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]

-(b*(B + I*C)*x)/(2*a^2) - ((I*b^2*(B + I*C) + a^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 = 0.0775119, antiderivative size = 87, normalized size of antiderivative = 0.95, number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.04, 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*}

Mathematica [B]  time = 0.296928, 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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Maple [B]  time = 0.079, size = 212, normalized size = 2.3 \begin{align*} -{\frac{C}{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}^{2}}\ln \left ( ia+ib+a\tan \left ({\frac{x}{2}} \right ) -b\tan \left ({\frac{x}{2}} \right ) \right ) }-{\frac{{\frac{i}{2}}B}{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}^{2}}\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 ) }+{\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 ) } \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]

-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+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

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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 = 2.08525, size = 178, normalized size = 1.93 \begin{align*} -\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} \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)*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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Sympy [A]  time = 1.93881, size = 75, normalized size = 0.82 \begin{align*} \frac{i B a e^{- i x} - B b x - C a e^{- i x} - i C b x}{2 a^{2}} + \frac{\left (- i B a^{2} - i B b^{2} - C a^{2} + C b^{2}\right ) \log{\left (\frac{a}{b} + e^{i x} \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]

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

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Giac [B]  time = 1.16486, size = 238, normalized size = 2.59 \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))