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

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Rubi [A]  time = 0.0405832, antiderivative size = 47, normalized size of antiderivative = 1., 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*}

Mathematica [A]  time = 0.119037, 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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Maple [B]  time = 0.059, size = 111, normalized size = 2.4 \begin{align*}{\frac{\ln \left ( c\tan \left ( x \right ) +b \right ) Bc}{{b}^{2}+{c}^{2}}}-{\frac{\ln \left ( c\tan \left ( x \right ) +b \right ) bC}{{b}^{2}+{c}^{2}}}-{\frac{\ln \left ( 1+ \left ( \tan \left ( x \right ) \right ) ^{2} \right ) Bc}{2\,{b}^{2}+2\,{c}^{2}}}+{\frac{\ln \left ( 1+ \left ( \tan \left ( x \right ) \right ) ^{2} \right ) bC}{2\,{b}^{2}+2\,{c}^{2}}}+{\frac{B\arctan \left ( \tan \left ( x \right ) \right ) b}{{b}^{2}+{c}^{2}}}+{\frac{C\arctan \left ( \tan \left ( x \right ) \right ) c}{{b}^{2}+{c}^{2}}} \end{align*}

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 = 1.51117, size = 244, normalized size = 5.19 \begin{align*} 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 )} \end{align*}

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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Fricas [A]  time = 1.98651, size = 139, normalized size = 2.96 \begin{align*} \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 )}} \end{align*}

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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Sympy [A]  time = 2.45975, size = 360, normalized size = 7.66 \begin{align*} \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 x - C \log{\left (\cos{\left (x \right )} \right )}}{b} & \text{for}\: c = 0 \\- \frac{i B x \sin{\left (x \right )}}{- 2 c \sin{\left (x \right )} + 2 i c \cos{\left (x \right )}} - \frac{B x \cos{\left (x \right )}}{- 2 c \sin{\left (x \right )} + 2 i c \cos{\left (x \right )}} - \frac{B \sin{\left (x \right )}}{- 2 c \sin{\left (x \right )} + 2 i c \cos{\left (x \right )}} - \frac{C x \sin{\left (x \right )}}{- 2 c \sin{\left (x \right )} + 2 i c \cos{\left (x \right )}} + \frac{i C x \cos{\left (x \right )}}{- 2 c \sin{\left (x \right )} + 2 i c \cos{\left (x \right )}} - \frac{i C \sin{\left (x \right )}}{- 2 c \sin{\left (x \right )} + 2 i c \cos{\left (x \right )}} & \text{for}\: b = - i c \\- \frac{i B x \sin{\left (x \right )}}{2 c \sin{\left (x \right )} + 2 i c \cos{\left (x \right )}} + \frac{B x \cos{\left (x \right )}}{2 c \sin{\left (x \right )} + 2 i c \cos{\left (x \right )}} + \frac{B \sin{\left (x \right )}}{2 c \sin{\left (x \right )} + 2 i c \cos{\left (x \right )}} + \frac{C x \sin{\left (x \right )}}{2 c \sin{\left (x \right )} + 2 i c \cos{\left (x \right )}} + \frac{i C x \cos{\left (x \right )}}{2 c \sin{\left (x \right )} + 2 i c \cos{\left (x \right )}} - \frac{i C \sin{\left (x \right )}}{2 c \sin{\left (x \right )} + 2 i c \cos{\left (x \right )}} & \text{for}\: b = i c \\\frac{B b x}{b^{2} + c^{2}} + \frac{B c \log{\left (\frac{b \cos{\left (x \right )}}{c} + \sin{\left (x \right )} \right )}}{b^{2} + c^{2}} - \frac{C b \log{\left (\frac{b \cos{\left (x \right )}}{c} + \sin{\left (x \right )} \right )}}{b^{2} + c^{2}} + \frac{C c x}{b^{2} + c^{2}} & \text{otherwise} \end{cases} \end{align*}

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*x - C*log(cos(x)))/b, Eq(c, 0)), (-I*B*x*sin(x
)/(-2*c*sin(x) + 2*I*c*cos(x)) - B*x*cos(x)/(-2*c*sin(x) + 2*I*c*cos(x)) - B*sin(x)/(-2*c*sin(x) + 2*I*c*cos(x
)) - C*x*sin(x)/(-2*c*sin(x) + 2*I*c*cos(x)) + I*C*x*cos(x)/(-2*c*sin(x) + 2*I*c*cos(x)) - I*C*sin(x)/(-2*c*si
n(x) + 2*I*c*cos(x)), Eq(b, -I*c)), (-I*B*x*sin(x)/(2*c*sin(x) + 2*I*c*cos(x)) + B*x*cos(x)/(2*c*sin(x) + 2*I*
c*cos(x)) + B*sin(x)/(2*c*sin(x) + 2*I*c*cos(x)) + C*x*sin(x)/(2*c*sin(x) + 2*I*c*cos(x)) + I*C*x*cos(x)/(2*c*
sin(x) + 2*I*c*cos(x)) - I*C*sin(x)/(2*c*sin(x) + 2*I*c*cos(x)), Eq(b, I*c)), (B*b*x/(b**2 + c**2) + B*c*log(b
*cos(x)/c + sin(x))/(b**2 + c**2) - C*b*log(b*cos(x)/c + sin(x))/(b**2 + c**2) + C*c*x/(b**2 + c**2), True))

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Giac [A]  time = 1.16139, size = 104, normalized size = 2.21 \begin{align*} \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}} \end{align*}

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)