3.804 \(\int \frac{A+B \cosh (x)+C \sinh (x)}{a+b \cosh (x)+b \sinh (x)} \, dx\)

Optimal. Leaf size=86 \[ -\frac{\left (a^2 (-(B+C))+2 a A b-b^2 (B-C)\right ) \log (a+b \sinh (x)+b \cosh (x))}{2 a^2 b}+\frac{x (2 a A-b (B-C))}{2 a^2}-\frac{(B-C) (\cosh (x)-\sinh (x))}{2 a} \]

[Out]

((2*a*A - b*(B - C))*x)/(2*a^2) - ((2*a*A*b - b^2*(B - C) - a^2*(B + C))*Log[a + b*Cosh[x] + b*Sinh[x]])/(2*a^
2*b) - ((B - C)*(Cosh[x] - Sinh[x]))/(2*a)

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Rubi [A]  time = 0.0830199, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.043, Rules used = {3130} \[ -\frac{\left (a^2 (-(B+C))+2 a A b-b^2 (B-C)\right ) \log (a+b \sinh (x)+b \cosh (x))}{2 a^2 b}+\frac{x (2 a A-b (B-C))}{2 a^2}-\frac{(B-C) (\cosh (x)-\sinh (x))}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cosh[x] + C*Sinh[x])/(a + b*Cosh[x] + b*Sinh[x]),x]

[Out]

((2*a*A - b*(B - C))*x)/(2*a^2) - ((2*a*A*b - b^2*(B - C) - a^2*(B + C))*Log[a + b*Cosh[x] + b*Sinh[x]])/(2*a^
2*b) - ((B - C)*(Cosh[x] - Sinh[x]))/(2*a)

Rule 3130

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

Rubi steps

\begin{align*} \int \frac{A+B \cosh (x)+C \sinh (x)}{a+b \cosh (x)+b \sinh (x)} \, dx &=\frac{(2 a A-b (B-C)) x}{2 a^2}-\frac{\left (2 a A b-b^2 (B-C)-a^2 (B+C)\right ) \log (a+b \cosh (x)+b \sinh (x))}{2 a^2 b}-\frac{(B-C) (\cosh (x)-\sinh (x))}{2 a}\\ \end{align*}

Mathematica [A]  time = 0.283727, size = 103, normalized size = 1.2 \[ \frac{\frac{2 \left (a^2 (B+C)-2 a A b+b^2 (B-C)\right ) \log \left ((b-a) \sinh \left (\frac{x}{2}\right )+(a+b) \cosh \left (\frac{x}{2}\right )\right )}{b}+x \left (\frac{a^2 (B+C)}{b}+2 a A+b (C-B)\right )+2 a (B-C) \sinh (x)-2 a (B-C) \cosh (x)}{4 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Cosh[x] + C*Sinh[x])/(a + b*Cosh[x] + b*Sinh[x]),x]

[Out]

((2*a*A + b*(-B + C) + (a^2*(B + C))/b)*x - 2*a*(B - C)*Cosh[x] + (2*(-2*a*A*b + b^2*(B - C) + a^2*(B + C))*Lo
g[(a + b)*Cosh[x/2] + (-a + b)*Sinh[x/2]])/b + 2*a*(B - C)*Sinh[x])/(4*a^2)

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Maple [B]  time = 0.055, size = 232, normalized size = 2.7 \begin{align*} -{\frac{B}{a} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}+{\frac{C}{a} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}+{\frac{A}{a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{Bb}{2\,{a}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{bC}{2\,{a}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{B}{2\,b}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }-{\frac{C}{2\,b}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }-{\frac{A}{a}\ln \left ( a\tanh \left ({\frac{x}{2}} \right ) -\tanh \left ({\frac{x}{2}} \right ) b-a-b \right ) }+{\frac{B}{2\,b}\ln \left ( a\tanh \left ({\frac{x}{2}} \right ) -\tanh \left ({\frac{x}{2}} \right ) b-a-b \right ) }+{\frac{Bb}{2\,{a}^{2}}\ln \left ( a\tanh \left ({\frac{x}{2}} \right ) -\tanh \left ({\frac{x}{2}} \right ) b-a-b \right ) }+{\frac{C}{2\,b}\ln \left ( a\tanh \left ({\frac{x}{2}} \right ) -\tanh \left ({\frac{x}{2}} \right ) b-a-b \right ) }-{\frac{bC}{2\,{a}^{2}}\ln \left ( a\tanh \left ({\frac{x}{2}} \right ) -\tanh \left ({\frac{x}{2}} \right ) b-a-b \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cosh(x)+C*sinh(x))/(a+b*cosh(x)+b*sinh(x)),x)

[Out]

-B/a/(tanh(1/2*x)+1)+C/a/(tanh(1/2*x)+1)+1/a*ln(tanh(1/2*x)+1)*A-1/2/a^2*ln(tanh(1/2*x)+1)*B*b+1/2/a^2*ln(tanh
(1/2*x)+1)*b*C-1/2*B/b*ln(tanh(1/2*x)-1)-1/2*C/b*ln(tanh(1/2*x)-1)-1/a*ln(a*tanh(1/2*x)-tanh(1/2*x)*b-a-b)*A+1
/2/b*ln(a*tanh(1/2*x)-tanh(1/2*x)*b-a-b)*B+1/2/a^2*b*ln(a*tanh(1/2*x)-tanh(1/2*x)*b-a-b)*B+1/2/b*ln(a*tanh(1/2
*x)-tanh(1/2*x)*b-a-b)*C-1/2/a^2*b*ln(a*tanh(1/2*x)-tanh(1/2*x)*b-a-b)*C

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Maxima [A]  time = 1.26165, size = 134, normalized size = 1.56 \begin{align*} \frac{1}{2} \, C{\left (\frac{x}{b} + \frac{e^{\left (-x\right )}}{a} + \frac{{\left (a^{2} - b^{2}\right )} \log \left (a e^{\left (-x\right )} + b\right )}{a^{2} b}\right )} + \frac{1}{2} \, B{\left (\frac{x}{b} - \frac{e^{\left (-x\right )}}{a} + \frac{{\left (a^{2} + b^{2}\right )} \log \left (a e^{\left (-x\right )} + b\right )}{a^{2} b}\right )} - \frac{A \log \left (a e^{\left (-x\right )} + b\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(x)+C*sinh(x))/(a+b*cosh(x)+b*sinh(x)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.46166, size = 342, normalized size = 3.98 \begin{align*} -\frac{{\left (B - C\right )} a b -{\left (2 \, A a b -{\left (B - C\right )} b^{2}\right )} x \cosh \left (x\right ) -{\left (2 \, A a b -{\left (B - C\right )} b^{2}\right )} x \sinh \left (x\right ) -{\left ({\left ({\left (B + C\right )} a^{2} - 2 \, A a b +{\left (B - C\right )} b^{2}\right )} \cosh \left (x\right ) +{\left ({\left (B + C\right )} a^{2} - 2 \, A a b +{\left (B - C\right )} b^{2}\right )} \sinh \left (x\right )\right )} \log \left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}{2 \,{\left (a^{2} b \cosh \left (x\right ) + a^{2} b \sinh \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(x)+C*sinh(x))/(a+b*cosh(x)+b*sinh(x)),x, algorithm="fricas")

[Out]

-1/2*((B - C)*a*b - (2*A*a*b - (B - C)*b^2)*x*cosh(x) - (2*A*a*b - (B - C)*b^2)*x*sinh(x) - (((B + C)*a^2 - 2*
A*a*b + (B - C)*b^2)*cosh(x) + ((B + C)*a^2 - 2*A*a*b + (B - C)*b^2)*sinh(x))*log(b*cosh(x) + b*sinh(x) + a))/
(a^2*b*cosh(x) + a^2*b*sinh(x))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(x)+C*sinh(x))/(a+b*cosh(x)+b*sinh(x)),x)

[Out]

Timed out

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Giac [A]  time = 1.11438, size = 107, normalized size = 1.24 \begin{align*} \frac{{\left (2 \, A a - B b + C b\right )} x}{2 \, a^{2}} - \frac{{\left (B a - C a\right )} e^{\left (-x\right )}}{2 \, a^{2}} + \frac{{\left (B a^{2} + C a^{2} - 2 \, A a b + B b^{2} - C b^{2}\right )} \log \left ({\left | b e^{x} + a \right |}\right )}{2 \, a^{2} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(x)+C*sinh(x))/(a+b*cosh(x)+b*sinh(x)),x, algorithm="giac")

[Out]

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