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

Optimal. Leaf size=84 \[ -\frac {i \left (a^2 (-B)+2 a A b-b^2 B\right ) \log (a-i b \sin (x)+b \cos (x))}{2 a^2 b}+\frac {x (2 a A-b B)}{2 a^2}+\frac {B \sin (x)}{2 a}-\frac {i B \cos (x)}{2 a} \]

[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

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Rubi [A]  time = 0.04, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {3132} \[ -\frac {i \left (a^2 (-B)+2 a A b-b^2 B\right ) \log (a-i b \sin (x)+b \cos (x))}{2 a^2 b}+\frac {x (2 a A-b B)}{2 a^2}+\frac {B \sin (x)}{2 a}-\frac {i B \cos (x)}{2 a} \]

Antiderivative was successfully verified.

[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 3132

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*} \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}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 147, normalized size = 1.75 \[ \frac {2 \left (a^2 B-2 a A b+b^2 B\right ) \tan ^{-1}\left (\frac {(a+b) \cot \left (\frac {x}{2}\right )}{a-b}\right )-2 i a A b \log \left (a^2+2 a b \cos (x)+b^2\right )+i a^2 B \log \left (a^2+2 a b \cos (x)+b^2\right )+i b^2 B \log \left (a^2+2 a b \cos (x)+b^2\right )+a^2 B x+2 a A b x+2 a b B \sin (x)-2 i a b B \cos (x)-b^2 B x}{4 a^2 b} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 2.14, size = 56, normalized size = 0.67 \[ \frac {B a^{2} x - i \, B a b e^{\left (i \, x\right )} + {\left (i \, B a^{2} - 2 i \, A a b + i \, B b^{2}\right )} \log \left (\frac {a e^{\left (i \, x\right )} + b}{a}\right )}{2 \, a^{2} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.16, size = 168, normalized size = 2.00 \[ -\frac {{\left (2 i \, A a - i \, B b\right )} \log \left (-a \tan \left (\frac {1}{2} \, x\right )^{2} + b \tan \left (\frac {1}{2} \, x\right )^{2} + 2 i \, a \tan \left (\frac {1}{2} \, x\right ) + a + b\right )}{4 \, a^{2}} - \frac {{\left (-2 i \, A a + i \, B b\right )} \log \left (\tan \left (\frac {1}{2} \, x\right ) + i\right )}{2 \, a^{2}} + \frac {{\left (2 \, B a^{2} - 2 \, A a b + B b^{2}\right )} {\left (x + 2 \, \arctan \left (\frac {i \, a \cos \relax (x) - a \sin \relax (x) + i \, a}{a \cos \relax (x) + i \, a \sin \relax (x) - a + 2 \, b}\right )\right )}}{4 \, a^{2} b} - \frac {2 i \, A a \tan \left (\frac {1}{2} \, x\right ) - i \, B b \tan \left (\frac {1}{2} \, x\right ) - 2 \, A a - 2 \, B a + B b}{2 \, a^{2} {\left (\tan \left (\frac {1}{2} \, x\right ) + i\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [B]  time = 0.18, size = 284, normalized size = 3.38 \[ \frac {i \ln \left (\tan \left (\frac {x}{2}\right )+i\right ) A}{a}-\frac {i \ln \left (\tan \left (\frac {x}{2}\right )+i\right ) b B}{2 a^{2}}+\frac {B}{a \left (\tan \left (\frac {x}{2}\right )+i\right )}-\frac {i B \ln \left (\tan \left (\frac {x}{2}\right )-i\right )}{2 b}+\frac {i \ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right ) A}{-a +b}-\frac {i b \ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right ) A}{a \left (-a +b \right )}-\frac {i a \ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right ) B}{2 b \left (-a +b \right )}+\frac {i \ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right ) B}{-2 a +2 b}-\frac {i b \ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right ) B}{2 a \left (-a +b \right )}+\frac {i b^{2} \ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right ) B}{2 a^{2} \left (-a +b \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I/a*ln(tan(1/2*x)+I)*A-1/2*I/a^2*ln(tan(1/2*x)+I)*b*B+B/a/(tan(1/2*x)+I)-1/2*I*B/b*ln(tan(1/2*x)-I)+I/(-a+b)*l
n(I*a+I*b-a*tan(1/2*x)+b*tan(1/2*x))*A-I/a*b/(-a+b)*ln(I*a+I*b-a*tan(1/2*x)+b*tan(1/2*x))*A-1/2*I*a/b/(-a+b)*l
n(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)*l
n(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((A+B*cos(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 = 8.39, size = 584, normalized size = 6.95 \[ \left (\sum _{k=1}^3\ln \left (-{\left (a-b\right )}^2\,\left (4\,A^2\,a^2-4\,A\,B\,a\,b-B^2\,a^2+B^2\,b^2\right )\,1{}\mathrm {i}-\mathrm {root}\left (a^4\,b^2\,d^3\,64{}\mathrm {i}-A\,B\,a\,b^3\,d\,64{}\mathrm {i}-A\,B\,a^3\,b\,d\,32{}\mathrm {i}+B^2\,a^2\,b^2\,d\,16{}\mathrm {i}+A^2\,a^2\,b^2\,d\,64{}\mathrm {i}+B^2\,b^4\,d\,16{}\mathrm {i}+B^2\,a^4\,d\,16{}\mathrm {i}-32\,A^2\,B\,a^2\,b+32\,A\,B^2\,a\,b^2-8\,B^3\,a^2\,b+16\,A\,B^2\,a^3-8\,B^3\,b^3,d,k\right )\,\left (4\,A\,a^3\,{\left (a-b\right )}^2-\mathrm {root}\left (a^4\,b^2\,d^3\,64{}\mathrm {i}-A\,B\,a\,b^3\,d\,64{}\mathrm {i}-A\,B\,a^3\,b\,d\,32{}\mathrm {i}+B^2\,a^2\,b^2\,d\,16{}\mathrm {i}+A^2\,a^2\,b^2\,d\,64{}\mathrm {i}+B^2\,b^4\,d\,16{}\mathrm {i}+B^2\,a^4\,d\,16{}\mathrm {i}-32\,A^2\,B\,a^2\,b+32\,A\,B^2\,a\,b^2-8\,B^3\,a^2\,b+16\,A\,B^2\,a^3-8\,B^3\,b^3,d,k\right )\,a^2\,{\left (a-b\right )}^2\,\left (a^2\,\mathrm {tan}\left (\frac {x}{2}\right )+b^2\,\mathrm {tan}\left (\frac {x}{2}\right )-a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )-a^2\,1{}\mathrm {i}+b^2\,1{}\mathrm {i}\right )\,8+4\,a\,\mathrm {tan}\left (\frac {x}{2}\right )\,{\left (a-b\right )}^2\,\left (A\,a^2\,1{}\mathrm {i}+B\,a^2\,1{}\mathrm {i}+B\,b^2\,1{}\mathrm {i}-A\,a\,b\,2{}\mathrm {i}-B\,a\,b\,1{}\mathrm {i}\right )\right )+\mathrm {tan}\left (\frac {x}{2}\right )\,{\left (a-b\right )}^2\,{\left (B\,a-2\,A\,a+B\,b\right )}^2\right )\,\mathrm {root}\left (a^4\,b^2\,d^3\,64{}\mathrm {i}-A\,B\,a\,b^3\,d\,64{}\mathrm {i}-A\,B\,a^3\,b\,d\,32{}\mathrm {i}+B^2\,a^2\,b^2\,d\,16{}\mathrm {i}+A^2\,a^2\,b^2\,d\,64{}\mathrm {i}+B^2\,b^4\,d\,16{}\mathrm {i}+B^2\,a^4\,d\,16{}\mathrm {i}-32\,A^2\,B\,a^2\,b+32\,A\,B^2\,a\,b^2-8\,B^3\,a^2\,b+16\,A\,B^2\,a^3-8\,B^3\,b^3,d,k\right )\right )+\frac {B}{a\,\left (\mathrm {tan}\left (\frac {x}{2}\right )+1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

symsum(log(tan(x/2)*(a - b)^2*(B*a - 2*A*a + B*b)^2 - root(a^4*b^2*d^3*64i - A*B*a*b^3*d*64i - A*B*a^3*b*d*32i
+ B^2*a^2*b^2*d*16i + A^2*a^2*b^2*d*64i + B^2*b^4*d*16i + B^2*a^4*d*16i - 32*A^2*B*a^2*b + 32*A*B^2*a*b^2 - 8
*B^3*a^2*b + 16*A*B^2*a^3 - 8*B^3*b^3, d, k)*(4*A*a^3*(a - b)^2 - 8*root(a^4*b^2*d^3*64i - A*B*a*b^3*d*64i - A
*B*a^3*b*d*32i + B^2*a^2*b^2*d*16i + A^2*a^2*b^2*d*64i + B^2*b^4*d*16i + B^2*a^4*d*16i - 32*A^2*B*a^2*b + 32*A
*B^2*a*b^2 - 8*B^3*a^2*b + 16*A*B^2*a^3 - 8*B^3*b^3, d, k)*a^2*(a - b)^2*(a^2*tan(x/2) + b^2*tan(x/2) - a^2*1i
+ b^2*1i - a*b*tan(x/2)) + 4*a*tan(x/2)*(a - b)^2*(A*a^2*1i + B*a^2*1i + B*b^2*1i - A*a*b*2i - B*a*b*1i)) - (
a - b)^2*(4*A^2*a^2 - B^2*a^2 + B^2*b^2 - 4*A*B*a*b)*1i)*root(a^4*b^2*d^3*64i - A*B*a*b^3*d*64i - A*B*a^3*b*d*
32i + B^2*a^2*b^2*d*16i + A^2*a^2*b^2*d*64i + B^2*b^4*d*16i + B^2*a^4*d*16i - 32*A^2*B*a^2*b + 32*A*B^2*a*b^2
- 8*B^3*a^2*b + 16*A*B^2*a^3 - 8*B^3*b^3, d, k), k, 1, 3) + B/(a*(tan(x/2) + 1i))

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sympy [A]  time = 0.58, size = 75, normalized size = 0.89 \[ \frac {B x}{2 b} + \begin {cases} - \frac {i B e^{i x}}{2 a} & \text {for}\: 2 a \neq 0 \\x \left (- \frac {B}{2 b} + \frac {B a + B b}{2 a b}\right ) & \text {otherwise} \end {cases} + \frac {i \left (- 2 A a b + B a^{2} + B b^{2}\right ) \log {\left (e^{i x} + \frac {b}{a} \right )}}{2 a^{2} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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