\(\int \frac {A+B \cos (x)}{a+b \cos (x)+i b \sin (x)} \, dx\) [538]

Optimal result

Integrand size = 22, antiderivative size = 84 \[ \int \frac {A+B \cos (x)}{a+b \cos (x)+i b \sin (x)} \, dx=\frac {(2 a A-b B) x}{2 a^2}+\frac {i B \cos (x)}{2 a}+\frac {i \left (2 a A b-a^2 B-b^2 B\right ) \log (a+b \cos (x)+i b \sin (x))}{2 a^2 b}+\frac {B \sin (x)}{2 a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {3211} \[ \int \frac {A+B \cos (x)}{a+b \cos (x)+i b \sin (x)} \, dx=\frac {i \left (a^2 (-B)+2 a A b-b^2 B\right ) \log (a+i b \sin (x)+b \cos (x))}{2 a^2 b}+\frac {x (2 a A-b B)}{2 a^2}+\frac {B \sin (x)}{2 a}+\frac {i B \cos (x)}{2 a} \]

[In]

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

[Out]

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

Rule 3211

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.))/(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x
_)]), x_Symbol] :> Simp[(2*a*A - b*B)*(x/(2*a^2)), x] + (Simp[B*(Sin[d + e*x]/(2*a*e)), x] - Simp[b*B*(Cos[d +
e*x]/(2*a*c*e)), x] + Simp[(a^2*B - 2*a*b*A + b^2*B)*(Log[RemoveContent[a + b*Cos[d + e*x] + c*Sin[d + e*x],
x]]/(2*a^2*c*e)), x]) /; FreeQ[{a, b, c, d, e, A, B}, x] && EqQ[b^2 + c^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(2 a A-b B) x}{2 a^2}+\frac {i B \cos (x)}{2 a}+\frac {i \left (2 a A b-a^2 B-b^2 B\right ) \log (a+b \cos (x)+i b \sin (x))}{2 a^2 b}+\frac {B \sin (x)}{2 a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.75 \[ \int \frac {A+B \cos (x)}{a+b \cos (x)+i b \sin (x)} \, dx=\frac {2 a A b x+a^2 B x-b^2 B x+2 \left (-2 a A b+a^2 B+b^2 B\right ) \arctan \left (\frac {(a+b) \cot \left (\frac {x}{2}\right )}{a-b}\right )+2 i a b B \cos (x)+2 i a A b \log \left (a^2+b^2+2 a b \cos (x)\right )-i a^2 B \log \left (a^2+b^2+2 a b \cos (x)\right )-i b^2 B \log \left (a^2+b^2+2 a b \cos (x)\right )+2 a b B \sin (x)}{4 a^2 b} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.21 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.02

method	result	size
risch	\(\frac {i B \,{\mathrm e}^{-i x}}{2 a}+\frac {x A}{a}-\frac {b x B}{2 a^{2}}+\frac {i \ln \left ({\mathrm e}^{i x}+\frac {a}{b}\right ) A}{a}-\frac {i \ln \left ({\mathrm e}^{i x}+\frac {a}{b}\right ) B}{2 b}-\frac {i b \ln \left ({\mathrm e}^{i x}+\frac {a}{b}\right ) B}{2 a^{2}}\)	\(86\)
default	\(-\frac {i \left (2 A a -B b \right ) \ln \left (-i+\tan \left (\frac {x}{2}\right )\right )}{2 a^{2}}+\frac {B}{a \left (-i+\tan \left (\frac {x}{2}\right )\right )}+\frac {i \left (2 A a b -B \,a^{2}-B \,b^{2}\right ) \ln \left (i a +i b +a \tan \left (\frac {x}{2}\right )-b \tan \left (\frac {x}{2}\right )\right )}{2 a^{2} b}+\frac {i B \ln \left (\tan \left (\frac {x}{2}\right )+i\right )}{2 b}\)	\(104\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.86 \[ \int \frac {A+B \cos (x)}{a+b \cos (x)+i b \sin (x)} \, dx=\frac {{\left (i \, B a b + {\left (2 \, A a b - B b^{2}\right )} x e^{\left (i \, x\right )} + {\left (-i \, B a^{2} + 2 i \, A a b - i \, B b^{2}\right )} e^{\left (i \, x\right )} \log \left (\frac {b e^{\left (i \, x\right )} + a}{b}\right )\right )} e^{\left (-i \, x\right )}}{2 \, a^{2} b} \]

[In]

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

[Out]

1/2*(I*B*a*b + (2*A*a*b - B*b^2)*x*e^(I*x) + (-I*B*a^2 + 2*I*A*a*b - I*B*b^2)*e^(I*x)*log((b*e^(I*x) + a)/b))*
e^(-I*x)/(a^2*b)

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.12 \[ \int \frac {A+B \cos (x)}{a+b \cos (x)+i b \sin (x)} \, dx=\begin {cases} \frac {i B e^{- i x}}{2 a} & \text {for}\: a \neq 0 \\x \left (- \frac {2 A a - B b}{2 a^{2}} + \frac {2 A a + B a - B b}{2 a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (2 A a - B b\right )}{2 a^{2}} - \frac {i \left (- 2 A a b + B a^{2} + B b^{2}\right ) \log {\left (\frac {a}{b} + e^{i x} \right )}}{2 a^{2} b} \]

[In]

integrate((A+B*cos(x))/(a+b*cos(x)+I*b*sin(x)),x)

[Out]

Piecewise((I*B*exp(-I*x)/(2*a), Ne(a, 0)), (x*(-(2*A*a - B*b)/(2*a**2) + (2*A*a + B*a - B*b)/(2*a**2)), True))
+ x*(2*A*a - B*b)/(2*a**2) - I*(-2*A*a*b + B*a**2 + B*b**2)*log(a/b + exp(I*x))/(2*a**2*b)

Maxima [F(-2)]

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

[In]

integrate((A+B*cos(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. 168 vs. \(2 (70) = 140\).

Time = 0.27 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.00 \[ \int \frac {A+B \cos (x)}{a+b \cos (x)+i b \sin (x)} \, dx=-\frac {{\left (-2 i \, A a + i \, B b\right )} \log \left (-a \tan \left (\frac {1}{2} \, x\right )^{2} + b \tan \left (\frac {1}{2} \, x\right )^{2} - 2 i \, a \tan \left (\frac {1}{2} \, x\right ) + a + b\right )}{4 \, a^{2}} - \frac {{\left (2 i \, A a - i \, B b\right )} \log \left (\tan \left (\frac {1}{2} \, x\right ) - i\right )}{2 \, a^{2}} + \frac {{\left (2 \, B a^{2} - 2 \, A a b + B b^{2}\right )} {\left (x + 2 \, \arctan \left (\frac {-i \, a \cos \left (x\right ) - a \sin \left (x\right ) - i \, a}{a \cos \left (x\right ) - i \, a \sin \left (x\right ) - a + 2 \, b}\right )\right )}}{4 \, a^{2} b} - \frac {-2 i \, A a \tan \left (\frac {1}{2} \, x\right ) + i \, B b \tan \left (\frac {1}{2} \, x\right ) - 2 \, A a - 2 \, B a + B b}{2 \, a^{2} {\left (\tan \left (\frac {1}{2} \, x\right ) - i\right )}} \]

[In]

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

[Out]

-1/4*(-2*I*A*a + I*B*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)*log(tan(1/2*x) - I)/a^2 + 1/4*(2*B*a^2 - 2*A*a*b + B*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*(-2*I*A*a*tan(1/2*x) + I*B*b*tan(1/2*x) - 2*A*a - 2*B*a + B*
b)/(a^2*(tan(1/2*x) - I))

Mupad [B] (verification not implemented)

Time = 28.31 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.18 \[ \int \frac {A+B \cos (x)}{a+b \cos (x)+i b \sin (x)} \, dx=\frac {B}{a\,\left (\mathrm {tan}\left (\frac {x}{2}\right )-\mathrm {i}\right )}+\frac {B\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,b}-\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )-\mathrm {i}\right )\,\left (A\,a\,1{}\mathrm {i}-\frac {B\,b\,1{}\mathrm {i}}{2}\right )}{a^2}-\frac {\ln \left (a+b-a\,\mathrm {tan}\left (\frac {x}{2}\right )\,1{}\mathrm {i}+b\,\mathrm {tan}\left (\frac {x}{2}\right )\,1{}\mathrm {i}\right )\,\left (B\,a^2-2\,A\,a\,b+B\,b^2\right )\,1{}\mathrm {i}}{2\,a^2\,b} \]

[In]

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

[Out]

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