3.85.16 \(\int \frac {-12 e^4+e^{\frac {2 (15 \log (x)+3 \log ^2(x))}{e^4}} (90-3 e^4+120 x+(36+48 x) \log (x))}{e^4 x^2} \, dx\)

Optimal. Leaf size=21 \[ \left (4+\frac {3}{x}\right ) \left (4+x^{\frac {6 (5+\log (x))}{e^4}}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 41, normalized size of antiderivative = 1.95, number of steps used = 4, number of rules used = 3, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {12, 14, 2288} \begin {gather*} \frac {x^{\frac {30}{e^4}-1} e^{\frac {6 \log ^2(x)}{e^4}} (4 x \log (x)+3 \log (x))}{\log (x)}+\frac {12}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-12*E^4 + E^((2*(15*Log[x] + 3*Log[x]^2))/E^4)*(90 - 3*E^4 + 120*x + (36 + 48*x)*Log[x]))/(E^4*x^2),x]

[Out]

12/x + (E^((6*Log[x]^2)/E^4)*x^(-1 + 30/E^4)*(3*Log[x] + 4*x*Log[x]))/Log[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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-12 e^4+e^{\frac {2 \left (15 \log (x)+3 \log ^2(x)\right )}{e^4}} \left (90-3 e^4+120 x+(36+48 x) \log (x)\right )}{x^2} \, dx}{e^4}\\ &=\frac {\int \left (-\frac {12 e^4}{x^2}+3 e^{\frac {6 \log ^2(x)}{e^4}} x^{-2+\frac {30}{e^4}} \left (30 \left (1-\frac {e^4}{30}\right )+40 x+12 \log (x)+16 x \log (x)\right )\right ) \, dx}{e^4}\\ &=\frac {12}{x}+\frac {3 \int e^{\frac {6 \log ^2(x)}{e^4}} x^{-2+\frac {30}{e^4}} \left (30 \left (1-\frac {e^4}{30}\right )+40 x+12 \log (x)+16 x \log (x)\right ) \, dx}{e^4}\\ &=\frac {12}{x}+\frac {e^{\frac {6 \log ^2(x)}{e^4}} x^{-1+\frac {30}{e^4}} (3 \log (x)+4 x \log (x))}{\log (x)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.23, size = 32, normalized size = 1.52 \begin {gather*} \frac {12}{x}+e^{\frac {6 \log ^2(x)}{e^4}} x^{-1+\frac {30}{e^4}} (3+4 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-12*E^4 + E^((2*(15*Log[x] + 3*Log[x]^2))/E^4)*(90 - 3*E^4 + 120*x + (36 + 48*x)*Log[x]))/(E^4*x^2)
,x]

[Out]

12/x + E^((6*Log[x]^2)/E^4)*x^(-1 + 30/E^4)*(3 + 4*x)

________________________________________________________________________________________

fricas [A]  time = 1.21, size = 26, normalized size = 1.24 \begin {gather*} \frac {{\left (4 \, x + 3\right )} e^{\left (6 \, {\left (\log \relax (x)^{2} + 5 \, \log \relax (x)\right )} e^{\left (-4\right )}\right )} + 12}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((48*x+36)*log(x)-3*exp(4)+120*x+90)*exp((3*log(x)^2+15*log(x))/exp(4))^2-12*exp(4))/x^2/exp(4),x,
algorithm="fricas")

[Out]

((4*x + 3)*e^(6*(log(x)^2 + 5*log(x))*e^(-4)) + 12)/x

________________________________________________________________________________________

giac [B]  time = 0.27, size = 48, normalized size = 2.29 \begin {gather*} \frac {{\left (4 \, x e^{\left (6 \, {\left (\log \relax (x)^{2} + 5 \, \log \relax (x)\right )} e^{\left (-4\right )} + 4\right )} + 12 \, e^{4} + 3 \, e^{\left (6 \, {\left (\log \relax (x)^{2} + 5 \, \log \relax (x)\right )} e^{\left (-4\right )} + 4\right )}\right )} e^{\left (-4\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((48*x+36)*log(x)-3*exp(4)+120*x+90)*exp((3*log(x)^2+15*log(x))/exp(4))^2-12*exp(4))/x^2/exp(4),x,
algorithm="giac")

[Out]

(4*x*e^(6*(log(x)^2 + 5*log(x))*e^(-4) + 4) + 12*e^4 + 3*e^(6*(log(x)^2 + 5*log(x))*e^(-4) + 4))*e^(-4)/x

________________________________________________________________________________________

maple [A]  time = 0.21, size = 28, normalized size = 1.33

method	result	size
risch	\(\frac {12}{x}+\frac {\left (3+4 x \right ) x^{2 \left (15+3 \ln \relax (x )\right ) {\mathrm e}^{-4}}}{x}\)	\(28\)
norman	\(\frac {12+3 \,{\mathrm e}^{2 \left (3 \ln \relax (x )^{2}+15 \ln \relax (x )\right ) {\mathrm e}^{-4}}+4 x \,{\mathrm e}^{2 \left (3 \ln \relax (x )^{2}+15 \ln \relax (x )\right ) {\mathrm e}^{-4}}}{x}\)	\(50\)
default	\({\mathrm e}^{-4} \left (\frac {3 \,{\mathrm e}^{4} {\mathrm e}^{2 \left (3 \ln \relax (x )^{2}+15 \ln \relax (x )\right ) {\mathrm e}^{-4}}+4 x \,{\mathrm e}^{4} {\mathrm e}^{2 \left (3 \ln \relax (x )^{2}+15 \ln \relax (x )\right ) {\mathrm e}^{-4}}}{x}+\frac {12 \,{\mathrm e}^{4}}{x}\right )\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((48*x+36)*ln(x)-3*exp(4)+120*x+90)*exp((3*ln(x)^2+15*ln(x))/exp(4))^2-12*exp(4))/x^2/exp(4),x,method=_RE
TURNVERBOSE)

[Out]

12/x+(3+4*x)/x*(x^(3*(5+ln(x))*exp(-4)))^2

________________________________________________________________________________________

maxima [C]  time = 0.53, size = 361, normalized size = 17.19 \begin {gather*} -\frac {1}{12} \, {\left (-3 i \, \sqrt {6} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{12} i \, \sqrt {6} {\left (e^{4} - 30\right )} e^{\left (-2\right )} + i \, \sqrt {6} e^{\left (-2\right )} \log \relax (x)\right ) e^{\left (-\frac {1}{24} \, {\left (e^{4} - 30\right )}^{2} e^{\left (-4\right )} + 6\right )} + 90 i \, \sqrt {6} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{12} i \, \sqrt {6} {\left (e^{4} - 30\right )} e^{\left (-2\right )} + i \, \sqrt {6} e^{\left (-2\right )} \log \relax (x)\right ) e^{\left (-\frac {1}{24} \, {\left (e^{4} - 30\right )}^{2} e^{\left (-4\right )} + 2\right )} + 120 i \, \sqrt {6} \sqrt {\pi } \operatorname {erf}\left (i \, \sqrt {6} e^{\left (-2\right )} \log \relax (x) + \frac {5}{2} i \, \sqrt {6} e^{\left (-2\right )}\right ) e^{\left (-\frac {75}{2} \, e^{\left (-4\right )} + 2\right )} + 3 \, \sqrt {6} {\left (\frac {\sqrt {6} \sqrt {\frac {1}{6}} \sqrt {\pi } {\left ({\left (e^{4} - 30\right )} e^{\left (-4\right )} - 12 \, e^{\left (-4\right )} \log \relax (x)\right )} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {\frac {1}{6}} \sqrt {-{\left ({\left (e^{4} - 30\right )} e^{\left (-4\right )} - 12 \, e^{\left (-4\right )} \log \relax (x)\right )}^{2} e^{4}}\right ) - 1\right )} {\left (e^{4} - 30\right )} e^{2}}{\sqrt {-{\left ({\left (e^{4} - 30\right )} e^{\left (-4\right )} - 12 \, e^{\left (-4\right )} \log \relax (x)\right )}^{2} e^{4}}} - 2 \, \sqrt {6} e^{\left (\frac {1}{24} \, {\left ({\left (e^{4} - 30\right )} e^{\left (-4\right )} - 12 \, e^{\left (-4\right )} \log \relax (x)\right )}^{2} e^{4} + 2\right )}\right )} e^{\left (-\frac {1}{24} \, {\left (e^{4} - 30\right )}^{2} e^{\left (-4\right )} + 2\right )} + 8 \, \sqrt {6} {\left (\frac {15 \, \sqrt {\pi } {\left (2 \, e^{\left (-4\right )} \log \relax (x) + 5 \, e^{\left (-4\right )}\right )} {\left (\operatorname {erf}\left (\sqrt {\frac {3}{2}} \sqrt {-{\left (2 \, e^{\left (-4\right )} \log \relax (x) + 5 \, e^{\left (-4\right )}\right )}^{2} e^{4}}\right ) - 1\right )} e^{2}}{\sqrt {-{\left (2 \, e^{\left (-4\right )} \log \relax (x) + 5 \, e^{\left (-4\right )}\right )}^{2} e^{4}}} - \sqrt {6} e^{\left (\frac {3}{2} \, {\left (2 \, e^{\left (-4\right )} \log \relax (x) + 5 \, e^{\left (-4\right )}\right )}^{2} e^{4} + 2\right )}\right )} e^{\left (-\frac {75}{2} \, e^{\left (-4\right )} + 2\right )} - \frac {144 \, e^{4}}{x}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((48*x+36)*log(x)-3*exp(4)+120*x+90)*exp((3*log(x)^2+15*log(x))/exp(4))^2-12*exp(4))/x^2/exp(4),x,
algorithm="maxima")

[Out]

-1/12*(-3*I*sqrt(6)*sqrt(pi)*erf(-1/12*I*sqrt(6)*(e^4 - 30)*e^(-2) + I*sqrt(6)*e^(-2)*log(x))*e^(-1/24*(e^4 -
30)^2*e^(-4) + 6) + 90*I*sqrt(6)*sqrt(pi)*erf(-1/12*I*sqrt(6)*(e^4 - 30)*e^(-2) + I*sqrt(6)*e^(-2)*log(x))*e^(
-1/24*(e^4 - 30)^2*e^(-4) + 2) + 120*I*sqrt(6)*sqrt(pi)*erf(I*sqrt(6)*e^(-2)*log(x) + 5/2*I*sqrt(6)*e^(-2))*e^
(-75/2*e^(-4) + 2) + 3*sqrt(6)*(sqrt(6)*sqrt(1/6)*sqrt(pi)*((e^4 - 30)*e^(-4) - 12*e^(-4)*log(x))*(erf(1/2*sqr
t(1/6)*sqrt(-((e^4 - 30)*e^(-4) - 12*e^(-4)*log(x))^2*e^4)) - 1)*(e^4 - 30)*e^2/sqrt(-((e^4 - 30)*e^(-4) - 12*
e^(-4)*log(x))^2*e^4) - 2*sqrt(6)*e^(1/24*((e^4 - 30)*e^(-4) - 12*e^(-4)*log(x))^2*e^4 + 2))*e^(-1/24*(e^4 - 3
0)^2*e^(-4) + 2) + 8*sqrt(6)*(15*sqrt(pi)*(2*e^(-4)*log(x) + 5*e^(-4))*(erf(sqrt(3/2)*sqrt(-(2*e^(-4)*log(x) +
5*e^(-4))^2*e^4)) - 1)*e^2/sqrt(-(2*e^(-4)*log(x) + 5*e^(-4))^2*e^4) - sqrt(6)*e^(3/2*(2*e^(-4)*log(x) + 5*e^
(-4))^2*e^4 + 2))*e^(-75/2*e^(-4) + 2) - 144*e^4/x)*e^(-4)

________________________________________________________________________________________

mupad [B]  time = 5.28, size = 41, normalized size = 1.95 \begin {gather*} \frac {3\,x^{30\,{\mathrm {e}}^{-4}}\,{\mathrm {e}}^{6\,{\mathrm {e}}^{-4}\,{\ln \relax (x)}^2}+12}{x}+4\,x^{30\,{\mathrm {e}}^{-4}}\,{\mathrm {e}}^{6\,{\mathrm {e}}^{-4}\,{\ln \relax (x)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-4)*(12*exp(4) - exp(2*exp(-4)*(15*log(x) + 3*log(x)^2))*(120*x - 3*exp(4) + log(x)*(48*x + 36) + 90
)))/x^2,x)

[Out]

(3*x^(30*exp(-4))*exp(6*exp(-4)*log(x)^2) + 12)/x + 4*x^(30*exp(-4))*exp(6*exp(-4)*log(x)^2)

________________________________________________________________________________________

sympy [A]  time = 0.30, size = 26, normalized size = 1.24 \begin {gather*} \frac {\left (4 x + 3\right ) e^{\frac {2 \left (3 \log {\relax (x )}^{2} + 15 \log {\relax (x )}\right )}{e^{4}}}}{x} + \frac {12}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((48*x+36)*ln(x)-3*exp(4)+120*x+90)*exp((3*ln(x)**2+15*ln(x))/exp(4))**2-12*exp(4))/x**2/exp(4),x)

[Out]

(4*x + 3)*exp(2*(3*log(x)**2 + 15*log(x))*exp(-4))/x + 12/x

________________________________________________________________________________________