3.67.40 \(\int \frac {25 \sqrt [35]{e}+(1-10 \sqrt [35]{e}) e^x+e^{\frac {1}{35}+2 x}}{25 \sqrt [35]{e}-10 e^{\frac {1}{35}+x}+e^{\frac {1}{35}+2 x}} \, dx\)

Optimal. Leaf size=17 \[ \frac {1}{\sqrt [35]{e} \left (5-e^x\right )}+x \]

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Rubi [A]  time = 0.09, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {2282, 12, 893} \begin {gather*} x+\frac {1}{\sqrt [35]{e} \left (5-e^x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(25*E^(1/35) + (1 - 10*E^(1/35))*E^x + E^(1/35 + 2*x))/(25*E^(1/35) - 10*E^(1/35 + x) + E^(1/35 + 2*x)),x]

[Out]

1/(E^(1/35)*(5 - 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 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 2282

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]]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\operatorname {Subst}\left (\int \frac {25 \sqrt [35]{e}+\left (1-10 \sqrt [35]{e}\right ) x+\sqrt [35]{e} x^2}{\sqrt [35]{e} (5-x)^2 x} \, dx,x,e^x\right )\\ &=\frac {\operatorname {Subst}\left (\int \frac {25 \sqrt [35]{e}+\left (1-10 \sqrt [35]{e}\right ) x+\sqrt [35]{e} x^2}{(5-x)^2 x} \, dx,x,e^x\right )}{\sqrt [35]{e}}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {1}{(-5+x)^2}+\frac {\sqrt [35]{e}}{x}\right ) \, dx,x,e^x\right )}{\sqrt [35]{e}}\\ &=\frac {1}{\sqrt [35]{e} \left (5-e^x\right )}+x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 17, normalized size = 1.00 \begin {gather*} \frac {1}{\sqrt [35]{e} \left (5-e^x\right )}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25*E^(1/35) + (1 - 10*E^(1/35))*E^x + E^(1/35 + 2*x))/(25*E^(1/35) - 10*E^(1/35 + x) + E^(1/35 + 2*
x)),x]

[Out]

1/(E^(1/35)*(5 - E^x)) + x

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fricas [B]  time = 0.59, size = 28, normalized size = 1.65 \begin {gather*} \frac {5 \, x e^{\frac {1}{35}} - x e^{\left (x + \frac {1}{35}\right )} + 1}{5 \, e^{\frac {1}{35}} - e^{\left (x + \frac {1}{35}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1/35)*exp(x)^2+(-10*exp(1/35)+1)*exp(x)+25*exp(1/35))/(exp(1/35)*exp(x)^2-10*exp(1/35)*exp(x)+2
5*exp(1/35)),x, algorithm="fricas")

[Out]

(5*x*e^(1/35) - x*e^(x + 1/35) + 1)/(5*e^(1/35) - e^(x + 1/35))

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giac [A]  time = 0.12, size = 12, normalized size = 0.71 \begin {gather*} x - \frac {e^{\left (-\frac {1}{35}\right )}}{e^{x} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1/35)*exp(x)^2+(-10*exp(1/35)+1)*exp(x)+25*exp(1/35))/(exp(1/35)*exp(x)^2-10*exp(1/35)*exp(x)+2
5*exp(1/35)),x, algorithm="giac")

[Out]

x - e^(-1/35)/(e^x - 5)

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maple [A]  time = 0.06, size = 13, normalized size = 0.76

method	result	size
risch	\(x -\frac {{\mathrm e}^{-\frac {1}{35}}}{{\mathrm e}^{x}-5}\)	\(13\)
derivativedivides	\({\mathrm e}^{-\frac {1}{35}} \left ({\mathrm e}^{\frac {1}{35}} \ln \left ({\mathrm e}^{x}\right )-\frac {1}{{\mathrm e}^{x}-5}\right )\)	\(21\)
default	\({\mathrm e}^{-\frac {1}{35}} \left ({\mathrm e}^{\frac {1}{35}} \ln \left ({\mathrm e}^{x}\right )-\frac {1}{{\mathrm e}^{x}-5}\right )\)	\(21\)
norman	\(\frac {{\mathrm e}^{x} x -5 x -{\mathrm e}^{-\frac {1}{35}}}{{\mathrm e}^{x}-5}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(1/35)*exp(x)^2+(-10*exp(1/35)+1)*exp(x)+25*exp(1/35))/(exp(1/35)*exp(x)^2-10*exp(1/35)*exp(x)+25*exp(
1/35)),x,method=_RETURNVERBOSE)

[Out]

x-1/(exp(x)-5)*exp(-1/35)

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maxima [A]  time = 0.39, size = 15, normalized size = 0.88 \begin {gather*} x + \frac {1}{5 \, e^{\frac {1}{35}} - e^{\left (x + \frac {1}{35}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1/35)*exp(x)^2+(-10*exp(1/35)+1)*exp(x)+25*exp(1/35))/(exp(1/35)*exp(x)^2-10*exp(1/35)*exp(x)+2
5*exp(1/35)),x, algorithm="maxima")

[Out]

x + 1/(5*e^(1/35) - e^(x + 1/35))

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mupad [B]  time = 4.31, size = 15, normalized size = 0.88 \begin {gather*} x-\frac {1}{{\mathrm {e}}^{x+\frac {1}{35}}-5\,{\mathrm {e}}^{1/35}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25*exp(1/35) + exp(2*x)*exp(1/35) - exp(x)*(10*exp(1/35) - 1))/(25*exp(1/35) + exp(2*x)*exp(1/35) - 10*ex
p(1/35)*exp(x)),x)

[Out]

x - 1/(exp(x + 1/35) - 5*exp(1/35))

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sympy [A]  time = 0.10, size = 17, normalized size = 1.00 \begin {gather*} x - \frac {1}{e^{\frac {1}{35}} e^{x} - 5 e^{\frac {1}{35}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1/35)*exp(x)**2+(-10*exp(1/35)+1)*exp(x)+25*exp(1/35))/(exp(1/35)*exp(x)**2-10*exp(1/35)*exp(x)
+25*exp(1/35)),x)

[Out]

x - 1/(exp(1/35)*exp(x) - 5*exp(1/35))

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