3.82.16 \(\int e^{-\log ^2(25 x)} (e^x (-2 x+29 x^2+10 x^3)+e^x (2 x-20 x^2) \log (25 x)) \, dx\)

Optimal. Leaf size=21 \[ e^{x-\log ^2(25 x)} x^2 (-1+10 x) \]

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Rubi [A]  time = 0.05, antiderivative size = 22, normalized size of antiderivative = 1.05, number of steps used = 1, number of rules used = 1, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {2288} \begin {gather*} x \left (x-10 x^2\right ) \left (-e^{x-\log ^2(25 x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(-2*x + 29*x^2 + 10*x^3) + E^x*(2*x - 20*x^2)*Log[25*x])/E^Log[25*x]^2,x]

[Out]

-(E^(x - Log[25*x]^2)*x*(x - 10*x^2))

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} &=-e^{x-\log ^2(25 x)} x \left (x-10 x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.35, size = 21, normalized size = 1.00 \begin {gather*} e^{x-\log ^2(25 x)} x^2 (-1+10 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-2*x + 29*x^2 + 10*x^3) + E^x*(2*x - 20*x^2)*Log[25*x])/E^Log[25*x]^2,x]

[Out]

E^(x - Log[25*x]^2)*x^2*(-1 + 10*x)

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fricas [A]  time = 0.66, size = 23, normalized size = 1.10 \begin {gather*} {\left (10 \, x^{3} - x^{2}\right )} e^{\left (-\log \left (25 \, x\right )^{2} + x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x^2+2*x)*exp(x)*log(25*x)+(10*x^3+29*x^2-2*x)*exp(x))/exp(log(25*x)^2),x, algorithm="fricas")

[Out]

(10*x^3 - x^2)*e^(-log(25*x)^2 + x)

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giac [A]  time = 0.15, size = 33, normalized size = 1.57 \begin {gather*} 10 \, x^{3} e^{\left (-\log \left (25 \, x\right )^{2} + x\right )} - x^{2} e^{\left (-\log \left (25 \, x\right )^{2} + x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x^2+2*x)*exp(x)*log(25*x)+(10*x^3+29*x^2-2*x)*exp(x))/exp(log(25*x)^2),x, algorithm="giac")

[Out]

10*x^3*e^(-log(25*x)^2 + x) - x^2*e^(-log(25*x)^2 + x)

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maple [A]  time = 0.03, size = 21, normalized size = 1.00

method	result	size
risch	\(\left (10 x -1\right ) x^{2} {\mathrm e}^{-\ln \left (25 x \right )^{2}+x}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-20*x^2+2*x)*exp(x)*ln(25*x)+(10*x^3+29*x^2-2*x)*exp(x))/exp(ln(25*x)^2),x,method=_RETURNVERBOSE)

[Out]

(10*x-1)*x^2*exp(-ln(25*x)^2+x)

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maxima [A]  time = 0.52, size = 33, normalized size = 1.57 \begin {gather*} {\left (10 \, x^{3} - x^{2}\right )} e^{\left (-4 \, \log \relax (5)^{2} - 4 \, \log \relax (5) \log \relax (x) - \log \relax (x)^{2} + x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x^2+2*x)*exp(x)*log(25*x)+(10*x^3+29*x^2-2*x)*exp(x))/exp(log(25*x)^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.94, size = 38, normalized size = 1.81 \begin {gather*} -\frac {{\mathrm {e}}^{-{\ln \relax (x)}^2-4\,{\ln \relax (5)}^2}\,\left (x^2\,{\mathrm {e}}^x-10\,x^3\,{\mathrm {e}}^x\right )}{x^{4\,\ln \relax (5)}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-log(25*x)^2)*(exp(x)*(29*x^2 - 2*x + 10*x^3) + log(25*x)*exp(x)*(2*x - 20*x^2)),x)

[Out]

-(exp(- log(x)^2 - 4*log(5)^2)*(x^2*exp(x) - 10*x^3*exp(x)))/x^(4*log(5))

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sympy [A]  time = 13.94, size = 27, normalized size = 1.29 \begin {gather*} \left (10 x^{3} e^{- \log {\left (25 x \right )}^{2}} - x^{2} e^{- \log {\left (25 x \right )}^{2}}\right ) e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x**2+2*x)*exp(x)*ln(25*x)+(10*x**3+29*x**2-2*x)*exp(x))/exp(ln(25*x)**2),x)

[Out]

(10*x**3*exp(-log(25*x)**2) - x**2*exp(-log(25*x)**2))*exp(x)

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