3.13.38 \(\int \frac {1}{5} (-30+10 x-2 e^3 x+15 x^2+10 x \log (2 x^2)) \, dx\)

Optimal. Leaf size=30 \[ (-3+x)^2-x^2+x^2 \left (-\frac {e^3}{5}+x+\log \left (2 x^2\right )\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 36, normalized size of antiderivative = 1.20, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {6, 12, 2304} \begin {gather*} x^3+\frac {1}{5} \left (5-e^3\right ) x^2-x^2+x^2 \log \left (2 x^2\right )-6 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-30 + 10*x - 2*E^3*x + 15*x^2 + 10*x*Log[2*x^2])/5,x]

[Out]

-6*x - x^2 + ((5 - E^3)*x^2)/5 + x^3 + x^2*Log[2*x^2]

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 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
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1}{5} \left (-30+\left (10-2 e^3\right ) x+15 x^2+10 x \log \left (2 x^2\right )\right ) \, dx\\ &=\frac {1}{5} \int \left (-30+\left (10-2 e^3\right ) x+15 x^2+10 x \log \left (2 x^2\right )\right ) \, dx\\ &=-6 x+\frac {1}{5} \left (5-e^3\right ) x^2+x^3+2 \int x \log \left (2 x^2\right ) \, dx\\ &=-6 x-x^2+\frac {1}{5} \left (5-e^3\right ) x^2+x^3+x^2 \log \left (2 x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 27, normalized size = 0.90 \begin {gather*} -6 x-\frac {e^3 x^2}{5}+x^3+x^2 \log \left (2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-30 + 10*x - 2*E^3*x + 15*x^2 + 10*x*Log[2*x^2])/5,x]

[Out]

-6*x - (E^3*x^2)/5 + x^3 + x^2*Log[2*x^2]

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fricas [A]  time = 0.78, size = 24, normalized size = 0.80 \begin {gather*} x^{3} - \frac {1}{5} \, x^{2} e^{3} + x^{2} \log \left (2 \, x^{2}\right ) - 6 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*x*log(2*x^2)-2/5*x*exp(3)+3*x^2+2*x-6,x, algorithm="fricas")

[Out]

x^3 - 1/5*x^2*e^3 + x^2*log(2*x^2) - 6*x

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giac [A]  time = 0.36, size = 24, normalized size = 0.80 \begin {gather*} x^{3} - \frac {1}{5} \, x^{2} e^{3} + x^{2} \log \left (2 \, x^{2}\right ) - 6 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*x*log(2*x^2)-2/5*x*exp(3)+3*x^2+2*x-6,x, algorithm="giac")

[Out]

x^3 - 1/5*x^2*e^3 + x^2*log(2*x^2) - 6*x

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maple [A]  time = 0.02, size = 25, normalized size = 0.83

method	result	size
default	\(x^{3}-6 x -\frac {x^{2} {\mathrm e}^{3}}{5}+x^{2} \ln \left (2 x^{2}\right )\)	\(25\)
norman	\(x^{3}-6 x -\frac {x^{2} {\mathrm e}^{3}}{5}+x^{2} \ln \left (2 x^{2}\right )\)	\(25\)
risch	\(x^{3}-6 x -\frac {x^{2} {\mathrm e}^{3}}{5}+x^{2} \ln \left (2 x^{2}\right )\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x*ln(2*x^2)-2/5*x*exp(3)+3*x^2+2*x-6,x,method=_RETURNVERBOSE)

[Out]

x^3-6*x-1/5*x^2*exp(3)+x^2*ln(2*x^2)

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maxima [A]  time = 0.44, size = 24, normalized size = 0.80 \begin {gather*} x^{3} - \frac {1}{5} \, x^{2} e^{3} + x^{2} \log \left (2 \, x^{2}\right ) - 6 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*x*log(2*x^2)-2/5*x*exp(3)+3*x^2+2*x-6,x, algorithm="maxima")

[Out]

x^3 - 1/5*x^2*e^3 + x^2*log(2*x^2) - 6*x

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mupad [B]  time = 0.87, size = 28, normalized size = 0.93 \begin {gather*} x^2\,\ln \left (x^2\right )-6\,x-\frac {x^2\,{\mathrm {e}}^3}{5}+x^2\,\ln \relax (2)+x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x - (2*x*exp(3))/5 + 2*x*log(2*x^2) + 3*x^2 - 6,x)

[Out]

x^2*log(x^2) - 6*x - (x^2*exp(3))/5 + x^2*log(2) + x^3

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sympy [A]  time = 0.10, size = 24, normalized size = 0.80 \begin {gather*} x^{3} + x^{2} \log {\left (2 x^{2} \right )} - \frac {x^{2} e^{3}}{5} - 6 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*x*ln(2*x**2)-2/5*x*exp(3)+3*x**2+2*x-6,x)

[Out]

x**3 + x**2*log(2*x**2) - x**2*exp(3)/5 - 6*x

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