3.33.7 $\int \frac{e(3+3x)}{25e^3{x}^2+10e^4{x}^2\log (\frac{8e^{\frac{1}{x}}}{x})+e^5{x}^2{\log }^2(\frac{8e^{\frac{1}{x}}}{x})}dx$

Optimal. Leaf size=22 $$\frac{3}{e^3(5+e\log \left(\frac{8e^{\frac{1}{x}}}{x}\right))}$$

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Rubi [A]  time = 0.16, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 57, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.053, Rules used = {12, 6688, 6686} $$\begin{array}{c}\frac{3}{e^3(e\log \left(\frac{8e^{\frac{1}{x}}}{x}\right)+5)}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(E*(3 + 3*x))/(25*E^3*x^2 + 10*E^4*x^2*Log[(8*E^x^(-1))/x] + E^5*x^2*Log[(8*E^x^(-1))/x]^2),x]

[Out]

3/(E^3*(5 + E*Log[(8*E^x^(-1))/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 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =e\int \frac{3+3x}{25e^3{x}^2+10e^4{x}^2\log \left(\frac{8e^{\frac{1}{x}}}{x}\right)+e^5{x}^2{\log }^2\left(\frac{8e^{\frac{1}{x}}}{x}\right)}dx\\ & =e\int \frac{3(1+x)}{e^3{x}^2{(5+e\log \left(\frac{8e^{\frac{1}{x}}}{x}\right))}^2}dx\\ & =\frac{3\int \frac{1+x}{x^2{(5+e\log \left(\frac{8e^{\frac{1}{x}}}{x}\right))}^2}dx}{e^2}\\ & =\frac{3}{e^3(5+e\log \left(\frac{8e^{\frac{1}{x}}}{x}\right))}\end{array}\end{array}$$

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Mathematica [A]  time = 0.02, size = 22, normalized size = 1.00 $$\begin{array}{c}\frac{3}{e^3(5+e\log \left(\frac{8e^{\frac{1}{x}}}{x}\right))}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E*(3 + 3*x))/(25*E^3*x^2 + 10*E^4*x^2*Log[(8*E^x^(-1))/x] + E^5*x^2*Log[(8*E^x^(-1))/x]^2),x]

[Out]

3/(E^3*(5 + E*Log[(8*E^x^(-1))/x]))

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fricas [A]  time = 0.55, size = 22, normalized size = 1.00 $$\begin{array}{c}\frac{3}{e^4\log \left(\frac{8e^{\frac{1}{x}}}{x}\right)+5e^3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x+3)*exp(1)/(x^2*exp(1)^2*exp(3)*log(8*exp(1/x)/x)^2+10*x^2*exp(1)*exp(3)*log(8*exp(1/x)/x)+25*x^
2*exp(3)),x, algorithm="fricas")

[Out]

3/(e^4*log(8*e^(1/x)/x) + 5*e^3)

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giac [A]  time = 0.25, size = 29, normalized size = 1.32 $$\begin{array}{c}\frac{3xe}{3x{e}^5\log (2)-x{e}^5\log (x)+5x{e}^4+e^5}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x+3)*exp(1)/(x^2*exp(1)^2*exp(3)*log(8*exp(1/x)/x)^2+10*x^2*exp(1)*exp(3)*log(8*exp(1/x)/x)+25*x^
2*exp(3)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.52, size = 36, normalized size = 1.64

method	result	size
norman	$-\frac{3{\mathrm{e}}^{-3}\mathrm{e}\ln \left(\frac{8{\mathrm{e}}^{\frac{1}{x}}}{x}\right)}{5(5+\ln \left(\frac{8{\mathrm{e}}^{\frac{1}{x}}}{x}\right)\mathrm{e})}$	$36$
default	$-\frac{3{\mathrm{e}}^{-3}x}{x\mathrm{e}\ln (x)-\mathrm{e}x(\ln \left(\frac{8{\mathrm{e}}^{\frac{1}{x}}}{x}\right)-\frac{1}{x}+\ln (x))-\mathrm{e}-5x}$	$47$
risch	$\frac{6i{\mathrm{e}}^{-3}}{\pi \mathrm{e}\mathrm{csgn}\left(\frac{i}{x}\right)\mathrm{csgn}\left(i{\mathrm{e}}^{\frac{1}{x}}\right)\mathrm{csgn}\left(\frac{i{\mathrm{e}}^{\frac{1}{x}}}{x}\right)-\pi \mathrm{e}\mathrm{csgn}\left(\frac{i}{x}\right)\mathrm{csgn}{\left(\frac{i{\mathrm{e}}^{\frac{1}{x}}}{x}\right)}^2-\pi \mathrm{e}\mathrm{csgn}\left(i{\mathrm{e}}^{\frac{1}{x}}\right)\mathrm{csgn}{\left(\frac{i{\mathrm{e}}^{\frac{1}{x}}}{x}\right)}^2+\pi \mathrm{e}\mathrm{csgn}{\left(\frac{i{\mathrm{e}}^{\frac{1}{x}}}{x}\right)}^3+6i\mathrm{e}\ln (2)-2i\mathrm{e}\ln (x)+2i\mathrm{e}\ln \left({\mathrm{e}}^{\frac{1}{x}}\right)+10i}$	$133$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x+3)*exp(1)/(x^2*exp(1)^2*exp(3)*ln(8*exp(1/x)/x)^2+10*x^2*exp(1)*exp(3)*ln(8*exp(1/x)/x)+25*x^2*exp(3)
),x,method=_RETURNVERBOSE)

[Out]

-3/5/exp(3)*exp(1)*ln(8*exp(1/x)/x)/(5+ln(8*exp(1/x)/x)*exp(1))

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maxima [A]  time = 0.58, size = 32, normalized size = 1.45 $$\begin{array}{c}-\frac{3xe}{x{e}^5\log (x)-(3e^5\log (2)+5e^4)x-e^5}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x+3)*exp(1)/(x^2*exp(1)^2*exp(3)*log(8*exp(1/x)/x)^2+10*x^2*exp(1)*exp(3)*log(8*exp(1/x)/x)+25*x^
2*exp(3)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.09, size = 20, normalized size = 0.91 $$\begin{array}{c}\frac{3{\mathrm{e}}^{-4}}{5{\mathrm{e}}^{-1}+\ln \left(\frac{8}{x}\right)+\frac{1}{x}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(1)*(3*x + 3))/(25*x^2*exp(3) + 10*x^2*exp(4)*log((8*exp(1/x))/x) + x^2*exp(5)*log((8*exp(1/x))/x)^2),
x)

[Out]

(3*exp(-4))/(5*exp(-1) + log(8/x) + 1/x)

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sympy [A]  time = 0.31, size = 19, normalized size = 0.86 $$\begin{array}{c}\frac{3}{e^4\log \left(\frac{8e^{\frac{1}{x}}}{x}\right)+5e^3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x+3)*exp(1)/(x**2*exp(1)**2*exp(3)*ln(8*exp(1/x)/x)**2+10*x**2*exp(1)*exp(3)*ln(8*exp(1/x)/x)+25*
x**2*exp(3)),x)

[Out]

3/(exp(4)*log(8*exp(1/x)/x) + 5*exp(3))

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