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) - (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)

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Rubi [A]  time = 0.0964729, antiderivative size = 101, normalized size of antiderivative = 1., 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 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]

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 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 \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*}

Mathematica [A]  time = 0.224024, 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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Maple [B]  time = 0.048, size = 438, normalized size = 4.3 \begin{align*} -2\,{\frac{\ln \left ( a \left ( \tan \left ( x/2 \right ) \right ) ^{2}-b \left ( \tan \left ( x/2 \right ) \right ) ^{2}+2\,c\tan \left ( x/2 \right ) +a+b \right ) ab}{ \left ( 2\,{b}^{2}+2\,{c}^{2} \right ) \left ( a-b \right ) }}+2\,{\frac{\ln \left ( a \left ( \tan \left ( x/2 \right ) \right ) ^{2}-b \left ( \tan \left ( x/2 \right ) \right ) ^{2}+2\,c\tan \left ( x/2 \right ) +a+b \right ){b}^{2}}{ \left ( 2\,{b}^{2}+2\,{c}^{2} \right ) \left ( a-b \right ) }}-4\,{\frac{ac}{ \left ( 2\,{b}^{2}+2\,{c}^{2} \right ) \sqrt{{a}^{2}-{b}^{2}-{c}^{2}}}\arctan \left ( 1/2\,{\frac{2\, \left ( a-b \right ) \tan \left ( x/2 \right ) +2\,c}{\sqrt{{a}^{2}-{b}^{2}-{c}^{2}}}} \right ) }-4\,{\frac{cb}{ \left ( 2\,{b}^{2}+2\,{c}^{2} \right ) \sqrt{{a}^{2}-{b}^{2}-{c}^{2}}}\arctan \left ( 1/2\,{\frac{2\, \left ( a-b \right ) \tan \left ( x/2 \right ) +2\,c}{\sqrt{{a}^{2}-{b}^{2}-{c}^{2}}}} \right ) }+4\,{\frac{abc}{ \left ( 2\,{b}^{2}+2\,{c}^{2} \right ) \sqrt{{a}^{2}-{b}^{2}-{c}^{2}} \left ( a-b \right ) }\arctan \left ( 1/2\,{\frac{2\, \left ( a-b \right ) \tan \left ( x/2 \right ) +2\,c}{\sqrt{{a}^{2}-{b}^{2}-{c}^{2}}}} \right ) }-4\,{\frac{c{b}^{2}}{ \left ( 2\,{b}^{2}+2\,{c}^{2} \right ) \sqrt{{a}^{2}-{b}^{2}-{c}^{2}} \left ( a-b \right ) }\arctan \left ( 1/2\,{\frac{2\, \left ( a-b \right ) \tan \left ( x/2 \right ) +2\,c}{\sqrt{{a}^{2}-{b}^{2}-{c}^{2}}}} \right ) }+2\,{\frac{b\ln \left ( 1+ \left ( \tan \left ( x/2 \right ) \right ) ^{2} \right ) }{2\,{b}^{2}+2\,{c}^{2}}}+4\,{\frac{c\arctan \left ( \tan \left ( x/2 \right ) \right ) }{2\,{b}^{2}+2\,{c}^{2}}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [B]  time = 2.14762, size = 1291, normalized size = 12.78 \begin{align*} \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 \left (x\right )^{2} - 2 \,{\left (a b^{3} + a b c^{2}\right )} \cos \left (x\right ) - 2 \,{\left (a b^{2} c + a c^{3} -{\left (b c^{3} -{\left (2 \, a^{2} b - b^{3}\right )} c\right )} \cos \left (x\right )\right )} \sin \left (x\right ) - 2 \,{\left (2 \, a b c \cos \left (x\right )^{2} - a b c +{\left (b^{2} c + c^{3}\right )} \cos \left (x\right ) -{\left (b^{3} + b c^{2} +{\left (a b^{2} - a c^{2}\right )} \cos \left (x\right )\right )} \sin \left (x\right )\right )} \sqrt{-a^{2} + b^{2} + c^{2}}}{2 \, a b \cos \left (x\right ) +{\left (b^{2} - c^{2}\right )} \cos \left (x\right )^{2} + a^{2} + c^{2} + 2 \,{\left (b c \cos \left (x\right ) + a c\right )} \sin \left (x\right )}\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 \left (x\right ) +{\left (b^{2} - c^{2}\right )} \cos \left (x\right )^{2} + a^{2} + c^{2} + 2 \,{\left (b c \cos \left (x\right ) + a c\right )} \sin \left (x\right )\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 \left (x\right ) + a c \sin \left (x\right ) + b^{2} + c^{2}\right )} \sqrt{a^{2} - b^{2} - c^{2}}}{{\left (c^{3} -{\left (a^{2} - b^{2}\right )} c\right )} \cos \left (x\right ) +{\left (a^{2} b - b^{3} - b c^{2}\right )} \sin \left (x\right )}\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 \left (x\right ) +{\left (b^{2} - c^{2}\right )} \cos \left (x\right )^{2} + a^{2} + c^{2} + 2 \,{\left (b c \cos \left (x\right ) + a c\right )} \sin \left (x\right )\right )}{2 \,{\left (a^{2} b^{2} - b^{4} - c^{4} +{\left (a^{2} - 2 \, b^{2}\right )} c^{2}\right )}}\right ] \end{align*}

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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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*cos(x)+c*sin(x)),x)

[Out]

Timed out

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Giac [A]  time = 1.15365, size = 216, normalized size = 2.14 \begin{align*} \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}} \end{align*}

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)