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

Optimal. Leaf size=115 \[ -\frac{2 \left (a b B-A \left (b^2+c^2\right )\right ) \tan ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )+c}{\sqrt{a^2-b^2-c^2}}\right )}{\left (b^2+c^2\right ) \sqrt{a^2-b^2-c^2}}+\frac{B c \log (a+b \cos (x)+c \sin (x))}{b^2+c^2}+\frac{b B x}{b^2+c^2} \]

[Out]

(b*B*x)/(b^2 + c^2) - (2*(a*b*B - A*(b^2 + c^2))*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*Log[a + b*Cos[x] + c*Sin[x]])/(b^2 + c^2)

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Rubi [A]  time = 0.129708, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {3138, 3124, 618, 204} \[ -\frac{2 \left (a b B-A \left (b^2+c^2\right )\right ) \tan ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )+c}{\sqrt{a^2-b^2-c^2}}\right )}{\left (b^2+c^2\right ) \sqrt{a^2-b^2-c^2}}+\frac{B c \log (a+b \cos (x)+c \sin (x))}{b^2+c^2}+\frac{b B x}{b^2+c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*B*x)/(b^2 + c^2) - (2*(a*b*B - A*(b^2 + c^2))*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*Log[a + b*Cos[x] + c*Sin[x]])/(b^2 + c^2)

Rule 3138

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.))/((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(
x_)]), x_Symbol] :> Simp[(b*B*(d + e*x))/(e*(b^2 + c^2)), x] + (Dist[(A*(b^2 + c^2) - a*b*B)/(b^2 + c^2), Int[
1/(a + b*Cos[d + e*x] + c*Sin[d + e*x]), x], x] + Simp[(c*B*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}, x] && NeQ[b^2 + c^2, 0] && NeQ[A*(b^2 + c^2) - a*b*B, 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)}{a+b \cos (x)+c \sin (x)} \, dx &=\frac{b B x}{b^2+c^2}+\frac{B c \log (a+b \cos (x)+c \sin (x))}{b^2+c^2}+\left (A-\frac{a b B}{b^2+c^2}\right ) \int \frac{1}{a+b \cos (x)+c \sin (x)} \, dx\\ &=\frac{b B x}{b^2+c^2}+\frac{B c \log (a+b \cos (x)+c \sin (x))}{b^2+c^2}+\left (2 \left (A-\frac{a b B}{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 x}{b^2+c^2}+\frac{B c \log (a+b \cos (x)+c \sin (x))}{b^2+c^2}-\left (4 \left (A-\frac{a b B}{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 x}{b^2+c^2}+\frac{2 \left (A-\frac{a b B}{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 \log (a+b \cos (x)+c \sin (x))}{b^2+c^2}\\ \end{align*}

Mathematica [A]  time = 0.269046, size = 95, normalized size = 0.83 \[ \frac{B (c \log (a+b \cos (x)+c \sin (x))+b x)-\frac{2 \left (A \left (b^2+c^2\right )-a b B\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )+c}{\sqrt{-a^2+b^2+c^2}}\right )}{\sqrt{-a^2+b^2+c^2}}}{b^2+c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.053, size = 544, normalized size = 4.7 \begin{align*}{\frac{aBc}{ \left ({b}^{2}+{c}^{2} \right ) \left ( a-b \right ) }\ln \left ( a \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}-b \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}+2\,c\tan \left ( x/2 \right ) +a+b \right ) }-{\frac{bBc}{ \left ({b}^{2}+{c}^{2} \right ) \left ( a-b \right ) }\ln \left ( a \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}-b \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}+2\,c\tan \left ( x/2 \right ) +a+b \right ) }+2\,{\frac{A{b}^{2}}{ \left ({b}^{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 ) }+2\,{\frac{A{c}^{2}}{ \left ({b}^{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 ) }-2\,{\frac{abB}{ \left ({b}^{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 ) }+2\,{\frac{B{c}^{2}}{ \left ({b}^{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 ) }-2\,{\frac{a{c}^{2}B}{ \left ({b}^{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{c}^{2}B}{ \left ({b}^{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 ) }-{\frac{Bc}{{b}^{2}+{c}^{2}}\ln \left ( 1+ \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2} \right ) }+2\,{\frac{bB\arctan \left ( \tan \left ( x/2 \right ) \right ) }{{b}^{2}+{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(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+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)*arctan(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))*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-B/(b^2+c^2)*c*ln(1+tan(1/2*x)^
2)+2*B/(b^2+c^2)*b*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))/(a+b*cos(x)+c*sin(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.79789, size = 1377, normalized size = 11.97 \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))/(a+b*cos(x)+c*sin(x)),x, algorithm="fricas")

[Out]

[-1/2*((B*a*b - A*b^2 - 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)*c
os(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)*x + (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 - A*c^2)*sqrt(a^2 - b^2 - c^2)*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*(B*a^2*b - B*b^3 - B*b*c^2)*x + (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))/(a+b*cos(x)+c*sin(x)),x)

[Out]

Timed out

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Giac [A]  time = 1.16656, size = 240, normalized size = 2.09 \begin{align*} \frac{B b x}{b^{2} + c^{2}} + \frac{B c \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 c \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} - 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))/(a+b*cos(x)+c*sin(x)),x, algorithm="giac")

[Out]

B*b*x/(b^2 + c^2) + 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) - B*c*log(t
an(1/2*x)^2 + 1)/(b^2 + c^2) + 2*(B*a*b - A*b^2 - 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))