3.12 \(\int \frac{\sin ^4(x)}{(a+a \sin (x))^2} \, dx\)

Optimal. Leaf size=66 \[ \frac{7 x}{2 a^2}+\frac{16 \cos (x)}{3 a^2}+\frac{8 \sin ^2(x) \cos (x)}{3 a^2 (\sin (x)+1)}-\frac{7 \sin (x) \cos (x)}{2 a^2}+\frac{\sin ^3(x) \cos (x)}{3 (a \sin (x)+a)^2} \]

[Out]

(7*x)/(2*a^2) + (16*Cos[x])/(3*a^2) - (7*Cos[x]*Sin[x])/(2*a^2) + (8*Cos[x]*Sin[x]^2)/(3*a^2*(1 + Sin[x])) + (
Cos[x]*Sin[x]^3)/(3*(a + a*Sin[x])^2)

________________________________________________________________________________________

Rubi [A]  time = 0.120751, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2765, 2977, 2734} \[ \frac{7 x}{2 a^2}+\frac{16 \cos (x)}{3 a^2}+\frac{8 \sin ^2(x) \cos (x)}{3 a^2 (\sin (x)+1)}-\frac{7 \sin (x) \cos (x)}{2 a^2}+\frac{\sin ^3(x) \cos (x)}{3 (a \sin (x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^4/(a + a*Sin[x])^2,x]

[Out]

(7*x)/(2*a^2) + (16*Cos[x])/(3*a^2) - (7*Cos[x]*Sin[x])/(2*a^2) + (8*Cos[x]*Sin[x]^2)/(3*a^2*(1 + Sin[x])) + (
Cos[x]*Sin[x]^3)/(3*(a + a*Sin[x])^2)

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 2734

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

Rubi steps

\begin{align*} \int \frac{\sin ^4(x)}{(a+a \sin (x))^2} \, dx &=\frac{\cos (x) \sin ^3(x)}{3 (a+a \sin (x))^2}-\frac{\int \frac{\sin ^2(x) (3 a-5 a \sin (x))}{a+a \sin (x)} \, dx}{3 a^2}\\ &=\frac{8 \cos (x) \sin ^2(x)}{3 a^2 (1+\sin (x))}+\frac{\cos (x) \sin ^3(x)}{3 (a+a \sin (x))^2}-\frac{\int \sin (x) \left (16 a^2-21 a^2 \sin (x)\right ) \, dx}{3 a^4}\\ &=\frac{7 x}{2 a^2}+\frac{16 \cos (x)}{3 a^2}-\frac{7 \cos (x) \sin (x)}{2 a^2}+\frac{8 \cos (x) \sin ^2(x)}{3 a^2 (1+\sin (x))}+\frac{\cos (x) \sin ^3(x)}{3 (a+a \sin (x))^2}\\ \end{align*}

Mathematica [A]  time = 0.239894, size = 100, normalized size = 1.52 \[ \frac{\left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right ) \left (21 (12 x-7) \cos \left (\frac{x}{2}\right )+(239-84 x) \cos \left (\frac{3 x}{2}\right )+3 \left (-5 \cos \left (\frac{5 x}{2}\right )+\cos \left (\frac{7 x}{2}\right )+2 \sin \left (\frac{x}{2}\right ) (56 x+(28 x+27) \cos (x)+6 \cos (2 x)+\cos (3 x)-50)\right )\right )}{48 a^2 (\sin (x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^4/(a + a*Sin[x])^2,x]

[Out]

((Cos[x/2] + Sin[x/2])*(21*(-7 + 12*x)*Cos[x/2] + (239 - 84*x)*Cos[(3*x)/2] + 3*(-5*Cos[(5*x)/2] + Cos[(7*x)/2
] + 2*(-50 + 56*x + (27 + 28*x)*Cos[x] + 6*Cos[2*x] + Cos[3*x])*Sin[x/2])))/(48*a^2*(1 + Sin[x])^2)

________________________________________________________________________________________

Maple [B]  time = 0.042, size = 126, normalized size = 1.9 \begin{align*}{\frac{1}{{a}^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{3} \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}+4\,{\frac{ \left ( \tan \left ( x/2 \right ) \right ) ^{2}}{{a}^{2} \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{1}{{a}^{2}}\tan \left ({\frac{x}{2}} \right ) \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}+4\,{\frac{1}{{a}^{2} \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+7\,{\frac{\arctan \left ( \tan \left ( x/2 \right ) \right ) }{{a}^{2}}}-{\frac{4}{3\,{a}^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) ^{-3}}+2\,{\frac{1}{{a}^{2} \left ( \tan \left ( x/2 \right ) +1 \right ) ^{2}}}+6\,{\frac{1}{{a}^{2} \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))^2,x)

[Out]

1/a^2/(tan(1/2*x)^2+1)^2*tan(1/2*x)^3+4/a^2/(tan(1/2*x)^2+1)^2*tan(1/2*x)^2-1/a^2/(tan(1/2*x)^2+1)^2*tan(1/2*x
)+4/a^2/(tan(1/2*x)^2+1)^2+7/a^2*arctan(tan(1/2*x))-4/3/a^2/(tan(1/2*x)+1)^3+2/a^2/(tan(1/2*x)+1)^2+6/a^2/(tan
(1/2*x)+1)

________________________________________________________________________________________

Maxima [B]  time = 2.4889, size = 267, normalized size = 4.05 \begin{align*} \frac{\frac{75 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac{97 \, \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac{126 \, \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac{98 \, \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + \frac{63 \, \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}} + \frac{21 \, \sin \left (x\right )^{6}}{{\left (\cos \left (x\right ) + 1\right )}^{6}} + 32}{3 \,{\left (a^{2} + \frac{3 \, a^{2} \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac{5 \, a^{2} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac{7 \, a^{2} \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac{7 \, a^{2} \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + \frac{5 \, a^{2} \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}} + \frac{3 \, a^{2} \sin \left (x\right )^{6}}{{\left (\cos \left (x\right ) + 1\right )}^{6}} + \frac{a^{2} \sin \left (x\right )^{7}}{{\left (\cos \left (x\right ) + 1\right )}^{7}}\right )}} + \frac{7 \, \arctan \left (\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1}\right )}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^4/(a+a*sin(x))^2,x, algorithm="maxima")

[Out]

1/3*(75*sin(x)/(cos(x) + 1) + 97*sin(x)^2/(cos(x) + 1)^2 + 126*sin(x)^3/(cos(x) + 1)^3 + 98*sin(x)^4/(cos(x) +
1)^4 + 63*sin(x)^5/(cos(x) + 1)^5 + 21*sin(x)^6/(cos(x) + 1)^6 + 32)/(a^2 + 3*a^2*sin(x)/(cos(x) + 1) + 5*a^2
*sin(x)^2/(cos(x) + 1)^2 + 7*a^2*sin(x)^3/(cos(x) + 1)^3 + 7*a^2*sin(x)^4/(cos(x) + 1)^4 + 5*a^2*sin(x)^5/(cos
(x) + 1)^5 + 3*a^2*sin(x)^6/(cos(x) + 1)^6 + a^2*sin(x)^7/(cos(x) + 1)^7) + 7*arctan(sin(x)/(cos(x) + 1))/a^2

________________________________________________________________________________________

Fricas [A]  time = 1.3966, size = 297, normalized size = 4.5 \begin{align*} -\frac{3 \, \cos \left (x\right )^{4} -{\left (21 \, x - 31\right )} \cos \left (x\right )^{2} - 6 \, \cos \left (x\right )^{3} +{\left (21 \, x + 38\right )} \cos \left (x\right ) +{\left (3 \, \cos \left (x\right )^{3} +{\left (21 \, x + 40\right )} \cos \left (x\right ) + 9 \, \cos \left (x\right )^{2} + 42 \, x + 2\right )} \sin \left (x\right ) + 42 \, x - 2}{6 \,{\left (a^{2} \cos \left (x\right )^{2} - a^{2} \cos \left (x\right ) - 2 \, a^{2} -{\left (a^{2} \cos \left (x\right ) + 2 \, a^{2}\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))^2,x, algorithm="fricas")

[Out]

-1/6*(3*cos(x)^4 - (21*x - 31)*cos(x)^2 - 6*cos(x)^3 + (21*x + 38)*cos(x) + (3*cos(x)^3 + (21*x + 40)*cos(x) +
9*cos(x)^2 + 42*x + 2)*sin(x) + 42*x - 2)/(a^2*cos(x)^2 - a^2*cos(x) - 2*a^2 - (a^2*cos(x) + 2*a^2)*sin(x))

________________________________________________________________________________________

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))**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 2.38438, size = 97, normalized size = 1.47 \begin{align*} \frac{7 \, x}{2 \, a^{2}} + \frac{\tan \left (\frac{1}{2} \, x\right )^{3} + 4 \, \tan \left (\frac{1}{2} \, x\right )^{2} - \tan \left (\frac{1}{2} \, x\right ) + 4}{{\left (\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right )}^{2} a^{2}} + \frac{2 \,{\left (9 \, \tan \left (\frac{1}{2} \, x\right )^{2} + 21 \, \tan \left (\frac{1}{2} \, x\right ) + 10\right )}}{3 \, a^{2}{\left (\tan \left (\frac{1}{2} \, x\right ) + 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^4/(a+a*sin(x))^2,x, algorithm="giac")

[Out]

7/2*x/a^2 + (tan(1/2*x)^3 + 4*tan(1/2*x)^2 - tan(1/2*x) + 4)/((tan(1/2*x)^2 + 1)^2*a^2) + 2/3*(9*tan(1/2*x)^2
+ 21*tan(1/2*x) + 10)/(a^2*(tan(1/2*x) + 1)^3)