3.49.25 \(\int \frac {4 x^2-9 x^4+4 x^5+e^3 (-2+2 x^2)+(-4 x+4 x^2+5 x^3-3 x^4) \log (-1-x)}{e^3 (x+x^2)} \, dx\)

Optimal. Leaf size=28 \[ x \left (2+\frac {(-2+x)^2 (x-\log (-1-x))}{e^3}\right )-2 \log (x) \]

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Rubi [B]  time = 0.35, antiderivative size = 98, normalized size of antiderivative = 3.50, number of steps used = 16, number of rules used = 9, integrand size = 64, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.141, Rules used = {12, 1593, 6742, 1620, 2417, 2389, 2295, 2395, 43} \begin {gather*} \frac {x^4}{e^3}-\frac {4 x^3}{e^3}-\frac {x^3 \log (-x-1)}{e^3}+\frac {4 x^2}{e^3}+\frac {4 x^2 \log (-x-1)}{e^3}-\frac {\left (9-2 e^3\right ) x}{e^3}+\frac {9 x}{e^3}-\frac {4 (x+1) \log (-x-1)}{e^3}-2 \log (x)+\frac {4 \log (x+1)}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*x^2 - 9*x^4 + 4*x^5 + E^3*(-2 + 2*x^2) + (-4*x + 4*x^2 + 5*x^3 - 3*x^4)*Log[-1 - x])/(E^3*(x + x^2)),x]

[Out]

(9*x)/E^3 - ((9 - 2*E^3)*x)/E^3 + (4*x^2)/E^3 - (4*x^3)/E^3 + x^4/E^3 + (4*x^2*Log[-1 - x])/E^3 - (x^3*Log[-1
- x])/E^3 - (4*(1 + x)*Log[-1 - x])/E^3 - 2*Log[x] + (4*Log[1 + x])/E^3

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

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 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 2417

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(Polyx_), x_Symbol] :> Int[ExpandIntegrand[Poly
x*(a + b*Log[c*(d + e*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, n, p}, x] && PolynomialQ[Polyx, 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} &=\frac {\int \frac {4 x^2-9 x^4+4 x^5+e^3 \left (-2+2 x^2\right )+\left (-4 x+4 x^2+5 x^3-3 x^4\right ) \log (-1-x)}{x+x^2} \, dx}{e^3}\\ &=\frac {\int \frac {4 x^2-9 x^4+4 x^5+e^3 \left (-2+2 x^2\right )+\left (-4 x+4 x^2+5 x^3-3 x^4\right ) \log (-1-x)}{x (1+x)} \, dx}{e^3}\\ &=\frac {\int \left (\frac {-2 e^3+4 \left (1+\frac {e^3}{2}\right ) x^2-9 x^4+4 x^5}{x (1+x)}-(-2+x) (-2+3 x) \log (-1-x)\right ) \, dx}{e^3}\\ &=\frac {\int \frac {-2 e^3+4 \left (1+\frac {e^3}{2}\right ) x^2-9 x^4+4 x^5}{x (1+x)} \, dx}{e^3}-\frac {\int (-2+x) (-2+3 x) \log (-1-x) \, dx}{e^3}\\ &=\frac {\int \left (-9 \left (1-\frac {2 e^3}{9}\right )-\frac {2 e^3}{x}+13 x-13 x^2+4 x^3+\frac {9}{1+x}\right ) \, dx}{e^3}-\frac {\int \left (4 \log (-1-x)-8 x \log (-1-x)+3 x^2 \log (-1-x)\right ) \, dx}{e^3}\\ &=-\frac {\left (9-2 e^3\right ) x}{e^3}+\frac {13 x^2}{2 e^3}-\frac {13 x^3}{3 e^3}+\frac {x^4}{e^3}-2 \log (x)+\frac {9 \log (1+x)}{e^3}-\frac {3 \int x^2 \log (-1-x) \, dx}{e^3}-\frac {4 \int \log (-1-x) \, dx}{e^3}+\frac {8 \int x \log (-1-x) \, dx}{e^3}\\ &=-\frac {\left (9-2 e^3\right ) x}{e^3}+\frac {13 x^2}{2 e^3}-\frac {13 x^3}{3 e^3}+\frac {x^4}{e^3}+\frac {4 x^2 \log (-1-x)}{e^3}-\frac {x^3 \log (-1-x)}{e^3}-2 \log (x)+\frac {9 \log (1+x)}{e^3}-\frac {\int \frac {x^3}{-1-x} \, dx}{e^3}+\frac {4 \int \frac {x^2}{-1-x} \, dx}{e^3}+\frac {4 \operatorname {Subst}(\int \log (x) \, dx,x,-1-x)}{e^3}\\ &=\frac {4 x}{e^3}-\frac {\left (9-2 e^3\right ) x}{e^3}+\frac {13 x^2}{2 e^3}-\frac {13 x^3}{3 e^3}+\frac {x^4}{e^3}+\frac {4 x^2 \log (-1-x)}{e^3}-\frac {x^3 \log (-1-x)}{e^3}-\frac {4 (1+x) \log (-1-x)}{e^3}-2 \log (x)+\frac {9 \log (1+x)}{e^3}-\frac {\int \left (-1+x-x^2+\frac {1}{1+x}\right ) \, dx}{e^3}+\frac {4 \int \left (1+\frac {1}{-1-x}-x\right ) \, dx}{e^3}\\ &=\frac {9 x}{e^3}-\frac {\left (9-2 e^3\right ) x}{e^3}+\frac {4 x^2}{e^3}-\frac {4 x^3}{e^3}+\frac {x^4}{e^3}+\frac {4 x^2 \log (-1-x)}{e^3}-\frac {x^3 \log (-1-x)}{e^3}-\frac {4 (1+x) \log (-1-x)}{e^3}-2 \log (x)+\frac {4 \log (1+x)}{e^3}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.08, size = 76, normalized size = 2.71 \begin {gather*} \frac {2 e^3 x+4 x^2-4 x^3+x^4-4 \log (-1-x)-4 x \log (-1-x)+4 x^2 \log (-1-x)-x^3 \log (-1-x)-2 e^3 \log (x)+4 \log (1+x)}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*x^2 - 9*x^4 + 4*x^5 + E^3*(-2 + 2*x^2) + (-4*x + 4*x^2 + 5*x^3 - 3*x^4)*Log[-1 - x])/(E^3*(x + x^
2)),x]

[Out]

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

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fricas [A]  time = 0.48, size = 48, normalized size = 1.71 \begin {gather*} {\left (x^{4} - 4 \, x^{3} + 4 \, x^{2} + 2 \, x e^{3} - 2 \, e^{3} \log \relax (x) - {\left (x^{3} - 4 \, x^{2} + 4 \, x\right )} \log \left (-x - 1\right )\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^4+5*x^3+4*x^2-4*x)*log(-x-1)+(2*x^2-2)*exp(3)+4*x^5-9*x^4+4*x^2)/(x^2+x)/exp(3),x, algorithm=
"fricas")

[Out]

(x^4 - 4*x^3 + 4*x^2 + 2*x*e^3 - 2*e^3*log(x) - (x^3 - 4*x^2 + 4*x)*log(-x - 1))*e^(-3)

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giac [B]  time = 0.15, size = 59, normalized size = 2.11 \begin {gather*} {\left (x^{4} - x^{3} \log \left (-x - 1\right ) - 4 \, x^{3} + 4 \, x^{2} \log \left (-x - 1\right ) + 4 \, x^{2} + 2 \, x e^{3} - 2 \, e^{3} \log \relax (x) - 4 \, x \log \left (-x - 1\right )\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^4+5*x^3+4*x^2-4*x)*log(-x-1)+(2*x^2-2)*exp(3)+4*x^5-9*x^4+4*x^2)/(x^2+x)/exp(3),x, algorithm=
"giac")

[Out]

(x^4 - x^3*log(-x - 1) - 4*x^3 + 4*x^2*log(-x - 1) + 4*x^2 + 2*x*e^3 - 2*e^3*log(x) - 4*x*log(-x - 1))*e^(-3)

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maple [A]  time = 0.12, size = 52, normalized size = 1.86

method	result	size
risch	\({\mathrm e}^{-3} \left (-x^{3}+4 x^{2}-4 x \right ) \ln \left (-x -1\right )+{\mathrm e}^{-3} x^{4}-4 \,{\mathrm e}^{-3} x^{3}+2 x +4 x^{2} {\mathrm e}^{-3}-2 \ln \relax (x )\)	\(52\)
norman	\({\mathrm e}^{-3} x^{4}+2 x +4 x^{2} {\mathrm e}^{-3}-4 \,{\mathrm e}^{-3} x^{3}-4 x \,{\mathrm e}^{-3} \ln \left (-x -1\right )+4 x^{2} {\mathrm e}^{-3} \ln \left (-x -1\right )-{\mathrm e}^{-3} x^{3} \ln \left (-x -1\right )-2 \ln \relax (x )\)	\(78\)
default	\({\mathrm e}^{-3} \left (\ln \left (-x -1\right ) \left (-x -1\right )^{3}+8 \left (-x -1\right )^{3}+7 \ln \left (-x -1\right ) \left (-x -1\right )^{2}+22 \left (-x -1\right )^{2}+15 \ln \left (-x -1\right ) \left (-x -1\right )-24 x -24+\left (-x -1\right )^{4}-2 \left (-x -1\right ) {\mathrm e}^{3}-2 \,{\mathrm e}^{3} \ln \left (-x \right )+9 \ln \left (-x -1\right )\right )\)	\(103\)
derivativedivides	\(-{\mathrm e}^{-3} \left (-\ln \left (-x -1\right ) \left (-x -1\right )^{3}-8 \left (-x -1\right )^{3}-7 \ln \left (-x -1\right ) \left (-x -1\right )^{2}-22 \left (-x -1\right )^{2}-15 \ln \left (-x -1\right ) \left (-x -1\right )+24 x +24-\left (-x -1\right )^{4}+2 \left (-x -1\right ) {\mathrm e}^{3}+2 \,{\mathrm e}^{3} \ln \left (-x \right )-9 \ln \left (-x -1\right )\right )\)	\(107\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*x^4+5*x^3+4*x^2-4*x)*ln(-x-1)+(2*x^2-2)*exp(3)+4*x^5-9*x^4+4*x^2)/(x^2+x)/exp(3),x,method=_RETURNVERB
OSE)

[Out]

exp(-3)*(-x^3+4*x^2-4*x)*ln(-x-1)+exp(-3)*x^4-4*exp(-3)*x^3+2*x+4*x^2*exp(-3)-2*ln(x)

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maxima [B]  time = 0.37, size = 128, normalized size = 4.57 \begin {gather*} \frac {1}{2} \, {\left (2 \, x^{4} - 8 \, x^{3} + 8 \, x^{2} + 4 \, {\left (x - \log \left (x + 1\right )\right )} e^{3} + 4 \, {\left (\log \left (x + 1\right ) - \log \relax (x)\right )} e^{3} - 4 \, \log \left (x + 1\right )^{2} - {\left (2 \, x^{3} - 3 \, x^{2} + 6 \, x - 6 \, \log \left (x + 1\right )\right )} \log \left (-x - 1\right ) + 5 \, {\left (x^{2} - 2 \, x + 2 \, \log \left (x + 1\right )\right )} \log \left (-x - 1\right ) + 8 \, {\left (x - \log \left (x + 1\right )\right )} \log \left (-x - 1\right ) - 4 \, \log \left (-x - 1\right )^{2}\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^4+5*x^3+4*x^2-4*x)*log(-x-1)+(2*x^2-2)*exp(3)+4*x^5-9*x^4+4*x^2)/(x^2+x)/exp(3),x, algorithm=
"maxima")

[Out]

1/2*(2*x^4 - 8*x^3 + 8*x^2 + 4*(x - log(x + 1))*e^3 + 4*(log(x + 1) - log(x))*e^3 - 4*log(x + 1)^2 - (2*x^3 -
3*x^2 + 6*x - 6*log(x + 1))*log(-x - 1) + 5*(x^2 - 2*x + 2*log(x + 1))*log(-x - 1) + 8*(x - log(x + 1))*log(-x
- 1) - 4*log(-x - 1)^2)*e^(-3)

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mupad [B]  time = 3.54, size = 65, normalized size = 2.32 \begin {gather*} 2\,x-2\,\ln \relax (x)+4\,x^2\,{\mathrm {e}}^{-3}-4\,x^3\,{\mathrm {e}}^{-3}+x^4\,{\mathrm {e}}^{-3}+4\,x^2\,{\mathrm {e}}^{-3}\,\ln \left (-x-1\right )-x^3\,{\mathrm {e}}^{-3}\,\ln \left (-x-1\right )-4\,x\,{\mathrm {e}}^{-3}\,\ln \left (-x-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-3)*(exp(3)*(2*x^2 - 2) + 4*x^2 - 9*x^4 + 4*x^5 - log(- x - 1)*(4*x - 4*x^2 - 5*x^3 + 3*x^4)))/(x + x
^2),x)

[Out]

2*x - 2*log(x) + 4*x^2*exp(-3) - 4*x^3*exp(-3) + x^4*exp(-3) + 4*x^2*exp(-3)*log(- x - 1) - x^3*exp(-3)*log(-
x - 1) - 4*x*exp(-3)*log(- x - 1)

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sympy [B]  time = 0.21, size = 53, normalized size = 1.89 \begin {gather*} \frac {\left (- x^{3} + 4 x^{2} - 4 x\right ) \log {\left (- x - 1 \right )}}{e^{3}} + \frac {x^{4} - 4 x^{3} + 4 x^{2} + 2 x e^{3} - 2 e^{3} \log {\relax (x )}}{e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x**4+5*x**3+4*x**2-4*x)*ln(-x-1)+(2*x**2-2)*exp(3)+4*x**5-9*x**4+4*x**2)/(x**2+x)/exp(3),x)

[Out]

(-x**3 + 4*x**2 - 4*x)*exp(-3)*log(-x - 1) + (x**4 - 4*x**3 + 4*x**2 + 2*x*exp(3) - 2*exp(3)*log(x))*exp(-3)

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