3.46.97 \(\int \frac {e^x \log (2)+(-1-20 x) \log (2)+(20 x \log (2)-e^x x \log (2)) \log (x)}{x \log ^2(x)} \, dx\)

Optimal. Leaf size=21 \[ \frac {\left (5-e^x+4 (-1+5 x)\right ) \log (2)}{\log (x)} \]

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Rubi [A]  time = 0.56, antiderivative size = 28, normalized size of antiderivative = 1.33, number of steps used = 14, number of rules used = 9, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.225, Rules used = {6688, 12, 6742, 2202, 2353, 2297, 2298, 2302, 30} \begin {gather*} \frac {20 x \log (2)}{\log (x)}-\frac {e^x \log (2)}{\log (x)}+\frac {\log (2)}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*Log[2] + (-1 - 20*x)*Log[2] + (20*x*Log[2] - E^x*x*Log[2])*Log[x])/(x*Log[x]^2),x]

[Out]

Log[2]/Log[x] - (E^x*Log[2])/Log[x] + (20*x*Log[2])/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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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} &=\int \frac {\log (2) \left (-1+e^x-20 x-\left (-20+e^x\right ) x \log (x)\right )}{x \log ^2(x)} \, dx\\ &=\log (2) \int \frac {-1+e^x-20 x-\left (-20+e^x\right ) x \log (x)}{x \log ^2(x)} \, dx\\ &=\log (2) \int \left (-\frac {e^x (-1+x \log (x))}{x \log ^2(x)}+\frac {-1-20 x+20 x \log (x)}{x \log ^2(x)}\right ) \, dx\\ &=-\left (\log (2) \int \frac {e^x (-1+x \log (x))}{x \log ^2(x)} \, dx\right )+\log (2) \int \frac {-1-20 x+20 x \log (x)}{x \log ^2(x)} \, dx\\ &=-\frac {e^x \log (2)}{\log (x)}+\log (2) \int \left (\frac {-1-20 x}{x \log ^2(x)}+\frac {20}{\log (x)}\right ) \, dx\\ &=-\frac {e^x \log (2)}{\log (x)}+\log (2) \int \frac {-1-20 x}{x \log ^2(x)} \, dx+(20 \log (2)) \int \frac {1}{\log (x)} \, dx\\ &=-\frac {e^x \log (2)}{\log (x)}+20 \log (2) \text {li}(x)+\log (2) \int \left (-\frac {20}{\log ^2(x)}-\frac {1}{x \log ^2(x)}\right ) \, dx\\ &=-\frac {e^x \log (2)}{\log (x)}+20 \log (2) \text {li}(x)-\log (2) \int \frac {1}{x \log ^2(x)} \, dx-(20 \log (2)) \int \frac {1}{\log ^2(x)} \, dx\\ &=-\frac {e^x \log (2)}{\log (x)}+\frac {20 x \log (2)}{\log (x)}+20 \log (2) \text {li}(x)-\log (2) \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )-(20 \log (2)) \int \frac {1}{\log (x)} \, dx\\ &=\frac {\log (2)}{\log (x)}-\frac {e^x \log (2)}{\log (x)}+\frac {20 x \log (2)}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 16, normalized size = 0.76 \begin {gather*} -\frac {\left (-1+e^x-20 x\right ) \log (2)}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*Log[2] + (-1 - 20*x)*Log[2] + (20*x*Log[2] - E^x*x*Log[2])*Log[x])/(x*Log[x]^2),x]

[Out]

-(((-1 + E^x - 20*x)*Log[2])/Log[x])

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fricas [A]  time = 0.56, size = 20, normalized size = 0.95 \begin {gather*} \frac {{\left (20 \, x + 1\right )} \log \relax (2) - e^{x} \log \relax (2)}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*log(2)*exp(x)+20*x*log(2))*log(x)+exp(x)*log(2)+(-20*x-1)*log(2))/x/log(x)^2,x, algorithm="fric
as")

[Out]

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

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giac [A]  time = 0.21, size = 19, normalized size = 0.90 \begin {gather*} \frac {20 \, x \log \relax (2) - e^{x} \log \relax (2) + \log \relax (2)}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*log(2)*exp(x)+20*x*log(2))*log(x)+exp(x)*log(2)+(-20*x-1)*log(2))/x/log(x)^2,x, algorithm="giac
")

[Out]

(20*x*log(2) - e^x*log(2) + log(2))/log(x)

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maple [A]  time = 0.06, size = 17, normalized size = 0.81

method	result	size
risch	\(\frac {\ln \relax (2) \left (20 x +1-{\mathrm e}^{x}\right )}{\ln \relax (x )}\)	\(17\)
norman	\(\frac {20 x \ln \relax (2)-{\mathrm e}^{x} \ln \relax (2)+\ln \relax (2)}{\ln \relax (x )}\)	\(20\)
default	\(-\frac {\ln \relax (2) {\mathrm e}^{x}}{\ln \relax (x )}+\frac {20 \ln \relax (2) x}{\ln \relax (x )}+\frac {\ln \relax (2)}{\ln \relax (x )}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x*ln(2)*exp(x)+20*x*ln(2))*ln(x)+exp(x)*ln(2)+(-20*x-1)*ln(2))/x/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

ln(2)*(20*x+1-exp(x))/ln(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -20 \, \Gamma \left (-1, -\log \relax (x)\right ) \log \relax (2) + 20 \, \int \frac {1}{\log \relax (x)}\,{d x} \log \relax (2) - \frac {e^{x} \log \relax (2)}{\log \relax (x)} + \frac {\log \relax (2)}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*log(2)*exp(x)+20*x*log(2))*log(x)+exp(x)*log(2)+(-20*x-1)*log(2))/x/log(x)^2,x, algorithm="maxi
ma")

[Out]

-20*gamma(-1, -log(x))*log(2) + 20*integrate(1/log(x), x)*log(2) - e^x*log(2)/log(x) + log(2)/log(x)

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mupad [B]  time = 3.47, size = 16, normalized size = 0.76 \begin {gather*} \frac {\ln \relax (2)\,\left (20\,x-{\mathrm {e}}^x+1\right )}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(20*x*log(2) - x*exp(x)*log(2)) - log(2)*(20*x + 1) + exp(x)*log(2))/(x*log(x)^2),x)

[Out]

(log(2)*(20*x - exp(x) + 1))/log(x)

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sympy [A]  time = 0.28, size = 22, normalized size = 1.05 \begin {gather*} \frac {20 x \log {\relax (2 )} + \log {\relax (2 )}}{\log {\relax (x )}} - \frac {e^{x} \log {\relax (2 )}}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*ln(2)*exp(x)+20*x*ln(2))*ln(x)+exp(x)*ln(2)+(-20*x-1)*ln(2))/x/ln(x)**2,x)

[Out]

(20*x*log(2) + log(2))/log(x) - exp(x)*log(2)/log(x)

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