3.443 \(\int \frac {\sin (x)}{a+b \cos (x)+c \sin (x)} \, dx\)

Optimal. Leaf size=101 \[ -\frac {2 a c \tan ^{-1}\left (\frac {(a-b) \tan \left (\frac {x}{2}\right )+c}{\sqrt {a^2-b^2-c^2}}\right )}{\left (b^2+c^2\right ) \sqrt {a^2-b^2-c^2}}-\frac {b \log (a+b \cos (x)+c \sin (x))}{b^2+c^2}+\frac {c x}{b^2+c^2} \]

[Out]

c*x/(b^2+c^2)-b*ln(a+b*cos(x)+c*sin(x))/(b^2+c^2)-2*a*c*arctan((c+(a-b)*tan(1/2*x))/(a^2-b^2-c^2)^(1/2))/(b^2+
c^2)/(a^2-b^2-c^2)^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3137, 3124, 618, 204} \[ -\frac {2 a c \tan ^{-1}\left (\frac {(a-b) \tan \left (\frac {x}{2}\right )+c}{\sqrt {a^2-b^2-c^2}}\right )}{\left (b^2+c^2\right ) \sqrt {a^2-b^2-c^2}}-\frac {b \log (a+b \cos (x)+c \sin (x))}{b^2+c^2}+\frac {c x}{b^2+c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x)/(b^2 + c^2) - (2*a*c*ArcTan[(c + (a - b)*Tan[x/2])/Sqrt[a^2 - b^2 - c^2]])/(Sqrt[a^2 - b^2 - c^2]*(b^2 +
c^2)) - (b*Log[a + b*Cos[x] + c*Sin[x]])/(b^2 + c^2)

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 3124

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(-1), x_Symbol] :> Module[{f = Free
Factors[Tan[(d + e*x)/2], x]}, Dist[(2*f)/e, Subst[Int[1/(a + b + 2*c*f*x + (a - b)*f^2*x^2), x], x, Tan[(d +
e*x)/2]/f], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0]

Rule 3137

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

Rubi steps

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

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Mathematica [A]  time = 0.22, size = 80, normalized size = 0.79 \[ \frac {\frac {2 a c \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {x}{2}\right )+c}{\sqrt {-a^2+b^2+c^2}}\right )}{\sqrt {-a^2+b^2+c^2}}-b \log (a+b \cos (x)+c \sin (x))+c x}{b^2+c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x + (2*a*c*ArcTanh[(c + (a - b)*Tan[x/2])/Sqrt[-a^2 + b^2 + c^2]])/Sqrt[-a^2 + b^2 + c^2] - b*Log[a + b*Cos
[x] + c*Sin[x]])/(b^2 + c^2)

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fricas [B]  time = 1.21, size = 579, normalized size = 5.73 \[ \left [-\frac {\sqrt {-a^{2} + b^{2} + c^{2}} a c \log \left (\frac {a^{2} b^{2} - 2 \, b^{4} - c^{4} - {\left (a^{2} + 3 \, b^{2}\right )} c^{2} - {\left (2 \, a^{2} b^{2} - b^{4} - 2 \, a^{2} c^{2} + c^{4}\right )} \cos \relax (x)^{2} - 2 \, {\left (a b^{3} + a b c^{2}\right )} \cos \relax (x) - 2 \, {\left (a b^{2} c + a c^{3} - {\left (b c^{3} - {\left (2 \, a^{2} b - b^{3}\right )} c\right )} \cos \relax (x)\right )} \sin \relax (x) - 2 \, {\left (2 \, a b c \cos \relax (x)^{2} - a b c + {\left (b^{2} c + c^{3}\right )} \cos \relax (x) - {\left (b^{3} + b c^{2} + {\left (a b^{2} - a c^{2}\right )} \cos \relax (x)\right )} \sin \relax (x)\right )} \sqrt {-a^{2} + b^{2} + c^{2}}}{2 \, a b \cos \relax (x) + {\left (b^{2} - c^{2}\right )} \cos \relax (x)^{2} + a^{2} + c^{2} + 2 \, {\left (b c \cos \relax (x) + a c\right )} \sin \relax (x)}\right ) + 2 \, {\left (c^{3} - {\left (a^{2} - b^{2}\right )} c\right )} x + {\left (a^{2} b - b^{3} - b c^{2}\right )} \log \left (2 \, a b \cos \relax (x) + {\left (b^{2} - c^{2}\right )} \cos \relax (x)^{2} + a^{2} + c^{2} + 2 \, {\left (b c \cos \relax (x) + a c\right )} \sin \relax (x)\right )}{2 \, {\left (a^{2} b^{2} - b^{4} - c^{4} + {\left (a^{2} - 2 \, b^{2}\right )} c^{2}\right )}}, -\frac {2 \, \sqrt {a^{2} - b^{2} - c^{2}} a c \arctan \left (-\frac {{\left (a b \cos \relax (x) + a c \sin \relax (x) + b^{2} + c^{2}\right )} \sqrt {a^{2} - b^{2} - c^{2}}}{{\left (c^{3} - {\left (a^{2} - b^{2}\right )} c\right )} \cos \relax (x) + {\left (a^{2} b - b^{3} - b c^{2}\right )} \sin \relax (x)}\right ) + 2 \, {\left (c^{3} - {\left (a^{2} - b^{2}\right )} c\right )} x + {\left (a^{2} b - b^{3} - b c^{2}\right )} \log \left (2 \, a b \cos \relax (x) + {\left (b^{2} - c^{2}\right )} \cos \relax (x)^{2} + a^{2} + c^{2} + 2 \, {\left (b c \cos \relax (x) + a c\right )} \sin \relax (x)\right )}{2 \, {\left (a^{2} b^{2} - b^{4} - c^{4} + {\left (a^{2} - 2 \, b^{2}\right )} c^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 160, normalized size = 1.58 \[ \frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - b \tan \left (\frac {1}{2} \, x\right ) + c}{\sqrt {a^{2} - b^{2} - c^{2}}}\right )\right )} a c}{\sqrt {a^{2} - b^{2} - c^{2}} {\left (b^{2} + c^{2}\right )}} + \frac {c x}{b^{2} + c^{2}} - \frac {b \log \left (-a \tan \left (\frac {1}{2} \, x\right )^{2} + b \tan \left (\frac {1}{2} \, x\right )^{2} - 2 \, c \tan \left (\frac {1}{2} \, x\right ) - a - b\right )}{b^{2} + c^{2}} + \frac {b \log \left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}{b^{2} + c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(pi*floor(1/2*x/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*x) - b*tan(1/2*x) + c)/sqrt(a^2 - b^2 - c^2))
)*a*c/(sqrt(a^2 - b^2 - c^2)*(b^2 + c^2)) + c*x/(b^2 + c^2) - b*log(-a*tan(1/2*x)^2 + b*tan(1/2*x)^2 - 2*c*tan
(1/2*x) - a - b)/(b^2 + c^2) + b*log(tan(1/2*x)^2 + 1)/(b^2 + c^2)

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maple [B]  time = 0.12, size = 438, normalized size = 4.34 \[ -\frac {2 \ln \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+2 c \tan \left (\frac {x}{2}\right )+a +b \right ) a b}{\left (2 b^{2}+2 c^{2}\right ) \left (a -b \right )}+\frac {2 \ln \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+2 c \tan \left (\frac {x}{2}\right )+a +b \right ) b^{2}}{\left (2 b^{2}+2 c^{2}\right ) \left (a -b \right )}-\frac {4 \arctan \left (\frac {2 \left (a -b \right ) \tan \left (\frac {x}{2}\right )+2 c}{2 \sqrt {a^{2}-b^{2}-c^{2}}}\right ) a c}{\left (2 b^{2}+2 c^{2}\right ) \sqrt {a^{2}-b^{2}-c^{2}}}-\frac {4 \arctan \left (\frac {2 \left (a -b \right ) \tan \left (\frac {x}{2}\right )+2 c}{2 \sqrt {a^{2}-b^{2}-c^{2}}}\right ) c b}{\left (2 b^{2}+2 c^{2}\right ) \sqrt {a^{2}-b^{2}-c^{2}}}+\frac {4 \arctan \left (\frac {2 \left (a -b \right ) \tan \left (\frac {x}{2}\right )+2 c}{2 \sqrt {a^{2}-b^{2}-c^{2}}}\right ) c a b}{\left (2 b^{2}+2 c^{2}\right ) \sqrt {a^{2}-b^{2}-c^{2}}\, \left (a -b \right )}-\frac {4 \arctan \left (\frac {2 \left (a -b \right ) \tan \left (\frac {x}{2}\right )+2 c}{2 \sqrt {a^{2}-b^{2}-c^{2}}}\right ) c \,b^{2}}{\left (2 b^{2}+2 c^{2}\right ) \sqrt {a^{2}-b^{2}-c^{2}}\, \left (a -b \right )}+\frac {2 b \ln \left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )}{2 b^{2}+2 c^{2}}+\frac {4 c \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{2 b^{2}+2 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(2*b^2+2*c^2)/(a-b)*ln(a*tan(1/2*x)^2-b*tan(1/2*x)^2+2*c*tan(1/2*x)+a+b)*a*b+2/(2*b^2+2*c^2)/(a-b)*ln(a*tan
(1/2*x)^2-b*tan(1/2*x)^2+2*c*tan(1/2*x)+a+b)*b^2-4/(2*b^2+2*c^2)/(a^2-b^2-c^2)^(1/2)*arctan(1/2*(2*(a-b)*tan(1
/2*x)+2*c)/(a^2-b^2-c^2)^(1/2))*a*c-4/(2*b^2+2*c^2)/(a^2-b^2-c^2)^(1/2)*arctan(1/2*(2*(a-b)*tan(1/2*x)+2*c)/(a
^2-b^2-c^2)^(1/2))*c*b+4/(2*b^2+2*c^2)/(a^2-b^2-c^2)^(1/2)*arctan(1/2*(2*(a-b)*tan(1/2*x)+2*c)/(a^2-b^2-c^2)^(
1/2))*c/(a-b)*a*b-4/(2*b^2+2*c^2)/(a^2-b^2-c^2)^(1/2)*arctan(1/2*(2*(a-b)*tan(1/2*x)+2*c)/(a^2-b^2-c^2)^(1/2))
*c/(a-b)*b^2+2/(2*b^2+2*c^2)*b*ln(1+tan(1/2*x)^2)+4/(2*b^2+2*c^2)*c*arctan(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(sin(x)/(a+b*cos(x)+c*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(c^2+b^2-a^2>0)', see `assume?`
for more details)Is c^2+b^2-a^2 positive or negative?

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mupad [B]  time = 11.44, size = 950, normalized size = 9.41 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )+1{}\mathrm {i}\right )}{b-c\,1{}\mathrm {i}}+\frac {\ln \left (64\,\mathrm {tan}\left (\frac {x}{2}\right )\,{\left (a-b\right )}^2-\frac {\left (a^2\,b-b\,c^2-b^3+a\,c\,\sqrt {-a^2+b^2+c^2}\right )\,\left (32\,a^2\,c+32\,b^2\,c-64\,a\,b\,c+64\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a-b\right )\,\left (-a^2+b\,a+c^2\right )+\frac {\left (a^2\,b-b\,c^2-b^3+a\,c\,\sqrt {-a^2+b^2+c^2}\right )\,\left (32\,b\,c^3-32\,a\,c^3-64\,b^3\,c+32\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a-b\right )\,\left (-2\,b^3+2\,a\,b^2+b\,c^2-2\,a\,c^2\right )+128\,a\,b^2\,c-64\,a^2\,b\,c+\frac {32\,\left (a-b\right )\,\left (a^2\,b-b\,c^2-b^3+a\,c\,\sqrt {-a^2+b^2+c^2}\right )\,\left (2\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,b^2-4\,a^2\,b\,c-2\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,c^2-2\,\mathrm {tan}\left (\frac {x}{2}\right )\,a\,b^3+a\,b^2\,c-2\,\mathrm {tan}\left (\frac {x}{2}\right )\,a\,b\,c^2+a\,c^3+3\,b^3\,c+3\,\mathrm {tan}\left (\frac {x}{2}\right )\,b^2\,c^2+3\,b\,c^3+3\,\mathrm {tan}\left (\frac {x}{2}\right )\,c^4\right )}{\left (b^2+c^2\right )\,\left (-a^2+b^2+c^2\right )}\right )}{\left (b^2+c^2\right )\,\left (-a^2+b^2+c^2\right )}\right )}{\left (b^2+c^2\right )\,\left (-a^2+b^2+c^2\right )}\right )\,\left (b\,\left (a^2-c^2\right )-b^3+a\,c\,\sqrt {-a^2+b^2+c^2}\right )}{\left (b^2+c^2\right )\,\left (-a^2+b^2+c^2\right )}-\frac {\ln \left (64\,\mathrm {tan}\left (\frac {x}{2}\right )\,{\left (a-b\right )}^2+\frac {\left (b\,c^2-a^2\,b+b^3+a\,c\,\sqrt {-a^2+b^2+c^2}\right )\,\left (32\,a^2\,c+32\,b^2\,c-64\,a\,b\,c+64\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a-b\right )\,\left (-a^2+b\,a+c^2\right )+\frac {\left (b\,c^2-a^2\,b+b^3+a\,c\,\sqrt {-a^2+b^2+c^2}\right )\,\left (32\,a\,c^3-32\,b\,c^3+64\,b^3\,c-32\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a-b\right )\,\left (-2\,b^3+2\,a\,b^2+b\,c^2-2\,a\,c^2\right )-128\,a\,b^2\,c+64\,a^2\,b\,c+\frac {32\,\left (a-b\right )\,\left (b\,c^2-a^2\,b+b^3+a\,c\,\sqrt {-a^2+b^2+c^2}\right )\,\left (2\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,b^2-4\,a^2\,b\,c-2\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,c^2-2\,\mathrm {tan}\left (\frac {x}{2}\right )\,a\,b^3+a\,b^2\,c-2\,\mathrm {tan}\left (\frac {x}{2}\right )\,a\,b\,c^2+a\,c^3+3\,b^3\,c+3\,\mathrm {tan}\left (\frac {x}{2}\right )\,b^2\,c^2+3\,b\,c^3+3\,\mathrm {tan}\left (\frac {x}{2}\right )\,c^4\right )}{\left (b^2+c^2\right )\,\left (-a^2+b^2+c^2\right )}\right )}{\left (b^2+c^2\right )\,\left (-a^2+b^2+c^2\right )}\right )}{\left (b^2+c^2\right )\,\left (-a^2+b^2+c^2\right )}\right )\,\left (b^3-b\,\left (a^2-c^2\right )+a\,c\,\sqrt {-a^2+b^2+c^2}\right )}{\left (b^2+c^2\right )\,\left (-a^2+b^2+c^2\right )}+\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )-\mathrm {i}\right )\,1{}\mathrm {i}}{-c+b\,1{}\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(tan(x/2) + 1i)/(b - c*1i) + (log(tan(x/2) - 1i)*1i)/(b*1i - c) + (log(64*tan(x/2)*(a - b)^2 - ((a^2*b - b*
c^2 - b^3 + a*c*(b^2 - a^2 + c^2)^(1/2))*(32*a^2*c + 32*b^2*c - 64*a*b*c + 64*tan(x/2)*(a - b)*(a*b - a^2 + c^
2) + ((a^2*b - b*c^2 - b^3 + a*c*(b^2 - a^2 + c^2)^(1/2))*(32*b*c^3 - 32*a*c^3 - 64*b^3*c + 32*tan(x/2)*(a - b
)*(2*a*b^2 - 2*a*c^2 + b*c^2 - 2*b^3) + 128*a*b^2*c - 64*a^2*b*c + (32*(a - b)*(a^2*b - b*c^2 - b^3 + a*c*(b^2
- a^2 + c^2)^(1/2))*(3*c^4*tan(x/2) + a*c^3 + 3*b*c^3 + 3*b^3*c + 2*a^2*b^2*tan(x/2) - 2*a^2*c^2*tan(x/2) + 3
*b^2*c^2*tan(x/2) - 2*a*b^3*tan(x/2) + a*b^2*c - 4*a^2*b*c - 2*a*b*c^2*tan(x/2)))/((b^2 + c^2)*(b^2 - a^2 + c^
2))))/((b^2 + c^2)*(b^2 - a^2 + c^2))))/((b^2 + c^2)*(b^2 - a^2 + c^2)))*(b*(a^2 - c^2) - b^3 + a*c*(b^2 - a^2
+ c^2)^(1/2)))/((b^2 + c^2)*(b^2 - a^2 + c^2)) - (log(64*tan(x/2)*(a - b)^2 + ((b*c^2 - a^2*b + b^3 + a*c*(b^
2 - a^2 + c^2)^(1/2))*(32*a^2*c + 32*b^2*c - 64*a*b*c + 64*tan(x/2)*(a - b)*(a*b - a^2 + c^2) + ((b*c^2 - a^2*
b + b^3 + a*c*(b^2 - a^2 + c^2)^(1/2))*(32*a*c^3 - 32*b*c^3 + 64*b^3*c - 32*tan(x/2)*(a - b)*(2*a*b^2 - 2*a*c^
2 + b*c^2 - 2*b^3) - 128*a*b^2*c + 64*a^2*b*c + (32*(a - b)*(b*c^2 - a^2*b + b^3 + a*c*(b^2 - a^2 + c^2)^(1/2)
)*(3*c^4*tan(x/2) + a*c^3 + 3*b*c^3 + 3*b^3*c + 2*a^2*b^2*tan(x/2) - 2*a^2*c^2*tan(x/2) + 3*b^2*c^2*tan(x/2) -
2*a*b^3*tan(x/2) + a*b^2*c - 4*a^2*b*c - 2*a*b*c^2*tan(x/2)))/((b^2 + c^2)*(b^2 - a^2 + c^2))))/((b^2 + c^2)*
(b^2 - a^2 + c^2))))/((b^2 + c^2)*(b^2 - a^2 + c^2)))*(b^3 - b*(a^2 - c^2) + a*c*(b^2 - a^2 + c^2)^(1/2)))/((b
^2 + c^2)*(b^2 - a^2 + c^2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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