3.45.67 \(\int \frac {-250 e^4 x+500 e^4 x \log (x)-500 e^4 x \log ^2(x)+(25 e^7+250 e^4 x) \log ^3(x)}{\log ^3(x)} \, dx\)

Optimal. Leaf size=29 \[ 5 \left (-2+e^4 \left (2+5 e^3 x+25 \left (x-\frac {x}{\log (x)}\right )^2\right )\right ) \]

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Rubi [A]  time = 0.21, antiderivative size = 39, normalized size of antiderivative = 1.34, number of steps used = 13, number of rules used = 6, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {6741, 12, 6742, 2306, 2309, 2178} \begin {gather*} 125 e^4 x^2+\frac {125 e^4 x^2}{\log ^2(x)}-\frac {250 e^4 x^2}{\log (x)}+25 e^7 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-250*E^4*x + 500*E^4*x*Log[x] - 500*E^4*x*Log[x]^2 + (25*E^7 + 250*E^4*x)*Log[x]^3)/Log[x]^3,x]

[Out]

25*E^7*x + 125*E^4*x^2 + (125*E^4*x^2)/Log[x]^2 - (250*E^4*x^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 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 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 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 {25 e^4 \left (-10 x+20 x \log (x)-20 x \log ^2(x)+e^3 \log ^3(x)+10 x \log ^3(x)\right )}{\log ^3(x)} \, dx\\ &=\left (25 e^4\right ) \int \frac {-10 x+20 x \log (x)-20 x \log ^2(x)+e^3 \log ^3(x)+10 x \log ^3(x)}{\log ^3(x)} \, dx\\ &=\left (25 e^4\right ) \int \left (e^3+10 x-\frac {10 x}{\log ^3(x)}+\frac {20 x}{\log ^2(x)}-\frac {20 x}{\log (x)}\right ) \, dx\\ &=25 e^7 x+125 e^4 x^2-\left (250 e^4\right ) \int \frac {x}{\log ^3(x)} \, dx+\left (500 e^4\right ) \int \frac {x}{\log ^2(x)} \, dx-\left (500 e^4\right ) \int \frac {x}{\log (x)} \, dx\\ &=25 e^7 x+125 e^4 x^2+\frac {125 e^4 x^2}{\log ^2(x)}-\frac {500 e^4 x^2}{\log (x)}-\left (250 e^4\right ) \int \frac {x}{\log ^2(x)} \, dx-\left (500 e^4\right ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )+\left (1000 e^4\right ) \int \frac {x}{\log (x)} \, dx\\ &=25 e^7 x+125 e^4 x^2-500 e^4 \text {Ei}(2 \log (x))+\frac {125 e^4 x^2}{\log ^2(x)}-\frac {250 e^4 x^2}{\log (x)}-\left (500 e^4\right ) \int \frac {x}{\log (x)} \, dx+\left (1000 e^4\right ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=25 e^7 x+125 e^4 x^2+500 e^4 \text {Ei}(2 \log (x))+\frac {125 e^4 x^2}{\log ^2(x)}-\frac {250 e^4 x^2}{\log (x)}-\left (500 e^4\right ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=25 e^7 x+125 e^4 x^2+\frac {125 e^4 x^2}{\log ^2(x)}-\frac {250 e^4 x^2}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 39, normalized size = 1.34 \begin {gather*} 25 e^7 x+125 e^4 x^2+\frac {125 e^4 x^2}{\log ^2(x)}-\frac {250 e^4 x^2}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-250*E^4*x + 500*E^4*x*Log[x] - 500*E^4*x*Log[x]^2 + (25*E^7 + 250*E^4*x)*Log[x]^3)/Log[x]^3,x]

[Out]

25*E^7*x + 125*E^4*x^2 + (125*E^4*x^2)/Log[x]^2 - (250*E^4*x^2)/Log[x]

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fricas [A]  time = 0.89, size = 41, normalized size = 1.41 \begin {gather*} -\frac {25 \, {\left (10 \, x^{2} e^{4} \log \relax (x) - 5 \, x^{2} e^{4} - {\left (5 \, x^{2} e^{4} + x e^{7}\right )} \log \relax (x)^{2}\right )}}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(4)*exp(log(5)+3)+250*x*exp(4))*log(x)^3-500*x*exp(4)*log(x)^2+500*x*exp(4)*log(x)-250*x*exp(
4))/log(x)^3,x, algorithm="fricas")

[Out]

-25*(10*x^2*e^4*log(x) - 5*x^2*e^4 - (5*x^2*e^4 + x*e^7)*log(x)^2)/log(x)^2

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giac [A]  time = 0.18, size = 35, normalized size = 1.21 \begin {gather*} 125 \, x^{2} e^{4} + 25 \, x e^{7} - \frac {250 \, x^{2} e^{4}}{\log \relax (x)} + \frac {125 \, x^{2} e^{4}}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(4)*exp(log(5)+3)+250*x*exp(4))*log(x)^3-500*x*exp(4)*log(x)^2+500*x*exp(4)*log(x)-250*x*exp(
4))/log(x)^3,x, algorithm="giac")

[Out]

125*x^2*e^4 + 25*x*e^7 - 250*x^2*e^4/log(x) + 125*x^2*e^4/log(x)^2

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maple [A]  time = 0.03, size = 30, normalized size = 1.03

method	result	size
risch	\(25 \,{\mathrm e}^{4} x \left ({\mathrm e}^{3}+5 x \right )-\frac {125 x^{2} {\mathrm e}^{4} \left (2 \ln \relax (x )-1\right )}{\ln \relax (x )^{2}}\)	\(30\)
norman	\(\frac {125 x^{2} {\mathrm e}^{4}-250 x^{2} {\mathrm e}^{4} \ln \relax (x )+125 x^{2} {\mathrm e}^{4} \ln \relax (x )^{2}+25 \,{\mathrm e}^{3} {\mathrm e}^{4} x \ln \relax (x )^{2}}{\ln \relax (x )^{2}}\)	\(45\)
default	\(25 \,{\mathrm e}^{3} {\mathrm e}^{4} x +125 x^{2} {\mathrm e}^{4}+500 \,{\mathrm e}^{4} \expIntegralEi \left (1, -2 \ln \relax (x )\right )+500 \,{\mathrm e}^{4} \left (-\frac {x^{2}}{\ln \relax (x )}-2 \expIntegralEi \left (1, -2 \ln \relax (x )\right )\right )-250 \,{\mathrm e}^{4} \left (-\frac {x^{2}}{2 \ln \relax (x )^{2}}-\frac {x^{2}}{\ln \relax (x )}-2 \expIntegralEi \left (1, -2 \ln \relax (x )\right )\right )\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*exp(4)*exp(ln(5)+3)+250*x*exp(4))*ln(x)^3-500*x*exp(4)*ln(x)^2+500*x*exp(4)*ln(x)-250*x*exp(4))/ln(x)^
3,x,method=_RETURNVERBOSE)

[Out]

25*exp(4)*x*(exp(3)+5*x)-125*x^2*exp(4)*(2*ln(x)-1)/ln(x)^2

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maxima [C]  time = 0.39, size = 42, normalized size = 1.45 \begin {gather*} 125 \, x^{2} e^{4} + 25 \, x e^{7} - 500 \, {\rm Ei}\left (2 \, \log \relax (x)\right ) e^{4} + 1000 \, e^{4} \Gamma \left (-1, -2 \, \log \relax (x)\right ) + 1000 \, e^{4} \Gamma \left (-2, -2 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(4)*exp(log(5)+3)+250*x*exp(4))*log(x)^3-500*x*exp(4)*log(x)^2+500*x*exp(4)*log(x)-250*x*exp(
4))/log(x)^3,x, algorithm="maxima")

[Out]

125*x^2*e^4 + 25*x*e^7 - 500*Ei(2*log(x))*e^4 + 1000*e^4*gamma(-1, -2*log(x)) + 1000*e^4*gamma(-2, -2*log(x))

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mupad [B]  time = 3.30, size = 34, normalized size = 1.17 \begin {gather*} \frac {125\,x^2\,{\mathrm {e}}^4-250\,x^2\,{\mathrm {e}}^4\,\ln \relax (x)}{{\ln \relax (x)}^2}+25\,x\,{\mathrm {e}}^4\,\left (5\,x+{\mathrm {e}}^3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)^3*(250*x*exp(4) + 5*exp(log(5) + 3)*exp(4)) - 250*x*exp(4) + 500*x*exp(4)*log(x) - 500*x*exp(4)*lo
g(x)^2)/log(x)^3,x)

[Out]

(125*x^2*exp(4) - 250*x^2*exp(4)*log(x))/log(x)^2 + 25*x*exp(4)*(5*x + exp(3))

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sympy [A]  time = 0.10, size = 39, normalized size = 1.34 \begin {gather*} 125 x^{2} e^{4} + 25 x e^{7} + \frac {- 250 x^{2} e^{4} \log {\relax (x )} + 125 x^{2} e^{4}}{\log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(4)*exp(ln(5)+3)+250*x*exp(4))*ln(x)**3-500*x*exp(4)*ln(x)**2+500*x*exp(4)*ln(x)-250*x*exp(4)
)/ln(x)**3,x)

[Out]

125*x**2*exp(4) + 25*x*exp(7) + (-250*x**2*exp(4)*log(x) + 125*x**2*exp(4))/log(x)**2

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