3.6.98 \(\int \frac {x+\cosh (x)+\sinh (x)}{\cosh (x)-\sinh (x)} \, dx\) [598]

Optimal. Leaf size=20 \[ -e^x+\frac {e^{2 x}}{2}+e^x x \]

[Out]

-exp(x)+1/2*exp(2*x)+exp(x)*x

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Rubi [A]
time = 0.04, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {5767, 6874, 2207, 2225, 2320, 12, 14} \begin {gather*} e^x x-e^x+\frac {e^{2 x}}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x + Cosh[x] + Sinh[x])/(Cosh[x] - Sinh[x]),x]

[Out]

-E^x + E^(2*x)/2 + E^x*x

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 5767

Int[(u_.)*(Cosh[v_]*(a_.) + (b_.)*Sinh[v_])^(n_.), x_Symbol] :> Int[u*(a*E^((a/b)*v))^n, x] /; FreeQ[{a, b, n}
, x] && EqQ[a^2 - b^2, 0]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 23, normalized size = 1.15 \begin {gather*} \frac {1}{2} \cosh (2 x)+(-1+x) \sinh (x)+\cosh (x) (-1+x+\sinh (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x + Cosh[x] + Sinh[x])/(Cosh[x] - Sinh[x]),x]

[Out]

Cosh[2*x]/2 + (-1 + x)*Sinh[x] + Cosh[x]*(-1 + x + Sinh[x])

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Maple [A]
time = 0.19, size = 16, normalized size = 0.80

method	result	size
risch	\(\left (-1+x \right ) {\mathrm e}^{x}+\frac {{\mathrm e}^{2 x}}{2}\)	\(14\)
default	\(\frac {\sinh \left (x \right )+x -1}{\cosh \left (x \right )-\sinh \left (x \right )}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x+cosh(x)+sinh(x))/(cosh(x)-sinh(x)),x,method=_RETURNVERBOSE)

[Out]

(sinh(x)+x-1)/(cosh(x)-sinh(x))

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Maxima [A]
time = 3.73, size = 13, normalized size = 0.65 \begin {gather*} {\left (x - 1\right )} e^{x} + \frac {1}{2} \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x+cosh(x)+sinh(x))/(cosh(x)-sinh(x)),x, algorithm="maxima")

[Out]

(x - 1)*e^x + 1/2*e^(2*x)

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Fricas [A]
time = 0.54, size = 20, normalized size = 1.00 \begin {gather*} \frac {2 \, x + \cosh \left (x\right ) + \sinh \left (x\right ) - 2}{2 \, {\left (\cosh \left (x\right ) - \sinh \left (x\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x+cosh(x)+sinh(x))/(cosh(x)-sinh(x)),x, algorithm="fricas")

[Out]

1/2*(2*x + cosh(x) + sinh(x) - 2)/(cosh(x) - sinh(x))

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Sympy [A]
time = 0.18, size = 26, normalized size = 1.30 \begin {gather*} \frac {x}{- \sinh {\left (x \right )} + \cosh {\left (x \right )}} + \frac {\sinh {\left (x \right )}}{- \sinh {\left (x \right )} + \cosh {\left (x \right )}} - \frac {1}{- \sinh {\left (x \right )} + \cosh {\left (x \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x+cosh(x)+sinh(x))/(cosh(x)-sinh(x)),x)

[Out]

x/(-sinh(x) + cosh(x)) + sinh(x)/(-sinh(x) + cosh(x)) - 1/(-sinh(x) + cosh(x))

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Giac [A]
time = 0.46, size = 11, normalized size = 0.55 \begin {gather*} \frac {1}{2} \, {\left (2 \, x + e^{x} - 2\right )} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x+cosh(x)+sinh(x))/(cosh(x)-sinh(x)),x, algorithm="giac")

[Out]

1/2*(2*x + e^x - 2)*e^x

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Mupad [B]
time = 0.06, size = 16, normalized size = 0.80 \begin {gather*} {\mathrm {e}}^x\,\left (x+\frac {{\mathrm {e}}^{-x}}{2}+\frac {{\mathrm {e}}^x}{2}-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + cosh(x) + sinh(x))/(cosh(x) - sinh(x)),x)

[Out]

exp(x)*(x + exp(-x)/2 + exp(x)/2 - 1)

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