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]

7/2*x/a^2+16/3*cos(x)/a^2-7/2*cos(x)*sin(x)/a^2+8/3*cos(x)*sin(x)^2/a^2/(1+sin(x))+1/3*cos(x)*sin(x)^3/(a+a*si
n(x))^2

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 66, normalized size of antiderivative = 1.00, 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 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]

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])

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 = 0.25, 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)

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 105, normalized size = 1.59 \[ -\frac {3 \, \cos \relax (x)^{4} - {\left (21 \, x - 31\right )} \cos \relax (x)^{2} - 6 \, \cos \relax (x)^{3} + {\left (21 \, x + 38\right )} \cos \relax (x) + {\left (3 \, \cos \relax (x)^{3} + {\left (21 \, x + 40\right )} \cos \relax (x) + 9 \, \cos \relax (x)^{2} + 42 \, x + 2\right )} \sin \relax (x) + 42 \, x - 2}{6 \, {\left (a^{2} \cos \relax (x)^{2} - a^{2} \cos \relax (x) - 2 \, a^{2} - {\left (a^{2} \cos \relax (x) + 2 \, a^{2}\right )} \sin \relax (x)\right )}} \]

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))

________________________________________________________________________________________

giac [A]  time = 0.33, size = 72, normalized size = 1.09 \[ \frac {7 \, x}{2 \, a^{2}} + \frac {\tan \left (\frac {1}{2} \, x\right )^{3} + 4 \, \tan \left (\frac {1}{2} \, x\right )^{2} - \tan \left (\frac {1}{2} \, x\right ) + 4}{{\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{2} a^{2}} + \frac {2 \, {\left (9 \, \tan \left (\frac {1}{2} \, x\right )^{2} + 21 \, \tan \left (\frac {1}{2} \, x\right ) + 10\right )}}{3 \, a^{2} {\left (\tan \left (\frac {1}{2} \, x\right ) + 1\right )}^{3}} \]

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)

________________________________________________________________________________________

maple [B]  time = 0.10, size = 126, normalized size = 1.91 \[ \frac {\tan ^{3}\left (\frac {x}{2}\right )}{a^{2} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {4 \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{a^{2} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}-\frac {\tan \left (\frac {x}{2}\right )}{a^{2} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {4}{a^{2} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {7 \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{a^{2}}-\frac {4}{3 a^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {2}{a^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {6}{a^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^4/(a+a*sin(x))^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 = 1.24, size = 198, normalized size = 3.00 \[ \frac {\frac {75 \, \sin \relax (x)}{\cos \relax (x) + 1} + \frac {97 \, \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {126 \, \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {98 \, \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + \frac {63 \, \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}} + \frac {21 \, \sin \relax (x)^{6}}{{\left (\cos \relax (x) + 1\right )}^{6}} + 32}{3 \, {\left (a^{2} + \frac {3 \, a^{2} \sin \relax (x)}{\cos \relax (x) + 1} + \frac {5 \, a^{2} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {7 \, a^{2} \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {7 \, a^{2} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + \frac {5 \, a^{2} \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}} + \frac {3 \, a^{2} \sin \relax (x)^{6}}{{\left (\cos \relax (x) + 1\right )}^{6}} + \frac {a^{2} \sin \relax (x)^{7}}{{\left (\cos \relax (x) + 1\right )}^{7}}\right )}} + \frac {7 \, \arctan \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )}{a^{2}} \]

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

________________________________________________________________________________________

mupad [B]  time = 6.81, size = 77, normalized size = 1.17 \[ \frac {7\,x}{2\,a^2}+\frac {7\,{\mathrm {tan}\left (\frac {x}{2}\right )}^6+21\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5+\frac {98\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4}{3}+42\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+\frac {97\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{3}+25\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {32}{3}}{a^2\,{\left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )}^2\,{\left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^4/(a + a*sin(x))^2,x)

[Out]

(7*x)/(2*a^2) + (25*tan(x/2) + (97*tan(x/2)^2)/3 + 42*tan(x/2)^3 + (98*tan(x/2)^4)/3 + 21*tan(x/2)^5 + 7*tan(x
/2)^6 + 32/3)/(a^2*(tan(x/2)^2 + 1)^2*(tan(x/2) + 1)^3)

________________________________________________________________________________________

sympy [B]  time = 10.92, size = 1423, normalized size = 21.56 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)**4/(a+a*sin(x))**2,x)

[Out]

21*x*tan(x/2)**7/(6*a**2*tan(x/2)**7 + 18*a**2*tan(x/2)**6 + 30*a**2*tan(x/2)**5 + 42*a**2*tan(x/2)**4 + 42*a*
*2*tan(x/2)**3 + 30*a**2*tan(x/2)**2 + 18*a**2*tan(x/2) + 6*a**2) + 63*x*tan(x/2)**6/(6*a**2*tan(x/2)**7 + 18*
a**2*tan(x/2)**6 + 30*a**2*tan(x/2)**5 + 42*a**2*tan(x/2)**4 + 42*a**2*tan(x/2)**3 + 30*a**2*tan(x/2)**2 + 18*
a**2*tan(x/2) + 6*a**2) + 105*x*tan(x/2)**5/(6*a**2*tan(x/2)**7 + 18*a**2*tan(x/2)**6 + 30*a**2*tan(x/2)**5 +
42*a**2*tan(x/2)**4 + 42*a**2*tan(x/2)**3 + 30*a**2*tan(x/2)**2 + 18*a**2*tan(x/2) + 6*a**2) + 147*x*tan(x/2)*
*4/(6*a**2*tan(x/2)**7 + 18*a**2*tan(x/2)**6 + 30*a**2*tan(x/2)**5 + 42*a**2*tan(x/2)**4 + 42*a**2*tan(x/2)**3
+ 30*a**2*tan(x/2)**2 + 18*a**2*tan(x/2) + 6*a**2) + 147*x*tan(x/2)**3/(6*a**2*tan(x/2)**7 + 18*a**2*tan(x/2)
**6 + 30*a**2*tan(x/2)**5 + 42*a**2*tan(x/2)**4 + 42*a**2*tan(x/2)**3 + 30*a**2*tan(x/2)**2 + 18*a**2*tan(x/2)
+ 6*a**2) + 105*x*tan(x/2)**2/(6*a**2*tan(x/2)**7 + 18*a**2*tan(x/2)**6 + 30*a**2*tan(x/2)**5 + 42*a**2*tan(x
/2)**4 + 42*a**2*tan(x/2)**3 + 30*a**2*tan(x/2)**2 + 18*a**2*tan(x/2) + 6*a**2) + 63*x*tan(x/2)/(6*a**2*tan(x/
2)**7 + 18*a**2*tan(x/2)**6 + 30*a**2*tan(x/2)**5 + 42*a**2*tan(x/2)**4 + 42*a**2*tan(x/2)**3 + 30*a**2*tan(x/
2)**2 + 18*a**2*tan(x/2) + 6*a**2) + 21*x/(6*a**2*tan(x/2)**7 + 18*a**2*tan(x/2)**6 + 30*a**2*tan(x/2)**5 + 42
*a**2*tan(x/2)**4 + 42*a**2*tan(x/2)**3 + 30*a**2*tan(x/2)**2 + 18*a**2*tan(x/2) + 6*a**2) + 42*tan(x/2)**6/(6
*a**2*tan(x/2)**7 + 18*a**2*tan(x/2)**6 + 30*a**2*tan(x/2)**5 + 42*a**2*tan(x/2)**4 + 42*a**2*tan(x/2)**3 + 30
*a**2*tan(x/2)**2 + 18*a**2*tan(x/2) + 6*a**2) + 126*tan(x/2)**5/(6*a**2*tan(x/2)**7 + 18*a**2*tan(x/2)**6 + 3
0*a**2*tan(x/2)**5 + 42*a**2*tan(x/2)**4 + 42*a**2*tan(x/2)**3 + 30*a**2*tan(x/2)**2 + 18*a**2*tan(x/2) + 6*a*
*2) + 196*tan(x/2)**4/(6*a**2*tan(x/2)**7 + 18*a**2*tan(x/2)**6 + 30*a**2*tan(x/2)**5 + 42*a**2*tan(x/2)**4 +
42*a**2*tan(x/2)**3 + 30*a**2*tan(x/2)**2 + 18*a**2*tan(x/2) + 6*a**2) + 252*tan(x/2)**3/(6*a**2*tan(x/2)**7 +
18*a**2*tan(x/2)**6 + 30*a**2*tan(x/2)**5 + 42*a**2*tan(x/2)**4 + 42*a**2*tan(x/2)**3 + 30*a**2*tan(x/2)**2 +
18*a**2*tan(x/2) + 6*a**2) + 194*tan(x/2)**2/(6*a**2*tan(x/2)**7 + 18*a**2*tan(x/2)**6 + 30*a**2*tan(x/2)**5
+ 42*a**2*tan(x/2)**4 + 42*a**2*tan(x/2)**3 + 30*a**2*tan(x/2)**2 + 18*a**2*tan(x/2) + 6*a**2) + 150*tan(x/2)/
(6*a**2*tan(x/2)**7 + 18*a**2*tan(x/2)**6 + 30*a**2*tan(x/2)**5 + 42*a**2*tan(x/2)**4 + 42*a**2*tan(x/2)**3 +
30*a**2*tan(x/2)**2 + 18*a**2*tan(x/2) + 6*a**2) + 64/(6*a**2*tan(x/2)**7 + 18*a**2*tan(x/2)**6 + 30*a**2*tan(
x/2)**5 + 42*a**2*tan(x/2)**4 + 42*a**2*tan(x/2)**3 + 30*a**2*tan(x/2)**2 + 18*a**2*tan(x/2) + 6*a**2)

________________________________________________________________________________________