Integrand size = 21, antiderivative size = 47 \[ \int \frac {B \cos (x)+C \sin (x)}{b \cos (x)+c \sin (x)} \, dx=\frac {(b B+c C) x}{b^2+c^2}+\frac {(B c-b C) \log (b \cos (x)+c \sin (x))}{b^2+c^2} \] Output:
(B*b+C*c)*x/(b^2+c^2)+(B*c-C*b)*ln(b*cos(x)+c*sin(x))/(b^2+c^2)
Time = 0.18 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.83 \[ \int \frac {B \cos (x)+C \sin (x)}{b \cos (x)+c \sin (x)} \, dx=\frac {(b B+c C) x+(B c-b C) \log (b \cos (x)+c \sin (x))}{b^2+c^2} \] Input:
Integrate[(B*Cos[x] + C*Sin[x])/(b*Cos[x] + c*Sin[x]),x]
Output:
((b*B + c*C)*x + (B*c - b*C)*Log[b*Cos[x] + c*Sin[x]])/(b^2 + c^2)
Time = 0.23 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3042, 3612}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {B \cos (x)+C \sin (x)}{b \cos (x)+c \sin (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {B \cos (x)+C \sin (x)}{b \cos (x)+c \sin (x)}dx\) |
\(\Big \downarrow \) 3612 |
\(\displaystyle \frac {x (b B+c C)}{b^2+c^2}+\frac {(B c-b C) \log (b \cos (x)+c \sin (x))}{b^2+c^2}\) |
Input:
Int[(B*Cos[x] + C*Sin[x])/(b*Cos[x] + c*Sin[x]),x]
Output:
((b*B + c*C)*x)/(b^2 + c^2) + ((B*c - b*C)*Log[b*Cos[x] + c*Sin[x]])/(b^2 + c^2)
Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)]) /((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]), x _Symbol] :> Simp[(b*B + c*C)*(x/(b^2 + c^2)), x] + Simp[(c*B - b*C)*(Log[a + b*Cos[d + e*x] + c*Sin[d + e*x]]/(e*(b^2 + c^2))), x] /; FreeQ[{a, b, c, d, e, A, B, C}, x] && NeQ[b^2 + c^2, 0] && EqQ[A*(b^2 + c^2) - a*(b*B + c*C ), 0]
Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.28
method | result | size |
parallelrisch | \(\frac {B x b +C x c -\left (\ln \left (\sec \left (\frac {x}{2}\right )^{2}\right )-\ln \left (b \tan \left (\frac {x}{2}\right )^{2}-2 c \tan \left (\frac {x}{2}\right )-b \right )\right ) \left (B c -b C \right )}{b^{2}+c^{2}}\) | \(60\) |
default | \(\frac {\frac {\left (-B c +b C \right ) \ln \left (1+\tan \left (x \right )^{2}\right )}{2}+\left (B b +C c \right ) \arctan \left (\tan \left (x \right )\right )}{b^{2}+c^{2}}+\frac {\left (B c -b C \right ) \ln \left (c \tan \left (x \right )+b \right )}{b^{2}+c^{2}}\) | \(66\) |
norman | \(\frac {\frac {\left (B b +C c \right ) x}{b^{2}+c^{2}}+\frac {\left (B b +C c \right ) x \tan \left (\frac {x}{2}\right )^{2}}{b^{2}+c^{2}}}{1+\tan \left (\frac {x}{2}\right )^{2}}+\frac {\left (B c -b C \right ) \ln \left (b \tan \left (\frac {x}{2}\right )^{2}-2 c \tan \left (\frac {x}{2}\right )-b \right )}{b^{2}+c^{2}}-\frac {\left (B c -b C \right ) \ln \left (1+\tan \left (\frac {x}{2}\right )^{2}\right )}{b^{2}+c^{2}}\) | \(122\) |
risch | \(\frac {i x C}{i c -b}-\frac {x B}{i c -b}-\frac {2 i x B c}{b^{2}+c^{2}}+\frac {2 i x b C}{b^{2}+c^{2}}+\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {i c +b}{i c -b}\right ) B c}{b^{2}+c^{2}}-\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {i c +b}{i c -b}\right ) b C}{b^{2}+c^{2}}\) | \(136\) |
Input:
int((B*cos(x)+C*sin(x))/(b*cos(x)+c*sin(x)),x,method=_RETURNVERBOSE)
Output:
1/(b^2+c^2)*(B*x*b+C*x*c-(ln(sec(1/2*x)^2)-ln(b*tan(1/2*x)^2-2*c*tan(1/2*x )-b))*(B*c-C*b))
Time = 0.08 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.26 \[ \int \frac {B \cos (x)+C \sin (x)}{b \cos (x)+c \sin (x)} \, dx=\frac {2 \, {\left (B b + C c\right )} x - {\left (C b - B c\right )} \log \left (2 \, b c \cos \left (x\right ) \sin \left (x\right ) + {\left (b^{2} - c^{2}\right )} \cos \left (x\right )^{2} + c^{2}\right )}{2 \, {\left (b^{2} + c^{2}\right )}} \] Input:
integrate((B*cos(x)+C*sin(x))/(b*cos(x)+c*sin(x)),x, algorithm="fricas")
Output:
1/2*(2*(B*b + C*c)*x - (C*b - B*c)*log(2*b*c*cos(x)*sin(x) + (b^2 - c^2)*c os(x)^2 + c^2))/(b^2 + c^2)
Result contains complex when optimal does not.
Time = 0.37 (sec) , antiderivative size = 360, normalized size of antiderivative = 7.66 \[ \int \frac {B \cos (x)+C \sin (x)}{b \cos (x)+c \sin (x)} \, dx=\begin {cases} \tilde {\infty } \left (B \log {\left (\sin {\left (x \right )} \right )} + C x\right ) & \text {for}\: b = 0 \wedge c = 0 \\\frac {B \log {\left (\sin {\left (x \right )} \right )} + C x}{c} & \text {for}\: b = 0 \\- \frac {B x \sin {\left (x \right )}}{2 i c \sin {\left (x \right )} + 2 c \cos {\left (x \right )}} + \frac {i B x \cos {\left (x \right )}}{2 i c \sin {\left (x \right )} + 2 c \cos {\left (x \right )}} - \frac {B \cos {\left (x \right )}}{2 i c \sin {\left (x \right )} + 2 c \cos {\left (x \right )}} + \frac {i C x \sin {\left (x \right )}}{2 i c \sin {\left (x \right )} + 2 c \cos {\left (x \right )}} + \frac {C x \cos {\left (x \right )}}{2 i c \sin {\left (x \right )} + 2 c \cos {\left (x \right )}} - \frac {i C \cos {\left (x \right )}}{2 i c \sin {\left (x \right )} + 2 c \cos {\left (x \right )}} & \text {for}\: b = - i c \\- \frac {B x \sin {\left (x \right )}}{- 2 i c \sin {\left (x \right )} + 2 c \cos {\left (x \right )}} - \frac {i B x \cos {\left (x \right )}}{- 2 i c \sin {\left (x \right )} + 2 c \cos {\left (x \right )}} - \frac {B \cos {\left (x \right )}}{- 2 i c \sin {\left (x \right )} + 2 c \cos {\left (x \right )}} - \frac {i C x \sin {\left (x \right )}}{- 2 i c \sin {\left (x \right )} + 2 c \cos {\left (x \right )}} + \frac {C x \cos {\left (x \right )}}{- 2 i c \sin {\left (x \right )} + 2 c \cos {\left (x \right )}} + \frac {i C \cos {\left (x \right )}}{- 2 i c \sin {\left (x \right )} + 2 c \cos {\left (x \right )}} & \text {for}\: b = i c \\\frac {B b x}{b^{2} + c^{2}} + \frac {B c \log {\left (\cos {\left (x \right )} + \frac {c \sin {\left (x \right )}}{b} \right )}}{b^{2} + c^{2}} - \frac {C b \log {\left (\cos {\left (x \right )} + \frac {c \sin {\left (x \right )}}{b} \right )}}{b^{2} + c^{2}} + \frac {C c x}{b^{2} + c^{2}} & \text {otherwise} \end {cases} \] Input:
integrate((B*cos(x)+C*sin(x))/(b*cos(x)+c*sin(x)),x)
Output:
Piecewise((zoo*(B*log(sin(x)) + C*x), Eq(b, 0) & Eq(c, 0)), ((B*log(sin(x) ) + C*x)/c, Eq(b, 0)), (-B*x*sin(x)/(2*I*c*sin(x) + 2*c*cos(x)) + I*B*x*co s(x)/(2*I*c*sin(x) + 2*c*cos(x)) - B*cos(x)/(2*I*c*sin(x) + 2*c*cos(x)) + I*C*x*sin(x)/(2*I*c*sin(x) + 2*c*cos(x)) + C*x*cos(x)/(2*I*c*sin(x) + 2*c* cos(x)) - I*C*cos(x)/(2*I*c*sin(x) + 2*c*cos(x)), Eq(b, -I*c)), (-B*x*sin( x)/(-2*I*c*sin(x) + 2*c*cos(x)) - I*B*x*cos(x)/(-2*I*c*sin(x) + 2*c*cos(x) ) - B*cos(x)/(-2*I*c*sin(x) + 2*c*cos(x)) - I*C*x*sin(x)/(-2*I*c*sin(x) + 2*c*cos(x)) + C*x*cos(x)/(-2*I*c*sin(x) + 2*c*cos(x)) + I*C*cos(x)/(-2*I*c *sin(x) + 2*c*cos(x)), Eq(b, I*c)), (B*b*x/(b**2 + c**2) + B*c*log(cos(x) + c*sin(x)/b)/(b**2 + c**2) - C*b*log(cos(x) + c*sin(x)/b)/(b**2 + c**2) + C*c*x/(b**2 + c**2), True))
Leaf count of result is larger than twice the leaf count of optimal. 181 vs. \(2 (48) = 96\).
Time = 0.11 (sec) , antiderivative size = 181, normalized size of antiderivative = 3.85 \[ \int \frac {B \cos (x)+C \sin (x)}{b \cos (x)+c \sin (x)} \, dx=B {\left (\frac {2 \, b \arctan \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1}\right )}{b^{2} + c^{2}} + \frac {c \log \left (-b - \frac {2 \, c \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {b \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}}\right )}{b^{2} + c^{2}} - \frac {c \log \left (\frac {\sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + 1\right )}{b^{2} + c^{2}}\right )} + C {\left (\frac {2 \, c \arctan \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1}\right )}{b^{2} + c^{2}} - \frac {b \log \left (-b - \frac {2 \, c \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {b \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}}\right )}{b^{2} + c^{2}} + \frac {b \log \left (\frac {\sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + 1\right )}{b^{2} + c^{2}}\right )} \] Input:
integrate((B*cos(x)+C*sin(x))/(b*cos(x)+c*sin(x)),x, algorithm="maxima")
Output:
B*(2*b*arctan(sin(x)/(cos(x) + 1))/(b^2 + c^2) + c*log(-b - 2*c*sin(x)/(co s(x) + 1) + b*sin(x)^2/(cos(x) + 1)^2)/(b^2 + c^2) - c*log(sin(x)^2/(cos(x ) + 1)^2 + 1)/(b^2 + c^2)) + C*(2*c*arctan(sin(x)/(cos(x) + 1))/(b^2 + c^2 ) - b*log(-b - 2*c*sin(x)/(cos(x) + 1) + b*sin(x)^2/(cos(x) + 1)^2)/(b^2 + c^2) + b*log(sin(x)^2/(cos(x) + 1)^2 + 1)/(b^2 + c^2))
Time = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.64 \[ \int \frac {B \cos (x)+C \sin (x)}{b \cos (x)+c \sin (x)} \, dx=\frac {{\left (B b + C c\right )} x}{b^{2} + c^{2}} + \frac {{\left (C b - B c\right )} \log \left (\tan \left (x\right )^{2} + 1\right )}{2 \, {\left (b^{2} + c^{2}\right )}} - \frac {{\left (C b c - B c^{2}\right )} \log \left ({\left | c \tan \left (x\right ) + b \right |}\right )}{b^{2} c + c^{3}} \] Input:
integrate((B*cos(x)+C*sin(x))/(b*cos(x)+c*sin(x)),x, algorithm="giac")
Output:
(B*b + C*c)*x/(b^2 + c^2) + 1/2*(C*b - B*c)*log(tan(x)^2 + 1)/(b^2 + c^2) - (C*b*c - B*c^2)*log(abs(c*tan(x) + b))/(b^2*c + c^3)
Time = 28.13 (sec) , antiderivative size = 1976, normalized size of antiderivative = 42.04 \[ \int \frac {B \cos (x)+C \sin (x)}{b \cos (x)+c \sin (x)} \, dx=\text {Too large to display} \] Input:
int((B*cos(x) + C*sin(x))/(b*cos(x) + c*sin(x)),x)
Output:
(log(b + 2*c*tan(x/2) - b*tan(x/2)^2)*(B*c - C*b))/(b^2 + c^2) - (log(1/(c os(x) + 1))*(2*B*c - 2*C*b))/(2*(b^2 + c^2)) + (2*atan(((((32*B*C^2*b^2 - ((2*B*c - 2*C*b)*(((2*B*c - 2*C*b)*(64*B*b^2*c^2 - 32*B*b^4 + 32*C*b*c^3 - 64*C*b^3*c + ((2*B*c - 2*C*b)*(96*b^4*c + 96*b^2*c^3))/(2*(b^2 + c^2))))/ (2*(b^2 + c^2)) - 32*C^2*b^2*c - 32*B^2*b^2*c + 64*B*C*b^3 + 64*B*C*b*c^2) )/(2*(b^2 + c^2)) + ((B*b + C*c)*(((B*b + C*c)*(64*B*b^2*c^2 - 32*B*b^4 + 32*C*b*c^3 - 64*C*b^3*c + ((2*B*c - 2*C*b)*(96*b^4*c + 96*b^2*c^3))/(2*(b^ 2 + c^2))))/(b^2 + c^2) + ((B*b + C*c)*(2*B*c - 2*C*b)*(96*b^4*c + 96*b^2* c^3))/(2*(b^2 + c^2)^2)))/(b^2 + c^2) - 32*B^2*C*b*c + ((B*b + C*c)^2*(2*B *c - 2*C*b)*(96*b^4*c + 96*b^2*c^3))/(2*(b^2 + c^2)^3))*(12*B^2*b*c^3 - 6* B^2*b^3*c - 6*C^2*b*c^3 + 12*C^2*b^3*c + 4*B*C*b^4 + 4*B*C*c^4 - 28*B*C*b^ 2*c^2))/((b^2 + c^2)^2*(B^2*b^2 + 4*B^2*c^2 + 4*C^2*b^2 + C^2*c^2 - 6*B*C* b*c)^2) - tan(x/2)*(((32*B^3*b*c - 32*B^2*C*b^2 - 64*C^3*b^2 + ((2*B*c - 2 *C*b)*(32*B^2*b^3 - 96*B^2*b*c^2 + 64*C^2*b*c^2 - ((2*B*c - 2*C*b)*(32*C*b ^2*c^2 - 64*C*b^4 + 32*B*b*c^3 + 128*B*b^3*c - ((2*B*c - 2*C*b)*(96*b*c^4 + 96*b^3*c^2))/(2*(b^2 + c^2))))/(2*(b^2 + c^2)) + 192*B*C*b^2*c))/(2*(b^2 + c^2)) + ((B*b + C*c)*(((B*b + C*c)*(32*C*b^2*c^2 - 64*C*b^4 + 32*B*b*c^ 3 + 128*B*b^3*c - ((2*B*c - 2*C*b)*(96*b*c^4 + 96*b^3*c^2))/(2*(b^2 + c^2) )))/(b^2 + c^2) - ((B*b + C*c)*(2*B*c - 2*C*b)*(96*b*c^4 + 96*b^3*c^2))/(2 *(b^2 + c^2)^2)))/(b^2 + c^2) + 64*B*C^2*b*c - ((B*b + C*c)^2*(2*B*c - ...
Time = 0.17 (sec) , antiderivative size = 1, normalized size of antiderivative = 0.02 \[ \int \frac {B \cos (x)+C \sin (x)}{b \cos (x)+c \sin (x)} \, dx=x \] Input:
int((B*cos(x)+C*sin(x))/(b*cos(x)+c*sin(x)),x)
Output:
x