3.2.85 \(\int \frac {\sin ^4(x)}{(a+b \sin (x))^2} \, dx\) [185]

Optimal. Leaf size=169 \[ \frac {\left (6 a^2+b^2\right ) x}{2 b^4}-\frac {2 a^3 \left (3 a^2-4 b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{b^4 \left (a^2-b^2\right )^{3/2}}+\frac {a \left (3 a^2-2 b^2\right ) \cos (x)}{b^3 \left (a^2-b^2\right )}-\frac {\left (3 a^2-b^2\right ) \cos (x) \sin (x)}{2 b^2 \left (a^2-b^2\right )}+\frac {a^2 \cos (x) \sin ^2(x)}{b \left (a^2-b^2\right ) (a+b \sin (x))} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.28, antiderivative size = 169, 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 = {2871, 3128, 3102, 2814, 2739, 632, 210} \begin {gather*} \frac {a^2 \sin ^2(x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\left (3 a^2-b^2\right ) \sin (x) \cos (x)}{2 b^2 \left (a^2-b^2\right )}+\frac {x \left (6 a^2+b^2\right )}{2 b^4}+\frac {a \left (3 a^2-2 b^2\right ) \cos (x)}{b^3 \left (a^2-b^2\right )}-\frac {2 a^3 \left (3 a^2-4 b^2\right ) \text {ArcTan}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{b^4 \left (a^2-b^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[x]^4/(a + b*Sin[x])^2,x]

[Out]

((6*a^2 + b^2)*x)/(2*b^4) - (2*a^3*(3*a^2 - 4*b^2)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(b^4*(a^2 - b^2)^
(3/2)) + (a*(3*a^2 - 2*b^2)*Cos[x])/(b^3*(a^2 - b^2)) - ((3*a^2 - b^2)*Cos[x]*Sin[x])/(2*b^2*(a^2 - b^2)) + (a
^2*Cos[x]*Sin[x]^2)/(b*(a^2 - b^2)*(a + b*Sin[x]))

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[2*(e/d), Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2814

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 2871

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/
(d*f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e
+ f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 +
c*d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 +
d*n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3102

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 3128

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

Rubi steps

\begin {align*} \int \frac {\sin ^4(x)}{(a+b \sin (x))^2} \, dx &=\frac {a^2 \cos (x) \sin ^2(x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\int \frac {\sin (x) \left (2 a^2-a b \sin (x)-\left (3 a^2-b^2\right ) \sin ^2(x)\right )}{a+b \sin (x)} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac {\left (3 a^2-b^2\right ) \cos (x) \sin (x)}{2 b^2 \left (a^2-b^2\right )}+\frac {a^2 \cos (x) \sin ^2(x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\int \frac {-a \left (3 a^2-b^2\right )+b \left (a^2+b^2\right ) \sin (x)+2 a \left (3 a^2-2 b^2\right ) \sin ^2(x)}{a+b \sin (x)} \, dx}{2 b^2 \left (a^2-b^2\right )}\\ &=\frac {a \left (3 a^2-2 b^2\right ) \cos (x)}{b^3 \left (a^2-b^2\right )}-\frac {\left (3 a^2-b^2\right ) \cos (x) \sin (x)}{2 b^2 \left (a^2-b^2\right )}+\frac {a^2 \cos (x) \sin ^2(x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\int \frac {-a b \left (3 a^2-b^2\right )-\left (a^2-b^2\right ) \left (6 a^2+b^2\right ) \sin (x)}{a+b \sin (x)} \, dx}{2 b^3 \left (a^2-b^2\right )}\\ &=\frac {\left (6 a^2+b^2\right ) x}{2 b^4}+\frac {a \left (3 a^2-2 b^2\right ) \cos (x)}{b^3 \left (a^2-b^2\right )}-\frac {\left (3 a^2-b^2\right ) \cos (x) \sin (x)}{2 b^2 \left (a^2-b^2\right )}+\frac {a^2 \cos (x) \sin ^2(x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\left (a^3 \left (3 a^2-4 b^2\right )\right ) \int \frac {1}{a+b \sin (x)} \, dx}{b^4 \left (a^2-b^2\right )}\\ &=\frac {\left (6 a^2+b^2\right ) x}{2 b^4}+\frac {a \left (3 a^2-2 b^2\right ) \cos (x)}{b^3 \left (a^2-b^2\right )}-\frac {\left (3 a^2-b^2\right ) \cos (x) \sin (x)}{2 b^2 \left (a^2-b^2\right )}+\frac {a^2 \cos (x) \sin ^2(x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\left (2 a^3 \left (3 a^2-4 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{b^4 \left (a^2-b^2\right )}\\ &=\frac {\left (6 a^2+b^2\right ) x}{2 b^4}+\frac {a \left (3 a^2-2 b^2\right ) \cos (x)}{b^3 \left (a^2-b^2\right )}-\frac {\left (3 a^2-b^2\right ) \cos (x) \sin (x)}{2 b^2 \left (a^2-b^2\right )}+\frac {a^2 \cos (x) \sin ^2(x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}+\frac {\left (4 a^3 \left (3 a^2-4 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {x}{2}\right )\right )}{b^4 \left (a^2-b^2\right )}\\ &=\frac {\left (6 a^2+b^2\right ) x}{2 b^4}-\frac {2 a^3 \left (3 a^2-4 b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{b^4 \left (a^2-b^2\right )^{3/2}}+\frac {a \left (3 a^2-2 b^2\right ) \cos (x)}{b^3 \left (a^2-b^2\right )}-\frac {\left (3 a^2-b^2\right ) \cos (x) \sin (x)}{2 b^2 \left (a^2-b^2\right )}+\frac {a^2 \cos (x) \sin ^2(x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.39, size = 115, normalized size = 0.68 \begin {gather*} \frac {12 a^2 x+2 b^2 x-\frac {8 a^3 \left (3 a^2-4 b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+4 a b \cos (x) \left (2+\frac {a^3}{(a-b) (a+b) (a+b \sin (x))}\right )-b^2 \sin (2 x)}{4 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(12*a^2*x + 2*b^2*x - (8*a^3*(3*a^2 - 4*b^2)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) + 4*a
*b*Cos[x]*(2 + a^3/((a - b)*(a + b)*(a + b*Sin[x]))) - b^2*Sin[2*x])/(4*b^4)

________________________________________________________________________________________

Maple [A]
time = 0.30, size = 183, normalized size = 1.08

method	result	size
default	\(\frac {\frac {2 \left (\frac {b^{2} \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{2}+2 a b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-\frac {b^{2} \tan \left (\frac {x}{2}\right )}{2}+2 a b \right )}{\left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\left (6 a^{2}+b^{2}\right ) \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{b^{4}}-\frac {2 a^{3} \left (\frac {-\frac {b^{2} \tan \left (\frac {x}{2}\right )}{a^{2}-b^{2}}-\frac {a b}{a^{2}-b^{2}}}{a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+2 b \tan \left (\frac {x}{2}\right )+a}+\frac {\left (3 a^{2}-4 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}\right )}{b^{4}}\)	\(183\)
risch	\(\frac {3 x \,a^{2}}{b^{4}}+\frac {x}{2 b^{2}}+\frac {i {\mathrm e}^{2 i x}}{8 b^{2}}+\frac {a \,{\mathrm e}^{i x}}{b^{3}}+\frac {a \,{\mathrm e}^{-i x}}{b^{3}}-\frac {i {\mathrm e}^{-2 i x}}{8 b^{2}}-\frac {2 i a^{4} \left (i b +a \,{\mathrm e}^{i x}\right )}{b^{4} \left (a^{2}-b^{2}\right ) \left (-i b \,{\mathrm e}^{2 i x}+i b +2 a \,{\mathrm e}^{i x}\right )}-\frac {3 a^{5} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) b^{4}}+\frac {4 a^{3} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) b^{2}}+\frac {3 a^{5} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) b^{4}}-\frac {4 a^{3} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) b^{2}}\)	\(423\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^4/(a+b*sin(x))^2,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

________________________________________________________________________________________

Fricas [A]
time = 0.46, size = 580, normalized size = 3.43 \begin {gather*} \left [\frac {{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \cos \left (x\right )^{3} - {\left (3 \, a^{6} - 4 \, a^{4} b^{2} + {\left (3 \, a^{5} b - 4 \, a^{3} b^{3}\right )} \sin \left (x\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) + {\left (6 \, a^{7} - 11 \, a^{5} b^{2} + 4 \, a^{3} b^{4} + a b^{6}\right )} x + {\left (6 \, a^{6} b - 11 \, a^{4} b^{3} + 6 \, a^{2} b^{5} - b^{7}\right )} \cos \left (x\right ) + {\left ({\left (6 \, a^{6} b - 11 \, a^{4} b^{3} + 4 \, a^{2} b^{5} + b^{7}\right )} x + 3 \, {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{2 \, {\left (a^{5} b^{4} - 2 \, a^{3} b^{6} + a b^{8} + {\left (a^{4} b^{5} - 2 \, a^{2} b^{7} + b^{9}\right )} \sin \left (x\right )\right )}}, \frac {{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \cos \left (x\right )^{3} + 2 \, {\left (3 \, a^{6} - 4 \, a^{4} b^{2} + {\left (3 \, a^{5} b - 4 \, a^{3} b^{3}\right )} \sin \left (x\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (x\right )}\right ) + {\left (6 \, a^{7} - 11 \, a^{5} b^{2} + 4 \, a^{3} b^{4} + a b^{6}\right )} x + {\left (6 \, a^{6} b - 11 \, a^{4} b^{3} + 6 \, a^{2} b^{5} - b^{7}\right )} \cos \left (x\right ) + {\left ({\left (6 \, a^{6} b - 11 \, a^{4} b^{3} + 4 \, a^{2} b^{5} + b^{7}\right )} x + 3 \, {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{2 \, {\left (a^{5} b^{4} - 2 \, a^{3} b^{6} + a b^{8} + {\left (a^{4} b^{5} - 2 \, a^{2} b^{7} + b^{9}\right )} \sin \left (x\right )\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((a^4*b^3 - 2*a^2*b^5 + b^7)*cos(x)^3 - (3*a^6 - 4*a^4*b^2 + (3*a^5*b - 4*a^3*b^3)*sin(x))*sqrt(-a^2 + b^
2)*log(-((2*a^2 - b^2)*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2 - 2*(a*cos(x)*sin(x) + b*cos(x))*sqrt(-a^2 + b^2))/
(b^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) + (6*a^7 - 11*a^5*b^2 + 4*a^3*b^4 + a*b^6)*x + (6*a^6*b - 11*a^4*b^
3 + 6*a^2*b^5 - b^7)*cos(x) + ((6*a^6*b - 11*a^4*b^3 + 4*a^2*b^5 + b^7)*x + 3*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*co
s(x))*sin(x))/(a^5*b^4 - 2*a^3*b^6 + a*b^8 + (a^4*b^5 - 2*a^2*b^7 + b^9)*sin(x)), 1/2*((a^4*b^3 - 2*a^2*b^5 +
b^7)*cos(x)^3 + 2*(3*a^6 - 4*a^4*b^2 + (3*a^5*b - 4*a^3*b^3)*sin(x))*sqrt(a^2 - b^2)*arctan(-(a*sin(x) + b)/(s
qrt(a^2 - b^2)*cos(x))) + (6*a^7 - 11*a^5*b^2 + 4*a^3*b^4 + a*b^6)*x + (6*a^6*b - 11*a^4*b^3 + 6*a^2*b^5 - b^7
)*cos(x) + ((6*a^6*b - 11*a^4*b^3 + 4*a^2*b^5 + b^7)*x + 3*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cos(x))*sin(x))/(a^5*
b^4 - 2*a^3*b^6 + a*b^8 + (a^4*b^5 - 2*a^2*b^7 + b^9)*sin(x))]

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]
time = 0.45, size = 184, normalized size = 1.09 \begin {gather*} -\frac {2 \, {\left (3 \, a^{5} - 4 \, a^{3} b^{2}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {2 \, {\left (a^{3} b \tan \left (\frac {1}{2} \, x\right ) + a^{4}\right )}}{{\left (a^{2} b^{3} - b^{5}\right )} {\left (a \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a\right )}} + \frac {{\left (6 \, a^{2} + b^{2}\right )} x}{2 \, b^{4}} + \frac {b \tan \left (\frac {1}{2} \, x\right )^{3} + 4 \, a \tan \left (\frac {1}{2} \, x\right )^{2} - b \tan \left (\frac {1}{2} \, x\right ) + 4 \, a}{{\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{2} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(3*a^5 - 4*a^3*b^2)*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - b^2)))/((a^2*b^
4 - b^6)*sqrt(a^2 - b^2)) + 2*(a^3*b*tan(1/2*x) + a^4)/((a^2*b^3 - b^5)*(a*tan(1/2*x)^2 + 2*b*tan(1/2*x) + a))
+ 1/2*(6*a^2 + b^2)*x/b^4 + (b*tan(1/2*x)^3 + 4*a*tan(1/2*x)^2 - b*tan(1/2*x) + 4*a)/((tan(1/2*x)^2 + 1)^2*b^
3)

________________________________________________________________________________________

Mupad [B]
time = 11.88, size = 2500, normalized size = 14.79 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^4/(a + b*sin(x))^2,x)

[Out]

(atan((((a^2*6i + b^2*1i)*((8*(a^2*b^11 + 10*a^4*b^9 + 13*a^6*b^7 - 60*a^8*b^5 + 36*a^10*b^3))/(b^12 - 2*a^2*b
^10 + a^4*b^8) + (8*tan(x/2)*(2*a*b^13 + 19*a^3*b^11 + 16*a^5*b^9 - 197*a^7*b^7 + 228*a^9*b^5 - 72*a^11*b^3))/
(b^13 - 2*a^2*b^11 + a^4*b^9) - ((a^2*6i + b^2*1i)*((8*(2*a*b^14 + 6*a^3*b^12 - 14*a^5*b^10 + 6*a^7*b^8))/(b^1
2 - 2*a^2*b^10 + a^4*b^8) - (((8*(4*a^2*b^15 - 8*a^4*b^13 + 4*a^6*b^11))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*ta
n(x/2)*(12*a*b^17 - 32*a^3*b^15 + 28*a^5*b^13 - 8*a^7*b^11))/(b^13 - 2*a^2*b^11 + a^4*b^9))*(a^2*6i + b^2*1i))
/(2*b^4) + (8*tan(x/2)*(32*a^4*b^12 - 56*a^6*b^10 + 24*a^8*b^8))/(b^13 - 2*a^2*b^11 + a^4*b^9)))/(2*b^4))*1i)/
(2*b^4) + ((a^2*6i + b^2*1i)*((8*(a^2*b^11 + 10*a^4*b^9 + 13*a^6*b^7 - 60*a^8*b^5 + 36*a^10*b^3))/(b^12 - 2*a^
2*b^10 + a^4*b^8) + (8*tan(x/2)*(2*a*b^13 + 19*a^3*b^11 + 16*a^5*b^9 - 197*a^7*b^7 + 228*a^9*b^5 - 72*a^11*b^3
))/(b^13 - 2*a^2*b^11 + a^4*b^9) + ((a^2*6i + b^2*1i)*((8*(2*a*b^14 + 6*a^3*b^12 - 14*a^5*b^10 + 6*a^7*b^8))/(
b^12 - 2*a^2*b^10 + a^4*b^8) + (((8*(4*a^2*b^15 - 8*a^4*b^13 + 4*a^6*b^11))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8
*tan(x/2)*(12*a*b^17 - 32*a^3*b^15 + 28*a^5*b^13 - 8*a^7*b^11))/(b^13 - 2*a^2*b^11 + a^4*b^9))*(a^2*6i + b^2*1
i))/(2*b^4) + (8*tan(x/2)*(32*a^4*b^12 - 56*a^6*b^10 + 24*a^8*b^8))/(b^13 - 2*a^2*b^11 + a^4*b^9)))/(2*b^4))*1
i)/(2*b^4))/((16*(54*a^11 + 4*a^5*b^6 + 9*a^7*b^4 - 81*a^9*b^2))/(b^12 - 2*a^2*b^10 + a^4*b^8) - ((a^2*6i + b^
2*1i)*((8*(a^2*b^11 + 10*a^4*b^9 + 13*a^6*b^7 - 60*a^8*b^5 + 36*a^10*b^3))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*
tan(x/2)*(2*a*b^13 + 19*a^3*b^11 + 16*a^5*b^9 - 197*a^7*b^7 + 228*a^9*b^5 - 72*a^11*b^3))/(b^13 - 2*a^2*b^11 +
a^4*b^9) - ((a^2*6i + b^2*1i)*((8*(2*a*b^14 + 6*a^3*b^12 - 14*a^5*b^10 + 6*a^7*b^8))/(b^12 - 2*a^2*b^10 + a^4
*b^8) - (((8*(4*a^2*b^15 - 8*a^4*b^13 + 4*a^6*b^11))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*tan(x/2)*(12*a*b^17 -
32*a^3*b^15 + 28*a^5*b^13 - 8*a^7*b^11))/(b^13 - 2*a^2*b^11 + a^4*b^9))*(a^2*6i + b^2*1i))/(2*b^4) + (8*tan(x/
2)*(32*a^4*b^12 - 56*a^6*b^10 + 24*a^8*b^8))/(b^13 - 2*a^2*b^11 + a^4*b^9)))/(2*b^4)))/(2*b^4) + ((a^2*6i + b^
2*1i)*((8*(a^2*b^11 + 10*a^4*b^9 + 13*a^6*b^7 - 60*a^8*b^5 + 36*a^10*b^3))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*
tan(x/2)*(2*a*b^13 + 19*a^3*b^11 + 16*a^5*b^9 - 197*a^7*b^7 + 228*a^9*b^5 - 72*a^11*b^3))/(b^13 - 2*a^2*b^11 +
a^4*b^9) + ((a^2*6i + b^2*1i)*((8*(2*a*b^14 + 6*a^3*b^12 - 14*a^5*b^10 + 6*a^7*b^8))/(b^12 - 2*a^2*b^10 + a^4
*b^8) + (((8*(4*a^2*b^15 - 8*a^4*b^13 + 4*a^6*b^11))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*tan(x/2)*(12*a*b^17 -
32*a^3*b^15 + 28*a^5*b^13 - 8*a^7*b^11))/(b^13 - 2*a^2*b^11 + a^4*b^9))*(a^2*6i + b^2*1i))/(2*b^4) + (8*tan(x/
2)*(32*a^4*b^12 - 56*a^6*b^10 + 24*a^8*b^8))/(b^13 - 2*a^2*b^11 + a^4*b^9)))/(2*b^4)))/(2*b^4) + (16*tan(x/2)*
(216*a^12 + 8*a^4*b^8 + 82*a^6*b^6 + 126*a^8*b^4 - 432*a^10*b^2))/(b^13 - 2*a^2*b^11 + a^4*b^9)))*(a^2*6i + b^
2*1i)*1i)/b^4 - ((tan(x/2)*(7*a*b^2 - 9*a^3))/(b^2*(a^2 - b^2)) - (2*(3*a^4 - 2*a^2*b^2))/(b^3*(a^2 - b^2)) +
(4*tan(x/2)^3*(2*a*b^2 - 3*a^3))/(b^2*(a^2 - b^2)) + (tan(x/2)^5*(a*b^2 - 3*a^3))/(b^2*(a^2 - b^2)) + (2*tan(x
/2)^4*(b^4 - 3*a^4 + a^2*b^2))/(b^3*(a^2 - b^2)) - (2*tan(x/2)^2*(6*a^4 + b^4 - 5*a^2*b^2))/(b^3*(a^2 - b^2)))
/(a + 2*b*tan(x/2) + 3*a*tan(x/2)^2 + 3*a*tan(x/2)^4 + a*tan(x/2)^6 + 4*b*tan(x/2)^3 + 2*b*tan(x/2)^5) + (a^3*
atan(((a^3*(3*a^2 - 4*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((8*(a^2*b^11 + 10*a^4*b^9 + 13*a^6*b^7 - 60*a^8*b^5 +
36*a^10*b^3))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*tan(x/2)*(2*a*b^13 + 19*a^3*b^11 + 16*a^5*b^9 - 197*a^7*b^7
+ 228*a^9*b^5 - 72*a^11*b^3))/(b^13 - 2*a^2*b^11 + a^4*b^9) + (a^3*(3*a^2 - 4*b^2)*(-(a + b)^3*(a - b)^3)^(1/2
)*((8*(2*a*b^14 + 6*a^3*b^12 - 14*a^5*b^10 + 6*a^7*b^8))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*tan(x/2)*(32*a^4*b
^12 - 56*a^6*b^10 + 24*a^8*b^8))/(b^13 - 2*a^2*b^11 + a^4*b^9) + (a^3*((8*(4*a^2*b^15 - 8*a^4*b^13 + 4*a^6*b^1
1))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*tan(x/2)*(12*a*b^17 - 32*a^3*b^15 + 28*a^5*b^13 - 8*a^7*b^11))/(b^13 -
2*a^2*b^11 + a^4*b^9))*(3*a^2 - 4*b^2)*(-(a + b)^3*(a - b)^3)^(1/2))/(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4))
)/(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4))*1i)/(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4) + (a^3*(3*a^2 - 4*b^2
)*(-(a + b)^3*(a - b)^3)^(1/2)*((8*(a^2*b^11 + 10*a^4*b^9 + 13*a^6*b^7 - 60*a^8*b^5 + 36*a^10*b^3))/(b^12 - 2*
a^2*b^10 + a^4*b^8) + (8*tan(x/2)*(2*a*b^13 + 19*a^3*b^11 + 16*a^5*b^9 - 197*a^7*b^7 + 228*a^9*b^5 - 72*a^11*b
^3))/(b^13 - 2*a^2*b^11 + a^4*b^9) - (a^3*(3*a^2 - 4*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((8*(2*a*b^14 + 6*a^3*b
^12 - 14*a^5*b^10 + 6*a^7*b^8))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*tan(x/2)*(32*a^4*b^12 - 56*a^6*b^10 + 24*a^
8*b^8))/(b^13 - 2*a^2*b^11 + a^4*b^9) - (a^3*((8*(4*a^2*b^15 - 8*a^4*b^13 + 4*a^6*b^11))/(b^12 - 2*a^2*b^10 +
a^4*b^8) + (8*tan(x/2)*(12*a*b^17 - 32*a^3*b^15 + 28*a^5*b^13 - 8*a^7*b^11))/(b^13 - 2*a^2*b^11 + a^4*b^9))*(3
*a^2 - 4*b^2)*(-(a + b)^3*(a - b)^3)^(1/2))/(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4)))/(b^10 - 3*a^2*b^8 + 3*a
^4*b^6 - a^6*b^4))*1i)/(b^10 - 3*a^2*b^8 + 3*a^...

________________________________________________________________________________________