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

Optimal. Leaf size=24 \[ \left (1+x+\frac {e^{2+x+x^2}+\log (3)}{5 x}\right ) \log (x) \]

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Rubi [B]  time = 0.17, antiderivative size = 53, normalized size of antiderivative = 2.21, number of steps used = 10, number of rules used = 4, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {12, 14, 2288, 2334} \begin {gather*} \frac {e^{x^2+x+2} \left (2 x^2 \log (x)+x \log (x)\right )}{5 x^2 (2 x+1)}+\frac {1}{5} \left (5 x+\frac {\log (3)}{x}\right ) \log (x)+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[x] + ((5*x + Log[3]/x)*Log[x])/5 + (E^(2 + x + x^2)*(x*Log[x] + 2*x^2*Log[x]))/(5*x^2*(1 + 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 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 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 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]
] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 28, normalized size = 1.17 \begin {gather*} \frac {\left (e^{2+x+x^2}+5 x+5 x^2+\log (3)\right ) \log (x)}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((E^(2 + x + x^2) + 5*x + 5*x^2 + Log[3])*Log[x])/(5*x)

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fricas [A]  time = 1.68, size = 25, normalized size = 1.04 \begin {gather*} \frac {{\left (5 \, x^{2} + 5 \, x + e^{\left (x^{2} + x + 2\right )} + \log \relax (3)\right )} \log \relax (x)}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(5*x^2 + 5*x + e^(x^2 + x + 2) + log(3))*log(x)/x

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giac [A]  time = 0.17, size = 33, normalized size = 1.38 \begin {gather*} \frac {5 \, x^{2} \log \relax (x) + 5 \, x \log \relax (x) + e^{\left (x^{2} + x + 2\right )} \log \relax (x) + \log \relax (3) \log \relax (x)}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(5*x^2*log(x) + 5*x*log(x) + e^(x^2 + x + 2)*log(x) + log(3)*log(x))/x

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maple [A]  time = 0.07, size = 26, normalized size = 1.08

method	result	size
risch	\(\frac {\left (5 x^{2}+\ln \relax (3)+{\mathrm e}^{x^{2}+x +2}\right ) \ln \relax (x )}{5 x}+\ln \relax (x )\)	\(26\)
default	\(\frac {{\mathrm e}^{x^{2}+x +2} \ln \relax (x )}{5 x}+\ln \relax (x )+x \ln \relax (x )+\frac {\ln \relax (3) \ln \relax (x )}{5 x}\)	\(31\)
norman	\(\frac {x^{2} \ln \relax (x )+x \ln \relax (x )+\frac {\ln \relax (3) \ln \relax (x )}{5}+\frac {\ln \relax (x ) {\mathrm e}^{x^{2}+x +2}}{5}}{x}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(5*x^2+ln(3)+exp(x^2+x+2))/x*ln(x)+ln(x)

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maxima [B]  time = 0.50, size = 49, normalized size = 2.04 \begin {gather*} x - \frac {5 \, x^{2} - {\left (5 \, x^{2} + \log \relax (3)\right )} \log \relax (x) - e^{\left (x^{2} + x + 2\right )} \log \relax (x) - \log \relax (3)}{5 \, x} - \frac {\log \relax (3)}{5 \, x} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 6.08, size = 25, normalized size = 1.04 \begin {gather*} \frac {\ln \relax (x)\,\left (5\,x+{\mathrm {e}}^{x^2+x+2}+\ln \relax (3)+5\,x^2\right )}{5\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(x)*(5*x + exp(x + x^2 + 2) + log(3) + 5*x^2))/(5*x)

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sympy [A]  time = 0.36, size = 32, normalized size = 1.33 \begin {gather*} \log {\relax (x )} + \frac {\left (5 x^{2} + \log {\relax (3 )}\right ) \log {\relax (x )}}{5 x} + \frac {e^{x^{2} + x + 2} \log {\relax (x )}}{5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x) + (5*x**2 + log(3))*log(x)/(5*x) + exp(x**2 + x + 2)*log(x)/(5*x)

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