3.93.93 \(\int \frac {(-6 x^3+4 x^4) \log (25)+e^{\frac {-1+\log (25)}{x^2 \log (25)}} (6-2 x+(-6+2 x-x^3) \log (25))}{x^3 \log (25)} \, dx\)

Optimal. Leaf size=24 \[ (-3+x) \left (-e^{\frac {1-\frac {1}{\log (25)}}{x^2}}+2 x\right ) \]

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Rubi [A]  time = 0.13, antiderivative size = 32, normalized size of antiderivative = 1.33, number of steps used = 4, number of rules used = 3, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {12, 14, 2288} \begin {gather*} (3-x) e^{\frac {1-\frac {1}{\log (25)}}{x^2}}+\frac {1}{2} (3-2 x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-6*x^3 + 4*x^4)*Log[25] + E^((-1 + Log[25])/(x^2*Log[25]))*(6 - 2*x + (-6 + 2*x - x^3)*Log[25]))/(x^3*Lo
g[25]),x]

[Out]

(3 - 2*x)^2/2 + E^((1 - Log[25]^(-1))/x^2)*(3 - 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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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} &=\frac {\int \frac {\left (-6 x^3+4 x^4\right ) \log (25)+e^{\frac {-1+\log (25)}{x^2 \log (25)}} \left (6-2 x+\left (-6+2 x-x^3\right ) \log (25)\right )}{x^3} \, dx}{\log (25)}\\ &=\frac {\int \left (2 (-3+2 x) \log (25)+\frac {e^{\frac {1-\frac {1}{\log (25)}}{x^2}} \left (6 (1-\log (25))-2 x (1-\log (25))-x^3 \log (25)\right )}{x^3}\right ) \, dx}{\log (25)}\\ &=\frac {1}{2} (3-2 x)^2+\frac {\int \frac {e^{\frac {1-\frac {1}{\log (25)}}{x^2}} \left (6 (1-\log (25))-2 x (1-\log (25))-x^3 \log (25)\right )}{x^3} \, dx}{\log (25)}\\ &=\frac {1}{2} (3-2 x)^2+e^{\frac {1-\frac {1}{\log (25)}}{x^2}} (3-x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 47, normalized size = 1.96 \begin {gather*} -6 x+2 x^2-\frac {e^{\frac {-1+\log (25)}{x^2 \log (25)}} (6-6 \log (25)+x (-2+2 \log (25)))}{2 (-1+\log (25))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-6*x^3 + 4*x^4)*Log[25] + E^((-1 + Log[25])/(x^2*Log[25]))*(6 - 2*x + (-6 + 2*x - x^3)*Log[25]))/(
x^3*Log[25]),x]

[Out]

-6*x + 2*x^2 - (E^((-1 + Log[25])/(x^2*Log[25]))*(6 - 6*Log[25] + x*(-2 + 2*Log[25])))/(2*(-1 + Log[25]))

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fricas [A]  time = 0.57, size = 30, normalized size = 1.25 \begin {gather*} 2 \, x^{2} - {\left (x - 3\right )} e^{\left (\frac {2 \, \log \relax (5) - 1}{2 \, x^{2} \log \relax (5)}\right )} - 6 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((2*(-x^3+2*x-6)*log(5)+6-2*x)*exp(1/2*(2*log(5)-1)/x^2/log(5))+2*(4*x^4-6*x^3)*log(5))/x^3/log(
5),x, algorithm="fricas")

[Out]

2*x^2 - (x - 3)*e^(1/2*(2*log(5) - 1)/(x^2*log(5))) - 6*x

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giac [B]  time = 0.26, size = 59, normalized size = 2.46 \begin {gather*} \frac {2 \, x^{2} \log \relax (5) - x e^{\left (\frac {2 \, \log \relax (5) - 1}{2 \, x^{2} \log \relax (5)}\right )} \log \relax (5) - 6 \, x \log \relax (5) + 3 \, e^{\left (\frac {2 \, \log \relax (5) - 1}{2 \, x^{2} \log \relax (5)}\right )} \log \relax (5)}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((2*(-x^3+2*x-6)*log(5)+6-2*x)*exp(1/2*(2*log(5)-1)/x^2/log(5))+2*(4*x^4-6*x^3)*log(5))/x^3/log(
5),x, algorithm="giac")

[Out]

(2*x^2*log(5) - x*e^(1/2*(2*log(5) - 1)/(x^2*log(5)))*log(5) - 6*x*log(5) + 3*e^(1/2*(2*log(5) - 1)/(x^2*log(5
)))*log(5))/log(5)

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maple [A]  time = 0.23, size = 42, normalized size = 1.75

method	result	size
risch	\(2 x^{2}-6 x +\frac {\left (-2 x \ln \relax (5)+6 \ln \relax (5)\right ) {\mathrm e}^{\frac {2 \ln \relax (5)-1}{2 x^{2} \ln \relax (5)}}}{2 \ln \relax (5)}\)	\(42\)
norman	\(\frac {-6 x^{3}+2 x^{4}+3 x^{2} {\mathrm e}^{\frac {2 \ln \relax (5)-1}{2 x^{2} \ln \relax (5)}}-x^{3} {\mathrm e}^{\frac {2 \ln \relax (5)-1}{2 x^{2} \ln \relax (5)}}}{x^{2}}\)	\(58\)
derivativedivides	\(-\frac {-4 x^{2} \ln \relax (5)+12 x \ln \relax (5)+\frac {3 \,{\mathrm e}^{\frac {1-\frac {1}{2 \ln \relax (5)}}{x^{2}}}}{1-\frac {1}{2 \ln \relax (5)}}-\frac {6 \ln \relax (5) {\mathrm e}^{\frac {1-\frac {1}{2 \ln \relax (5)}}{x^{2}}}}{1-\frac {1}{2 \ln \relax (5)}}+2 \ln \relax (5) x \,{\mathrm e}^{\frac {1-\frac {1}{2 \ln \relax (5)}}{x^{2}}}}{2 \ln \relax (5)}\)	\(90\)
default	\(\frac {4 x^{2} \ln \relax (5)-12 x \ln \relax (5)-\frac {3 \,{\mathrm e}^{\frac {1-\frac {1}{2 \ln \relax (5)}}{x^{2}}}}{1-\frac {1}{2 \ln \relax (5)}}+\frac {6 \ln \relax (5) {\mathrm e}^{\frac {1-\frac {1}{2 \ln \relax (5)}}{x^{2}}}}{1-\frac {1}{2 \ln \relax (5)}}-2 \ln \relax (5) x \,{\mathrm e}^{\frac {1-\frac {1}{2 \ln \relax (5)}}{x^{2}}}}{2 \ln \relax (5)}\)	\(90\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((2*(-x^3+2*x-6)*ln(5)+6-2*x)*exp(1/2*(2*ln(5)-1)/x^2/ln(5))+2*(4*x^4-6*x^3)*ln(5))/x^3/ln(5),x,method
=_RETURNVERBOSE)

[Out]

2*x^2-6*x+1/2/ln(5)*(-2*x*ln(5)+6*ln(5))*exp(1/2*(2*ln(5)-1)/x^2/ln(5))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {2 \, x^{2} \log \relax (5) - x e^{\left (\frac {2 \, \log \relax (5) - 1}{2 \, x^{2} \log \relax (5)}\right )} \log \relax (5) - 6 \, x \log \relax (5) + 3 \, e^{\left (\frac {2 \, \log \relax (5) - 1}{2 \, x^{2} \log \relax (5)}\right )} \log \relax (5)}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((2*(-x^3+2*x-6)*log(5)+6-2*x)*exp(1/2*(2*log(5)-1)/x^2/log(5))+2*(4*x^4-6*x^3)*log(5))/x^3/log(
5),x, algorithm="maxima")

[Out]

(2*x^2*log(5) - 6*x*log(5) - integrate((x^3*log(5) - x*(2*log(5) - 1) + 6*log(5) - 3)*e^(1/x^2 - 1/2/(x^2*log(
5)))/x^3, x))/log(5)

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mupad [B]  time = 7.19, size = 52, normalized size = 2.17 \begin {gather*} \frac {{\mathrm {e}}^{\frac {1}{x^2}-\frac {1}{2\,x^2\,\ln \relax (5)}}\,\ln \left (125\right )}{\ln \relax (5)}-x\,{\mathrm {e}}^{\frac {1}{x^2}-\frac {1}{2\,x^2\,\ln \relax (5)}}-6\,x+\frac {x^2\,\ln \left (25\right )}{\ln \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(5)*(6*x^3 - 4*x^4) + (exp((log(5) - 1/2)/(x^2*log(5)))*(2*x + 2*log(5)*(x^3 - 2*x + 6) - 6))/2)/(x^3
*log(5)),x)

[Out]

(exp(1/x^2 - 1/(2*x^2*log(5)))*log(125))/log(5) - x*exp(1/x^2 - 1/(2*x^2*log(5))) - 6*x + (x^2*log(25))/log(5)

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sympy [A]  time = 0.29, size = 26, normalized size = 1.08 \begin {gather*} 2 x^{2} - 6 x + \left (3 - x\right ) e^{\frac {- \frac {1}{2} + \log {\relax (5 )}}{x^{2} \log {\relax (5 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((2*(-x**3+2*x-6)*ln(5)+6-2*x)*exp(1/2*(2*ln(5)-1)/x**2/ln(5))+2*(4*x**4-6*x**3)*ln(5))/x**3/ln(
5),x)

[Out]

2*x**2 - 6*x + (3 - x)*exp((-1/2 + log(5))/(x**2*log(5)))

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