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^3-9/5*cos(x)/a^3+1/5*cos(x)*sin(x)^3/(a+a*sin(x))^3+3/5*cos(x)*sin(x)^2/a/(a+a*sin(x))^2-3*cos(x)/(a^3+
a^3*sin(x))

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Rubi [A]  time = 0.22, antiderivative size = 71, normalized size of antiderivative = 1.00, 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 12

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

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]

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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.08, 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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fricas [B]  time = 0.49, size = 132, normalized size = 1.86 \[ -\frac {3 \, {\left (5 \, x + 13\right )} \cos \relax (x)^{3} + 5 \, \cos \relax (x)^{4} + {\left (45 \, x - 28\right )} \cos \relax (x)^{2} - 3 \, {\left (10 \, x + 21\right )} \cos \relax (x) + {\left ({\left (15 \, x - 34\right )} \cos \relax (x)^{2} + 5 \, \cos \relax (x)^{3} - 2 \, {\left (15 \, x + 31\right )} \cos \relax (x) - 60 \, x + 1\right )} \sin \relax (x) - 60 \, x - 1}{5 \, {\left (a^{3} \cos \relax (x)^{3} + 3 \, a^{3} \cos \relax (x)^{2} - 2 \, a^{3} \cos \relax (x) - 4 \, a^{3} + {\left (a^{3} \cos \relax (x)^{2} - 2 \, a^{3} \cos \relax (x) - 4 \, a^{3}\right )} \sin \relax (x)\right )}} \]

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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giac [A]  time = 0.88, size = 67, normalized size = 0.94 \[ -\frac {3 \, x}{a^{3}} - \frac {2}{{\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )} a^{3}} - \frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, x\right )^{4} + 70 \, \tan \left (\frac {1}{2} \, x\right )^{3} + 120 \, \tan \left (\frac {1}{2} \, x\right )^{2} + 80 \, \tan \left (\frac {1}{2} \, x\right ) + 19\right )}}{5 \, a^{3} {\left (\tan \left (\frac {1}{2} \, x\right ) + 1\right )}^{5}} \]

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)

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maple [A]  time = 0.10, size = 79, normalized size = 1.11 \[ -\frac {2}{a^{3} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )}-\frac {6 \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{a^{3}}-\frac {8}{5 a^{3} \left (\tan \left (\frac {x}{2}\right )+1\right )^{5}}+\frac {4}{a^{3} \left (\tan \left (\frac {x}{2}\right )+1\right )^{4}}-\frac {4}{a^{3} \left (\tan \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {6}{a^{3} \left (\tan \left (\frac {x}{2}\right )+1\right )} \]

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 = 0.93, size = 198, normalized size = 2.79 \[ -\frac {2 \, {\left (\frac {105 \, \sin \relax (x)}{\cos \relax (x) + 1} + \frac {189 \, \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {200 \, \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {160 \, \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + \frac {75 \, \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}} + \frac {15 \, \sin \relax (x)^{6}}{{\left (\cos \relax (x) + 1\right )}^{6}} + 24\right )}}{5 \, {\left (a^{3} + \frac {5 \, a^{3} \sin \relax (x)}{\cos \relax (x) + 1} + \frac {11 \, a^{3} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {15 \, a^{3} \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {15 \, a^{3} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + \frac {11 \, a^{3} \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}} + \frac {5 \, a^{3} \sin \relax (x)^{6}}{{\left (\cos \relax (x) + 1\right )}^{6}} + \frac {a^{3} \sin \relax (x)^{7}}{{\left (\cos \relax (x) + 1\right )}^{7}}\right )}} - \frac {6 \, \arctan \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )}{a^{3}} \]

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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mupad [B]  time = 6.89, size = 78, normalized size = 1.10 \[ -\frac {3\,x}{a^3}-\frac {6\,{\mathrm {tan}\left (\frac {x}{2}\right )}^6+30\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5+64\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+80\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+\frac {378\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{5}+42\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {48}{5}}{a^3\,\left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )\,{\left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^4/(a + a*sin(x))^3,x)

[Out]

- (3*x)/a^3 - (42*tan(x/2) + (378*tan(x/2)^2)/5 + 80*tan(x/2)^3 + 64*tan(x/2)^4 + 30*tan(x/2)^5 + 6*tan(x/2)^6
+ 48/5)/(a^3*(tan(x/2)^2 + 1)*(tan(x/2) + 1)^5)

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sympy [B]  time = 19.64, size = 1425, normalized size = 20.07 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)**4/(a+a*sin(x))**3,x)

[Out]

-15*x*tan(x/2)**7/(5*a**3*tan(x/2)**7 + 25*a**3*tan(x/2)**6 + 55*a**3*tan(x/2)**5 + 75*a**3*tan(x/2)**4 + 75*a
**3*tan(x/2)**3 + 55*a**3*tan(x/2)**2 + 25*a**3*tan(x/2) + 5*a**3) - 75*x*tan(x/2)**6/(5*a**3*tan(x/2)**7 + 25
*a**3*tan(x/2)**6 + 55*a**3*tan(x/2)**5 + 75*a**3*tan(x/2)**4 + 75*a**3*tan(x/2)**3 + 55*a**3*tan(x/2)**2 + 25
*a**3*tan(x/2) + 5*a**3) - 165*x*tan(x/2)**5/(5*a**3*tan(x/2)**7 + 25*a**3*tan(x/2)**6 + 55*a**3*tan(x/2)**5 +
75*a**3*tan(x/2)**4 + 75*a**3*tan(x/2)**3 + 55*a**3*tan(x/2)**2 + 25*a**3*tan(x/2) + 5*a**3) - 225*x*tan(x/2)
**4/(5*a**3*tan(x/2)**7 + 25*a**3*tan(x/2)**6 + 55*a**3*tan(x/2)**5 + 75*a**3*tan(x/2)**4 + 75*a**3*tan(x/2)**
3 + 55*a**3*tan(x/2)**2 + 25*a**3*tan(x/2) + 5*a**3) - 225*x*tan(x/2)**3/(5*a**3*tan(x/2)**7 + 25*a**3*tan(x/2
)**6 + 55*a**3*tan(x/2)**5 + 75*a**3*tan(x/2)**4 + 75*a**3*tan(x/2)**3 + 55*a**3*tan(x/2)**2 + 25*a**3*tan(x/2
) + 5*a**3) - 165*x*tan(x/2)**2/(5*a**3*tan(x/2)**7 + 25*a**3*tan(x/2)**6 + 55*a**3*tan(x/2)**5 + 75*a**3*tan(
x/2)**4 + 75*a**3*tan(x/2)**3 + 55*a**3*tan(x/2)**2 + 25*a**3*tan(x/2) + 5*a**3) - 75*x*tan(x/2)/(5*a**3*tan(x
/2)**7 + 25*a**3*tan(x/2)**6 + 55*a**3*tan(x/2)**5 + 75*a**3*tan(x/2)**4 + 75*a**3*tan(x/2)**3 + 55*a**3*tan(x
/2)**2 + 25*a**3*tan(x/2) + 5*a**3) - 15*x/(5*a**3*tan(x/2)**7 + 25*a**3*tan(x/2)**6 + 55*a**3*tan(x/2)**5 + 7
5*a**3*tan(x/2)**4 + 75*a**3*tan(x/2)**3 + 55*a**3*tan(x/2)**2 + 25*a**3*tan(x/2) + 5*a**3) - 30*tan(x/2)**6/(
5*a**3*tan(x/2)**7 + 25*a**3*tan(x/2)**6 + 55*a**3*tan(x/2)**5 + 75*a**3*tan(x/2)**4 + 75*a**3*tan(x/2)**3 + 5
5*a**3*tan(x/2)**2 + 25*a**3*tan(x/2) + 5*a**3) - 150*tan(x/2)**5/(5*a**3*tan(x/2)**7 + 25*a**3*tan(x/2)**6 +
55*a**3*tan(x/2)**5 + 75*a**3*tan(x/2)**4 + 75*a**3*tan(x/2)**3 + 55*a**3*tan(x/2)**2 + 25*a**3*tan(x/2) + 5*a
**3) - 320*tan(x/2)**4/(5*a**3*tan(x/2)**7 + 25*a**3*tan(x/2)**6 + 55*a**3*tan(x/2)**5 + 75*a**3*tan(x/2)**4 +
75*a**3*tan(x/2)**3 + 55*a**3*tan(x/2)**2 + 25*a**3*tan(x/2) + 5*a**3) - 400*tan(x/2)**3/(5*a**3*tan(x/2)**7
+ 25*a**3*tan(x/2)**6 + 55*a**3*tan(x/2)**5 + 75*a**3*tan(x/2)**4 + 75*a**3*tan(x/2)**3 + 55*a**3*tan(x/2)**2
+ 25*a**3*tan(x/2) + 5*a**3) - 378*tan(x/2)**2/(5*a**3*tan(x/2)**7 + 25*a**3*tan(x/2)**6 + 55*a**3*tan(x/2)**5
+ 75*a**3*tan(x/2)**4 + 75*a**3*tan(x/2)**3 + 55*a**3*tan(x/2)**2 + 25*a**3*tan(x/2) + 5*a**3) - 210*tan(x/2)
/(5*a**3*tan(x/2)**7 + 25*a**3*tan(x/2)**6 + 55*a**3*tan(x/2)**5 + 75*a**3*tan(x/2)**4 + 75*a**3*tan(x/2)**3 +
55*a**3*tan(x/2)**2 + 25*a**3*tan(x/2) + 5*a**3) - 48/(5*a**3*tan(x/2)**7 + 25*a**3*tan(x/2)**6 + 55*a**3*tan
(x/2)**5 + 75*a**3*tan(x/2)**4 + 75*a**3*tan(x/2)**3 + 55*a**3*tan(x/2)**2 + 25*a**3*tan(x/2) + 5*a**3)

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