3.43.3 \(\int \frac {-900 e^{225 \log ^2(x)} \log (x)+540 e^{450 \log ^2(x)} \log (x)}{x} \, dx\)

Optimal. Leaf size=18 \[ \frac {3}{5} \left (-\frac {5}{3}+e^{225 \log ^2(x)}\right )^2 \]

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Rubi [A]  time = 0.13, antiderivative size = 23, normalized size of antiderivative = 1.28, number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {14, 2276, 2204, 2209} \begin {gather*} \frac {3}{5} e^{450 \log ^2(x)}-2 e^{225 \log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-900*E^(225*Log[x]^2)*Log[x] + 540*E^(450*Log[x]^2)*Log[x])/x,x]

[Out]

-2*E^(225*Log[x]^2) + (3*E^(450*Log[x]^2))/5

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

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 2276

Int[(F_)^(((a_.) + Log[(c_.)*(x_)^(n_.)]^2*(b_.))*(d_.))*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[(e*x)^(m + 1)/(
e*n*(c*x^n)^((m + 1)/n)), Subst[Int[E^(a*d*Log[F] + ((m + 1)*x)/n + b*d*Log[F]*x^2), x], x, Log[c*x^n]], x] /;
FreeQ[{F, a, b, c, d, e, m, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {900 e^{225 \log ^2(x)} \log (x)}{x}+\frac {540 e^{450 \log ^2(x)} \log (x)}{x}\right ) \, dx\\ &=540 \int \frac {e^{450 \log ^2(x)} \log (x)}{x} \, dx-900 \int \frac {e^{225 \log ^2(x)} \log (x)}{x} \, dx\\ &=540 \operatorname {Subst}\left (\int e^{450 x^2} x \, dx,x,\log (x)\right )-900 \operatorname {Subst}\left (\int e^{225 x^2} x \, dx,x,\log (x)\right )\\ &=-2 e^{225 \log ^2(x)}+\frac {3}{5} e^{450 \log ^2(x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 18, normalized size = 1.00 \begin {gather*} \frac {1}{15} \left (-5+3 e^{225 \log ^2(x)}\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-900*E^(225*Log[x]^2)*Log[x] + 540*E^(450*Log[x]^2)*Log[x])/x,x]

[Out]

(-5 + 3*E^(225*Log[x]^2))^2/15

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fricas [A]  time = 0.67, size = 19, normalized size = 1.06 \begin {gather*} \frac {3}{5} \, e^{\left (450 \, \log \relax (x)^{2}\right )} - 2 \, e^{\left (225 \, \log \relax (x)^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((540*log(x)*exp(225*log(x)^2)^2-900*log(x)*exp(225*log(x)^2))/x,x, algorithm="fricas")

[Out]

3/5*e^(450*log(x)^2) - 2*e^(225*log(x)^2)

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giac [A]  time = 0.23, size = 19, normalized size = 1.06 \begin {gather*} \frac {3}{5} \, e^{\left (450 \, \log \relax (x)^{2}\right )} - 2 \, e^{\left (225 \, \log \relax (x)^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((540*log(x)*exp(225*log(x)^2)^2-900*log(x)*exp(225*log(x)^2))/x,x, algorithm="giac")

[Out]

3/5*e^(450*log(x)^2) - 2*e^(225*log(x)^2)

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maple [A]  time = 0.07, size = 20, normalized size = 1.11

method	result	size
risch	\(-2 \,{\mathrm e}^{225 \ln \relax (x )^{2}}+\frac {3 \,{\mathrm e}^{450 \ln \relax (x )^{2}}}{5}\)	\(20\)
derivativedivides	\(-2 \,{\mathrm e}^{225 \ln \relax (x )^{2}}+\frac {3 \,{\mathrm e}^{450 \ln \relax (x )^{2}}}{5}\)	\(22\)
default	\(-2 \,{\mathrm e}^{225 \ln \relax (x )^{2}}+\frac {3 \,{\mathrm e}^{450 \ln \relax (x )^{2}}}{5}\)	\(22\)
norman	\(-2 \,{\mathrm e}^{225 \ln \relax (x )^{2}}+\frac {3 \,{\mathrm e}^{450 \ln \relax (x )^{2}}}{5}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((540*ln(x)*exp(225*ln(x)^2)^2-900*ln(x)*exp(225*ln(x)^2))/x,x,method=_RETURNVERBOSE)

[Out]

-2*exp(225*ln(x)^2)+3/5*exp(450*ln(x)^2)

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maxima [A]  time = 1.18, size = 19, normalized size = 1.06 \begin {gather*} \frac {3}{5} \, e^{\left (450 \, \log \relax (x)^{2}\right )} - 2 \, e^{\left (225 \, \log \relax (x)^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((540*log(x)*exp(225*log(x)^2)^2-900*log(x)*exp(225*log(x)^2))/x,x, algorithm="maxima")

[Out]

3/5*e^(450*log(x)^2) - 2*e^(225*log(x)^2)

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mupad [B]  time = 2.97, size = 20, normalized size = 1.11 \begin {gather*} \frac {{\mathrm {e}}^{225\,{\ln \relax (x)}^2}\,\left (3\,{\mathrm {e}}^{225\,{\ln \relax (x)}^2}-10\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(900*exp(225*log(x)^2)*log(x) - 540*exp(450*log(x)^2)*log(x))/x,x)

[Out]

(exp(225*log(x)^2)*(3*exp(225*log(x)^2) - 10))/5

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sympy [A]  time = 0.28, size = 20, normalized size = 1.11 \begin {gather*} \frac {3 e^{450 \log {\relax (x )}^{2}}}{5} - 2 e^{225 \log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((540*ln(x)*exp(225*ln(x)**2)**2-900*ln(x)*exp(225*ln(x)**2))/x,x)

[Out]

3*exp(450*log(x)**2)/5 - 2*exp(225*log(x)**2)

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