3.44.17 \(\int \frac {1}{8} (-16 e^4+(16 e^4-e^8 x) \log (3)+e^8 x \log ^2(3)+(16 e^4-e^8 x+2 e^8 x \log (3)) \log (\frac {3}{x})+e^8 x \log ^2(\frac {3}{x})) \, dx\)

Optimal. Leaf size=29 \[ \frac {\left (x+x \left (3+\frac {1}{4} e^4 x \left (\log (3)+\log \left (\frac {3}{x}\right )\right )\right )\right )^2}{x^2} \]

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Rubi [B]  time = 0.08, antiderivative size = 140, normalized size of antiderivative = 4.83, number of steps used = 7, number of rules used = 6, integrand size = 74, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {12, 6, 2313, 9, 2305, 2304} \begin {gather*} \frac {e^8 x^2}{32}+\frac {1}{16} e^8 x^2 \log ^2\left (\frac {3}{x}\right )+\frac {1}{16} e^8 x^2 \log ^2(3)+\frac {1}{16} e^8 x^2 \log \left (\frac {3}{x}\right )+\frac {1}{16} \left (32 e^4 x-e^8 x^2 (1-\log (9))\right ) \log \left (\frac {3}{x}\right )-2 e^4 x-\frac {\left (32-e^4 x (1-\log (9))\right )^2}{32 (1-\log (9))}-\frac {1}{16} \left (16-e^4 x\right )^2 \log (3) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16*E^4 + (16*E^4 - E^8*x)*Log[3] + E^8*x*Log[3]^2 + (16*E^4 - E^8*x + 2*E^8*x*Log[3])*Log[3/x] + E^8*x*L
og[3/x]^2)/8,x]

[Out]

-2*E^4*x + (E^8*x^2)/32 - ((16 - E^4*x)^2*Log[3])/16 + (E^8*x^2*Log[3]^2)/16 - (32 - E^4*x*(1 - Log[9]))^2/(32
*(1 - Log[9])) + (E^8*x^2*Log[3/x])/16 + ((32*E^4*x - E^8*x^2*(1 - Log[9]))*Log[3/x])/16 + (E^8*x^2*Log[3/x]^2
)/16

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 9

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 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] && GtQ[p, 0]

Rule 2313

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(d +
e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]] /; FreeQ[{a,
b, c, d, e, n, r}, x] && IGtQ[q, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{8} \int \left (-16 e^4+\left (16 e^4-e^8 x\right ) \log (3)+e^8 x \log ^2(3)+\left (16 e^4-e^8 x+2 e^8 x \log (3)\right ) \log \left (\frac {3}{x}\right )+e^8 x \log ^2\left (\frac {3}{x}\right )\right ) \, dx\\ &=-2 e^4 x-\frac {1}{16} \left (16-e^4 x\right )^2 \log (3)+\frac {1}{16} e^8 x^2 \log ^2(3)+\frac {1}{8} \int \left (16 e^4-e^8 x+2 e^8 x \log (3)\right ) \log \left (\frac {3}{x}\right ) \, dx+\frac {1}{8} e^8 \int x \log ^2\left (\frac {3}{x}\right ) \, dx\\ &=-2 e^4 x-\frac {1}{16} \left (16-e^4 x\right )^2 \log (3)+\frac {1}{16} e^8 x^2 \log ^2(3)+\frac {1}{16} e^8 x^2 \log ^2\left (\frac {3}{x}\right )+\frac {1}{8} \int \left (16 e^4+e^8 x (-1+2 \log (3))\right ) \log \left (\frac {3}{x}\right ) \, dx+\frac {1}{8} e^8 \int x \log \left (\frac {3}{x}\right ) \, dx\\ &=-2 e^4 x+\frac {e^8 x^2}{32}-\frac {1}{16} \left (16-e^4 x\right )^2 \log (3)+\frac {1}{16} e^8 x^2 \log ^2(3)+\frac {1}{16} e^8 x^2 \log \left (\frac {3}{x}\right )+\frac {1}{16} \left (32 e^4 x-e^8 x^2 (1-\log (9))\right ) \log \left (\frac {3}{x}\right )+\frac {1}{16} e^8 x^2 \log ^2\left (\frac {3}{x}\right )+\frac {1}{8} \int \frac {1}{2} e^4 \left (32-e^4 x (1-\log (9))\right ) \, dx\\ &=-2 e^4 x+\frac {e^8 x^2}{32}-\frac {1}{16} \left (16-e^4 x\right )^2 \log (3)+\frac {1}{16} e^8 x^2 \log ^2(3)-\frac {\left (32-e^4 x (1-\log (9))\right )^2}{32 (1-\log (9))}+\frac {1}{16} e^8 x^2 \log \left (\frac {3}{x}\right )+\frac {1}{16} \left (32 e^4 x-e^8 x^2 (1-\log (9))\right ) \log \left (\frac {3}{x}\right )+\frac {1}{16} e^8 x^2 \log ^2\left (\frac {3}{x}\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.10, size = 76, normalized size = 2.62 \begin {gather*} \frac {1}{32} e^4 x \left (64 \log (3)+e^4 x \log (3)-e^4 x \log (9)+e^4 x \log ^2(9)+\left (64+e^4 x (1+\log (81))\right ) \log \left (\frac {3}{x}\right )+e^4 x \log \left (\frac {1}{x}\right ) \left (-1+\log (9)+2 \log \left (\frac {3}{x}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16*E^4 + (16*E^4 - E^8*x)*Log[3] + E^8*x*Log[3]^2 + (16*E^4 - E^8*x + 2*E^8*x*Log[3])*Log[3/x] + E
^8*x*Log[3/x]^2)/8,x]

[Out]

(E^4*x*(64*Log[3] + E^4*x*Log[3] - E^4*x*Log[9] + E^4*x*Log[9]^2 + (64 + E^4*x*(1 + Log[81]))*Log[3/x] + E^4*x
*Log[x^(-1)]*(-1 + Log[9] + 2*Log[3/x])))/32

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fricas [A]  time = 0.81, size = 56, normalized size = 1.93 \begin {gather*} \frac {1}{16} \, x^{2} e^{8} \log \relax (3)^{2} + \frac {1}{16} \, x^{2} e^{8} \log \left (\frac {3}{x}\right )^{2} + 2 \, x e^{4} \log \relax (3) + \frac {1}{8} \, {\left (x^{2} e^{8} \log \relax (3) + 16 \, x e^{4}\right )} \log \left (\frac {3}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*x*exp(4)^2*log(3/x)^2+1/8*(2*x*exp(4)^2*log(3)-x*exp(4)^2+16*exp(4))*log(3/x)+1/8*x*exp(4)^2*log
(3)^2+1/8*(-x*exp(4)^2+16*exp(4))*log(3)-2*exp(4),x, algorithm="fricas")

[Out]

1/16*x^2*e^8*log(3)^2 + 1/16*x^2*e^8*log(3/x)^2 + 2*x*e^4*log(3) + 1/8*(x^2*e^8*log(3) + 16*x*e^4)*log(3/x)

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giac [B]  time = 0.19, size = 118, normalized size = 4.07 \begin {gather*} \frac {1}{16} \, x^{2} e^{8} \log \relax (3)^{2} + \frac {1}{32} \, {\left (2 \, e^{8} \log \relax (3) + \frac {64 \, e^{4}}{x} - e^{8}\right )} x^{2} + \frac {1}{32} \, {\left (2 \, x^{2} \log \left (\frac {3}{x}\right )^{2} + 2 \, x^{2} \log \left (\frac {3}{x}\right ) + x^{2}\right )} e^{8} - 2 \, x e^{4} - \frac {1}{16} \, {\left (x^{2} e^{8} - 32 \, x e^{4}\right )} \log \relax (3) + \frac {1}{16} \, {\left (2 \, x^{2} e^{8} \log \relax (3) - x^{2} e^{8} + 32 \, x e^{4}\right )} \log \left (\frac {3}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*x*exp(4)^2*log(3/x)^2+1/8*(2*x*exp(4)^2*log(3)-x*exp(4)^2+16*exp(4))*log(3/x)+1/8*x*exp(4)^2*log
(3)^2+1/8*(-x*exp(4)^2+16*exp(4))*log(3)-2*exp(4),x, algorithm="giac")

[Out]

1/16*x^2*e^8*log(3)^2 + 1/32*(2*e^8*log(3) + 64*e^4/x - e^8)*x^2 + 1/32*(2*x^2*log(3/x)^2 + 2*x^2*log(3/x) + x
^2)*e^8 - 2*x*e^4 - 1/16*(x^2*e^8 - 32*x*e^4)*log(3) + 1/16*(2*x^2*e^8*log(3) - x^2*e^8 + 32*x*e^4)*log(3/x)

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maple [B]  time = 0.06, size = 61, normalized size = 2.10

method	result	size
risch	\(2 \,{\mathrm e}^{4} x \ln \relax (3)+\frac {{\mathrm e}^{8} \ln \relax (3) x^{2} \ln \left (\frac {3}{x}\right )}{8}+2 \,{\mathrm e}^{4} \ln \left (\frac {3}{x}\right ) x +\frac {{\mathrm e}^{8} \ln \relax (3)^{2} x^{2}}{16}+\frac {{\mathrm e}^{8} x^{2} \ln \left (\frac {3}{x}\right )^{2}}{16}\)	\(61\)
default	\(2 \,{\mathrm e}^{4} x \ln \relax (3)+\frac {{\mathrm e}^{8} \ln \relax (3) x^{2} \ln \left (\frac {3}{x}\right )}{8}+2 \,{\mathrm e}^{4} \ln \left (\frac {3}{x}\right ) x +\frac {{\mathrm e}^{8} \ln \relax (3)^{2} x^{2}}{16}+\frac {{\mathrm e}^{8} x^{2} \ln \left (\frac {3}{x}\right )^{2}}{16}\)	\(67\)
norman	\(2 \,{\mathrm e}^{4} x \ln \relax (3)+\frac {{\mathrm e}^{8} \ln \relax (3) x^{2} \ln \left (\frac {3}{x}\right )}{8}+2 \,{\mathrm e}^{4} \ln \left (\frac {3}{x}\right ) x +\frac {{\mathrm e}^{8} \ln \relax (3)^{2} x^{2}}{16}+\frac {{\mathrm e}^{8} x^{2} \ln \left (\frac {3}{x}\right )^{2}}{16}\)	\(67\)
derivativedivides	\(\frac {{\mathrm e}^{8} \ln \relax (3)^{2} x^{2}}{16}-\frac {9 \,{\mathrm e}^{8} \ln \relax (3) \left (-\frac {x^{2} \ln \left (\frac {3}{x}\right )}{18}-\frac {x^{2}}{36}\right )}{4}-\frac {9 \,{\mathrm e}^{8} \left (-\frac {x^{2} \ln \left (\frac {3}{x}\right )^{2}}{18}-\frac {x^{2} \ln \left (\frac {3}{x}\right )}{18}-\frac {x^{2}}{36}\right )}{8}-\frac {{\mathrm e}^{8} \ln \relax (3) x^{2}}{16}+2 \,{\mathrm e}^{4} x \ln \relax (3)+\frac {9 \,{\mathrm e}^{8} \left (-\frac {x^{2} \ln \left (\frac {3}{x}\right )}{18}-\frac {x^{2}}{36}\right )}{8}-6 \,{\mathrm e}^{4} \left (-\frac {x \ln \left (\frac {3}{x}\right )}{3}-\frac {x}{3}\right )-2 x \,{\mathrm e}^{4}\)	\(139\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/8*x*exp(4)^2*ln(3/x)^2+1/8*(2*x*exp(4)^2*ln(3)-x*exp(4)^2+16*exp(4))*ln(3/x)+1/8*x*exp(4)^2*ln(3)^2+1/8*
(-x*exp(4)^2+16*exp(4))*ln(3)-2*exp(4),x,method=_RETURNVERBOSE)

[Out]

2*exp(4)*x*ln(3)+1/8*exp(8)*ln(3)*x^2*ln(3/x)+2*exp(4)*ln(3/x)*x+1/16*exp(8)*ln(3)^2*x^2+1/16*exp(8)*x^2*ln(3/
x)^2

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maxima [B]  time = 0.36, size = 108, normalized size = 3.72 \begin {gather*} \frac {1}{16} \, x^{2} e^{8} \log \relax (3)^{2} + \frac {1}{16} \, x^{2} e^{8} \log \left (\frac {3}{x}\right )^{2} + \frac {1}{32} \, {\left (2 \, e^{8} \log \relax (3) - e^{8}\right )} x^{2} + \frac {1}{32} \, {\left (2 \, x^{2} \log \left (\frac {3}{x}\right ) + x^{2}\right )} e^{8} - \frac {1}{16} \, {\left (x^{2} e^{8} - 32 \, x e^{4}\right )} \log \relax (3) + \frac {1}{16} \, {\left (2 \, x^{2} e^{8} \log \relax (3) - x^{2} e^{8} + 32 \, x e^{4}\right )} \log \left (\frac {3}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*x*exp(4)^2*log(3/x)^2+1/8*(2*x*exp(4)^2*log(3)-x*exp(4)^2+16*exp(4))*log(3/x)+1/8*x*exp(4)^2*log
(3)^2+1/8*(-x*exp(4)^2+16*exp(4))*log(3)-2*exp(4),x, algorithm="maxima")

[Out]

1/16*x^2*e^8*log(3)^2 + 1/16*x^2*e^8*log(3/x)^2 + 1/32*(2*e^8*log(3) - e^8)*x^2 + 1/32*(2*x^2*log(3/x) + x^2)*
e^8 - 1/16*(x^2*e^8 - 32*x*e^4)*log(3) + 1/16*(2*x^2*e^8*log(3) - x^2*e^8 + 32*x*e^4)*log(3/x)

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mupad [B]  time = 3.24, size = 31, normalized size = 1.07 \begin {gather*} \frac {x\,{\mathrm {e}}^4\,\left (\ln \left (\frac {1}{x}\right )+2\,\ln \relax (3)\right )\,\left (x\,\ln \left (\frac {1}{x}\right )\,{\mathrm {e}}^4+2\,x\,{\mathrm {e}}^4\,\ln \relax (3)+32\right )}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(3/x)*(16*exp(4) - x*exp(8) + 2*x*exp(8)*log(3)))/8 - 2*exp(4) + (log(3)*(16*exp(4) - x*exp(8)))/8 + (
x*exp(8)*log(3)^2)/8 + (x*exp(8)*log(3/x)^2)/8,x)

[Out]

(x*exp(4)*(log(1/x) + 2*log(3))*(x*log(1/x)*exp(4) + 2*x*exp(4)*log(3) + 32))/16

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sympy [B]  time = 0.18, size = 61, normalized size = 2.10 \begin {gather*} \frac {x^{2} e^{8} \log {\left (\frac {3}{x} \right )}^{2}}{16} + \frac {x^{2} e^{8} \log {\relax (3 )}^{2}}{16} + 2 x e^{4} \log {\relax (3 )} + \left (\frac {x^{2} e^{8} \log {\relax (3 )}}{8} + 2 x e^{4}\right ) \log {\left (\frac {3}{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*x*exp(4)**2*ln(3/x)**2+1/8*(2*x*exp(4)**2*ln(3)-x*exp(4)**2+16*exp(4))*ln(3/x)+1/8*x*exp(4)**2*l
n(3)**2+1/8*(-x*exp(4)**2+16*exp(4))*ln(3)-2*exp(4),x)

[Out]

x**2*exp(8)*log(3/x)**2/16 + x**2*exp(8)*log(3)**2/16 + 2*x*exp(4)*log(3) + (x**2*exp(8)*log(3)/8 + 2*x*exp(4)
)*log(3/x)

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