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

Optimal. Leaf size=144 \[ \frac {b \sin ^2(x) \cos (x)}{3 a^2}+\frac {2 b^5 \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a^5 \sqrt {a^2-b^2}}+\frac {b \left (2 a^2+3 b^2\right ) \cos (x)}{3 a^4}-\frac {\left (3 a^2+4 b^2\right ) \sin (x) \cos (x)}{8 a^3}+\frac {x \left (3 a^4+4 a^2 b^2+8 b^4\right )}{8 a^5}-\frac {\sin ^3(x) \cos (x)}{4 a} \]

[Out]

1/8*(3*a^4+4*a^2*b^2+8*b^4)*x/a^5+1/3*b*(2*a^2+3*b^2)*cos(x)/a^4-1/8*(3*a^2+4*b^2)*cos(x)*sin(x)/a^3+1/3*b*cos
(x)*sin(x)^2/a^2-1/4*cos(x)*sin(x)^3/a+2*b^5*arctanh((a+b*tan(1/2*x))/(a^2-b^2)^(1/2))/a^5/(a^2-b^2)^(1/2)

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Rubi [A]  time = 0.59, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {3853, 4104, 3919, 3831, 2660, 618, 206} \[ \frac {x \left (4 a^2 b^2+3 a^4+8 b^4\right )}{8 a^5}+\frac {b \left (2 a^2+3 b^2\right ) \cos (x)}{3 a^4}-\frac {\left (3 a^2+4 b^2\right ) \sin (x) \cos (x)}{8 a^3}+\frac {2 b^5 \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a^5 \sqrt {a^2-b^2}}+\frac {b \sin ^2(x) \cos (x)}{3 a^2}-\frac {\sin ^3(x) \cos (x)}{4 a} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^4/(a + b*Csc[x]),x]

[Out]

((3*a^4 + 4*a^2*b^2 + 8*b^4)*x)/(8*a^5) + (2*b^5*ArcTanh[(a + b*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^5*Sqrt[a^2 - b^
2]) + (b*(2*a^2 + 3*b^2)*Cos[x])/(3*a^4) - ((3*a^2 + 4*b^2)*Cos[x]*Sin[x])/(8*a^3) + (b*Cos[x]*Sin[x]^2)/(3*a^
2) - (Cos[x]*Sin[x]^3)/(4*a)

Rule 206

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

Rule 618

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 2660

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 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3853

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(Cot[e + f*
x]*(d*Csc[e + f*x])^n)/(a*f*n), x] - Dist[1/(a*d*n), Int[((d*Csc[e + f*x])^(n + 1)*Simp[b*n - a*(n + 1)*Csc[e
+ f*x] - b*(n + 1)*Csc[e + f*x]^2, x])/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 -
b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]

Rule 3919

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

Rule 4104

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n)/(a*f*n), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rubi steps

\begin {align*} \int \frac {\sin ^4(x)}{a+b \csc (x)} \, dx &=-\frac {\cos (x) \sin ^3(x)}{4 a}+\frac {\int \frac {\left (-4 b+3 a \csc (x)+3 b \csc ^2(x)\right ) \sin ^3(x)}{a+b \csc (x)} \, dx}{4 a}\\ &=\frac {b \cos (x) \sin ^2(x)}{3 a^2}-\frac {\cos (x) \sin ^3(x)}{4 a}-\frac {\int \frac {\left (-3 \left (3 a^2+4 b^2\right )-a b \csc (x)+8 b^2 \csc ^2(x)\right ) \sin ^2(x)}{a+b \csc (x)} \, dx}{12 a^2}\\ &=-\frac {\left (3 a^2+4 b^2\right ) \cos (x) \sin (x)}{8 a^3}+\frac {b \cos (x) \sin ^2(x)}{3 a^2}-\frac {\cos (x) \sin ^3(x)}{4 a}+\frac {\int \frac {\left (-8 b \left (2 a^2+3 b^2\right )+a \left (9 a^2-4 b^2\right ) \csc (x)+3 b \left (3 a^2+4 b^2\right ) \csc ^2(x)\right ) \sin (x)}{a+b \csc (x)} \, dx}{24 a^3}\\ &=\frac {b \left (2 a^2+3 b^2\right ) \cos (x)}{3 a^4}-\frac {\left (3 a^2+4 b^2\right ) \cos (x) \sin (x)}{8 a^3}+\frac {b \cos (x) \sin ^2(x)}{3 a^2}-\frac {\cos (x) \sin ^3(x)}{4 a}-\frac {\int \frac {-3 \left (3 a^4+4 a^2 b^2+8 b^4\right )-3 a b \left (3 a^2+4 b^2\right ) \csc (x)}{a+b \csc (x)} \, dx}{24 a^4}\\ &=\frac {\left (3 a^4+4 a^2 b^2+8 b^4\right ) x}{8 a^5}+\frac {b \left (2 a^2+3 b^2\right ) \cos (x)}{3 a^4}-\frac {\left (3 a^2+4 b^2\right ) \cos (x) \sin (x)}{8 a^3}+\frac {b \cos (x) \sin ^2(x)}{3 a^2}-\frac {\cos (x) \sin ^3(x)}{4 a}-\frac {b^5 \int \frac {\csc (x)}{a+b \csc (x)} \, dx}{a^5}\\ &=\frac {\left (3 a^4+4 a^2 b^2+8 b^4\right ) x}{8 a^5}+\frac {b \left (2 a^2+3 b^2\right ) \cos (x)}{3 a^4}-\frac {\left (3 a^2+4 b^2\right ) \cos (x) \sin (x)}{8 a^3}+\frac {b \cos (x) \sin ^2(x)}{3 a^2}-\frac {\cos (x) \sin ^3(x)}{4 a}-\frac {b^4 \int \frac {1}{1+\frac {a \sin (x)}{b}} \, dx}{a^5}\\ &=\frac {\left (3 a^4+4 a^2 b^2+8 b^4\right ) x}{8 a^5}+\frac {b \left (2 a^2+3 b^2\right ) \cos (x)}{3 a^4}-\frac {\left (3 a^2+4 b^2\right ) \cos (x) \sin (x)}{8 a^3}+\frac {b \cos (x) \sin ^2(x)}{3 a^2}-\frac {\cos (x) \sin ^3(x)}{4 a}-\frac {\left (2 b^4\right ) \operatorname {Subst}\left (\int \frac {1}{1+\frac {2 a x}{b}+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{a^5}\\ &=\frac {\left (3 a^4+4 a^2 b^2+8 b^4\right ) x}{8 a^5}+\frac {b \left (2 a^2+3 b^2\right ) \cos (x)}{3 a^4}-\frac {\left (3 a^2+4 b^2\right ) \cos (x) \sin (x)}{8 a^3}+\frac {b \cos (x) \sin ^2(x)}{3 a^2}-\frac {\cos (x) \sin ^3(x)}{4 a}+\frac {\left (4 b^4\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (1-\frac {a^2}{b^2}\right )-x^2} \, dx,x,\frac {2 a}{b}+2 \tan \left (\frac {x}{2}\right )\right )}{a^5}\\ &=\frac {\left (3 a^4+4 a^2 b^2+8 b^4\right ) x}{8 a^5}+\frac {2 b^5 \tanh ^{-1}\left (\frac {b \left (\frac {a}{b}+\tan \left (\frac {x}{2}\right )\right )}{\sqrt {a^2-b^2}}\right )}{a^5 \sqrt {a^2-b^2}}+\frac {b \left (2 a^2+3 b^2\right ) \cos (x)}{3 a^4}-\frac {\left (3 a^2+4 b^2\right ) \cos (x) \sin (x)}{8 a^3}+\frac {b \cos (x) \sin ^2(x)}{3 a^2}-\frac {\cos (x) \sin ^3(x)}{4 a}\\ \end {align*}

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Mathematica [A]  time = 0.31, size = 129, normalized size = 0.90 \[ \frac {36 a^4 x-24 a^4 \sin (2 x)+3 a^4 \sin (4 x)-8 a^3 b \cos (3 x)+48 a^2 b^2 x-24 a^2 b^2 \sin (2 x)+24 a b \left (3 a^2+4 b^2\right ) \cos (x)-\frac {192 b^5 \tan ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}+96 b^4 x}{96 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^4/(a + b*Csc[x]),x]

[Out]

(36*a^4*x + 48*a^2*b^2*x + 96*b^4*x - (192*b^5*ArcTan[(a + b*Tan[x/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + 2
4*a*b*(3*a^2 + 4*b^2)*Cos[x] - 8*a^3*b*Cos[3*x] - 24*a^4*Sin[2*x] - 24*a^2*b^2*Sin[2*x] + 3*a^4*Sin[4*x])/(96*
a^5)

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fricas [A]  time = 1.13, size = 410, normalized size = 2.85 \[ \left [\frac {12 \, \sqrt {a^{2} - b^{2}} b^{5} \log \left (\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \relax (x)^{2} + 2 \, a b \sin \relax (x) + a^{2} + b^{2} + 2 \, {\left (b \cos \relax (x) \sin \relax (x) + a \cos \relax (x)\right )} \sqrt {a^{2} - b^{2}}}{a^{2} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2}}\right ) - 8 \, {\left (a^{5} b - a^{3} b^{3}\right )} \cos \relax (x)^{3} + 3 \, {\left (3 \, a^{6} + a^{4} b^{2} + 4 \, a^{2} b^{4} - 8 \, b^{6}\right )} x + 24 \, {\left (a^{5} b - a b^{5}\right )} \cos \relax (x) + 3 \, {\left (2 \, {\left (a^{6} - a^{4} b^{2}\right )} \cos \relax (x)^{3} - {\left (5 \, a^{6} - a^{4} b^{2} - 4 \, a^{2} b^{4}\right )} \cos \relax (x)\right )} \sin \relax (x)}{24 \, {\left (a^{7} - a^{5} b^{2}\right )}}, \frac {24 \, \sqrt {-a^{2} + b^{2}} b^{5} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \sin \relax (x) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \relax (x)}\right ) - 8 \, {\left (a^{5} b - a^{3} b^{3}\right )} \cos \relax (x)^{3} + 3 \, {\left (3 \, a^{6} + a^{4} b^{2} + 4 \, a^{2} b^{4} - 8 \, b^{6}\right )} x + 24 \, {\left (a^{5} b - a b^{5}\right )} \cos \relax (x) + 3 \, {\left (2 \, {\left (a^{6} - a^{4} b^{2}\right )} \cos \relax (x)^{3} - {\left (5 \, a^{6} - a^{4} b^{2} - 4 \, a^{2} b^{4}\right )} \cos \relax (x)\right )} \sin \relax (x)}{24 \, {\left (a^{7} - a^{5} b^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*(12*sqrt(a^2 - b^2)*b^5*log(((a^2 - 2*b^2)*cos(x)^2 + 2*a*b*sin(x) + a^2 + b^2 + 2*(b*cos(x)*sin(x) + a*
cos(x))*sqrt(a^2 - b^2))/(a^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) - 8*(a^5*b - a^3*b^3)*cos(x)^3 + 3*(3*a^6
+ a^4*b^2 + 4*a^2*b^4 - 8*b^6)*x + 24*(a^5*b - a*b^5)*cos(x) + 3*(2*(a^6 - a^4*b^2)*cos(x)^3 - (5*a^6 - a^4*b^
2 - 4*a^2*b^4)*cos(x))*sin(x))/(a^7 - a^5*b^2), 1/24*(24*sqrt(-a^2 + b^2)*b^5*arctan(-sqrt(-a^2 + b^2)*(b*sin(
x) + a)/((a^2 - b^2)*cos(x))) - 8*(a^5*b - a^3*b^3)*cos(x)^3 + 3*(3*a^6 + a^4*b^2 + 4*a^2*b^4 - 8*b^6)*x + 24*
(a^5*b - a*b^5)*cos(x) + 3*(2*(a^6 - a^4*b^2)*cos(x)^3 - (5*a^6 - a^4*b^2 - 4*a^2*b^4)*cos(x))*sin(x))/(a^7 -
a^5*b^2)]

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giac [A]  time = 0.66, size = 252, normalized size = 1.75 \[ -\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, x\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )} b^{5}}{\sqrt {-a^{2} + b^{2}} a^{5}} + \frac {{\left (3 \, a^{4} + 4 \, a^{2} b^{2} + 8 \, b^{4}\right )} x}{8 \, a^{5}} + \frac {9 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{7} + 12 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{7} + 24 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{6} + 33 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{5} + 12 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{5} + 48 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{4} + 72 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{4} - 33 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{3} - 12 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} + 64 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{2} + 72 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{2} - 9 \, a^{3} \tan \left (\frac {1}{2} \, x\right ) - 12 \, a b^{2} \tan \left (\frac {1}{2} \, x\right ) + 16 \, a^{2} b + 24 \, b^{3}}{12 \, {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{4} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(pi*floor(1/2*x/pi + 1/2)*sgn(b) + arctan((b*tan(1/2*x) + a)/sqrt(-a^2 + b^2)))*b^5/(sqrt(-a^2 + b^2)*a^5)
+ 1/8*(3*a^4 + 4*a^2*b^2 + 8*b^4)*x/a^5 + 1/12*(9*a^3*tan(1/2*x)^7 + 12*a*b^2*tan(1/2*x)^7 + 24*b^3*tan(1/2*x)
^6 + 33*a^3*tan(1/2*x)^5 + 12*a*b^2*tan(1/2*x)^5 + 48*a^2*b*tan(1/2*x)^4 + 72*b^3*tan(1/2*x)^4 - 33*a^3*tan(1/
2*x)^3 - 12*a*b^2*tan(1/2*x)^3 + 64*a^2*b*tan(1/2*x)^2 + 72*b^3*tan(1/2*x)^2 - 9*a^3*tan(1/2*x) - 12*a*b^2*tan
(1/2*x) + 16*a^2*b + 24*b^3)/((tan(1/2*x)^2 + 1)^4*a^4)

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maple [B]  time = 0.38, size = 405, normalized size = 2.81 \[ -\frac {2 b^{5} \arctan \left (\frac {2 \tan \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{a^{5} \sqrt {-a^{2}+b^{2}}}+\frac {3 \left (\tan ^{7}\left (\frac {x}{2}\right )\right )}{4 a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {\left (\tan ^{7}\left (\frac {x}{2}\right )\right ) b^{2}}{a^{3} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {2 b^{3} \left (\tan ^{6}\left (\frac {x}{2}\right )\right )}{a^{4} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {\left (\tan ^{5}\left (\frac {x}{2}\right )\right ) b^{2}}{a^{3} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {11 \left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{4 a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {4 \left (\tan ^{4}\left (\frac {x}{2}\right )\right ) b}{a^{2} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {6 \left (\tan ^{4}\left (\frac {x}{2}\right )\right ) b^{3}}{a^{4} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}-\frac {\left (\tan ^{3}\left (\frac {x}{2}\right )\right ) b^{2}}{a^{3} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}-\frac {11 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{4 a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {6 \left (\tan ^{2}\left (\frac {x}{2}\right )\right ) b^{3}}{a^{4} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {16 \left (\tan ^{2}\left (\frac {x}{2}\right )\right ) b}{3 a^{2} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}-\frac {3 \tan \left (\frac {x}{2}\right )}{4 a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}-\frac {\tan \left (\frac {x}{2}\right ) b^{2}}{a^{3} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {4 b}{3 a^{2} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {2 b^{3}}{a^{4} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {2 \arctan \left (\tan \left (\frac {x}{2}\right )\right ) b^{4}}{a^{5}}+\frac {3 \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{4 a}+\frac {\arctan \left (\tan \left (\frac {x}{2}\right )\right ) b^{2}}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*b^5/a^5/(-a^2+b^2)^(1/2)*arctan(1/2*(2*tan(1/2*x)*b+2*a)/(-a^2+b^2)^(1/2))+3/4/a/(tan(1/2*x)^2+1)^4*tan(1/2
*x)^7+1/a^3/(tan(1/2*x)^2+1)^4*tan(1/2*x)^7*b^2+2/a^4/(tan(1/2*x)^2+1)^4*b^3*tan(1/2*x)^6+1/a^3/(tan(1/2*x)^2+
1)^4*tan(1/2*x)^5*b^2+11/4/a/(tan(1/2*x)^2+1)^4*tan(1/2*x)^5+4/a^2/(tan(1/2*x)^2+1)^4*tan(1/2*x)^4*b+6/a^4/(ta
n(1/2*x)^2+1)^4*tan(1/2*x)^4*b^3-1/a^3/(tan(1/2*x)^2+1)^4*tan(1/2*x)^3*b^2-11/4/a/(tan(1/2*x)^2+1)^4*tan(1/2*x
)^3+6/a^4/(tan(1/2*x)^2+1)^4*tan(1/2*x)^2*b^3+16/3/a^2/(tan(1/2*x)^2+1)^4*tan(1/2*x)^2*b-3/4/a/(tan(1/2*x)^2+1
)^4*tan(1/2*x)-1/a^3/(tan(1/2*x)^2+1)^4*tan(1/2*x)*b^2+4/3/a^2/(tan(1/2*x)^2+1)^4*b+2/a^4/(tan(1/2*x)^2+1)^4*b
^3+2/a^5*arctan(tan(1/2*x))*b^4+3/4/a*arctan(tan(1/2*x))+1/a^3*arctan(tan(1/2*x))*b^2

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^4/(a+b*csc(x)),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*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 1.26, size = 1639, normalized size = 11.38 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^4/(a + b/sin(x)),x)

[Out]

((2*(2*a^2*b + 3*b^3))/(3*a^4) - (tan(x/2)*(3*a^2 + 4*b^2))/(4*a^3) + (tan(x/2)^7*(3*a^2 + 4*b^2))/(4*a^3) - (
tan(x/2)^3*(11*a^2 + 4*b^2))/(4*a^3) + (tan(x/2)^5*(11*a^2 + 4*b^2))/(4*a^3) + (2*b^3*tan(x/2)^6)/a^4 + (2*tan
(x/2)^4*(2*a^2*b + 3*b^3))/a^4 + (2*tan(x/2)^2*(8*a^2*b + 9*b^3))/(3*a^4))/(4*tan(x/2)^2 + 6*tan(x/2)^4 + 4*ta
n(x/2)^6 + tan(x/2)^8 + 1) - (atan((81*b^3*tan(x/2))/(8*((27*a^2*b)/8 + (81*b^3)/8 + (63*b^5)/(2*a^2) + (35*b^
7)/a^4 + (40*b^9)/a^6)) + (63*b^5*tan(x/2))/(2*((27*a^4*b)/8 + (63*b^5)/2 + (81*a^2*b^3)/8 + (35*b^7)/a^2 + (4
0*b^9)/a^4)) + (35*b^7*tan(x/2))/((27*a^6*b)/8 + 35*b^7 + (63*a^2*b^5)/2 + (81*a^4*b^3)/8 + (40*b^9)/a^2) + (4
0*b^9*tan(x/2))/((27*a^8*b)/8 + 40*b^9 + 35*a^2*b^7 + (63*a^4*b^5)/2 + (81*a^6*b^3)/8) + (27*a*b*tan(x/2))/(8*
((27*a*b)/8 + (81*b^3)/(8*a) + (63*b^5)/(2*a^3) + (35*b^7)/a^5 + (40*b^9)/a^7)))*(a^4*3i + b^4*8i + a^2*b^2*4i
)*1i)/(4*a^5) + (b^5*atan(((b^5*(a^2 - b^2)^(1/2)*((32*a^4*b^10 + 32*a^6*b^8 + 32*a^8*b^6 + 12*a^10*b^4 + (9*a
^12*b^2)/2)/a^11 + (tan(x/2)*(18*a^14*b - 128*a^4*b^11 + 64*a^6*b^9 + 64*a^8*b^7 + 104*a^10*b^5 + 39*a^12*b^3)
)/(2*a^12) + (b^5*(a^2 - b^2)^(1/2)*((12*a^14*b + 16*a^10*b^5 + 4*a^12*b^3)/a^11 + (64*b^6*tan(x/2))/a^2 + (b^
5*(a^2 - b^2)^(1/2)*(32*a^3*b^2 + (tan(x/2)*(192*a^16*b - 128*a^14*b^3))/(2*a^12)))/(a^7 - a^5*b^2)))/(a^7 - a
^5*b^2))*1i)/(a^7 - a^5*b^2) + (b^5*(a^2 - b^2)^(1/2)*((32*a^4*b^10 + 32*a^6*b^8 + 32*a^8*b^6 + 12*a^10*b^4 +
(9*a^12*b^2)/2)/a^11 + (tan(x/2)*(18*a^14*b - 128*a^4*b^11 + 64*a^6*b^9 + 64*a^8*b^7 + 104*a^10*b^5 + 39*a^12*
b^3))/(2*a^12) - (b^5*(a^2 - b^2)^(1/2)*((12*a^14*b + 16*a^10*b^5 + 4*a^12*b^3)/a^11 + (64*b^6*tan(x/2))/a^2 -
(b^5*(a^2 - b^2)^(1/2)*(32*a^3*b^2 + (tan(x/2)*(192*a^16*b - 128*a^14*b^3))/(2*a^12)))/(a^7 - a^5*b^2)))/(a^7
- a^5*b^2))*1i)/(a^7 - a^5*b^2))/((32*b^13 + 40*a^2*b^11 + 24*a^4*b^9 + 9*a^6*b^7)/a^11 + (tan(x/2)*(128*b^14
+ 128*a^2*b^12 + 128*a^4*b^10 + 48*a^6*b^8 + 18*a^8*b^6))/a^12 + (b^5*(a^2 - b^2)^(1/2)*((32*a^4*b^10 + 32*a^
6*b^8 + 32*a^8*b^6 + 12*a^10*b^4 + (9*a^12*b^2)/2)/a^11 + (tan(x/2)*(18*a^14*b - 128*a^4*b^11 + 64*a^6*b^9 + 6
4*a^8*b^7 + 104*a^10*b^5 + 39*a^12*b^3))/(2*a^12) + (b^5*(a^2 - b^2)^(1/2)*((12*a^14*b + 16*a^10*b^5 + 4*a^12*
b^3)/a^11 + (64*b^6*tan(x/2))/a^2 + (b^5*(a^2 - b^2)^(1/2)*(32*a^3*b^2 + (tan(x/2)*(192*a^16*b - 128*a^14*b^3)
)/(2*a^12)))/(a^7 - a^5*b^2)))/(a^7 - a^5*b^2)))/(a^7 - a^5*b^2) - (b^5*(a^2 - b^2)^(1/2)*((32*a^4*b^10 + 32*a
^6*b^8 + 32*a^8*b^6 + 12*a^10*b^4 + (9*a^12*b^2)/2)/a^11 + (tan(x/2)*(18*a^14*b - 128*a^4*b^11 + 64*a^6*b^9 +
64*a^8*b^7 + 104*a^10*b^5 + 39*a^12*b^3))/(2*a^12) - (b^5*(a^2 - b^2)^(1/2)*((12*a^14*b + 16*a^10*b^5 + 4*a^12
*b^3)/a^11 + (64*b^6*tan(x/2))/a^2 - (b^5*(a^2 - b^2)^(1/2)*(32*a^3*b^2 + (tan(x/2)*(192*a^16*b - 128*a^14*b^3
))/(2*a^12)))/(a^7 - a^5*b^2)))/(a^7 - a^5*b^2)))/(a^7 - a^5*b^2)))*(a^2 - b^2)^(1/2)*2i)/(a^7 - a^5*b^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{4}{\relax (x )}}{a + b \csc {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(x)**4/(a + b*csc(x)), x)

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