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

Optimal. Leaf size=24 \[ -16+x+\frac {\left (\frac {e^x}{x}-x\right ) \log (\log (5))}{5 x} \]

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Rubi [A]  time = 0.03, antiderivative size = 15, normalized size of antiderivative = 0.62, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {12, 14, 2197} \begin {gather*} \frac {e^x \log (\log (5))}{5 x^2}+x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x + (E^x*Log[Log[5]])/(5*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 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 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 15, normalized size = 0.62 \begin {gather*} x+\frac {e^x \log (\log (5))}{5 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x + (E^x*Log[Log[5]])/(5*x^2)

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fricas [A]  time = 0.61, size = 17, normalized size = 0.71 \begin {gather*} \frac {5 \, x^{3} + e^{x} \log \left (\log \relax (5)\right )}{5 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(5*x^3 + e^x*log(log(5)))/x^2

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giac [A]  time = 0.24, size = 17, normalized size = 0.71 \begin {gather*} \frac {5 \, x^{3} + e^{x} \log \left (\log \relax (5)\right )}{5 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(5*x^3 + e^x*log(log(5)))/x^2

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maple [A]  time = 0.10, size = 13, normalized size = 0.54

method	result	size
default	\(x +\frac {{\mathrm e}^{x} \ln \left (\ln \relax (5)\right )}{5 x^{2}}\)	\(13\)
risch	\(x +\frac {{\mathrm e}^{x} \ln \left (\ln \relax (5)\right )}{5 x^{2}}\)	\(13\)
norman	\(\frac {x^{3}+\frac {{\mathrm e}^{x} \ln \left (\ln \relax (5)\right )}{5}}{x^{2}}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.49, size = 22, normalized size = 0.92 \begin {gather*} \frac {1}{5} \, \Gamma \left (-1, -x\right ) \log \left (\log \relax (5)\right ) + \frac {2}{5} \, \Gamma \left (-2, -x\right ) \log \left (\log \relax (5)\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*gamma(-1, -x)*log(log(5)) + 2/5*gamma(-2, -x)*log(log(5)) + x

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mupad [B]  time = 0.06, size = 12, normalized size = 0.50 \begin {gather*} x+\frac {{\mathrm {e}}^x\,\ln \left (\ln \relax (5)\right )}{5\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + (exp(x)*log(log(5)))/(5*x^2)

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sympy [A]  time = 0.10, size = 14, normalized size = 0.58 \begin {gather*} x + \frac {e^{x} \log {\left (\log {\relax (5 )} \right )}}{5 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + exp(x)*log(log(5))/(5*x**2)

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