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

Optimal. Leaf size=103 \[ -\frac {i \left (-\left (a^2 (B+i C)\right )+2 a A b-b^2 (B-i C)\right ) \log (a-i b \sin (x)+b \cos (x))}{2 a^2 b}+\frac {x (2 a A-b B+i b C)}{2 a^2}-\frac {(C+i B) (\cos (x)+i \sin (x))}{2 a} \]

[Out]

1/2*(2*a*A-b*B+I*b*C)*x/a^2-1/2*I*(2*a*A*b-b^2*(B-I*C)-a^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 = 103, 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} \[ -\frac {i \left (a^2 (-(B+i C))+2 a A b-b^2 (B-i C)\right ) \log (a-i b \sin (x)+b \cos (x))}{2 a^2 b}+\frac {x (2 a A-b B+i b C)}{2 a^2}-\frac {(C+i B) (\cos (x)+i \sin (x))}{2 a} \]

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*b*C)*x)/(2*a^2) - ((I/2)*(2*a*A*b - b^2*(B - I*C) - a^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 b C) x}{2 a^2}-\frac {i \left (2 a A b-b^2 (B-i C)-a^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 = 0.45, size = 167, normalized size = 1.62 \[ \frac {\frac {\left (i a^2 (B+i C)-2 i a A b+b^2 (C+i B)\right ) \log \left (a^2+2 a b \cos (x)+b^2\right )}{b}+\frac {2 \left (a^2 (B+i C)-2 a A b+b^2 (B-i C)\right ) \tan ^{-1}\left (\frac {(a+b) \cot \left (\frac {x}{2}\right )}{a-b}\right )}{b}+x \left (\frac {a^2 (B+i C)}{b}+2 a A-b (B-i C)\right )+2 a (B-i C) \sin (x)-2 i a (B-i C) \cos (x)}{4 a^2} \]

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

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fricas [A]  time = 1.09, size = 73, normalized size = 0.71 \[ \frac {{\left (B + i \, C\right )} a^{2} x + {\left (-i \, B - C\right )} a b e^{\left (i \, x\right )} + {\left ({\left (i \, B - C\right )} a^{2} - 2 i \, A a b + {\left (i \, B + C\right )} b^{2}\right )} \log \left (\frac {a e^{\left (i \, x\right )} + b}{a}\right )}{2 \, a^{2} b} \]

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

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giac [B]  time = 0.15, size = 203, normalized size = 1.97 \[ -\frac {{\left (2 i \, A a - i \, B b - C b\right )} \log \left (-a \tan \left (\frac {1}{2} \, x\right )^{2} + b \tan \left (\frac {1}{2} \, x\right )^{2} + 2 i \, a \tan \left (\frac {1}{2} \, x\right ) + a + b\right )}{4 \, a^{2}} - \frac {{\left (-2 i \, A a + i \, B b + C b\right )} \log \left (\tan \left (\frac {1}{2} \, x\right ) + i\right )}{2 \, a^{2}} + \frac {{\left (2 \, B a^{2} + 2 i \, C a^{2} - 2 \, A a b + B b^{2} - i \, C b^{2}\right )} {\left (x + 2 \, \arctan \left (\frac {i \, a \cos \relax (x) - a \sin \relax (x) + i \, a}{a \cos \relax (x) + i \, a \sin \relax (x) - a + 2 \, b}\right )\right )}}{4 \, a^{2} b} - \frac {2 i \, A a \tan \left (\frac {1}{2} \, x\right ) - i \, B b \tan \left (\frac {1}{2} \, x\right ) - C b \tan \left (\frac {1}{2} \, x\right ) - 2 \, A a - 2 \, B a + 2 i \, C a + B b - i \, C b}{2 \, a^{2} {\left (\tan \left (\frac {1}{2} \, x\right ) + i\right )}} \]

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 = 0.19, size = 475, normalized size = 4.61 \[ \frac {i \ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right ) B}{-2 a +2 b}+\frac {B}{a \left (\tan \left (\frac {x}{2}\right )+i\right )}-\frac {i b \ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right ) A}{a \left (-a +b \right )}-\frac {i B \ln \left (\tan \left (\frac {x}{2}\right )-i\right )}{2 b}-\frac {\ln \left (\tan \left (\frac {x}{2}\right )+i\right ) b C}{2 a^{2}}+\frac {C \ln \left (\tan \left (\frac {x}{2}\right )-i\right )}{2 b}-\frac {i C}{a \left (\tan \left (\frac {x}{2}\right )+i\right )}+\frac {a \ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right ) C}{2 b \left (-a +b \right )}-\frac {\ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right ) C}{2 \left (-a +b \right )}-\frac {b \ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right ) C}{2 a \left (-a +b \right )}+\frac {b^{2} \ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right ) C}{2 a^{2} \left (-a +b \right )}-\frac {i \ln \left (\tan \left (\frac {x}{2}\right )+i\right ) b B}{2 a^{2}}-\frac {i a \ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right ) B}{2 b \left (-a +b \right )}+\frac {i \ln \left (\tan \left (\frac {x}{2}\right )+i\right ) A}{a}-\frac {i b \ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right ) B}{2 a \left (-a +b \right )}+\frac {i \ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right ) A}{-a +b}+\frac {i b^{2} \ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right ) B}{2 a^{2} \left (-a +b \right )} \]

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

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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 = 6.86, size = 133, normalized size = 1.29 \[ \ln \left (a+b+a\,\mathrm {tan}\left (\frac {x}{2}\right )\,1{}\mathrm {i}-b\,\mathrm {tan}\left (\frac {x}{2}\right )\,1{}\mathrm {i}\right )\,\left (\frac {-\frac {C}{2}+\frac {B\,1{}\mathrm {i}}{2}}{b}+\frac {\frac {B\,b^2\,1{}\mathrm {i}}{2}+\frac {C\,b^2}{2}-A\,a\,b\,1{}\mathrm {i}}{a^2\,b}\right )+\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )+1{}\mathrm {i}\right )\,\left (2\,A\,a-B\,b+C\,b\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,a^2}+\frac {5\,B-C\,5{}\mathrm {i}}{5\,a\,\left (\mathrm {tan}\left (\frac {x}{2}\right )+1{}\mathrm {i}\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )-\mathrm {i}\right )\,\left (-C+B\,1{}\mathrm {i}\right )}{2\,b} \]

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

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sympy [A]  time = 1.27, size = 109, normalized size = 1.06 \[ \begin {cases} - \frac {\left (i B + C\right ) e^{i x}}{2 a} & \text {for}\: 2 a \neq 0 \\x \left (- \frac {B + i C}{2 b} + \frac {B a + B b + i C a - i C b}{2 a b}\right ) & \text {otherwise} \end {cases} - \frac {x \left (- B - i C\right )}{2 b} + \frac {i \left (- 2 A a b + B a^{2} + B b^{2} + i C a^{2} - i C b^{2}\right ) \log {\left (e^{i x} + \frac {b}{a} \right )}}{2 a^{2} b} \]

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

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