3.8 \(\int \frac{\csc ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx\)

Optimal. Leaf size=331 \[ -\frac{\sqrt{2} c \left (\sqrt{b^2-4 a c} \left (b^2-a c\right )-3 a b c+b^3\right ) \tan ^{-1}\left (\frac{\tan \left (\frac{x}{2}\right ) \left (b-\sqrt{b^2-4 a c}\right )+2 c}{\sqrt{2} \sqrt{-b \sqrt{b^2-4 a c}-2 c (a+c)+b^2}}\right )}{a^3 \sqrt{b^2-4 a c} \sqrt{-b \sqrt{b^2-4 a c}-2 c (a+c)+b^2}}+\frac{\sqrt{2} c \left (-\sqrt{b^2-4 a c} \left (b^2-a c\right )-3 a b c+b^3\right ) \tan ^{-1}\left (\frac{\tan \left (\frac{x}{2}\right ) \left (\sqrt{b^2-4 a c}+b\right )+2 c}{\sqrt{2} \sqrt{b \sqrt{b^2-4 a c}-2 c (a+c)+b^2}}\right )}{a^3 \sqrt{b^2-4 a c} \sqrt{b \sqrt{b^2-4 a c}-2 c (a+c)+b^2}}-\frac{\left (b^2-a c\right ) \tanh ^{-1}(\cos (x))}{a^3}+\frac{b \cot (x)}{a^2}-\frac{\tanh ^{-1}(\cos (x))}{2 a}-\frac{\cot (x) \csc (x)}{2 a} \]

[Out]

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

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Rubi [A]  time = 3.21015, antiderivative size = 331, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.474, Rules used = {3256, 3770, 3767, 8, 3768, 3292, 2660, 618, 204} \[ -\frac{\sqrt{2} c \left (\sqrt{b^2-4 a c} \left (b^2-a c\right )-3 a b c+b^3\right ) \tan ^{-1}\left (\frac{\tan \left (\frac{x}{2}\right ) \left (b-\sqrt{b^2-4 a c}\right )+2 c}{\sqrt{2} \sqrt{-b \sqrt{b^2-4 a c}-2 c (a+c)+b^2}}\right )}{a^3 \sqrt{b^2-4 a c} \sqrt{-b \sqrt{b^2-4 a c}-2 c (a+c)+b^2}}+\frac{\sqrt{2} c \left (-\sqrt{b^2-4 a c} \left (b^2-a c\right )-3 a b c+b^3\right ) \tan ^{-1}\left (\frac{\tan \left (\frac{x}{2}\right ) \left (\sqrt{b^2-4 a c}+b\right )+2 c}{\sqrt{2} \sqrt{b \sqrt{b^2-4 a c}-2 c (a+c)+b^2}}\right )}{a^3 \sqrt{b^2-4 a c} \sqrt{b \sqrt{b^2-4 a c}-2 c (a+c)+b^2}}-\frac{\left (b^2-a c\right ) \tanh ^{-1}(\cos (x))}{a^3}+\frac{b \cot (x)}{a^2}-\frac{\tanh ^{-1}(\cos (x))}{2 a}-\frac{\cot (x) \csc (x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^3/(a + b*Sin[x] + c*Sin[x]^2),x]

[Out]

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

Rule 3256

Int[sin[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*sin[(d_.) + (e_.)*(x_)]^(n_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]
^(n2_.))^(p_), x_Symbol] :> Int[ExpandTrig[sin[d + e*x]^m*(a + b*sin[d + e*x]^n + c*sin[d + e*x]^(2*n))^p, x],
x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IntegersQ[m, n, p]

Rule 3770

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 3292

Int[((A_) + (B_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + (b_.)*sin[(d_.) + (e_.)*(x_)] + (c_.)*sin[(d_.) + (e_.)*(x
_)]^2), x_Symbol] :> Module[{q = Rt[b^2 - 4*a*c, 2]}, Dist[B + (b*B - 2*A*c)/q, Int[1/(b + q + 2*c*Sin[d + e*x
]), x], x] + Dist[B - (b*B - 2*A*c)/q, Int[1/(b - q + 2*c*Sin[d + e*x]), x], x]] /; FreeQ[{a, b, c, d, e, A, B
}, 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 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 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])

Rubi steps

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

Mathematica [C]  time = 1.63299, size = 481, normalized size = 1.45 \[ \frac{\csc ^2(x) (-2 a-2 b \sin (x)+c \cos (2 x)-c) \left (-4 \left (a^2-2 a c+2 b^2\right ) \log \left (\sin \left (\frac{x}{2}\right )\right )+4 \left (a^2-2 a c+2 b^2\right ) \log \left (\cos \left (\frac{x}{2}\right )\right )+a^2 \csc ^2\left (\frac{x}{2}\right )-a^2 \sec ^2\left (\frac{x}{2}\right )+\frac{8 c \left (b^2 \sqrt{4 a c-b^2}-a c \sqrt{4 a c-b^2}+3 i a b c-i b^3\right ) \tan ^{-1}\left (\frac{2 c+\tan \left (\frac{x}{2}\right ) \left (b-i \sqrt{4 a c-b^2}\right )}{\sqrt{2} \sqrt{-i b \sqrt{4 a c-b^2}-2 c (a+c)+b^2}}\right )}{\sqrt{2 a c-\frac{b^2}{2}} \sqrt{-i b \sqrt{4 a c-b^2}-2 c (a+c)+b^2}}+\frac{8 c \left (b^2 \sqrt{4 a c-b^2}-a c \sqrt{4 a c-b^2}-3 i a b c+i b^3\right ) \tan ^{-1}\left (\frac{2 c+\tan \left (\frac{x}{2}\right ) \left (b+i \sqrt{4 a c-b^2}\right )}{\sqrt{2} \sqrt{i b \sqrt{4 a c-b^2}-2 c (a+c)+b^2}}\right )}{\sqrt{2 a c-\frac{b^2}{2}} \sqrt{i b \sqrt{4 a c-b^2}-2 c (a+c)+b^2}}+4 a b \tan \left (\frac{x}{2}\right )-4 a b \cot \left (\frac{x}{2}\right )\right )}{16 a^3 \left (a \csc ^2(x)+b \csc (x)+c\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^3/(a + b*Sin[x] + c*Sin[x]^2),x]

[Out]

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

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Maple [B]  time = 0.184, size = 1369, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)^3/(a+b*sin(x)+c*sin(x)^2),x)

[Out]

1/8/a*tan(1/2*x)^2-1/2/a^2*tan(1/2*x)*b+4/a/(4*a*c-b^2)/(4*c*a-2*b^2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)*arcta
n((-2*a*tan(1/2*x)+(-4*a*c+b^2)^(1/2)-b)/(4*c*a-2*b^2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2))*(-4*a*c+b^2)^(1/2)*
c^2-8/a^2/(4*a*c-b^2)/(4*c*a-2*b^2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)*arctan((-2*a*tan(1/2*x)+(-4*a*c+b^2)^(1
/2)-b)/(4*c*a-2*b^2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2))*(-4*a*c+b^2)^(1/2)*b^2*c+2/a^3/(4*a*c-b^2)/(4*c*a-2*b
^2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)*arctan((-2*a*tan(1/2*x)+(-4*a*c+b^2)^(1/2)-b)/(4*c*a-2*b^2+2*b*(-4*a*c+
b^2)^(1/2)+4*a^2)^(1/2))*(-4*a*c+b^2)^(1/2)*b^4-16/a/(4*a*c-b^2)/(4*c*a-2*b^2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1
/2)*arctan((-2*a*tan(1/2*x)+(-4*a*c+b^2)^(1/2)-b)/(4*c*a-2*b^2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2))*b*c^2+12/a
^2/(4*a*c-b^2)/(4*c*a-2*b^2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)*arctan((-2*a*tan(1/2*x)+(-4*a*c+b^2)^(1/2)-b)/
(4*c*a-2*b^2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2))*b^3*c-2/a^3/(4*a*c-b^2)/(4*c*a-2*b^2+2*b*(-4*a*c+b^2)^(1/2)+
4*a^2)^(1/2)*arctan((-2*a*tan(1/2*x)+(-4*a*c+b^2)^(1/2)-b)/(4*c*a-2*b^2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2))*b
^5+4/a/(4*a*c-b^2)/(4*c*a-2*b^2-2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)*arctan((2*a*tan(1/2*x)+b+(-4*a*c+b^2)^(1/2
))/(4*c*a-2*b^2-2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2))*(-4*a*c+b^2)^(1/2)*c^2-8/a^2/(4*a*c-b^2)/(4*c*a-2*b^2-2*b
*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)*arctan((2*a*tan(1/2*x)+b+(-4*a*c+b^2)^(1/2))/(4*c*a-2*b^2-2*b*(-4*a*c+b^2)^(1
/2)+4*a^2)^(1/2))*(-4*a*c+b^2)^(1/2)*b^2*c+2/a^3/(4*a*c-b^2)/(4*c*a-2*b^2-2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)*
arctan((2*a*tan(1/2*x)+b+(-4*a*c+b^2)^(1/2))/(4*c*a-2*b^2-2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2))*(-4*a*c+b^2)^(1
/2)*b^4+16/a/(4*a*c-b^2)/(4*c*a-2*b^2-2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)*arctan((2*a*tan(1/2*x)+b+(-4*a*c+b^2
)^(1/2))/(4*c*a-2*b^2-2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2))*b*c^2-12/a^2/(4*a*c-b^2)/(4*c*a-2*b^2-2*b*(-4*a*c+b
^2)^(1/2)+4*a^2)^(1/2)*arctan((2*a*tan(1/2*x)+b+(-4*a*c+b^2)^(1/2))/(4*c*a-2*b^2-2*b*(-4*a*c+b^2)^(1/2)+4*a^2)
^(1/2))*b^3*c+2/a^3/(4*a*c-b^2)/(4*c*a-2*b^2-2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)*arctan((2*a*tan(1/2*x)+b+(-4*
a*c+b^2)^(1/2))/(4*c*a-2*b^2-2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2))*b^5-1/8/a/tan(1/2*x)^2+1/2/a*ln(tan(1/2*x))-
1/a^2*ln(tan(1/2*x))*c+1/a^3*ln(tan(1/2*x))*b^2+1/2*b/a^2/tan(1/2*x)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^3/(a+b*sin(x)+c*sin(x)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^3/(a+b*sin(x)+c*sin(x)^2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{3}{\left (x \right )}}{a + b \sin{\left (x \right )} + c \sin ^{2}{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)**3/(a+b*sin(x)+c*sin(x)**2),x)

[Out]

Integral(csc(x)**3/(a + b*sin(x) + c*sin(x)**2), x)

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^3/(a+b*sin(x)+c*sin(x)^2),x, algorithm="giac")

[Out]

Timed out