3.40.29 \(\int \frac {-1+16 x^2+4 x^3-4 x^4+(8 x+2 x^2-6 x^3) \log (x)-2 x^2 \log ^2(x)}{x} \, dx\)

Optimal. Leaf size=22 \[ 11+\log (5)-\log (x)-(4+x-x (x+\log (x)))^2 \]

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Rubi [B]  time = 0.07, antiderivative size = 51, normalized size of antiderivative = 2.32, number of steps used = 11, number of rules used = 5, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.106, Rules used = {14, 2356, 2295, 2304, 2305} \begin {gather*} -x^4+2 x^3-2 x^3 \log (x)+7 x^2-x^2 \log ^2(x)+2 x^2 \log (x)-8 x+8 x \log (x)-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + 16*x^2 + 4*x^3 - 4*x^4 + (8*x + 2*x^2 - 6*x^3)*Log[x] - 2*x^2*Log[x]^2)/x,x]

[Out]

-8*x + 7*x^2 + 2*x^3 - x^4 - Log[x] + 8*x*Log[x] + 2*x^2*Log[x] - 2*x^3*Log[x] - x^2*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 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 2356

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Polyx_), x_Symbol] :> Int[ExpandIntegrand[Polyx*(a + b*Log[c*
x^n])^p, x], x] /; FreeQ[{a, b, c, n, p}, x] && PolynomialQ[Polyx, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {-1+16 x^2+4 x^3-4 x^4}{x}-2 (1+x) (-4+3 x) \log (x)-2 x \log ^2(x)\right ) \, dx\\ &=-(2 \int (1+x) (-4+3 x) \log (x) \, dx)-2 \int x \log ^2(x) \, dx+\int \frac {-1+16 x^2+4 x^3-4 x^4}{x} \, dx\\ &=-x^2 \log ^2(x)+2 \int x \log (x) \, dx-2 \int \left (-4 \log (x)-x \log (x)+3 x^2 \log (x)\right ) \, dx+\int \left (-\frac {1}{x}+16 x+4 x^2-4 x^3\right ) \, dx\\ &=\frac {15 x^2}{2}+\frac {4 x^3}{3}-x^4-\log (x)+x^2 \log (x)-x^2 \log ^2(x)+2 \int x \log (x) \, dx-6 \int x^2 \log (x) \, dx+8 \int \log (x) \, dx\\ &=-8 x+7 x^2+2 x^3-x^4-\log (x)+8 x \log (x)+2 x^2 \log (x)-2 x^3 \log (x)-x^2 \log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.01, size = 51, normalized size = 2.32 \begin {gather*} -8 x+7 x^2+2 x^3-x^4-\log (x)+8 x \log (x)+2 x^2 \log (x)-2 x^3 \log (x)-x^2 \log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + 16*x^2 + 4*x^3 - 4*x^4 + (8*x + 2*x^2 - 6*x^3)*Log[x] - 2*x^2*Log[x]^2)/x,x]

[Out]

-8*x + 7*x^2 + 2*x^3 - x^4 - Log[x] + 8*x*Log[x] + 2*x^2*Log[x] - 2*x^3*Log[x] - x^2*Log[x]^2

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fricas [B]  time = 0.60, size = 47, normalized size = 2.14 \begin {gather*} -x^{4} - x^{2} \log \relax (x)^{2} + 2 \, x^{3} + 7 \, x^{2} - {\left (2 \, x^{3} - 2 \, x^{2} - 8 \, x + 1\right )} \log \relax (x) - 8 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^2*log(x)^2+(-6*x^3+2*x^2+8*x)*log(x)-4*x^4+4*x^3+16*x^2-1)/x,x, algorithm="fricas")

[Out]

-x^4 - x^2*log(x)^2 + 2*x^3 + 7*x^2 - (2*x^3 - 2*x^2 - 8*x + 1)*log(x) - 8*x

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giac [B]  time = 0.16, size = 48, normalized size = 2.18 \begin {gather*} -x^{4} - x^{2} \log \relax (x)^{2} + 2 \, x^{3} + 7 \, x^{2} - 2 \, {\left (x^{3} - x^{2} - 4 \, x\right )} \log \relax (x) - 8 \, x - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^2*log(x)^2+(-6*x^3+2*x^2+8*x)*log(x)-4*x^4+4*x^3+16*x^2-1)/x,x, algorithm="giac")

[Out]

-x^4 - x^2*log(x)^2 + 2*x^3 + 7*x^2 - 2*(x^3 - x^2 - 4*x)*log(x) - 8*x - log(x)

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maple [B]  time = 0.02, size = 50, normalized size = 2.27

method	result	size
risch	\(-x^{2} \ln \relax (x )^{2}+\left (-2 x^{3}+2 x^{2}+8 x \right ) \ln \relax (x )-x^{4}+2 x^{3}+7 x^{2}-8 x -\ln \relax (x )\)	\(50\)
default	\(-x^{2} \ln \relax (x )^{2}+2 x^{2} \ln \relax (x )+7 x^{2}-2 x^{3} \ln \relax (x )+2 x^{3}-x^{4}+8 x \ln \relax (x )-8 x -\ln \relax (x )\)	\(52\)
norman	\(-x^{2} \ln \relax (x )^{2}+2 x^{2} \ln \relax (x )+7 x^{2}-2 x^{3} \ln \relax (x )+2 x^{3}-x^{4}+8 x \ln \relax (x )-8 x -\ln \relax (x )\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x^2*ln(x)^2+(-6*x^3+2*x^2+8*x)*ln(x)-4*x^4+4*x^3+16*x^2-1)/x,x,method=_RETURNVERBOSE)

[Out]

-x^2*ln(x)^2+(-2*x^3+2*x^2+8*x)*ln(x)-x^4+2*x^3+7*x^2-8*x-ln(x)

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maxima [B]  time = 0.34, size = 58, normalized size = 2.64 \begin {gather*} -x^{4} - 2 \, x^{3} \log \relax (x) - \frac {1}{2} \, {\left (2 \, \log \relax (x)^{2} - 2 \, \log \relax (x) + 1\right )} x^{2} + 2 \, x^{3} + x^{2} \log \relax (x) + \frac {15}{2} \, x^{2} + 8 \, x \log \relax (x) - 8 \, x - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^2*log(x)^2+(-6*x^3+2*x^2+8*x)*log(x)-4*x^4+4*x^3+16*x^2-1)/x,x, algorithm="maxima")

[Out]

-x^4 - 2*x^3*log(x) - 1/2*(2*log(x)^2 - 2*log(x) + 1)*x^2 + 2*x^3 + x^2*log(x) + 15/2*x^2 + 8*x*log(x) - 8*x -
log(x)

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mupad [B]  time = 2.24, size = 51, normalized size = 2.32 \begin {gather*} -x^4-2\,x^3\,\ln \relax (x)+2\,x^3-x^2\,{\ln \relax (x)}^2+2\,x^2\,\ln \relax (x)+7\,x^2+8\,x\,\ln \relax (x)-8\,x-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x^2*log(x)^2 - 16*x^2 - 4*x^3 + 4*x^4 - log(x)*(8*x + 2*x^2 - 6*x^3) + 1)/x,x)

[Out]

2*x^2*log(x) - log(x) - 8*x - 2*x^3*log(x) - x^2*log(x)^2 + 8*x*log(x) + 7*x^2 + 2*x^3 - x^4

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sympy [B]  time = 0.18, size = 44, normalized size = 2.00 \begin {gather*} - x^{4} + 2 x^{3} - x^{2} \log {\relax (x )}^{2} + 7 x^{2} - 8 x + \left (- 2 x^{3} + 2 x^{2} + 8 x\right ) \log {\relax (x )} - \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x**2*ln(x)**2+(-6*x**3+2*x**2+8*x)*ln(x)-4*x**4+4*x**3+16*x**2-1)/x,x)

[Out]

-x**4 + 2*x**3 - x**2*log(x)**2 + 7*x**2 - 8*x + (-2*x**3 + 2*x**2 + 8*x)*log(x) - log(x)

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