3.94.76 \(\int \frac {-9025-190 x-x^2+(190 x+2 x^2) \log (x)+(-950-10 x) \log ^2(x)-10 x \log ^3(x)+75 \log ^4(x)}{25 x \log ^2(x)} \, dx\)

Optimal. Leaf size=20 \[ \frac {\left (19+\frac {x}{5}-\log ^2(x)\right )^2}{\log (x)} \]

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Rubi [B]  time = 0.35, antiderivative size = 42, normalized size of antiderivative = 2.10, number of steps used = 22, number of rules used = 13, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.232, Rules used = {12, 6742, 43, 2353, 2297, 2298, 2302, 30, 2306, 2309, 2178, 2330, 2295} \begin {gather*} \frac {x^2}{25 \log (x)}+\log ^3(x)-\frac {2}{5} x \log (x)-38 \log (x)+\frac {38 x}{5 \log (x)}+\frac {361}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-9025 - 190*x - x^2 + (190*x + 2*x^2)*Log[x] + (-950 - 10*x)*Log[x]^2 - 10*x*Log[x]^3 + 75*Log[x]^4)/(25*
x*Log[x]^2),x]

[Out]

361/Log[x] + (38*x)/(5*Log[x]) + x^2/(25*Log[x]) - 38*Log[x] - (2*x*Log[x])/5 + Log[x]^3

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

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 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 2330

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

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 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{25} \int \frac {-9025-190 x-x^2+\left (190 x+2 x^2\right ) \log (x)+(-950-10 x) \log ^2(x)-10 x \log ^3(x)+75 \log ^4(x)}{x \log ^2(x)} \, dx\\ &=\frac {1}{25} \int \left (-\frac {10 (95+x)}{x}-\frac {(95+x)^2}{x \log ^2(x)}+\frac {2 (95+x)}{\log (x)}-10 \log (x)+\frac {75 \log ^2(x)}{x}\right ) \, dx\\ &=-\left (\frac {1}{25} \int \frac {(95+x)^2}{x \log ^2(x)} \, dx\right )+\frac {2}{25} \int \frac {95+x}{\log (x)} \, dx-\frac {2}{5} \int \frac {95+x}{x} \, dx-\frac {2}{5} \int \log (x) \, dx+3 \int \frac {\log ^2(x)}{x} \, dx\\ &=\frac {2 x}{5}-\frac {2}{5} x \log (x)-\frac {1}{25} \int \left (\frac {190}{\log ^2(x)}+\frac {9025}{x \log ^2(x)}+\frac {x}{\log ^2(x)}\right ) \, dx+\frac {2}{25} \int \left (\frac {95}{\log (x)}+\frac {x}{\log (x)}\right ) \, dx-\frac {2}{5} \int \left (1+\frac {95}{x}\right ) \, dx+3 \operatorname {Subst}\left (\int x^2 \, dx,x,\log (x)\right )\\ &=-38 \log (x)-\frac {2}{5} x \log (x)+\log ^3(x)-\frac {1}{25} \int \frac {x}{\log ^2(x)} \, dx+\frac {2}{25} \int \frac {x}{\log (x)} \, dx-\frac {38}{5} \int \frac {1}{\log ^2(x)} \, dx+\frac {38}{5} \int \frac {1}{\log (x)} \, dx-361 \int \frac {1}{x \log ^2(x)} \, dx\\ &=\frac {38 x}{5 \log (x)}+\frac {x^2}{25 \log (x)}-38 \log (x)-\frac {2}{5} x \log (x)+\log ^3(x)+\frac {38 \text {li}(x)}{5}-\frac {2}{25} \int \frac {x}{\log (x)} \, dx+\frac {2}{25} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )-\frac {38}{5} \int \frac {1}{\log (x)} \, dx-361 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )\\ &=\frac {2}{25} \text {Ei}(2 \log (x))+\frac {361}{\log (x)}+\frac {38 x}{5 \log (x)}+\frac {x^2}{25 \log (x)}-38 \log (x)-\frac {2}{5} x \log (x)+\log ^3(x)-\frac {2}{25} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=\frac {361}{\log (x)}+\frac {38 x}{5 \log (x)}+\frac {x^2}{25 \log (x)}-38 \log (x)-\frac {2}{5} x \log (x)+\log ^3(x)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.02, size = 42, normalized size = 2.10 \begin {gather*} \frac {361}{\log (x)}+\frac {38 x}{5 \log (x)}+\frac {x^2}{25 \log (x)}-38 \log (x)-\frac {2}{5} x \log (x)+\log ^3(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-9025 - 190*x - x^2 + (190*x + 2*x^2)*Log[x] + (-950 - 10*x)*Log[x]^2 - 10*x*Log[x]^3 + 75*Log[x]^4
)/(25*x*Log[x]^2),x]

[Out]

361/Log[x] + (38*x)/(5*Log[x]) + x^2/(25*Log[x]) - 38*Log[x] - (2*x*Log[x])/5 + Log[x]^3

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fricas [A]  time = 0.88, size = 29, normalized size = 1.45 \begin {gather*} \frac {25 \, \log \relax (x)^{4} - 10 \, {\left (x + 95\right )} \log \relax (x)^{2} + x^{2} + 190 \, x + 9025}{25 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(75*log(x)^4-10*x*log(x)^3+(-10*x-950)*log(x)^2+(2*x^2+190*x)*log(x)-x^2-190*x-9025)/x/log(x)^2
,x, algorithm="fricas")

[Out]

1/25*(25*log(x)^4 - 10*(x + 95)*log(x)^2 + x^2 + 190*x + 9025)/log(x)

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giac [A]  time = 0.24, size = 28, normalized size = 1.40 \begin {gather*} \log \relax (x)^{3} - \frac {2}{5} \, x \log \relax (x) + \frac {x^{2} + 190 \, x + 9025}{25 \, \log \relax (x)} - 38 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(75*log(x)^4-10*x*log(x)^3+(-10*x-950)*log(x)^2+(2*x^2+190*x)*log(x)-x^2-190*x-9025)/x/log(x)^2
,x, algorithm="giac")

[Out]

log(x)^3 - 2/5*x*log(x) + 1/25*(x^2 + 190*x + 9025)/log(x) - 38*log(x)

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maple [A]  time = 0.03, size = 29, normalized size = 1.45

method	result	size
risch	\(\ln \relax (x )^{3}-\frac {2 x \ln \relax (x )}{5}-38 \ln \relax (x )+\frac {x^{2}+190 x +9025}{25 \ln \relax (x )}\)	\(29\)
norman	\(\frac {361+\ln \relax (x )^{4}-38 \ln \relax (x )^{2}+\frac {38 x}{5}+\frac {x^{2}}{25}-\frac {2 x \ln \relax (x )^{2}}{5}}{\ln \relax (x )}\)	\(33\)
default	\(\ln \relax (x )^{3}-\frac {2 x \ln \relax (x )}{5}-38 \ln \relax (x )+\frac {x^{2}}{25 \ln \relax (x )}+\frac {38 x}{5 \ln \relax (x )}+\frac {361}{\ln \relax (x )}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/25*(75*ln(x)^4-10*x*ln(x)^3+(-10*x-950)*ln(x)^2+(2*x^2+190*x)*ln(x)-x^2-190*x-9025)/x/ln(x)^2,x,method=_
RETURNVERBOSE)

[Out]

ln(x)^3-2/5*x*ln(x)-38*ln(x)+1/25*(x^2+190*x+9025)/ln(x)

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maxima [C]  time = 0.39, size = 48, normalized size = 2.40 \begin {gather*} \log \relax (x)^{3} - \frac {2}{5} \, x \log \relax (x) + \frac {361}{\log \relax (x)} + \frac {2}{25} \, {\rm Ei}\left (2 \, \log \relax (x)\right ) + \frac {38}{5} \, {\rm Ei}\left (\log \relax (x)\right ) - \frac {38}{5} \, \Gamma \left (-1, -\log \relax (x)\right ) - \frac {2}{25} \, \Gamma \left (-1, -2 \, \log \relax (x)\right ) - 38 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(75*log(x)^4-10*x*log(x)^3+(-10*x-950)*log(x)^2+(2*x^2+190*x)*log(x)-x^2-190*x-9025)/x/log(x)^2
,x, algorithm="maxima")

[Out]

log(x)^3 - 2/5*x*log(x) + 361/log(x) + 2/25*Ei(2*log(x)) + 38/5*Ei(log(x)) - 38/5*gamma(-1, -log(x)) - 2/25*ga
mma(-1, -2*log(x)) - 38*log(x)

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mupad [B]  time = 8.86, size = 17, normalized size = 0.85 \begin {gather*} \frac {{\left (-5\,{\ln \relax (x)}^2+x+95\right )}^2}{25\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((38*x)/5 + (2*x*log(x)^3)/5 - 3*log(x)^4 - (log(x)*(190*x + 2*x^2))/25 + x^2/25 + (log(x)^2*(10*x + 950)
)/25 + 361)/(x*log(x)^2),x)

[Out]

(x - 5*log(x)^2 + 95)^2/(25*log(x))

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sympy [B]  time = 0.13, size = 31, normalized size = 1.55 \begin {gather*} - \frac {2 x \log {\relax (x )}}{5} + \frac {x^{2} + 190 x + 9025}{25 \log {\relax (x )}} + \log {\relax (x )}^{3} - 38 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(75*ln(x)**4-10*x*ln(x)**3+(-10*x-950)*ln(x)**2+(2*x**2+190*x)*ln(x)-x**2-190*x-9025)/x/ln(x)**
2,x)

[Out]

-2*x*log(x)/5 + (x**2 + 190*x + 9025)/(25*log(x)) + log(x)**3 - 38*log(x)

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