3.75.17 \(\int \frac {-4 x^2+e^{\frac {25+1250 x+15675 x^2+1250 x^3+25 x^4}{x^2}} (-50-1250 x+1250 x^3+50 x^4) \log ^2(x)}{x^3 \log ^2(x)} \, dx\)

Optimal. Leaf size=19 \[ e^{25 \left (25+\frac {1}{x}+x\right )^2}+\frac {4}{\log (x)} \]

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Rubi [A]  time = 2.25, antiderivative size = 24, normalized size of antiderivative = 1.26, number of steps used = 6, number of rules used = 5, integrand size = 60, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6688, 14, 6706, 2302, 30} \begin {gather*} e^{\frac {25 \left (x^2+25 x+1\right )^2}{x^2}}+\frac {4}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4*x^2 + E^((25 + 1250*x + 15675*x^2 + 1250*x^3 + 25*x^4)/x^2)*(-50 - 1250*x + 1250*x^3 + 50*x^4)*Log[x]^
2)/(x^3*Log[x]^2),x]

[Out]

E^((25*(1 + 25*x + x^2)^2)/x^2) + 4/Log[x]

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 30

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

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]]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {50 e^{\frac {25 \left (1+25 x+x^2\right )^2}{x^2}} \left (-1-25 x+25 x^3+x^4\right )-\frac {4 x^2}{\log ^2(x)}}{x^3} \, dx\\ &=\int \left (\frac {50 e^{\frac {25 \left (1+25 x+x^2\right )^2}{x^2}} (-1+x) (1+x) \left (1+25 x+x^2\right )}{x^3}-\frac {4}{x \log ^2(x)}\right ) \, dx\\ &=-\left (4 \int \frac {1}{x \log ^2(x)} \, dx\right )+50 \int \frac {e^{\frac {25 \left (1+25 x+x^2\right )^2}{x^2}} (-1+x) (1+x) \left (1+25 x+x^2\right )}{x^3} \, dx\\ &=e^{\frac {25 \left (1+25 x+x^2\right )^2}{x^2}}-4 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )\\ &=e^{\frac {25 \left (1+25 x+x^2\right )^2}{x^2}}+\frac {4}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.52, size = 24, normalized size = 1.26 \begin {gather*} e^{\frac {25 \left (1+25 x+x^2\right )^2}{x^2}}+\frac {4}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*x^2 + E^((25 + 1250*x + 15675*x^2 + 1250*x^3 + 25*x^4)/x^2)*(-50 - 1250*x + 1250*x^3 + 50*x^4)*L
og[x]^2)/(x^3*Log[x]^2),x]

[Out]

E^((25*(1 + 25*x + x^2)^2)/x^2) + 4/Log[x]

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fricas [A]  time = 0.83, size = 34, normalized size = 1.79 \begin {gather*} \frac {e^{\left (\frac {25 \, {\left (x^{4} + 50 \, x^{3} + 627 \, x^{2} + 50 \, x + 1\right )}}{x^{2}}\right )} \log \relax (x) + 4}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x^4+1250*x^3-1250*x-50)*exp((25*x^4+1250*x^3+15675*x^2+1250*x+25)/x^2)*log(x)^2-4*x^2)/x^3/log(
x)^2,x, algorithm="fricas")

[Out]

(e^(25*(x^4 + 50*x^3 + 627*x^2 + 50*x + 1)/x^2)*log(x) + 4)/log(x)

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giac [A]  time = 0.27, size = 34, normalized size = 1.79 \begin {gather*} \frac {e^{\left (\frac {25 \, {\left (x^{4} + 50 \, x^{3} + 627 \, x^{2} + 50 \, x + 1\right )}}{x^{2}}\right )} \log \relax (x) + 4}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x^4+1250*x^3-1250*x-50)*exp((25*x^4+1250*x^3+15675*x^2+1250*x+25)/x^2)*log(x)^2-4*x^2)/x^3/log(
x)^2,x, algorithm="giac")

[Out]

(e^(25*(x^4 + 50*x^3 + 627*x^2 + 50*x + 1)/x^2)*log(x) + 4)/log(x)

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maple [A]  time = 0.11, size = 24, normalized size = 1.26

method	result	size
risch	\(\frac {4}{\ln \relax (x )}+{\mathrm e}^{\frac {25 \left (x^{2}+25 x +1\right )^{2}}{x^{2}}}\)	\(24\)
default	\(\frac {4}{\ln \relax (x )}+{\mathrm e}^{\frac {25 x^{4}+1250 x^{3}+15675 x^{2}+1250 x +25}{x^{2}}}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((50*x^4+1250*x^3-1250*x-50)*exp((25*x^4+1250*x^3+15675*x^2+1250*x+25)/x^2)*ln(x)^2-4*x^2)/x^3/ln(x)^2,x,m
ethod=_RETURNVERBOSE)

[Out]

4/ln(x)+exp(25*(x^2+25*x+1)^2/x^2)

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maxima [A]  time = 0.50, size = 31, normalized size = 1.63 \begin {gather*} \frac {e^{\left (25 \, x^{2} + 1250 \, x + \frac {1250}{x} + \frac {25}{x^{2}} + 15675\right )} \log \relax (x) + 4}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x^4+1250*x^3-1250*x-50)*exp((25*x^4+1250*x^3+15675*x^2+1250*x+25)/x^2)*log(x)^2-4*x^2)/x^3/log(
x)^2,x, algorithm="maxima")

[Out]

(e^(25*x^2 + 1250*x + 1250/x + 25/x^2 + 15675)*log(x) + 4)/log(x)

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mupad [B]  time = 4.87, size = 32, normalized size = 1.68 \begin {gather*} \frac {4}{\ln \relax (x)}+{\mathrm {e}}^{1250\,x}\,{\mathrm {e}}^{15675}\,{\mathrm {e}}^{\frac {25}{x^2}}\,{\mathrm {e}}^{25\,x^2}\,{\mathrm {e}}^{1250/x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x^2 + exp((1250*x + 15675*x^2 + 1250*x^3 + 25*x^4 + 25)/x^2)*log(x)^2*(1250*x - 1250*x^3 - 50*x^4 + 50
))/(x^3*log(x)^2),x)

[Out]

4/log(x) + exp(1250*x)*exp(15675)*exp(25/x^2)*exp(25*x^2)*exp(1250/x)

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sympy [A]  time = 0.38, size = 29, normalized size = 1.53 \begin {gather*} e^{\frac {25 x^{4} + 1250 x^{3} + 15675 x^{2} + 1250 x + 25}{x^{2}}} + \frac {4}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x**4+1250*x**3-1250*x-50)*exp((25*x**4+1250*x**3+15675*x**2+1250*x+25)/x**2)*ln(x)**2-4*x**2)/x
**3/ln(x)**2,x)

[Out]

exp((25*x**4 + 1250*x**3 + 15675*x**2 + 1250*x + 25)/x**2) + 4/log(x)

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