3.20.74 \(\int \frac {4 e^{1+x}-4 e^{1+x} x \log (x)+(12 x-4 x \log (\frac {7}{4})) \log ^2(x)}{x \log ^2(x)} \, dx\)

Optimal. Leaf size=24 \[ 4 \left (x \left (3-\log \left (\frac {7}{4}\right )\right )-\frac {e^{1+x}}{\log (x)}\right ) \]

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Rubi [A]  time = 0.73, antiderivative size = 23, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {6741, 12, 6742, 2202} \begin {gather*} 4 x \left (3-\log \left (\frac {7}{4}\right )\right )-\frac {4 e^{x+1}}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*E^(1 + x) - 4*E^(1 + x)*x*Log[x] + (12*x - 4*x*Log[7/4])*Log[x]^2)/(x*Log[x]^2),x]

[Out]

4*x*(3 - Log[7/4]) - (4*E^(1 + x))/Log[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 2202

Int[Log[(d_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*(x_)^(m_.)*((e_) + Log[(d_.)*(x_)]*(h_.)*((f_.) +
(g_.)*(x_))), x_Symbol] :> Simp[(e*x^(m + 1)*F^(c*(a + b*x))*Log[d*x]^(n + 1))/(n + 1), x] /; FreeQ[{F, a, b,
c, d, e, f, g, h, m, n}, x] && EqQ[e*(m + 1) - f*h*(n + 1), 0] && EqQ[g*h*(n + 1) - b*c*e*Log[F], 0] && NeQ[n,
-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} &=\int \frac {4 \left (e^{1+x}-e^{1+x} x \log (x)+3 x \left (1-\frac {1}{3} \log \left (\frac {7}{4}\right )\right ) \log ^2(x)\right )}{x \log ^2(x)} \, dx\\ &=4 \int \frac {e^{1+x}-e^{1+x} x \log (x)+3 x \left (1-\frac {1}{3} \log \left (\frac {7}{4}\right )\right ) \log ^2(x)}{x \log ^2(x)} \, dx\\ &=4 \int \left (3 \left (1-\frac {1}{3} \log \left (\frac {7}{4}\right )\right )-\frac {e^{1+x} (-1+x \log (x))}{x \log ^2(x)}\right ) \, dx\\ &=4 x \left (3-\log \left (\frac {7}{4}\right )\right )-4 \int \frac {e^{1+x} (-1+x \log (x))}{x \log ^2(x)} \, dx\\ &=4 x \left (3-\log \left (\frac {7}{4}\right )\right )-\frac {4 e^{1+x}}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 22, normalized size = 0.92 \begin {gather*} 12 x-4 x \log \left (\frac {7}{4}\right )-\frac {4 e^{1+x}}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*E^(1 + x) - 4*E^(1 + x)*x*Log[x] + (12*x - 4*x*Log[7/4])*Log[x]^2)/(x*Log[x]^2),x]

[Out]

12*x - 4*x*Log[7/4] - (4*E^(1 + x))/Log[x]

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fricas [A]  time = 0.78, size = 22, normalized size = 0.92 \begin {gather*} -\frac {4 \, {\left ({\left (x \log \left (\frac {7}{4}\right ) - 3 \, x\right )} \log \relax (x) + e^{\left (x + 1\right )}\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*log(7/4)+12*x)*log(x)^2-4*x*exp(x+1)*log(x)+4*exp(x+1))/x/log(x)^2,x, algorithm="fricas")

[Out]

-4*((x*log(7/4) - 3*x)*log(x) + e^(x + 1))/log(x)

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giac [A]  time = 0.26, size = 29, normalized size = 1.21 \begin {gather*} -\frac {4 \, {\left (x \log \relax (7) \log \relax (x) - 2 \, x \log \relax (2) \log \relax (x) - 3 \, x \log \relax (x) + e^{\left (x + 1\right )}\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*log(7/4)+12*x)*log(x)^2-4*x*exp(x+1)*log(x)+4*exp(x+1))/x/log(x)^2,x, algorithm="giac")

[Out]

-4*(x*log(7)*log(x) - 2*x*log(2)*log(x) - 3*x*log(x) + e^(x + 1))/log(x)

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maple [A]  time = 0.05, size = 20, normalized size = 0.83

method	result	size
default	\(12 x -\frac {4 \,{\mathrm e}^{x +1}}{\ln \relax (x )}-4 x \ln \left (\frac {7}{4}\right )\)	\(20\)
risch	\(8 x \ln \relax (2)-4 x \ln \relax (7)+12 x -\frac {4 \,{\mathrm e}^{x +1}}{\ln \relax (x )}\)	\(25\)
norman	\(\frac {\left (-4 \ln \relax (7)+8 \ln \relax (2)+12\right ) x \ln \relax (x )-4 \,{\mathrm e}^{x +1}}{\ln \relax (x )}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x*ln(7/4)+12*x)*ln(x)^2-4*x*exp(x+1)*ln(x)+4*exp(x+1))/x/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

12*x-4*exp(x+1)/ln(x)-4*x*ln(7/4)

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maxima [A]  time = 0.49, size = 19, normalized size = 0.79 \begin {gather*} -4 \, x \log \left (\frac {7}{4}\right ) + 12 \, x - \frac {4 \, e^{\left (x + 1\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*log(7/4)+12*x)*log(x)^2-4*x*exp(x+1)*log(x)+4*exp(x+1))/x/log(x)^2,x, algorithm="maxima")

[Out]

-4*x*log(7/4) + 12*x - 4*e^(x + 1)/log(x)

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mupad [B]  time = 1.18, size = 20, normalized size = 0.83 \begin {gather*} -x\,\left (4\,\ln \left (\frac {7}{4}\right )-12\right )-\frac {4\,{\mathrm {e}}^{x+1}}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*exp(x + 1) + log(x)^2*(12*x - 4*x*log(7/4)) - 4*x*exp(x + 1)*log(x))/(x*log(x)^2),x)

[Out]

- x*(4*log(7/4) - 12) - (4*exp(x + 1))/log(x)

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sympy [A]  time = 0.26, size = 22, normalized size = 0.92 \begin {gather*} x \left (- 4 \log {\relax (7 )} + 8 \log {\relax (2 )} + 12\right ) - \frac {4 e^{x + 1}}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*ln(7/4)+12*x)*ln(x)**2-4*x*exp(x+1)*ln(x)+4*exp(x+1))/x/ln(x)**2,x)

[Out]

x*(-4*log(7) + 8*log(2) + 12) - 4*exp(x + 1)/log(x)

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