3.23 \(\int \frac{\sin ^4(x)}{(a+a \sin (x))^3} \, dx\)

Optimal. Leaf size=71 \[ -\frac{3 x}{a^3}-\frac{9 \cos (x)}{5 a^3}-\frac{3 \cos (x)}{a^3 \sin (x)+a^3}+\frac{\sin ^3(x) \cos (x)}{5 (a \sin (x)+a)^3}+\frac{3 \sin ^2(x) \cos (x)}{5 a (a \sin (x)+a)^2} \]

[Out]

(-3*x)/a^3 - (9*Cos[x])/(5*a^3) + (Cos[x]*Sin[x]^3)/(5*(a + a*Sin[x])^3) + (3*Cos[x]*Sin[x]^2)/(5*a*(a + a*Sin
[x])^2) - (3*Cos[x])/(a^3 + a^3*Sin[x])

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Rubi [A]  time = 0.221008, antiderivative size = 71, 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 = {2765, 2977, 2968, 3023, 12, 2735, 2648} \[ -\frac{3 x}{a^3}-\frac{9 \cos (x)}{5 a^3}-\frac{3 \cos (x)}{a^3 \sin (x)+a^3}+\frac{\sin ^3(x) \cos (x)}{5 (a \sin (x)+a)^3}+\frac{3 \sin ^2(x) \cos (x)}{5 a (a \sin (x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^4/(a + a*Sin[x])^3,x]

[Out]

(-3*x)/a^3 - (9*Cos[x])/(5*a^3) + (Cos[x]*Sin[x]^3)/(5*(a + a*Sin[x])^3) + (3*Cos[x]*Sin[x]^2)/(5*a*(a + a*Sin
[x])^2) - (3*Cos[x])/(a^3 + a^3*Sin[x])

Rule 2765

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

Rule 2977

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

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{\sin ^4(x)}{(a+a \sin (x))^3} \, dx &=\frac{\cos (x) \sin ^3(x)}{5 (a+a \sin (x))^3}-\frac{\int \frac{\sin ^2(x) (3 a-6 a \sin (x))}{(a+a \sin (x))^2} \, dx}{5 a^2}\\ &=\frac{\cos (x) \sin ^3(x)}{5 (a+a \sin (x))^3}+\frac{3 \cos (x) \sin ^2(x)}{5 a (a+a \sin (x))^2}-\frac{\int \frac{\sin (x) \left (18 a^2-27 a^2 \sin (x)\right )}{a+a \sin (x)} \, dx}{15 a^4}\\ &=\frac{\cos (x) \sin ^3(x)}{5 (a+a \sin (x))^3}+\frac{3 \cos (x) \sin ^2(x)}{5 a (a+a \sin (x))^2}-\frac{\int \frac{18 a^2 \sin (x)-27 a^2 \sin ^2(x)}{a+a \sin (x)} \, dx}{15 a^4}\\ &=-\frac{9 \cos (x)}{5 a^3}+\frac{\cos (x) \sin ^3(x)}{5 (a+a \sin (x))^3}+\frac{3 \cos (x) \sin ^2(x)}{5 a (a+a \sin (x))^2}-\frac{\int \frac{45 a^3 \sin (x)}{a+a \sin (x)} \, dx}{15 a^5}\\ &=-\frac{9 \cos (x)}{5 a^3}+\frac{\cos (x) \sin ^3(x)}{5 (a+a \sin (x))^3}+\frac{3 \cos (x) \sin ^2(x)}{5 a (a+a \sin (x))^2}-\frac{3 \int \frac{\sin (x)}{a+a \sin (x)} \, dx}{a^2}\\ &=-\frac{3 x}{a^3}-\frac{9 \cos (x)}{5 a^3}+\frac{\cos (x) \sin ^3(x)}{5 (a+a \sin (x))^3}+\frac{3 \cos (x) \sin ^2(x)}{5 a (a+a \sin (x))^2}+\frac{3 \int \frac{1}{a+a \sin (x)} \, dx}{a^2}\\ &=-\frac{3 x}{a^3}-\frac{9 \cos (x)}{5 a^3}+\frac{\cos (x) \sin ^3(x)}{5 (a+a \sin (x))^3}+\frac{3 \cos (x) \sin ^2(x)}{5 a (a+a \sin (x))^2}-\frac{3 \cos (x)}{a^3+a^3 \sin (x)}\\ \end{align*}

Mathematica [A]  time = 0.0765933, size = 140, normalized size = 1.97 \[ \frac{\left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right ) \left (\sin \left (\frac{x}{2}\right )-\cos \left (\frac{x}{2}\right )-15 x \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )^5-5 \cos (x) \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )^5+48 \sin \left (\frac{x}{2}\right ) \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )^4+6 \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )^3-12 \sin \left (\frac{x}{2}\right ) \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )^2\right )}{5 (a \sin (x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^4/(a + a*Sin[x])^3,x]

[Out]

((Cos[x/2] + Sin[x/2])*(-Cos[x/2] + Sin[x/2] - 12*Sin[x/2]*(Cos[x/2] + Sin[x/2])^2 + 6*(Cos[x/2] + Sin[x/2])^3
+ 48*Sin[x/2]*(Cos[x/2] + Sin[x/2])^4 - 15*x*(Cos[x/2] + Sin[x/2])^5 - 5*Cos[x]*(Cos[x/2] + Sin[x/2])^5))/(5*
(a + a*Sin[x])^3)

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Maple [A]  time = 0.046, size = 79, normalized size = 1.1 \begin{align*} -2\,{\frac{1}{{a}^{3} \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}+1 \right ) }}-6\,{\frac{\arctan \left ( \tan \left ( x/2 \right ) \right ) }{{a}^{3}}}-{\frac{8}{5\,{a}^{3}} \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) ^{-5}}+4\,{\frac{1}{{a}^{3} \left ( \tan \left ( x/2 \right ) +1 \right ) ^{4}}}-4\,{\frac{1}{{a}^{3} \left ( \tan \left ( x/2 \right ) +1 \right ) ^{2}}}-6\,{\frac{1}{{a}^{3} \left ( \tan \left ( x/2 \right ) +1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^4/(a+a*sin(x))^3,x)

[Out]

-2/a^3/(tan(1/2*x)^2+1)-6/a^3*arctan(tan(1/2*x))-8/5/a^3/(tan(1/2*x)+1)^5+4/a^3/(tan(1/2*x)+1)^4-4/a^3/(tan(1/
2*x)+1)^2-6/a^3/(tan(1/2*x)+1)

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Maxima [B]  time = 2.08608, size = 267, normalized size = 3.76 \begin{align*} -\frac{2 \,{\left (\frac{105 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac{189 \, \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac{200 \, \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac{160 \, \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + \frac{75 \, \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}} + \frac{15 \, \sin \left (x\right )^{6}}{{\left (\cos \left (x\right ) + 1\right )}^{6}} + 24\right )}}{5 \,{\left (a^{3} + \frac{5 \, a^{3} \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac{11 \, a^{3} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac{15 \, a^{3} \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac{15 \, a^{3} \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + \frac{11 \, a^{3} \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}} + \frac{5 \, a^{3} \sin \left (x\right )^{6}}{{\left (\cos \left (x\right ) + 1\right )}^{6}} + \frac{a^{3} \sin \left (x\right )^{7}}{{\left (\cos \left (x\right ) + 1\right )}^{7}}\right )}} - \frac{6 \, \arctan \left (\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1}\right )}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^4/(a+a*sin(x))^3,x, algorithm="maxima")

[Out]

-2/5*(105*sin(x)/(cos(x) + 1) + 189*sin(x)^2/(cos(x) + 1)^2 + 200*sin(x)^3/(cos(x) + 1)^3 + 160*sin(x)^4/(cos(
x) + 1)^4 + 75*sin(x)^5/(cos(x) + 1)^5 + 15*sin(x)^6/(cos(x) + 1)^6 + 24)/(a^3 + 5*a^3*sin(x)/(cos(x) + 1) + 1
1*a^3*sin(x)^2/(cos(x) + 1)^2 + 15*a^3*sin(x)^3/(cos(x) + 1)^3 + 15*a^3*sin(x)^4/(cos(x) + 1)^4 + 11*a^3*sin(x
)^5/(cos(x) + 1)^5 + 5*a^3*sin(x)^6/(cos(x) + 1)^6 + a^3*sin(x)^7/(cos(x) + 1)^7) - 6*arctan(sin(x)/(cos(x) +
1))/a^3

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Fricas [B]  time = 1.39637, size = 379, normalized size = 5.34 \begin{align*} -\frac{3 \,{\left (5 \, x + 13\right )} \cos \left (x\right )^{3} + 5 \, \cos \left (x\right )^{4} +{\left (45 \, x - 28\right )} \cos \left (x\right )^{2} - 3 \,{\left (10 \, x + 21\right )} \cos \left (x\right ) +{\left ({\left (15 \, x - 34\right )} \cos \left (x\right )^{2} + 5 \, \cos \left (x\right )^{3} - 2 \,{\left (15 \, x + 31\right )} \cos \left (x\right ) - 60 \, x + 1\right )} \sin \left (x\right ) - 60 \, x - 1}{5 \,{\left (a^{3} \cos \left (x\right )^{3} + 3 \, a^{3} \cos \left (x\right )^{2} - 2 \, a^{3} \cos \left (x\right ) - 4 \, a^{3} +{\left (a^{3} \cos \left (x\right )^{2} - 2 \, a^{3} \cos \left (x\right ) - 4 \, a^{3}\right )} \sin \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^4/(a+a*sin(x))^3,x, algorithm="fricas")

[Out]

-1/5*(3*(5*x + 13)*cos(x)^3 + 5*cos(x)^4 + (45*x - 28)*cos(x)^2 - 3*(10*x + 21)*cos(x) + ((15*x - 34)*cos(x)^2
+ 5*cos(x)^3 - 2*(15*x + 31)*cos(x) - 60*x + 1)*sin(x) - 60*x - 1)/(a^3*cos(x)^3 + 3*a^3*cos(x)^2 - 2*a^3*cos
(x) - 4*a^3 + (a^3*cos(x)^2 - 2*a^3*cos(x) - 4*a^3)*sin(x))

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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+a*sin(x))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.31993, size = 90, normalized size = 1.27 \begin{align*} -\frac{3 \, x}{a^{3}} - \frac{2}{{\left (\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right )} a^{3}} - \frac{2 \,{\left (15 \, \tan \left (\frac{1}{2} \, x\right )^{4} + 70 \, \tan \left (\frac{1}{2} \, x\right )^{3} + 120 \, \tan \left (\frac{1}{2} \, x\right )^{2} + 80 \, \tan \left (\frac{1}{2} \, x\right ) + 19\right )}}{5 \, a^{3}{\left (\tan \left (\frac{1}{2} \, x\right ) + 1\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^4/(a+a*sin(x))^3,x, algorithm="giac")

[Out]

-3*x/a^3 - 2/((tan(1/2*x)^2 + 1)*a^3) - 2/5*(15*tan(1/2*x)^4 + 70*tan(1/2*x)^3 + 120*tan(1/2*x)^2 + 80*tan(1/2
*x) + 19)/(a^3*(tan(1/2*x) + 1)^5)