3.81.69 \(\int \frac {-128+144 x-28 x^2+10 x^3+2 x^4+(16 x-4 x^2+2 x^3) \log (3)+(-16 x+4 x^2-2 x^3) \log (2 x^2)}{x^3} \, dx\)

Optimal. Leaf size=25 \[ \left (5-\frac {2 (4-x)}{x}+x+\log (3)-\log \left (2 x^2\right )\right )^2 \]

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Rubi [B]  time = 0.12, antiderivative size = 69, normalized size of antiderivative = 2.76, 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 = {14, 1628, 2357, 2295, 2304, 2301} \begin {gather*} x^2+\frac {64}{x^2}+\log ^2\left (2 x^2\right )-2 x \log \left (2 x^2\right )+\frac {16 \log \left (2 x^2\right )}{x}+4 x+\frac {32}{x}+2 x (5+\log (3))-4 (7+\log (3)) \log (x)-\frac {16 (9+\log (3))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-128 + 144*x - 28*x^2 + 10*x^3 + 2*x^4 + (16*x - 4*x^2 + 2*x^3)*Log[3] + (-16*x + 4*x^2 - 2*x^3)*Log[2*x^
2])/x^3,x]

[Out]

64/x^2 + 32/x + 4*x + x^2 + 2*x*(5 + Log[3]) - (16*(9 + Log[3]))/x - 4*(7 + Log[3])*Log[x] + (16*Log[2*x^2])/x
- 2*x*Log[2*x^2] + Log[2*x^2]^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 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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 (8-2 x+x^2\right ) \left (-8+x^2+x (7+\log (3))\right )}{x^3}-\frac {2 \left (8-2 x+x^2\right ) \log \left (2 x^2\right )}{x^2}\right ) \, dx\\ &=2 \int \frac {\left (8-2 x+x^2\right ) \left (-8+x^2+x (7+\log (3))\right )}{x^3} \, dx-2 \int \frac {\left (8-2 x+x^2\right ) \log \left (2 x^2\right )}{x^2} \, dx\\ &=2 \int \left (-\frac {64}{x^3}+x+5 \left (1+\frac {\log (3)}{5}\right )-\frac {2 (7+\log (3))}{x}+\frac {8 (9+\log (3))}{x^2}\right ) \, dx-2 \int \left (\log \left (2 x^2\right )+\frac {8 \log \left (2 x^2\right )}{x^2}-\frac {2 \log \left (2 x^2\right )}{x}\right ) \, dx\\ &=\frac {64}{x^2}+x^2+2 x (5+\log (3))-\frac {16 (9+\log (3))}{x}-4 (7+\log (3)) \log (x)-2 \int \log \left (2 x^2\right ) \, dx+4 \int \frac {\log \left (2 x^2\right )}{x} \, dx-16 \int \frac {\log \left (2 x^2\right )}{x^2} \, dx\\ &=\frac {64}{x^2}+\frac {32}{x}+4 x+x^2+2 x (5+\log (3))-\frac {16 (9+\log (3))}{x}-4 (7+\log (3)) \log (x)+\frac {16 \log \left (2 x^2\right )}{x}-2 x \log \left (2 x^2\right )+\log ^2\left (2 x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.07, size = 60, normalized size = 2.40 \begin {gather*} \frac {64}{x^2}+\frac {32}{x}+4 x+x^2+2 x (5+\log (3))+\frac {16 \left (-9+\log \left (\frac {2 x^2}{3}\right )\right )}{x}+\left (-7+\log \left (\frac {2 x^2}{3}\right )\right )^2-2 x \log \left (2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-128 + 144*x - 28*x^2 + 10*x^3 + 2*x^4 + (16*x - 4*x^2 + 2*x^3)*Log[3] + (-16*x + 4*x^2 - 2*x^3)*Lo
g[2*x^2])/x^3,x]

[Out]

64/x^2 + 32/x + 4*x + x^2 + 2*x*(5 + Log[3]) + (16*(-9 + Log[(2*x^2)/3]))/x + (-7 + Log[(2*x^2)/3])^2 - 2*x*Lo
g[2*x^2]

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fricas [B]  time = 0.98, size = 66, normalized size = 2.64 \begin {gather*} \frac {x^{4} + x^{2} \log \left (2 \, x^{2}\right )^{2} + 14 \, x^{3} + 2 \, {\left (x^{3} - 8 \, x\right )} \log \relax (3) - 2 \, {\left (x^{3} + x^{2} \log \relax (3) + 7 \, x^{2} - 8 \, x\right )} \log \left (2 \, x^{2}\right ) - 112 \, x + 64}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3+4*x^2-16*x)*log(2*x^2)+(2*x^3-4*x^2+16*x)*log(3)+2*x^4+10*x^3-28*x^2+144*x-128)/x^3,x, algo
rithm="fricas")

[Out]

(x^4 + x^2*log(2*x^2)^2 + 14*x^3 + 2*(x^3 - 8*x)*log(3) - 2*(x^3 + x^2*log(3) + 7*x^2 - 8*x)*log(2*x^2) - 112*
x + 64)/x^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left (x^{4} + 5 \, x^{3} - 14 \, x^{2} + {\left (x^{3} - 2 \, x^{2} + 8 \, x\right )} \log \relax (3) - {\left (x^{3} - 2 \, x^{2} + 8 \, x\right )} \log \left (2 \, x^{2}\right ) + 72 \, x - 64\right )}}{x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3+4*x^2-16*x)*log(2*x^2)+(2*x^3-4*x^2+16*x)*log(3)+2*x^4+10*x^3-28*x^2+144*x-128)/x^3,x, algo
rithm="giac")

[Out]

integrate(2*(x^4 + 5*x^3 - 14*x^2 + (x^3 - 2*x^2 + 8*x)*log(3) - (x^3 - 2*x^2 + 8*x)*log(2*x^2) + 72*x - 64)/x
^3, x)

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maple [B]  time = 0.05, size = 76, normalized size = 3.04

method	result	size
norman	\(\frac {64+x^{4}+x^{2} \ln \left (2 x^{2}\right )^{2}+\left (-112-16 \ln \relax (3)\right ) x +\left (14+2 \ln \relax (3)\right ) x^{3}+\left (-14-2 \ln \relax (3)\right ) x^{2} \ln \left (2 x^{2}\right )+16 x \ln \left (2 x^{2}\right )-2 x^{3} \ln \left (2 x^{2}\right )}{x^{2}}\)	\(76\)
default	\(14 x +x^{2}-28 \ln \relax (x )-\frac {112}{x}+\frac {64}{x^{2}}-2 x \ln \relax (2)+4 \ln \relax (2) \ln \relax (x )+\frac {16 \ln \relax (2)}{x}+2 x \ln \relax (3)-4 \ln \relax (3) \ln \relax (x )-\frac {16 \ln \relax (3)}{x}-2 x \ln \left (x^{2}\right )+4 \ln \relax (x ) \ln \left (x^{2}\right )-4 \ln \relax (x )^{2}+\frac {16 \ln \left (x^{2}\right )}{x}\)	\(88\)
risch	\(14 x +x^{2}-28 \ln \relax (x )-\frac {112}{x}+\frac {64}{x^{2}}-2 x \ln \relax (2)+4 \ln \relax (2) \ln \relax (x )+\frac {16 \ln \relax (2)}{x}+2 x \ln \relax (3)-4 \ln \relax (3) \ln \relax (x )-\frac {16 \ln \relax (3)}{x}-2 x \ln \left (x^{2}\right )+4 \ln \relax (x ) \ln \left (x^{2}\right )-4 \ln \relax (x )^{2}+\frac {16 \ln \left (x^{2}\right )}{x}\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^3+4*x^2-16*x)*ln(2*x^2)+(2*x^3-4*x^2+16*x)*ln(3)+2*x^4+10*x^3-28*x^2+144*x-128)/x^3,x,method=_RETUR
NVERBOSE)

[Out]

(64+x^4+x^2*ln(2*x^2)^2+(-112-16*ln(3))*x+(14+2*ln(3))*x^3+(-14-2*ln(3))*x^2*ln(2*x^2)+16*x*ln(2*x^2)-2*x^3*ln
(2*x^2))/x^2

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maxima [B]  time = 0.36, size = 67, normalized size = 2.68 \begin {gather*} x^{2} + 2 \, x \log \relax (3) - 2 \, x \log \left (2 \, x^{2}\right ) + \log \left (2 \, x^{2}\right )^{2} - 4 \, \log \relax (3) \log \relax (x) + 14 \, x - \frac {16 \, \log \relax (3)}{x} + \frac {16 \, \log \left (2 \, x^{2}\right )}{x} - \frac {112}{x} + \frac {64}{x^{2}} - 28 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3+4*x^2-16*x)*log(2*x^2)+(2*x^3-4*x^2+16*x)*log(3)+2*x^4+10*x^3-28*x^2+144*x-128)/x^3,x, algo
rithm="maxima")

[Out]

x^2 + 2*x*log(3) - 2*x*log(2*x^2) + log(2*x^2)^2 - 4*log(3)*log(x) + 14*x - 16*log(3)/x + 16*log(2*x^2)/x - 11
2/x + 64/x^2 - 28*log(x)

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mupad [B]  time = 5.13, size = 63, normalized size = 2.52 \begin {gather*} {\ln \left (2\,x^2\right )}^2-\ln \left (x^2\right )\,\left (\ln \relax (9)+14\right )+x\,\left (\ln \relax (9)-2\,\ln \left (2\,x^2\right )+14\right )+\frac {64\,x-x^2\,\left (16\,\ln \relax (3)-16\,\ln \left (2\,x^2\right )+112\right )}{x^3}+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((144*x + log(3)*(16*x - 4*x^2 + 2*x^3) - log(2*x^2)*(16*x - 4*x^2 + 2*x^3) - 28*x^2 + 10*x^3 + 2*x^4 - 128
)/x^3,x)

[Out]

log(2*x^2)^2 - log(x^2)*(log(9) + 14) + x*(log(9) - 2*log(2*x^2) + 14) + (64*x - x^2*(16*log(3) - 16*log(2*x^2
) + 112))/x^3 + x^2

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sympy [B]  time = 0.33, size = 60, normalized size = 2.40 \begin {gather*} x^{2} + x \left (2 \log {\relax (3 )} + 14\right ) - 4 \left (\log {\relax (3 )} + 7\right ) \log {\relax (x )} + \log {\left (2 x^{2} \right )}^{2} + \frac {\left (16 - 2 x^{2}\right ) \log {\left (2 x^{2} \right )}}{x} + \frac {x \left (-112 - 16 \log {\relax (3 )}\right ) + 64}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**3+4*x**2-16*x)*ln(2*x**2)+(2*x**3-4*x**2+16*x)*ln(3)+2*x**4+10*x**3-28*x**2+144*x-128)/x**3,
x)

[Out]

x**2 + x*(2*log(3) + 14) - 4*(log(3) + 7)*log(x) + log(2*x**2)**2 + (16 - 2*x**2)*log(2*x**2)/x + (x*(-112 - 1
6*log(3)) + 64)/x**2

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