3.12.11 \(\int \frac {e^{1+x+x^2} (-1-x-x^2-2 x^3)+e^{1+x+x^2} (-x-2 x^2) \log (x)}{x \log (5)} \, dx\) [1111]

Optimal. Leaf size=20 \[ 6-\frac {e^{1+x+x^2} (x+\log (x))}{\log (5)} \]

[Out]

6-(x+ln(x))*exp(x^2+x+1)/ln(5)

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Rubi [A]
time = 0.17, antiderivative size = 32, normalized size of antiderivative = 1.60, number of steps used = 20, number of rules used = 7, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {12, 14, 2266, 2235, 2272, 2273, 2634} \begin {gather*} -\frac {e^{x^2+x+1} x}{\log (5)}-\frac {e^{x^2+x+1} \log (x)}{\log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

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 2235

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2
]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2266

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[F^(a - b^2/(4*c)), Int[F^((b + 2*c*x)^2/(4*c))
, x], x] /; FreeQ[{F, a, b, c}, x]

Rule 2272

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[e*(F^(a + b*x + c*x^2)/(2
*c*Log[F])), x] - Dist[(b*e - 2*c*d)/(2*c), Int[F^(a + b*x + c*x^2), x], x] /; FreeQ[{F, a, b, c, d, e}, x] &&
NeQ[b*e - 2*c*d, 0]

Rule 2273

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_))^(m_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
(F^(a + b*x + c*x^2)/(2*c*Log[F])), x] + (-Dist[(b*e - 2*c*d)/(2*c), Int[(d + e*x)^(m - 1)*F^(a + b*x + c*x^2)
, x], x] - Dist[(m - 1)*(e^2/(2*c*Log[F])), Int[(d + e*x)^(m - 2)*F^(a + b*x + c*x^2), x], x]) /; FreeQ[{F, a,
b, c, d, e}, x] && NeQ[b*e - 2*c*d, 0] && GtQ[m, 1]

Rule 2634

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^{1+x+x^2} \left (-1-x-x^2-2 x^3\right )+e^{1+x+x^2} \left (-x-2 x^2\right ) \log (x)}{x} \, dx}{\log (5)}\\ &=\frac {\int \left (-e^{1+x+x^2}-\frac {e^{1+x+x^2}}{x}-e^{1+x+x^2} x-2 e^{1+x+x^2} x^2-e^{1+x+x^2} \log (x)-2 e^{1+x+x^2} x \log (x)\right ) \, dx}{\log (5)}\\ &=-\frac {\int e^{1+x+x^2} \, dx}{\log (5)}-\frac {\int \frac {e^{1+x+x^2}}{x} \, dx}{\log (5)}-\frac {\int e^{1+x+x^2} x \, dx}{\log (5)}-\frac {\int e^{1+x+x^2} \log (x) \, dx}{\log (5)}-\frac {2 \int e^{1+x+x^2} x^2 \, dx}{\log (5)}-\frac {2 \int e^{1+x+x^2} x \log (x) \, dx}{\log (5)}\\ &=-\frac {e^{1+x+x^2}}{2 \log (5)}-\frac {e^{1+x+x^2} x}{\log (5)}-\frac {e^{1+x+x^2} \log (x)}{\log (5)}+\frac {\int e^{1+x+x^2} \, dx}{2 \log (5)}+\frac {\int e^{1+x+x^2} \, dx}{\log (5)}-\frac {\int \frac {e^{1+x+x^2}}{x} \, dx}{\log (5)}+\frac {\int e^{1+x+x^2} x \, dx}{\log (5)}+\frac {\int \frac {e^{3/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2}+x\right )}{2 x} \, dx}{\log (5)}+\frac {2 \int \frac {2 e^{1+x+x^2}-e^{3/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2}+x\right )}{4 x} \, dx}{\log (5)}-\frac {e^{3/4} \int e^{\frac {1}{4} (1+2 x)^2} \, dx}{\log (5)}\\ &=-\frac {e^{1+x+x^2} x}{\log (5)}-\frac {e^{3/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (1+2 x)\right )}{2 \log (5)}-\frac {e^{1+x+x^2} \log (x)}{\log (5)}-\frac {\int e^{1+x+x^2} \, dx}{2 \log (5)}+\frac {\int \frac {2 e^{1+x+x^2}-e^{3/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2}+x\right )}{x} \, dx}{2 \log (5)}-\frac {\int \frac {e^{1+x+x^2}}{x} \, dx}{\log (5)}+\frac {e^{3/4} \int e^{\frac {1}{4} (1+2 x)^2} \, dx}{2 \log (5)}+\frac {e^{3/4} \int e^{\frac {1}{4} (1+2 x)^2} \, dx}{\log (5)}+\frac {\left (e^{3/4} \sqrt {\pi }\right ) \int \frac {\text {erfi}\left (\frac {1}{2}+x\right )}{x} \, dx}{2 \log (5)}\\ &=-\frac {e^{1+x+x^2} x}{\log (5)}+\frac {e^{3/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (1+2 x)\right )}{4 \log (5)}-\frac {e^{1+x+x^2} \log (x)}{\log (5)}+\frac {\int \left (\frac {2 e^{1+x+x^2}}{x}-\frac {e^{3/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2}+x\right )}{x}\right ) \, dx}{2 \log (5)}-\frac {\int \frac {e^{1+x+x^2}}{x} \, dx}{\log (5)}-\frac {e^{3/4} \int e^{\frac {1}{4} (1+2 x)^2} \, dx}{2 \log (5)}+\frac {\left (e^{3/4} \sqrt {\pi }\right ) \int \frac {\text {erfi}\left (\frac {1}{2}+x\right )}{x} \, dx}{2 \log (5)}\\ &=-\frac {e^{1+x+x^2} x}{\log (5)}-\frac {e^{1+x+x^2} \log (x)}{\log (5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.03, size = 18, normalized size = 0.90 \begin {gather*} -\frac {e^{1+x+x^2} (x+\log (x))}{\log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.07, size = 31, normalized size = 1.55

method	result	size
norman	\(-\frac {x \,{\mathrm e}^{x^{2}+x +1}}{\ln \left (5\right )}-\frac {{\mathrm e}^{x^{2}+x +1} \ln \left (x \right )}{\ln \left (5\right )}\)	\(31\)
risch	\(-\frac {x \,{\mathrm e}^{x^{2}+x +1}}{\ln \left (5\right )}-\frac {{\mathrm e}^{x^{2}+x +1} \ln \left (x \right )}{\ln \left (5\right )}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^2-x)*exp(x^2+x+1)*ln(x)+(-2*x^3-x^2-x-1)*exp(x^2+x+1))/x/ln(5),x,method=_RETURNVERBOSE)

[Out]

-x/ln(5)*exp(x^2+x+1)-1/ln(5)*exp(x^2+x+1)*ln(x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(I*sqrt(pi)*erf(I*x + 1/2*I)*e^(3/4) - 2*integrate((2*x^3*e + x^2*e + (2*x^2*e + x*e)*log(x) + e)*e^(x^2 +
x)/x, x))/log(5)

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Fricas [A]
time = 0.36, size = 26, normalized size = 1.30 \begin {gather*} -\frac {x e^{\left (x^{2} + x + 1\right )} + e^{\left (x^{2} + x + 1\right )} \log \left (x\right )}{\log \left (5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.13, size = 17, normalized size = 0.85 \begin {gather*} \frac {\left (- x - \log {\left (x \right )}\right ) e^{x^{2} + x + 1}}{\log {\left (5 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-x - log(x))*exp(x**2 + x + 1)/log(5)

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Giac [A]
time = 0.39, size = 26, normalized size = 1.30 \begin {gather*} -\frac {x e^{\left (x^{2} + x + 1\right )} + e^{\left (x^{2} + x + 1\right )} \log \left (x\right )}{\log \left (5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.94, size = 18, normalized size = 0.90 \begin {gather*} -\frac {{\mathrm {e}}^{x^2}\,\mathrm {e}\,{\mathrm {e}}^x\,\left (x+\ln \left (x\right )\right )}{\ln \left (5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(exp(x^2)*exp(1)*exp(x)*(x + log(x)))/log(5)

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