3.42.30 \(\int \frac {-253-2 x-625 x^2+(253+4 x+1875 x^2) \log (x)}{\log ^2(x)} \, dx\)

Optimal. Leaf size=16 \[ \frac {x \left (253+2 x+625 x^2\right )}{\log (x)} \]

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Rubi [A]  time = 0.19, antiderivative size = 26, normalized size of antiderivative = 1.62, number of steps used = 19, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6742, 2356, 2297, 2298, 2306, 2309, 2178} \begin {gather*} \frac {625 x^3}{\log (x)}+\frac {2 x^2}{\log (x)}+\frac {253 x}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-253 - 2*x - 625*x^2 + (253 + 4*x + 1875*x^2)*Log[x])/Log[x]^2,x]

[Out]

(253*x)/Log[x] + (2*x^2)/Log[x] + (625*x^3)/Log[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 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 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 2356

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Polyx_), x_Symbol] :> Int[ExpandIntegrand[Polyx*(a + b*Log[c*
x^n])^p, x], x] /; FreeQ[{a, b, c, n, p}, x] && PolynomialQ[Polyx, 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 \left (\frac {-253-2 x-625 x^2}{\log ^2(x)}+\frac {253+4 x+1875 x^2}{\log (x)}\right ) \, dx\\ &=\int \frac {-253-2 x-625 x^2}{\log ^2(x)} \, dx+\int \frac {253+4 x+1875 x^2}{\log (x)} \, dx\\ &=\int \left (-\frac {253}{\log ^2(x)}-\frac {2 x}{\log ^2(x)}-\frac {625 x^2}{\log ^2(x)}\right ) \, dx+\int \left (\frac {253}{\log (x)}+\frac {4 x}{\log (x)}+\frac {1875 x^2}{\log (x)}\right ) \, dx\\ &=-\left (2 \int \frac {x}{\log ^2(x)} \, dx\right )+4 \int \frac {x}{\log (x)} \, dx-253 \int \frac {1}{\log ^2(x)} \, dx+253 \int \frac {1}{\log (x)} \, dx-625 \int \frac {x^2}{\log ^2(x)} \, dx+1875 \int \frac {x^2}{\log (x)} \, dx\\ &=\frac {253 x}{\log (x)}+\frac {2 x^2}{\log (x)}+\frac {625 x^3}{\log (x)}+253 \text {li}(x)-4 \int \frac {x}{\log (x)} \, dx+4 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )-253 \int \frac {1}{\log (x)} \, dx-1875 \int \frac {x^2}{\log (x)} \, dx+1875 \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )\\ &=4 \text {Ei}(2 \log (x))+1875 \text {Ei}(3 \log (x))+\frac {253 x}{\log (x)}+\frac {2 x^2}{\log (x)}+\frac {625 x^3}{\log (x)}-4 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )-1875 \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )\\ &=\frac {253 x}{\log (x)}+\frac {2 x^2}{\log (x)}+\frac {625 x^3}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 16, normalized size = 1.00 \begin {gather*} \frac {x \left (253+2 x+625 x^2\right )}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-253 - 2*x - 625*x^2 + (253 + 4*x + 1875*x^2)*Log[x])/Log[x]^2,x]

[Out]

(x*(253 + 2*x + 625*x^2))/Log[x]

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fricas [A]  time = 1.18, size = 19, normalized size = 1.19 \begin {gather*} \frac {625 \, x^{3} + 2 \, x^{2} + 253 \, x}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1875*x^2+4*x+253)*log(x)-625*x^2-2*x-253)/log(x)^2,x, algorithm="fricas")

[Out]

(625*x^3 + 2*x^2 + 253*x)/log(x)

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giac [A]  time = 0.20, size = 26, normalized size = 1.62 \begin {gather*} \frac {625 \, x^{3}}{\log \relax (x)} + \frac {2 \, x^{2}}{\log \relax (x)} + \frac {253 \, x}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1875*x^2+4*x+253)*log(x)-625*x^2-2*x-253)/log(x)^2,x, algorithm="giac")

[Out]

625*x^3/log(x) + 2*x^2/log(x) + 253*x/log(x)

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maple [A]  time = 0.02, size = 17, normalized size = 1.06

method	result	size
risch	\(\frac {x \left (625 x^{2}+2 x +253\right )}{\ln \relax (x )}\)	\(17\)
norman	\(\frac {625 x^{3}+2 x^{2}+253 x}{\ln \relax (x )}\)	\(20\)
default	\(\frac {625 x^{3}}{\ln \relax (x )}+\frac {2 x^{2}}{\ln \relax (x )}+\frac {253 x}{\ln \relax (x )}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1875*x^2+4*x+253)*ln(x)-625*x^2-2*x-253)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

x*(625*x^2+2*x+253)/ln(x)

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maxima [C]  time = 0.42, size = 44, normalized size = 2.75 \begin {gather*} 1875 \, {\rm Ei}\left (3 \, \log \relax (x)\right ) + 4 \, {\rm Ei}\left (2 \, \log \relax (x)\right ) + 253 \, {\rm Ei}\left (\log \relax (x)\right ) - 253 \, \Gamma \left (-1, -\log \relax (x)\right ) - 4 \, \Gamma \left (-1, -2 \, \log \relax (x)\right ) - 1875 \, \Gamma \left (-1, -3 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1875*x^2+4*x+253)*log(x)-625*x^2-2*x-253)/log(x)^2,x, algorithm="maxima")

[Out]

1875*Ei(3*log(x)) + 4*Ei(2*log(x)) + 253*Ei(log(x)) - 253*gamma(-1, -log(x)) - 4*gamma(-1, -2*log(x)) - 1875*g
amma(-1, -3*log(x))

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mupad [B]  time = 3.75, size = 16, normalized size = 1.00 \begin {gather*} \frac {x\,\left (625\,x^2+2\,x+253\right )}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x - log(x)*(4*x + 1875*x^2 + 253) + 625*x^2 + 253)/log(x)^2,x)

[Out]

(x*(2*x + 625*x^2 + 253))/log(x)

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sympy [A]  time = 0.09, size = 15, normalized size = 0.94 \begin {gather*} \frac {625 x^{3} + 2 x^{2} + 253 x}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1875*x**2+4*x+253)*ln(x)-625*x**2-2*x-253)/ln(x)**2,x)

[Out]

(625*x**3 + 2*x**2 + 253*x)/log(x)

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