\(\int \frac {A+C \sin (x)}{a+b \cos (x)-i b \sin (x)} \, dx\) [544]

Optimal result

Integrand size = 22, antiderivative size = 85 \[ \int \frac {A+C \sin (x)}{a+b \cos (x)-i b \sin (x)} \, dx=\frac {(2 a A+i b C) x}{2 a^2}-\frac {C \cos (x)}{2 a}-\frac {\left (2 i a A b+a^2 C-b^2 C\right ) \log (a+b \cos (x)-i b \sin (x))}{2 a^2 b}-\frac {i C \sin (x)}{2 a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 85, 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 = {3210} \[ \int \frac {A+C \sin (x)}{a+b \cos (x)-i b \sin (x)} \, dx=-\frac {\left (a^2 C+2 i a A b-b^2 C\right ) \log (a-i b \sin (x)+b \cos (x))}{2 a^2 b}+\frac {x (2 a A+i b C)}{2 a^2}-\frac {i C \sin (x)}{2 a}-\frac {C \cos (x)}{2 a} \]

[In]

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

[Out]

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

Rule 3210

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

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

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.79 \[ \int \frac {A+C \sin (x)}{a+b \cos (x)-i b \sin (x)} \, dx=\frac {2 a A b x+i a^2 C x+i b^2 C x+2 i \left (2 i a A b+a^2 C-b^2 C\right ) \arctan \left (\frac {(a+b) \cot \left (\frac {x}{2}\right )}{a-b}\right )-2 a b C \cos (x)-2 i a A b \log \left (a^2+b^2+2 a b \cos (x)\right )-a^2 C \log \left (a^2+b^2+2 a b \cos (x)\right )+b^2 C \log \left (a^2+b^2+2 a b \cos (x)\right )-2 i a b C \sin (x)}{4 a^2 b} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.30 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.91

method	result	size
risch	\(-\frac {C \,{\mathrm e}^{i x}}{2 a}+\frac {i x C}{2 b}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {b}{a}\right ) C}{2 b}+\frac {b \ln \left ({\mathrm e}^{i x}+\frac {b}{a}\right ) C}{2 a^{2}}-\frac {i \ln \left ({\mathrm e}^{i x}+\frac {b}{a}\right ) A}{a}\)	\(77\)
default	\(\frac {C \ln \left (-i+\tan \left (\frac {x}{2}\right )\right )}{2 b}+\frac {i \left (-i C \,a^{2}+i C \,b^{2}+2 A a b \right ) \left (a -b \right ) \ln \left (i a +i b -a \tan \left (\frac {x}{2}\right )+b \tan \left (\frac {x}{2}\right )\right )}{2 a^{2} b \left (-a +b \right )}-\frac {i C}{a \left (\tan \left (\frac {x}{2}\right )+i\right )}+\frac {\left (2 i A a -b C \right ) \ln \left (\tan \left (\frac {x}{2}\right )+i\right )}{2 a^{2}}\)	\(119\)

[In]

int((A+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*x*C-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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.67 \[ \int \frac {A+C \sin (x)}{a+b \cos (x)-i b \sin (x)} \, dx=\frac {i \, C a^{2} x - C a b e^{\left (i \, x\right )} - {\left (C a^{2} + 2 i \, A a b - C b^{2}\right )} \log \left (\frac {a e^{\left (i \, x\right )} + b}{a}\right )}{2 \, a^{2} b} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.92 \[ \int \frac {A+C \sin (x)}{a+b \cos (x)-i b \sin (x)} \, dx=\frac {i C x}{2 b} + \begin {cases} - \frac {C e^{i x}}{2 a} & \text {for}\: a \neq 0 \\x \left (- \frac {i C}{2 b} + \frac {i C a - i C b}{2 a b}\right ) & \text {otherwise} \end {cases} - \frac {\left (2 i A a b + C a^{2} - C b^{2}\right ) \log {\left (e^{i x} + \frac {b}{a} \right )}}{2 a^{2} b} \]

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate((A+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. 169 vs. \(2 (69) = 138\).

Time = 0.27 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.99 \[ \int \frac {A+C \sin (x)}{a+b \cos (x)-i b \sin (x)} \, dx=-\frac {{\left (2 i \, A a - C b\right )} \log \left (-a \tan \left (\frac {1}{2} \, x\right )^{2} + b \tan \left (\frac {1}{2} \, x\right )^{2} + 2 i \, a \tan \left (\frac {1}{2} \, x\right ) + a + b\right )}{4 \, a^{2}} - \frac {{\left (-2 i \, A a + C b\right )} \log \left (\tan \left (\frac {1}{2} \, x\right ) + i\right )}{2 \, a^{2}} - \frac {{\left (-2 i \, C a^{2} + 2 \, A a b + i \, C b^{2}\right )} {\left (x + 2 \, \arctan \left (\frac {i \, a \cos \left (x\right ) - a \sin \left (x\right ) + i \, a}{a \cos \left (x\right ) + i \, a \sin \left (x\right ) - a + 2 \, b}\right )\right )}}{4 \, a^{2} b} - \frac {2 i \, A a \tan \left (\frac {1}{2} \, x\right ) - C b \tan \left (\frac {1}{2} \, x\right ) - 2 \, A a + 2 i \, C a - i \, C b}{2 \, a^{2} {\left (\tan \left (\frac {1}{2} \, x\right ) + i\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 28.45 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.13 \[ \int \frac {A+C \sin (x)}{a+b \cos (x)-i b \sin (x)} \, dx=-\ln \left (a+b+a\,\mathrm {tan}\left (\frac {x}{2}\right )\,1{}\mathrm {i}-b\,\mathrm {tan}\left (\frac {x}{2}\right )\,1{}\mathrm {i}\right )\,\left (\frac {C}{2\,b}-\frac {C\,b}{2\,a^2}+\frac {A\,1{}\mathrm {i}}{a}\right )-\frac {C\,1{}\mathrm {i}}{a\,\left (\mathrm {tan}\left (\frac {x}{2}\right )+1{}\mathrm {i}\right )}+\frac {C\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )-\mathrm {i}\right )}{2\,b}+\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )+1{}\mathrm {i}\right )\,\left (-\frac {C\,b}{2}+A\,a\,1{}\mathrm {i}\right )}{a^2} \]

[In]

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

[Out]

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