3.25.45 \(\int \frac {-9+(-25 x-50 e^{x^2} x^2) \log ^2(2 x)}{3 x \log ^2(2 x)} \, dx\)

Optimal. Leaf size=25 \[ \frac {25}{3} \left (1-e^{x^2}-x\right )+\frac {3}{\log (2 x)} \]

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Rubi [A]  time = 0.24, antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {12, 6688, 2209, 2302, 30} \begin {gather*} -\frac {25 e^{x^2}}{3}-\frac {25 x}{3}+\frac {3}{\log (2 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-9 + (-25*x - 50*E^x^2*x^2)*Log[2*x]^2)/(3*x*Log[2*x]^2),x]

[Out]

(-25*E^x^2)/3 - (25*x)/3 + 3/Log[2*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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \frac {-9+\left (-25 x-50 e^{x^2} x^2\right ) \log ^2(2 x)}{x \log ^2(2 x)} \, dx\\ &=\frac {1}{3} \int \left (-25-50 e^{x^2} x-\frac {9}{x \log ^2(2 x)}\right ) \, dx\\ &=-\frac {25 x}{3}-3 \int \frac {1}{x \log ^2(2 x)} \, dx-\frac {50}{3} \int e^{x^2} x \, dx\\ &=-\frac {25 e^{x^2}}{3}-\frac {25 x}{3}-3 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (2 x)\right )\\ &=-\frac {25 e^{x^2}}{3}-\frac {25 x}{3}+\frac {3}{\log (2 x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 23, normalized size = 0.92 \begin {gather*} -\frac {25 e^{x^2}}{3}-\frac {25 x}{3}+\frac {3}{\log (2 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-9 + (-25*x - 50*E^x^2*x^2)*Log[2*x]^2)/(3*x*Log[2*x]^2),x]

[Out]

(-25*E^x^2)/3 - (25*x)/3 + 3/Log[2*x]

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fricas [A]  time = 0.99, size = 22, normalized size = 0.88 \begin {gather*} -\frac {25 \, {\left (x + e^{\left (x^{2}\right )}\right )} \log \left (2 \, x\right ) - 9}{3 \, \log \left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-50*x^2*exp(x^2)-25*x)*log(2*x)^2-9)/x/log(2*x)^2,x, algorithm="fricas")

[Out]

-1/3*(25*(x + e^(x^2))*log(2*x) - 9)/log(2*x)

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giac [A]  time = 0.40, size = 27, normalized size = 1.08 \begin {gather*} -\frac {25 \, x \log \left (2 \, x\right ) + 25 \, e^{\left (x^{2}\right )} \log \left (2 \, x\right ) - 9}{3 \, \log \left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-50*x^2*exp(x^2)-25*x)*log(2*x)^2-9)/x/log(2*x)^2,x, algorithm="giac")

[Out]

-1/3*(25*x*log(2*x) + 25*e^(x^2)*log(2*x) - 9)/log(2*x)

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maple [A]  time = 0.02, size = 19, normalized size = 0.76

method	result	size
default	\(-\frac {25 x}{3}+\frac {3}{\ln \left (2 x \right )}-\frac {25 \,{\mathrm e}^{x^{2}}}{3}\)	\(19\)
risch	\(-\frac {25 x}{3}+\frac {3}{\ln \left (2 x \right )}-\frac {25 \,{\mathrm e}^{x^{2}}}{3}\)	\(19\)
norman	\(\frac {3-\frac {25 x \ln \left (2 x \right )}{3}-\frac {25 \,{\mathrm e}^{x^{2}} \ln \left (2 x \right )}{3}}{\ln \left (2 x \right )}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((-50*x^2*exp(x^2)-25*x)*ln(2*x)^2-9)/x/ln(2*x)^2,x,method=_RETURNVERBOSE)

[Out]

-25/3*x+3/ln(2*x)-25/3*exp(x^2)

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maxima [A]  time = 0.39, size = 18, normalized size = 0.72 \begin {gather*} -\frac {25}{3} \, x + \frac {3}{\log \left (2 \, x\right )} - \frac {25}{3} \, e^{\left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-50*x^2*exp(x^2)-25*x)*log(2*x)^2-9)/x/log(2*x)^2,x, algorithm="maxima")

[Out]

-25/3*x + 3/log(2*x) - 25/3*e^(x^2)

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mupad [B]  time = 1.40, size = 18, normalized size = 0.72 \begin {gather*} \frac {3}{\ln \left (2\,x\right )}-\frac {25\,{\mathrm {e}}^{x^2}}{3}-\frac {25\,x}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((log(2*x)^2*(25*x + 50*x^2*exp(x^2)))/3 + 3)/(x*log(2*x)^2),x)

[Out]

3/log(2*x) - (25*exp(x^2))/3 - (25*x)/3

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sympy [A]  time = 0.32, size = 19, normalized size = 0.76 \begin {gather*} - \frac {25 x}{3} - \frac {25 e^{x^{2}}}{3} + \frac {3}{\log {\left (2 x \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-50*x**2*exp(x**2)-25*x)*ln(2*x)**2-9)/x/ln(2*x)**2,x)

[Out]

-25*x/3 - 25*exp(x**2)/3 + 3/log(2*x)

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