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

Optimal. Leaf size=53 \[ -\frac {3 x}{2 a}+\frac {4 \cos ^3(x)}{3 a}-\frac {4 \cos (x)}{a}+\frac {\sin ^3(x) \cos (x)}{a \sin (x)+a}+\frac {3 \sin (x) \cos (x)}{2 a} \]

[Out]

-3/2*x/a-4*cos(x)/a+4/3*cos(x)^3/a+3/2*cos(x)*sin(x)/a+cos(x)*sin(x)^3/(a+a*sin(x))

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Rubi [A]  time = 0.07, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2767, 2748, 2635, 8, 2633} \[ -\frac {3 x}{2 a}+\frac {4 \cos ^3(x)}{3 a}-\frac {4 \cos (x)}{a}+\frac {\sin ^3(x) \cos (x)}{a \sin (x)+a}+\frac {3 \sin (x) \cos (x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^4/(a + a*Sin[x]),x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2767

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

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 101, normalized size = 1.91 \[ \frac {\left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right ) \left (-36 x \sin \left (\frac {x}{2}\right )+69 \sin \left (\frac {x}{2}\right )-18 \sin \left (\frac {3 x}{2}\right )+2 \sin \left (\frac {5 x}{2}\right )+\sin \left (\frac {7 x}{2}\right )-3 (12 x+7) \cos \left (\frac {x}{2}\right )-18 \cos \left (\frac {3 x}{2}\right )-2 \cos \left (\frac {5 x}{2}\right )+\cos \left (\frac {7 x}{2}\right )\right )}{24 a (\sin (x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^4/(a + a*Sin[x]),x]

[Out]

((Cos[x/2] + Sin[x/2])*(-3*(7 + 12*x)*Cos[x/2] - 18*Cos[(3*x)/2] - 2*Cos[(5*x)/2] + Cos[(7*x)/2] + 69*Sin[x/2]
- 36*x*Sin[x/2] - 18*Sin[(3*x)/2] + 2*Sin[(5*x)/2] + Sin[(7*x)/2]))/(24*a*(1 + Sin[x]))

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fricas [A]  time = 0.45, size = 70, normalized size = 1.32 \[ \frac {2 \, \cos \relax (x)^{4} - \cos \relax (x)^{3} - 3 \, {\left (3 \, x + 5\right )} \cos \relax (x) - 12 \, \cos \relax (x)^{2} + {\left (2 \, \cos \relax (x)^{3} + 3 \, \cos \relax (x)^{2} - 9 \, x - 9 \, \cos \relax (x) + 6\right )} \sin \relax (x) - 9 \, x - 6}{6 \, {\left (a \cos \relax (x) + a \sin \relax (x) + a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*cos(x)^4 - cos(x)^3 - 3*(3*x + 5)*cos(x) - 12*cos(x)^2 + (2*cos(x)^3 + 3*cos(x)^2 - 9*x - 9*cos(x) + 6)
*sin(x) - 9*x - 6)/(a*cos(x) + a*sin(x) + a)

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giac [A]  time = 0.91, size = 67, normalized size = 1.26 \[ -\frac {3 \, x}{2 \, a} - \frac {2}{a {\left (\tan \left (\frac {1}{2} \, x\right ) + 1\right )}} - \frac {3 \, \tan \left (\frac {1}{2} \, x\right )^{5} + 6 \, \tan \left (\frac {1}{2} \, x\right )^{4} + 24 \, \tan \left (\frac {1}{2} \, x\right )^{2} - 3 \, \tan \left (\frac {1}{2} \, x\right ) + 10}{3 \, {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{3} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*x/a - 2/(a*(tan(1/2*x) + 1)) - 1/3*(3*tan(1/2*x)^5 + 6*tan(1/2*x)^4 + 24*tan(1/2*x)^2 - 3*tan(1/2*x) + 10
)/((tan(1/2*x)^2 + 1)^3*a)

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maple [B]  time = 0.08, size = 121, normalized size = 2.28 \[ -\frac {\tan ^{5}\left (\frac {x}{2}\right )}{a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}-\frac {2 \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}-\frac {8 \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}+\frac {\tan \left (\frac {x}{2}\right )}{a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}-\frac {10}{3 a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}-\frac {3 \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{a}-\frac {2}{a \left (\tan \left (\frac {x}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a/(tan(1/2*x)^2+1)^3*tan(1/2*x)^5-2/a/(tan(1/2*x)^2+1)^3*tan(1/2*x)^4-8/a/(tan(1/2*x)^2+1)^3*tan(1/2*x)^2+1
/a/(tan(1/2*x)^2+1)^3*tan(1/2*x)-10/3/a/(tan(1/2*x)^2+1)^3-3/a*arctan(tan(1/2*x))-2/a/(tan(1/2*x)+1)

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maxima [B]  time = 0.83, size = 180, normalized size = 3.40 \[ -\frac {\frac {7 \, \sin \relax (x)}{\cos \relax (x) + 1} + \frac {39 \, \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {24 \, \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {24 \, \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + \frac {9 \, \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}} + \frac {9 \, \sin \relax (x)^{6}}{{\left (\cos \relax (x) + 1\right )}^{6}} + 16}{3 \, {\left (a + \frac {a \sin \relax (x)}{\cos \relax (x) + 1} + \frac {3 \, a \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {3 \, a \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {3 \, a \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + \frac {3 \, a \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}} + \frac {a \sin \relax (x)^{6}}{{\left (\cos \relax (x) + 1\right )}^{6}} + \frac {a \sin \relax (x)^{7}}{{\left (\cos \relax (x) + 1\right )}^{7}}\right )}} - \frac {3 \, \arctan \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(7*sin(x)/(cos(x) + 1) + 39*sin(x)^2/(cos(x) + 1)^2 + 24*sin(x)^3/(cos(x) + 1)^3 + 24*sin(x)^4/(cos(x) +
1)^4 + 9*sin(x)^5/(cos(x) + 1)^5 + 9*sin(x)^6/(cos(x) + 1)^6 + 16)/(a + a*sin(x)/(cos(x) + 1) + 3*a*sin(x)^2/(
cos(x) + 1)^2 + 3*a*sin(x)^3/(cos(x) + 1)^3 + 3*a*sin(x)^4/(cos(x) + 1)^4 + 3*a*sin(x)^5/(cos(x) + 1)^5 + a*si
n(x)^6/(cos(x) + 1)^6 + a*sin(x)^7/(cos(x) + 1)^7) - 3*arctan(sin(x)/(cos(x) + 1))/a

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mupad [B]  time = 6.82, size = 78, normalized size = 1.47 \[ -\frac {3\,x}{2\,a}-\frac {3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5+8\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+8\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+13\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+\frac {7\,\mathrm {tan}\left (\frac {x}{2}\right )}{3}+\frac {16}{3}}{a\,{\left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )}^3\,\left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^4/(a + a*sin(x)),x)

[Out]

- (3*x)/(2*a) - ((7*tan(x/2))/3 + 13*tan(x/2)^2 + 8*tan(x/2)^3 + 8*tan(x/2)^4 + 3*tan(x/2)^5 + 3*tan(x/2)^6 +
16/3)/(a*(tan(x/2)^2 + 1)^3*(tan(x/2) + 1))

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sympy [B]  time = 5.91, size = 1221, normalized size = 23.04 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9*x*tan(x/2)**7/(6*a*tan(x/2)**7 + 6*a*tan(x/2)**6 + 18*a*tan(x/2)**5 + 18*a*tan(x/2)**4 + 18*a*tan(x/2)**3 +
18*a*tan(x/2)**2 + 6*a*tan(x/2) + 6*a) - 9*x*tan(x/2)**6/(6*a*tan(x/2)**7 + 6*a*tan(x/2)**6 + 18*a*tan(x/2)**
5 + 18*a*tan(x/2)**4 + 18*a*tan(x/2)**3 + 18*a*tan(x/2)**2 + 6*a*tan(x/2) + 6*a) - 27*x*tan(x/2)**5/(6*a*tan(x
/2)**7 + 6*a*tan(x/2)**6 + 18*a*tan(x/2)**5 + 18*a*tan(x/2)**4 + 18*a*tan(x/2)**3 + 18*a*tan(x/2)**2 + 6*a*tan
(x/2) + 6*a) - 27*x*tan(x/2)**4/(6*a*tan(x/2)**7 + 6*a*tan(x/2)**6 + 18*a*tan(x/2)**5 + 18*a*tan(x/2)**4 + 18*
a*tan(x/2)**3 + 18*a*tan(x/2)**2 + 6*a*tan(x/2) + 6*a) - 27*x*tan(x/2)**3/(6*a*tan(x/2)**7 + 6*a*tan(x/2)**6 +
18*a*tan(x/2)**5 + 18*a*tan(x/2)**4 + 18*a*tan(x/2)**3 + 18*a*tan(x/2)**2 + 6*a*tan(x/2) + 6*a) - 27*x*tan(x/
2)**2/(6*a*tan(x/2)**7 + 6*a*tan(x/2)**6 + 18*a*tan(x/2)**5 + 18*a*tan(x/2)**4 + 18*a*tan(x/2)**3 + 18*a*tan(x
/2)**2 + 6*a*tan(x/2) + 6*a) - 9*x*tan(x/2)/(6*a*tan(x/2)**7 + 6*a*tan(x/2)**6 + 18*a*tan(x/2)**5 + 18*a*tan(x
/2)**4 + 18*a*tan(x/2)**3 + 18*a*tan(x/2)**2 + 6*a*tan(x/2) + 6*a) - 9*x/(6*a*tan(x/2)**7 + 6*a*tan(x/2)**6 +
18*a*tan(x/2)**5 + 18*a*tan(x/2)**4 + 18*a*tan(x/2)**3 + 18*a*tan(x/2)**2 + 6*a*tan(x/2) + 6*a) - 18*tan(x/2)*
*6/(6*a*tan(x/2)**7 + 6*a*tan(x/2)**6 + 18*a*tan(x/2)**5 + 18*a*tan(x/2)**4 + 18*a*tan(x/2)**3 + 18*a*tan(x/2)
**2 + 6*a*tan(x/2) + 6*a) - 18*tan(x/2)**5/(6*a*tan(x/2)**7 + 6*a*tan(x/2)**6 + 18*a*tan(x/2)**5 + 18*a*tan(x/
2)**4 + 18*a*tan(x/2)**3 + 18*a*tan(x/2)**2 + 6*a*tan(x/2) + 6*a) - 48*tan(x/2)**4/(6*a*tan(x/2)**7 + 6*a*tan(
x/2)**6 + 18*a*tan(x/2)**5 + 18*a*tan(x/2)**4 + 18*a*tan(x/2)**3 + 18*a*tan(x/2)**2 + 6*a*tan(x/2) + 6*a) - 48
*tan(x/2)**3/(6*a*tan(x/2)**7 + 6*a*tan(x/2)**6 + 18*a*tan(x/2)**5 + 18*a*tan(x/2)**4 + 18*a*tan(x/2)**3 + 18*
a*tan(x/2)**2 + 6*a*tan(x/2) + 6*a) - 78*tan(x/2)**2/(6*a*tan(x/2)**7 + 6*a*tan(x/2)**6 + 18*a*tan(x/2)**5 + 1
8*a*tan(x/2)**4 + 18*a*tan(x/2)**3 + 18*a*tan(x/2)**2 + 6*a*tan(x/2) + 6*a) - 14*tan(x/2)/(6*a*tan(x/2)**7 + 6
*a*tan(x/2)**6 + 18*a*tan(x/2)**5 + 18*a*tan(x/2)**4 + 18*a*tan(x/2)**3 + 18*a*tan(x/2)**2 + 6*a*tan(x/2) + 6*
a) - 32/(6*a*tan(x/2)**7 + 6*a*tan(x/2)**6 + 18*a*tan(x/2)**5 + 18*a*tan(x/2)**4 + 18*a*tan(x/2)**3 + 18*a*tan
(x/2)**2 + 6*a*tan(x/2) + 6*a)

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