3.85.6 \(\int \frac {9 x^2+3 x^3+(-3 x^2+3 x^3) \log (-1+x)+e^{e^x x} (-25 x+(-25+25 x+e^x (25 x-25 x^3)) \log (-1+x))}{-3 x^2+3 x^3} \, dx\)

Optimal. Leaf size=22 \[ \left (3-\frac {25 e^{e^x x}}{3 x}+x\right ) \log (-1+x) \]

________________________________________________________________________________________

Rubi [B]  time = 1.34, antiderivative size = 74, normalized size of antiderivative = 3.36, number of steps used = 10, number of rules used = 6, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {1593, 6742, 43, 2389, 2295, 2288} \begin {gather*} -\frac {25 e^{e^x x} \left (e^x x \log (x-1)-e^x x^3 \log (x-1)\right )}{3 (1-x) x^2 \left (e^x x+e^x\right )}+4 \log (1-x)-(1-x) \log (x-1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(9*x^2 + 3*x^3 + (-3*x^2 + 3*x^3)*Log[-1 + x] + E^(E^x*x)*(-25*x + (-25 + 25*x + E^x*(25*x - 25*x^3))*Log[
-1 + x]))/(-3*x^2 + 3*x^3),x]

[Out]

4*Log[1 - x] - (1 - x)*Log[-1 + x] - (25*E^(E^x*x)*(E^x*x*Log[-1 + x] - E^x*x^3*Log[-1 + x]))/(3*(1 - x)*x^2*(
E^x + E^x*x))

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 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, 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 {9 x^2+3 x^3+\left (-3 x^2+3 x^3\right ) \log (-1+x)+e^{e^x x} \left (-25 x+\left (-25+25 x+e^x \left (25 x-25 x^3\right )\right ) \log (-1+x)\right )}{x^2 (-3+3 x)} \, dx\\ &=\int \left (\frac {3+x-\log (-1+x)+x \log (-1+x)}{-1+x}-\frac {25 e^{e^x x} \left (x+\log (-1+x)-x \log (-1+x)-e^x x \log (-1+x)+e^x x^3 \log (-1+x)\right )}{3 (-1+x) x^2}\right ) \, dx\\ &=-\left (\frac {25}{3} \int \frac {e^{e^x x} \left (x+\log (-1+x)-x \log (-1+x)-e^x x \log (-1+x)+e^x x^3 \log (-1+x)\right )}{(-1+x) x^2} \, dx\right )+\int \frac {3+x-\log (-1+x)+x \log (-1+x)}{-1+x} \, dx\\ &=-\frac {25 e^{e^x x} \left (e^x x \log (-1+x)-e^x x^3 \log (-1+x)\right )}{3 (1-x) x^2 \left (e^x+e^x x\right )}+\int \left (\frac {3+x}{-1+x}+\log (-1+x)\right ) \, dx\\ &=-\frac {25 e^{e^x x} \left (e^x x \log (-1+x)-e^x x^3 \log (-1+x)\right )}{3 (1-x) x^2 \left (e^x+e^x x\right )}+\int \frac {3+x}{-1+x} \, dx+\int \log (-1+x) \, dx\\ &=-\frac {25 e^{e^x x} \left (e^x x \log (-1+x)-e^x x^3 \log (-1+x)\right )}{3 (1-x) x^2 \left (e^x+e^x x\right )}+\int \left (1+\frac {4}{-1+x}\right ) \, dx+\operatorname {Subst}(\int \log (x) \, dx,x,-1+x)\\ &=4 \log (1-x)+(-1+x) \log (-1+x)-\frac {25 e^{e^x x} \left (e^x x \log (-1+x)-e^x x^3 \log (-1+x)\right )}{3 (1-x) x^2 \left (e^x+e^x x\right )}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.58, size = 37, normalized size = 1.68 \begin {gather*} \frac {1}{3} \left (12 \log (1-x)+\frac {\left (-25 e^{e^x x}+3 (-1+x) x\right ) \log (-1+x)}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(9*x^2 + 3*x^3 + (-3*x^2 + 3*x^3)*Log[-1 + x] + E^(E^x*x)*(-25*x + (-25 + 25*x + E^x*(25*x - 25*x^3)
)*Log[-1 + x]))/(-3*x^2 + 3*x^3),x]

[Out]

(12*Log[1 - x] + ((-25*E^(E^x*x) + 3*(-1 + x)*x)*Log[-1 + x])/x)/3

________________________________________________________________________________________

fricas [A]  time = 0.80, size = 30, normalized size = 1.36 \begin {gather*} \frac {3 \, {\left (x^{2} + 3 \, x\right )} \log \left (x - 1\right ) - 25 \, e^{\left (x e^{x}\right )} \log \left (x - 1\right )}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-25*x^3+25*x)*exp(x)+25*x-25)*log(x-1)-25*x)*exp(exp(x)*x)+(3*x^3-3*x^2)*log(x-1)+3*x^3+9*x^2)/(
3*x^3-3*x^2),x, algorithm="fricas")

[Out]

1/3*(3*(x^2 + 3*x)*log(x - 1) - 25*e^(x*e^x)*log(x - 1))/x

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {3 \, x^{3} + 9 \, x^{2} - 25 \, {\left ({\left ({\left (x^{3} - x\right )} e^{x} - x + 1\right )} \log \left (x - 1\right ) + x\right )} e^{\left (x e^{x}\right )} + 3 \, {\left (x^{3} - x^{2}\right )} \log \left (x - 1\right )}{3 \, {\left (x^{3} - x^{2}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-25*x^3+25*x)*exp(x)+25*x-25)*log(x-1)-25*x)*exp(exp(x)*x)+(3*x^3-3*x^2)*log(x-1)+3*x^3+9*x^2)/(
3*x^3-3*x^2),x, algorithm="giac")

[Out]

integrate(1/3*(3*x^3 + 9*x^2 - 25*(((x^3 - x)*e^x - x + 1)*log(x - 1) + x)*e^(x*e^x) + 3*(x^3 - x^2)*log(x - 1
))/(x^3 - x^2), x)

________________________________________________________________________________________

maple [A]  time = 0.38, size = 28, normalized size = 1.27

method	result	size
risch	\(\ln \left (x -1\right ) x +3 \ln \left (x -1\right )-\frac {25 \ln \left (x -1\right ) {\mathrm e}^{{\mathrm e}^{x} x}}{3 x}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((-25*x^3+25*x)*exp(x)+25*x-25)*ln(x-1)-25*x)*exp(exp(x)*x)+(3*x^3-3*x^2)*ln(x-1)+3*x^3+9*x^2)/(3*x^3-3*
x^2),x,method=_RETURNVERBOSE)

[Out]

ln(x-1)*x+3*ln(x-1)-25/3*ln(x-1)/x*exp(exp(x)*x)

________________________________________________________________________________________

maxima [A]  time = 0.43, size = 40, normalized size = 1.82 \begin {gather*} {\left (x + \log \left (x - 1\right )\right )} \log \left (x - 1\right ) - \log \left (x - 1\right )^{2} - \frac {25 \, e^{\left (x e^{x}\right )} \log \left (x - 1\right )}{3 \, x} + 3 \, \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-25*x^3+25*x)*exp(x)+25*x-25)*log(x-1)-25*x)*exp(exp(x)*x)+(3*x^3-3*x^2)*log(x-1)+3*x^3+9*x^2)/(
3*x^3-3*x^2),x, algorithm="maxima")

[Out]

(x + log(x - 1))*log(x - 1) - log(x - 1)^2 - 25/3*e^(x*e^x)*log(x - 1)/x + 3*log(x - 1)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {{\mathrm {e}}^{x\,{\mathrm {e}}^x}\,\left (25\,x-\ln \left (x-1\right )\,\left (25\,x+{\mathrm {e}}^x\,\left (25\,x-25\,x^3\right )-25\right )\right )+\ln \left (x-1\right )\,\left (3\,x^2-3\,x^3\right )-9\,x^2-3\,x^3}{3\,x^2-3\,x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x*exp(x))*(25*x - log(x - 1)*(25*x + exp(x)*(25*x - 25*x^3) - 25)) + log(x - 1)*(3*x^2 - 3*x^3) - 9*x
^2 - 3*x^3)/(3*x^2 - 3*x^3),x)

[Out]

int((exp(x*exp(x))*(25*x - log(x - 1)*(25*x + exp(x)*(25*x - 25*x^3) - 25)) + log(x - 1)*(3*x^2 - 3*x^3) - 9*x
^2 - 3*x^3)/(3*x^2 - 3*x^3), x)

________________________________________________________________________________________

sympy [A]  time = 0.40, size = 29, normalized size = 1.32 \begin {gather*} x \log {\left (x - 1 \right )} + 3 \log {\left (x - 1 \right )} - \frac {25 e^{x e^{x}} \log {\left (x - 1 \right )}}{3 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-25*x**3+25*x)*exp(x)+25*x-25)*ln(x-1)-25*x)*exp(exp(x)*x)+(3*x**3-3*x**2)*ln(x-1)+3*x**3+9*x**2
)/(3*x**3-3*x**2),x)

[Out]

x*log(x - 1) + 3*log(x - 1) - 25*exp(x*exp(x))*log(x - 1)/(3*x)

________________________________________________________________________________________