3.95.52 \(\int \frac {e^{2 x}+4 x^2+(2 e^{2 x} x+2 x^2) \log (x)+(-3 x^2-6 x^2 \log (x)) \log (x^2)}{x} \, dx\)

Optimal. Leaf size=23 \[ \log (x) \left (e^{2 x}+4 x^2-3 x^2 \log \left (x^2\right )\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 40, normalized size of antiderivative = 1.74, number of steps used = 8, number of rules used = 5, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {14, 2288, 2304, 2303, 2366} \begin {gather*} -2 x^2+3 x^2 \log (x)+x^2 (\log (x)+2)-3 x^2 \log (x) \log \left (x^2\right )+e^{2 x} \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(2*x) + 4*x^2 + (2*E^(2*x)*x + 2*x^2)*Log[x] + (-3*x^2 - 6*x^2*Log[x])*Log[x^2])/x,x]

[Out]

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

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 2303

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(b*(d*x)^(m + 1)*Log[c*x^n])/(
d*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && EqQ[a*(m + 1) - b*n, 0]

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 2366

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.))*((g_.)*(x_))^(m_.), x_Sy
mbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[Sim
plifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] &&  !(EqQ[p, 1] && EqQ[a, 0] &&
NeQ[d, 0])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {e^{2 x} (1+2 x \log (x))}{x}-x \left (-4-2 \log (x)+3 \log \left (x^2\right )+6 \log (x) \log \left (x^2\right )\right )\right ) \, dx\\ &=\int \frac {e^{2 x} (1+2 x \log (x))}{x} \, dx-\int x \left (-4-2 \log (x)+3 \log \left (x^2\right )+6 \log (x) \log \left (x^2\right )\right ) \, dx\\ &=e^{2 x} \log (x)-\int \left (-2 x (2+\log (x))+3 x (1+2 \log (x)) \log \left (x^2\right )\right ) \, dx\\ &=e^{2 x} \log (x)+2 \int x (2+\log (x)) \, dx-3 \int x (1+2 \log (x)) \log \left (x^2\right ) \, dx\\ &=-\frac {x^2}{2}+e^{2 x} \log (x)+x^2 (2+\log (x))-3 x^2 \log (x) \log \left (x^2\right )+6 \int x \log (x) \, dx\\ &=-2 x^2+e^{2 x} \log (x)+3 x^2 \log (x)+x^2 (2+\log (x))-3 x^2 \log (x) \log \left (x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 23, normalized size = 1.00 \begin {gather*} \log (x) \left (e^{2 x}+4 x^2-3 x^2 \log \left (x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*x) + 4*x^2 + (2*E^(2*x)*x + 2*x^2)*Log[x] + (-3*x^2 - 6*x^2*Log[x])*Log[x^2])/x,x]

[Out]

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

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fricas [A]  time = 0.49, size = 23, normalized size = 1.00 \begin {gather*} -6 \, x^{2} \log \relax (x)^{2} + {\left (4 \, x^{2} + e^{\left (2 \, x\right )}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x^2*log(x)-3*x^2)*log(x^2)+(2*x*exp(2*x)+2*x^2)*log(x)+exp(2*x)+4*x^2)/x,x, algorithm="fricas")

[Out]

-6*x^2*log(x)^2 + (4*x^2 + e^(2*x))*log(x)

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giac [A]  time = 0.15, size = 24, normalized size = 1.04 \begin {gather*} -6 \, x^{2} \log \relax (x)^{2} + 4 \, x^{2} \log \relax (x) + e^{\left (2 \, x\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x^2*log(x)-3*x^2)*log(x^2)+(2*x*exp(2*x)+2*x^2)*log(x)+exp(2*x)+4*x^2)/x,x, algorithm="giac")

[Out]

-6*x^2*log(x)^2 + 4*x^2*log(x) + e^(2*x)*log(x)

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maple [A]  time = 0.10, size = 27, normalized size = 1.17

method	result	size
default	\(4 x^{2} \ln \relax (x )-3 \ln \relax (x ) \ln \left (x^{2}\right ) x^{2}+\ln \relax (x ) {\mathrm e}^{2 x}\)	\(27\)
risch	\(-6 x^{2} \ln \relax (x )^{2}+\left (\frac {3 i \pi \,x^{2} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )}{2}-3 i \pi \,x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\frac {3 i \pi \,x^{2} \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}+4 x^{2}+{\mathrm e}^{2 x}\right ) \ln \relax (x )\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-6*x^2*ln(x)-3*x^2)*ln(x^2)+(2*x*exp(2*x)+2*x^2)*ln(x)+exp(2*x)+4*x^2)/x,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.39, size = 42, normalized size = 1.83 \begin {gather*} -\frac {3}{2} \, x^{2} \log \left (x^{2}\right ) + 4 \, x^{2} \log \relax (x) - 3 \, {\left (x^{2} \log \left (x^{2}\right ) - x^{2}\right )} \log \relax (x) + e^{\left (2 \, x\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x^2*log(x)-3*x^2)*log(x^2)+(2*x*exp(2*x)+2*x^2)*log(x)+exp(2*x)+4*x^2)/x,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 7.24, size = 22, normalized size = 0.96 \begin {gather*} \ln \relax (x)\,\left ({\mathrm {e}}^{2\,x}-3\,x^2\,\ln \left (x^2\right )+4\,x^2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x) - log(x^2)*(6*x^2*log(x) + 3*x^2) + 4*x^2 + log(x)*(2*x*exp(2*x) + 2*x^2))/x,x)

[Out]

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

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sympy [A]  time = 0.37, size = 26, normalized size = 1.13 \begin {gather*} - 6 x^{2} \log {\relax (x )}^{2} + 4 x^{2} \log {\relax (x )} + e^{2 x} \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x**2*ln(x)-3*x**2)*ln(x**2)+(2*x*exp(2*x)+2*x**2)*ln(x)+exp(2*x)+4*x**2)/x,x)

[Out]

-6*x**2*log(x)**2 + 4*x**2*log(x) + exp(2*x)*log(x)

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