3.19.55 \(\int \frac {e^{1+x} (1-x)-5 e x^2+(e^{1+x}-6 x+5 e x^2) \log (\frac {-e^{1+x}+6 x-5 e x^2}{3 x})}{e^{1+x} x^2-6 x^3+5 e x^4+(2 e^{1+x} x-12 x^2+10 e x^3) \log (\frac {-e^{1+x}+6 x-5 e x^2}{3 x})+(e^{1+x}-6 x+5 e x^2) \log ^2(\frac {-e^{1+x}+6 x-5 e x^2}{3 x})} \, dx\)

Optimal. Leaf size=26 \[ \frac {x}{x+\log \left (2-\frac {e \left (e^x+5 x^2\right )}{3 x}\right )} \]

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Rubi [A]  time = 0.77, antiderivative size = 31, normalized size of antiderivative = 1.19, number of steps used = 3, number of rules used = 3, integrand size = 172, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {6688, 6711, 32} \begin {gather*} -\frac {1}{\frac {x}{\log \left (-\frac {5 e x}{3}-\frac {e^{x+1}}{3 x}+2\right )}+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(1 + x)*(1 - x) - 5*E*x^2 + (E^(1 + x) - 6*x + 5*E*x^2)*Log[(-E^(1 + x) + 6*x - 5*E*x^2)/(3*x)])/(E^(1
+ x)*x^2 - 6*x^3 + 5*E*x^4 + (2*E^(1 + x)*x - 12*x^2 + 10*E*x^3)*Log[(-E^(1 + x) + 6*x - 5*E*x^2)/(3*x)] + (E^
(1 + x) - 6*x + 5*E*x^2)*Log[(-E^(1 + x) + 6*x - 5*E*x^2)/(3*x)]^2),x]

[Out]

-(1 + x/Log[2 - E^(1 + x)/(3*x) - (5*E*x)/3])^(-1)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 6688

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

Rule 6711

Int[(u_)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x] - q*v*D[w
, x])]}, Dist[c*p, Subst[Int[(b + a*x^p)^m, x], x, v*w^(m*q + 1)], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q}
, x] && EqQ[p + q*(m*p + 1), 0] && IntegerQ[p] && IntegerQ[m]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-e \left (e^x (-1+x)+5 x^2\right )+\left (e^{1+x}-6 x+5 e x^2\right ) \log \left (2-\frac {e^{1+x}}{3 x}-\frac {5 e x}{3}\right )}{\left (e^{1+x}-6 x+5 e x^2\right ) \left (x+\log \left (2-\frac {e^{1+x}}{3 x}-\frac {5 e x}{3}\right )\right )^2} \, dx\\ &=\operatorname {Subst}\left (\int \frac {1}{(1+x)^2} \, dx,x,\frac {x}{\log \left (2-\frac {e^{1+x}}{3 x}-\frac {5 e x}{3}\right )}\right )\\ &=-\frac {1}{1+\frac {x}{\log \left (2-\frac {e^{1+x}}{3 x}-\frac {5 e x}{3}\right )}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 27, normalized size = 1.04 \begin {gather*} \frac {x}{x+\log \left (2-\frac {e^{1+x}}{3 x}-\frac {5 e x}{3}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(1 + x)*(1 - x) - 5*E*x^2 + (E^(1 + x) - 6*x + 5*E*x^2)*Log[(-E^(1 + x) + 6*x - 5*E*x^2)/(3*x)])/
(E^(1 + x)*x^2 - 6*x^3 + 5*E*x^4 + (2*E^(1 + x)*x - 12*x^2 + 10*E*x^3)*Log[(-E^(1 + x) + 6*x - 5*E*x^2)/(3*x)]
+ (E^(1 + x) - 6*x + 5*E*x^2)*Log[(-E^(1 + x) + 6*x - 5*E*x^2)/(3*x)]^2),x]

[Out]

x/(x + Log[2 - E^(1 + x)/(3*x) - (5*E*x)/3])

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fricas [A]  time = 0.63, size = 27, normalized size = 1.04 \begin {gather*} \frac {x}{x + \log \left (-\frac {5 \, x^{2} e - 6 \, x + e^{\left (x + 1\right )}}{3 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(1)*exp(x)+5*x^2*exp(1)-6*x)*log(1/3*(-exp(1)*exp(x)-5*x^2*exp(1)+6*x)/x)+(-x+1)*exp(1)*exp(x)-
5*x^2*exp(1))/((exp(1)*exp(x)+5*x^2*exp(1)-6*x)*log(1/3*(-exp(1)*exp(x)-5*x^2*exp(1)+6*x)/x)^2+(2*x*exp(1)*exp
(x)+10*x^3*exp(1)-12*x^2)*log(1/3*(-exp(1)*exp(x)-5*x^2*exp(1)+6*x)/x)+x^2*exp(1)*exp(x)+5*x^4*exp(1)-6*x^3),x
, algorithm="fricas")

[Out]

x/(x + log(-1/3*(5*x^2*e - 6*x + e^(x + 1))/x))

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giac [A]  time = 1.35, size = 27, normalized size = 1.04 \begin {gather*} \frac {x}{x + \log \left (-\frac {5 \, x^{2} e - 6 \, x + e^{\left (x + 1\right )}}{3 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(1)*exp(x)+5*x^2*exp(1)-6*x)*log(1/3*(-exp(1)*exp(x)-5*x^2*exp(1)+6*x)/x)+(-x+1)*exp(1)*exp(x)-
5*x^2*exp(1))/((exp(1)*exp(x)+5*x^2*exp(1)-6*x)*log(1/3*(-exp(1)*exp(x)-5*x^2*exp(1)+6*x)/x)^2+(2*x*exp(1)*exp
(x)+10*x^3*exp(1)-12*x^2)*log(1/3*(-exp(1)*exp(x)-5*x^2*exp(1)+6*x)/x)+x^2*exp(1)*exp(x)+5*x^4*exp(1)-6*x^3),x
, algorithm="giac")

[Out]

x/(x + log(-1/3*(5*x^2*e - 6*x + e^(x + 1))/x))

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maple [C]  time = 0.22, size = 234, normalized size = 9.00

method	result	size
risch	\(\frac {2 x}{-i \pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (\left (x^{2}+\frac {{\mathrm e}^{x}}{5}\right ) {\mathrm e}-\frac {6 x}{5}\right )\right ) \mathrm {csgn}\left (\frac {i \left (\left (x^{2}+\frac {{\mathrm e}^{x}}{5}\right ) {\mathrm e}-\frac {6 x}{5}\right )}{x}\right )+i \pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (\left (x^{2}+\frac {{\mathrm e}^{x}}{5}\right ) {\mathrm e}-\frac {6 x}{5}\right )}{x}\right )^{2}-2 i \pi \mathrm {csgn}\left (\frac {i \left (\left (x^{2}+\frac {{\mathrm e}^{x}}{5}\right ) {\mathrm e}-\frac {6 x}{5}\right )}{x}\right )^{2}+i \pi \,\mathrm {csgn}\left (i \left (\left (x^{2}+\frac {{\mathrm e}^{x}}{5}\right ) {\mathrm e}-\frac {6 x}{5}\right )\right ) \mathrm {csgn}\left (\frac {i \left (\left (x^{2}+\frac {{\mathrm e}^{x}}{5}\right ) {\mathrm e}-\frac {6 x}{5}\right )}{x}\right )^{2}+i \pi \mathrm {csgn}\left (\frac {i \left (\left (x^{2}+\frac {{\mathrm e}^{x}}{5}\right ) {\mathrm e}-\frac {6 x}{5}\right )}{x}\right )^{3}+2 i \pi +2 \ln \relax (5)-2 \ln \relax (3)+2 x -2 \ln \relax (x )+2 \ln \left (\left (x^{2}+\frac {{\mathrm e}^{x}}{5}\right ) {\mathrm e}-\frac {6 x}{5}\right )}\)	\(234\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(1)*exp(x)+5*x^2*exp(1)-6*x)*ln(1/3*(-exp(1)*exp(x)-5*x^2*exp(1)+6*x)/x)+(1-x)*exp(1)*exp(x)-5*x^2*ex
p(1))/((exp(1)*exp(x)+5*x^2*exp(1)-6*x)*ln(1/3*(-exp(1)*exp(x)-5*x^2*exp(1)+6*x)/x)^2+(2*x*exp(1)*exp(x)+10*x^
3*exp(1)-12*x^2)*ln(1/3*(-exp(1)*exp(x)-5*x^2*exp(1)+6*x)/x)+x^2*exp(1)*exp(x)+5*x^4*exp(1)-6*x^3),x,method=_R
ETURNVERBOSE)

[Out]

2*x/(-I*Pi*csgn(I/x)*csgn(I*((x^2+1/5*exp(x))*exp(1)-6/5*x))*csgn(I/x*((x^2+1/5*exp(x))*exp(1)-6/5*x))+I*Pi*cs
gn(I/x)*csgn(I/x*((x^2+1/5*exp(x))*exp(1)-6/5*x))^2-2*I*Pi*csgn(I/x*((x^2+1/5*exp(x))*exp(1)-6/5*x))^2+I*Pi*cs
gn(I*((x^2+1/5*exp(x))*exp(1)-6/5*x))*csgn(I/x*((x^2+1/5*exp(x))*exp(1)-6/5*x))^2+I*Pi*csgn(I/x*((x^2+1/5*exp(
x))*exp(1)-6/5*x))^3+2*I*Pi+2*ln(5)-2*ln(3)+2*x-2*ln(x)+2*ln((x^2+1/5*exp(x))*exp(1)-6/5*x))

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maxima [A]  time = 1.06, size = 32, normalized size = 1.23 \begin {gather*} \frac {x}{x - \log \relax (3) + \log \left (-5 \, x^{2} e + 6 \, x - e^{\left (x + 1\right )}\right ) - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(1)*exp(x)+5*x^2*exp(1)-6*x)*log(1/3*(-exp(1)*exp(x)-5*x^2*exp(1)+6*x)/x)+(-x+1)*exp(1)*exp(x)-
5*x^2*exp(1))/((exp(1)*exp(x)+5*x^2*exp(1)-6*x)*log(1/3*(-exp(1)*exp(x)-5*x^2*exp(1)+6*x)/x)^2+(2*x*exp(1)*exp
(x)+10*x^3*exp(1)-12*x^2)*log(1/3*(-exp(1)*exp(x)-5*x^2*exp(1)+6*x)/x)+x^2*exp(1)*exp(x)+5*x^4*exp(1)-6*x^3),x
, algorithm="maxima")

[Out]

x/(x - log(3) + log(-5*x^2*e + 6*x - e^(x + 1)) - log(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {5\,x^2\,\mathrm {e}-\ln \left (-\frac {\frac {5\,x^2\,\mathrm {e}}{3}-2\,x+\frac {\mathrm {e}\,{\mathrm {e}}^x}{3}}{x}\right )\,\left (5\,x^2\,\mathrm {e}-6\,x+\mathrm {e}\,{\mathrm {e}}^x\right )+\mathrm {e}\,{\mathrm {e}}^x\,\left (x-1\right )}{{\ln \left (-\frac {\frac {5\,x^2\,\mathrm {e}}{3}-2\,x+\frac {\mathrm {e}\,{\mathrm {e}}^x}{3}}{x}\right )}^2\,\left (5\,x^2\,\mathrm {e}-6\,x+\mathrm {e}\,{\mathrm {e}}^x\right )+5\,x^4\,\mathrm {e}+\ln \left (-\frac {\frac {5\,x^2\,\mathrm {e}}{3}-2\,x+\frac {\mathrm {e}\,{\mathrm {e}}^x}{3}}{x}\right )\,\left (10\,x^3\,\mathrm {e}-12\,x^2+2\,x\,\mathrm {e}\,{\mathrm {e}}^x\right )-6\,x^3+x^2\,\mathrm {e}\,{\mathrm {e}}^x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x^2*exp(1) - log(-((5*x^2*exp(1))/3 - 2*x + (exp(1)*exp(x))/3)/x)*(5*x^2*exp(1) - 6*x + exp(1)*exp(x))
+ exp(1)*exp(x)*(x - 1))/(log(-((5*x^2*exp(1))/3 - 2*x + (exp(1)*exp(x))/3)/x)^2*(5*x^2*exp(1) - 6*x + exp(1)
*exp(x)) + 5*x^4*exp(1) + log(-((5*x^2*exp(1))/3 - 2*x + (exp(1)*exp(x))/3)/x)*(10*x^3*exp(1) - 12*x^2 + 2*x*e
xp(1)*exp(x)) - 6*x^3 + x^2*exp(1)*exp(x)),x)

[Out]

int(-(5*x^2*exp(1) - log(-((5*x^2*exp(1))/3 - 2*x + (exp(1)*exp(x))/3)/x)*(5*x^2*exp(1) - 6*x + exp(1)*exp(x))
+ exp(1)*exp(x)*(x - 1))/(log(-((5*x^2*exp(1))/3 - 2*x + (exp(1)*exp(x))/3)/x)^2*(5*x^2*exp(1) - 6*x + exp(1)
*exp(x)) + 5*x^4*exp(1) + log(-((5*x^2*exp(1))/3 - 2*x + (exp(1)*exp(x))/3)/x)*(10*x^3*exp(1) - 12*x^2 + 2*x*e
xp(1)*exp(x)) - 6*x^3 + x^2*exp(1)*exp(x)), x)

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sympy [A]  time = 0.29, size = 27, normalized size = 1.04 \begin {gather*} \frac {x}{x + \log {\left (\frac {- \frac {5 e x^{2}}{3} + 2 x - \frac {e e^{x}}{3}}{x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(1)*exp(x)+5*x**2*exp(1)-6*x)*ln(1/3*(-exp(1)*exp(x)-5*x**2*exp(1)+6*x)/x)+(-x+1)*exp(1)*exp(x)
-5*x**2*exp(1))/((exp(1)*exp(x)+5*x**2*exp(1)-6*x)*ln(1/3*(-exp(1)*exp(x)-5*x**2*exp(1)+6*x)/x)**2+(2*x*exp(1)
*exp(x)+10*x**3*exp(1)-12*x**2)*ln(1/3*(-exp(1)*exp(x)-5*x**2*exp(1)+6*x)/x)+x**2*exp(1)*exp(x)+5*x**4*exp(1)-
6*x**3),x)

[Out]

x/(x + log((-5*E*x**2/3 + 2*x - E*exp(x)/3)/x))

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