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]

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

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Rubi [A]  time = 0.07837, antiderivative size = 85, normalized size of antiderivative = 0.94, number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.04, 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*}

Mathematica [B]  time = 0.276705, 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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Maple [B]  time = 0.085, size = 388, normalized size = 4.3 \begin{align*}{\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{{\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 ) }+{\frac{Ca}{2\,b \left ( -a+b \right ) }\ln \left ( ia+ib-a\tan \left ({\frac{x}{2}} \right ) +b\tan \left ({\frac{x}{2}} \right ) \right ) }-{\frac{C}{-2\,a+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 \left ( -a+b \right ) }\ln \left ( ia+ib-a\tan \left ({\frac{x}{2}} \right ) +b\tan \left ({\frac{x}{2}} \right ) \right ) }+{\frac{{b}^{2}C}{2\,{a}^{2} \left ( -a+b \right ) }\ln \left ( ia+ib-a\tan \left ({\frac{x}{2}} \right ) +b\tan \left ({\frac{x}{2}} \right ) \right ) }-{\frac{{\frac{i}{2}}aB}{b \left ( -a+b \right ) }\ln \left ( ia+ib-a\tan \left ({\frac{x}{2}} \right ) +b\tan \left ({\frac{x}{2}} \right ) \right ) }+{\frac{{\frac{i}{2}}B}{-a+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 \left ( -a+b \right ) }\ln \left ( ia+ib-a\tan \left ({\frac{x}{2}} \right ) +b\tan \left ({\frac{x}{2}} \right ) \right ) }+{\frac{{\frac{i}{2}}{b}^{2}B}{{a}^{2} \left ( -a+b \right ) }\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 ) } \end{align*}

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

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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((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 = 2.05515, size = 153, normalized size = 1.7 \begin{align*} \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} \end{align*}

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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Sympy [A]  time = 1.06271, size = 75, normalized size = 0.83 \begin{align*} \frac{B a x - i B b e^{i x} + i C a x - C b e^{i x}}{2 a b} + \frac{\left (i B a^{2} + i B b^{2} - C a^{2} + C b^{2}\right ) \log{\left (e^{i x} + \frac{b}{a} \right )}}{2 a^{2} b} \end{align*}

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]

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

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Giac [B]  time = 1.16851, size = 238, normalized size = 2.64 \begin{align*} -\frac{2 \,{\left (B a^{3} + i \, C a^{3} - B a^{2} b - i \, C a^{2} b + 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 (i \, B b + C b\right )} \log \left (\tan \left (\frac{1}{2} \, x\right ) + i\right )}{2 \, a^{2}} - \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 )}} \end{align*}

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]

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