3.9.44 \(\int \frac {6 e^5+400 x^2+e^x (-x^2+x^3)-6 e^5 \log (x)}{6 e^5 x^2} \, dx\)

Optimal. Leaf size=22 \[ \frac {\left (400+e^x\right ) (-2+x)}{6 e^5}+\frac {\log (x)}{x} \]

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Rubi [A]  time = 0.06, antiderivative size = 38, normalized size of antiderivative = 1.73, number of steps used = 10, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {12, 14, 2176, 2194, 2304} \begin {gather*} -\frac {1}{6} e^{x-5} (1-x)-\frac {e^{x-5}}{6}+\frac {200 x}{3 e^5}+\frac {\log (x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6*E^5 + 400*x^2 + E^x*(-x^2 + x^3) - 6*E^5*Log[x])/(6*E^5*x^2),x]

[Out]

-1/6*E^(-5 + x) - (E^(-5 + x)*(1 - x))/6 + (200*x)/(3*E^5) + Log[x]/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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {6 e^5+400 x^2+e^x \left (-x^2+x^3\right )-6 e^5 \log (x)}{x^2} \, dx}{6 e^5}\\ &=\frac {\int \left (e^x (-1+x)-\frac {2 \left (-3 e^5-200 x^2+3 e^5 \log (x)\right )}{x^2}\right ) \, dx}{6 e^5}\\ &=\frac {\int e^x (-1+x) \, dx}{6 e^5}-\frac {\int \frac {-3 e^5-200 x^2+3 e^5 \log (x)}{x^2} \, dx}{3 e^5}\\ &=-\frac {1}{6} e^{-5+x} (1-x)-\frac {\int e^x \, dx}{6 e^5}-\frac {\int \left (\frac {-3 e^5-200 x^2}{x^2}+\frac {3 e^5 \log (x)}{x^2}\right ) \, dx}{3 e^5}\\ &=-\frac {1}{6} e^{-5+x}-\frac {1}{6} e^{-5+x} (1-x)-\frac {\int \frac {-3 e^5-200 x^2}{x^2} \, dx}{3 e^5}-\int \frac {\log (x)}{x^2} \, dx\\ &=-\frac {1}{6} e^{-5+x}-\frac {1}{6} e^{-5+x} (1-x)+\frac {1}{x}+\frac {\log (x)}{x}-\frac {\int \left (-200-\frac {3 e^5}{x^2}\right ) \, dx}{3 e^5}\\ &=-\frac {1}{6} e^{-5+x}-\frac {1}{6} e^{-5+x} (1-x)+\frac {200 x}{3 e^5}+\frac {\log (x)}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 33, normalized size = 1.50 \begin {gather*} \frac {200 x}{3 e^5}+\frac {1}{6} e^x \left (-\frac {2}{e^5}+\frac {x}{e^5}\right )+\frac {\log (x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6*E^5 + 400*x^2 + E^x*(-x^2 + x^3) - 6*E^5*Log[x])/(6*E^5*x^2),x]

[Out]

(200*x)/(3*E^5) + (E^x*(-2/E^5 + x/E^5))/6 + Log[x]/x

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fricas [A]  time = 0.64, size = 29, normalized size = 1.32 \begin {gather*} \frac {{\left (400 \, x^{2} + {\left (x^{2} - 2 \, x\right )} e^{x} + 6 \, e^{5} \log \relax (x)\right )} e^{\left (-5\right )}}{6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(-6*exp(5)*log(x)+(x^3-x^2)*exp(x)+6*exp(5)+400*x^2)/x^2/exp(5),x, algorithm="fricas")

[Out]

1/6*(400*x^2 + (x^2 - 2*x)*e^x + 6*e^5*log(x))*e^(-5)/x

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giac [A]  time = 0.43, size = 30, normalized size = 1.36 \begin {gather*} \frac {{\left (x^{2} e^{x} + 400 \, x^{2} - 2 \, x e^{x} + 6 \, e^{5} \log \relax (x)\right )} e^{\left (-5\right )}}{6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(-6*exp(5)*log(x)+(x^3-x^2)*exp(x)+6*exp(5)+400*x^2)/x^2/exp(5),x, algorithm="giac")

[Out]

1/6*(x^2*e^x + 400*x^2 - 2*x*e^x + 6*e^5*log(x))*e^(-5)/x

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maple [A]  time = 0.08, size = 26, normalized size = 1.18

method	result	size
risch	\(\frac {x \,{\mathrm e}^{x -5}}{6}-\frac {{\mathrm e}^{x -5}}{3}+\frac {200 x \,{\mathrm e}^{-5}}{3}+\frac {\ln \relax (x )}{x}\)	\(26\)
default	\(\frac {{\mathrm e}^{-5} \left ({\mathrm e}^{x} x -2 \,{\mathrm e}^{x}+400 x +\frac {6 \,{\mathrm e}^{5} \ln \relax (x )}{x}\right )}{6}\)	\(28\)
norman	\(\frac {\frac {200 \,{\mathrm e}^{-5} x^{2}}{3}-\frac {x \,{\mathrm e}^{-5} {\mathrm e}^{x}}{3}+\frac {x^{2} {\mathrm e}^{-5} {\mathrm e}^{x}}{6}+\ln \relax (x )}{x}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/6*(-6*exp(5)*ln(x)+(x^3-x^2)*exp(x)+6*exp(5)+400*x^2)/x^2/exp(5),x,method=_RETURNVERBOSE)

[Out]

1/6*x*exp(x-5)-1/3*exp(x-5)+200/3*x*exp(-5)+ln(x)/x

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maxima [B]  time = 0.61, size = 39, normalized size = 1.77 \begin {gather*} \frac {1}{6} \, {\left (6 \, {\left (\frac {\log \relax (x)}{x} + \frac {1}{x}\right )} e^{5} + {\left (x - 1\right )} e^{x} + 400 \, x - \frac {6 \, e^{5}}{x} - e^{x}\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(-6*exp(5)*log(x)+(x^3-x^2)*exp(x)+6*exp(5)+400*x^2)/x^2/exp(5),x, algorithm="maxima")

[Out]

1/6*(6*(log(x)/x + 1/x)*e^5 + (x - 1)*e^x + 400*x - 6*e^5/x - e^x)*e^(-5)

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mupad [B]  time = 0.69, size = 25, normalized size = 1.14 \begin {gather*} \frac {\ln \relax (x)}{x}+\frac {200\,x\,{\mathrm {e}}^{-5}}{3}-\frac {{\mathrm {e}}^{-5}\,{\mathrm {e}}^x}{3}+\frac {x\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^x}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-5)*(exp(5) - exp(5)*log(x) - (exp(x)*(x^2 - x^3))/6 + (200*x^2)/3))/x^2,x)

[Out]

log(x)/x + (200*x*exp(-5))/3 - (exp(-5)*exp(x))/3 + (x*exp(-5)*exp(x))/6

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sympy [A]  time = 0.30, size = 24, normalized size = 1.09 \begin {gather*} \frac {200 x}{3 e^{5}} + \frac {\left (x - 2\right ) e^{x}}{6 e^{5}} + \frac {\log {\relax (x )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(-6*exp(5)*ln(x)+(x**3-x**2)*exp(x)+6*exp(5)+400*x**2)/x**2/exp(5),x)

[Out]

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

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