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

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

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 105, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.44, 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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fricas [A]  time = 1.46, size = 89, normalized size = 0.85 \[ \frac {{\left ({\left (i \, B - C\right )} a b + {\left (2 \, A a b - {\left (B + i \, C\right )} b^{2}\right )} x e^{\left (i \, x\right )} + {\left ({\left (-i \, B - C\right )} a^{2} + 2 i \, A a b + {\left (-i \, B + C\right )} b^{2}\right )} e^{\left (i \, x\right )} \log \left (\frac {b e^{\left (i \, x\right )} + a}{b}\right )\right )} e^{\left (-i \, x\right )}}{2 \, a^{2} b} \]

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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giac [B]  time = 0.15, size = 203, normalized size = 1.93 \[ -\frac {{\left (-2 i \, A a + i \, B b - C b\right )} \log \left (-a \tan \left (\frac {1}{2} \, x\right )^{2} + b \tan \left (\frac {1}{2} \, x\right )^{2} - 2 i \, a \tan \left (\frac {1}{2} \, x\right ) + a + b\right )}{4 \, a^{2}} - \frac {{\left (2 i \, A a - i \, B b + C b\right )} \log \left (\tan \left (\frac {1}{2} \, x\right ) - i\right )}{2 \, a^{2}} + \frac {{\left (2 \, B a^{2} - 2 i \, C a^{2} - 2 \, A a b + B b^{2} + i \, C b^{2}\right )} {\left (x + 2 \, \arctan \left (\frac {-i \, a \cos \relax (x) - a \sin \relax (x) - i \, a}{a \cos \relax (x) - i \, a \sin \relax (x) - a + 2 \, b}\right )\right )}}{4 \, a^{2} b} - \frac {-2 i \, A a \tan \left (\frac {1}{2} \, x\right ) + i \, B b \tan \left (\frac {1}{2} \, x\right ) - C b \tan \left (\frac {1}{2} \, x\right ) - 2 \, A a - 2 \, B a - 2 i \, C a + B b + i \, C b}{2 \, a^{2} {\left (\tan \left (\frac {1}{2} \, x\right ) - i\right )}} \]

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]

-1/4*(-2*I*A*a + I*B*b - C*b)*log(-a*tan(1/2*x)^2 + b*tan(1/2*x)^2 - 2*I*a*tan(1/2*x) + a + b)/a^2 - 1/2*(2*I*
A*a - I*B*b + C*b)*log(tan(1/2*x) - I)/a^2 + 1/4*(2*B*a^2 - 2*I*C*a^2 - 2*A*a*b + B*b^2 + I*C*b^2)*(x + 2*arct
an((-I*a*cos(x) - a*sin(x) - I*a)/(a*cos(x) - I*a*sin(x) - a + 2*b)))/(a^2*b) - 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))

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maple [B]  time = 0.18, size = 257, normalized size = 2.45 \[ \frac {C \ln \left (\tan \left (\frac {x}{2}\right )+i\right )}{2 b}+\frac {i B \ln \left (\tan \left (\frac {x}{2}\right )+i\right )}{2 b}+\frac {i C}{a \left (\tan \left (\frac {x}{2}\right )-i\right )}+\frac {B}{a \left (\tan \left (\frac {x}{2}\right )-i\right )}-\frac {i \ln \left (\tan \left (\frac {x}{2}\right )-i\right ) A}{a}+\frac {i \ln \left (\tan \left (\frac {x}{2}\right )-i\right ) b B}{2 a^{2}}-\frac {\ln \left (\tan \left (\frac {x}{2}\right )-i\right ) b C}{2 a^{2}}-\frac {\ln \left (i a +i b +a \tan \left (\frac {x}{2}\right )-b \tan \left (\frac {x}{2}\right )\right ) C}{2 b}+\frac {b \ln \left (i a +i b +a \tan \left (\frac {x}{2}\right )-b \tan \left (\frac {x}{2}\right )\right ) C}{2 a^{2}}+\frac {i \ln \left (i a +i b +a \tan \left (\frac {x}{2}\right )-b \tan \left (\frac {x}{2}\right )\right ) A}{a}-\frac {i \ln \left (i a +i b +a \tan \left (\frac {x}{2}\right )-b \tan \left (\frac {x}{2}\right )\right ) B}{2 b}-\frac {i b \ln \left (i a +i b +a \tan \left (\frac {x}{2}\right )-b \tan \left (\frac {x}{2}\right )\right ) B}{2 a^{2}} \]

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*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-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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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 >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [B]  time = 6.93, size = 132, normalized size = 1.26 \[ -\ln \left (a+b-a\,\mathrm {tan}\left (\frac {x}{2}\right )\,1{}\mathrm {i}+b\,\mathrm {tan}\left (\frac {x}{2}\right )\,1{}\mathrm {i}\right )\,\left (\frac {\frac {C}{2}+\frac {B\,1{}\mathrm {i}}{2}}{b}-\frac {\frac {C\,b^2}{2}-\frac {B\,b^2\,1{}\mathrm {i}}{2}+A\,a\,b\,1{}\mathrm {i}}{a^2\,b}\right )+\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )+1{}\mathrm {i}\right )\,\left (C+B\,1{}\mathrm {i}\right )}{2\,b}+\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )-\mathrm {i}\right )\,\left (B\,b-2\,A\,a+C\,b\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,a^2}+\frac {5\,B+C\,5{}\mathrm {i}}{5\,a\,\left (\mathrm {tan}\left (\frac {x}{2}\right )-\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cos(x) + C*sin(x))/(a + b*cos(x) + b*sin(x)*1i),x)

[Out]

(log(tan(x/2) + 1i)*(B*1i + C))/(2*b) - log(a + b - a*tan(x/2)*1i + b*tan(x/2)*1i)*(((B*1i)/2 + C/2)/b - ((C*b
^2)/2 - (B*b^2*1i)/2 + A*a*b*1i)/(a^2*b)) + (log(tan(x/2) - 1i)*(B*b - 2*A*a + C*b*1i)*1i)/(2*a^2) + (5*B + C*
5i)/(5*a*(tan(x/2) - 1i))

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sympy [A]  time = 1.37, size = 138, normalized size = 1.31 \[ \begin {cases} - \frac {\left (- i B + C\right ) e^{- i x}}{2 a} & \text {for}\: 2 a \neq 0 \\x \left (- \frac {2 A a - B b - i C b}{2 a^{2}} - \frac {i \left (2 i A a + i B a - i B b - C a + C b\right )}{2 a^{2}}\right ) & \text {otherwise} \end {cases} - \frac {x \left (- 2 A a + B b + i C b\right )}{2 a^{2}} - \frac {i \left (- 2 A a b + B a^{2} + B b^{2} - i C a^{2} + i C b^{2}\right ) \log {\left (\frac {a}{b} + e^{i x} \right )}}{2 a^{2} b} \]

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]

Piecewise((-(-I*B + C)*exp(-I*x)/(2*a), Ne(2*a, 0)), (x*(-(2*A*a - B*b - I*C*b)/(2*a**2) - I*(2*I*A*a + I*B*a
- I*B*b - C*a + C*b)/(2*a**2)), True)) - x*(-2*A*a + B*b + I*C*b)/(2*a**2) - I*(-2*A*a*b + B*a**2 + B*b**2 - I
*C*a**2 + I*C*b**2)*log(a/b + exp(I*x))/(2*a**2*b)

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