3.60.18 \(\int \frac {e^x (3-15 x+15 x^2+625 x^3)+e^x (-3 x-250 x^2) \log (x)+25 e^x x \log ^2(x)}{25 x^3 \log (9)-10 x^2 \log (9) \log (x)+x \log (9) \log ^2(x)} \, dx\)

Optimal. Leaf size=22 \[ \frac {e^x \left (25+\frac {3}{5 x-\log (x)}\right )}{\log (9)} \]

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Rubi [B]  time = 0.35, antiderivative size = 49, normalized size of antiderivative = 2.23, number of steps used = 3, number of rules used = 3, integrand size = 73, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {6688, 12, 2288} \begin {gather*} \frac {e^x \left (625 x^3+15 x^2+25 x \log ^2(x)-(250 x+3) x \log (x)\right )}{x \log (9) (5 x-\log (x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(3 - 15*x + 15*x^2 + 625*x^3) + E^x*(-3*x - 250*x^2)*Log[x] + 25*E^x*x*Log[x]^2)/(25*x^3*Log[9] - 10*
x^2*Log[9]*Log[x] + x*Log[9]*Log[x]^2),x]

[Out]

(E^x*(15*x^2 + 625*x^3 - x*(3 + 250*x)*Log[x] + 25*x*Log[x]^2))/(x*Log[9]*(5*x - Log[x])^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 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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^x \left (3-15 x+15 x^2+625 x^3-x (3+250 x) \log (x)+25 x \log ^2(x)\right )}{x \log (9) (5 x-\log (x))^2} \, dx\\ &=\frac {\int \frac {e^x \left (3-15 x+15 x^2+625 x^3-x (3+250 x) \log (x)+25 x \log ^2(x)\right )}{x (5 x-\log (x))^2} \, dx}{\log (9)}\\ &=\frac {e^x \left (15 x^2+625 x^3-x (3+250 x) \log (x)+25 x \log ^2(x)\right )}{x \log (9) (5 x-\log (x))^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 27, normalized size = 1.23 \begin {gather*} \frac {e^x (3+125 x-25 \log (x))}{\log (9) (5 x-\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(3 - 15*x + 15*x^2 + 625*x^3) + E^x*(-3*x - 250*x^2)*Log[x] + 25*E^x*x*Log[x]^2)/(25*x^3*Log[9]
- 10*x^2*Log[9]*Log[x] + x*Log[9]*Log[x]^2),x]

[Out]

(E^x*(3 + 125*x - 25*Log[x]))/(Log[9]*(5*x - Log[x]))

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fricas [A]  time = 0.73, size = 31, normalized size = 1.41 \begin {gather*} \frac {{\left (125 \, x + 3\right )} e^{x} - 25 \, e^{x} \log \relax (x)}{2 \, {\left (5 \, x \log \relax (3) - \log \relax (3) \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*x*exp(x)*log(x)^2+(-250*x^2-3*x)*exp(x)*log(x)+(625*x^3+15*x^2-15*x+3)*exp(x))/(2*x*log(3)*log(x
)^2-20*x^2*log(3)*log(x)+50*x^3*log(3)),x, algorithm="fricas")

[Out]

1/2*((125*x + 3)*e^x - 25*e^x*log(x))/(5*x*log(3) - log(3)*log(x))

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giac [A]  time = 0.17, size = 32, normalized size = 1.45 \begin {gather*} \frac {125 \, x e^{x} - 25 \, e^{x} \log \relax (x) + 3 \, e^{x}}{2 \, {\left (5 \, x \log \relax (3) - \log \relax (3) \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*x*exp(x)*log(x)^2+(-250*x^2-3*x)*exp(x)*log(x)+(625*x^3+15*x^2-15*x+3)*exp(x))/(2*x*log(3)*log(x
)^2-20*x^2*log(3)*log(x)+50*x^3*log(3)),x, algorithm="giac")

[Out]

1/2*(125*x*e^x - 25*e^x*log(x) + 3*e^x)/(5*x*log(3) - log(3)*log(x))

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maple [A]  time = 0.04, size = 28, normalized size = 1.27

method	result	size
risch	\(\frac {25 \,{\mathrm e}^{x}}{2 \ln \relax (3)}+\frac {3 \,{\mathrm e}^{x}}{2 \ln \relax (3) \left (-\ln \relax (x )+5 x \right )}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25*x*exp(x)*ln(x)^2+(-250*x^2-3*x)*exp(x)*ln(x)+(625*x^3+15*x^2-15*x+3)*exp(x))/(2*x*ln(3)*ln(x)^2-20*x^2
*ln(3)*ln(x)+50*x^3*ln(3)),x,method=_RETURNVERBOSE)

[Out]

25/2*exp(x)/ln(3)+3/2*exp(x)/ln(3)/(-ln(x)+5*x)

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maxima [A]  time = 0.50, size = 27, normalized size = 1.23 \begin {gather*} \frac {{\left (125 \, x - 25 \, \log \relax (x) + 3\right )} e^{x}}{2 \, {\left (5 \, x \log \relax (3) - \log \relax (3) \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*x*exp(x)*log(x)^2+(-250*x^2-3*x)*exp(x)*log(x)+(625*x^3+15*x^2-15*x+3)*exp(x))/(2*x*log(3)*log(x
)^2-20*x^2*log(3)*log(x)+50*x^3*log(3)),x, algorithm="maxima")

[Out]

1/2*(125*x - 25*log(x) + 3)*e^x/(5*x*log(3) - log(3)*log(x))

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mupad [B]  time = 4.71, size = 27, normalized size = 1.23 \begin {gather*} \frac {{\mathrm {e}}^x\,\left (125\,x-25\,\ln \relax (x)+3\right )}{2\,\ln \relax (3)\,\left (5\,x-\ln \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(15*x^2 - 15*x + 625*x^3 + 3) - exp(x)*log(x)*(3*x + 250*x^2) + 25*x*exp(x)*log(x)^2)/(50*x^3*log(
3) + 2*x*log(3)*log(x)^2 - 20*x^2*log(3)*log(x)),x)

[Out]

(exp(x)*(125*x - 25*log(x) + 3))/(2*log(3)*(5*x - log(x)))

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sympy [A]  time = 0.31, size = 27, normalized size = 1.23 \begin {gather*} \frac {\left (125 x - 25 \log {\relax (x )} + 3\right ) e^{x}}{10 x \log {\relax (3 )} - 2 \log {\relax (3 )} \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*x*exp(x)*ln(x)**2+(-250*x**2-3*x)*exp(x)*ln(x)+(625*x**3+15*x**2-15*x+3)*exp(x))/(2*x*ln(3)*ln(x
)**2-20*x**2*ln(3)*ln(x)+50*x**3*ln(3)),x)

[Out]

(125*x - 25*log(x) + 3)*exp(x)/(10*x*log(3) - 2*log(3)*log(x))

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