3.62.84 \(\int \frac {-11 x-3 e^x x+(22 x-22 x^2+e^x (6 x-3 x^2-3 x^3)) \log (\frac {-1+x}{x})}{-1+x} \, dx\)

Optimal. Leaf size=22 \[ -\left (\left (2+3 \left (3+e^x\right )\right ) x^2 \log \left (\frac {-1+x}{x}\right )\right ) \]

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Rubi [A]  time = 0.87, antiderivative size = 30, normalized size of antiderivative = 1.36, number of steps used = 24, number of rules used = 12, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {6688, 6742, 2199, 2194, 2178, 2196, 2176, 2554, 14, 43, 2455, 193} \begin {gather*} -3 e^x x^2 \log \left (1-\frac {1}{x}\right )-11 x^2 \log \left (1-\frac {1}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-11*x - 3*E^x*x + (22*x - 22*x^2 + E^x*(6*x - 3*x^2 - 3*x^3))*Log[(-1 + x)/x])/(-1 + x),x]

[Out]

-11*x^2*Log[1 - x^(-1)] - 3*E^x*x^2*Log[1 - x^(-1)]

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 193

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b}, x] && LtQ[n, 0]
&& IntegerQ[p]

Rule 2176

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] &&  !$UseGamma === True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2194

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 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rule 2455

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.)*(x_))^(m_.), x_Symbol] :> Simp[((f*x)^(m
+ 1)*(a + b*Log[c*(d + e*x^n)^p]))/(f*(m + 1)), x] - Dist[(b*e*n*p)/(f*(m + 1)), Int[(x^(n - 1)*(f*x)^(m + 1))
/(d + e*x^n), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && NeQ[m, -1]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x \left (11+3 e^x+(-1+x) \left (22+3 e^x (2+x)\right ) \log \left (\frac {-1+x}{x}\right )\right )}{1-x} \, dx\\ &=\int \left (-\frac {3 e^x x \left (1-2 \log \left (\frac {-1+x}{x}\right )+x \log \left (\frac {-1+x}{x}\right )+x^2 \log \left (\frac {-1+x}{x}\right )\right )}{-1+x}-\frac {11 \left (x-2 x \log \left (\frac {-1+x}{x}\right )+2 x^2 \log \left (\frac {-1+x}{x}\right )\right )}{-1+x}\right ) \, dx\\ &=-\left (3 \int \frac {e^x x \left (1-2 \log \left (\frac {-1+x}{x}\right )+x \log \left (\frac {-1+x}{x}\right )+x^2 \log \left (\frac {-1+x}{x}\right )\right )}{-1+x} \, dx\right )-11 \int \frac {x-2 x \log \left (\frac {-1+x}{x}\right )+2 x^2 \log \left (\frac {-1+x}{x}\right )}{-1+x} \, dx\\ &=-\left (3 \int \frac {e^x x \left (-1-\left (-2+x+x^2\right ) \log \left (\frac {-1+x}{x}\right )\right )}{1-x} \, dx\right )-11 \int x \left (\frac {1}{-1+x}+2 \log \left (\frac {-1+x}{x}\right )\right ) \, dx\\ &=-\left (3 \int \left (\frac {e^x x}{-1+x}+e^x x (2+x) \log \left (1-\frac {1}{x}\right )\right ) \, dx\right )-11 \int \left (\frac {x}{-1+x}+2 x \log \left (1-\frac {1}{x}\right )\right ) \, dx\\ &=-\left (3 \int \frac {e^x x}{-1+x} \, dx\right )-3 \int e^x x (2+x) \log \left (1-\frac {1}{x}\right ) \, dx-11 \int \frac {x}{-1+x} \, dx-22 \int x \log \left (1-\frac {1}{x}\right ) \, dx\\ &=-11 x^2 \log \left (1-\frac {1}{x}\right )-3 e^x x^2 \log \left (1-\frac {1}{x}\right )-3 \int \left (e^x+\frac {e^x}{-1+x}\right ) \, dx+3 \int \frac {e^x x}{-1+x} \, dx-11 \int \left (1+\frac {1}{-1+x}\right ) \, dx+11 \int \frac {1}{1-\frac {1}{x}} \, dx\\ &=-11 x-11 x^2 \log \left (1-\frac {1}{x}\right )-3 e^x x^2 \log \left (1-\frac {1}{x}\right )-11 \log (1-x)-3 \int e^x \, dx+3 \int \left (e^x+\frac {e^x}{-1+x}\right ) \, dx-3 \int \frac {e^x}{-1+x} \, dx+11 \int \frac {x}{-1+x} \, dx\\ &=-3 e^x-11 x-3 e \text {Ei}(-1+x)-11 x^2 \log \left (1-\frac {1}{x}\right )-3 e^x x^2 \log \left (1-\frac {1}{x}\right )-11 \log (1-x)+3 \int e^x \, dx+3 \int \frac {e^x}{-1+x} \, dx+11 \int \left (1+\frac {1}{-1+x}\right ) \, dx\\ &=-11 x^2 \log \left (1-\frac {1}{x}\right )-3 e^x x^2 \log \left (1-\frac {1}{x}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 5.03, size = 20, normalized size = 0.91 \begin {gather*} -\left (\left (11+3 e^x\right ) x^2 \log \left (\frac {-1+x}{x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-11*x - 3*E^x*x + (22*x - 22*x^2 + E^x*(6*x - 3*x^2 - 3*x^3))*Log[(-1 + x)/x])/(-1 + x),x]

[Out]

-((11 + 3*E^x)*x^2*Log[(-1 + x)/x])

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fricas [A]  time = 0.58, size = 23, normalized size = 1.05 \begin {gather*} -{\left (3 \, x^{2} e^{x} + 11 \, x^{2}\right )} \log \left (\frac {x - 1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x^3-3*x^2+6*x)*exp(x)-22*x^2+22*x)*log((x-1)/x)-3*exp(x)*x-11*x)/(x-1),x, algorithm="fricas")

[Out]

-(3*x^2*e^x + 11*x^2)*log((x - 1)/x)

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giac [A]  time = 0.18, size = 29, normalized size = 1.32 \begin {gather*} -3 \, x^{2} e^{x} \log \left (\frac {x - 1}{x}\right ) - 11 \, x^{2} \log \left (\frac {x - 1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x^3-3*x^2+6*x)*exp(x)-22*x^2+22*x)*log((x-1)/x)-3*exp(x)*x-11*x)/(x-1),x, algorithm="giac")

[Out]

-3*x^2*e^x*log((x - 1)/x) - 11*x^2*log((x - 1)/x)

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maple [A]  time = 0.44, size = 30, normalized size = 1.36

method	result	size
norman	\(-11 \ln \left (\frac {x -1}{x}\right ) x^{2}-3 \ln \left (\frac {x -1}{x}\right ) {\mathrm e}^{x} x^{2}\)	\(30\)
default	\(-3 \ln \left (\frac {x -1}{x}\right ) {\mathrm e}^{x} x^{2}-11 \ln \left (x -1\right )-11 \ln \left (-\frac {1}{x}\right )+11 \ln \left (1-\frac {1}{x}\right ) \left (1-\frac {1}{x}\right ) \left (-1-\frac {1}{x}\right ) x^{2}\)	\(58\)
risch	\(\left (-3 \,{\mathrm e}^{x} x^{2}-11 x^{2}\right ) \ln \left (x -1\right )+3 x^{2} {\mathrm e}^{x} \ln \relax (x )+11 x^{2} \ln \relax (x )+\frac {11 i \pi \,x^{2} \mathrm {csgn}\left (\frac {i \left (x -1\right )}{x}\right )^{3}}{2}-\frac {11 i \pi \,x^{2} \mathrm {csgn}\left (\frac {i \left (x -1\right )}{x}\right )^{2} \mathrm {csgn}\left (\frac {i}{x}\right )}{2}-\frac {11 i \pi \,x^{2} \mathrm {csgn}\left (\frac {i \left (x -1\right )}{x}\right )^{2} \mathrm {csgn}\left (i \left (x -1\right )\right )}{2}+\frac {11 i \pi \,x^{2} \mathrm {csgn}\left (\frac {i \left (x -1\right )}{x}\right ) \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (x -1\right )\right )}{2}+\frac {3 i \pi \,x^{2} \mathrm {csgn}\left (\frac {i \left (x -1\right )}{x}\right )^{3} {\mathrm e}^{x}}{2}-\frac {3 i \pi \,x^{2} \mathrm {csgn}\left (\frac {i \left (x -1\right )}{x}\right )^{2} \mathrm {csgn}\left (\frac {i}{x}\right ) {\mathrm e}^{x}}{2}-\frac {3 i \pi \,x^{2} \mathrm {csgn}\left (\frac {i \left (x -1\right )}{x}\right )^{2} \mathrm {csgn}\left (i \left (x -1\right )\right ) {\mathrm e}^{x}}{2}+\frac {3 i \pi \,x^{2} \mathrm {csgn}\left (\frac {i \left (x -1\right )}{x}\right ) \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (x -1\right )\right ) {\mathrm e}^{x}}{2}\)	\(248\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-3*x^3-3*x^2+6*x)*exp(x)-22*x^2+22*x)*ln((x-1)/x)-3*exp(x)*x-11*x)/(x-1),x,method=_RETURNVERBOSE)

[Out]

-11*ln((x-1)/x)*x^2-3*ln((x-1)/x)*exp(x)*x^2

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maxima [B]  time = 0.39, size = 43, normalized size = 1.95 \begin {gather*} 3 \, x^{2} e^{x} \log \relax (x) + 11 \, x^{2} \log \relax (x) - {\left (3 \, x^{2} e^{x} + 11 \, x^{2} - 11\right )} \log \left (x - 1\right ) - 11 \, \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x^3-3*x^2+6*x)*exp(x)-22*x^2+22*x)*log((x-1)/x)-3*exp(x)*x-11*x)/(x-1),x, algorithm="maxima")

[Out]

3*x^2*e^x*log(x) + 11*x^2*log(x) - (3*x^2*e^x + 11*x^2 - 11)*log(x - 1) - 11*log(x - 1)

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mupad [B]  time = 4.37, size = 19, normalized size = 0.86 \begin {gather*} -x^2\,\ln \left (\frac {x-1}{x}\right )\,\left (3\,{\mathrm {e}}^x+11\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(11*x + log((x - 1)/x)*(22*x^2 - 22*x + exp(x)*(3*x^2 - 6*x + 3*x^3)) + 3*x*exp(x))/(x - 1),x)

[Out]

-x^2*log((x - 1)/x)*(3*exp(x) + 11)

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sympy [A]  time = 0.41, size = 27, normalized size = 1.23 \begin {gather*} - 3 x^{2} e^{x} \log {\left (\frac {x - 1}{x} \right )} - 11 x^{2} \log {\left (\frac {x - 1}{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x**3-3*x**2+6*x)*exp(x)-22*x**2+22*x)*ln((x-1)/x)-3*exp(x)*x-11*x)/(x-1),x)

[Out]

-3*x**2*exp(x)*log((x - 1)/x) - 11*x**2*log((x - 1)/x)

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