3.18 \(\int \frac{A+B \cos (d+e x)+C \sin (d+e x)}{a+b \cos (d+e x)} \, dx\)

Optimal. Leaf size=87 \[ \frac{2 (A b-a B) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a+b}}\right )}{b e \sqrt{a-b} \sqrt{a+b}}-\frac{C \log (a+b \cos (d+e x))}{b e}+\frac{B x}{b} \]

[Out]

(B*x)/b + (2*(A*b - a*B)*ArcTan[(Sqrt[a - b]*Tan[(d + e*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]*e) - (
C*Log[a + b*Cos[d + e*x]])/(b*e)

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Rubi [A]  time = 0.143702, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4377, 2735, 2659, 205, 2668, 31} \[ \frac{2 (A b-a B) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a+b}}\right )}{b e \sqrt{a-b} \sqrt{a+b}}-\frac{C \log (a+b \cos (d+e x))}{b e}+\frac{B x}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*x)/b + (2*(A*b - a*B)*ArcTan[(Sqrt[a - b]*Tan[(d + e*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]*e) - (
C*Log[a + b*Cos[d + e*x]])/(b*e)

Rule 4377

Int[(u_)*((v_) + (d_.)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_.)), x_Symbol] :> With[{e = FreeFactors[Cos[c*(a +
b*x)], x]}, Int[ActivateTrig[u*v], x] + Dist[d, Int[ActivateTrig[u]*Sin[c*(a + b*x)]^n, x], x] /; FunctionOfQ[
Cos[c*(a + b*x)]/e, u, x]] /; FreeQ[{a, b, c, d}, x] &&  !FreeQ[v, x] && IntegerQ[(n - 1)/2] && NonsumQ[u] &&
(EqQ[F, Sin] || EqQ[F, sin])

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.224411, size = 82, normalized size = 0.94 \[ \frac{\frac{2 (a B-A b) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}-C \log (a+b \cos (d+e x))+B (d+e x)}{b e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*(d + e*x) + (2*(-(A*b) + a*B)*ArcTanh[((a - b)*Tan[(d + e*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] - C*Lo
g[a + b*Cos[d + e*x]])/(b*e)

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Maple [B]  time = 0.043, size = 226, normalized size = 2.6 \begin{align*}{\frac{C}{eb}\ln \left ( \left ( \tan \left ({\frac{ex}{2}}+{\frac{d}{2}} \right ) \right ) ^{2}+1 \right ) }+2\,{\frac{B\arctan \left ( \tan \left ( 1/2\,ex+d/2 \right ) \right ) }{eb}}-{\frac{Ca}{eb \left ( a-b \right ) }\ln \left ( \left ( \tan \left ({\frac{ex}{2}}+{\frac{d}{2}} \right ) \right ) ^{2}a- \left ( \tan \left ({\frac{ex}{2}}+{\frac{d}{2}} \right ) \right ) ^{2}b+a+b \right ) }+{\frac{C}{e \left ( a-b \right ) }\ln \left ( \left ( \tan \left ({\frac{ex}{2}}+{\frac{d}{2}} \right ) \right ) ^{2}a- \left ( \tan \left ({\frac{ex}{2}}+{\frac{d}{2}} \right ) \right ) ^{2}b+a+b \right ) }+2\,{\frac{A}{e\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,ex+d/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-2\,{\frac{Ba}{eb\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,ex+d/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d)),x)

[Out]

1/e/b*C*ln(tan(1/2*e*x+1/2*d)^2+1)+2/e/b*B*arctan(tan(1/2*e*x+1/2*d))-1/e/b/(a-b)*ln(tan(1/2*e*x+1/2*d)^2*a-ta
n(1/2*e*x+1/2*d)^2*b+a+b)*C*a+1/e/(a-b)*ln(tan(1/2*e*x+1/2*d)^2*a-tan(1/2*e*x+1/2*d)^2*b+a+b)*C+2/e/((a+b)*(a-
b))^(1/2)*arctan((a-b)*tan(1/2*e*x+1/2*d)/((a+b)*(a-b))^(1/2))*A-2/e/b/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/
2*e*x+1/2*d)/((a+b)*(a-b))^(1/2))*B*a

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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(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.9228, size = 721, normalized size = 8.29 \begin{align*} \left [\frac{2 \,{\left (B a^{2} - B b^{2}\right )} e x +{\left (B a - A b\right )} \sqrt{-a^{2} + b^{2}} \log \left (\frac{2 \, a b \cos \left (e x + d\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (e x + d\right )^{2} + 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (e x + d\right ) + b\right )} \sin \left (e x + d\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (e x + d\right )^{2} + 2 \, a b \cos \left (e x + d\right ) + a^{2}}\right ) -{\left (C a^{2} - C b^{2}\right )} \log \left (b^{2} \cos \left (e x + d\right )^{2} + 2 \, a b \cos \left (e x + d\right ) + a^{2}\right )}{2 \,{\left (a^{2} b - b^{3}\right )} e}, \frac{2 \,{\left (B a^{2} - B b^{2}\right )} e x - 2 \,{\left (B a - A b\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \cos \left (e x + d\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (e x + d\right )}\right ) -{\left (C a^{2} - C b^{2}\right )} \log \left (b^{2} \cos \left (e x + d\right )^{2} + 2 \, a b \cos \left (e x + d\right ) + a^{2}\right )}{2 \,{\left (a^{2} b - b^{3}\right )} e}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d)),x, algorithm="fricas")

[Out]

[1/2*(2*(B*a^2 - B*b^2)*e*x + (B*a - A*b)*sqrt(-a^2 + b^2)*log((2*a*b*cos(e*x + d) + (2*a^2 - b^2)*cos(e*x + d
)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(e*x + d) + b)*sin(e*x + d) - a^2 + 2*b^2)/(b^2*cos(e*x + d)^2 + 2*a*b*cos(e*x
+ d) + a^2)) - (C*a^2 - C*b^2)*log(b^2*cos(e*x + d)^2 + 2*a*b*cos(e*x + d) + a^2))/((a^2*b - b^3)*e), 1/2*(2*(
B*a^2 - B*b^2)*e*x - 2*(B*a - A*b)*sqrt(a^2 - b^2)*arctan(-(a*cos(e*x + d) + b)/(sqrt(a^2 - b^2)*sin(e*x + d))
) - (C*a^2 - C*b^2)*log(b^2*cos(e*x + d)^2 + 2*a*b*cos(e*x + d) + a^2))/((a^2*b - b^3)*e)]

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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(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d)),x)

[Out]

Timed out

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Giac [B]  time = 1.188, size = 225, normalized size = 2.59 \begin{align*}{\left (\frac{{\left (x e + d\right )} B}{b} - \frac{C \log \left (-a \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + b \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} - a - b\right )}{b} + \frac{C \log \left (\tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + 1\right )}{b} + \frac{2 \,{\left (\pi \left \lfloor \frac{x e + d}{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 e + \frac{1}{2} \, d\right ) - b \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )}{\sqrt{a^{2} - b^{2}}}\right )\right )}{\left (B a - A b\right )}}{\sqrt{a^{2} - b^{2}} b}\right )} e^{\left (-1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d)),x, algorithm="giac")

[Out]

((x*e + d)*B/b - C*log(-a*tan(1/2*x*e + 1/2*d)^2 + b*tan(1/2*x*e + 1/2*d)^2 - a - b)/b + C*log(tan(1/2*x*e + 1
/2*d)^2 + 1)/b + 2*(pi*floor(1/2*(x*e + d)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*x*e + 1/2*d) - b*tan
(1/2*x*e + 1/2*d))/sqrt(a^2 - b^2)))*(B*a - A*b)/(sqrt(a^2 - b^2)*b))*e^(-1)