3.31.65 \(\int \frac {1}{5} (-55-4 x+16 x^2-2 x^3+(48+32 x-2 x^2-8 x^3+2 x^4) \log (x)+(64-48 x^2+5 x^4) \log ^2(x)) \, dx\)

Optimal. Leaf size=24 \[ \frac {1}{5} x \left (x^2+\left (5-x+\left (-8+x^2\right ) \log (x)\right )^2\right ) \]

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Rubi [B]  time = 0.13, antiderivative size = 77, normalized size of antiderivative = 3.21, number of steps used = 17, number of rules used = 6, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {12, 2356, 2295, 2304, 2296, 2305} \begin {gather*} \frac {1}{5} x^5 \log ^2(x)-\frac {2}{5} x^4 \log (x)+\frac {2 x^3}{5}-\frac {16}{5} x^3 \log ^2(x)+2 x^3 \log (x)-2 x^2+\frac {16}{5} x^2 \log (x)+5 x+\frac {64}{5} x \log ^2(x)-16 x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-55 - 4*x + 16*x^2 - 2*x^3 + (48 + 32*x - 2*x^2 - 8*x^3 + 2*x^4)*Log[x] + (64 - 48*x^2 + 5*x^4)*Log[x]^2)
/5,x]

[Out]

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

Rule 12

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

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 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

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} &=\frac {1}{5} \int \left (-55-4 x+16 x^2-2 x^3+\left (48+32 x-2 x^2-8 x^3+2 x^4\right ) \log (x)+\left (64-48 x^2+5 x^4\right ) \log ^2(x)\right ) \, dx\\ &=-11 x-\frac {2 x^2}{5}+\frac {16 x^3}{15}-\frac {x^4}{10}+\frac {1}{5} \int \left (48+32 x-2 x^2-8 x^3+2 x^4\right ) \log (x) \, dx+\frac {1}{5} \int \left (64-48 x^2+5 x^4\right ) \log ^2(x) \, dx\\ &=-11 x-\frac {2 x^2}{5}+\frac {16 x^3}{15}-\frac {x^4}{10}+\frac {1}{5} \int \left (48 \log (x)+32 x \log (x)-2 x^2 \log (x)-8 x^3 \log (x)+2 x^4 \log (x)\right ) \, dx+\frac {1}{5} \int \left (64 \log ^2(x)-48 x^2 \log ^2(x)+5 x^4 \log ^2(x)\right ) \, dx\\ &=-11 x-\frac {2 x^2}{5}+\frac {16 x^3}{15}-\frac {x^4}{10}-\frac {2}{5} \int x^2 \log (x) \, dx+\frac {2}{5} \int x^4 \log (x) \, dx-\frac {8}{5} \int x^3 \log (x) \, dx+\frac {32}{5} \int x \log (x) \, dx+\frac {48}{5} \int \log (x) \, dx-\frac {48}{5} \int x^2 \log ^2(x) \, dx+\frac {64}{5} \int \log ^2(x) \, dx+\int x^4 \log ^2(x) \, dx\\ &=-\frac {103 x}{5}-2 x^2+\frac {10 x^3}{9}-\frac {2 x^5}{125}+\frac {48}{5} x \log (x)+\frac {16}{5} x^2 \log (x)-\frac {2}{15} x^3 \log (x)-\frac {2}{5} x^4 \log (x)+\frac {2}{25} x^5 \log (x)+\frac {64}{5} x \log ^2(x)-\frac {16}{5} x^3 \log ^2(x)+\frac {1}{5} x^5 \log ^2(x)-\frac {2}{5} \int x^4 \log (x) \, dx+\frac {32}{5} \int x^2 \log (x) \, dx-\frac {128}{5} \int \log (x) \, dx\\ &=5 x-2 x^2+\frac {2 x^3}{5}-16 x \log (x)+\frac {16}{5} x^2 \log (x)+2 x^3 \log (x)-\frac {2}{5} x^4 \log (x)+\frac {64}{5} x \log ^2(x)-\frac {16}{5} x^3 \log ^2(x)+\frac {1}{5} x^5 \log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.01, size = 77, normalized size = 3.21 \begin {gather*} 5 x-2 x^2+\frac {2 x^3}{5}-16 x \log (x)+\frac {16}{5} x^2 \log (x)+2 x^3 \log (x)-\frac {2}{5} x^4 \log (x)+\frac {64}{5} x \log ^2(x)-\frac {16}{5} x^3 \log ^2(x)+\frac {1}{5} x^5 \log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-55 - 4*x + 16*x^2 - 2*x^3 + (48 + 32*x - 2*x^2 - 8*x^3 + 2*x^4)*Log[x] + (64 - 48*x^2 + 5*x^4)*Log
[x]^2)/5,x]

[Out]

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

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fricas [B]  time = 0.56, size = 53, normalized size = 2.21 \begin {gather*} \frac {2}{5} \, x^{3} + \frac {1}{5} \, {\left (x^{5} - 16 \, x^{3} + 64 \, x\right )} \log \relax (x)^{2} - 2 \, x^{2} - \frac {2}{5} \, {\left (x^{4} - 5 \, x^{3} - 8 \, x^{2} + 40 \, x\right )} \log \relax (x) + 5 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(5*x^4-48*x^2+64)*log(x)^2+1/5*(2*x^4-8*x^3-2*x^2+32*x+48)*log(x)-2/5*x^3+16/5*x^2-4/5*x-11,x, a
lgorithm="fricas")

[Out]

2/5*x^3 + 1/5*(x^5 - 16*x^3 + 64*x)*log(x)^2 - 2*x^2 - 2/5*(x^4 - 5*x^3 - 8*x^2 + 40*x)*log(x) + 5*x

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giac [B]  time = 0.23, size = 65, normalized size = 2.71 \begin {gather*} \frac {1}{5} \, x^{5} \log \relax (x)^{2} - \frac {2}{5} \, x^{4} \log \relax (x) - \frac {16}{5} \, x^{3} \log \relax (x)^{2} + 2 \, x^{3} \log \relax (x) + \frac {2}{5} \, x^{3} + \frac {16}{5} \, x^{2} \log \relax (x) + \frac {64}{5} \, x \log \relax (x)^{2} - 2 \, x^{2} - 16 \, x \log \relax (x) + 5 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(5*x^4-48*x^2+64)*log(x)^2+1/5*(2*x^4-8*x^3-2*x^2+32*x+48)*log(x)-2/5*x^3+16/5*x^2-4/5*x-11,x, a
lgorithm="giac")

[Out]

1/5*x^5*log(x)^2 - 2/5*x^4*log(x) - 16/5*x^3*log(x)^2 + 2*x^3*log(x) + 2/5*x^3 + 16/5*x^2*log(x) + 64/5*x*log(
x)^2 - 2*x^2 - 16*x*log(x) + 5*x

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maple [B]  time = 0.04, size = 66, normalized size = 2.75

method	result	size
default	\(5 x -2 x^{2}+\frac {2 x^{3}}{5}+\frac {x^{5} \ln \relax (x )^{2}}{5}-\frac {16 x^{3} \ln \relax (x )^{2}}{5}+2 x^{3} \ln \relax (x )+\frac {64 x \ln \relax (x )^{2}}{5}-16 x \ln \relax (x )-\frac {2 x^{4} \ln \relax (x )}{5}+\frac {16 x^{2} \ln \relax (x )}{5}\)	\(66\)
norman	\(5 x -2 x^{2}+\frac {2 x^{3}}{5}+\frac {x^{5} \ln \relax (x )^{2}}{5}-\frac {16 x^{3} \ln \relax (x )^{2}}{5}+2 x^{3} \ln \relax (x )+\frac {64 x \ln \relax (x )^{2}}{5}-16 x \ln \relax (x )-\frac {2 x^{4} \ln \relax (x )}{5}+\frac {16 x^{2} \ln \relax (x )}{5}\)	\(66\)
risch	\(5 x -2 x^{2}+\frac {2 x^{3}}{5}+\frac {x^{5} \ln \relax (x )^{2}}{5}-\frac {16 x^{3} \ln \relax (x )^{2}}{5}+2 x^{3} \ln \relax (x )+\frac {64 x \ln \relax (x )^{2}}{5}-16 x \ln \relax (x )-\frac {2 x^{4} \ln \relax (x )}{5}+\frac {16 x^{2} \ln \relax (x )}{5}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(5*x^4-48*x^2+64)*ln(x)^2+1/5*(2*x^4-8*x^3-2*x^2+32*x+48)*ln(x)-2/5*x^3+16/5*x^2-4/5*x-11,x,method=_RE
TURNVERBOSE)

[Out]

5*x-2*x^2+2/5*x^3+1/5*x^5*ln(x)^2-16/5*x^3*ln(x)^2+2*x^3*ln(x)+64/5*x*ln(x)^2-16*x*ln(x)-2/5*x^4*ln(x)+16/5*x^
2*ln(x)

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maxima [B]  time = 0.70, size = 94, normalized size = 3.92 \begin {gather*} \frac {1}{125} \, {\left (25 \, \log \relax (x)^{2} - 10 \, \log \relax (x) + 2\right )} x^{5} - \frac {2}{125} \, x^{5} - \frac {16}{45} \, {\left (9 \, \log \relax (x)^{2} - 6 \, \log \relax (x) + 2\right )} x^{3} + \frac {10}{9} \, x^{3} + \frac {64}{5} \, {\left (\log \relax (x)^{2} - 2 \, \log \relax (x) + 2\right )} x - 2 \, x^{2} + \frac {2}{75} \, {\left (3 \, x^{5} - 15 \, x^{4} - 5 \, x^{3} + 120 \, x^{2} + 360 \, x\right )} \log \relax (x) - \frac {103}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(5*x^4-48*x^2+64)*log(x)^2+1/5*(2*x^4-8*x^3-2*x^2+32*x+48)*log(x)-2/5*x^3+16/5*x^2-4/5*x-11,x, a
lgorithm="maxima")

[Out]

1/125*(25*log(x)^2 - 10*log(x) + 2)*x^5 - 2/125*x^5 - 16/45*(9*log(x)^2 - 6*log(x) + 2)*x^3 + 10/9*x^3 + 64/5*
(log(x)^2 - 2*log(x) + 2)*x - 2*x^2 + 2/75*(3*x^5 - 15*x^4 - 5*x^3 + 120*x^2 + 360*x)*log(x) - 103/5*x

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mupad [B]  time = 1.84, size = 59, normalized size = 2.46 \begin {gather*} \frac {x\,\left (x^4\,{\ln \relax (x)}^2-2\,x^3\,\ln \relax (x)-16\,x^2\,{\ln \relax (x)}^2+10\,x^2\,\ln \relax (x)+2\,x^2+16\,x\,\ln \relax (x)-10\,x+64\,{\ln \relax (x)}^2-80\,\ln \relax (x)+25\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(32*x - 2*x^2 - 8*x^3 + 2*x^4 + 48))/5 - (4*x)/5 + (log(x)^2*(5*x^4 - 48*x^2 + 64))/5 + (16*x^2)/5
- (2*x^3)/5 - 11,x)

[Out]

(x*(10*x^2*log(x) - 80*log(x) - 10*x - 2*x^3*log(x) + 64*log(x)^2 - 16*x^2*log(x)^2 + x^4*log(x)^2 + 16*x*log(
x) + 2*x^2 + 25))/5

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sympy [B]  time = 0.16, size = 61, normalized size = 2.54 \begin {gather*} \frac {2 x^{3}}{5} - 2 x^{2} + 5 x + \left (\frac {x^{5}}{5} - \frac {16 x^{3}}{5} + \frac {64 x}{5}\right ) \log {\relax (x )}^{2} + \left (- \frac {2 x^{4}}{5} + 2 x^{3} + \frac {16 x^{2}}{5} - 16 x\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(5*x**4-48*x**2+64)*ln(x)**2+1/5*(2*x**4-8*x**3-2*x**2+32*x+48)*ln(x)-2/5*x**3+16/5*x**2-4/5*x-1
1,x)

[Out]

2*x**3/5 - 2*x**2 + 5*x + (x**5/5 - 16*x**3/5 + 64*x/5)*log(x)**2 + (-2*x**4/5 + 2*x**3 + 16*x**2/5 - 16*x)*lo
g(x)

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