3.67.73 \(\int \frac {-3 x^2+8 x^3+e^{\frac {6 e^{25}}{x^2}} (-12 e^{25} x+2 x^3)+(2 x-4 x^2) \log (4)+e^{\frac {3 e^{25}}{x^2}} (-24 e^{25} x-2 x^2+8 x^3+(12 e^{25}-2 x^2) \log (4))+(2 x-4 x^2+e^{\frac {3 e^{25}}{x^2}} (12 e^{25}-2 x^2)) \log (x)}{x^2} \, dx\)

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

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Rubi [B]  time = 0.33, antiderivative size = 80, normalized size of antiderivative = 3.20, number of steps used = 11, number of rules used = 5, integrand size = 126, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {14, 2288, 2346, 2301, 2295} \begin {gather*} e^{\frac {6 e^{25}}{x^2}} x^2+4 x^2+2 e^{\frac {3 e^{25}}{x^2}-25} x \left (2 e^{25} x-e^{25} \log (x)-e^{25} \log (4)\right )+4 x+\log ^2(x)-4 x \log (x)-x (3+\log (256))+\log (16) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3*x^2 + 8*x^3 + E^((6*E^25)/x^2)*(-12*E^25*x + 2*x^3) + (2*x - 4*x^2)*Log[4] + E^((3*E^25)/x^2)*(-24*E^2
5*x - 2*x^2 + 8*x^3 + (12*E^25 - 2*x^2)*Log[4]) + (2*x - 4*x^2 + E^((3*E^25)/x^2)*(12*E^25 - 2*x^2))*Log[x])/x
^2,x]

[Out]

4*x + 4*x^2 + E^((6*E^25)/x^2)*x^2 - x*(3 + Log[256]) - 4*x*Log[x] + Log[16]*Log[x] + Log[x]^2 + 2*E^(-25 + (3
*E^25)/x^2)*x*(2*E^25*x - E^25*Log[4] - E^25*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 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 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2346

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.))/(x_), x_Symbol] :> Dist[d, Int[((d
+ e*x)^(q - 1)*(a + b*Log[c*x^n])^p)/x, x], x] + Dist[e, Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^p, x], x] /
; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && GtQ[q, 0] && IntegerQ[2*q]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 e^{\frac {6 e^{25}}{x^2}} \left (-6 e^{25}+x^2\right )}{x}+\frac {8 x^2-3 x \left (1+\frac {4 \log (4)}{3}\right )+\log (16)+2 \log (x)-4 x \log (x)}{x}+\frac {2 e^{\frac {3 e^{25}}{x^2}} \left (-12 e^{25} x+4 x^3+6 e^{25} \log (4)-x^2 (1+\log (4))+6 e^{25} \log (x)-x^2 \log (x)\right )}{x^2}\right ) \, dx\\ &=2 \int \frac {e^{\frac {6 e^{25}}{x^2}} \left (-6 e^{25}+x^2\right )}{x} \, dx+2 \int \frac {e^{\frac {3 e^{25}}{x^2}} \left (-12 e^{25} x+4 x^3+6 e^{25} \log (4)-x^2 (1+\log (4))+6 e^{25} \log (x)-x^2 \log (x)\right )}{x^2} \, dx+\int \frac {8 x^2-3 x \left (1+\frac {4 \log (4)}{3}\right )+\log (16)+2 \log (x)-4 x \log (x)}{x} \, dx\\ &=e^{\frac {6 e^{25}}{x^2}} x^2+2 e^{-25+\frac {3 e^{25}}{x^2}} x \left (2 e^{25} x-e^{25} \log (4)-e^{25} \log (x)\right )+\int \left (\frac {8 x^2+\log (16)-x (3+\log (256))}{x}-\frac {2 (-1+2 x) \log (x)}{x}\right ) \, dx\\ &=e^{\frac {6 e^{25}}{x^2}} x^2+2 e^{-25+\frac {3 e^{25}}{x^2}} x \left (2 e^{25} x-e^{25} \log (4)-e^{25} \log (x)\right )-2 \int \frac {(-1+2 x) \log (x)}{x} \, dx+\int \frac {8 x^2+\log (16)-x (3+\log (256))}{x} \, dx\\ &=e^{\frac {6 e^{25}}{x^2}} x^2+2 e^{-25+\frac {3 e^{25}}{x^2}} x \left (2 e^{25} x-e^{25} \log (4)-e^{25} \log (x)\right )+2 \int \frac {\log (x)}{x} \, dx-4 \int \log (x) \, dx+\int \left (-3+8 x+\frac {\log (16)}{x}-\log (256)\right ) \, dx\\ &=4 x+4 x^2+e^{\frac {6 e^{25}}{x^2}} x^2-x (3+\log (256))-4 x \log (x)+\log (16) \log (x)+\log ^2(x)+2 e^{-25+\frac {3 e^{25}}{x^2}} x \left (2 e^{25} x-e^{25} \log (4)-e^{25} \log (x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.15, size = 64, normalized size = 2.56 \begin {gather*} x \left (1+\left (2+e^{\frac {3 e^{25}}{x^2}}\right )^2 x-4 \log (4)-2 e^{\frac {3 e^{25}}{x^2}} \log (4)\right )+\left (-2 \left (2+e^{\frac {3 e^{25}}{x^2}}\right ) x+\log (16)\right ) \log (x)+\log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*x^2 + 8*x^3 + E^((6*E^25)/x^2)*(-12*E^25*x + 2*x^3) + (2*x - 4*x^2)*Log[4] + E^((3*E^25)/x^2)*(-
24*E^25*x - 2*x^2 + 8*x^3 + (12*E^25 - 2*x^2)*Log[4]) + (2*x - 4*x^2 + E^((3*E^25)/x^2)*(12*E^25 - 2*x^2))*Log
[x])/x^2,x]

[Out]

x*(1 + (2 + E^((3*E^25)/x^2))^2*x - 4*Log[4] - 2*E^((3*E^25)/x^2)*Log[4]) + (-2*(2 + E^((3*E^25)/x^2))*x + Log
[16])*Log[x] + Log[x]^2

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fricas [B]  time = 0.48, size = 69, normalized size = 2.76 \begin {gather*} x^{2} e^{\left (\frac {6 \, e^{25}}{x^{2}}\right )} + 4 \, x^{2} + 4 \, {\left (x^{2} - x \log \relax (2)\right )} e^{\left (\frac {3 \, e^{25}}{x^{2}}\right )} - 8 \, x \log \relax (2) - 2 \, {\left (x e^{\left (\frac {3 \, e^{25}}{x^{2}}\right )} + 2 \, x - 2 \, \log \relax (2)\right )} \log \relax (x) + \log \relax (x)^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*exp(25)-2*x^2)*exp(3*exp(25)/x^2)-4*x^2+2*x)*log(x)+(-12*x*exp(25)+2*x^3)*exp(3*exp(25)/x^2)^2
+(2*(12*exp(25)-2*x^2)*log(2)-24*x*exp(25)+8*x^3-2*x^2)*exp(3*exp(25)/x^2)+2*(-4*x^2+2*x)*log(2)+8*x^3-3*x^2)/
x^2,x, algorithm="fricas")

[Out]

x^2*e^(6*e^25/x^2) + 4*x^2 + 4*(x^2 - x*log(2))*e^(3*e^25/x^2) - 8*x*log(2) - 2*(x*e^(3*e^25/x^2) + 2*x - 2*lo
g(2))*log(x) + log(x)^2 + x

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giac [B]  time = 0.17, size = 126, normalized size = 5.04 \begin {gather*} x^{2} e^{\left (\frac {25 \, x^{2} + 6 \, e^{25}}{x^{2}} - 25\right )} + {\left (4 \, x^{2} e^{25} + 4 \, x^{2} e^{\left (\frac {25 \, x^{2} + 3 \, e^{25}}{x^{2}}\right )} - 8 \, x e^{25} \log \relax (2) - 4 \, x e^{\left (\frac {25 \, x^{2} + 3 \, e^{25}}{x^{2}}\right )} \log \relax (2) - 4 \, x e^{25} \log \relax (x) - 2 \, x e^{\left (\frac {25 \, x^{2} + 3 \, e^{25}}{x^{2}}\right )} \log \relax (x) + 4 \, e^{25} \log \relax (2) \log \relax (x) + e^{25} \log \relax (x)^{2} + x e^{25}\right )} e^{\left (-25\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*exp(25)-2*x^2)*exp(3*exp(25)/x^2)-4*x^2+2*x)*log(x)+(-12*x*exp(25)+2*x^3)*exp(3*exp(25)/x^2)^2
+(2*(12*exp(25)-2*x^2)*log(2)-24*x*exp(25)+8*x^3-2*x^2)*exp(3*exp(25)/x^2)+2*(-4*x^2+2*x)*log(2)+8*x^3-3*x^2)/
x^2,x, algorithm="giac")

[Out]

x^2*e^((25*x^2 + 6*e^25)/x^2 - 25) + (4*x^2*e^25 + 4*x^2*e^((25*x^2 + 3*e^25)/x^2) - 8*x*e^25*log(2) - 4*x*e^(
(25*x^2 + 3*e^25)/x^2)*log(2) - 4*x*e^25*log(x) - 2*x*e^((25*x^2 + 3*e^25)/x^2)*log(x) + 4*e^25*log(2)*log(x)
+ e^25*log(x)^2 + x*e^25)*e^(-25)

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maple [B]  time = 0.12, size = 79, normalized size = 3.16

method	result	size
risch	\(\ln \relax (x )^{2}+\left (-2 \,{\mathrm e}^{\frac {3 \,{\mathrm e}^{25}}{x^{2}}} x -4 x \right ) \ln \relax (x )+x^{2} {\mathrm e}^{\frac {6 \,{\mathrm e}^{25}}{x^{2}}}-4 \ln \relax (2) x \,{\mathrm e}^{\frac {3 \,{\mathrm e}^{25}}{x^{2}}}+4 \,{\mathrm e}^{\frac {3 \,{\mathrm e}^{25}}{x^{2}}} x^{2}+4 \ln \relax (2) \ln \relax (x )-8 x \ln \relax (2)+4 x^{2}+x\)	\(79\)
default	\(\frac {4 \,{\mathrm e}^{\frac {3 \,{\mathrm e}^{25}}{x^{2}}} x^{3}-2 \ln \relax (x ) {\mathrm e}^{\frac {3 \,{\mathrm e}^{25}}{x^{2}}} x^{2}-4 \ln \relax (2) {\mathrm e}^{\frac {3 \,{\mathrm e}^{25}}{x^{2}}} x^{2}}{x}+4 x^{2}-8 x \ln \relax (2)+x +4 \ln \relax (2) \ln \relax (x )+x^{2} {\mathrm e}^{\frac {6 \,{\mathrm e}^{25}}{x^{2}}}-4 x \ln \relax (x )+\ln \relax (x )^{2}\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((12*exp(25)-2*x^2)*exp(3*exp(25)/x^2)-4*x^2+2*x)*ln(x)+(-12*x*exp(25)+2*x^3)*exp(3*exp(25)/x^2)^2+(2*(12
*exp(25)-2*x^2)*ln(2)-24*x*exp(25)+8*x^3-2*x^2)*exp(3*exp(25)/x^2)+2*(-4*x^2+2*x)*ln(2)+8*x^3-3*x^2)/x^2,x,met
hod=_RETURNVERBOSE)

[Out]

ln(x)^2+(-2*exp(3*exp(25)/x^2)*x-4*x)*ln(x)+x^2*exp(6*exp(25)/x^2)-4*ln(2)*x*exp(3*exp(25)/x^2)+4*exp(3*exp(25
)/x^2)*x^2+4*ln(2)*ln(x)-8*x*ln(2)+4*x^2+x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -2 \, \sqrt {3} x \sqrt {-\frac {e^{25}}{x^{2}}} \Gamma \left (-\frac {1}{2}, -\frac {3 \, e^{25}}{x^{2}}\right ) \log \relax (2) - \sqrt {3} x \sqrt {-\frac {e^{25}}{x^{2}}} \Gamma \left (-\frac {1}{2}, -\frac {3 \, e^{25}}{x^{2}}\right ) - 2 \, x e^{\left (\frac {3 \, e^{25}}{x^{2}}\right )} \log \relax (x) - \frac {4 \, \sqrt {3} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {3} \sqrt {-\frac {e^{25}}{x^{2}}}\right ) - 1\right )} e^{25} \log \relax (2)}{x \sqrt {-\frac {e^{25}}{x^{2}}}} + 4 \, x^{2} + 6 \, {\rm Ei}\left (\frac {6 \, e^{25}}{x^{2}}\right ) e^{25} + 12 \, {\rm Ei}\left (\frac {3 \, e^{25}}{x^{2}}\right ) e^{25} - 12 \, e^{25} \Gamma \left (-1, -\frac {3 \, e^{25}}{x^{2}}\right ) - 6 \, e^{25} \Gamma \left (-1, -\frac {6 \, e^{25}}{x^{2}}\right ) - 8 \, x \log \relax (2) - 4 \, x \log \relax (x) + 4 \, \log \relax (2) \log \relax (x) + \log \relax (x)^{2} + x + 2 \, \int e^{\left (\frac {3 \, e^{25}}{x^{2}}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*exp(25)-2*x^2)*exp(3*exp(25)/x^2)-4*x^2+2*x)*log(x)+(-12*x*exp(25)+2*x^3)*exp(3*exp(25)/x^2)^2
+(2*(12*exp(25)-2*x^2)*log(2)-24*x*exp(25)+8*x^3-2*x^2)*exp(3*exp(25)/x^2)+2*(-4*x^2+2*x)*log(2)+8*x^3-3*x^2)/
x^2,x, algorithm="maxima")

[Out]

-2*sqrt(3)*x*sqrt(-e^25/x^2)*gamma(-1/2, -3*e^25/x^2)*log(2) - sqrt(3)*x*sqrt(-e^25/x^2)*gamma(-1/2, -3*e^25/x
^2) - 2*x*e^(3*e^25/x^2)*log(x) - 4*sqrt(3)*sqrt(pi)*(erf(sqrt(3)*sqrt(-e^25/x^2)) - 1)*e^25*log(2)/(x*sqrt(-e
^25/x^2)) + 4*x^2 + 6*Ei(6*e^25/x^2)*e^25 + 12*Ei(3*e^25/x^2)*e^25 - 12*e^25*gamma(-1, -3*e^25/x^2) - 6*e^25*g
amma(-1, -6*e^25/x^2) - 8*x*log(2) - 4*x*log(x) + 4*log(2)*log(x) + log(x)^2 + x + 2*integrate(e^(3*e^25/x^2),
x)

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mupad [B]  time = 4.37, size = 78, normalized size = 3.12 \begin {gather*} x+4\,x^2\,{\mathrm {e}}^{\frac {3\,{\mathrm {e}}^{25}}{x^2}}+x^2\,{\mathrm {e}}^{\frac {6\,{\mathrm {e}}^{25}}{x^2}}-8\,x\,\ln \relax (2)+{\ln \relax (x)}^2+4\,\ln \relax (2)\,\ln \relax (x)-4\,x\,\ln \relax (x)+4\,x^2-4\,x\,{\mathrm {e}}^{\frac {3\,{\mathrm {e}}^{25}}{x^2}}\,\ln \relax (2)-2\,x\,{\mathrm {e}}^{\frac {3\,{\mathrm {e}}^{25}}{x^2}}\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*log(2)*(2*x - 4*x^2) - exp((6*exp(25))/x^2)*(12*x*exp(25) - 2*x^3) + log(x)*(2*x + exp((3*exp(25))/x^2)
*(12*exp(25) - 2*x^2) - 4*x^2) + exp((3*exp(25))/x^2)*(2*log(2)*(12*exp(25) - 2*x^2) - 24*x*exp(25) - 2*x^2 +
8*x^3) - 3*x^2 + 8*x^3)/x^2,x)

[Out]

x + 4*x^2*exp((3*exp(25))/x^2) + x^2*exp((6*exp(25))/x^2) - 8*x*log(2) + log(x)^2 + 4*log(2)*log(x) - 4*x*log(
x) + 4*x^2 - 4*x*exp((3*exp(25))/x^2)*log(2) - 2*x*exp((3*exp(25))/x^2)*log(x)

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sympy [B]  time = 0.68, size = 75, normalized size = 3.00 \begin {gather*} x^{2} e^{\frac {6 e^{25}}{x^{2}}} + 4 x^{2} - 4 x \log {\relax (x )} + x \left (1 - 8 \log {\relax (2 )}\right ) + \left (4 x^{2} - 2 x \log {\relax (x )} - 4 x \log {\relax (2 )}\right ) e^{\frac {3 e^{25}}{x^{2}}} + \log {\relax (x )}^{2} + 4 \log {\relax (2 )} \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*exp(25)-2*x**2)*exp(3*exp(25)/x**2)-4*x**2+2*x)*ln(x)+(-12*x*exp(25)+2*x**3)*exp(3*exp(25)/x**
2)**2+(2*(12*exp(25)-2*x**2)*ln(2)-24*x*exp(25)+8*x**3-2*x**2)*exp(3*exp(25)/x**2)+2*(-4*x**2+2*x)*ln(2)+8*x**
3-3*x**2)/x**2,x)

[Out]

x**2*exp(6*exp(25)/x**2) + 4*x**2 - 4*x*log(x) + x*(1 - 8*log(2)) + (4*x**2 - 2*x*log(x) - 4*x*log(2))*exp(3*e
xp(25)/x**2) + log(x)**2 + 4*log(2)*log(x)

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