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]

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

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Rubi [A]  time = 0.05, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, 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*}

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Mathematica [A]  time = 0.25, 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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fricas [A]  time = 0.45, size = 59, normalized size = 0.77 \[ -\frac {C a^{2} x - C a b \cosh \relax (x) - C a b \sinh \relax (x) - {\left (C a^{2} + 2 \, A a b - C b^{2}\right )} \log \left (a \cosh \relax (x) + a \sinh \relax (x) + b\right )}{2 \, a^{2} b} \]

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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giac [A]  time = 0.13, size = 49, normalized size = 0.64 \[ -\frac {C x}{2 \, b} + \frac {C e^{x}}{2 \, a} + \frac {{\left (C a^{2} + 2 \, A a b - C b^{2}\right )} \log \left ({\left | a e^{x} + b \right |}\right )}{2 \, a^{2} b} \]

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)

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maple [A]  time = 0.19, size = 125, normalized size = 1.62 \[ -\frac {C}{a \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) A}{a}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) b C}{2 a^{2}}-\frac {C \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 b}+\frac {\ln \left (a \tanh \left (\frac {x}{2}\right )-\tanh \left (\frac {x}{2}\right ) b +a +b \right ) A}{a}+\frac {\ln \left (a \tanh \left (\frac {x}{2}\right )-\tanh \left (\frac {x}{2}\right ) b +a +b \right ) C}{2 b}-\frac {b \ln \left (a \tanh \left (\frac {x}{2}\right )-\tanh \left (\frac {x}{2}\right ) b +a +b \right ) C}{2 a^{2}} \]

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]

-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-1/2*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)*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 = 0.33, size = 65, normalized size = 0.84 \[ A {\left (\frac {x}{a} + \frac {\log \left (b e^{\left (-x\right )} + a\right )}{a}\right )} - \frac {1}{2} \, C {\left (\frac {b x}{a^{2}} - \frac {e^{x}}{a} - \frac {{\left (a^{2} - b^{2}\right )} \log \left (b e^{\left (-x\right )} + a\right )}{a^{2} b}\right )} \]

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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mupad [B]  time = 0.12, size = 48, normalized size = 0.62 \[ \frac {C\,{\mathrm {e}}^x}{2\,a}-\frac {C\,x}{2\,b}+\frac {\ln \left (b+a\,{\mathrm {e}}^x\right )\,\left (C\,a^2+2\,A\,a\,b-C\,b^2\right )}{2\,a^2\,b} \]

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]

(C*exp(x))/(2*a) - (C*x)/(2*b) + (log(b + a*exp(x))*(C*a^2 - C*b^2 + 2*A*a*b))/(2*a^2*b)

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sympy [A]  time = 5.26, size = 852, normalized size = 11.06 \[ \text {result too large to display} \]

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]

Piecewise((zoo*(A*x + C*cosh(x)), Eq(a, 0) & Eq(b, 0)), (2*A/(-2*b*sinh(x) + 2*b*cosh(x)) + C*x*sinh(x)/(-2*b*
sinh(x) + 2*b*cosh(x)) - C*x*cosh(x)/(-2*b*sinh(x) + 2*b*cosh(x)) + C*cosh(x)/(-2*b*sinh(x) + 2*b*cosh(x)), Eq
(a, 0)), ((A*x + C*cosh(x))/a, Eq(b, 0)), (2*A*x*tanh(x/2)/(2*b*tanh(x/2) - 2*b) - 2*A*x/(2*b*tanh(x/2) - 2*b)
- 2*A*log(tanh(x/2) + 1)*tanh(x/2)/(2*b*tanh(x/2) - 2*b) + 2*A*log(tanh(x/2) + 1)/(2*b*tanh(x/2) - 2*b) - C*x
*tanh(x/2)/(2*b*tanh(x/2) - 2*b) + C*x/(2*b*tanh(x/2) - 2*b) - 2*C/(2*b*tanh(x/2) - 2*b), Eq(a, b)), (2*A*a*b*
x*tanh(x/2)/(2*a**2*b*tanh(x/2) - 2*a**2*b) - 2*A*a*b*x/(2*a**2*b*tanh(x/2) - 2*a**2*b) - 2*A*a*b*log(tanh(x/2
) + 1)*tanh(x/2)/(2*a**2*b*tanh(x/2) - 2*a**2*b) + 2*A*a*b*log(tanh(x/2) + 1)/(2*a**2*b*tanh(x/2) - 2*a**2*b)
+ 2*A*a*b*log(a/(a - b) + b/(a - b) + tanh(x/2))*tanh(x/2)/(2*a**2*b*tanh(x/2) - 2*a**2*b) - 2*A*a*b*log(a/(a
- b) + b/(a - b) + tanh(x/2))/(2*a**2*b*tanh(x/2) - 2*a**2*b) - C*a**2*log(tanh(x/2) + 1)*tanh(x/2)/(2*a**2*b*
tanh(x/2) - 2*a**2*b) + C*a**2*log(tanh(x/2) + 1)/(2*a**2*b*tanh(x/2) - 2*a**2*b) + C*a**2*log(a/(a - b) + b/(
a - b) + tanh(x/2))*tanh(x/2)/(2*a**2*b*tanh(x/2) - 2*a**2*b) - C*a**2*log(a/(a - b) + b/(a - b) + tanh(x/2))/
(2*a**2*b*tanh(x/2) - 2*a**2*b) - 2*C*a*b/(2*a**2*b*tanh(x/2) - 2*a**2*b) - C*b**2*x*tanh(x/2)/(2*a**2*b*tanh(
x/2) - 2*a**2*b) + C*b**2*x/(2*a**2*b*tanh(x/2) - 2*a**2*b) + C*b**2*log(tanh(x/2) + 1)*tanh(x/2)/(2*a**2*b*ta
nh(x/2) - 2*a**2*b) - C*b**2*log(tanh(x/2) + 1)/(2*a**2*b*tanh(x/2) - 2*a**2*b) - C*b**2*log(a/(a - b) + b/(a
- b) + tanh(x/2))*tanh(x/2)/(2*a**2*b*tanh(x/2) - 2*a**2*b) + C*b**2*log(a/(a - b) + b/(a - b) + tanh(x/2))/(2
*a**2*b*tanh(x/2) - 2*a**2*b), True))

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