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

Optimal. Leaf size=105 \[ \frac{i \left (a^2 (-(B-i C))+2 a A b-b^2 (B+i C)\right ) \log (a+i b \sin (x)+b \cos (x))}{2 a^2 b}+\frac{x (2 a A-b (B+i C))}{2 a^2}+\frac{(-C+i B) (\cos (x)-i \sin (x))}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.0736377, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.038, Rules used = {3130} \[ \frac{i \left (a^2 (-(B-i C))+2 a A b-b^2 (B+i C)\right ) \log (a+i b \sin (x)+b \cos (x))}{2 a^2 b}+\frac{x (2 a A-b (B+i C))}{2 a^2}+\frac{(-C+i B) (\cos (x)-i \sin (x))}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3130

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

Rubi steps

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

Mathematica [A]  time = 0.412856, size = 165, normalized size = 1.57 \[ \frac{x \left (a^2 (B-i C)+2 a A b-b^2 (B+i C)\right )+\left (a^2 (-C-i B)+2 i a A b+b^2 (C-i B)\right ) \log \left (a^2+2 a b \cos (x)+b^2\right )+2 \left (a^2 (B-i C)-2 a A b+b^2 (B+i C)\right ) \tan ^{-1}\left (\frac{(a+b) \cot \left (\frac{x}{2}\right )}{a-b}\right )+2 a b (B+i C) \sin (x)+2 i a b (B+i C) \cos (x)}{4 a^2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.08, size = 257, normalized size = 2.5 \begin{align*} -{\frac{C}{2\,b}\ln \left ( ia+ib+a\tan \left ({\frac{x}{2}} \right ) -b\tan \left ({\frac{x}{2}} \right ) \right ) }+{\frac{bC}{2\,{a}^{2}}\ln \left ( ia+ib+a\tan \left ({\frac{x}{2}} \right ) -b\tan \left ({\frac{x}{2}} \right ) \right ) }+{\frac{iA}{a}\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 ( ia+ib+a\tan \left ({\frac{x}{2}} \right ) -b\tan \left ({\frac{x}{2}} \right ) \right ) }-{\frac{{\frac{i}{2}}bB}{{a}^{2}}\ln \left ( ia+ib+a\tan \left ({\frac{x}{2}} \right ) -b\tan \left ({\frac{x}{2}} \right ) \right ) }+{\frac{C}{2\,b}\ln \left ( \tan \left ({\frac{x}{2}} \right ) +i \right ) }+{\frac{{\frac{i}{2}}B}{b}\ln \left ( \tan \left ({\frac{x}{2}} \right ) +i \right ) }+{\frac{iC}{a} \left ( \tan \left ({\frac{x}{2}} \right ) -i \right ) ^{-1}}+{\frac{B}{a} \left ( \tan \left ({\frac{x}{2}} \right ) -i \right ) ^{-1}}-{\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{bC}{2\,{a}^{2}}\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)+C*sin(x))/(a+b*cos(x)+I*b*sin(x)),x)

[Out]

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

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.14681, size = 209, normalized size = 1.99 \begin{align*} \frac{{\left ({\left (i \, B - C\right )} a b +{\left (2 \, A a b -{\left (B + i \, C\right )} b^{2}\right )} x e^{\left (i \, x\right )} +{\left ({\left (-i \, B - C\right )} a^{2} + 2 i \, A a b +{\left (-i \, B + C\right )} b^{2}\right )} e^{\left (i \, x\right )} \log \left (\frac{b e^{\left (i \, x\right )} + a}{b}\right )\right )} e^{\left (-i \, x\right )}}{2 \, a^{2} b} \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)+I*b*sin(x)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.33136, size = 87, normalized size = 0.83 \begin{align*} \left (\frac{i A}{a} - \frac{i B}{2 b} - \frac{i B b}{2 a^{2}} - \frac{C}{2 b} + \frac{C b}{2 a^{2}}\right ) \log{\left (\frac{a}{b} + e^{i x} \right )} + \frac{2 A a x + i B a e^{- i x} - B b x - C a e^{- i x} - i C b x}{2 a^{2}} \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)+I*b*sin(x)),x)

[Out]

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

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Giac [B]  time = 1.12666, size = 278, normalized size = 2.65 \begin{align*} -\frac{2 \,{\left (B a^{3} - i \, C a^{3} - 2 \, A a^{2} b - B a^{2} b + i \, C a^{2} b + 2 \, A a b^{2} + B a b^{2} + i \, C a b^{2} - B b^{3} - i \, C 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{{\left (-i \, B - C\right )} \log \left (\tan \left (\frac{1}{2} \, x\right ) + i\right )}{2 \, b} - \frac{{\left (2 i \, A a - i \, B b + C 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 ) - C b \tan \left (\frac{1}{2} \, x\right ) - 2 \, A a - 2 \, B a - 2 i \, C a + B b + i \, C 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)+C*sin(x))/(a+b*cos(x)+I*b*sin(x)),x, algorithm="giac")

[Out]

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