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

Optimal. Leaf size=90 \[ \frac {\left (i a^2 (B+i C)+b^2 (C+i B)\right ) \log (a-i b \sin (x)+b \cos (x))}{2 a^2 b}-\frac {b x (B-i C)}{2 a^2}-\frac {(C+i B) (\cos (x)+i \sin (x))}{2 a} \]

[Out]

-1/2*b*(B-I*C)*x/a^2+1/2*(I*a^2*(B+I*C)+b^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 = 85, normalized size of antiderivative = 0.94, 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} \[ -\frac {b x (B-i C)}{2 a^2}+\frac {1}{2} \left (\frac {b (C+i B)}{a^2}+\frac {i (B+i C)}{b}\right ) \log (a-i b \sin (x)+b \cos (x))-\frac {(C+i B) (\cos (x)+i \sin (x))}{2 a} \]

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*C))/b + (b*(I*B + C))/a^2)*Log[a + b*Cos[x] - I*b*Sin[x]])/2 - ((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 {1}{2} \left (\frac {i (B+i C)}{b}+\frac {b (i B+C)}{a^2}\right ) \log (a+b \cos (x)-i b \sin (x))-\frac {(i B+C) (\cos (x)+i \sin (x))}{2 a}\\ \end {align*}

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Mathematica [B]  time = 0.29, size = 195, normalized size = 2.17 \[ \frac {x \left (a^2 B+i a^2 C-b^2 B+i b^2 C\right )}{4 a^2 b}+\frac {i \left (a^2 B+i a^2 C+b^2 B-i b^2 C\right ) \log \left (a^2+2 a b \cos (x)+b^2\right )}{4 a^2 b}+\frac {\left (a^2 B+i a^2 C+b^2 B-i b^2 C\right ) \tan ^{-1}\left (\frac {(a+b) \cos \left (\frac {x}{2}\right )}{a \sin \left (\frac {x}{2}\right )-b \sin \left (\frac {x}{2}\right )}\right )}{2 a^2 b}+\frac {(B-i C) \sin (x)}{2 a}-\frac {i (B-i C) \cos (x)}{2 a} \]

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 = 0.96, size = 68, normalized size = 0.76 \[ \frac {{\left (B + i \, C\right )} a^{2} x + {\left (-i \, B - C\right )} a b e^{\left (i \, x\right )} + {\left ({\left (i \, B - C\right )} a^{2} + {\left (i \, B + C\right )} b^{2}\right )} \log \left (\frac {a e^{\left (i \, x\right )} + b}{a}\right )}{2 \, a^{2} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [B]  time = 0.16, size = 178, normalized size = 1.98 \[ -\frac {{\left (-i \, B b - C b\right )} \log \left (-a \tan \left (\frac {1}{2} \, x\right )^{2} + b \tan \left (\frac {1}{2} \, x\right )^{2} + 2 i \, a \tan \left (\frac {1}{2} \, x\right ) + a + b\right )}{4 \, a^{2}} - \frac {{\left (i \, B b + C b\right )} \log \left (\tan \left (\frac {1}{2} \, x\right ) + i\right )}{2 \, a^{2}} + \frac {{\left (2 \, B a^{2} + 2 i \, C a^{2} + B b^{2} - i \, C b^{2}\right )} {\left (x + 2 \, \arctan \left (\frac {i \, a \cos \relax (x) - a \sin \relax (x) + i \, a}{a \cos \relax (x) + i \, a \sin \relax (x) - a + 2 \, b}\right )\right )}}{4 \, a^{2} b} - \frac {-i \, B b \tan \left (\frac {1}{2} \, x\right ) - C b \tan \left (\frac {1}{2} \, x\right ) - 2 \, B a + 2 i \, C a + B b - i \, C b}{2 \, a^{2} {\left (\tan \left (\frac {1}{2} \, x\right ) + i\right )}} \]

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 = 0.19, size = 388, normalized size = 4.31 \[ -\frac {i C}{a \left (\tan \left (\frac {x}{2}\right )+i\right )}+\frac {B}{a \left (\tan \left (\frac {x}{2}\right )+i\right )}-\frac {i \ln \left (\tan \left (\frac {x}{2}\right )+i\right ) b B}{2 a^{2}}-\frac {\ln \left (\tan \left (\frac {x}{2}\right )+i\right ) b C}{2 a^{2}}+\frac {C \ln \left (\tan \left (\frac {x}{2}\right )-i\right )}{2 b}-\frac {i B \ln \left (\tan \left (\frac {x}{2}\right )-i\right )}{2 b}+\frac {a \ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right ) C}{2 b \left (-a +b \right )}-\frac {\ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right ) C}{2 \left (-a +b \right )}-\frac {b \ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right ) C}{2 a \left (-a +b \right )}+\frac {b^{2} \ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right ) C}{2 a^{2} \left (-a +b \right )}-\frac {i a \ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right ) B}{2 b \left (-a +b \right )}+\frac {i \ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right ) B}{-2 a +2 b}-\frac {i b \ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right ) B}{2 a \left (-a +b \right )}+\frac {i b^{2} \ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right ) B}{2 a^{2} \left (-a +b \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 = 4.51, size = 118, normalized size = 1.31 \[ \ln \left (a+b+a\,\mathrm {tan}\left (\frac {x}{2}\right )\,1{}\mathrm {i}-b\,\mathrm {tan}\left (\frac {x}{2}\right )\,1{}\mathrm {i}\right )\,\left (\frac {-\frac {C}{2}+\frac {B\,1{}\mathrm {i}}{2}}{b}+\frac {\frac {C\,b^2}{2}+\frac {B\,b^2\,1{}\mathrm {i}}{2}}{a^2\,b}\right )+\frac {B-C\,1{}\mathrm {i}}{a\,\left (\mathrm {tan}\left (\frac {x}{2}\right )+1{}\mathrm {i}\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )+1{}\mathrm {i}\right )\,\left (C\,b+B\,b\,1{}\mathrm {i}\right )}{2\,a^2}-\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )-\mathrm {i}\right )\,\left (-C+B\,1{}\mathrm {i}\right )}{2\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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sympy [A]  time = 0.81, size = 102, normalized size = 1.13 \[ \begin {cases} - \frac {\left (i B + C\right ) e^{i x}}{2 a} & \text {for}\: 2 a \neq 0 \\x \left (- \frac {B + i C}{2 b} + \frac {B a + B b + i C a - i C b}{2 a b}\right ) & \text {otherwise} \end {cases} - \frac {x \left (- B - i C\right )}{2 b} + \frac {i \left (B a^{2} + B b^{2} + i C a^{2} - i C b^{2}\right ) \log {\left (e^{i x} + \frac {b}{a} \right )}}{2 a^{2} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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*(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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