3.17.34 $\int \frac{160e^{1+\frac{5e}{4x\log (3)}}}{-125x^2\log (3)+75e^{\frac{5e}{4x\log (3)}}{x}^2\log (3)-15e^{\frac{5e}{2x\log (3)}}{x}^2\log (3)+e^{\frac{15e}{4x\log (3)}}{x}^2\log (3)}dx$

Optimal. Leaf size=20 $$\frac{64}{{(-5+e^{\frac{5e}{4x\log (3)}})}^2}$$

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Rubi [A]  time = 0.35, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 90, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.033, Rules used = {12, 6688, 6686} $$\begin{array}{c}\frac{64}{{(5-e^{\frac{5e}{x\log (81)}})}^2}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(160*E^(1 + (5*E)/(4*x*Log[3])))/(-125*x^2*Log[3] + 75*E^((5*E)/(4*x*Log[3]))*x^2*Log[3] - 15*E^((5*E)/(2*
x*Log[3]))*x^2*Log[3] + E^((15*E)/(4*x*Log[3]))*x^2*Log[3]),x]

[Out]

64/(5 - E^((5*E)/(x*Log[81])))^2

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =160\int \frac{e^{1+\frac{5e}{4x\log (3)}}}{-125x^2\log (3)+75e^{\frac{5e}{4x\log (3)}}{x}^2\log (3)-15e^{\frac{5e}{2x\log (3)}}{x}^2\log (3)+e^{\frac{15e}{4x\log (3)}}{x}^2\log (3)}dx\\ & =160\int \frac{e^{1+\frac{5e}{x\log (81)}}}{{(-5+e^{\frac{5e}{x\log (81)}})}^3{x}^2\log (3)}dx\\ & =\frac{160\int \frac{e^{1+\frac{5e}{x\log (81)}}}{{(-5+e^{\frac{5e}{x\log (81)}})}^3{x}^2}dx}{\log (3)}\\ & =\frac{64}{{(5-e^{\frac{5e}{x\log (81)}})}^2}\end{array}\end{array}$$

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Mathematica [A]  time = 0.04, size = 18, normalized size = 0.90 $$\begin{array}{c}\frac{64}{{(-5+e^{\frac{5e}{x\log (81)}})}^2}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(160*E^(1 + (5*E)/(4*x*Log[3])))/(-125*x^2*Log[3] + 75*E^((5*E)/(4*x*Log[3]))*x^2*Log[3] - 15*E^((5*
E)/(2*x*Log[3]))*x^2*Log[3] + E^((15*E)/(4*x*Log[3]))*x^2*Log[3]),x]

[Out]

64/(-5 + E^((5*E)/(x*Log[81])))^2

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fricas [B]  time = 0.77, size = 85, normalized size = 4.25 $$\begin{array}{c}-\frac{64e^{(2\log (5)+2)}}{10e^{(\frac{4x\log (5)\log (3)+4x\log (3)+e^{(\log (5)+1)}}{4x\log (3)}+\log (5)+1)}-e^{\left(\frac{4x\log (5)\log (3)+4x\log (3)+e^{(\log (5)+1)}}{2x\log (3)}\right)}-25e^{(2\log (5)+2)}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(32*exp(log(5)+1)*exp(1/4*exp(log(5)+1)/x/log(3))/(x^2*log(3)*exp(1/4*exp(log(5)+1)/x/log(3))^3-15*x^
2*log(3)*exp(1/4*exp(log(5)+1)/x/log(3))^2+75*x^2*log(3)*exp(1/4*exp(log(5)+1)/x/log(3))-125*x^2*log(3)),x, al
gorithm="fricas")

[Out]

-64*e^(2*log(5) + 2)/(10*e^(1/4*(4*x*log(5)*log(3) + 4*x*log(3) + e^(log(5) + 1))/(x*log(3)) + log(5) + 1) - e
^(1/2*(4*x*log(5)*log(3) + 4*x*log(3) + e^(log(5) + 1))/(x*log(3))) - 25*e^(2*log(5) + 2))

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giac [A]  time = 0.55, size = 41, normalized size = 2.05 $$\begin{array}{c}\frac{64e}{25e+e^{(\frac{5e}{2x\log (3)}+1)}-10e^{(\frac{5e}{4x\log (3)}+1)}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(32*exp(log(5)+1)*exp(1/4*exp(log(5)+1)/x/log(3))/(x^2*log(3)*exp(1/4*exp(log(5)+1)/x/log(3))^3-15*x^
2*log(3)*exp(1/4*exp(log(5)+1)/x/log(3))^2+75*x^2*log(3)*exp(1/4*exp(log(5)+1)/x/log(3))-125*x^2*log(3)),x, al
gorithm="giac")

[Out]

64*e/(25*e + e^(5/2*e/(x*log(3)) + 1) - 10*e^(5/4*e/(x*log(3)) + 1))

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maple [A]  time = 0.18, size = 19, normalized size = 0.95

method	result	size
risch	$\frac{64}{{({\mathrm{e}}^{\frac{5\mathrm{e}}{4x\ln (3)}}-5)}^2}$	$19$
norman	$\frac{64}{{({\mathrm{e}}^{\frac{{\mathrm{e}}^{\ln (5)+1}}{4x\ln (3)}}-5)}^2}$	$22$
derivativedivides	$\frac{64{\mathrm{e}}^2{\mathrm{e}}^{-2}}{{({\mathrm{e}}^{\frac{{\mathrm{e}}^{\ln (5)+1}}{4x\ln (3)}}-5)}^2}$	$33$
default	$\frac{64{\mathrm{e}}^2{\mathrm{e}}^{-2}}{{({\mathrm{e}}^{\frac{{\mathrm{e}}^{\ln (5)+1}}{4x\ln (3)}}-5)}^2}$	$33$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(32*exp(ln(5)+1)*exp(1/4*exp(ln(5)+1)/x/ln(3))/(x^2*ln(3)*exp(1/4*exp(ln(5)+1)/x/ln(3))^3-15*x^2*ln(3)*exp(
1/4*exp(ln(5)+1)/x/ln(3))^2+75*x^2*ln(3)*exp(1/4*exp(ln(5)+1)/x/ln(3))-125*x^2*ln(3)),x,method=_RETURNVERBOSE)

[Out]

64/(exp(5/4*exp(1)/x/ln(3))-5)^2

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maxima [B]  time = 0.54, size = 59, normalized size = 2.95 $$\begin{array}{c}-\frac{64(e^{\left(\frac{5e}{2x\log (3)}\right)}-10e^{\left(\frac{5e}{4x\log (3)}\right)})}{25(e^{\left(\frac{5e}{2x\log (3)}\right)}-10e^{\left(\frac{5e}{4x\log (3)}\right)}+25)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(32*exp(log(5)+1)*exp(1/4*exp(log(5)+1)/x/log(3))/(x^2*log(3)*exp(1/4*exp(log(5)+1)/x/log(3))^3-15*x^
2*log(3)*exp(1/4*exp(log(5)+1)/x/log(3))^2+75*x^2*log(3)*exp(1/4*exp(log(5)+1)/x/log(3))-125*x^2*log(3)),x, al
gorithm="maxima")

[Out]

-64/25*(e^(5/2*e/(x*log(3))) - 10*e^(5/4*e/(x*log(3))))/(e^(5/2*e/(x*log(3))) - 10*e^(5/4*e/(x*log(3))) + 25)

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mupad [B]  time = 1.28, size = 18, normalized size = 0.90 $$\begin{array}{c}\frac{64}{{({\mathrm{e}}^{\frac{5\mathrm{e}}{4x\ln (3)}}-5)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(32*exp(log(5) + 1)*exp(exp(log(5) + 1)/(4*x*log(3))))/(125*x^2*log(3) + 15*x^2*exp(exp(log(5) + 1)/(2*x*
log(3)))*log(3) - 75*x^2*exp(exp(log(5) + 1)/(4*x*log(3)))*log(3) - x^2*exp((3*exp(log(5) + 1))/(4*x*log(3)))*
log(3)),x)

[Out]

64/(exp((5*exp(1))/(4*x*log(3))) - 5)^2

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sympy [A]  time = 0.14, size = 31, normalized size = 1.55 $$\begin{array}{c}\frac{64}{-10e^{\frac{5e}{4x\log (3)}}+e^{\frac{5e}{2x\log (3)}}+25}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(32*exp(ln(5)+1)*exp(1/4*exp(ln(5)+1)/x/ln(3))/(x**2*ln(3)*exp(1/4*exp(ln(5)+1)/x/ln(3))**3-15*x**2*l
n(3)*exp(1/4*exp(ln(5)+1)/x/ln(3))**2+75*x**2*ln(3)*exp(1/4*exp(ln(5)+1)/x/ln(3))-125*x**2*ln(3)),x)

[Out]

64/(-10*exp(5*E/(4*x*log(3))) + exp(5*E/(2*x*log(3))) + 25)

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