3.13.29 \(\int \frac {e^x (-24+24 x)-32 \log (\frac {5}{2 x})-16 \log ^2(\frac {5}{2 x})}{15 x^2} \, dx\)

Optimal. Leaf size=25 \[ \frac {8 \left (e^x+\frac {2}{3} \log ^2\left (\frac {5}{2 x}\right )\right )}{5 x} \]

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Rubi [B]  time = 0.12, antiderivative size = 51, normalized size of antiderivative = 2.04, number of steps used = 6, number of rules used = 6, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {12, 14, 2197, 2304, 2366, 2303} \begin {gather*} \frac {8 e^x}{5 x}+\frac {16 \left (\log \left (\frac {5}{2 x}\right )+2\right ) \log \left (\frac {5}{2 x}\right )}{15 x}-\frac {32 \log \left (\frac {5}{2 x}\right )}{15 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(-24 + 24*x) - 32*Log[5/(2*x)] - 16*Log[5/(2*x)]^2)/(15*x^2),x]

[Out]

(8*E^x)/(5*x) - (32*Log[5/(2*x)])/(15*x) + (16*Log[5/(2*x)]*(2 + Log[5/(2*x)]))/(15*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 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
]

Rule 2303

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(b*(d*x)^(m + 1)*Log[c*x^n])/(
d*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && EqQ[a*(m + 1) - b*n, 0]

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
]

Rule 2366

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.))*((g_.)*(x_))^(m_.), x_Sy
mbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[Sim
plifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] &&  !(EqQ[p, 1] && EqQ[a, 0] &&
NeQ[d, 0])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{15} \int \frac {e^x (-24+24 x)-32 \log \left (\frac {5}{2 x}\right )-16 \log ^2\left (\frac {5}{2 x}\right )}{x^2} \, dx\\ &=\frac {1}{15} \int \left (\frac {24 e^x (-1+x)}{x^2}-\frac {16 \log \left (\frac {5}{2 x}\right ) \left (2+\log \left (\frac {5}{2 x}\right )\right )}{x^2}\right ) \, dx\\ &=-\left (\frac {16}{15} \int \frac {\log \left (\frac {5}{2 x}\right ) \left (2+\log \left (\frac {5}{2 x}\right )\right )}{x^2} \, dx\right )+\frac {8}{5} \int \frac {e^x (-1+x)}{x^2} \, dx\\ &=\frac {8 e^x}{5 x}-\frac {16 \log \left (\frac {5}{2 x}\right )}{15 x}+\frac {16 \log \left (\frac {5}{2 x}\right ) \left (2+\log \left (\frac {5}{2 x}\right )\right )}{15 x}-\frac {16}{15} \int \frac {-1-\log \left (\frac {5}{2 x}\right )}{x^2} \, dx\\ &=\frac {8 e^x}{5 x}-\frac {32 \log \left (\frac {5}{2 x}\right )}{15 x}+\frac {16 \log \left (\frac {5}{2 x}\right ) \left (2+\log \left (\frac {5}{2 x}\right )\right )}{15 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 28, normalized size = 1.12 \begin {gather*} \frac {8}{15} \left (\frac {3 e^x}{x}+\frac {2 \log ^2\left (\frac {5}{2 x}\right )}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-24 + 24*x) - 32*Log[5/(2*x)] - 16*Log[5/(2*x)]^2)/(15*x^2),x]

[Out]

(8*((3*E^x)/x + (2*Log[5/(2*x)]^2)/x))/15

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fricas [A]  time = 0.71, size = 20, normalized size = 0.80 \begin {gather*} \frac {8 \, {\left (2 \, \log \left (\frac {5}{2 \, x}\right )^{2} + 3 \, e^{x}\right )}}{15 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*(-16*log(5/2/x)^2-32*log(5/2/x)+(24*x-24)*exp(x))/x^2,x, algorithm="fricas")

[Out]

8/15*(2*log(5/2/x)^2 + 3*e^x)/x

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giac [A]  time = 0.41, size = 20, normalized size = 0.80 \begin {gather*} \frac {8 \, {\left (2 \, \log \left (\frac {5}{2 \, x}\right )^{2} + 3 \, e^{x}\right )}}{15 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*(-16*log(5/2/x)^2-32*log(5/2/x)+(24*x-24)*exp(x))/x^2,x, algorithm="giac")

[Out]

8/15*(2*log(5/2/x)^2 + 3*e^x)/x

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maple [A]  time = 0.09, size = 20, normalized size = 0.80

method	result	size
norman	\(\frac {\frac {16 \ln \left (\frac {5}{2 x}\right )^{2}}{15}+\frac {8 \,{\mathrm e}^{x}}{5}}{x}\)	\(20\)
default	\(\frac {16 \ln \left (\frac {5}{2 x}\right )^{2}}{15 x}+\frac {8 \,{\mathrm e}^{x}}{5 x}\)	\(22\)
risch	\(\frac {16 \ln \relax (x )^{2}}{15 x}+\frac {16 \left (2 \ln \relax (2)-2 \ln \relax (5)\right ) \ln \relax (x )}{15 x}+\frac {\frac {16 \ln \relax (2)^{2}}{15}-\frac {32 \ln \relax (2) \ln \relax (5)}{15}+\frac {16 \ln \relax (5)^{2}}{15}+\frac {8 \,{\mathrm e}^{x}}{5}}{x}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/15*(-16*ln(5/2/x)^2-32*ln(5/2/x)+(24*x-24)*exp(x))/x^2,x,method=_RETURNVERBOSE)

[Out]

(16/15*ln(5/2/x)^2+8/5*exp(x))/x

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maxima [C]  time = 0.55, size = 25, normalized size = 1.00 \begin {gather*} \frac {16 \, \log \left (\frac {5}{2 \, x}\right )^{2}}{15 \, x} + \frac {8}{5} \, {\rm Ei}\relax (x) - \frac {8}{5} \, \Gamma \left (-1, -x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*(-16*log(5/2/x)^2-32*log(5/2/x)+(24*x-24)*exp(x))/x^2,x, algorithm="maxima")

[Out]

16/15*log(5/2/x)^2/x + 8/5*Ei(x) - 8/5*gamma(-1, -x)

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mupad [B]  time = 0.92, size = 20, normalized size = 0.80 \begin {gather*} \frac {8\,\left (2\,{\ln \left (\frac {5}{2\,x}\right )}^2+3\,{\mathrm {e}}^x\right )}{15\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((32*log(5/(2*x)))/15 + (16*log(5/(2*x))^2)/15 - (exp(x)*(24*x - 24))/15)/x^2,x)

[Out]

(8*(3*exp(x) + 2*log(5/(2*x))^2))/(15*x)

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sympy [A]  time = 0.30, size = 20, normalized size = 0.80 \begin {gather*} \frac {8 e^{x}}{5 x} + \frac {16 \log {\left (\frac {5}{2 x} \right )}^{2}}{15 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*(-16*ln(5/2/x)**2-32*ln(5/2/x)+(24*x-24)*exp(x))/x**2,x)

[Out]

8*exp(x)/(5*x) + 16*log(5/(2*x))**2/(15*x)

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