3.84.86 \(\int \frac {e^{13+e-x+2 x^2+\log ^2(2 x)} (-4-x+4 x^2+2 \log (2 x))}{16 x^5} \, dx\)

Optimal. Leaf size=27 \[ e^{5+e-x+x \left (\frac {4}{x}+2 x\right )+(-2+\log (2 x))^2} \]

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Rubi [B]  time = 0.30, antiderivative size = 55, normalized size of antiderivative = 2.04, number of steps used = 2, number of rules used = 2, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {12, 2288} \begin {gather*} \frac {e^{2 x^2-x+\log ^2(2 x)+e+13} \left (-4 x^2+x-2 \log (2 x)\right )}{16 x^5 \left (-4 x-\frac {2 \log (2 x)}{x}+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(13 + E - x + 2*x^2 + Log[2*x]^2)*(-4 - x + 4*x^2 + 2*Log[2*x]))/(16*x^5),x]

[Out]

(E^(13 + E - x + 2*x^2 + Log[2*x]^2)*(x - 4*x^2 - 2*Log[2*x]))/(16*x^5*(1 - 4*x - (2*Log[2*x])/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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{16} \int \frac {e^{13+e-x+2 x^2+\log ^2(2 x)} \left (-4-x+4 x^2+2 \log (2 x)\right )}{x^5} \, dx\\ &=\frac {e^{13+e-x+2 x^2+\log ^2(2 x)} \left (x-4 x^2-2 \log (2 x)\right )}{16 x^5 \left (1-4 x-\frac {2 \log (2 x)}{x}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 26, normalized size = 0.96 \begin {gather*} \frac {e^{13+e-x+2 x^2+\log ^2(2 x)}}{16 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(13 + E - x + 2*x^2 + Log[2*x]^2)*(-4 - x + 4*x^2 + 2*Log[2*x]))/(16*x^5),x]

[Out]

E^(13 + E - x + 2*x^2 + Log[2*x]^2)/(16*x^4)

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fricas [A]  time = 0.97, size = 25, normalized size = 0.93 \begin {gather*} e^{\left (2 \, x^{2} + \log \left (2 \, x\right )^{2} - x + e - 4 \, \log \left (2 \, x\right ) + 13\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(2*x)+4*x^2-x-4)*exp(log(2*x)^2-4*log(2*x)+exp(1)+2*x^2-x+13)/x,x, algorithm="fricas")

[Out]

e^(2*x^2 + log(2*x)^2 - x + e - 4*log(2*x) + 13)

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giac [A]  time = 0.23, size = 25, normalized size = 0.93 \begin {gather*} e^{\left (2 \, x^{2} + \log \left (2 \, x\right )^{2} - x + e - 4 \, \log \left (2 \, x\right ) + 13\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(2*x)+4*x^2-x-4)*exp(log(2*x)^2-4*log(2*x)+exp(1)+2*x^2-x+13)/x,x, algorithm="giac")

[Out]

e^(2*x^2 + log(2*x)^2 - x + e - 4*log(2*x) + 13)

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maple [A]  time = 0.03, size = 25, normalized size = 0.93

method	result	size
risch	\(\frac {{\mathrm e}^{\ln \left (2 x \right )^{2}+13+{\mathrm e}+2 x^{2}-x}}{16 x^{4}}\)	\(25\)
norman	\({\mathrm e}^{\ln \left (2 x \right )^{2}-4 \ln \left (2 x \right )+{\mathrm e}+2 x^{2}-x +13}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*ln(2*x)+4*x^2-x-4)*exp(ln(2*x)^2-4*ln(2*x)+exp(1)+2*x^2-x+13)/x,x,method=_RETURNVERBOSE)

[Out]

1/16/x^4*exp(ln(2*x)^2+13+exp(1)+2*x^2-x)

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maxima [A]  time = 0.55, size = 32, normalized size = 1.19 \begin {gather*} \frac {e^{\left (2 \, x^{2} + \log \relax (2)^{2} + 2 \, \log \relax (2) \log \relax (x) + \log \relax (x)^{2} - x + e + 13\right )}}{16 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(2*x)+4*x^2-x-4)*exp(log(2*x)^2-4*log(2*x)+exp(1)+2*x^2-x+13)/x,x, algorithm="maxima")

[Out]

1/16*e^(2*x^2 + log(2)^2 + 2*log(2)*log(x) + log(x)^2 - x + e + 13)/x^4

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mupad [B]  time = 5.44, size = 35, normalized size = 1.30 \begin {gather*} \frac {x^{2\,\ln \relax (2)-4}\,{\mathrm {e}}^{{\ln \relax (2)}^2}\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{13}\,{\mathrm {e}}^{2\,x^2}\,{\mathrm {e}}^{{\ln \relax (x)}^2}\,{\mathrm {e}}^{\mathrm {e}}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(1) - 4*log(2*x) - x + log(2*x)^2 + 2*x^2 + 13)*(x - 2*log(2*x) - 4*x^2 + 4))/x,x)

[Out]

(x^(2*log(2) - 4)*exp(log(2)^2)*exp(-x)*exp(13)*exp(2*x^2)*exp(log(x)^2)*exp(exp(1)))/16

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sympy [A]  time = 0.30, size = 24, normalized size = 0.89 \begin {gather*} \frac {e^{2 x^{2} - x + \log {\left (2 x \right )}^{2} + e + 13}}{16 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*ln(2*x)+4*x**2-x-4)*exp(ln(2*x)**2-4*ln(2*x)+exp(1)+2*x**2-x+13)/x,x)

[Out]

exp(2*x**2 - x + log(2*x)**2 + E + 13)/(16*x**4)

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