3.550 \(\int \frac{A+B \cos (x)+C \sin (x)}{a+b \cos (x)+c \sin (x)} \, dx\)
Optimal. Leaf size=131 \[ \frac{2 \tan ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )+c}{\sqrt{a^2-b^2-c^2}}\right ) \left (A \left (b^2+c^2\right )-a (b B+c C)\right )}{\left (b^2+c^2\right ) \sqrt{a^2-b^2-c^2}}+\frac{(B c-b C) \log (a+b \cos (x)+c \sin (x))}{b^2+c^2}+\frac{x (b B+c C)}{b^2+c^2} \]
[Out]
((b*B + c*C)*x)/(b^2 + c^2) + (2*(A*(b^2 + c^2) - a*(b*B + c*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*c - b*C)*Log[a + b*Cos[x] + c*Sin[x]])/(b^2 + c^2)
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Rubi [A] time = 0.126317, antiderivative size = 131, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used =
{3136, 3124, 618, 204} \[ \frac{2 \tan ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )+c}{\sqrt{a^2-b^2-c^2}}\right ) \left (A \left (b^2+c^2\right )-a (b B+c C)\right )}{\left (b^2+c^2\right ) \sqrt{a^2-b^2-c^2}}+\frac{(B c-b C) \log (a+b \cos (x)+c \sin (x))}{b^2+c^2}+\frac{x (b B+c C)}{b^2+c^2} \]
Antiderivative was successfully verified.
[In]
Int[(A + B*Cos[x] + C*Sin[x])/(a + b*Cos[x] + c*Sin[x]),x]
[Out]
((b*B + c*C)*x)/(b^2 + c^2) + (2*(A*(b^2 + c^2) - a*(b*B + c*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*c - b*C)*Log[a + b*Cos[x] + c*Sin[x]])/(b^2 + c^2)
Rule 3136
Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_)]*(
b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]), x_Symbol] :> Simp[((b*B + c*C)*x)/(b^2 + c^2), x] + (Dist[(A*(b^2 + c^2
) - a*(b*B + c*C))/(b^2 + c^2), Int[1/(a + b*Cos[d + e*x] + c*Sin[d + e*x]), x], x] + Simp[((c*B - 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, B, C}, x] && NeQ[b^2 + c^
2, 0] && NeQ[A*(b^2 + c^2) - a*(b*B + 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{A+B \cos (x)+C \sin (x)}{a+b \cos (x)+c \sin (x)} \, dx &=\frac{(b B+c C) x}{b^2+c^2}+\frac{(B c-b C) \log (a+b \cos (x)+c \sin (x))}{b^2+c^2}+\left (A-\frac{a (b B+c C)}{b^2+c^2}\right ) \int \frac{1}{a+b \cos (x)+c \sin (x)} \, dx\\ &=\frac{(b B+c C) x}{b^2+c^2}+\frac{(B c-b C) \log (a+b \cos (x)+c \sin (x))}{b^2+c^2}+\left (2 \left (A-\frac{a (b B+c C)}{b^2+c^2}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+2 c x+(a-b) x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )\\ &=\frac{(b B+c C) x}{b^2+c^2}+\frac{(B c-b C) \log (a+b \cos (x)+c \sin (x))}{b^2+c^2}-\left (4 \left (A-\frac{a (b B+c C)}{b^2+c^2}\right )\right ) \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 )\\ &=\frac{(b B+c C) x}{b^2+c^2}+\frac{2 \left (A-\frac{a (b B+c C)}{b^2+c^2}\right ) \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}}+\frac{(B c-b C) \log (a+b \cos (x)+c \sin (x))}{b^2+c^2}\\ \end{align*}
Mathematica [A] time = 0.333339, size = 110, normalized size = 0.84 \[ \frac{\frac{2 \left (a (b B+c C)-A \left (b^2+c^2\right )\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )+c}{\sqrt{-a^2+b^2+c^2}}\right )}{\sqrt{-a^2+b^2+c^2}}+(B c-b C) \log (a+b \cos (x)+c \sin (x))+x (b B+c C)}{b^2+c^2} \]
Antiderivative was successfully verified.
[In]
Integrate[(A + B*Cos[x] + C*Sin[x])/(a + b*Cos[x] + c*Sin[x]),x]
[Out]
((b*B + c*C)*x + (2*(-(A*(b^2 + c^2)) + a*(b*B + c*C))*ArcTanh[(c + (a - b)*Tan[x/2])/Sqrt[-a^2 + b^2 + c^2]])
/Sqrt[-a^2 + b^2 + c^2] + (B*c - b*C)*Log[a + b*Cos[x] + c*Sin[x]])/(b^2 + c^2)
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Maple [B] time = 0.056, size = 954, normalized size = 7.3 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*cos(x)+C*sin(x))/(a+b*cos(x)+c*sin(x)),x)
[Out]
1/(b^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*c-1/(b^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*B*c-1/(b^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*C+1/(b^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*C+2/(b^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*b^2+2/(b^2+c^2)/(a^2-b^2-c^2)^(1/2)*arcta
n(1/2*(2*(a-b)*tan(1/2*x)+2*c)/(a^2-b^2-c^2)^(1/2))*A*c^2-2/(b^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*b*B+2/(b^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))*B*c^2-2/(b^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*C-2/(b^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*c-2/
(b^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^2/(a-b)*a*B+2/(b^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^2/(a-b)*b*B+2/(b^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*C-2/(b^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*C-B/(b^2+c^2)*c*ln(1+tan(1/2*x)^
2)+C/(b^2+c^2)*b*ln(1+tan(1/2*x)^2)+2*B/(b^2+c^2)*b*arctan(tan(1/2*x))+2*C/(b^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((A+B*cos(x)+C*sin(x))/(a+b*cos(x)+c*sin(x)),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 2.98718, size = 1550, normalized size = 11.83 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*cos(x)+C*sin(x))/(a+b*cos(x)+c*sin(x)),x, algorithm="fricas")
[Out]
[-1/2*((B*a*b - A*b^2 + C*a*c - A*c^2)*sqrt(-a^2 + b^2 + c^2)*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*(B*a^2*b - B*b^3 - B*b*c^2 - C*c^3 + (C*a^2 - C*b^2)*c)*x + (C*a^2*b - C*b^3 - C*b*c^2 +
B*c^3 - (B*a^2 - B*b^2)*c)*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*(B*a*b - A*b^2 + C*a*c - A*c^2)*sqrt(a^2 - b^2 - c^2)*arc
tan(-(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*(B*a^2*b - B*b^3 - B*b*c^2 - C*c^3 + (C*a^2 - C*b^2)*c)*x + (C*a^2*b - C*b^3 - C*b*c^2 +
B*c^3 - (B*a^2 - B*b^2)*c)*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((A+B*cos(x)+C*sin(x))/(a+b*cos(x)+c*sin(x)),x)
[Out]
Timed out
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Giac [A] time = 1.1498, size = 269, normalized size = 2.05 \begin{align*} \frac{{\left (B b + C c\right )} x}{b^{2} + c^{2}} - \frac{{\left (C b - B c\right )} \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{{\left (C b - B c\right )} \log \left (\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right )}{b^{2} + c^{2}} + \frac{2 \,{\left (B a b - A b^{2} + C a c - A c^{2}\right )}{\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 )}}{\sqrt{a^{2} - b^{2} - c^{2}}{\left (b^{2} + c^{2}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*cos(x)+C*sin(x))/(a+b*cos(x)+c*sin(x)),x, algorithm="giac")
[Out]
(B*b + C*c)*x/(b^2 + c^2) - (C*b - B*c)*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) + (C*b - B*c)*log(tan(1/2*x)^2 + 1)/(b^2 + c^2) + 2*(B*a*b - A*b^2 + C*a*c - A*c^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)))/(sqrt(a^2 - b^2 - c^2
)*(b^2 + c^2))