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

Optimal. Leaf size=101 \[ \frac {a \left (a^2+b^2\right ) \tanh ^{-1}(\cos (x))}{b^4}+\frac {\left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac {\sin (x) (b-a \cot (x))}{\sqrt {a^2+b^2}}\right )}{b^4}-\frac {\left (a^2+b^2\right ) \csc (x)}{b^3}+\frac {a \tanh ^{-1}(\cos (x))}{2 b^2}+\frac {a \cot (x) \csc (x)}{2 b^2}-\frac {\csc ^3(x)}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3510, 3486, 3768, 3770, 3509, 206} \[ -\frac {\left (a^2+b^2\right ) \csc (x)}{b^3}+\frac {a \left (a^2+b^2\right ) \tanh ^{-1}(\cos (x))}{b^4}+\frac {\left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac {\sin (x) (b-a \cot (x))}{\sqrt {a^2+b^2}}\right )}{b^4}+\frac {a \tanh ^{-1}(\cos (x))}{2 b^2}+\frac {a \cot (x) \csc (x)}{2 b^2}-\frac {\csc ^3(x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^5/(a + b*Cot[x]),x]

[Out]

(a*ArcTanh[Cos[x]])/(2*b^2) + (a*(a^2 + b^2)*ArcTanh[Cos[x]])/b^4 + ((a^2 + b^2)^(3/2)*ArcTanh[((b - a*Cot[x])
*Sin[x])/Sqrt[a^2 + b^2]])/b^4 - ((a^2 + b^2)*Csc[x])/b^3 + (a*Cot[x]*Csc[x])/(2*b^2) - Csc[x]^3/(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 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3509

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

Rule 3510

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Dist[d^2/b^2, I
nt[(d*Sec[e + f*x])^(m - 2)*(a - b*Tan[e + f*x]), x], x] + Dist[(d^2*(a^2 + b^2))/b^2, Int[(d*Sec[e + f*x])^(m
- 2)/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 + b^2, 0] && IGtQ[m, 1]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

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

Rubi steps

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

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Mathematica [A]  time = 1.27, size = 198, normalized size = 1.96 \[ -\frac {48 a^3 \log \left (\sin \left (\frac {x}{2}\right )\right )-48 a^3 \log \left (\cos \left (\frac {x}{2}\right )\right )+4 b \left (6 a^2+7 b^2\right ) \cot \left (\frac {x}{2}\right )-96 \left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac {b \tan \left (\frac {x}{2}\right )-a}{\sqrt {a^2+b^2}}\right )+24 a^2 b \tan \left (\frac {x}{2}\right )-6 a b^2 \csc ^2\left (\frac {x}{2}\right )+6 a b^2 \sec ^2\left (\frac {x}{2}\right )+72 a b^2 \log \left (\sin \left (\frac {x}{2}\right )\right )-72 a b^2 \log \left (\cos \left (\frac {x}{2}\right )\right )+28 b^3 \tan \left (\frac {x}{2}\right )+b^3 \sin (x) \csc ^4\left (\frac {x}{2}\right )+16 b^3 \sin ^4\left (\frac {x}{2}\right ) \csc ^3(x)}{48 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^5/(a + b*Cot[x]),x]

[Out]

-1/48*(-96*(a^2 + b^2)^(3/2)*ArcTanh[(-a + b*Tan[x/2])/Sqrt[a^2 + b^2]] + 4*b*(6*a^2 + 7*b^2)*Cot[x/2] - 6*a*b
^2*Csc[x/2]^2 - 48*a^3*Log[Cos[x/2]] - 72*a*b^2*Log[Cos[x/2]] + 48*a^3*Log[Sin[x/2]] + 72*a*b^2*Log[Sin[x/2]]
+ 6*a*b^2*Sec[x/2]^2 + 16*b^3*Csc[x]^3*Sin[x/2]^4 + b^3*Csc[x/2]^4*Sin[x] + 24*a^2*b*Tan[x/2] + 28*b^3*Tan[x/2
])/b^4

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fricas [B]  time = 0.61, size = 264, normalized size = 2.61 \[ -\frac {6 \, a b^{2} \cos \relax (x) \sin \relax (x) - 6 \, {\left ({\left (a^{2} + b^{2}\right )} \cos \relax (x)^{2} - a^{2} - b^{2}\right )} \sqrt {a^{2} + b^{2}} \log \left (-\frac {2 \, a b \cos \relax (x) \sin \relax (x) - {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} - a^{2} - 2 \, b^{2} + 2 \, \sqrt {a^{2} + b^{2}} {\left (a \cos \relax (x) - b \sin \relax (x)\right )}}{2 \, a b \cos \relax (x) \sin \relax (x) - {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + a^{2}}\right ) \sin \relax (x) - 12 \, a^{2} b - 16 \, b^{3} + 12 \, {\left (a^{2} b + b^{3}\right )} \cos \relax (x)^{2} + 3 \, {\left (2 \, a^{3} + 3 \, a b^{2} - {\left (2 \, a^{3} + 3 \, a b^{2}\right )} \cos \relax (x)^{2}\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) \sin \relax (x) - 3 \, {\left (2 \, a^{3} + 3 \, a b^{2} - {\left (2 \, a^{3} + 3 \, a b^{2}\right )} \cos \relax (x)^{2}\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) \sin \relax (x)}{12 \, {\left (b^{4} \cos \relax (x)^{2} - b^{4}\right )} \sin \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(6*a*b^2*cos(x)*sin(x) - 6*((a^2 + b^2)*cos(x)^2 - a^2 - b^2)*sqrt(a^2 + b^2)*log(-(2*a*b*cos(x)*sin(x)
- (a^2 - b^2)*cos(x)^2 - a^2 - 2*b^2 + 2*sqrt(a^2 + b^2)*(a*cos(x) - b*sin(x)))/(2*a*b*cos(x)*sin(x) - (a^2 -
b^2)*cos(x)^2 + a^2))*sin(x) - 12*a^2*b - 16*b^3 + 12*(a^2*b + b^3)*cos(x)^2 + 3*(2*a^3 + 3*a*b^2 - (2*a^3 + 3
*a*b^2)*cos(x)^2)*log(1/2*cos(x) + 1/2)*sin(x) - 3*(2*a^3 + 3*a*b^2 - (2*a^3 + 3*a*b^2)*cos(x)^2)*log(-1/2*cos
(x) + 1/2)*sin(x))/((b^4*cos(x)^2 - b^4)*sin(x))

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giac [B]  time = 0.76, size = 220, normalized size = 2.18 \[ -\frac {b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} + 3 \, a b \tan \left (\frac {1}{2} \, x\right )^{2} + 12 \, a^{2} \tan \left (\frac {1}{2} \, x\right ) + 15 \, b^{2} \tan \left (\frac {1}{2} \, x\right )}{24 \, b^{3}} - \frac {{\left (2 \, a^{3} + 3 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{2 \, b^{4}} - \frac {{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (\frac {{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b^{4}} + \frac {44 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{3} + 66 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{2} - 15 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{2} + 3 \, a b^{2} \tan \left (\frac {1}{2} \, x\right ) - b^{3}}{24 \, b^{4} \tan \left (\frac {1}{2} \, x\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(b^2*tan(1/2*x)^3 + 3*a*b*tan(1/2*x)^2 + 12*a^2*tan(1/2*x) + 15*b^2*tan(1/2*x))/b^3 - 1/2*(2*a^3 + 3*a*b
^2)*log(abs(tan(1/2*x)))/b^4 - (a^4 + 2*a^2*b^2 + b^4)*log(abs(2*b*tan(1/2*x) - 2*a - 2*sqrt(a^2 + b^2))/abs(2
*b*tan(1/2*x) - 2*a + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*b^4) + 1/24*(44*a^3*tan(1/2*x)^3 + 66*a*b^2*tan(1/2
*x)^3 - 12*a^2*b*tan(1/2*x)^2 - 15*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 [B]  time = 0.26, size = 232, normalized size = 2.30 \[ -\frac {\tan ^{3}\left (\frac {x}{2}\right )}{24 b}-\frac {a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{8 b^{2}}-\frac {a^{2} \tan \left (\frac {x}{2}\right )}{2 b^{3}}-\frac {5 \tan \left (\frac {x}{2}\right )}{8 b}+\frac {2 \arctanh \left (\frac {2 \tan \left (\frac {x}{2}\right ) b -2 a}{2 \sqrt {a^{2}+b^{2}}}\right ) a^{4}}{b^{4} \sqrt {a^{2}+b^{2}}}+\frac {4 \arctanh \left (\frac {2 \tan \left (\frac {x}{2}\right ) b -2 a}{2 \sqrt {a^{2}+b^{2}}}\right ) a^{2}}{b^{2} \sqrt {a^{2}+b^{2}}}+\frac {2 \arctanh \left (\frac {2 \tan \left (\frac {x}{2}\right ) b -2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}-\frac {1}{24 b \tan \left (\frac {x}{2}\right )^{3}}-\frac {a^{2}}{2 b^{3} \tan \left (\frac {x}{2}\right )}-\frac {5}{8 b \tan \left (\frac {x}{2}\right )}+\frac {a}{8 b^{2} \tan \left (\frac {x}{2}\right )^{2}}-\frac {a^{3} \ln \left (\tan \left (\frac {x}{2}\right )\right )}{b^{4}}-\frac {3 a \ln \left (\tan \left (\frac {x}{2}\right )\right )}{2 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)^5/(a+b*cot(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)-5/8/b*tan(1/2*x)+2/b^4/(a^2+b^2)^(1/2)*arct
anh(1/2*(2*tan(1/2*x)*b-2*a)/(a^2+b^2)^(1/2))*a^4+4/b^2/(a^2+b^2)^(1/2)*arctanh(1/2*(2*tan(1/2*x)*b-2*a)/(a^2+
b^2)^(1/2))*a^2+2/(a^2+b^2)^(1/2)*arctanh(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^2-5/8/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))-3/2/b^2*a*ln(tan(1/2*x))

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maxima [B]  time = 0.61, size = 216, normalized size = 2.14 \[ -\frac {\frac {3 \, a b \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {b^{2} \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {3 \, {\left (4 \, a^{2} + 5 \, b^{2}\right )} \sin \relax (x)}{\cos \relax (x) + 1}}{24 \, b^{3}} - \frac {{\left (2 \, a^{3} + 3 \, a b^{2}\right )} \log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )}{2 \, b^{4}} - \frac {{\left (b^{2} - \frac {3 \, a b \sin \relax (x)}{\cos \relax (x) + 1} + \frac {3 \, {\left (4 \, a^{2} + 5 \, b^{2}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}\right )} {\left (\cos \relax (x) + 1\right )}^{3}}{24 \, b^{3} \sin \relax (x)^{3}} - \frac {{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (\frac {a - \frac {b \sin \relax (x)}{\cos \relax (x) + 1} + \sqrt {a^{2} + b^{2}}}{a - \frac {b \sin \relax (x)}{\cos \relax (x) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(3*a*b*sin(x)^2/(cos(x) + 1)^2 + b^2*sin(x)^3/(cos(x) + 1)^3 + 3*(4*a^2 + 5*b^2)*sin(x)/(cos(x) + 1))/b^
3 - 1/2*(2*a^3 + 3*a*b^2)*log(sin(x)/(cos(x) + 1))/b^4 - 1/24*(b^2 - 3*a*b*sin(x)/(cos(x) + 1) + 3*(4*a^2 + 5*
b^2)*sin(x)^2/(cos(x) + 1)^2)*(cos(x) + 1)^3/(b^3*sin(x)^3) - (a^4 + 2*a^2*b^2 + b^4)*log((a - b*sin(x)/(cos(x
) + 1) + sqrt(a^2 + b^2))/(a - b*sin(x)/(cos(x) + 1) - sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*b^4)

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mupad [B]  time = 0.62, size = 674, normalized size = 6.67 \[ -\mathrm {tan}\left (\frac {x}{2}\right )\,\left (\frac {5}{8\,b}+\frac {a^2}{2\,b^3}\right )-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{24\,b}-\frac {a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{8\,b^2}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (4\,a^2+5\,b^2\right )+\frac {b^2}{3}-a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}{8\,b^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}-\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )\,\left (a^3+\frac {3\,a\,b^2}{2}\right )}{b^4}-\frac {\mathrm {atan}\left (\frac {\frac {\sqrt {{\left (a^2+b^2\right )}^3}\,\left (\frac {4\,a^4\,b^4+7\,a^2\,b^6+2\,b^8}{b^6}+\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (8\,a^5\,b^2+16\,a^3\,b^4+7\,a\,b^6\right )}{b^5}+\frac {\sqrt {{\left (a^2+b^2\right )}^3}\,\left (2\,a\,b^2+\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (8\,a^2\,b^6+6\,b^8\right )}{b^5}\right )}{b^4}\right )\,1{}\mathrm {i}}{b^4}+\frac {\sqrt {{\left (a^2+b^2\right )}^3}\,\left (\frac {4\,a^4\,b^4+7\,a^2\,b^6+2\,b^8}{b^6}+\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (8\,a^5\,b^2+16\,a^3\,b^4+7\,a\,b^6\right )}{b^5}-\frac {\sqrt {{\left (a^2+b^2\right )}^3}\,\left (2\,a\,b^2+\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (8\,a^2\,b^6+6\,b^8\right )}{b^5}\right )}{b^4}\right )\,1{}\mathrm {i}}{b^4}}{\frac {2\,\left (2\,a^7+7\,a^5\,b^2+8\,a^3\,b^4+3\,a\,b^6\right )}{b^6}+\frac {\sqrt {{\left (a^2+b^2\right )}^3}\,\left (\frac {4\,a^4\,b^4+7\,a^2\,b^6+2\,b^8}{b^6}+\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (8\,a^5\,b^2+16\,a^3\,b^4+7\,a\,b^6\right )}{b^5}+\frac {\sqrt {{\left (a^2+b^2\right )}^3}\,\left (2\,a\,b^2+\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (8\,a^2\,b^6+6\,b^8\right )}{b^5}\right )}{b^4}\right )}{b^4}-\frac {\sqrt {{\left (a^2+b^2\right )}^3}\,\left (\frac {4\,a^4\,b^4+7\,a^2\,b^6+2\,b^8}{b^6}+\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (8\,a^5\,b^2+16\,a^3\,b^4+7\,a\,b^6\right )}{b^5}-\frac {\sqrt {{\left (a^2+b^2\right )}^3}\,\left (2\,a\,b^2+\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (8\,a^2\,b^6+6\,b^8\right )}{b^5}\right )}{b^4}\right )}{b^4}-\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (2\,a^6+8\,a^4\,b^2+10\,a^2\,b^4+4\,b^6\right )}{b^5}}\right )\,\sqrt {{\left (a^2+b^2\right )}^3}\,2{}\mathrm {i}}{b^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)^5*(a + b*cot(x))),x)

[Out]

- tan(x/2)*(5/(8*b) + a^2/(2*b^3)) - tan(x/2)^3/(24*b) - (a*tan(x/2)^2)/(8*b^2) - (tan(x/2)^2*(4*a^2 + 5*b^2)
+ b^2/3 - a*b*tan(x/2))/(8*b^3*tan(x/2)^3) - (atan(((((a^2 + b^2)^3)^(1/2)*((2*b^8 + 7*a^2*b^6 + 4*a^4*b^4)/b^
6 + (tan(x/2)*(7*a*b^6 + 16*a^3*b^4 + 8*a^5*b^2))/b^5 + (((a^2 + b^2)^3)^(1/2)*(2*a*b^2 + (tan(x/2)*(6*b^8 + 8
*a^2*b^6))/b^5))/b^4)*1i)/b^4 + (((a^2 + b^2)^3)^(1/2)*((2*b^8 + 7*a^2*b^6 + 4*a^4*b^4)/b^6 + (tan(x/2)*(7*a*b
^6 + 16*a^3*b^4 + 8*a^5*b^2))/b^5 - (((a^2 + b^2)^3)^(1/2)*(2*a*b^2 + (tan(x/2)*(6*b^8 + 8*a^2*b^6))/b^5))/b^4
)*1i)/b^4)/((2*(3*a*b^6 + 2*a^7 + 8*a^3*b^4 + 7*a^5*b^2))/b^6 + (((a^2 + b^2)^3)^(1/2)*((2*b^8 + 7*a^2*b^6 + 4
*a^4*b^4)/b^6 + (tan(x/2)*(7*a*b^6 + 16*a^3*b^4 + 8*a^5*b^2))/b^5 + (((a^2 + b^2)^3)^(1/2)*(2*a*b^2 + (tan(x/2
)*(6*b^8 + 8*a^2*b^6))/b^5))/b^4))/b^4 - (((a^2 + b^2)^3)^(1/2)*((2*b^8 + 7*a^2*b^6 + 4*a^4*b^4)/b^6 + (tan(x/
2)*(7*a*b^6 + 16*a^3*b^4 + 8*a^5*b^2))/b^5 - (((a^2 + b^2)^3)^(1/2)*(2*a*b^2 + (tan(x/2)*(6*b^8 + 8*a^2*b^6))/
b^5))/b^4))/b^4 - (2*tan(x/2)*(2*a^6 + 4*b^6 + 10*a^2*b^4 + 8*a^4*b^2))/b^5))*((a^2 + b^2)^3)^(1/2)*2i)/b^4 -
(log(tan(x/2))*((3*a*b^2)/2 + a^3))/b^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(x)**5/(a + b*cot(x)), x)

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