\(\int \frac {A+B \cos (x)+C \sin (x)}{a+b \cos (x)-i b \sin (x)} \, dx\) [554]

Optimal result

Integrand size = 26, antiderivative size = 103 \[ \int \frac {A+B \cos (x)+C \sin (x)}{a+b \cos (x)-i b \sin (x)} \, dx=\frac {(2 a A-b B+i b C) x}{2 a^2}-\frac {i \left (2 a A b-b^2 (B-i C)-a^2 (B+i C)\right ) \log (a+b \cos (x)-i b \sin (x))}{2 a^2 b}-\frac {(i B+C) (\cos (x)+i \sin (x))}{2 a} \]

[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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {3209} \[ \int \frac {A+B \cos (x)+C \sin (x)}{a+b \cos (x)-i b \sin (x)} \, dx=-\frac {i \left (-\left (a^2 (B+i C)\right )+2 a A b-b^2 (B-i C)\right ) \log (a-i b \sin (x)+b \cos (x))}{2 a^2 b}+\frac {x (2 a A-b B+i b C)}{2 a^2}-\frac {(C+i B) (\cos (x)+i \sin (x))}{2 a} \]

[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 3209

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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.62 \[ \int \frac {A+B \cos (x)+C \sin (x)}{a+b \cos (x)-i b \sin (x)} \, dx=\frac {\left (2 a A-b (B-i C)+\frac {a^2 (B+i C)}{b}\right ) x+\frac {2 \left (-2 a A b+b^2 (B-i C)+a^2 (B+i C)\right ) \arctan \left (\frac {(a+b) \cot \left (\frac {x}{2}\right )}{a-b}\right )}{b}-2 i a (B-i C) \cos (x)+\frac {\left (-2 i a A b+i a^2 (B+i C)+b^2 (i B+C)\right ) \log \left (a^2+b^2+2 a b \cos (x)\right )}{b}+2 a (B-i C) \sin (x)}{4 a^2} \]

[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)

Maple [A] (verified)

Time = 1.32 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.31

method	result	size
risch	\(-\frac {C \,{\mathrm e}^{i x}}{2 a}-\frac {i B \,{\mathrm e}^{i x}}{2 a}+\frac {i x C}{2 b}+\frac {B x}{2 b}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {b}{a}\right ) C}{2 b}+\frac {b \ln \left ({\mathrm e}^{i x}+\frac {b}{a}\right ) C}{2 a^{2}}-\frac {i \ln \left ({\mathrm e}^{i x}+\frac {b}{a}\right ) A}{a}+\frac {i \ln \left ({\mathrm e}^{i x}+\frac {b}{a}\right ) B}{2 b}+\frac {i b \ln \left ({\mathrm e}^{i x}+\frac {b}{a}\right ) B}{2 a^{2}}\)	\(135\)
default	\(\frac {i \left (-i C \,a^{2}+i C \,b^{2}+2 A a b -B \,a^{2}-B \,b^{2}\right ) \left (a -b \right ) \ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right )}{2 a^{2} b \left (-a +b \right )}-\frac {i \left (i C +B \right ) \ln \left (-i+\tan \left (\frac {x}{2}\right )\right )}{2 b}-\frac {i C -B}{a \left (\tan \left (\frac {x}{2}\right )+i\right )}+\frac {\left (2 i A a -i B b -b C \right ) \ln \left (\tan \left (\frac {x}{2}\right )+i\right )}{2 a^{2}}\)	\(148\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.71 \[ \int \frac {A+B \cos (x)+C \sin (x)}{a+b \cos (x)-i b \sin (x)} \, dx=\frac {{\left (B + i \, C\right )} a^{2} x + {\left (-i \, B - C\right )} a b e^{\left (i \, x\right )} + {\left ({\left (i \, B - C\right )} a^{2} - 2 i \, A a b + {\left (i \, B + C\right )} b^{2}\right )} \log \left (\frac {a e^{\left (i \, x\right )} + b}{a}\right )}{2 \, a^{2} b} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.69 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.02 \[ \int \frac {A+B \cos (x)+C \sin (x)}{a+b \cos (x)-i b \sin (x)} \, dx=\begin {cases} \frac {\left (- i B - C\right ) e^{i x}}{2 a} & \text {for}\: a \neq 0 \\x \left (- \frac {B + i C}{2 b} + \frac {B a + B b + i C a - i C b}{2 a b}\right ) & \text {otherwise} \end {cases} + \frac {x \left (B + i C\right )}{2 b} + \frac {i \left (- 2 A a b + B a^{2} + B b^{2} + i C a^{2} - i C b^{2}\right ) \log {\left (e^{i x} + \frac {b}{a} \right )}}{2 a^{2} b} \]

[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(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)

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \cos (x)+C \sin (x)}{a+b \cos (x)-i b \sin (x)} \, dx=\text {Exception raised: RuntimeError} \]

[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: expt: undefined: 0 to a negative exponent.

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (85) = 170\).

Time = 0.27 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.97 \[ \int \frac {A+B \cos (x)+C \sin (x)}{a+b \cos (x)-i b \sin (x)} \, dx=-\frac {{\left (2 i \, A a - i \, B b - C b\right )} \log \left (-a \tan \left (\frac {1}{2} \, x\right )^{2} + b \tan \left (\frac {1}{2} \, x\right )^{2} + 2 i \, a \tan \left (\frac {1}{2} \, x\right ) + a + b\right )}{4 \, a^{2}} - \frac {{\left (-2 i \, A a + i \, B b + C b\right )} \log \left (\tan \left (\frac {1}{2} \, x\right ) + i\right )}{2 \, a^{2}} + \frac {{\left (2 \, B a^{2} + 2 i \, C a^{2} - 2 \, A a b + B b^{2} - i \, C b^{2}\right )} {\left (x + 2 \, \arctan \left (\frac {i \, a \cos \left (x\right ) - a \sin \left (x\right ) + i \, a}{a \cos \left (x\right ) + i \, a \sin \left (x\right ) - a + 2 \, b}\right )\right )}}{4 \, a^{2} b} - \frac {2 i \, A a \tan \left (\frac {1}{2} \, x\right ) - i \, B b \tan \left (\frac {1}{2} \, x\right ) - C b \tan \left (\frac {1}{2} \, x\right ) - 2 \, A a - 2 \, B a + 2 i \, C a + B b - i \, C b}{2 \, a^{2} {\left (\tan \left (\frac {1}{2} \, x\right ) + i\right )}} \]

[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))

Mupad [B] (verification not implemented)

Time = 31.92 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.29 \[ \int \frac {A+B \cos (x)+C \sin (x)}{a+b \cos (x)-i b \sin (x)} \, dx=\ln \left (a+b+a\,\mathrm {tan}\left (\frac {x}{2}\right )\,1{}\mathrm {i}-b\,\mathrm {tan}\left (\frac {x}{2}\right )\,1{}\mathrm {i}\right )\,\left (\frac {-\frac {C}{2}+\frac {B\,1{}\mathrm {i}}{2}}{b}+\frac {\frac {B\,b^2\,1{}\mathrm {i}}{2}+\frac {C\,b^2}{2}-A\,a\,b\,1{}\mathrm {i}}{a^2\,b}\right )+\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )+1{}\mathrm {i}\right )\,\left (2\,A\,a-B\,b+C\,b\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,a^2}+\frac {5\,B-C\,5{}\mathrm {i}}{5\,a\,\left (\mathrm {tan}\left (\frac {x}{2}\right )+1{}\mathrm {i}\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )-\mathrm {i}\right )\,\left (-C+B\,1{}\mathrm {i}\right )}{2\,b} \]

[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)