\(\int \frac {\sin ^4(x)}{a+b \sin (x)} \, dx\) [176]

Optimal result

Integrand size = 13, antiderivative size = 110 \[ \int \frac {\sin ^4(x)}{a+b \sin (x)} \, dx=-\frac {a \left (2 a^2+b^2\right ) x}{2 b^4}+\frac {2 a^4 \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{b^4 \sqrt {a^2-b^2}}-\frac {\left (3 a^2+2 b^2\right ) \cos (x)}{3 b^3}+\frac {a \cos (x) \sin (x)}{2 b^2}-\frac {\cos (x) \sin ^2(x)}{3 b} \] Output:

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

Mathematica [A] (verified)

Time = 0.83 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.89 \[ \int \frac {\sin ^4(x)}{a+b \sin (x)} \, dx=\frac {-6 a \left (2 a^2+b^2\right ) x+\frac {24 a^4 \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-3 b \left (4 a^2+3 b^2\right ) \cos (x)+b^3 \cos (3 x)+3 a b^2 \sin (2 x)}{12 b^4} \] Input:

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

Output:

(-6*a*(2*a^2 + b^2)*x + (24*a^4*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/ 
Sqrt[a^2 - b^2] - 3*b*(4*a^2 + 3*b^2)*Cos[x] + b^3*Cos[3*x] + 3*a*b^2*Sin[ 
2*x])/(12*b^4)
 

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.21, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.077, Rules used = {3042, 3272, 3042, 3528, 25, 3042, 3502, 27, 3042, 3214, 3042, 3139, 1083, 217}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (x)^4}{a+b \sin (x)}dx\)

\(\Big \downarrow \) 3272

\(\displaystyle \frac {\int \frac {\sin (x) \left (-3 a \sin ^2(x)+2 b \sin (x)+2 a\right )}{a+b \sin (x)}dx}{3 b}-\frac {\sin ^2(x) \cos (x)}{3 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sin (x) \left (-3 a \sin (x)^2+2 b \sin (x)+2 a\right )}{a+b \sin (x)}dx}{3 b}-\frac {\sin ^2(x) \cos (x)}{3 b}\)

\(\Big \downarrow \) 3528

\(\displaystyle \frac {\frac {\int -\frac {3 a^2-b \sin (x) a-2 \left (3 a^2+2 b^2\right ) \sin ^2(x)}{a+b \sin (x)}dx}{2 b}+\frac {3 a \sin (x) \cos (x)}{2 b}}{3 b}-\frac {\sin ^2(x) \cos (x)}{3 b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {3 a \sin (x) \cos (x)}{2 b}-\frac {\int \frac {3 a^2-b \sin (x) a-2 \left (3 a^2+2 b^2\right ) \sin ^2(x)}{a+b \sin (x)}dx}{2 b}}{3 b}-\frac {\sin ^2(x) \cos (x)}{3 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 a \sin (x) \cos (x)}{2 b}-\frac {\int \frac {3 a^2-b \sin (x) a-2 \left (3 a^2+2 b^2\right ) \sin (x)^2}{a+b \sin (x)}dx}{2 b}}{3 b}-\frac {\sin ^2(x) \cos (x)}{3 b}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {3 a \sin (x) \cos (x)}{2 b}-\frac {\frac {\int \frac {3 \left (b a^2+\left (2 a^2+b^2\right ) \sin (x) a\right )}{a+b \sin (x)}dx}{b}+\frac {2 \left (3 a^2+2 b^2\right ) \cos (x)}{b}}{2 b}}{3 b}-\frac {\sin ^2(x) \cos (x)}{3 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 a \sin (x) \cos (x)}{2 b}-\frac {\frac {3 \int \frac {b a^2+\left (2 a^2+b^2\right ) \sin (x) a}{a+b \sin (x)}dx}{b}+\frac {2 \left (3 a^2+2 b^2\right ) \cos (x)}{b}}{2 b}}{3 b}-\frac {\sin ^2(x) \cos (x)}{3 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 a \sin (x) \cos (x)}{2 b}-\frac {\frac {3 \int \frac {b a^2+\left (2 a^2+b^2\right ) \sin (x) a}{a+b \sin (x)}dx}{b}+\frac {2 \left (3 a^2+2 b^2\right ) \cos (x)}{b}}{2 b}}{3 b}-\frac {\sin ^2(x) \cos (x)}{3 b}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {\frac {3 a \sin (x) \cos (x)}{2 b}-\frac {\frac {3 \left (\frac {a x \left (2 a^2+b^2\right )}{b}-\frac {2 a^4 \int \frac {1}{a+b \sin (x)}dx}{b}\right )}{b}+\frac {2 \left (3 a^2+2 b^2\right ) \cos (x)}{b}}{2 b}}{3 b}-\frac {\sin ^2(x) \cos (x)}{3 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 a \sin (x) \cos (x)}{2 b}-\frac {\frac {3 \left (\frac {a x \left (2 a^2+b^2\right )}{b}-\frac {2 a^4 \int \frac {1}{a+b \sin (x)}dx}{b}\right )}{b}+\frac {2 \left (3 a^2+2 b^2\right ) \cos (x)}{b}}{2 b}}{3 b}-\frac {\sin ^2(x) \cos (x)}{3 b}\)

\(\Big \downarrow \) 3139

\(\displaystyle \frac {\frac {3 a \sin (x) \cos (x)}{2 b}-\frac {\frac {3 \left (\frac {a x \left (2 a^2+b^2\right )}{b}-\frac {4 a^4 \int \frac {1}{a \tan ^2\left (\frac {x}{2}\right )+2 b \tan \left (\frac {x}{2}\right )+a}d\tan \left (\frac {x}{2}\right )}{b}\right )}{b}+\frac {2 \left (3 a^2+2 b^2\right ) \cos (x)}{b}}{2 b}}{3 b}-\frac {\sin ^2(x) \cos (x)}{3 b}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {\frac {3 a \sin (x) \cos (x)}{2 b}-\frac {\frac {3 \left (\frac {8 a^4 \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {x}{2}\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {x}{2}\right )\right )}{b}+\frac {a x \left (2 a^2+b^2\right )}{b}\right )}{b}+\frac {2 \left (3 a^2+2 b^2\right ) \cos (x)}{b}}{2 b}}{3 b}-\frac {\sin ^2(x) \cos (x)}{3 b}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {3 a \sin (x) \cos (x)}{2 b}-\frac {\frac {2 \left (3 a^2+2 b^2\right ) \cos (x)}{b}+\frac {3 \left (\frac {a x \left (2 a^2+b^2\right )}{b}-\frac {4 a^4 \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2}}\right )}{b}}{2 b}}{3 b}-\frac {\sin ^2(x) \cos (x)}{3 b}\)

Input:

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

Output:

-1/3*(Cos[x]*Sin[x]^2)/b + (-1/2*((3*((a*(2*a^2 + b^2)*x)/b - (4*a^4*ArcTa 
n[(2*b + 2*a*Tan[x/2])/(2*Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2])))/b + (2* 
(3*a^2 + 2*b^2)*Cos[x])/b)/b + (3*a*Cos[x]*Sin[x])/(2*b))/(3*b)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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 1083

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3139

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre 
eFactors[Tan[(c + d*x)/2], x]}, Simp[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 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(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 3272

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

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[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 3528

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] + Simp[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])))
 

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.30

method	result	size
default	\(\frac {2 a^{4} \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{4} \sqrt {a^{2}-b^{2}}}-\frac {2 \left (\frac {\frac {a \,b^{2} \tan \left (\frac {x}{2}\right )^{5}}{2}+a^{2} b \tan \left (\frac {x}{2}\right )^{4}+\left (2 a^{2} b +2 b^{3}\right ) \tan \left (\frac {x}{2}\right )^{2}-\frac {a \,b^{2} \tan \left (\frac {x}{2}\right )}{2}+a^{2} b +\frac {2 b^{3}}{3}}{\left (\tan \left (\frac {x}{2}\right )^{2}+1\right )^{3}}+\frac {a \left (2 a^{2}+b^{2}\right ) \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{2}\right )}{b^{4}}\)	\(143\)
risch	\(-\frac {a^{3} x}{b^{4}}-\frac {a x}{2 b^{2}}-\frac {{\mathrm e}^{i x} a^{2}}{2 b^{3}}-\frac {3 \,{\mathrm e}^{i x}}{8 b}-\frac {{\mathrm e}^{-i x} a^{2}}{2 b^{3}}-\frac {3 \,{\mathrm e}^{-i x}}{8 b}-\frac {a^{4} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, b^{4}}+\frac {a^{4} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, b^{4}}+\frac {\cos \left (3 x \right )}{12 b}+\frac {a \sin \left (2 x \right )}{4 b^{2}}\)	\(212\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 333, normalized size of antiderivative = 3.03 \[ \int \frac {\sin ^4(x)}{a+b \sin (x)} \, dx=\left [-\frac {3 \, \sqrt {-a^{2} + b^{2}} a^{4} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) - 2 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (x\right )^{3} - 3 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (x\right ) \sin \left (x\right ) + 3 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} x + 6 \, {\left (a^{4} b - b^{5}\right )} \cos \left (x\right )}{6 \, {\left (a^{2} b^{4} - b^{6}\right )}}, -\frac {6 \, \sqrt {a^{2} - b^{2}} a^{4} \arctan \left (-\frac {a \sin \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (x\right )}\right ) - 2 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (x\right )^{3} - 3 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (x\right ) \sin \left (x\right ) + 3 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} x + 6 \, {\left (a^{4} b - b^{5}\right )} \cos \left (x\right )}{6 \, {\left (a^{2} b^{4} - b^{6}\right )}}\right ] \] Input:

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

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^4(x)}{a+b \sin (x)} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sin ^4(x)}{a+b \sin (x)} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.35 \[ \int \frac {\sin ^4(x)}{a+b \sin (x)} \, dx=\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{4}}{\sqrt {a^{2} - b^{2}} b^{4}} - \frac {{\left (2 \, a^{3} + a b^{2}\right )} x}{2 \, b^{4}} - \frac {3 \, a b \tan \left (\frac {1}{2} \, x\right )^{5} + 6 \, a^{2} \tan \left (\frac {1}{2} \, x\right )^{4} + 12 \, a^{2} \tan \left (\frac {1}{2} \, x\right )^{2} + 12 \, b^{2} \tan \left (\frac {1}{2} \, x\right )^{2} - 3 \, a b \tan \left (\frac {1}{2} \, x\right ) + 6 \, a^{2} + 4 \, b^{2}}{3 \, {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{3} b^{3}} \] Input:

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

Output:

2*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - 
b^2)))*a^4/(sqrt(a^2 - b^2)*b^4) - 1/2*(2*a^3 + a*b^2)*x/b^4 - 1/3*(3*a*b* 
tan(1/2*x)^5 + 6*a^2*tan(1/2*x)^4 + 12*a^2*tan(1/2*x)^2 + 12*b^2*tan(1/2*x 
)^2 - 3*a*b*tan(1/2*x) + 6*a^2 + 4*b^2)/((tan(1/2*x)^2 + 1)^3*b^3)
 

Mupad [B] (verification not implemented)

Time = 17.43 (sec) , antiderivative size = 1075, normalized size of antiderivative = 9.77 \[ \int \frac {\sin ^4(x)}{a+b \sin (x)} \, dx=\text {Too large to display} \] Input:

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

Output:

- ((2*(3*a^2 + 2*b^2))/(3*b^3) + (a*tan(x/2)^5)/b^2 + (2*a^2*tan(x/2)^4)/b 
^3 + (4*tan(x/2)^2*(a^2 + b^2))/b^3 - (a*tan(x/2))/b^2)/(3*tan(x/2)^2 + 3* 
tan(x/2)^4 + tan(x/2)^6 + 1) - (a*atan((8*a^4*tan(x/2))/(8*a^4 + (40*a^6)/ 
b^2 + (48*a^8)/b^4) + (40*a^6*tan(x/2))/(40*a^6 + 8*a^4*b^2 + (48*a^8)/b^2 
) + (48*a^8*tan(x/2))/(48*a^8 + 8*a^4*b^4 + 40*a^6*b^2))*(2*a^2 + b^2))/b^ 
4 - (a^4*atan(((a^4*((8*(a^4*b^7 + 4*a^6*b^5 + 4*a^8*b^3))/b^8 + (8*tan(x/ 
2)*(2*a^3*b^9 + 7*a^5*b^7 + 4*a^7*b^5 - 8*a^9*b^3))/b^9 + (a^4*((8*(2*a^2* 
b^10 + 2*a^4*b^8))/b^8 + (64*a^5*tan(x/2))/b + (a^4*(32*a^2*b^3 + (8*tan(x 
/2)*(12*a*b^13 - 8*a^3*b^11))/b^9))/(b^4*(b^2 - a^2)^(1/2))))/(b^4*(b^2 - 
a^2)^(1/2)))*1i)/(b^4*(b^2 - a^2)^(1/2)) + (a^4*((8*(a^4*b^7 + 4*a^6*b^5 + 
 4*a^8*b^3))/b^8 + (8*tan(x/2)*(2*a^3*b^9 + 7*a^5*b^7 + 4*a^7*b^5 - 8*a^9* 
b^3))/b^9 - (a^4*((8*(2*a^2*b^10 + 2*a^4*b^8))/b^8 + (64*a^5*tan(x/2))/b - 
 (a^4*(32*a^2*b^3 + (8*tan(x/2)*(12*a*b^13 - 8*a^3*b^11))/b^9))/(b^4*(b^2 
- a^2)^(1/2))))/(b^4*(b^2 - a^2)^(1/2)))*1i)/(b^4*(b^2 - a^2)^(1/2)))/((16 
*(2*a^10 + a^8*b^2))/b^8 + (16*tan(x/2)*(8*a^11 + 2*a^7*b^4 + 8*a^9*b^2))/ 
b^9 + (a^4*((8*(a^4*b^7 + 4*a^6*b^5 + 4*a^8*b^3))/b^8 + (8*tan(x/2)*(2*a^3 
*b^9 + 7*a^5*b^7 + 4*a^7*b^5 - 8*a^9*b^3))/b^9 + (a^4*((8*(2*a^2*b^10 + 2* 
a^4*b^8))/b^8 + (64*a^5*tan(x/2))/b + (a^4*(32*a^2*b^3 + (8*tan(x/2)*(12*a 
*b^13 - 8*a^3*b^11))/b^9))/(b^4*(b^2 - a^2)^(1/2))))/(b^4*(b^2 - a^2)^(1/2 
))))/(b^4*(b^2 - a^2)^(1/2)) - (a^4*((8*(a^4*b^7 + 4*a^6*b^5 + 4*a^8*b^...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.52 \[ \int \frac {\sin ^4(x)}{a+b \sin (x)} \, dx=\frac {12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{4}-2 \cos \left (x \right ) \sin \left (x \right )^{2} a^{2} b^{3}+2 \cos \left (x \right ) \sin \left (x \right )^{2} b^{5}+3 \cos \left (x \right ) \sin \left (x \right ) a^{3} b^{2}-3 \cos \left (x \right ) \sin \left (x \right ) a \,b^{4}-6 \cos \left (x \right ) a^{4} b +2 \cos \left (x \right ) a^{2} b^{3}+4 \cos \left (x \right ) b^{5}-6 a^{5} x -2 a^{4} b +3 a^{3} b^{2} x -2 a^{2} b^{3}+3 a \,b^{4} x +4 b^{5}}{6 b^{4} \left (a^{2}-b^{2}\right )} \] Input:

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

Output:

(12*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2))*a**4 - 2*co 
s(x)*sin(x)**2*a**2*b**3 + 2*cos(x)*sin(x)**2*b**5 + 3*cos(x)*sin(x)*a**3* 
b**2 - 3*cos(x)*sin(x)*a*b**4 - 6*cos(x)*a**4*b + 2*cos(x)*a**2*b**3 + 4*c 
os(x)*b**5 - 6*a**5*x - 2*a**4*b + 3*a**3*b**2*x - 2*a**2*b**3 + 3*a*b**4* 
x + 4*b**5)/(6*b**4*(a**2 - b**2))