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

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

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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} $$\begin{array}{c}\frac{x^{\frac{30}{e^4}-1}{e}^{\frac{6{\log }^2(x)}{e^4}}(4x\log (x)+3\log (x))}{\log (x)}+\frac{12}{x}\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{\int \frac{-12e^4+e^{\frac{2(15\log (x)+3{\log }^2(x))}{e^4}}(90-3e^4+120x+(36+48x)\log (x))}{x^2}dx}{e^4}\\ & =\frac{\int (-\frac{12e^4}{x^2}+3e^{\frac{6{\log }^2(x)}{e^4}}{x}^{-2+\frac{30}{e^4}}(30(1-\frac{e^4}{30})+40x+12\log (x)+16x\log (x)))dx}{e^4}\\ & =\frac{12}{x}+\frac{3\int e^{\frac{6{\log }^2(x)}{e^4}}{x}^{-2+\frac{30}{e^4}}(30(1-\frac{e^4}{30})+40x+12\log (x)+16x\log (x))dx}{e^4}\\ & =\frac{12}{x}+\frac{e^{\frac{6{\log }^2(x)}{e^4}}{x}^{-1+\frac{30}{e^4}}(3\log (x)+4x\log (x))}{\log (x)}\end{array}\end{array}$$

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

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)

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

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

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

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

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

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

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

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

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)

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

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)

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sympy [A]  time = 0.30, size = 26, normalized size = 1.24 $$\begin{array}{c}\frac{(4x+3)e^{\frac{2(3\log {(x)}^2+15\log (x))}{e^4}}}{x}+\frac{12}{x}\end{array}$$

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

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