3.39 \(\int \frac {\csc ^5(x)}{a+b \csc (x)} \, dx\)

Optimal. Leaf size=112 \[ \frac {a \left (2 a^2+b^2\right ) \tanh ^{-1}(\cos (x))}{2 b^4}-\frac {\left (3 a^2+2 b^2\right ) \cot (x)}{3 b^3}-\frac {2 a^4 \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{b^4 \sqrt {a^2-b^2}}+\frac {a \cot (x) \csc (x)}{2 b^2}-\frac {\cot (x) \csc ^2(x)}{3 b} \]

[Out]

1/2*a*(2*a^2+b^2)*arctanh(cos(x))/b^4-1/3*(3*a^2+2*b^2)*cot(x)/b^3+1/2*a*cot(x)*csc(x)/b^2-1/3*cot(x)*csc(x)^2
/b-2*a^4*arctanh((a+b*tan(1/2*x))/(a^2-b^2)^(1/2))/b^4/(a^2-b^2)^(1/2)

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Rubi [A]  time = 0.41, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {3851, 4092, 4082, 3998, 3770, 3831, 2660, 618, 206} \[ -\frac {\left (3 a^2+2 b^2\right ) \cot (x)}{3 b^3}+\frac {a \left (2 a^2+b^2\right ) \tanh ^{-1}(\cos (x))}{2 b^4}-\frac {2 a^4 \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{b^4 \sqrt {a^2-b^2}}+\frac {a \cot (x) \csc (x)}{2 b^2}-\frac {\cot (x) \csc ^2(x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^5/(a + b*Csc[x]),x]

[Out]

(a*(2*a^2 + b^2)*ArcTanh[Cos[x]])/(2*b^4) - (2*a^4*ArcTanh[(a + b*Tan[x/2])/Sqrt[a^2 - b^2]])/(b^4*Sqrt[a^2 -
b^2]) - ((3*a^2 + 2*b^2)*Cot[x])/(3*b^3) + (a*Cot[x]*Csc[x])/(2*b^2) - (Cot[x]*Csc[x]^2)/(3*b)

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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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 3851

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[(d^3*Cot[e
+ f*x]*(d*Csc[e + f*x])^(n - 3))/(b*f*(n - 2)), x] + Dist[d^3/(b*(n - 2)), Int[((d*Csc[e + f*x])^(n - 3)*Simp
[a*(n - 3) + b*(n - 3)*Csc[e + f*x] - a*(n - 2)*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] && GtQ[n, 3]

Rule 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 4082

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

Rule 4092

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(
e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(C*Csc[e + f*x]*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m + 2)
+ A*(m + 3))*Csc[e + f*x] - (2*a*C - b*B*(m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m
}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\csc ^5(x)}{a+b \csc (x)} \, dx &=-\frac {\cot (x) \csc ^2(x)}{3 b}+\frac {\int \frac {\csc ^2(x) \left (2 a+2 b \csc (x)-3 a \csc ^2(x)\right )}{a+b \csc (x)} \, dx}{3 b}\\ &=\frac {a \cot (x) \csc (x)}{2 b^2}-\frac {\cot (x) \csc ^2(x)}{3 b}+\frac {\int \frac {\csc (x) \left (-3 a^2+a b \csc (x)+2 \left (3 a^2+2 b^2\right ) \csc ^2(x)\right )}{a+b \csc (x)} \, dx}{6 b^2}\\ &=-\frac {\left (3 a^2+2 b^2\right ) \cot (x)}{3 b^3}+\frac {a \cot (x) \csc (x)}{2 b^2}-\frac {\cot (x) \csc ^2(x)}{3 b}+\frac {\int \frac {\csc (x) \left (-3 a^2 b-3 a \left (2 a^2+b^2\right ) \csc (x)\right )}{a+b \csc (x)} \, dx}{6 b^3}\\ &=-\frac {\left (3 a^2+2 b^2\right ) \cot (x)}{3 b^3}+\frac {a \cot (x) \csc (x)}{2 b^2}-\frac {\cot (x) \csc ^2(x)}{3 b}+\frac {a^4 \int \frac {\csc (x)}{a+b \csc (x)} \, dx}{b^4}-\frac {\left (a \left (2 a^2+b^2\right )\right ) \int \csc (x) \, dx}{2 b^4}\\ &=\frac {a \left (2 a^2+b^2\right ) \tanh ^{-1}(\cos (x))}{2 b^4}-\frac {\left (3 a^2+2 b^2\right ) \cot (x)}{3 b^3}+\frac {a \cot (x) \csc (x)}{2 b^2}-\frac {\cot (x) \csc ^2(x)}{3 b}+\frac {a^4 \int \frac {1}{1+\frac {a \sin (x)}{b}} \, dx}{b^5}\\ &=\frac {a \left (2 a^2+b^2\right ) \tanh ^{-1}(\cos (x))}{2 b^4}-\frac {\left (3 a^2+2 b^2\right ) \cot (x)}{3 b^3}+\frac {a \cot (x) \csc (x)}{2 b^2}-\frac {\cot (x) \csc ^2(x)}{3 b}+\frac {\left (2 a^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 )}{b^5}\\ &=\frac {a \left (2 a^2+b^2\right ) \tanh ^{-1}(\cos (x))}{2 b^4}-\frac {\left (3 a^2+2 b^2\right ) \cot (x)}{3 b^3}+\frac {a \cot (x) \csc (x)}{2 b^2}-\frac {\cot (x) \csc ^2(x)}{3 b}-\frac {\left (4 a^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 )}{b^5}\\ &=\frac {a \left (2 a^2+b^2\right ) \tanh ^{-1}(\cos (x))}{2 b^4}-\frac {2 a^4 \tanh ^{-1}\left (\frac {b \left (\frac {a}{b}+\tan \left (\frac {x}{2}\right )\right )}{\sqrt {a^2-b^2}}\right )}{b^4 \sqrt {a^2-b^2}}-\frac {\left (3 a^2+2 b^2\right ) \cot (x)}{3 b^3}+\frac {a \cot (x) \csc (x)}{2 b^2}-\frac {\cot (x) \csc ^2(x)}{3 b}\\ \end {align*}

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Mathematica [A]  time = 1.73, size = 125, normalized size = 1.12 \[ \frac {b \left (3 a^2+2 b^2\right ) \cos (3 x) \csc ^3(x)-3 b \cot (x) \csc (x) \left (\left (a^2+2 b^2\right ) \csc (x)-2 a b\right )+6 a \left (2 a^2+b^2\right ) \left (\log \left (\cos \left (\frac {x}{2}\right )\right )-\log \left (\sin \left (\frac {x}{2}\right )\right )\right )+\frac {24 a^4 \tan ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}}{12 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^5/(a + b*Csc[x]),x]

[Out]

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

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fricas [B]  time = 0.74, size = 607, normalized size = 5.42 \[ \left [\frac {4 \, {\left (3 \, a^{4} b - a^{2} b^{3} - 2 \, b^{5}\right )} \cos \relax (x)^{3} - 6 \, {\left (a^{4} \cos \relax (x)^{2} - a^{4}\right )} \sqrt {a^{2} - b^{2}} \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 ) \sin \relax (x) + 6 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \relax (x) \sin \relax (x) + 3 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4} - {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \relax (x)^{2}\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) \sin \relax (x) - 3 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4} - {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \relax (x)^{2}\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) \sin \relax (x) - 12 \, {\left (a^{4} b - b^{5}\right )} \cos \relax (x)}{12 \, {\left (a^{2} b^{4} - b^{6} - {\left (a^{2} b^{4} - b^{6}\right )} \cos \relax (x)^{2}\right )} \sin \relax (x)}, \frac {4 \, {\left (3 \, a^{4} b - a^{2} b^{3} - 2 \, b^{5}\right )} \cos \relax (x)^{3} + 12 \, {\left (a^{4} \cos \relax (x)^{2} - a^{4}\right )} \sqrt {-a^{2} + b^{2}} \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 ) \sin \relax (x) + 6 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \relax (x) \sin \relax (x) + 3 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4} - {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \relax (x)^{2}\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) \sin \relax (x) - 3 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4} - {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \relax (x)^{2}\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) \sin \relax (x) - 12 \, {\left (a^{4} b - b^{5}\right )} \cos \relax (x)}{12 \, {\left (a^{2} b^{4} - b^{6} - {\left (a^{2} b^{4} - b^{6}\right )} \cos \relax (x)^{2}\right )} \sin \relax (x)}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^5/(a+b*csc(x)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.66, size = 194, normalized size = 1.73 \[ \frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, x\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )} a^{4}}{\sqrt {-a^{2} + b^{2}} b^{4}} + \frac {b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} - 3 \, a b \tan \left (\frac {1}{2} \, x\right )^{2} + 12 \, a^{2} \tan \left (\frac {1}{2} \, x\right ) + 9 \, b^{2} \tan \left (\frac {1}{2} \, x\right )}{24 \, b^{3}} - \frac {{\left (2 \, a^{3} + a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{2 \, b^{4}} + \frac {44 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{3} + 22 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{2} - 9 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{2} + 3 \, a b^{2} \tan \left (\frac {1}{2} \, x\right ) - b^{3}}{24 \, b^{4} \tan \left (\frac {1}{2} \, x\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^5/(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)))*a^4/(sqrt(-a^2 + b^2)*b^4) +
1/24*(b^2*tan(1/2*x)^3 - 3*a*b*tan(1/2*x)^2 + 12*a^2*tan(1/2*x) + 9*b^2*tan(1/2*x))/b^3 - 1/2*(2*a^3 + a*b^2)
*log(abs(tan(1/2*x)))/b^4 + 1/24*(44*a^3*tan(1/2*x)^3 + 22*a*b^2*tan(1/2*x)^3 - 12*a^2*b*tan(1/2*x)^2 - 9*b^3*
tan(1/2*x)^2 + 3*a*b^2*tan(1/2*x) - b^3)/(b^4*tan(1/2*x)^3)

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maple [A]  time = 0.27, size = 162, normalized size = 1.45 \[ \frac {\tan ^{3}\left (\frac {x}{2}\right )}{24 b}-\frac {a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{8 b^{2}}+\frac {a^{2} \tan \left (\frac {x}{2}\right )}{2 b^{3}}+\frac {3 \tan \left (\frac {x}{2}\right )}{8 b}+\frac {2 a^{4} \arctan \left (\frac {2 \tan \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{b^{4} \sqrt {-a^{2}+b^{2}}}-\frac {1}{24 b \tan \left (\frac {x}{2}\right )^{3}}-\frac {a^{2}}{2 b^{3} \tan \left (\frac {x}{2}\right )}-\frac {3}{8 b \tan \left (\frac {x}{2}\right )}+\frac {a}{8 b^{2} \tan \left (\frac {x}{2}\right )^{2}}-\frac {a^{3} \ln \left (\tan \left (\frac {x}{2}\right )\right )}{b^{4}}-\frac {a \ln \left (\tan \left (\frac {x}{2}\right )\right )}{2 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)^5/(a+b*csc(x)),x)

[Out]

1/24/b*tan(1/2*x)^3-1/8/b^2*a*tan(1/2*x)^2+1/2/b^3*a^2*tan(1/2*x)+3/8/b*tan(1/2*x)+2/b^4*a^4/(-a^2+b^2)^(1/2)*
arctan(1/2*(2*tan(1/2*x)*b+2*a)/(-a^2+b^2)^(1/2))-1/24/b/tan(1/2*x)^3-1/2/b^3/tan(1/2*x)*a^2-3/8/b/tan(1/2*x)+
1/8*a/b^2/tan(1/2*x)^2-1/b^4*a^3*ln(tan(1/2*x))-1/2/b^2*a*ln(tan(1/2*x))

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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(csc(x)^5/(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 = 0.79, size = 588, normalized size = 5.25 \[ -\frac {b^2\,\left (\frac {3\,a\,\sin \left (2\,x\right )\,\sqrt {a^2-b^2}}{4}+\frac {3\,a\,\sin \left (3\,x\right )\,\ln \left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )\,\sqrt {a^2-b^2}}{8}-\frac {9\,a\,\ln \left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )\,\sin \relax (x)\,\sqrt {a^2-b^2}}{8}\right )+b^3\,\left (\frac {\cos \left (3\,x\right )\,\sqrt {a^2-b^2}}{2}-\frac {3\,\cos \relax (x)\,\sqrt {a^2-b^2}}{2}\right )-b\,\left (\frac {3\,a^2\,\cos \relax (x)\,\sqrt {a^2-b^2}}{4}-\frac {3\,a^2\,\cos \left (3\,x\right )\,\sqrt {a^2-b^2}}{4}\right )-\frac {9\,a^3\,\ln \left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )\,\sin \relax (x)\,\sqrt {a^2-b^2}}{4}+\frac {3\,a^3\,\sin \left (3\,x\right )\,\ln \left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )\,\sqrt {a^2-b^2}}{4}+\frac {a^4\,\mathrm {atan}\left (\frac {a^4\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,8{}\mathrm {i}-b^4\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,1{}\mathrm {i}+a\,b^3\,\cos \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,1{}\mathrm {i}+a^3\,b\,\cos \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,4{}\mathrm {i}}{-8\,\sin \left (\frac {x}{2}\right )\,a^5-4\,\cos \left (\frac {x}{2}\right )\,a^4\,b+4\,\sin \left (\frac {x}{2}\right )\,a^3\,b^2+\cos \left (\frac {x}{2}\right )\,a^2\,b^3+2\,\sin \left (\frac {x}{2}\right )\,a\,b^4+\cos \left (\frac {x}{2}\right )\,b^5}\right )\,\sin \relax (x)\,9{}\mathrm {i}}{2}-\frac {a^4\,\mathrm {atan}\left (\frac {a^4\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,8{}\mathrm {i}-b^4\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,1{}\mathrm {i}+a\,b^3\,\cos \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,1{}\mathrm {i}+a^3\,b\,\cos \left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,4{}\mathrm {i}}{-8\,\sin \left (\frac {x}{2}\right )\,a^5-4\,\cos \left (\frac {x}{2}\right )\,a^4\,b+4\,\sin \left (\frac {x}{2}\right )\,a^3\,b^2+\cos \left (\frac {x}{2}\right )\,a^2\,b^3+2\,\sin \left (\frac {x}{2}\right )\,a\,b^4+\cos \left (\frac {x}{2}\right )\,b^5}\right )\,\sin \left (3\,x\right )\,3{}\mathrm {i}}{2}}{\frac {3\,b^4\,\sin \left (3\,x\right )\,\sqrt {a^2-b^2}}{4}-\frac {9\,b^4\,\sin \relax (x)\,\sqrt {a^2-b^2}}{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)^5*(a + b/sin(x))),x)

[Out]

-(b^2*((3*a*sin(2*x)*(a^2 - b^2)^(1/2))/4 + (3*a*sin(3*x)*log(sin(x/2)/cos(x/2))*(a^2 - b^2)^(1/2))/8 - (9*a*l
og(sin(x/2)/cos(x/2))*sin(x)*(a^2 - b^2)^(1/2))/8) + b^3*((cos(3*x)*(a^2 - b^2)^(1/2))/2 - (3*cos(x)*(a^2 - b^
2)^(1/2))/2) - b*((3*a^2*cos(x)*(a^2 - b^2)^(1/2))/4 - (3*a^2*cos(3*x)*(a^2 - b^2)^(1/2))/4) + (a^4*atan((a^4*
sin(x/2)*(a^2 - b^2)^(1/2)*8i - b^4*sin(x/2)*(a^2 - b^2)^(1/2)*1i + a*b^3*cos(x/2)*(a^2 - b^2)^(1/2)*1i + a^3*
b*cos(x/2)*(a^2 - b^2)^(1/2)*4i)/(b^5*cos(x/2) - 8*a^5*sin(x/2) + a^2*b^3*cos(x/2) + 4*a^3*b^2*sin(x/2) - 4*a^
4*b*cos(x/2) + 2*a*b^4*sin(x/2)))*sin(x)*9i)/2 - (a^4*atan((a^4*sin(x/2)*(a^2 - b^2)^(1/2)*8i - b^4*sin(x/2)*(
a^2 - b^2)^(1/2)*1i + a*b^3*cos(x/2)*(a^2 - b^2)^(1/2)*1i + a^3*b*cos(x/2)*(a^2 - b^2)^(1/2)*4i)/(b^5*cos(x/2)
- 8*a^5*sin(x/2) + a^2*b^3*cos(x/2) + 4*a^3*b^2*sin(x/2) - 4*a^4*b*cos(x/2) + 2*a*b^4*sin(x/2)))*sin(3*x)*3i)
/2 - (9*a^3*log(sin(x/2)/cos(x/2))*sin(x)*(a^2 - b^2)^(1/2))/4 + (3*a^3*sin(3*x)*log(sin(x/2)/cos(x/2))*(a^2 -
b^2)^(1/2))/4)/((3*b^4*sin(3*x)*(a^2 - b^2)^(1/2))/4 - (9*b^4*sin(x)*(a^2 - b^2)^(1/2))/4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{5}{\relax (x )}}{a + b \csc {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)**5/(a+b*csc(x)),x)

[Out]

Integral(csc(x)**5/(a + b*csc(x)), x)

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