3.102.54 \(\int \frac {e^{x/5} (e^4 (-5+x)-10 x+5 x^2+x^3)-e^{x/5} x^2 \log (\frac {x^2}{4096})}{5 x^2} \, dx\)

Optimal. Leaf size=30 \[ \frac {e^{x/5} \left (e^4-x \left (-x+\log \left (\frac {x^2}{4096}\right )\right )\right )}{x} \]

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Rubi [C]  time = 0.12, antiderivative size = 78, normalized size of antiderivative = 2.60, number of steps used = 12, number of rules used = 7, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.127, Rules used = {12, 14, 2194, 2177, 2178, 2176, 2554} \begin {gather*} -\frac {1}{5} \left (10-e^4\right ) \text {Ei}\left (\frac {x}{5}\right )-\frac {1}{5} e^4 \text {Ei}\left (\frac {x}{5}\right )+2 \text {Ei}\left (\frac {x}{5}\right )-e^{x/5} \log \left (\frac {x^2}{4096}\right )+e^{x/5} x+\frac {e^{\frac {x}{5}+4}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(x/5)*(E^4*(-5 + x) - 10*x + 5*x^2 + x^3) - E^(x/5)*x^2*Log[x^2/4096])/(5*x^2),x]

[Out]

E^(4 + x/5)/x + E^(x/5)*x + 2*ExpIntegralEi[x/5] - (E^4*ExpIntegralEi[x/5])/5 - ((10 - E^4)*ExpIntegralEi[x/5]
)/5 - E^(x/5)*Log[x^2/4096]

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 2177

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

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$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 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {e^{x/5} \left (e^4 (-5+x)-10 x+5 x^2+x^3\right )-e^{x/5} x^2 \log \left (\frac {x^2}{4096}\right )}{x^2} \, dx\\ &=\frac {1}{5} \int \left (5 e^{x/5}-\frac {5 e^{4+\frac {x}{5}}}{x^2}+\frac {\left (1-\frac {10}{e^4}\right ) e^{4+\frac {x}{5}}}{x}+e^{x/5} x-e^{x/5} \log \left (\frac {x^2}{4096}\right )\right ) \, dx\\ &=\frac {1}{5} \int e^{x/5} x \, dx-\frac {1}{5} \int e^{x/5} \log \left (\frac {x^2}{4096}\right ) \, dx+\frac {1}{5} \left (1-\frac {10}{e^4}\right ) \int \frac {e^{4+\frac {x}{5}}}{x} \, dx+\int e^{x/5} \, dx-\int \frac {e^{4+\frac {x}{5}}}{x^2} \, dx\\ &=5 e^{x/5}+\frac {e^{4+\frac {x}{5}}}{x}+e^{x/5} x-\frac {1}{5} \left (10-e^4\right ) \text {Ei}\left (\frac {x}{5}\right )-e^{x/5} \log \left (\frac {x^2}{4096}\right )-\frac {1}{5} \int \frac {e^{4+\frac {x}{5}}}{x} \, dx+\frac {1}{5} \int \frac {10 e^{x/5}}{x} \, dx-\int e^{x/5} \, dx\\ &=\frac {e^{4+\frac {x}{5}}}{x}+e^{x/5} x-\frac {1}{5} e^4 \text {Ei}\left (\frac {x}{5}\right )-\frac {1}{5} \left (10-e^4\right ) \text {Ei}\left (\frac {x}{5}\right )-e^{x/5} \log \left (\frac {x^2}{4096}\right )+2 \int \frac {e^{x/5}}{x} \, dx\\ &=\frac {e^{4+\frac {x}{5}}}{x}+e^{x/5} x+2 \text {Ei}\left (\frac {x}{5}\right )-\frac {1}{5} e^4 \text {Ei}\left (\frac {x}{5}\right )-\frac {1}{5} \left (10-e^4\right ) \text {Ei}\left (\frac {x}{5}\right )-e^{x/5} \log \left (\frac {x^2}{4096}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 29, normalized size = 0.97 \begin {gather*} \frac {e^{x/5} \left (e^4+x^2-x \log \left (\frac {x^2}{4096}\right )\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(x/5)*(E^4*(-5 + x) - 10*x + 5*x^2 + x^3) - E^(x/5)*x^2*Log[x^2/4096])/(5*x^2),x]

[Out]

(E^(x/5)*(E^4 + x^2 - x*Log[x^2/4096]))/x

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fricas [A]  time = 1.39, size = 30, normalized size = 1.00 \begin {gather*} -\frac {x e^{\left (\frac {1}{5} \, x\right )} \log \left (\frac {1}{4096} \, x^{2}\right ) - {\left (x^{2} + e^{4}\right )} e^{\left (\frac {1}{5} \, x\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-x^2*exp(1/5*x)*log(1/4096*x^2)+((x-5)*exp(4)+x^3+5*x^2-10*x)*exp(1/5*x))/x^2,x, algorithm="fri
cas")

[Out]

-(x*e^(1/5*x)*log(1/4096*x^2) - (x^2 + e^4)*e^(1/5*x))/x

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giac [A]  time = 0.17, size = 32, normalized size = 1.07 \begin {gather*} \frac {x^{2} e^{\left (\frac {1}{5} \, x\right )} - x e^{\left (\frac {1}{5} \, x\right )} \log \left (\frac {1}{4096} \, x^{2}\right ) + e^{\left (\frac {1}{5} \, x + 4\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-x^2*exp(1/5*x)*log(1/4096*x^2)+((x-5)*exp(4)+x^3+5*x^2-10*x)*exp(1/5*x))/x^2,x, algorithm="gia
c")

[Out]

(x^2*e^(1/5*x) - x*e^(1/5*x)*log(1/4096*x^2) + e^(1/5*x + 4))/x

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maple [A]  time = 0.14, size = 34, normalized size = 1.13

method	result	size
norman	\(\frac {x^{2} {\mathrm e}^{\frac {x}{5}}+{\mathrm e}^{4} {\mathrm e}^{\frac {x}{5}}-x \,{\mathrm e}^{\frac {x}{5}} \ln \left (\frac {x^{2}}{4096}\right )}{x}\)	\(34\)
default	\(\frac {5 x^{2} {\mathrm e}^{\frac {x}{5}}+5 \,{\mathrm e}^{4} {\mathrm e}^{\frac {x}{5}}-5 \left (\ln \left (\frac {x^{2}}{4096}\right )-2 \ln \relax (x )\right ) x \,{\mathrm e}^{\frac {x}{5}}-10 \ln \relax (x ) {\mathrm e}^{\frac {x}{5}} x}{5 x}\)	\(51\)
risch	\(-2 \,{\mathrm e}^{\frac {x}{5}} \ln \relax (x )+\frac {\left (i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right ) x -2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2} x +i \pi \mathrm {csgn}\left (i x^{2}\right )^{3} x +24 x \ln \relax (2)+2 x^{2}+2 \,{\mathrm e}^{4}\right ) {\mathrm e}^{\frac {x}{5}}}{2 x}\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-x^2*exp(1/5*x)*ln(1/4096*x^2)+((x-5)*exp(4)+x^3+5*x^2-10*x)*exp(1/5*x))/x^2,x,method=_RETURNVERBOSE)

[Out]

(x^2*exp(1/5*x)+exp(4)*exp(1/5*x)-x*exp(1/5*x)*ln(1/4096*x^2))/x

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maxima [C]  time = 0.39, size = 44, normalized size = 1.47 \begin {gather*} \frac {1}{5} \, {\rm Ei}\left (\frac {1}{5} \, x\right ) e^{4} + {\left (x - 5\right )} e^{\left (\frac {1}{5} \, x\right )} - \frac {1}{5} \, e^{4} \Gamma \left (-1, -\frac {1}{5} \, x\right ) - e^{\left (\frac {1}{5} \, x\right )} \log \left (\frac {1}{4096} \, x^{2}\right ) + 5 \, e^{\left (\frac {1}{5} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-x^2*exp(1/5*x)*log(1/4096*x^2)+((x-5)*exp(4)+x^3+5*x^2-10*x)*exp(1/5*x))/x^2,x, algorithm="max
ima")

[Out]

1/5*Ei(1/5*x)*e^4 + (x - 5)*e^(1/5*x) - 1/5*e^4*gamma(-1, -1/5*x) - e^(1/5*x)*log(1/4096*x^2) + 5*e^(1/5*x)

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mupad [B]  time = 6.75, size = 26, normalized size = 0.87 \begin {gather*} {\mathrm {e}}^{x/5}\,\left (x-\ln \left (\frac {x^2}{4096}\right )\right )+\frac {{\mathrm {e}}^{\frac {x}{5}+4}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x/5)*(exp(4)*(x - 5) - 10*x + 5*x^2 + x^3))/5 - (x^2*exp(x/5)*log(x^2/4096))/5)/x^2,x)

[Out]

exp(x/5)*(x - log(x^2/4096)) + exp(x/5 + 4)/x

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sympy [A]  time = 0.31, size = 20, normalized size = 0.67 \begin {gather*} \frac {\left (x^{2} - x \log {\left (\frac {x^{2}}{4096} \right )} + e^{4}\right ) e^{\frac {x}{5}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-x**2*exp(1/5*x)*ln(1/4096*x**2)+((x-5)*exp(4)+x**3+5*x**2-10*x)*exp(1/5*x))/x**2,x)

[Out]

(x**2 - x*log(x**2/4096) + exp(4))*exp(x/5)/x

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