3.27.69 \(\int (15-2 x+e^{-27+24 x+3 x^2} (1+24 x+6 x^2)-8 \log (x)+\log ^2(x)) \, dx\)

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

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Rubi [A]  time = 0.04, antiderivative size = 45, normalized size of antiderivative = 1.80, number of steps used = 5, number of rules used = 3, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2288, 2295, 2296} \begin {gather*} -x^2+\frac {e^{3 x^2+24 x-27} \left (x^2+4 x\right )}{x+4}+25 x+x \log ^2(x)-10 x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[15 - 2*x + E^(-27 + 24*x + 3*x^2)*(1 + 24*x + 6*x^2) - 8*Log[x] + Log[x]^2,x]

[Out]

25*x - x^2 + (E^(-27 + 24*x + 3*x^2)*(4*x + x^2))/(4 + x) - 10*x*Log[x] + x*Log[x]^2

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=15 x-x^2-8 \int \log (x) \, dx+\int e^{-27+24 x+3 x^2} \left (1+24 x+6 x^2\right ) \, dx+\int \log ^2(x) \, dx\\ &=23 x-x^2+\frac {e^{-27+24 x+3 x^2} \left (4 x+x^2\right )}{4+x}-8 x \log (x)+x \log ^2(x)-2 \int \log (x) \, dx\\ &=25 x-x^2+\frac {e^{-27+24 x+3 x^2} \left (4 x+x^2\right )}{4+x}-10 x \log (x)+x \log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 25, normalized size = 1.00 \begin {gather*} x \left (25+e^{-27+3 x (8+x)}-x-10 \log (x)+\log ^2(x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[15 - 2*x + E^(-27 + 24*x + 3*x^2)*(1 + 24*x + 6*x^2) - 8*Log[x] + Log[x]^2,x]

[Out]

x*(25 + E^(-27 + 3*x*(8 + x)) - x - 10*Log[x] + Log[x]^2)

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fricas [A]  time = 0.47, size = 33, normalized size = 1.32 \begin {gather*} x \log \relax (x)^{2} - x^{2} + x e^{\left (3 \, x^{2} + 24 \, x - 27\right )} - 10 \, x \log \relax (x) + 25 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(x)^2-8*log(x)+(6*x^2+24*x+1)*exp(3*x^2+24*x-27)-2*x+15,x, algorithm="fricas")

[Out]

x*log(x)^2 - x^2 + x*e^(3*x^2 + 24*x - 27) - 10*x*log(x) + 25*x

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giac [A]  time = 0.23, size = 33, normalized size = 1.32 \begin {gather*} x \log \relax (x)^{2} - x^{2} + x e^{\left (3 \, x^{2} + 24 \, x - 27\right )} - 10 \, x \log \relax (x) + 25 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(x)^2-8*log(x)+(6*x^2+24*x+1)*exp(3*x^2+24*x-27)-2*x+15,x, algorithm="giac")

[Out]

x*log(x)^2 - x^2 + x*e^(3*x^2 + 24*x - 27) - 10*x*log(x) + 25*x

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maple [A]  time = 0.04, size = 32, normalized size = 1.28

method	result	size
risch	\(25 x +x \ln \relax (x )^{2}-10 x \ln \relax (x )+{\mathrm e}^{3 \left (x +9\right ) \left (x -1\right )} x -x^{2}\)	\(32\)
default	\(25 x +x \ln \relax (x )^{2}-10 x \ln \relax (x )+{\mathrm e}^{3 x^{2}+24 x -27} x -x^{2}\)	\(34\)
norman	\(25 x +x \ln \relax (x )^{2}-10 x \ln \relax (x )+{\mathrm e}^{3 x^{2}+24 x -27} x -x^{2}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(x)^2-8*ln(x)+(6*x^2+24*x+1)*exp(3*x^2+24*x-27)-2*x+15,x,method=_RETURNVERBOSE)

[Out]

25*x+x*ln(x)^2-10*x*ln(x)+exp(3*(x+9)*(x-1))*x-x^2

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maxima [A]  time = 0.46, size = 39, normalized size = 1.56 \begin {gather*} {\left (\log \relax (x)^{2} - 2 \, \log \relax (x) + 2\right )} x - x^{2} + x e^{\left (3 \, x^{2} + 24 \, x - 27\right )} - 8 \, x \log \relax (x) + 23 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(x)^2-8*log(x)+(6*x^2+24*x+1)*exp(3*x^2+24*x-27)-2*x+15,x, algorithm="maxima")

[Out]

(log(x)^2 - 2*log(x) + 2)*x - x^2 + x*e^(3*x^2 + 24*x - 27) - 8*x*log(x) + 23*x

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mupad [B]  time = 1.54, size = 37, normalized size = 1.48 \begin {gather*} x\,{\mathrm {e}}^{-27}\,\left ({\mathrm {e}}^{27}\,{\ln \relax (x)}^2-10\,{\mathrm {e}}^{27}\,\ln \relax (x)+25\,{\mathrm {e}}^{27}+{\mathrm {e}}^{3\,x^2+24\,x}-x\,{\mathrm {e}}^{27}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(x)^2 - 8*log(x) - 2*x + exp(24*x + 3*x^2 - 27)*(24*x + 6*x^2 + 1) + 15,x)

[Out]

x*exp(-27)*(25*exp(27) + exp(24*x + 3*x^2) + exp(27)*log(x)^2 - x*exp(27) - 10*exp(27)*log(x))

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sympy [A]  time = 0.28, size = 32, normalized size = 1.28 \begin {gather*} - x^{2} + x e^{3 x^{2} + 24 x - 27} + x \log {\relax (x )}^{2} - 10 x \log {\relax (x )} + 25 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(x)**2-8*ln(x)+(6*x**2+24*x+1)*exp(3*x**2+24*x-27)-2*x+15,x)

[Out]

-x**2 + x*exp(3*x**2 + 24*x - 27) + x*log(x)**2 - 10*x*log(x) + 25*x

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