3.45.24 \(\int \frac {1}{25} e^{\frac {1}{25} (x^3+x^2 \log (3)-25 x^2 \log (5 x))} (-25 x+3 x^2+2 x \log (3)-50 x \log (5 x)) \, dx\)

Optimal. Leaf size=19 \[ e^{\frac {1}{25} x^2 (x+\log (3)-25 \log (5 x))} \]

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Rubi [A]  time = 0.19, antiderivative size = 33, normalized size of antiderivative = 1.74, number of steps used = 3, number of rules used = 3, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {6, 12, 6706} \begin {gather*} 3^{\frac {x^2}{25}} 5^{-x^2} e^{\frac {x^3}{25}} x^{-x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((x^3 + x^2*Log[3] - 25*x^2*Log[5*x])/25)*(-25*x + 3*x^2 + 2*x*Log[3] - 50*x*Log[5*x]))/25,x]

[Out]

(3^(x^2/25)*E^(x^3/25))/(5^x^2*x^x^2)

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1}{25} e^{\frac {1}{25} \left (x^3+x^2 \log (3)-25 x^2 \log (5 x)\right )} \left (3 x^2+x (-25+2 \log (3))-50 x \log (5 x)\right ) \, dx\\ &=\frac {1}{25} \int e^{\frac {1}{25} \left (x^3+x^2 \log (3)-25 x^2 \log (5 x)\right )} \left (3 x^2+x (-25+2 \log (3))-50 x \log (5 x)\right ) \, dx\\ &=3^{\frac {x^2}{25}} 5^{-x^2} e^{\frac {x^3}{25}} x^{-x^2}\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.91, size = 53, normalized size = 2.79 \begin {gather*} \frac {1}{25} \int e^{\frac {1}{25} \left (x^3+x^2 \log (3)-25 x^2 \log (5 x)\right )} \left (-25 x+3 x^2+2 x \log (3)-50 x \log (5 x)\right ) \, dx \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((x^3 + x^2*Log[3] - 25*x^2*Log[5*x])/25)*(-25*x + 3*x^2 + 2*x*Log[3] - 50*x*Log[5*x]))/25,x]

[Out]

Integrate[E^((x^3 + x^2*Log[3] - 25*x^2*Log[5*x])/25)*(-25*x + 3*x^2 + 2*x*Log[3] - 50*x*Log[5*x]), x]/25

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fricas [A]  time = 0.59, size = 23, normalized size = 1.21 \begin {gather*} e^{\left (\frac {1}{25} \, x^{3} + \frac {1}{25} \, x^{2} \log \relax (3) - x^{2} \log \left (5 \, x\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(-50*x*log(5*x)+2*x*log(3)+3*x^2-25*x)*exp(-x^2*log(5*x)+1/25*x^2*log(3)+1/25*x^3),x, algorithm
="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 23, normalized size = 1.21 \begin {gather*} e^{\left (\frac {1}{25} \, x^{3} + \frac {1}{25} \, x^{2} \log \relax (3) - x^{2} \log \left (5 \, x\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(-50*x*log(5*x)+2*x*log(3)+3*x^2-25*x)*exp(-x^2*log(5*x)+1/25*x^2*log(3)+1/25*x^3),x, algorithm
="giac")

[Out]

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

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maple [A]  time = 0.05, size = 24, normalized size = 1.26

method	result	size
norman	\({\mathrm e}^{-x^{2} \ln \left (5 x \right )+\frac {x^{2} \ln \relax (3)}{25}+\frac {x^{3}}{25}}\)	\(24\)
risch	\(\left (5 x \right )^{-x^{2}} 3^{\frac {x^{2}}{25}} {\mathrm e}^{\frac {x^{3}}{25}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/25*(-50*x*ln(5*x)+2*x*ln(3)+3*x^2-25*x)*exp(-x^2*ln(5*x)+1/25*x^2*ln(3)+1/25*x^3),x,method=_RETURNVERBOS
E)

[Out]

exp(-x^2*ln(5*x)+1/25*x^2*ln(3)+1/25*x^3)

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maxima [A]  time = 0.37, size = 23, normalized size = 1.21 \begin {gather*} e^{\left (\frac {1}{25} \, x^{3} + \frac {1}{25} \, x^{2} \log \relax (3) - x^{2} \log \left (5 \, x\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(-50*x*log(5*x)+2*x*log(3)+3*x^2-25*x)*exp(-x^2*log(5*x)+1/25*x^2*log(3)+1/25*x^3),x, algorithm
="maxima")

[Out]

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

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mupad [B]  time = 3.17, size = 26, normalized size = 1.37 \begin {gather*} \frac {{\left (\frac {1}{5}\right )}^{x^2}\,3^{\frac {x^2}{25}}\,{\mathrm {e}}^{\frac {x^3}{25}}}{x^{x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((x^2*log(3))/25 - x^2*log(5*x) + x^3/25)*(25*x + 50*x*log(5*x) - 2*x*log(3) - 3*x^2))/25,x)

[Out]

((1/5)^(x^2)*3^(x^2/25)*exp(x^3/25))/x^(x^2)

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sympy [A]  time = 0.33, size = 22, normalized size = 1.16 \begin {gather*} e^{\frac {x^{3}}{25} - x^{2} \log {\left (5 x \right )} + \frac {x^{2} \log {\relax (3 )}}{25}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(-50*x*ln(5*x)+2*x*ln(3)+3*x**2-25*x)*exp(-x**2*ln(5*x)+1/25*x**2*ln(3)+1/25*x**3),x)

[Out]

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

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