3.571 \(\int \frac {x (b+a \cosh (x))}{(a+b \cosh (x))^2} \, dx\)

Optimal. Leaf size=25 \[ \frac {x \sinh (x)}{a+b \cosh (x)}-\frac {\log (a+b \cosh (x))}{b} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {5637, 2668, 31} \[ \frac {x \sinh (x)}{a+b \cosh (x)}-\frac {\log (a+b \cosh (x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[(x*(b + a*Cosh[x]))/(a + b*Cosh[x])^2,x]

[Out]

-(Log[a + b*Cosh[x]]/b) + (x*Sinh[x])/(a + b*Cosh[x])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 5637

Int[((Cosh[(c_.) + (d_.)*(x_)]*(B_.) + (A_))*((e_.) + (f_.)*(x_)))/(Cosh[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^2,
x_Symbol] :> Simp[(B*(e + f*x)*Sinh[c + d*x])/(a*d*(a + b*Cosh[c + d*x])), x] - Dist[(B*f)/(a*d), Int[Sinh[c +
d*x]/(a + b*Cosh[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && EqQ[a*A - b*B, 0]

Rubi steps

\begin {align*} \int \frac {x (b+a \cosh (x))}{(a+b \cosh (x))^2} \, dx &=\frac {x \sinh (x)}{a+b \cosh (x)}-\int \frac {\sinh (x)}{a+b \cosh (x)} \, dx\\ &=\frac {x \sinh (x)}{a+b \cosh (x)}-\frac {\operatorname {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \cosh (x)\right )}{b}\\ &=-\frac {\log (a+b \cosh (x))}{b}+\frac {x \sinh (x)}{a+b \cosh (x)}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 25, normalized size = 1.00 \[ \frac {x \sinh (x)}{a+b \cosh (x)}-\frac {\log (a+b \cosh (x))}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*(b + a*Cosh[x]))/(a + b*Cosh[x])^2,x]

[Out]

-(Log[a + b*Cosh[x]]/b) + (x*Sinh[x])/(a + b*Cosh[x])

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fricas [B]  time = 0.44, size = 129, normalized size = 5.16 \[ \frac {2 \, b x \cosh \relax (x)^{2} + 2 \, b x \sinh \relax (x)^{2} + 2 \, a x \cosh \relax (x) - {\left (b \cosh \relax (x)^{2} + b \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (b \cosh \relax (x) + a\right )} \sinh \relax (x) + b\right )} \log \left (\frac {2 \, {\left (b \cosh \relax (x) + a\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) + 2 \, {\left (2 \, b x \cosh \relax (x) + a x\right )} \sinh \relax (x)}{b^{2} \cosh \relax (x)^{2} + b^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + b^{2} + 2 \, {\left (b^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b+a*cosh(x))/(a+b*cosh(x))^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.14, size = 100, normalized size = 4.00 \[ \frac {2 \, b x e^{\left (2 \, x\right )} - b e^{\left (2 \, x\right )} \log \left (-b e^{\left (2 \, x\right )} - 2 \, a e^{x} - b\right ) - 2 \, a e^{x} \log \left (-b e^{\left (2 \, x\right )} - 2 \, a e^{x} - b\right ) - 2 \, b x - b \log \left (-b e^{\left (2 \, x\right )} - 2 \, a e^{x} - b\right )}{b^{2} e^{\left (2 \, x\right )} + 2 \, a b e^{x} + b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b+a*cosh(x))/(a+b*cosh(x))^2,x, algorithm="giac")

[Out]

(2*b*x*e^(2*x) - b*e^(2*x)*log(-b*e^(2*x) - 2*a*e^x - b) - 2*a*e^x*log(-b*e^(2*x) - 2*a*e^x - b) - 2*b*x - b*l
og(-b*e^(2*x) - 2*a*e^x - b))/(b^2*e^(2*x) + 2*a*b*e^x + b^2)

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maple [B]  time = 0.48, size = 55, normalized size = 2.20 \[ \frac {2 x}{b}-\frac {2 x \left (a \,{\mathrm e}^{x}+b \right )}{b \left (b \,{\mathrm e}^{2 x}+2 a \,{\mathrm e}^{x}+b \right )}-\frac {\ln \left ({\mathrm e}^{2 x}+\frac {2 a \,{\mathrm e}^{x}}{b}+1\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b+a*cosh(x))/(a+b*cosh(x))^2,x)

[Out]

2*x/b-2*x*(a*exp(x)+b)/b/(b*exp(2*x)+2*a*exp(x)+b)-1/b*ln(exp(2*x)+2/b*a*exp(x)+1)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b+a*cosh(x))/(a+b*cosh(x))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 1.53, size = 105, normalized size = 4.20 \[ \frac {2\,x}{b}+\frac {\frac {2\,\left (b^3\,x-a^2\,b\,x\right )}{a^2\,b-b^3}+\frac {2\,{\mathrm {e}}^x\,\left (a\,b^3\,x-a^3\,b\,x\right )}{b\,\left (a^2\,b-b^3\right )}}{b+2\,a\,{\mathrm {e}}^x+b\,{\mathrm {e}}^{2\,x}}-\frac {\ln \left (b+2\,a\,{\mathrm {e}}^x+b\,{\mathrm {e}}^{2\,x}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(b + a*cosh(x)))/(a + b*cosh(x))^2,x)

[Out]

(2*x)/b + ((2*(b^3*x - a^2*b*x))/(a^2*b - b^3) + (2*exp(x)*(a*b^3*x - a^3*b*x))/(b*(a^2*b - b^3)))/(b + 2*a*ex
p(x) + b*exp(2*x)) - log(b + 2*a*exp(x) + b*exp(2*x))/b

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b+a*cosh(x))/(a+b*cosh(x))**2,x)

[Out]

Timed out

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