\frac{ \left ( \tan \left ( x/2 \right ) \right ) ^{2}}{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{{a}^{2}}{{b}^{3} \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}-{\frac{4}{3\,b} \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}}}-{\frac{a}{{b}^{2}}\arctan \left ( \tan \left ({\frac{x}{2}} \right ) \right ) }+2\,{\frac{{a}^{4}}{{b}^{4}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( x/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.85976, size = 728, normalized size = 6.62 \begin{align*} \left [-\frac{3 \, \sqrt{-a^{2} + b^{2}} a^{4} \log \left (\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} + 2 \,{\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) - 2 \,{\left (a^{2} b^{3} - b^{5}\right )} \cos \left (x\right )^{3} - 3 \,{\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (x\right ) \sin \left (x\right ) + 3 \,{\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} x + 6 \,{\left (a^{4} b - b^{5}\right )} \cos \left (x\right )}{6 \,{\left (a^{2} b^{4} - b^{6}\right )}}, -\frac{6 \, \sqrt{a^{2} - b^{2}} a^{4} \arctan \left (-\frac{a \sin \left (x\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (x\right )}\right ) - 2 \,{\left (a^{2} b^{3} - b^{5}\right )} \cos \left (x\right )^{3} - 3 \,{\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (x\right ) \sin \left (x\right ) + 3 \,{\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} x + 6 \,{\left (a^{4} b - b^{5}\right )} \cos \left (x\right )}{6 \,{\left (a^{2} b^{4} - b^{6}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(3*sqrt(-a^2 + b^2)*a^4*log(((2*a^2 - b^2)*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2 + 2*(a*cos(x)*sin(x) + b*
cos(x))*sqrt(-a^2 + b^2))/(b^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) - 2*(a^2*b^3 - b^5)*cos(x)^3 - 3*(a^3*b^2
- a*b^4)*cos(x)*sin(x) + 3*(2*a^5 - a^3*b^2 - a*b^4)*x + 6*(a^4*b - b^5)*cos(x))/(a^2*b^4 - b^6), -1/6*(6*sqr
t(a^2 - b^2)*a^4*arctan(-(a*sin(x) + b)/(sqrt(a^2 - b^2)*cos(x))) - 2*(a^2*b^3 - b^5)*cos(x)^3 - 3*(a^3*b^2 -
a*b^4)*cos(x)*sin(x) + 3*(2*a^5 - a^3*b^2 - a*b^4)*x + 6*(a^4*b - b^5)*cos(x))/(a^2*b^4 - b^6)]

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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*sin(x)),x)

[Out]

Timed out

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Giac [A]  time = 1.56705, size = 201, normalized size = 1.83 \begin{align*} \frac{2 \,{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, x\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )} a^{4}}{\sqrt{a^{2} - b^{2}} b^{4}} - \frac{{\left (2 \, a^{3} + a b^{2}\right )} x}{2 \, b^{4}} - \frac{3 \, a b \tan \left (\frac{1}{2} \, x\right )^{5} + 6 \, a^{2} \tan \left (\frac{1}{2} \, x\right )^{4} + 12 \, a^{2} \tan \left (\frac{1}{2} \, x\right )^{2} + 12 \, b^{2} \tan \left (\frac{1}{2} \, x\right )^{2} - 3 \, a b \tan \left (\frac{1}{2} \, x\right ) + 6 \, a^{2} + 4 \, b^{2}}{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*sin(x)),x, algorithm="giac")

[Out]

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