3.58.71 \(\int \frac {-32 e^x x^3+e^{2 x} (-448+128 x)+(8 x^6+e^x (128 x^3-32 x^4)) \log (x)-4 x^6 \log ^2(x)}{x^8} \, dx\)

Optimal. Leaf size=22 \[ \frac {4 \left (x+\left (\frac {4 e^x}{x^3}-\log (x)\right )^2\right )}{x} \]

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Rubi [A]  time = 0.14, antiderivative size = 41, normalized size of antiderivative = 1.86, number of steps used = 6, number of rules used = 6, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {14, 2197, 2304, 2366, 2303, 2288} \begin {gather*} \frac {64 e^{2 x}}{x^7}-\frac {32 e^x \log (x)}{x^4}-\frac {4 (2-\log (x)) \log (x)}{x}+\frac {8 \log (x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-32*E^x*x^3 + E^(2*x)*(-448 + 128*x) + (8*x^6 + E^x*(128*x^3 - 32*x^4))*Log[x] - 4*x^6*Log[x]^2)/x^8,x]

[Out]

(64*E^(2*x))/x^7 - (32*E^x*Log[x])/x^4 + (8*Log[x])/x - (4*(2 - Log[x])*Log[x])/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 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]

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} &=\int \left (\frac {64 e^{2 x} (-7+2 x)}{x^8}-\frac {4 (-2+\log (x)) \log (x)}{x^2}-\frac {32 e^x (1-4 \log (x)+x \log (x))}{x^5}\right ) \, dx\\ &=-\left (4 \int \frac {(-2+\log (x)) \log (x)}{x^2} \, dx\right )-32 \int \frac {e^x (1-4 \log (x)+x \log (x))}{x^5} \, dx+64 \int \frac {e^{2 x} (-7+2 x)}{x^8} \, dx\\ &=\frac {64 e^{2 x}}{x^7}-\frac {32 e^x \log (x)}{x^4}+\frac {4 \log (x)}{x}-\frac {4 (2-\log (x)) \log (x)}{x}+4 \int \frac {1-\log (x)}{x^2} \, dx\\ &=\frac {64 e^{2 x}}{x^7}-\frac {32 e^x \log (x)}{x^4}+\frac {8 \log (x)}{x}-\frac {4 (2-\log (x)) \log (x)}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 19, normalized size = 0.86 \begin {gather*} \frac {4 \left (-4 e^x+x^3 \log (x)\right )^2}{x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-32*E^x*x^3 + E^(2*x)*(-448 + 128*x) + (8*x^6 + E^x*(128*x^3 - 32*x^4))*Log[x] - 4*x^6*Log[x]^2)/x^
8,x]

[Out]

(4*(-4*E^x + x^3*Log[x])^2)/x^7

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fricas [A]  time = 1.24, size = 29, normalized size = 1.32 \begin {gather*} \frac {4 \, {\left (x^{6} \log \relax (x)^{2} - 8 \, x^{3} e^{x} \log \relax (x) + 16 \, e^{\left (2 \, x\right )}\right )}}{x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^6*log(x)^2+((-32*x^4+128*x^3)*exp(x)+8*x^6)*log(x)+(128*x-448)*exp(x)^2-32*exp(x)*x^3)/x^8,x,
algorithm="fricas")

[Out]

4*(x^6*log(x)^2 - 8*x^3*e^x*log(x) + 16*e^(2*x))/x^7

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giac [A]  time = 0.15, size = 29, normalized size = 1.32 \begin {gather*} \frac {4 \, {\left (x^{6} \log \relax (x)^{2} - 8 \, x^{3} e^{x} \log \relax (x) + 16 \, e^{\left (2 \, x\right )}\right )}}{x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^6*log(x)^2+((-32*x^4+128*x^3)*exp(x)+8*x^6)*log(x)+(128*x-448)*exp(x)^2-32*exp(x)*x^3)/x^8,x,
algorithm="giac")

[Out]

4*(x^6*log(x)^2 - 8*x^3*e^x*log(x) + 16*e^(2*x))/x^7

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maple [A]  time = 0.05, size = 29, normalized size = 1.32

method	result	size
risch	\(\frac {4 \ln \relax (x )^{2}}{x}-\frac {32 \,{\mathrm e}^{x} \ln \relax (x )}{x^{4}}+\frac {64 \,{\mathrm e}^{2 x}}{x^{7}}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*x^6*ln(x)^2+((-32*x^4+128*x^3)*exp(x)+8*x^6)*ln(x)+(128*x-448)*exp(x)^2-32*exp(x)*x^3)/x^8,x,method=_R
ETURNVERBOSE)

[Out]

4*ln(x)^2/x-32/x^4*exp(x)*ln(x)+64/x^7*exp(2*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {8 \, \log \relax (x)}{x} - \frac {8}{x} + \frac {4 \, {\left (x^{3} \log \relax (x)^{2} + 2 \, x^{3} \log \relax (x) + 2 \, x^{3} - 8 \, e^{x} \log \relax (x)\right )}}{x^{4}} + 32 \, \Gamma \left (-4, -x\right ) - 8192 \, \Gamma \left (-6, -2 \, x\right ) - 57344 \, \Gamma \left (-7, -2 \, x\right ) + 32 \, \int \frac {e^{x}}{x^{5}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^6*log(x)^2+((-32*x^4+128*x^3)*exp(x)+8*x^6)*log(x)+(128*x-448)*exp(x)^2-32*exp(x)*x^3)/x^8,x,
algorithm="maxima")

[Out]

-8*log(x)/x - 8/x + 4*(x^3*log(x)^2 + 2*x^3*log(x) + 2*x^3 - 8*e^x*log(x))/x^4 + 32*gamma(-4, -x) - 8192*gamma
(-6, -2*x) - 57344*gamma(-7, -2*x) + 32*integrate(e^x/x^5, x)

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mupad [B]  time = 4.41, size = 102, normalized size = 4.64 \begin {gather*} \frac {4\,\mathrm {ei}\relax (x)}{3}+\frac {4\,\mathrm {expint}\left (-x\right )}{3}+\frac {4\,\left ({\ln \relax (x)}^2+2\,\ln \relax (x)+2\right )}{x}+\frac {64\,{\mathrm {e}}^{2\,x}}{x^7}+32\,{\mathrm {e}}^x\,\left (\frac {1}{24\,x}+\frac {1}{24\,x^2}+\frac {1}{12\,x^3}+\frac {1}{4\,x^4}\right )-\frac {8\,{\mathrm {e}}^x+\frac {4\,x^2\,{\mathrm {e}}^x}{3}+x^3\,\left (\frac {4\,{\mathrm {e}}^x}{3}+8\,\ln \relax (x)+8\right )+32\,{\mathrm {e}}^x\,\ln \relax (x)+\frac {8\,x\,{\mathrm {e}}^x}{3}}{x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(32*x^3*exp(x) + 4*x^6*log(x)^2 - exp(2*x)*(128*x - 448) - log(x)*(exp(x)*(128*x^3 - 32*x^4) + 8*x^6))/x^
8,x)

[Out]

(4*ei(x))/3 + (4*expint(-x))/3 + (4*(2*log(x) + log(x)^2 + 2))/x + (64*exp(2*x))/x^7 + 32*exp(x)*(1/(24*x) + 1
/(24*x^2) + 1/(12*x^3) + 1/(4*x^4)) - (8*exp(x) + (4*x^2*exp(x))/3 + x^3*((4*exp(x))/3 + 8*log(x) + 8) + 32*ex
p(x)*log(x) + (8*x*exp(x))/3)/x^4

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sympy [A]  time = 0.32, size = 32, normalized size = 1.45 \begin {gather*} \frac {4 \log {\relax (x )}^{2}}{x} + \frac {- 32 x^{7} e^{x} \log {\relax (x )} + 64 x^{4} e^{2 x}}{x^{11}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x**6*ln(x)**2+((-32*x**4+128*x**3)*exp(x)+8*x**6)*ln(x)+(128*x-448)*exp(x)**2-32*exp(x)*x**3)/x*
*8,x)

[Out]

4*log(x)**2/x + (-32*x**7*exp(x)*log(x) + 64*x**4*exp(2*x))/x**11

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