3.183 \(\int \frac {\csc ^3(x)}{a+b \sin (x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {2802, 3055, 3001, 3770, 2660, 618, 204} \[ -\frac {2 b^3 \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{a^3 \sqrt {a^2-b^2}}-\frac {\left (a^2+2 b^2\right ) \tanh ^{-1}(\cos (x))}{2 a^3}+\frac {b \cot (x)}{a^2}-\frac {\cot (x) \csc (x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^3/(a + b*Sin[x]),x]

[Out]

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

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[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 2802

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 -
b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n
*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n
+ 3)*Sin[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] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !
(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*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] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

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

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Mathematica [A]  time = 0.52, size = 144, normalized size = 1.71 \[ \frac {-\frac {16 b^3 \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-a^2 \csc ^2\left (\frac {x}{2}\right )+a^2 \sec ^2\left (\frac {x}{2}\right )+4 a^2 \log \left (\sin \left (\frac {x}{2}\right )\right )-4 a^2 \log \left (\cos \left (\frac {x}{2}\right )\right )-4 a b \tan \left (\frac {x}{2}\right )+4 a b \cot \left (\frac {x}{2}\right )+8 b^2 \log \left (\sin \left (\frac {x}{2}\right )\right )-8 b^2 \log \left (\cos \left (\frac {x}{2}\right )\right )}{8 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^3/(a + b*Sin[x]),x]

[Out]

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

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fricas [B]  time = 0.69, size = 490, normalized size = 5.83 \[ \left [\frac {4 \, {\left (a^{3} b - a b^{3}\right )} \cos \relax (x) \sin \relax (x) + 2 \, {\left (b^{3} \cos \relax (x)^{2} - b^{3}\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2} - 2 \, {\left (a \cos \relax (x) \sin \relax (x) + b \cos \relax (x)\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2}}\right ) - 2 \, {\left (a^{4} - a^{2} b^{2}\right )} \cos \relax (x) - {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4} - {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} \cos \relax (x)^{2}\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4} - {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} \cos \relax (x)^{2}\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right )}{4 \, {\left (a^{5} - a^{3} b^{2} - {\left (a^{5} - a^{3} b^{2}\right )} \cos \relax (x)^{2}\right )}}, \frac {4 \, {\left (a^{3} b - a b^{3}\right )} \cos \relax (x) \sin \relax (x) - 4 \, {\left (b^{3} \cos \relax (x)^{2} - b^{3}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \relax (x) + b}{\sqrt {a^{2} - b^{2}} \cos \relax (x)}\right ) - 2 \, {\left (a^{4} - a^{2} b^{2}\right )} \cos \relax (x) - {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4} - {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} \cos \relax (x)^{2}\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4} - {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} \cos \relax (x)^{2}\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right )}{4 \, {\left (a^{5} - a^{3} b^{2} - {\left (a^{5} - a^{3} b^{2}\right )} \cos \relax (x)^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^3/(a+b*sin(x)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.26, size = 141, normalized size = 1.68 \[ -\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{3}}{\sqrt {a^{2} - b^{2}} a^{3}} + \frac {a \tan \left (\frac {1}{2} \, x\right )^{2} - 4 \, b \tan \left (\frac {1}{2} \, x\right )}{8 \, a^{2}} + \frac {{\left (a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{2 \, a^{3}} - \frac {6 \, a^{2} \tan \left (\frac {1}{2} \, x\right )^{2} + 12 \, b^{2} \tan \left (\frac {1}{2} \, x\right )^{2} - 4 \, a b \tan \left (\frac {1}{2} \, x\right ) + a^{2}}{8 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^3/(a+b*sin(x)),x, algorithm="giac")

[Out]

-2*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - b^2)))*b^3/(sqrt(a^2 - b^2)*a^3) +
1/8*(a*tan(1/2*x)^2 - 4*b*tan(1/2*x))/a^2 + 1/2*(a^2 + 2*b^2)*log(abs(tan(1/2*x)))/a^3 - 1/8*(6*a^2*tan(1/2*x)
^2 + 12*b^2*tan(1/2*x)^2 - 4*a*b*tan(1/2*x) + a^2)/(a^3*tan(1/2*x)^2)

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maple [A]  time = 0.12, size = 112, normalized size = 1.33 \[ \frac {\tan ^{2}\left (\frac {x}{2}\right )}{8 a}-\frac {\tan \left (\frac {x}{2}\right ) b}{2 a^{2}}-\frac {1}{8 a \tan \left (\frac {x}{2}\right )^{2}}+\frac {\ln \left (\tan \left (\frac {x}{2}\right )\right )}{2 a}+\frac {\ln \left (\tan \left (\frac {x}{2}\right )\right ) b^{2}}{a^{3}}+\frac {b}{2 a^{2} \tan \left (\frac {x}{2}\right )}-\frac {2 b^{3} \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{3} \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)^3/(a+b*sin(x)),x)

[Out]

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

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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(csc(x)^3/(a+b*sin(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*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 7.39, size = 531, normalized size = 6.32 \[ \frac {a^4\,\left (\frac {\cos \relax (x)}{2}-\frac {\ln \left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )}{4}+\frac {\cos \left (2\,x\right )\,\ln \left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )}{4}\right )-a^2\,\left (\frac {b^2\,\ln \left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )}{4}+\frac {b^2\,\cos \relax (x)}{2}-\frac {b^2\,\cos \left (2\,x\right )\,\ln \left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )}{4}\right )+\frac {b^4\,\ln \left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )}{2}-b^3\,\mathrm {atan}\left (\frac {-a^4\,\sin \left (\frac {x}{2}\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}+b^4\,\sin \left (\frac {x}{2}\right )\,\sqrt {b^2-a^2}\,8{}\mathrm {i}+a\,b^3\,\cos \left (\frac {x}{2}\right )\,\sqrt {b^2-a^2}\,4{}\mathrm {i}+a^3\,b\,\cos \left (\frac {x}{2}\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}}{\cos \left (\frac {x}{2}\right )\,a^5+2\,\sin \left (\frac {x}{2}\right )\,a^4\,b+\cos \left (\frac {x}{2}\right )\,a^3\,b^2+4\,\sin \left (\frac {x}{2}\right )\,a^2\,b^3-4\,\cos \left (\frac {x}{2}\right )\,a\,b^4-8\,\sin \left (\frac {x}{2}\right )\,b^5}\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}-\frac {b^4\,\cos \left (2\,x\right )\,\ln \left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )}{2}+\frac {a\,b^3\,\sin \left (2\,x\right )}{2}-\frac {a^3\,b\,\sin \left (2\,x\right )}{2}+b^3\,\cos \left (2\,x\right )\,\mathrm {atan}\left (\frac {-a^4\,\sin \left (\frac {x}{2}\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}+b^4\,\sin \left (\frac {x}{2}\right )\,\sqrt {b^2-a^2}\,8{}\mathrm {i}+a\,b^3\,\cos \left (\frac {x}{2}\right )\,\sqrt {b^2-a^2}\,4{}\mathrm {i}+a^3\,b\,\cos \left (\frac {x}{2}\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}}{\cos \left (\frac {x}{2}\right )\,a^5+2\,\sin \left (\frac {x}{2}\right )\,a^4\,b+\cos \left (\frac {x}{2}\right )\,a^3\,b^2+4\,\sin \left (\frac {x}{2}\right )\,a^2\,b^3-4\,\cos \left (\frac {x}{2}\right )\,a\,b^4-8\,\sin \left (\frac {x}{2}\right )\,b^5}\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}}{\frac {a^5\,\cos \left (2\,x\right )}{2}-\frac {a^5}{2}+\frac {a^3\,b^2}{2}-\frac {a^3\,b^2\,\cos \left (2\,x\right )}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)^3*(a + b*sin(x))),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{3}{\relax (x )}}{a + b \sin {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)**3/(a+b*sin(x)),x)

[Out]

Integral(csc(x)**3/(a + b*sin(x)), x)

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