3.539 \(\int \frac{A+B \cos (x)}{a+b \cos (x)-i b \sin (x)} \, dx\)

Optimal. Leaf size=84 \[ -\frac{i \left (a^2 (-B)+2 a A b-b^2 B\right ) \log (a-i b \sin (x)+b \cos (x))}{2 a^2 b}+\frac{x (2 a A-b B)}{2 a^2}+\frac{B \sin (x)}{2 a}-\frac{i B \cos (x)}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.0424567, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.045, Rules used = {3132} \[ -\frac{i \left (a^2 (-B)+2 a A b-b^2 B\right ) \log (a-i b \sin (x)+b \cos (x))}{2 a^2 b}+\frac{x (2 a A-b B)}{2 a^2}+\frac{B \sin (x)}{2 a}-\frac{i B \cos (x)}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3132

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

Rubi steps

\begin{align*} \int \frac{A+B \cos (x)}{a+b \cos (x)-i b \sin (x)} \, dx &=\frac{(2 a A-b B) x}{2 a^2}-\frac{i B \cos (x)}{2 a}-\frac{i \left (2 a A b-a^2 B-b^2 B\right ) \log (a+b \cos (x)-i b \sin (x))}{2 a^2 b}+\frac{B \sin (x)}{2 a}\\ \end{align*}

Mathematica [A]  time = 0.182585, size = 147, normalized size = 1.75 \[ \frac{2 \left (a^2 B-2 a A b+b^2 B\right ) \tan ^{-1}\left (\frac{(a+b) \cot \left (\frac{x}{2}\right )}{a-b}\right )-2 i a A b \log \left (a^2+2 a b \cos (x)+b^2\right )+i a^2 B \log \left (a^2+2 a b \cos (x)+b^2\right )+i b^2 B \log \left (a^2+2 a b \cos (x)+b^2\right )+a^2 B x+2 a A b x+2 a b B \sin (x)-2 i a b B \cos (x)-b^2 B x}{4 a^2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*A*b*x + a^2*B*x - b^2*B*x + 2*(-2*a*A*b + a^2*B + b^2*B)*ArcTan[((a + b)*Cot[x/2])/(a - b)] - (2*I)*a*b*B
*Cos[x] - (2*I)*a*A*b*Log[a^2 + b^2 + 2*a*b*Cos[x]] + I*a^2*B*Log[a^2 + b^2 + 2*a*b*Cos[x]] + I*b^2*B*Log[a^2
+ b^2 + 2*a*b*Cos[x]] + 2*a*b*B*Sin[x])/(4*a^2*b)

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Maple [B]  time = 0.08, size = 284, normalized size = 3.4 \begin{align*}{\frac{iA}{a}\ln \left ( \tan \left ({\frac{x}{2}} \right ) +i \right ) }-{\frac{{\frac{i}{2}}bB}{{a}^{2}}\ln \left ( \tan \left ({\frac{x}{2}} \right ) +i \right ) }+{\frac{B}{a} \left ( \tan \left ({\frac{x}{2}} \right ) +i \right ) ^{-1}}+{\frac{iA}{-a+b}\ln \left ( ia+ib-a\tan \left ({\frac{x}{2}} \right ) +b\tan \left ({\frac{x}{2}} \right ) \right ) }-{\frac{iAb}{a \left ( -a+b \right ) }\ln \left ( ia+ib-a\tan \left ({\frac{x}{2}} \right ) +b\tan \left ({\frac{x}{2}} \right ) \right ) }-{\frac{{\frac{i}{2}}aB}{b \left ( -a+b \right ) }\ln \left ( ia+ib-a\tan \left ({\frac{x}{2}} \right ) +b\tan \left ({\frac{x}{2}} \right ) \right ) }+{\frac{{\frac{i}{2}}B}{-a+b}\ln \left ( ia+ib-a\tan \left ({\frac{x}{2}} \right ) +b\tan \left ({\frac{x}{2}} \right ) \right ) }-{\frac{{\frac{i}{2}}bB}{a \left ( -a+b \right ) }\ln \left ( ia+ib-a\tan \left ({\frac{x}{2}} \right ) +b\tan \left ({\frac{x}{2}} \right ) \right ) }+{\frac{{\frac{i}{2}}{b}^{2}B}{{a}^{2} \left ( -a+b \right ) }\ln \left ( ia+ib-a\tan \left ({\frac{x}{2}} \right ) +b\tan \left ({\frac{x}{2}} \right ) \right ) }-{\frac{{\frac{i}{2}}B}{b}\ln \left ( \tan \left ({\frac{x}{2}} \right ) -i \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(x))/(a+b*cos(x)-I*b*sin(x)),x)

[Out]

I/a*ln(tan(1/2*x)+I)*A-1/2*I/a^2*ln(tan(1/2*x)+I)*b*B+B/a/(tan(1/2*x)+I)+I/(-a+b)*ln(I*a+I*b-a*tan(1/2*x)+b*ta
n(1/2*x))*A-I/a*b/(-a+b)*ln(I*a+I*b-a*tan(1/2*x)+b*tan(1/2*x))*A-1/2*I*a/b/(-a+b)*ln(I*a+I*b-a*tan(1/2*x)+b*ta
n(1/2*x))*B+1/2*I/(-a+b)*ln(I*a+I*b-a*tan(1/2*x)+b*tan(1/2*x))*B-1/2*I/a*b/(-a+b)*ln(I*a+I*b-a*tan(1/2*x)+b*ta
n(1/2*x))*B+1/2*I/a^2*b^2/(-a+b)*ln(I*a+I*b-a*tan(1/2*x)+b*tan(1/2*x))*B-1/2*I*B/b*ln(tan(1/2*x)-I)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(x))/(a+b*cos(x)-I*b*sin(x)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.89394, size = 132, normalized size = 1.57 \begin{align*} \frac{B a^{2} x - i \, B a b e^{\left (i \, x\right )} +{\left (i \, B a^{2} - 2 i \, A a b + i \, B b^{2}\right )} \log \left (\frac{a e^{\left (i \, x\right )} + b}{a}\right )}{2 \, a^{2} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(x))/(a+b*cos(x)-I*b*sin(x)),x, algorithm="fricas")

[Out]

1/2*(B*a^2*x - I*B*a*b*e^(I*x) + (I*B*a^2 - 2*I*A*a*b + I*B*b^2)*log((a*e^(I*x) + b)/a))/(a^2*b)

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Sympy [A]  time = 0.884295, size = 51, normalized size = 0.61 \begin{align*} \left (- \frac{i A}{a} + \frac{i B}{2 b} + \frac{i B b}{2 a^{2}}\right ) \log{\left (e^{i x} + \frac{b}{a} \right )} + \frac{B a x - i B b e^{i x}}{2 a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(x))/(a+b*cos(x)-I*b*sin(x)),x)

[Out]

(-I*A/a + I*B/(2*b) + I*B*b/(2*a**2))*log(exp(I*x) + b/a) + (B*a*x - I*B*b*exp(I*x))/(2*a*b)

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Giac [B]  time = 1.15591, size = 209, normalized size = 2.49 \begin{align*} -\frac{2 \,{\left (B a^{3} - 2 \, A a^{2} b - B a^{2} b + 2 \, A a b^{2} + B a b^{2} - B b^{3}\right )} \log \left (-a \tan \left (\frac{1}{2} \, x\right ) + b \tan \left (\frac{1}{2} \, x\right ) + i \, a + i \, b\right )}{4 i \, a^{3} b - 4 i \, a^{2} b^{2}} - \frac{i \, B \log \left (\tan \left (\frac{1}{2} \, x\right ) - i\right )}{2 \, b} - \frac{{\left (-2 i \, A a + i \, B b\right )} \log \left (\tan \left (\frac{1}{2} \, x\right ) + i\right )}{2 \, a^{2}} - \frac{2 i \, A a \tan \left (\frac{1}{2} \, x\right ) - i \, B b \tan \left (\frac{1}{2} \, x\right ) - 2 \, A a - 2 \, B a + B b}{2 \, a^{2}{\left (\tan \left (\frac{1}{2} \, x\right ) + i\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(x))/(a+b*cos(x)-I*b*sin(x)),x, algorithm="giac")

[Out]

-2*(B*a^3 - 2*A*a^2*b - B*a^2*b + 2*A*a*b^2 + B*a*b^2 - B*b^3)*log(-a*tan(1/2*x) + b*tan(1/2*x) + I*a + I*b)/(
4*I*a^3*b - 4*I*a^2*b^2) - 1/2*I*B*log(tan(1/2*x) - I)/b - 1/2*(-2*I*A*a + I*B*b)*log(tan(1/2*x) + I)/a^2 - 1/
2*(2*I*A*a*tan(1/2*x) - I*B*b*tan(1/2*x) - 2*A*a - 2*B*a + B*b)/(a^2*(tan(1/2*x) + I))