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

Optimal. Leaf size=78 \[ \frac{\left (a^2 (-B)+2 a A b-b^2 B\right ) \log (a-b \sinh (x)+b \cosh (x))}{2 a^2 b}+\frac{x (2 a A-b B)}{2 a^2}+\frac{B \sinh (x)}{2 a}+\frac{B \cosh (x)}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.0455736, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {3132} \[ \frac{\left (a^2 (-B)+2 a A b-b^2 B\right ) \log (a-b \sinh (x)+b \cosh (x))}{2 a^2 b}+\frac{x (2 a A-b B)}{2 a^2}+\frac{B \sinh (x)}{2 a}+\frac{B \cosh (x)}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3132

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

Rubi steps

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

Mathematica [A]  time = 0.157305, size = 86, normalized size = 1.1 \[ \frac{x \left (a^2 B+2 a A b-b^2 B\right )-2 \left (a^2 B-2 a A b+b^2 B\right ) \log \left ((a-b) \sinh \left (\frac{x}{2}\right )+(a+b) \cosh \left (\frac{x}{2}\right )\right )+2 a b B \sinh (x)+2 a b B \cosh (x)}{4 a^2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.044, size = 125, normalized size = 1.6 \begin{align*}{\frac{B}{2\,b}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{B}{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{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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*B/b*ln(tanh(1/2*x)+1)-B/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/a*ln(a*t
anh(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

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Maxima [A]  time = 1.07242, size = 84, normalized size = 1.08 \begin{align*} A{\left (\frac{x}{a} + \frac{\log \left (b e^{\left (-x\right )} + a\right )}{a}\right )} - \frac{1}{2} \, B{\left (\frac{b x}{a^{2}} - \frac{e^{x}}{a} + \frac{{\left (a^{2} + b^{2}\right )} \log \left (b e^{\left (-x\right )} + a\right )}{a^{2} b}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.30002, size = 154, normalized size = 1.97 \begin{align*} \frac{B a^{2} x + B a b \cosh \left (x\right ) + B a b \sinh \left (x\right ) -{\left (B a^{2} - 2 \, A a b + B b^{2}\right )} \log \left (a \cosh \left (x\right ) + a \sinh \left (x\right ) + b\right )}{2 \, a^{2} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))/(a+b*cosh(x)-b*sinh(x)),x)

[Out]

Timed out

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Giac [A]  time = 1.15686, size = 65, normalized size = 0.83 \begin{align*} \frac{B x}{2 \, b} + \frac{B e^{x}}{2 \, a} - \frac{{\left (B a^{2} - 2 \, A a b + B b^{2}\right )} \log \left ({\left | a e^{x} + b \right |}\right )}{2 \, a^{2} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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