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

Optimal. Leaf size=226 \[ \frac{\sqrt{2} \left (1-\frac{b}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\tan \left (\frac{x}{2}\right ) \left (b-\sqrt{b^2-4 a c}\right )+2 c}{\sqrt{2} \sqrt{-b \sqrt{b^2-4 a c}-2 c (a+c)+b^2}}\right )}{\sqrt{-b \sqrt{b^2-4 a c}-2 c (a+c)+b^2}}+\frac{\sqrt{2} \left (\frac{b}{\sqrt{b^2-4 a c}}+1\right ) \tan ^{-1}\left (\frac{\tan \left (\frac{x}{2}\right ) \left (\sqrt{b^2-4 a c}+b\right )+2 c}{\sqrt{2} \sqrt{b \sqrt{b^2-4 a c}-2 c (a+c)+b^2}}\right )}{\sqrt{b \sqrt{b^2-4 a c}-2 c (a+c)+b^2}} \]

[Out]

(Sqrt[2]*(1 - b/Sqrt[b^2 - 4*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]])])/Sqrt[b^2 - 2*c*(a + c) - b*Sqrt[b^2 - 4*a*c]] + (Sqrt[2]*(1 + b/Sqrt[b^2 - 4*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]])])/
Sqrt[b^2 - 2*c*(a + c) + b*Sqrt[b^2 - 4*a*c]]

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Rubi [A]  time = 0.548391, antiderivative size = 226, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {3256, 2660, 618, 204} \[ \frac{\sqrt{2} \left (1-\frac{b}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\tan \left (\frac{x}{2}\right ) \left (b-\sqrt{b^2-4 a c}\right )+2 c}{\sqrt{2} \sqrt{-b \sqrt{b^2-4 a c}-2 c (a+c)+b^2}}\right )}{\sqrt{-b \sqrt{b^2-4 a c}-2 c (a+c)+b^2}}+\frac{\sqrt{2} \left (\frac{b}{\sqrt{b^2-4 a c}}+1\right ) \tan ^{-1}\left (\frac{\tan \left (\frac{x}{2}\right ) \left (\sqrt{b^2-4 a c}+b\right )+2 c}{\sqrt{2} \sqrt{b \sqrt{b^2-4 a c}-2 c (a+c)+b^2}}\right )}{\sqrt{b \sqrt{b^2-4 a c}-2 c (a+c)+b^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*(1 - b/Sqrt[b^2 - 4*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]])])/Sqrt[b^2 - 2*c*(a + c) - b*Sqrt[b^2 - 4*a*c]] + (Sqrt[2]*(1 + b/Sqrt[b^2 - 4*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]])])/
Sqrt[b^2 - 2*c*(a + c) + b*Sqrt[b^2 - 4*a*c]]

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 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{\sin (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx &=\int \left (\frac{1-\frac{b}{\sqrt{b^2-4 a c}}}{b-\sqrt{b^2-4 a c}+2 c \sin (x)}+\frac{1+\frac{b}{\sqrt{b^2-4 a c}}}{b+\sqrt{b^2-4 a c}+2 c \sin (x)}\right ) \, dx\\ &=\left (1-\frac{b}{\sqrt{b^2-4 a c}}\right ) \int \frac{1}{b-\sqrt{b^2-4 a c}+2 c \sin (x)} \, dx+\left (1+\frac{b}{\sqrt{b^2-4 a c}}\right ) \int \frac{1}{b+\sqrt{b^2-4 a c}+2 c \sin (x)} \, dx\\ &=\left (2 \left (1-\frac{b}{\sqrt{b^2-4 a c}}\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 )+\left (2 \left (1+\frac{b}{\sqrt{b^2-4 a c}}\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 )\\ &=-\left (\left (4 \left (1-\frac{b}{\sqrt{b^2-4 a c}}\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 )\right )-\left (4 \left (1+\frac{b}{\sqrt{b^2-4 a c}}\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 )\\ &=\frac{\sqrt{2} \left (1-\frac{b}{\sqrt{b^2-4 a c}}\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 )}{\sqrt{b^2-2 c (a+c)-b \sqrt{b^2-4 a c}}}+\frac{\sqrt{2} \left (1+\frac{b}{\sqrt{b^2-4 a c}}\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 )}{\sqrt{b^2-2 c (a+c)+b \sqrt{b^2-4 a c}}}\\ \end{align*}

Mathematica [C]  time = 0.695455, size = 268, normalized size = 1.19 \[ \frac{\frac{\left (\sqrt{4 a c-b^2}+i b\right ) \tan ^{-1}\left (\frac{2 c+\tan \left (\frac{x}{2}\right ) \left (b-i \sqrt{4 a c-b^2}\right )}{\sqrt{2} \sqrt{-i b \sqrt{4 a c-b^2}-2 c (a+c)+b^2}}\right )}{\sqrt{-i b \sqrt{4 a c-b^2}-2 c (a+c)+b^2}}+\frac{\left (\sqrt{4 a c-b^2}-i b\right ) \tan ^{-1}\left (\frac{2 c+\tan \left (\frac{x}{2}\right ) \left (b+i \sqrt{4 a c-b^2}\right )}{\sqrt{2} \sqrt{i b \sqrt{4 a c-b^2}-2 c (a+c)+b^2}}\right )}{\sqrt{i b \sqrt{4 a c-b^2}-2 c (a+c)+b^2}}}{\sqrt{2 a c-\frac{b^2}{2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((I*b + 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*c*(a + c) - I*b*Sqrt[-b^2 + 4*a*c]] + (((-I)*b + 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*c*(a + c) + I*b*Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2/2 + 2*a*c]

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Maple [A]  time = 0.133, size = 216, normalized size = 1. \begin{align*} 8\,{\frac{a\sqrt{-4\,ca+{b}^{2}}}{ \left ( 8\,ca-2\,{b}^{2} \right ) \sqrt{4\,ca-2\,{b}^{2}+2\,b\sqrt{-4\,ca+{b}^{2}}+4\,{a}^{2}}}\arctan \left ({\frac{-2\,a\tan \left ( x/2 \right ) +\sqrt{-4\,ca+{b}^{2}}-b}{\sqrt{4\,ca-2\,{b}^{2}+2\,b\sqrt{-4\,ca+{b}^{2}}+4\,{a}^{2}}}} \right ) }+8\,{\frac{a\sqrt{-4\,ca+{b}^{2}}}{ \left ( 8\,ca-2\,{b}^{2} \right ) \sqrt{4\,ca-2\,{b}^{2}-2\,b\sqrt{-4\,ca+{b}^{2}}+4\,{a}^{2}}}\arctan \left ({\frac{2\,a\tan \left ( x/2 \right ) +b+\sqrt{-4\,ca+{b}^{2}}}{\sqrt{4\,ca-2\,{b}^{2}-2\,b\sqrt{-4\,ca+{b}^{2}}+4\,{a}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*a*(-4*a*c+b^2)^(1/2)/(8*a*c-2*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))+8*a*(-4*a*c+b^2)^(1/2)/(8*a*c-2*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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 4.92798, size = 7050, normalized size = 31.19 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(2)*sqrt(-(2*a^2 - b^2 + 2*a*c - (a^2*b^2 - b^4 - 4*a*c^3 - (8*a^2 - b^2)*c^2 - 2*(2*a^3 - 3*a*b^2)*c
)*sqrt(b^2/(a^4*b^2 - 2*a^2*b^4 + b^6 - 4*a*c^5 - (16*a^2 - b^2)*c^4 - 12*(2*a^3 - a*b^2)*c^3 - 2*(8*a^4 - 11*
a^2*b^2 + b^4)*c^2 - 4*(a^5 - 3*a^3*b^2 + 2*a*b^4)*c)))/(a^2*b^2 - b^4 - 4*a*c^3 - (8*a^2 - b^2)*c^2 - 2*(2*a^
3 - 3*a*b^2)*c))*log(2*a*b^2*sin(x) + 4*a*b*c + 2*(a^3*b^2 - a*b^4 - 4*a^2*c^3 - (8*a^3 - a*b^2)*c^2 - 2*(2*a^
4 - 3*a^2*b^2)*c)*sqrt(b^2/(a^4*b^2 - 2*a^2*b^4 + b^6 - 4*a*c^5 - (16*a^2 - b^2)*c^4 - 12*(2*a^3 - a*b^2)*c^3
- 2*(8*a^4 - 11*a^2*b^2 + b^4)*c^2 - 4*(a^5 - 3*a^3*b^2 + 2*a*b^4)*c))*sin(x) - sqrt(2)*((a^3*b^3 - a*b^5 + 4*
a*b*c^4 + (4*a^2*b - b^3)*c^3 - (4*a^3*b + 5*a*b^3)*c^2 - (4*a^4*b - 5*a^2*b^3 - b^5)*c)*sqrt(b^2/(a^4*b^2 - 2
*a^2*b^4 + b^6 - 4*a*c^5 - (16*a^2 - b^2)*c^4 - 12*(2*a^3 - a*b^2)*c^3 - 2*(8*a^4 - 11*a^2*b^2 + b^4)*c^2 - 4*
(a^5 - 3*a^3*b^2 + 2*a*b^4)*c))*cos(x) + (a*b^3 - 4*a*b*c^2 - (4*a^2*b - b^3)*c)*cos(x))*sqrt(-(2*a^2 - b^2 +
2*a*c - (a^2*b^2 - b^4 - 4*a*c^3 - (8*a^2 - b^2)*c^2 - 2*(2*a^3 - 3*a*b^2)*c)*sqrt(b^2/(a^4*b^2 - 2*a^2*b^4 +
b^6 - 4*a*c^5 - (16*a^2 - b^2)*c^4 - 12*(2*a^3 - a*b^2)*c^3 - 2*(8*a^4 - 11*a^2*b^2 + b^4)*c^2 - 4*(a^5 - 3*a^
3*b^2 + 2*a*b^4)*c)))/(a^2*b^2 - b^4 - 4*a*c^3 - (8*a^2 - b^2)*c^2 - 2*(2*a^3 - 3*a*b^2)*c))) + 1/4*sqrt(2)*sq
rt(-(2*a^2 - b^2 + 2*a*c + (a^2*b^2 - b^4 - 4*a*c^3 - (8*a^2 - b^2)*c^2 - 2*(2*a^3 - 3*a*b^2)*c)*sqrt(b^2/(a^4
*b^2 - 2*a^2*b^4 + b^6 - 4*a*c^5 - (16*a^2 - b^2)*c^4 - 12*(2*a^3 - a*b^2)*c^3 - 2*(8*a^4 - 11*a^2*b^2 + b^4)*
c^2 - 4*(a^5 - 3*a^3*b^2 + 2*a*b^4)*c)))/(a^2*b^2 - b^4 - 4*a*c^3 - (8*a^2 - b^2)*c^2 - 2*(2*a^3 - 3*a*b^2)*c)
)*log(2*a*b^2*sin(x) + 4*a*b*c - 2*(a^3*b^2 - a*b^4 - 4*a^2*c^3 - (8*a^3 - a*b^2)*c^2 - 2*(2*a^4 - 3*a^2*b^2)*
c)*sqrt(b^2/(a^4*b^2 - 2*a^2*b^4 + b^6 - 4*a*c^5 - (16*a^2 - b^2)*c^4 - 12*(2*a^3 - a*b^2)*c^3 - 2*(8*a^4 - 11
*a^2*b^2 + b^4)*c^2 - 4*(a^5 - 3*a^3*b^2 + 2*a*b^4)*c))*sin(x) - sqrt(2)*((a^3*b^3 - a*b^5 + 4*a*b*c^4 + (4*a^
2*b - b^3)*c^3 - (4*a^3*b + 5*a*b^3)*c^2 - (4*a^4*b - 5*a^2*b^3 - b^5)*c)*sqrt(b^2/(a^4*b^2 - 2*a^2*b^4 + b^6
- 4*a*c^5 - (16*a^2 - b^2)*c^4 - 12*(2*a^3 - a*b^2)*c^3 - 2*(8*a^4 - 11*a^2*b^2 + b^4)*c^2 - 4*(a^5 - 3*a^3*b^
2 + 2*a*b^4)*c))*cos(x) - (a*b^3 - 4*a*b*c^2 - (4*a^2*b - b^3)*c)*cos(x))*sqrt(-(2*a^2 - b^2 + 2*a*c + (a^2*b^
2 - b^4 - 4*a*c^3 - (8*a^2 - b^2)*c^2 - 2*(2*a^3 - 3*a*b^2)*c)*sqrt(b^2/(a^4*b^2 - 2*a^2*b^4 + b^6 - 4*a*c^5 -
(16*a^2 - b^2)*c^4 - 12*(2*a^3 - a*b^2)*c^3 - 2*(8*a^4 - 11*a^2*b^2 + b^4)*c^2 - 4*(a^5 - 3*a^3*b^2 + 2*a*b^4
)*c)))/(a^2*b^2 - b^4 - 4*a*c^3 - (8*a^2 - b^2)*c^2 - 2*(2*a^3 - 3*a*b^2)*c))) - 1/4*sqrt(2)*sqrt(-(2*a^2 - b^
2 + 2*a*c + (a^2*b^2 - b^4 - 4*a*c^3 - (8*a^2 - b^2)*c^2 - 2*(2*a^3 - 3*a*b^2)*c)*sqrt(b^2/(a^4*b^2 - 2*a^2*b^
4 + b^6 - 4*a*c^5 - (16*a^2 - b^2)*c^4 - 12*(2*a^3 - a*b^2)*c^3 - 2*(8*a^4 - 11*a^2*b^2 + b^4)*c^2 - 4*(a^5 -
3*a^3*b^2 + 2*a*b^4)*c)))/(a^2*b^2 - b^4 - 4*a*c^3 - (8*a^2 - b^2)*c^2 - 2*(2*a^3 - 3*a*b^2)*c))*log(-2*a*b^2*
sin(x) - 4*a*b*c + 2*(a^3*b^2 - a*b^4 - 4*a^2*c^3 - (8*a^3 - a*b^2)*c^2 - 2*(2*a^4 - 3*a^2*b^2)*c)*sqrt(b^2/(a
^4*b^2 - 2*a^2*b^4 + b^6 - 4*a*c^5 - (16*a^2 - b^2)*c^4 - 12*(2*a^3 - a*b^2)*c^3 - 2*(8*a^4 - 11*a^2*b^2 + b^4
)*c^2 - 4*(a^5 - 3*a^3*b^2 + 2*a*b^4)*c))*sin(x) - sqrt(2)*((a^3*b^3 - a*b^5 + 4*a*b*c^4 + (4*a^2*b - b^3)*c^3
- (4*a^3*b + 5*a*b^3)*c^2 - (4*a^4*b - 5*a^2*b^3 - b^5)*c)*sqrt(b^2/(a^4*b^2 - 2*a^2*b^4 + b^6 - 4*a*c^5 - (1
6*a^2 - b^2)*c^4 - 12*(2*a^3 - a*b^2)*c^3 - 2*(8*a^4 - 11*a^2*b^2 + b^4)*c^2 - 4*(a^5 - 3*a^3*b^2 + 2*a*b^4)*c
))*cos(x) - (a*b^3 - 4*a*b*c^2 - (4*a^2*b - b^3)*c)*cos(x))*sqrt(-(2*a^2 - b^2 + 2*a*c + (a^2*b^2 - b^4 - 4*a*
c^3 - (8*a^2 - b^2)*c^2 - 2*(2*a^3 - 3*a*b^2)*c)*sqrt(b^2/(a^4*b^2 - 2*a^2*b^4 + b^6 - 4*a*c^5 - (16*a^2 - b^2
)*c^4 - 12*(2*a^3 - a*b^2)*c^3 - 2*(8*a^4 - 11*a^2*b^2 + b^4)*c^2 - 4*(a^5 - 3*a^3*b^2 + 2*a*b^4)*c)))/(a^2*b^
2 - b^4 - 4*a*c^3 - (8*a^2 - b^2)*c^2 - 2*(2*a^3 - 3*a*b^2)*c))) + 1/4*sqrt(2)*sqrt(-(2*a^2 - b^2 + 2*a*c - (a
^2*b^2 - b^4 - 4*a*c^3 - (8*a^2 - b^2)*c^2 - 2*(2*a^3 - 3*a*b^2)*c)*sqrt(b^2/(a^4*b^2 - 2*a^2*b^4 + b^6 - 4*a*
c^5 - (16*a^2 - b^2)*c^4 - 12*(2*a^3 - a*b^2)*c^3 - 2*(8*a^4 - 11*a^2*b^2 + b^4)*c^2 - 4*(a^5 - 3*a^3*b^2 + 2*
a*b^4)*c)))/(a^2*b^2 - b^4 - 4*a*c^3 - (8*a^2 - b^2)*c^2 - 2*(2*a^3 - 3*a*b^2)*c))*log(-2*a*b^2*sin(x) - 4*a*b
*c - 2*(a^3*b^2 - a*b^4 - 4*a^2*c^3 - (8*a^3 - a*b^2)*c^2 - 2*(2*a^4 - 3*a^2*b^2)*c)*sqrt(b^2/(a^4*b^2 - 2*a^2
*b^4 + b^6 - 4*a*c^5 - (16*a^2 - b^2)*c^4 - 12*(2*a^3 - a*b^2)*c^3 - 2*(8*a^4 - 11*a^2*b^2 + b^4)*c^2 - 4*(a^5
- 3*a^3*b^2 + 2*a*b^4)*c))*sin(x) - sqrt(2)*((a^3*b^3 - a*b^5 + 4*a*b*c^4 + (4*a^2*b - b^3)*c^3 - (4*a^3*b +
5*a*b^3)*c^2 - (4*a^4*b - 5*a^2*b^3 - b^5)*c)*sqrt(b^2/(a^4*b^2 - 2*a^2*b^4 + b^6 - 4*a*c^5 - (16*a^2 - b^2)*c
^4 - 12*(2*a^3 - a*b^2)*c^3 - 2*(8*a^4 - 11*a^2*b^2 + b^4)*c^2 - 4*(a^5 - 3*a^3*b^2 + 2*a*b^4)*c))*cos(x) + (a
*b^3 - 4*a*b*c^2 - (4*a^2*b - b^3)*c)*cos(x))*sqrt(-(2*a^2 - b^2 + 2*a*c - (a^2*b^2 - b^4 - 4*a*c^3 - (8*a^2 -
b^2)*c^2 - 2*(2*a^3 - 3*a*b^2)*c)*sqrt(b^2/(a^4*b^2 - 2*a^2*b^4 + b^6 - 4*a*c^5 - (16*a^2 - b^2)*c^4 - 12*(2*
a^3 - a*b^2)*c^3 - 2*(8*a^4 - 11*a^2*b^2 + b^4)*c^2 - 4*(a^5 - 3*a^3*b^2 + 2*a*b^4)*c)))/(a^2*b^2 - b^4 - 4*a*
c^3 - (8*a^2 - b^2)*c^2 - 2*(2*a^3 - 3*a*b^2)*c)))

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Sympy [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(sin(x)/(a+b*sin(x)+c*sin(x)**2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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