3.14.7 \(\int \frac {100-40 e^2+4 e^4-x^3+(10 x^4+2 x^5) \log (x)+x^5 \log ^2(x)}{x^5} \, dx\)

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

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Rubi [A]  time = 0.06, antiderivative size = 30, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 5, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {14, 2346, 2301, 2295, 2296} \begin {gather*} -\frac {\left (5-e^2\right )^2}{x^4}+\frac {1}{x}+x \log ^2(x)+5 \log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(100 - 40*E^2 + 4*E^4 - x^3 + (10*x^4 + 2*x^5)*Log[x] + x^5*Log[x]^2)/x^5,x]

[Out]

-((5 - E^2)^2/x^4) + x^(-1) + 5*Log[x]^2 + x*Log[x]^2

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2346

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.))/(x_), x_Symbol] :> Dist[d, Int[((d
+ e*x)^(q - 1)*(a + b*Log[c*x^n])^p)/x, x], x] + Dist[e, Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^p, x], x] /
; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && GtQ[q, 0] && IntegerQ[2*q]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {100-40 e^2+4 e^4-x^3}{x^5}+\frac {2 (5+x) \log (x)}{x}+\log ^2(x)\right ) \, dx\\ &=2 \int \frac {(5+x) \log (x)}{x} \, dx+\int \frac {100-40 e^2+4 e^4-x^3}{x^5} \, dx+\int \log ^2(x) \, dx\\ &=x \log ^2(x)+10 \int \frac {\log (x)}{x} \, dx+\int \left (\frac {4 \left (-5+e^2\right )^2}{x^5}-\frac {1}{x^2}\right ) \, dx\\ &=-\frac {\left (5-e^2\right )^2}{x^4}+\frac {1}{x}+5 \log ^2(x)+x \log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 37, normalized size = 1.28 \begin {gather*} -\frac {25}{x^4}+\frac {10 e^2}{x^4}-\frac {e^4}{x^4}+\frac {1}{x}+5 \log ^2(x)+x \log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(100 - 40*E^2 + 4*E^4 - x^3 + (10*x^4 + 2*x^5)*Log[x] + x^5*Log[x]^2)/x^5,x]

[Out]

-25/x^4 + (10*E^2)/x^4 - E^4/x^4 + x^(-1) + 5*Log[x]^2 + x*Log[x]^2

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fricas [A]  time = 0.96, size = 31, normalized size = 1.07 \begin {gather*} \frac {x^{3} + {\left (x^{5} + 5 \, x^{4}\right )} \log \relax (x)^{2} - e^{4} + 10 \, e^{2} - 25}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5*log(x)^2+(2*x^5+10*x^4)*log(x)+4*exp(2)^2-40*exp(2)-x^3+100)/x^5,x, algorithm="fricas")

[Out]

(x^3 + (x^5 + 5*x^4)*log(x)^2 - e^4 + 10*e^2 - 25)/x^4

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giac [A]  time = 0.30, size = 34, normalized size = 1.17 \begin {gather*} \frac {x^{5} \log \relax (x)^{2} + 5 \, x^{4} \log \relax (x)^{2} + x^{3} - e^{4} + 10 \, e^{2} - 25}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5*log(x)^2+(2*x^5+10*x^4)*log(x)+4*exp(2)^2-40*exp(2)-x^3+100)/x^5,x, algorithm="giac")

[Out]

(x^5*log(x)^2 + 5*x^4*log(x)^2 + x^3 - e^4 + 10*e^2 - 25)/x^4

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maple [A]  time = 0.04, size = 28, normalized size = 0.97

method	result	size
risch	\(\left (5+x \right ) \ln \relax (x )^{2}-\frac {-x^{3}+{\mathrm e}^{4}-10 \,{\mathrm e}^{2}+25}{x^{4}}\)	\(28\)
norman	\(\frac {x^{3}+x^{5} \ln \relax (x )^{2}+5 x^{4} \ln \relax (x )^{2}-25-{\mathrm e}^{4}+10 \,{\mathrm e}^{2}}{x^{4}}\)	\(37\)
default	\(x \ln \relax (x )^{2}+5 \ln \relax (x )^{2}+\frac {1}{x}-\frac {{\mathrm e}^{4}}{x^{4}}+\frac {10 \,{\mathrm e}^{2}}{x^{4}}-\frac {25}{x^{4}}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*ln(x)^2+(2*x^5+10*x^4)*ln(x)+4*exp(2)^2-40*exp(2)-x^3+100)/x^5,x,method=_RETURNVERBOSE)

[Out]

(5+x)*ln(x)^2-(-x^3+exp(4)-10*exp(2)+25)/x^4

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maxima [A]  time = 0.43, size = 49, normalized size = 1.69 \begin {gather*} {\left (\log \relax (x)^{2} - 2 \, \log \relax (x) + 2\right )} x + 2 \, x \log \relax (x) + 5 \, \log \relax (x)^{2} - 2 \, x + \frac {1}{x} - \frac {e^{4}}{x^{4}} + \frac {10 \, e^{2}}{x^{4}} - \frac {25}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5*log(x)^2+(2*x^5+10*x^4)*log(x)+4*exp(2)^2-40*exp(2)-x^3+100)/x^5,x, algorithm="maxima")

[Out]

(log(x)^2 - 2*log(x) + 2)*x + 2*x*log(x) + 5*log(x)^2 - 2*x + 1/x - e^4/x^4 + 10*e^2/x^4 - 25/x^4

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mupad [B]  time = 0.94, size = 30, normalized size = 1.03 \begin {gather*} x\,{\ln \relax (x)}^2+5\,{\ln \relax (x)}^2+\frac {x^4-x\,{\left ({\mathrm {e}}^2-5\right )}^2}{x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*exp(4) - 40*exp(2) + log(x)*(10*x^4 + 2*x^5) + x^5*log(x)^2 - x^3 + 100)/x^5,x)

[Out]

x*log(x)^2 + 5*log(x)^2 + (x^4 - x*(exp(2) - 5)^2)/x^5

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sympy [A]  time = 0.25, size = 24, normalized size = 0.83 \begin {gather*} \left (x + 5\right ) \log {\relax (x )}^{2} - \frac {- x^{3} - 10 e^{2} + 25 + e^{4}}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**5*ln(x)**2+(2*x**5+10*x**4)*ln(x)+4*exp(2)**2-40*exp(2)-x**3+100)/x**5,x)

[Out]

(x + 5)*log(x)**2 - (-x**3 - 10*exp(2) + 25 + exp(4))/x**4

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