3.99.40 \(\int \frac {36 x^4+e^{32} (-2 x+2 x^2)+e^{16} (6 x^2-18 x^3)+(e^{32} (2-2 x)+12 e^{16} x^2) \log (x)}{9 x} \, dx\)

Optimal. Leaf size=26 \[ 1+x^2 \left (x+\frac {e^{16} (-x+\log (x))}{3 x}\right )^2 \]

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Rubi [B]  time = 0.10, antiderivative size = 72, normalized size of antiderivative = 2.77, number of steps used = 9, number of rules used = 6, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {12, 14, 2357, 2295, 2301, 2304} \begin {gather*} x^4-\frac {2 e^{16} x^3}{3}+\frac {1}{9} e^{16} \left (3+e^{16}\right ) x^2-\frac {e^{16} x^2}{3}+\frac {2}{3} e^{16} x^2 \log (x)+\frac {1}{9} e^{32} \log ^2(x)-\frac {2}{9} e^{32} x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(36*x^4 + E^32*(-2*x + 2*x^2) + E^16*(6*x^2 - 18*x^3) + (E^32*(2 - 2*x) + 12*E^16*x^2)*Log[x])/(9*x),x]

[Out]

-1/3*(E^16*x^2) + (E^16*(3 + E^16)*x^2)/9 - (2*E^16*x^3)/3 + x^4 - (2*E^32*x*Log[x])/9 + (2*E^16*x^2*Log[x])/3
+ (E^32*Log[x]^2)/9

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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} &=\frac {1}{9} \int \frac {36 x^4+e^{32} \left (-2 x+2 x^2\right )+e^{16} \left (6 x^2-18 x^3\right )+\left (e^{32} (2-2 x)+12 e^{16} x^2\right ) \log (x)}{x} \, dx\\ &=\frac {1}{9} \int \left (2 \left (-e^{32}+e^{16} \left (3+e^{16}\right ) x-9 e^{16} x^2+18 x^3\right )+\frac {2 e^{16} \left (e^{16}-e^{16} x+6 x^2\right ) \log (x)}{x}\right ) \, dx\\ &=\frac {2}{9} \int \left (-e^{32}+e^{16} \left (3+e^{16}\right ) x-9 e^{16} x^2+18 x^3\right ) \, dx+\frac {1}{9} \left (2 e^{16}\right ) \int \frac {\left (e^{16}-e^{16} x+6 x^2\right ) \log (x)}{x} \, dx\\ &=-\frac {2 e^{32} x}{9}+\frac {1}{9} e^{16} \left (3+e^{16}\right ) x^2-\frac {2 e^{16} x^3}{3}+x^4+\frac {1}{9} \left (2 e^{16}\right ) \int \left (-e^{16} \log (x)+\frac {e^{16} \log (x)}{x}+6 x \log (x)\right ) \, dx\\ &=-\frac {2 e^{32} x}{9}+\frac {1}{9} e^{16} \left (3+e^{16}\right ) x^2-\frac {2 e^{16} x^3}{3}+x^4+\frac {1}{3} \left (4 e^{16}\right ) \int x \log (x) \, dx-\frac {1}{9} \left (2 e^{32}\right ) \int \log (x) \, dx+\frac {1}{9} \left (2 e^{32}\right ) \int \frac {\log (x)}{x} \, dx\\ &=-\frac {1}{3} e^{16} x^2+\frac {1}{9} e^{16} \left (3+e^{16}\right ) x^2-\frac {2 e^{16} x^3}{3}+x^4-\frac {2}{9} e^{32} x \log (x)+\frac {2}{3} e^{16} x^2 \log (x)+\frac {1}{9} e^{32} \log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 23, normalized size = 0.88 \begin {gather*} \frac {1}{9} \left (\left (e^{16}-3 x\right ) x-e^{16} \log (x)\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(36*x^4 + E^32*(-2*x + 2*x^2) + E^16*(6*x^2 - 18*x^3) + (E^32*(2 - 2*x) + 12*E^16*x^2)*Log[x])/(9*x)
,x]

[Out]

((E^16 - 3*x)*x - E^16*Log[x])^2/9

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fricas [A]  time = 0.74, size = 43, normalized size = 1.65 \begin {gather*} x^{4} - \frac {2}{3} \, x^{3} e^{16} + \frac {1}{9} \, x^{2} e^{32} + \frac {1}{9} \, e^{32} \log \relax (x)^{2} + \frac {2}{9} \, {\left (3 \, x^{2} e^{16} - x e^{32}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(((-2*x+2)*exp(16)^2+12*x^2*exp(16))*log(x)+(2*x^2-2*x)*exp(16)^2+(-18*x^3+6*x^2)*exp(16)+36*x^4
)/x,x, algorithm="fricas")

[Out]

x^4 - 2/3*x^3*e^16 + 1/9*x^2*e^32 + 1/9*e^32*log(x)^2 + 2/9*(3*x^2*e^16 - x*e^32)*log(x)

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giac [A]  time = 0.20, size = 42, normalized size = 1.62 \begin {gather*} x^{4} - \frac {2}{3} \, x^{3} e^{16} + \frac {2}{3} \, x^{2} e^{16} \log \relax (x) + \frac {1}{9} \, x^{2} e^{32} - \frac {2}{9} \, x e^{32} \log \relax (x) + \frac {1}{9} \, e^{32} \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(((-2*x+2)*exp(16)^2+12*x^2*exp(16))*log(x)+(2*x^2-2*x)*exp(16)^2+(-18*x^3+6*x^2)*exp(16)+36*x^4
)/x,x, algorithm="giac")

[Out]

x^4 - 2/3*x^3*e^16 + 2/3*x^2*e^16*log(x) + 1/9*x^2*e^32 - 2/9*x*e^32*log(x) + 1/9*e^32*log(x)^2

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maple [A]  time = 0.04, size = 44, normalized size = 1.69

method	result	size
risch	\(\frac {{\mathrm e}^{32} \ln \relax (x )^{2}}{9}+\frac {\left (-2 x \,{\mathrm e}^{32}+6 x^{2} {\mathrm e}^{16}\right ) \ln \relax (x )}{9}+\frac {{\mathrm e}^{32} x^{2}}{9}-\frac {2 x^{3} {\mathrm e}^{16}}{3}+x^{4}\)	\(44\)
norman	\(x^{4}-\frac {2 x^{3} {\mathrm e}^{16}}{3}+\frac {{\mathrm e}^{32} x^{2}}{9}+\frac {{\mathrm e}^{32} \ln \relax (x )^{2}}{9}+\frac {2 x^{2} {\mathrm e}^{16} \ln \relax (x )}{3}-\frac {2 \,{\mathrm e}^{32} \ln \relax (x ) x}{9}\)	\(49\)
default	\(-\frac {2 \,{\mathrm e}^{32} \left (x \ln \relax (x )-x \right )}{9}+\frac {4 \,{\mathrm e}^{16} \left (\frac {x^{2} \ln \relax (x )}{2}-\frac {x^{2}}{4}\right )}{3}+\frac {{\mathrm e}^{32} x^{2}}{9}-\frac {2 x^{3} {\mathrm e}^{16}}{3}+x^{4}+\frac {{\mathrm e}^{32} \ln \relax (x )^{2}}{9}-\frac {2 x \,{\mathrm e}^{32}}{9}+\frac {x^{2} {\mathrm e}^{16}}{3}\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/9*(((-2*x+2)*exp(16)^2+12*x^2*exp(16))*ln(x)+(2*x^2-2*x)*exp(16)^2+(-18*x^3+6*x^2)*exp(16)+36*x^4)/x,x,m
ethod=_RETURNVERBOSE)

[Out]

1/9*exp(32)*ln(x)^2+1/9*(-2*x*exp(32)+6*x^2*exp(16))*ln(x)+1/9*exp(32)*x^2-2/3*x^3*exp(16)+x^4

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maxima [B]  time = 0.37, size = 67, normalized size = 2.58 \begin {gather*} x^{4} - \frac {2}{3} \, x^{3} e^{16} + \frac {1}{9} \, x^{2} e^{32} + \frac {1}{3} \, x^{2} e^{16} + \frac {1}{9} \, e^{32} \log \relax (x)^{2} - \frac {2}{9} \, {\left (x \log \relax (x) - x\right )} e^{32} - \frac {2}{9} \, x e^{32} + \frac {1}{3} \, {\left (2 \, x^{2} \log \relax (x) - x^{2}\right )} e^{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(((-2*x+2)*exp(16)^2+12*x^2*exp(16))*log(x)+(2*x^2-2*x)*exp(16)^2+(-18*x^3+6*x^2)*exp(16)+36*x^4
)/x,x, algorithm="maxima")

[Out]

x^4 - 2/3*x^3*e^16 + 1/9*x^2*e^32 + 1/3*x^2*e^16 + 1/9*e^32*log(x)^2 - 2/9*(x*log(x) - x)*e^32 - 2/9*x*e^32 +
1/3*(2*x^2*log(x) - x^2)*e^16

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mupad [B]  time = 5.86, size = 20, normalized size = 0.77 \begin {gather*} \frac {{\left ({\mathrm {e}}^{16}\,\ln \relax (x)-x\,{\mathrm {e}}^{16}+3\,x^2\right )}^2}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((log(x)*(12*x^2*exp(16) - exp(32)*(2*x - 2)))/9 - (exp(32)*(2*x - 2*x^2))/9 + (exp(16)*(6*x^2 - 18*x^3))/
9 + 4*x^4)/x,x)

[Out]

(exp(16)*log(x) - x*exp(16) + 3*x^2)^2/9

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sympy [B]  time = 0.16, size = 53, normalized size = 2.04 \begin {gather*} x^{4} - \frac {2 x^{3} e^{16}}{3} + \frac {x^{2} e^{32}}{9} + \left (\frac {2 x^{2} e^{16}}{3} - \frac {2 x e^{32}}{9}\right ) \log {\relax (x )} + \frac {e^{32} \log {\relax (x )}^{2}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(((-2*x+2)*exp(16)**2+12*x**2*exp(16))*ln(x)+(2*x**2-2*x)*exp(16)**2+(-18*x**3+6*x**2)*exp(16)+3
6*x**4)/x,x)

[Out]

x**4 - 2*x**3*exp(16)/3 + x**2*exp(32)/9 + (2*x**2*exp(16)/3 - 2*x*exp(32)/9)*log(x) + exp(32)*log(x)**2/9

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