3.609 \(\int x^m \log ^2(x) \, dx\)

Optimal. Leaf size=42 \[ \frac {2 x^{m+1}}{(m+1)^3}+\frac {x^{m+1} \log ^2(x)}{m+1}-\frac {2 x^{m+1} \log (x)}{(m+1)^2} \]

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Rubi [A]  time = 0.02, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2305, 2304} \[ \frac {2 x^{m+1}}{(m+1)^3}+\frac {x^{m+1} \log ^2(x)}{m+1}-\frac {2 x^{m+1} \log (x)}{(m+1)^2} \]

Antiderivative was successfully verified.

[In]

Int[x^m*Log[x]^2,x]

[Out]

(2*x^(1 + m))/(1 + m)^3 - (2*x^(1 + m)*Log[x])/(1 + m)^2 + (x^(1 + m)*Log[x]^2)/(1 + m)

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]

Rubi steps

\begin {align*} \int x^m \log ^2(x) \, dx &=\frac {x^{1+m} \log ^2(x)}{1+m}-\frac {2 \int x^m \log (x) \, dx}{1+m}\\ &=\frac {2 x^{1+m}}{(1+m)^3}-\frac {2 x^{1+m} \log (x)}{(1+m)^2}+\frac {x^{1+m} \log ^2(x)}{1+m}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 30, normalized size = 0.71 \[ \frac {x^{m+1} \left ((m+1)^2 \log ^2(x)-2 (m+1) \log (x)+2\right )}{(m+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*Log[x]^2,x]

[Out]

(x^(1 + m)*(2 - 2*(1 + m)*Log[x] + (1 + m)^2*Log[x]^2))/(1 + m)^3

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^m \log ^2(x) \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[x^m*Log[x]^2,x]

[Out]

Could not integrate

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fricas [A]  time = 0.92, size = 45, normalized size = 1.07 \[ \frac {{\left ({\left (m^{2} + 2 \, m + 1\right )} x \log \relax (x)^{2} - 2 \, {\left (m + 1\right )} x \log \relax (x) + 2 \, x\right )} x^{m}}{m^{3} + 3 \, m^{2} + 3 \, m + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*log(x)^2,x, algorithm="fricas")

[Out]

((m^2 + 2*m + 1)*x*log(x)^2 - 2*(m + 1)*x*log(x) + 2*x)*x^m/(m^3 + 3*m^2 + 3*m + 1)

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giac [A]  time = 0.64, size = 84, normalized size = 2.00 \[ -\frac {2 \, m x x^{m} \log \relax (x)}{{\left (m^{2} + 2 \, m + 1\right )} {\left (m + 1\right )}} + \frac {x^{m + 1} \log \relax (x)^{2}}{m + 1} - \frac {2 \, x x^{m} \log \relax (x)}{{\left (m^{2} + 2 \, m + 1\right )} {\left (m + 1\right )}} + \frac {2 \, x x^{m}}{{\left (m^{2} + 2 \, m + 1\right )} {\left (m + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*log(x)^2,x, algorithm="giac")

[Out]

-2*m*x*x^m*log(x)/((m^2 + 2*m + 1)*(m + 1)) + x^(m + 1)*log(x)^2/(m + 1) - 2*x*x^m*log(x)/((m^2 + 2*m + 1)*(m
+ 1)) + 2*x*x^m/((m^2 + 2*m + 1)*(m + 1))

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maple [A]  time = 0.04, size = 41, normalized size = 0.98

method	result	size
risch	\(\frac {x \left (m^{2} \ln \relax (x )^{2}+2 m \ln \relax (x )^{2}-2 m \ln \relax (x )+\ln \relax (x )^{2}-2 \ln \relax (x )+2\right ) x^{m}}{\left (1+m \right )^{3}}\)	\(41\)
norman	\(\frac {x \ln \relax (x )^{2} {\mathrm e}^{m \ln \relax (x )}}{1+m}+\frac {2 x \,{\mathrm e}^{m \ln \relax (x )}}{m^{3}+3 m^{2}+3 m +1}-\frac {2 x \ln \relax (x ) {\mathrm e}^{m \ln \relax (x )}}{m^{2}+2 m +1}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

x*(m^2*ln(x)^2+2*m*ln(x)^2-2*m*ln(x)+ln(x)^2-2*ln(x)+2)/(1+m)^3*x^m

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maxima [A]  time = 0.42, size = 42, normalized size = 1.00 \[ \frac {x^{m + 1} \log \relax (x)^{2}}{m + 1} - \frac {2 \, x^{m + 1} \log \relax (x)}{{\left (m + 1\right )}^{2}} + \frac {2 \, x^{m + 1}}{{\left (m + 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*log(x)^2,x, algorithm="maxima")

[Out]

x^(m + 1)*log(x)^2/(m + 1) - 2*x^(m + 1)*log(x)/(m + 1)^2 + 2*x^(m + 1)/(m + 1)^3

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mupad [B]  time = 0.35, size = 43, normalized size = 1.02 \[ \left \{\begin {array}{cl} \frac {{\ln \relax (x)}^3}{3} & \text {\ if\ \ }m=-1\\ \frac {x^{m+1}\,\left ({\ln \relax (x)}^2\,{\left (m+1\right )}^2-2\,\ln \relax (x)\,\left (m+1\right )+2\right )}{{\left (m+1\right )}^3} & \text {\ if\ \ }m\neq -1 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*log(x)^2,x)

[Out]

piecewise(m == -1, log(x)^3/3, m ~= -1, (x^(m + 1)*(- 2*log(x)*(m + 1) + log(x)^2*(m + 1)^2 + 2))/(m + 1)^3)

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sympy [A]  time = 1.57, size = 155, normalized size = 3.69 \[ \begin {cases} \frac {m^{2} x x^{m} \log {\relax (x )}^{2}}{m^{3} + 3 m^{2} + 3 m + 1} + \frac {2 m x x^{m} \log {\relax (x )}^{2}}{m^{3} + 3 m^{2} + 3 m + 1} - \frac {2 m x x^{m} \log {\relax (x )}}{m^{3} + 3 m^{2} + 3 m + 1} + \frac {x x^{m} \log {\relax (x )}^{2}}{m^{3} + 3 m^{2} + 3 m + 1} - \frac {2 x x^{m} \log {\relax (x )}}{m^{3} + 3 m^{2} + 3 m + 1} + \frac {2 x x^{m}}{m^{3} + 3 m^{2} + 3 m + 1} & \text {for}\: m \neq -1 \\\frac {\log {\relax (x )}^{3}}{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*ln(x)**2,x)

[Out]

Piecewise((m**2*x*x**m*log(x)**2/(m**3 + 3*m**2 + 3*m + 1) + 2*m*x*x**m*log(x)**2/(m**3 + 3*m**2 + 3*m + 1) -
2*m*x*x**m*log(x)/(m**3 + 3*m**2 + 3*m + 1) + x*x**m*log(x)**2/(m**3 + 3*m**2 + 3*m + 1) - 2*x*x**m*log(x)/(m*
*3 + 3*m**2 + 3*m + 1) + 2*x*x**m/(m**3 + 3*m**2 + 3*m + 1), Ne(m, -1)), (log(x)**3/3, True))

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