3.6.23 \(\int \frac {-12 x+42 x^2-54 x^3+36 x^4+e^5 (-4+6 x-6 x^2)+(2 e^{10}+e^5 (6 x-12 x^2)) \log (x)}{x} \, dx\)

Optimal. Leaf size=22 \[ \log ^2(30)+\left (-2+(3-3 x) x+e^5 \log (x)\right )^2 \]

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Rubi [B]  time = 0.11, antiderivative size = 80, normalized size of antiderivative = 3.64, number of steps used = 9, number of rules used = 5, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.085, Rules used = {14, 2357, 2295, 2301, 2304} \begin {gather*} 9 x^4-18 x^3+3 \left (7-e^5\right ) x^2+3 e^5 x^2-6 e^5 x^2 \log (x)-6 \left (2-e^5\right ) x-6 e^5 x+e^{10} \log ^2(x)+6 e^5 x \log (x)-4 e^5 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-12*x + 42*x^2 - 54*x^3 + 36*x^4 + E^5*(-4 + 6*x - 6*x^2) + (2*E^10 + E^5*(6*x - 12*x^2))*Log[x])/x,x]

[Out]

-6*E^5*x - 6*(2 - E^5)*x + 3*E^5*x^2 + 3*(7 - E^5)*x^2 - 18*x^3 + 9*x^4 - 4*E^5*Log[x] + 6*E^5*x*Log[x] - 6*E^
5*x^2*Log[x] + E^10*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 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 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 \left (-2 e^5-3 \left (2-e^5\right ) x+3 \left (7-e^5\right ) x^2-27 x^3+18 x^4\right )}{x}+\frac {2 e^5 \left (e^5+3 x-6 x^2\right ) \log (x)}{x}\right ) \, dx\\ &=2 \int \frac {-2 e^5-3 \left (2-e^5\right ) x+3 \left (7-e^5\right ) x^2-27 x^3+18 x^4}{x} \, dx+\left (2 e^5\right ) \int \frac {\left (e^5+3 x-6 x^2\right ) \log (x)}{x} \, dx\\ &=2 \int \left (-3 \left (2-e^5\right )-\frac {2 e^5}{x}+3 \left (7-e^5\right ) x-27 x^2+18 x^3\right ) \, dx+\left (2 e^5\right ) \int \left (3 \log (x)+\frac {e^5 \log (x)}{x}-6 x \log (x)\right ) \, dx\\ &=-6 \left (2-e^5\right ) x+3 \left (7-e^5\right ) x^2-18 x^3+9 x^4-4 e^5 \log (x)+\left (6 e^5\right ) \int \log (x) \, dx-\left (12 e^5\right ) \int x \log (x) \, dx+\left (2 e^{10}\right ) \int \frac {\log (x)}{x} \, dx\\ &=-6 e^5 x-6 \left (2-e^5\right ) x+3 e^5 x^2+3 \left (7-e^5\right ) x^2-18 x^3+9 x^4-4 e^5 \log (x)+6 e^5 x \log (x)-6 e^5 x^2 \log (x)+e^{10} \log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.01, size = 52, normalized size = 2.36 \begin {gather*} -12 x+21 x^2-18 x^3+9 x^4-4 e^5 \log (x)+6 e^5 x \log (x)-6 e^5 x^2 \log (x)+e^{10} \log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-12*x + 42*x^2 - 54*x^3 + 36*x^4 + E^5*(-4 + 6*x - 6*x^2) + (2*E^10 + E^5*(6*x - 12*x^2))*Log[x])/x
,x]

[Out]

-12*x + 21*x^2 - 18*x^3 + 9*x^4 - 4*E^5*Log[x] + 6*E^5*x*Log[x] - 6*E^5*x^2*Log[x] + E^10*Log[x]^2

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fricas [A]  time = 0.72, size = 42, normalized size = 1.91 \begin {gather*} 9 \, x^{4} - 18 \, x^{3} - 2 \, {\left (3 \, x^{2} - 3 \, x + 2\right )} e^{5} \log \relax (x) + e^{10} \log \relax (x)^{2} + 21 \, x^{2} - 12 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(5)^2+(-12*x^2+6*x)*exp(5))*log(x)+(-6*x^2+6*x-4)*exp(5)+36*x^4-54*x^3+42*x^2-12*x)/x,x, algo
rithm="fricas")

[Out]

9*x^4 - 18*x^3 - 2*(3*x^2 - 3*x + 2)*e^5*log(x) + e^10*log(x)^2 + 21*x^2 - 12*x

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giac [B]  time = 0.44, size = 48, normalized size = 2.18 \begin {gather*} 9 \, x^{4} - 6 \, x^{2} e^{5} \log \relax (x) - 18 \, x^{3} + 6 \, x e^{5} \log \relax (x) + e^{10} \log \relax (x)^{2} + 21 \, x^{2} - 4 \, e^{5} \log \relax (x) - 12 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(5)^2+(-12*x^2+6*x)*exp(5))*log(x)+(-6*x^2+6*x-4)*exp(5)+36*x^4-54*x^3+42*x^2-12*x)/x,x, algo
rithm="giac")

[Out]

9*x^4 - 6*x^2*e^5*log(x) - 18*x^3 + 6*x*e^5*log(x) + e^10*log(x)^2 + 21*x^2 - 4*e^5*log(x) - 12*x

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maple [B]  time = 0.08, size = 49, normalized size = 2.23

method	result	size
risch	\({\mathrm e}^{10} \ln \relax (x )^{2}+\left (-6 x^{2} {\mathrm e}^{5}+6 x \,{\mathrm e}^{5}\right ) \ln \relax (x )+9 x^{4}-18 x^{3}+21 x^{2}-12 x -4 \,{\mathrm e}^{5} \ln \relax (x )\)	\(49\)
norman	\(-4 \,{\mathrm e}^{5} \ln \relax (x )+{\mathrm e}^{10} \ln \relax (x )^{2}-12 x +21 x^{2}-18 x^{3}+9 x^{4}+6 x \,{\mathrm e}^{5} \ln \relax (x )-6 x^{2} {\mathrm e}^{5} \ln \relax (x )\)	\(51\)
default	\(-12 \,{\mathrm e}^{5} \left (\frac {x^{2} \ln \relax (x )}{2}-\frac {x^{2}}{4}\right )+9 x^{4}+{\mathrm e}^{10} \ln \relax (x )^{2}+6 \,{\mathrm e}^{5} \left (x \ln \relax (x )-x \right )-3 x^{2} {\mathrm e}^{5}-18 x^{3}+6 x \,{\mathrm e}^{5}+21 x^{2}-4 \,{\mathrm e}^{5} \ln \relax (x )-12 x\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(5)^2+(-12*x^2+6*x)*exp(5))*ln(x)+(-6*x^2+6*x-4)*exp(5)+36*x^4-54*x^3+42*x^2-12*x)/x,x,method=_RETU
RNVERBOSE)

[Out]

exp(10)*ln(x)^2+(-6*x^2*exp(5)+6*x*exp(5))*ln(x)+9*x^4-18*x^3+21*x^2-12*x-4*exp(5)*ln(x)

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maxima [B]  time = 0.43, size = 73, normalized size = 3.32 \begin {gather*} 9 \, x^{4} - 18 \, x^{3} - 3 \, x^{2} e^{5} + e^{10} \log \relax (x)^{2} + 21 \, x^{2} - 3 \, {\left (2 \, x^{2} \log \relax (x) - x^{2}\right )} e^{5} + 6 \, {\left (x \log \relax (x) - x\right )} e^{5} + 6 \, x e^{5} - 4 \, e^{5} \log \relax (x) - 12 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(5)^2+(-12*x^2+6*x)*exp(5))*log(x)+(-6*x^2+6*x-4)*exp(5)+36*x^4-54*x^3+42*x^2-12*x)/x,x, algo
rithm="maxima")

[Out]

9*x^4 - 18*x^3 - 3*x^2*e^5 + e^10*log(x)^2 + 21*x^2 - 3*(2*x^2*log(x) - x^2)*e^5 + 6*(x*log(x) - x)*e^5 + 6*x*
e^5 - 4*e^5*log(x) - 12*x

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mupad [B]  time = 0.56, size = 48, normalized size = 2.18 \begin {gather*} 9\,x^4-18\,x^3-6\,{\mathrm {e}}^5\,x^2\,\ln \relax (x)+21\,x^2+6\,{\mathrm {e}}^5\,x\,\ln \relax (x)-12\,x+{\mathrm {e}}^{10}\,{\ln \relax (x)}^2-4\,{\mathrm {e}}^5\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(12*x + exp(5)*(6*x^2 - 6*x + 4) - log(x)*(2*exp(10) + exp(5)*(6*x - 12*x^2)) - 42*x^2 + 54*x^3 - 36*x^4)
/x,x)

[Out]

exp(10)*log(x)^2 - 12*x - 4*exp(5)*log(x) + 21*x^2 - 18*x^3 + 9*x^4 + 6*x*exp(5)*log(x) - 6*x^2*exp(5)*log(x)

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sympy [B]  time = 0.23, size = 53, normalized size = 2.41 \begin {gather*} 9 x^{4} - 18 x^{3} + 21 x^{2} - 12 x + \left (- 6 x^{2} e^{5} + 6 x e^{5}\right ) \log {\relax (x )} + e^{10} \log {\relax (x )}^{2} - 4 e^{5} \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(5)**2+(-12*x**2+6*x)*exp(5))*ln(x)+(-6*x**2+6*x-4)*exp(5)+36*x**4-54*x**3+42*x**2-12*x)/x,x)

[Out]

9*x**4 - 18*x**3 + 21*x**2 - 12*x + (-6*x**2*exp(5) + 6*x*exp(5))*log(x) + exp(10)*log(x)**2 - 4*exp(5)*log(x)

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