3.90.7 \(\int \frac {e^5 (25+25 x) \log (4)+25 e^5 \log (4) \log (x)+25 e^5 \log ^2(x)}{x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=29 \[ \frac {25 e^5 (1+x) (\log (4)+\log (x))}{-x^2+x (x-\log (x))} \]

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Rubi [A]  time = 0.40, antiderivative size = 34, normalized size of antiderivative = 1.17, number of steps used = 13, number of rules used = 9, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.237, Rules used = {6688, 12, 6742, 2353, 2306, 2309, 2178, 2302, 30} \begin {gather*} -\frac {25 e^5}{x}-\frac {25 e^5 \log (4)}{\log (x)}-\frac {25 e^5 \log (4)}{x \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^5*(25 + 25*x)*Log[4] + 25*E^5*Log[4]*Log[x] + 25*E^5*Log[x]^2)/(x^2*Log[x]^2),x]

[Out]

(-25*E^5)/x - (25*E^5*Log[4])/Log[x] - (25*E^5*Log[4])/(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 30

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

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

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 2306

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

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

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 {25 e^5 \left ((1+x) \log (4)+\log (4) \log (x)+\log ^2(x)\right )}{x^2 \log ^2(x)} \, dx\\ &=\left (25 e^5\right ) \int \frac {(1+x) \log (4)+\log (4) \log (x)+\log ^2(x)}{x^2 \log ^2(x)} \, dx\\ &=\left (25 e^5\right ) \int \left (\frac {1}{x^2}+\frac {(1+x) \log (4)}{x^2 \log ^2(x)}+\frac {\log (4)}{x^2 \log (x)}\right ) \, dx\\ &=-\frac {25 e^5}{x}+\left (25 e^5 \log (4)\right ) \int \frac {1+x}{x^2 \log ^2(x)} \, dx+\left (25 e^5 \log (4)\right ) \int \frac {1}{x^2 \log (x)} \, dx\\ &=-\frac {25 e^5}{x}+\left (25 e^5 \log (4)\right ) \int \left (\frac {1}{x^2 \log ^2(x)}+\frac {1}{x \log ^2(x)}\right ) \, dx+\left (25 e^5 \log (4)\right ) \operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (x)\right )\\ &=-\frac {25 e^5}{x}+25 e^5 \text {Ei}(-\log (x)) \log (4)+\left (25 e^5 \log (4)\right ) \int \frac {1}{x^2 \log ^2(x)} \, dx+\left (25 e^5 \log (4)\right ) \int \frac {1}{x \log ^2(x)} \, dx\\ &=-\frac {25 e^5}{x}+25 e^5 \text {Ei}(-\log (x)) \log (4)-\frac {25 e^5 \log (4)}{x \log (x)}-\left (25 e^5 \log (4)\right ) \int \frac {1}{x^2 \log (x)} \, dx+\left (25 e^5 \log (4)\right ) \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )\\ &=-\frac {25 e^5}{x}+25 e^5 \text {Ei}(-\log (x)) \log (4)-\frac {25 e^5 \log (4)}{\log (x)}-\frac {25 e^5 \log (4)}{x \log (x)}-\left (25 e^5 \log (4)\right ) \operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (x)\right )\\ &=-\frac {25 e^5}{x}-\frac {25 e^5 \log (4)}{\log (x)}-\frac {25 e^5 \log (4)}{x \log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 25, normalized size = 0.86 \begin {gather*} 25 e^5 \left (-\frac {1}{x}-\frac {(1+x) \log (4)}{x \log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^5*(25 + 25*x)*Log[4] + 25*E^5*Log[4]*Log[x] + 25*E^5*Log[x]^2)/(x^2*Log[x]^2),x]

[Out]

25*E^5*(-x^(-1) - ((1 + x)*Log[4])/(x*Log[x]))

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fricas [A]  time = 0.44, size = 24, normalized size = 0.83 \begin {gather*} -\frac {25 \, {\left (2 \, {\left (x + 1\right )} e^{5} \log \relax (2) + e^{5} \log \relax (x)\right )}}{x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*exp(5)*log(x)^2+50*exp(5)*log(2)*log(x)+2*(25*x+25)*exp(5)*log(2))/x^2/log(x)^2,x, algorithm="fr
icas")

[Out]

-25*(2*(x + 1)*e^5*log(2) + e^5*log(x))/(x*log(x))

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giac [A]  time = 0.14, size = 28, normalized size = 0.97 \begin {gather*} -\frac {25 \, {\left (2 \, x e^{5} \log \relax (2) + 2 \, e^{5} \log \relax (2) + e^{5} \log \relax (x)\right )}}{x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*exp(5)*log(x)^2+50*exp(5)*log(2)*log(x)+2*(25*x+25)*exp(5)*log(2))/x^2/log(x)^2,x, algorithm="gi
ac")

[Out]

-25*(2*x*e^5*log(2) + 2*e^5*log(2) + e^5*log(x))/(x*log(x))

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maple [A]  time = 0.03, size = 25, normalized size = 0.86

method	result	size
risch	\(-\frac {25 \,{\mathrm e}^{5}}{x}-\frac {50 \,{\mathrm e}^{5} \ln \relax (2) \left (x +1\right )}{x \ln \relax (x )}\)	\(25\)
norman	\(\frac {-50 \,{\mathrm e}^{5} \ln \relax (2)-25 \,{\mathrm e}^{5} \ln \relax (x )-50 x \,{\mathrm e}^{5} \ln \relax (2)}{x \ln \relax (x )}\)	\(29\)
default	\(-\frac {25 \,{\mathrm e}^{5}}{x}-50 \,{\mathrm e}^{5} \ln \relax (2) \expIntegralEi \left (1, \ln \relax (x )\right )-\frac {50 \,{\mathrm e}^{5} \ln \relax (2)}{\ln \relax (x )}+50 \,{\mathrm e}^{5} \ln \relax (2) \left (-\frac {1}{x \ln \relax (x )}+\expIntegralEi \left (1, \ln \relax (x )\right )\right )\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25*exp(5)*ln(x)^2+50*exp(5)*ln(2)*ln(x)+2*(25*x+25)*exp(5)*ln(2))/x^2/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

-25*exp(5)/x-50/x*exp(5)*ln(2)*(x+1)/ln(x)

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maxima [C]  time = 0.39, size = 39, normalized size = 1.34 \begin {gather*} 50 \, {\rm Ei}\left (-\log \relax (x)\right ) e^{5} \log \relax (2) - 50 \, e^{5} \Gamma \left (-1, \log \relax (x)\right ) \log \relax (2) - \frac {50 \, e^{5} \log \relax (2)}{\log \relax (x)} - \frac {25 \, e^{5}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*exp(5)*log(x)^2+50*exp(5)*log(2)*log(x)+2*(25*x+25)*exp(5)*log(2))/x^2/log(x)^2,x, algorithm="ma
xima")

[Out]

50*Ei(-log(x))*e^5*log(2) - 50*e^5*gamma(-1, log(x))*log(2) - 50*e^5*log(2)/log(x) - 25*e^5/x

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mupad [B]  time = 6.08, size = 29, normalized size = 1.00 \begin {gather*} -\frac {25\,{\mathrm {e}}^5}{x}-\frac {25\,{\mathrm {e}}^5\,\left (2\,\ln \relax (2)+2\,x\,\ln \relax (2)\right )}{x\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25*exp(5)*log(x)^2 + 50*exp(5)*log(2)*log(x) + 2*exp(5)*log(2)*(25*x + 25))/(x^2*log(x)^2),x)

[Out]

- (25*exp(5))/x - (25*exp(5)*(2*log(2) + 2*x*log(2)))/(x*log(x))

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sympy [A]  time = 0.10, size = 31, normalized size = 1.07 \begin {gather*} \frac {- 50 x e^{5} \log {\relax (2 )} - 50 e^{5} \log {\relax (2 )}}{x \log {\relax (x )}} - \frac {25 e^{5}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*exp(5)*ln(x)**2+50*exp(5)*ln(2)*ln(x)+2*(25*x+25)*exp(5)*ln(2))/x**2/ln(x)**2,x)

[Out]

(-50*x*exp(5)*log(2) - 50*exp(5)*log(2))/(x*log(x)) - 25*exp(5)/x

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