3.529 \(\int \frac {B \cos (x)+C \sin (x)}{b \cos (x)+c \sin (x)} \, dx\)

Optimal. Leaf size=47 \[ \frac {x (b B+c C)}{b^2+c^2}+\frac {(B c-b C) \log (b \cos (x)+c \sin (x))}{b^2+c^2} \]

[Out]

(B*b+C*c)*x/(b^2+c^2)+(B*c-C*b)*ln(b*cos(x)+c*sin(x))/(b^2+c^2)

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Rubi [A]  time = 0.04, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {3133} \[ \frac {x (b B+c C)}{b^2+c^2}+\frac {(B c-b C) \log (b \cos (x)+c \sin (x))}{b^2+c^2} \]

Antiderivative was successfully verified.

[In]

Int[(B*Cos[x] + C*Sin[x])/(b*Cos[x] + c*Sin[x]),x]

[Out]

((b*B + c*C)*x)/(b^2 + c^2) + ((B*c - b*C)*Log[b*Cos[x] + c*Sin[x]])/(b^2 + c^2)

Rule 3133

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)*L
og[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]

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 39, normalized size = 0.83 \[ \frac {x (b B+c C)+(B c-b C) \log (b \cos (x)+c \sin (x))}{b^2+c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(B*Cos[x] + C*Sin[x])/(b*Cos[x] + c*Sin[x]),x]

[Out]

((b*B + c*C)*x + (B*c - b*C)*Log[b*Cos[x] + c*Sin[x]])/(b^2 + c^2)

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fricas [A]  time = 0.88, size = 59, normalized size = 1.26 \[ \frac {2 \, {\left (B b + C c\right )} x - {\left (C b - B c\right )} \log \left (2 \, b c \cos \relax (x) \sin \relax (x) + {\left (b^{2} - c^{2}\right )} \cos \relax (x)^{2} + c^{2}\right )}{2 \, {\left (b^{2} + c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(x)+C*sin(x))/(b*cos(x)+c*sin(x)),x, algorithm="fricas")

[Out]

1/2*(2*(B*b + C*c)*x - (C*b - B*c)*log(2*b*c*cos(x)*sin(x) + (b^2 - c^2)*cos(x)^2 + c^2))/(b^2 + c^2)

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giac [A]  time = 0.17, size = 77, normalized size = 1.64 \[ \frac {{\left (B b + C c\right )} x}{b^{2} + c^{2}} + \frac {{\left (C b - B c\right )} \log \left (\tan \relax (x)^{2} + 1\right )}{2 \, {\left (b^{2} + c^{2}\right )}} - \frac {{\left (C b c - B c^{2}\right )} \log \left ({\left | c \tan \relax (x) + b \right |}\right )}{b^{2} c + c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(x)+C*sin(x))/(b*cos(x)+c*sin(x)),x, algorithm="giac")

[Out]

(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)

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maple [B]  time = 0.14, size = 111, normalized size = 2.36 \[ \frac {\ln \left (c \tan \relax (x )+b \right ) B c}{b^{2}+c^{2}}-\frac {\ln \left (c \tan \relax (x )+b \right ) b C}{b^{2}+c^{2}}-\frac {\ln \left (1+\tan ^{2}\relax (x )\right ) B c}{2 \left (b^{2}+c^{2}\right )}+\frac {\ln \left (1+\tan ^{2}\relax (x )\right ) b C}{2 b^{2}+2 c^{2}}+\frac {B \arctan \left (\tan \relax (x )\right ) b}{b^{2}+c^{2}}+\frac {C \arctan \left (\tan \relax (x )\right ) c}{b^{2}+c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*cos(x)+C*sin(x))/(b*cos(x)+c*sin(x)),x)

[Out]

1/(b^2+c^2)*ln(c*tan(x)+b)*B*c-1/(b^2+c^2)*ln(c*tan(x)+b)*b*C-1/2/(b^2+c^2)*ln(1+tan(x)^2)*B*c+1/2/(b^2+c^2)*l
n(1+tan(x)^2)*b*C+1/(b^2+c^2)*B*arctan(tan(x))*b+1/(b^2+c^2)*C*arctan(tan(x))*c

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maxima [B]  time = 0.42, size = 181, normalized size = 3.85 \[ B {\left (\frac {2 \, b \arctan \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )}{b^{2} + c^{2}} + \frac {c \log \left (-b - \frac {2 \, c \sin \relax (x)}{\cos \relax (x) + 1} + \frac {b \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}\right )}{b^{2} + c^{2}} - \frac {c \log \left (\frac {\sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + 1\right )}{b^{2} + c^{2}}\right )} + C {\left (\frac {2 \, c \arctan \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )}{b^{2} + c^{2}} - \frac {b \log \left (-b - \frac {2 \, c \sin \relax (x)}{\cos \relax (x) + 1} + \frac {b \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}\right )}{b^{2} + c^{2}} + \frac {b \log \left (\frac {\sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + 1\right )}{b^{2} + c^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(x)+C*sin(x))/(b*cos(x)+c*sin(x)),x, algorithm="maxima")

[Out]

B*(2*b*arctan(sin(x)/(cos(x) + 1))/(b^2 + c^2) + c*log(-b - 2*c*sin(x)/(cos(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))

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mupad [B]  time = 12.82, size = 1976, normalized size = 42.04 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*cos(x) + C*sin(x))/(b*cos(x) + c*sin(x)),x)

[Out]

(log(b + 2*c*tan(x/2) - b*tan(x/2)^2)*(B*c - C*b))/(b^2 + c^2) - (log(1/(cos(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 - 3
2*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 - 2*C*b)*(96*b*c^4 + 96*b^3*c^2))/(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) + ((((B*b + C*c)^3*(96*b*c^4 + 96*b^3*c^2))/(b^2 + c^2)^3
- ((B*b + C*c)*(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))/(b^2 + c^2) + ((2*B*c - 2*C*b)*(((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)))/(2*(b^2 + c^2)))*(B^2*b^4 + 4*B^2*c^4 - 4*C^2*b^4 - C^2*c^4 - 13*B^2*b^2*c^2 +
13*C^2*b^2*c^2 - 18*B*C*b*c^3 + 18*B*C*b^3*c))/((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)) + ((((B*b + C*c)*(((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))/(b^2 + c^2) - ((B*b + C*c)^3*(96*b^4*c + 96*b^2*c^3))/(b^2 + c^2)^3 + ((2*B*c - 2*C*b)*(((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)))/(2*(b^2 + c^
2)))*(B^2*b^4 + 4*B^2*c^4 - 4*C^2*b^4 - C^2*c^4 - 13*B^2*b^2*c^2 + 13*C^2*b^2*c^2 - 18*B*C*b*c^3 + 18*B*C*b^3*
c))/((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))*(b^4 + c^4 + 2*b^2*c^2))/(32*B*
b^2 + 32*C*b*c))*(B*b + C*c))/(b^2 + c^2)

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sympy [A]  time = 0.88, size = 371, normalized size = 7.89 \[ \begin {cases} \tilde {\infty } \left (B \log {\left (\sin {\relax (x )} \right )} + C x\right ) & \text {for}\: b = 0 \wedge c = 0 \\\frac {B \log {\left (\sin {\relax (x )} \right )} + C x}{c} & \text {for}\: b = 0 \\\frac {B x \sin {\relax (x )}}{- 2 i c \sin {\relax (x )} - 2 c \cos {\relax (x )}} - \frac {i B x \cos {\relax (x )}}{- 2 i c \sin {\relax (x )} - 2 c \cos {\relax (x )}} - \frac {i B \sin {\relax (x )}}{- 2 i c \sin {\relax (x )} - 2 c \cos {\relax (x )}} - \frac {i C x \sin {\relax (x )}}{- 2 i c \sin {\relax (x )} - 2 c \cos {\relax (x )}} - \frac {C x \cos {\relax (x )}}{- 2 i c \sin {\relax (x )} - 2 c \cos {\relax (x )}} + \frac {C \sin {\relax (x )}}{- 2 i c \sin {\relax (x )} - 2 c \cos {\relax (x )}} & \text {for}\: b = - i c \\\frac {B x \sin {\relax (x )}}{2 i c \sin {\relax (x )} - 2 c \cos {\relax (x )}} + \frac {i B x \cos {\relax (x )}}{2 i c \sin {\relax (x )} - 2 c \cos {\relax (x )}} + \frac {i B \sin {\relax (x )}}{2 i c \sin {\relax (x )} - 2 c \cos {\relax (x )}} + \frac {i C x \sin {\relax (x )}}{2 i c \sin {\relax (x )} - 2 c \cos {\relax (x )}} - \frac {C x \cos {\relax (x )}}{2 i c \sin {\relax (x )} - 2 c \cos {\relax (x )}} + \frac {C \sin {\relax (x )}}{2 i c \sin {\relax (x )} - 2 c \cos {\relax (x )}} & \text {for}\: b = i c \\\frac {B b x}{b^{2} + c^{2}} + \frac {B c \log {\left (\cos {\relax (x )} + \frac {c \sin {\relax (x )}}{b} \right )}}{b^{2} + c^{2}} - \frac {C b \log {\left (\cos {\relax (x )} + \frac {c \sin {\relax (x )}}{b} \right )}}{b^{2} + c^{2}} + \frac {C c x}{b^{2} + c^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(x)+C*sin(x))/(b*cos(x)+c*sin(x)),x)

[Out]

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*cos(x)/(-2*I*c*sin(x) - 2*c*cos(x)) - I*B*sin(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)) + C*sin(x)/(-2*I*c*s
in(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)) + I*B*sin(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)) + C*sin(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))

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