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

Optimal. Leaf size=84 \[ -\frac {\left (2 a^2+b^2\right ) \tanh ^{-1}(\cos (x))}{2 b^3}+\frac {2 a^3 \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2}}+\frac {a \cot (x)}{b^2}-\frac {\cot (x) \csc (x)}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {3851, 4082, 3998, 3770, 3831, 2660, 618, 206} \[ -\frac {\left (2 a^2+b^2\right ) \tanh ^{-1}(\cos (x))}{2 b^3}+\frac {2 a^3 \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2}}+\frac {a \cot (x)}{b^2}-\frac {\cot (x) \csc (x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^4/(a + b*Csc[x]),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.50, size = 144, normalized size = 1.71 \[ \frac {8 a^2 \log \left (\sin \left (\frac {x}{2}\right )\right )-8 a^2 \log \left (\cos \left (\frac {x}{2}\right )\right )-\frac {16 a^3 \tan ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}-4 a b \tan \left (\frac {x}{2}\right )+4 a b \cot \left (\frac {x}{2}\right )-b^2 \csc ^2\left (\frac {x}{2}\right )+b^2 \sec ^2\left (\frac {x}{2}\right )+4 b^2 \log \left (\sin \left (\frac {x}{2}\right )\right )-4 b^2 \log \left (\cos \left (\frac {x}{2}\right )\right )}{8 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^4/(a + b*Csc[x]),x]

[Out]

((-16*a^3*ArcTan[(a + b*Tan[x/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + 4*a*b*Cot[x/2] - b^2*Csc[x/2]^2 - 8*a^
2*Log[Cos[x/2]] - 4*b^2*Log[Cos[x/2]] + 8*a^2*Log[Sin[x/2]] + 4*b^2*Log[Sin[x/2]] + b^2*Sec[x/2]^2 - 4*a*b*Tan
[x/2])/(8*b^3)

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fricas [B]  time = 0.87, size = 524, normalized size = 6.24 \[ \left [\frac {4 \, {\left (a^{3} b - a b^{3}\right )} \cos \relax (x) \sin \relax (x) - 2 \, {\left (a^{3} \cos \relax (x)^{2} - a^{3}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \relax (x)^{2} + 2 \, a b \sin \relax (x) + a^{2} + b^{2} + 2 \, {\left (b \cos \relax (x) \sin \relax (x) + a \cos \relax (x)\right )} \sqrt {a^{2} - b^{2}}}{a^{2} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2}}\right ) - 2 \, {\left (a^{2} b^{2} - b^{4}\right )} \cos \relax (x) - {\left (2 \, a^{4} - a^{2} b^{2} - b^{4} - {\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} \cos \relax (x)^{2}\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + {\left (2 \, a^{4} - a^{2} b^{2} - b^{4} - {\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} \cos \relax (x)^{2}\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right )}{4 \, {\left (a^{2} b^{3} - b^{5} - {\left (a^{2} b^{3} - b^{5}\right )} \cos \relax (x)^{2}\right )}}, \frac {4 \, {\left (a^{3} b - a b^{3}\right )} \cos \relax (x) \sin \relax (x) - 4 \, {\left (a^{3} \cos \relax (x)^{2} - a^{3}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \sin \relax (x) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \relax (x)}\right ) - 2 \, {\left (a^{2} b^{2} - b^{4}\right )} \cos \relax (x) - {\left (2 \, a^{4} - a^{2} b^{2} - b^{4} - {\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} \cos \relax (x)^{2}\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + {\left (2 \, a^{4} - a^{2} b^{2} - b^{4} - {\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} \cos \relax (x)^{2}\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right )}{4 \, {\left (a^{2} b^{3} - b^{5} - {\left (a^{2} b^{3} - b^{5}\right )} \cos \relax (x)^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.53, size = 141, normalized size = 1.68 \[ -\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, x\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )} a^{3}}{\sqrt {-a^{2} + b^{2}} b^{3}} + \frac {b \tan \left (\frac {1}{2} \, x\right )^{2} - 4 \, a \tan \left (\frac {1}{2} \, x\right )}{8 \, b^{2}} + \frac {{\left (2 \, a^{2} + b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{2 \, b^{3}} - \frac {12 \, a^{2} \tan \left (\frac {1}{2} \, x\right )^{2} + 6 \, b^{2} \tan \left (\frac {1}{2} \, x\right )^{2} - 4 \, a b \tan \left (\frac {1}{2} \, x\right ) + b^{2}}{8 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(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)))*a^3/(sqrt(-a^2 + b^2)*b^3)
+ 1/8*(b*tan(1/2*x)^2 - 4*a*tan(1/2*x))/b^2 + 1/2*(2*a^2 + b^2)*log(abs(tan(1/2*x)))/b^3 - 1/8*(12*a^2*tan(1/2
*x)^2 + 6*b^2*tan(1/2*x)^2 - 4*a*b*tan(1/2*x) + b^2)/(b^3*tan(1/2*x)^2)

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maple [A]  time = 0.23, size = 112, normalized size = 1.33 \[ \frac {\tan ^{2}\left (\frac {x}{2}\right )}{8 b}-\frac {a \tan \left (\frac {x}{2}\right )}{2 b^{2}}-\frac {2 a^{3} \arctan \left (\frac {2 \tan \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{b^{3} \sqrt {-a^{2}+b^{2}}}-\frac {1}{8 b \tan \left (\frac {x}{2}\right )^{2}}+\frac {\ln \left (\tan \left (\frac {x}{2}\right )\right ) a^{2}}{b^{3}}+\frac {\ln \left (\tan \left (\frac {x}{2}\right )\right )}{2 b}+\frac {a}{2 b^{2} \tan \left (\frac {x}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/b*tan(1/2*x)^2-1/2/b^2*a*tan(1/2*x)-2/b^3*a^3/(-a^2+b^2)^(1/2)*arctan(1/2*(2*tan(1/2*x)*b+2*a)/(-a^2+b^2)^
(1/2))-1/8/b/tan(1/2*x)^2+1/b^3*ln(tan(1/2*x))*a^2+1/2/b*ln(tan(1/2*x))+1/2*a/b^2/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)^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 = 0.58, size = 515, normalized size = 6.13 \[ -\frac {b^2\,\left (\frac {\cos \relax (x)\,\sqrt {a^2-b^2}}{2}-\frac {\ln \left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )\,\sqrt {a^2-b^2}}{4}+\frac {\cos \left (2\,x\right )\,\ln \left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )\,\sqrt {a^2-b^2}}{4}\right )-\frac {a^2\,\ln \left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )\,\sqrt {a^2-b^2}}{2}-\frac {a\,b\,\sin \left (2\,x\right )\,\sqrt {a^2-b^2}}{2}+\frac {a^2\,\cos \left (2\,x\right )\,\ln \left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )\,\sqrt {a^2-b^2}}{2}+a^3\,\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 )\,1{}\mathrm {i}-a^3\,\cos \left (2\,x\right )\,\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 )\,1{}\mathrm {i}}{\frac {b^3\,\sqrt {a^2-b^2}}{2}-\frac {b^3\,\cos \left (2\,x\right )\,\sqrt {a^2-b^2}}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)^4*(a + b/sin(x))),x)

[Out]

-(a^3*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)))*1i + b^2*((cos(x)*(a^2 - b^2)^(1/2))/2 - (log(sin(x/2)/cos(x
/2))*(a^2 - b^2)^(1/2))/4 + (cos(2*x)*log(sin(x/2)/cos(x/2))*(a^2 - b^2)^(1/2))/4) - (a^2*log(sin(x/2)/cos(x/2
))*(a^2 - b^2)^(1/2))/2 - a^3*cos(2*x)*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)))*1i - (a*b*sin(2*x)*(a^2 - b
^2)^(1/2))/2 + (a^2*cos(2*x)*log(sin(x/2)/cos(x/2))*(a^2 - b^2)^(1/2))/2)/((b^3*(a^2 - b^2)^(1/2))/2 - (b^3*co
s(2*x)*(a^2 - b^2)^(1/2))/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(x)**4/(a + b*csc(x)), x)

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