3.805 $$\int \frac{A+C \sinh (x)}{a+b \cosh (x)-b \sinh (x)} \, dx$$

Optimal. Leaf size=77 $\frac{\left (a^2 C+2 a A b-b^2 C\right ) \log (a-b \sinh (x)+b \cosh (x))}{2 a^2 b}+\frac{x (2 a A-b C)}{2 a^2}+\frac{C \sinh (x)}{2 a}+\frac{C \cosh (x)}{2 a}$

[Out]

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

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Rubi [A]  time = 0.0512851, antiderivative size = 77, 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 = {3131} $\frac{\left (a^2 C+2 a A b-b^2 C\right ) \log (a-b \sinh (x)+b \cosh (x))}{2 a^2 b}+\frac{x (2 a A-b C)}{2 a^2}+\frac{C \sinh (x)}{2 a}+\frac{C \cosh (x)}{2 a}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 3131

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

Rubi steps

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

Mathematica [A]  time = 0.233741, size = 86, normalized size = 1.12 $\frac{\frac{2 \left (a^2 C+2 a A b-b^2 C\right ) \log \left ((a-b) \sinh \left (\frac{x}{2}\right )+(a+b) \cosh \left (\frac{x}{2}\right )\right )}{b}+x \left (-\frac{a^2 C}{b}+2 a A-b C\right )+2 a C \sinh (x)+2 a C \cosh (x)}{4 a^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.048, size = 125, normalized size = 1.6 \begin{align*} -{\frac{C}{2\,b}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\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{bC}{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{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+C*sinh(x))/(a+b*cosh(x)-b*sinh(x)),x)

[Out]

-1/2*C/b*ln(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*C+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)*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 = 1.12609, size = 88, normalized size = 1.14 \begin{align*} A{\left (\frac{x}{a} + \frac{\log \left (b e^{\left (-x\right )} + a\right )}{a}\right )} - \frac{1}{2} \, C{\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+C*sinh(x))/(a+b*cosh(x)-b*sinh(x)),x, algorithm="maxima")

[Out]

A*(x/a + log(b*e^(-x) + a)/a) - 1/2*C*(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 = 2.3295, size = 155, normalized size = 2.01 \begin{align*} -\frac{C a^{2} x - C a b \cosh \left (x\right ) - C a b \sinh \left (x\right ) -{\left (C a^{2} + 2 \, A a b - C 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+C*sinh(x))/(a+b*cosh(x)-b*sinh(x)),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.13059, size = 66, normalized size = 0.86 \begin{align*} -\frac{C x}{2 \, b} + \frac{C e^{x}}{2 \, a} + \frac{{\left (C a^{2} + 2 \, A a b - C 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+C*sinh(x))/(a+b*cosh(x)-b*sinh(x)),x, algorithm="giac")

[Out]

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