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

Optimal. Leaf size=92 \[ -\frac {\left (a^2 (C+i B)+i b^2 (B+i C)\right ) \log (a+i b \sin (x)+b \cos (x))}{2 a^2 b}-\frac {b x (B+i C)}{2 a^2}+\frac {(-C+i B) (\cos (x)-i \sin (x))}{2 a} \]

[Out]

-1/2*b*(B+I*C)*x/a^2-1/2*(I*b^2*(B+I*C)+a^2*(I*B+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 = 0.08, antiderivative size = 87, normalized size of antiderivative = 0.95, number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {3130} \[ -\frac {\left (\frac {i b^2 (B+i C)}{a^2}+i B+C\right ) \log (a+i b \sin (x)+b \cos (x))}{2 b}-\frac {b x (B+i C)}{2 a^2}+\frac {(-C+i B) (\cos (x)-i \sin (x))}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(B + I*C)*x)/(2*a^2) - ((I*B + (I*b^2*(B + I*C))/a^2 + C)*Log[a + b*Cos[x] + I*b*Sin[x]])/(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 {B \cos (x)+C \sin (x)}{a+b \cos (x)+i b \sin (x)} \, dx &=-\frac {b (B+i C) x}{2 a^2}-\frac {\left (i B+\frac {i b^2 (B+i C)}{a^2}+C\right ) \log (a+b \cos (x)+i b \sin (x))}{2 b}+\frac {(i B-C) (\cos (x)-i \sin (x))}{2 a}\\ \end {align*}

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Mathematica [B]  time = 0.32, size = 195, normalized size = 2.12 \[ \frac {x \left (a^2 B-i a^2 C-b^2 B-i b^2 C\right )}{4 a^2 b}-\frac {i \left (a^2 B-i a^2 C+b^2 B+i b^2 C\right ) \log \left (a^2+2 a b \cos (x)+b^2\right )}{4 a^2 b}-\frac {\left (a^2 B-i a^2 C+b^2 B+i b^2 C\right ) \tan ^{-1}\left (\frac {(a+b) \cos \left (\frac {x}{2}\right )}{b \sin \left (\frac {x}{2}\right )-a \sin \left (\frac {x}{2}\right )}\right )}{2 a^2 b}+\frac {(B+i C) \sin (x)}{2 a}+\frac {i (B+i C) \cos (x)}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.05, size = 78, normalized size = 0.85 \[ -\frac {{\left ({\left (B + i \, C\right )} b^{2} x e^{\left (i \, x\right )} - {\left (i \, B - C\right )} a b - {\left ({\left (-i \, B - C\right )} a^{2} + {\left (-i \, B + C\right )} b^{2}\right )} e^{\left (i \, x\right )} \log \left (\frac {b e^{\left (i \, x\right )} + a}{b}\right )\right )} e^{\left (-i \, x\right )}}{2 \, a^{2} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((B + I*C)*b^2*x*e^(I*x) - (I*B - C)*a*b - ((-I*B - C)*a^2 + (-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 = 0.16, size = 178, normalized size = 1.93 \[ -\frac {{\left (i \, B b - C b\right )} \log \left (-a \tan \left (\frac {1}{2} \, x\right )^{2} + b \tan \left (\frac {1}{2} \, x\right )^{2} - 2 i \, a \tan \left (\frac {1}{2} \, x\right ) + a + b\right )}{4 \, a^{2}} - \frac {{\left (-i \, B b + C b\right )} \log \left (\tan \left (\frac {1}{2} \, x\right ) - i\right )}{2 \, a^{2}} + \frac {{\left (2 \, B a^{2} - 2 i \, C a^{2} + B b^{2} + i \, C b^{2}\right )} {\left (x + 2 \, \arctan \left (\frac {-i \, a \cos \relax (x) - a \sin \relax (x) - i \, a}{a \cos \relax (x) - i \, a \sin \relax (x) - a + 2 \, b}\right )\right )}}{4 \, a^{2} b} - \frac {i \, B b \tan \left (\frac {1}{2} \, x\right ) - C b \tan \left (\frac {1}{2} \, x\right ) - 2 \, B a - 2 i \, C a + B b + i \, C b}{2 \, a^{2} {\left (\tan \left (\frac {1}{2} \, x\right ) - i\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(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*(-I*B*b + C*b)*l
og(tan(1/2*x) - I)/a^2 + 1/4*(2*B*a^2 - 2*I*C*a^2 + B*b^2 + I*C*b^2)*(x + 2*arctan((-I*a*cos(x) - a*sin(x) - I
*a)/(a*cos(x) - I*a*sin(x) - a + 2*b)))/(a^2*b) - 1/2*(I*B*b*tan(1/2*x) - C*b*tan(1/2*x) - 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 = 0.20, size = 212, normalized size = 2.30 \[ \frac {C \ln \left (\tan \left (\frac {x}{2}\right )+i\right )}{2 b}+\frac {i B \ln \left (\tan \left (\frac {x}{2}\right )+i\right )}{2 b}+\frac {i C}{a \left (\tan \left (\frac {x}{2}\right )-i\right )}+\frac {B}{a \left (\tan \left (\frac {x}{2}\right )-i\right )}+\frac {i \ln \left (\tan \left (\frac {x}{2}\right )-i\right ) b B}{2 a^{2}}-\frac {\ln \left (\tan \left (\frac {x}{2}\right )-i\right ) b C}{2 a^{2}}-\frac {\ln \left (i a +i b +a \tan \left (\frac {x}{2}\right )-b \tan \left (\frac {x}{2}\right )\right ) C}{2 b}+\frac {b \ln \left (i a +i b +a \tan \left (\frac {x}{2}\right )-b \tan \left (\frac {x}{2}\right )\right ) C}{2 a^{2}}-\frac {i \ln \left (i a +i b +a \tan \left (\frac {x}{2}\right )-b \tan \left (\frac {x}{2}\right )\right ) B}{2 b}-\frac {i b \ln \left (i a +i b +a \tan \left (\frac {x}{2}\right )-b \tan \left (\frac {x}{2}\right )\right ) B}{2 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((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)+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-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((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 = 5.32, size = 118, normalized size = 1.28 \[ -\ln \left (a+b-a\,\mathrm {tan}\left (\frac {x}{2}\right )\,1{}\mathrm {i}+b\,\mathrm {tan}\left (\frac {x}{2}\right )\,1{}\mathrm {i}\right )\,\left (\frac {\frac {C}{2}+\frac {B\,1{}\mathrm {i}}{2}}{b}+\frac {-\frac {C\,b^2}{2}+\frac {B\,b^2\,1{}\mathrm {i}}{2}}{a^2\,b}\right )+\frac {B+C\,1{}\mathrm {i}}{a\,\left (\mathrm {tan}\left (\frac {x}{2}\right )-\mathrm {i}\right )}+\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )+1{}\mathrm {i}\right )\,\left (C+B\,1{}\mathrm {i}\right )}{2\,b}+\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )-\mathrm {i}\right )\,\left (-C\,b+B\,b\,1{}\mathrm {i}\right )}{2\,a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.85, size = 116, normalized size = 1.26 \[ \begin {cases} - \frac {\left (- i B + C\right ) e^{- i x}}{2 a} & \text {for}\: 2 a \neq 0 \\x \left (- \frac {- B b - i C b}{2 a^{2}} - \frac {i \left (i B a - i B b - C a + C b\right )}{2 a^{2}}\right ) & \text {otherwise} \end {cases} - \frac {x \left (B b + i C b\right )}{2 a^{2}} - \frac {i \left (B a^{2} + B b^{2} - i C a^{2} + i C b^{2}\right ) \log {\left (\frac {a}{b} + e^{i x} \right )}}{2 a^{2} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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*b - I*C*b)/(2*a**2) - I*(I*B*a - I*B*b - C*a + C
*b)/(2*a**2)), True)) - x*(B*b + I*C*b)/(2*a**2) - I*(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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