3.91.19 \(\int \frac {-7-14 x+(7+7 x) \log (-1-x)+(7+7 x) \log (x)}{e^2 (x^2+x^3)} \, dx\)

Optimal. Leaf size=17 \[ -\frac {7 (\log (-1-x)+\log (x))}{e^2 x} \]

________________________________________________________________________________________

Rubi [A]  time = 0.41, antiderivative size = 25, normalized size of antiderivative = 1.47, number of steps used = 15, number of rules used = 10, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {12, 1593, 6741, 6742, 77, 2395, 36, 31, 29, 2304} \begin {gather*} -\frac {7 \log (-x-1)}{e^2 x}-\frac {7 \log (x)}{e^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-7 - 14*x + (7 + 7*x)*Log[-1 - x] + (7 + 7*x)*Log[x])/(E^2*(x^2 + x^3)),x]

[Out]

(-7*Log[-1 - x])/(E^2*x) - (7*Log[x])/(E^2*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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-7-14 x+(7+7 x) \log (-1-x)+(7+7 x) \log (x)}{x^2+x^3} \, dx}{e^2}\\ &=\frac {\int \frac {-7-14 x+(7+7 x) \log (-1-x)+(7+7 x) \log (x)}{x^2 (1+x)} \, dx}{e^2}\\ &=\frac {\int \frac {7 (-1-2 x+\log (-1-x)+x \log (-1-x)+\log (x)+x \log (x))}{x^2 (1+x)} \, dx}{e^2}\\ &=\frac {7 \int \frac {-1-2 x+\log (-1-x)+x \log (-1-x)+\log (x)+x \log (x)}{x^2 (1+x)} \, dx}{e^2}\\ &=\frac {7 \int \left (\frac {-1-2 x+\log (-1-x)+x \log (-1-x)}{x^2 (1+x)}+\frac {\log (x)}{x^2}\right ) \, dx}{e^2}\\ &=\frac {7 \int \frac {-1-2 x+\log (-1-x)+x \log (-1-x)}{x^2 (1+x)} \, dx}{e^2}+\frac {7 \int \frac {\log (x)}{x^2} \, dx}{e^2}\\ &=-\frac {7}{e^2 x}-\frac {7 \log (x)}{e^2 x}+\frac {7 \int \left (\frac {-1-2 x}{x^2 (1+x)}+\frac {\log (-1-x)}{x^2}\right ) \, dx}{e^2}\\ &=-\frac {7}{e^2 x}-\frac {7 \log (x)}{e^2 x}+\frac {7 \int \frac {-1-2 x}{x^2 (1+x)} \, dx}{e^2}+\frac {7 \int \frac {\log (-1-x)}{x^2} \, dx}{e^2}\\ &=-\frac {7}{e^2 x}-\frac {7 \log (-1-x)}{e^2 x}-\frac {7 \log (x)}{e^2 x}-\frac {7 \int \frac {1}{(-1-x) x} \, dx}{e^2}+\frac {7 \int \left (-\frac {1}{x^2}-\frac {1}{x}+\frac {1}{1+x}\right ) \, dx}{e^2}\\ &=-\frac {7 \log (-1-x)}{e^2 x}-\frac {7 \log (x)}{e^2}-\frac {7 \log (x)}{e^2 x}+\frac {7 \log (1+x)}{e^2}+\frac {7 \int \frac {1}{-1-x} \, dx}{e^2}+\frac {7 \int \frac {1}{x} \, dx}{e^2}\\ &=-\frac {7 \log (-1-x)}{e^2 x}-\frac {7 \log (x)}{e^2 x}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 24, normalized size = 1.41 \begin {gather*} \frac {7 \left (-\frac {\log (-1-x)}{x}-\frac {\log (x)}{x}\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-7 - 14*x + (7 + 7*x)*Log[-1 - x] + (7 + 7*x)*Log[x])/(E^2*(x^2 + x^3)),x]

[Out]

(7*(-(Log[-1 - x]/x) - Log[x]/x))/E^2

________________________________________________________________________________________

fricas [A]  time = 0.51, size = 16, normalized size = 0.94 \begin {gather*} -\frac {7 \, {\left (\log \relax (x) + \log \left (-x - 1\right )\right )} e^{\left (-2\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((7*x+7)*log(x)+(7*x+7)*log(-x-1)-14*x-7)/(x^3+x^2)/exp(2),x, algorithm="fricas")

[Out]

-7*(log(x) + log(-x - 1))*e^(-2)/x

________________________________________________________________________________________

giac [A]  time = 0.16, size = 21, normalized size = 1.24 \begin {gather*} -7 \, {\left (\frac {\log \relax (x)}{x} + \frac {\log \left (-x - 1\right )}{x}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((7*x+7)*log(x)+(7*x+7)*log(-x-1)-14*x-7)/(x^3+x^2)/exp(2),x, algorithm="giac")

[Out]

-7*(log(x)/x + log(-x - 1)/x)*e^(-2)

________________________________________________________________________________________

maple [A]  time = 0.46, size = 24, normalized size = 1.41

method	result	size
risch	\(-\frac {7 \,{\mathrm e}^{-2} \ln \left (-x -1\right )}{x}-\frac {7 \,{\mathrm e}^{-2} \ln \relax (x )}{x}\)	\(24\)
default	\({\mathrm e}^{-2} \left (7 \ln \left (-x \right )+\frac {7 \ln \left (-x -1\right ) \left (-x -1\right )}{x}-\frac {7 \ln \relax (x )}{x}-7 \ln \relax (x )+7 \ln \left (x +1\right )\right )\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((7*x+7)*ln(x)+(7*x+7)*ln(-x-1)-14*x-7)/(x^3+x^2)/exp(2),x,method=_RETURNVERBOSE)

[Out]

-7*exp(-2)/x*ln(-x-1)-7*exp(-2)*ln(x)/x

________________________________________________________________________________________

maxima [B]  time = 0.40, size = 41, normalized size = 2.41 \begin {gather*} -7 \, {\left (\frac {{\left (x + 1\right )} \log \relax (x) - {\left (x - 1\right )} \log \left (-x - 1\right ) + 1}{x} - \frac {1}{x} + \log \left (x + 1\right ) - \log \relax (x)\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((7*x+7)*log(x)+(7*x+7)*log(-x-1)-14*x-7)/(x^3+x^2)/exp(2),x, algorithm="maxima")

[Out]

-7*(((x + 1)*log(x) - (x - 1)*log(-x - 1) + 1)/x - 1/x + log(x + 1) - log(x))*e^(-2)

________________________________________________________________________________________

mupad [B]  time = 7.23, size = 16, normalized size = 0.94 \begin {gather*} -\frac {7\,{\mathrm {e}}^{-2}\,\left (\ln \left (-x-1\right )+\ln \relax (x)\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-2)*(14*x - log(- x - 1)*(7*x + 7) - log(x)*(7*x + 7) + 7))/(x^2 + x^3),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.42, size = 24, normalized size = 1.41 \begin {gather*} - \frac {7 \log {\relax (x )}}{x e^{2}} - \frac {7 \log {\left (- x - 1 \right )}}{x e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((7*x+7)*ln(x)+(7*x+7)*ln(-x-1)-14*x-7)/(x**3+x**2)/exp(2),x)

[Out]

-7*exp(-2)*log(x)/x - 7*exp(-2)*log(-x - 1)/x

________________________________________________________________________________________