3.52.92 \(\int \frac {e^{\frac {4+2 x}{x}} (26 x+5 x^3)+e^{\frac {4+2 x}{x}} (-104-20 x^2+10 x^3) \log (x)}{5 x^2} \, dx\)

Optimal. Leaf size=23 \[ 5+e^{\frac {4+2 x}{x}} \left (\frac {26}{5}+x^2\right ) \log (x) \]

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Rubi [A]  time = 0.47, antiderivative size = 31, normalized size of antiderivative = 1.35, number of steps used = 30, number of rules used = 10, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {12, 14, 2210, 2214, 2206, 2554, 6483, 6475, 2209, 2226} \begin {gather*} e^{\frac {4}{x}+2} x^2 \log (x)+\frac {26}{5} e^{\frac {4}{x}+2} \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((4 + 2*x)/x)*(26*x + 5*x^3) + E^((4 + 2*x)/x)*(-104 - 20*x^2 + 10*x^3)*Log[x])/(5*x^2),x]

[Out]

(26*E^(2 + 4/x)*Log[x])/5 + E^(2 + 4/x)*x^2*Log[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 2206

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> Simp[((c + d*x)*F^(a + b*(c + d*x)^n))/d, x]
- Dist[b*n*Log[F], Int[(c + d*x)^n*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2/n]
&& ILtQ[n, 0]

Rule 2209

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

Rule 2210

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Simp[(F^a*ExpIntegralEi[
b*(c + d*x)^n*Log[F]])/(f*n), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]

Rule 2214

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*F^(a + b*(c + d*x)^n))/(d*(m + 1)), x] - Dist[(b*n*Log[F])/(m + 1), Int[(c + d*x)^(m + n)*F^(a + b*(c +
d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

Rule 2226

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, 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]

Rule 6475

Int[ExpIntegralE[1, (b_.)*(x_)]/(x_), x_Symbol] :> Simp[b*x*HypergeometricPFQ[{1, 1, 1}, {2, 2, 2}, -(b*x)], x
] + (-Simp[EulerGamma*Log[x], x] - Simp[(1*Log[b*x]^2)/2, x]) /; FreeQ[b, x]

Rule 6483

Int[ExpIntegralEi[(b_.)*(x_)]/(x_), x_Symbol] :> Simp[Log[x]*(ExpIntegralEi[b*x] + ExpIntegralE[1, -(b*x)]), x
] - Int[ExpIntegralE[1, -(b*x)]/x, x] /; FreeQ[b, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {e^{\frac {4+2 x}{x}} \left (26 x+5 x^3\right )+e^{\frac {4+2 x}{x}} \left (-104-20 x^2+10 x^3\right ) \log (x)}{x^2} \, dx\\ &=\frac {1}{5} \int \left (\frac {26 e^{2+\frac {4}{x}}}{x}+5 e^{2+\frac {4}{x}} x-20 e^{2+\frac {4}{x}} \log (x)-\frac {104 e^{2+\frac {4}{x}} \log (x)}{x^2}+10 e^{2+\frac {4}{x}} x \log (x)\right ) \, dx\\ &=2 \int e^{2+\frac {4}{x}} x \log (x) \, dx-4 \int e^{2+\frac {4}{x}} \log (x) \, dx+\frac {26}{5} \int \frac {e^{2+\frac {4}{x}}}{x} \, dx-\frac {104}{5} \int \frac {e^{2+\frac {4}{x}} \log (x)}{x^2} \, dx+\int e^{2+\frac {4}{x}} x \, dx\\ &=\frac {1}{2} e^{2+\frac {4}{x}} x^2-\frac {26}{5} e^2 \text {Ei}\left (\frac {4}{x}\right )+\frac {26}{5} e^{2+\frac {4}{x}} \log (x)+e^{2+\frac {4}{x}} x^2 \log (x)+2 \int e^{2+\frac {4}{x}} \, dx-2 \int \left (\frac {1}{2} e^{2+\frac {4}{x}} (4+x)-\frac {8 e^2 \text {Ei}\left (\frac {4}{x}\right )}{x}\right ) \, dx+4 \int e^2 \left (e^{4/x}-\frac {4 \text {Ei}\left (\frac {4}{x}\right )}{x}\right ) \, dx+\frac {104}{5} \int -\frac {e^{2+\frac {4}{x}}}{4 x} \, dx\\ &=2 e^{2+\frac {4}{x}} x+\frac {1}{2} e^{2+\frac {4}{x}} x^2-\frac {26}{5} e^2 \text {Ei}\left (\frac {4}{x}\right )+\frac {26}{5} e^{2+\frac {4}{x}} \log (x)+e^{2+\frac {4}{x}} x^2 \log (x)-\frac {26}{5} \int \frac {e^{2+\frac {4}{x}}}{x} \, dx+8 \int \frac {e^{2+\frac {4}{x}}}{x} \, dx+\left (4 e^2\right ) \int \left (e^{4/x}-\frac {4 \text {Ei}\left (\frac {4}{x}\right )}{x}\right ) \, dx+\left (16 e^2\right ) \int \frac {\text {Ei}\left (\frac {4}{x}\right )}{x} \, dx-\int e^{2+\frac {4}{x}} (4+x) \, dx\\ &=2 e^{2+\frac {4}{x}} x+\frac {1}{2} e^{2+\frac {4}{x}} x^2-8 e^2 \text {Ei}\left (\frac {4}{x}\right )+\frac {26}{5} e^{2+\frac {4}{x}} \log (x)+e^{2+\frac {4}{x}} x^2 \log (x)+\left (4 e^2\right ) \int e^{4/x} \, dx-\left (16 e^2\right ) \int \frac {\text {Ei}\left (\frac {4}{x}\right )}{x} \, dx-\left (16 e^2\right ) \operatorname {Subst}\left (\int \frac {\text {Ei}(4 x)}{x} \, dx,x,\frac {1}{x}\right )-\int \left (4 e^{2+\frac {4}{x}}+e^{2+\frac {4}{x}} x\right ) \, dx\\ &=6 e^{2+\frac {4}{x}} x+\frac {1}{2} e^{2+\frac {4}{x}} x^2-8 e^2 \text {Ei}\left (\frac {4}{x}\right )-16 e^2 \left (E_1\left (-\frac {4}{x}\right )+\text {Ei}\left (\frac {4}{x}\right )\right ) \log \left (\frac {1}{x}\right )+\frac {26}{5} e^{2+\frac {4}{x}} \log (x)+e^{2+\frac {4}{x}} x^2 \log (x)-4 \int e^{2+\frac {4}{x}} \, dx+\left (16 e^2\right ) \int \frac {e^{4/x}}{x} \, dx+\left (16 e^2\right ) \operatorname {Subst}\left (\int \frac {E_1(-4 x)}{x} \, dx,x,\frac {1}{x}\right )+\left (16 e^2\right ) \operatorname {Subst}\left (\int \frac {\text {Ei}(4 x)}{x} \, dx,x,\frac {1}{x}\right )-\int e^{2+\frac {4}{x}} x \, dx\\ &=2 e^{2+\frac {4}{x}} x-24 e^2 \text {Ei}\left (\frac {4}{x}\right )-\frac {64 e^2 \, _3F_3\left (1,1,1;2,2,2;\frac {4}{x}\right )}{x}-8 e^2 \log ^2\left (-\frac {4}{x}\right )+\frac {26}{5} e^{2+\frac {4}{x}} \log (x)+16 e^2 \gamma \log (x)+e^{2+\frac {4}{x}} x^2 \log (x)-2 \int e^{2+\frac {4}{x}} \, dx-16 \int \frac {e^{2+\frac {4}{x}}}{x} \, dx-\left (16 e^2\right ) \operatorname {Subst}\left (\int \frac {E_1(-4 x)}{x} \, dx,x,\frac {1}{x}\right )\\ &=-8 e^2 \text {Ei}\left (\frac {4}{x}\right )+\frac {26}{5} e^{2+\frac {4}{x}} \log (x)+e^{2+\frac {4}{x}} x^2 \log (x)-8 \int \frac {e^{2+\frac {4}{x}}}{x} \, dx\\ &=\frac {26}{5} e^{2+\frac {4}{x}} \log (x)+e^{2+\frac {4}{x}} x^2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 22, normalized size = 0.96 \begin {gather*} \frac {1}{5} e^{2+\frac {4}{x}} \left (26+5 x^2\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((4 + 2*x)/x)*(26*x + 5*x^3) + E^((4 + 2*x)/x)*(-104 - 20*x^2 + 10*x^3)*Log[x])/(5*x^2),x]

[Out]

(E^(2 + 4/x)*(26 + 5*x^2)*Log[x])/5

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fricas [A]  time = 0.73, size = 20, normalized size = 0.87 \begin {gather*} \frac {1}{5} \, {\left (5 \, x^{2} + 26\right )} e^{\left (\frac {2 \, {\left (x + 2\right )}}{x}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((10*x^3-20*x^2-104)*exp((2*x+4)/x)*log(x)+(5*x^3+26*x)*exp((2*x+4)/x))/x^2,x, algorithm="fricas
")

[Out]

1/5*(5*x^2 + 26)*e^(2*(x + 2)/x)*log(x)

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giac [A]  time = 0.21, size = 29, normalized size = 1.26 \begin {gather*} x^{2} e^{\left (\frac {2 \, {\left (x + 2\right )}}{x}\right )} \log \relax (x) + \frac {26}{5} \, e^{\left (\frac {2 \, {\left (x + 2\right )}}{x}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((10*x^3-20*x^2-104)*exp((2*x+4)/x)*log(x)+(5*x^3+26*x)*exp((2*x+4)/x))/x^2,x, algorithm="giac")

[Out]

x^2*e^(2*(x + 2)/x)*log(x) + 26/5*e^(2*(x + 2)/x)*log(x)

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maple [A]  time = 0.05, size = 21, normalized size = 0.91

method	result	size
risch	\(\frac {\left (5 x^{2}+26\right ) {\mathrm e}^{\frac {2 x +4}{x}} \ln \relax (x )}{5}\)	\(21\)
norman	\(\frac {x^{3} {\mathrm e}^{\frac {2 x +4}{x}} \ln \relax (x )+\frac {26 x \,{\mathrm e}^{\frac {2 x +4}{x}} \ln \relax (x )}{5}}{x}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((10*x^3-20*x^2-104)*exp((2*x+4)/x)*ln(x)+(5*x^3+26*x)*exp((2*x+4)/x))/x^2,x,method=_RETURNVERBOSE)

[Out]

1/5*(5*x^2+26)*exp(2*(2+x)/x)*ln(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {26}{5} \, {\rm Ei}\left (\frac {4}{x}\right ) e^{2} + 16 \, e^{2} \Gamma \left (-2, -\frac {4}{x}\right ) + \frac {1}{5} \, \int \frac {2 \, {\left (5 \, x^{3} e^{2} - 10 \, x^{2} e^{2} - 52 \, e^{2}\right )} e^{\frac {4}{x}} \log \relax (x)}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((10*x^3-20*x^2-104)*exp((2*x+4)/x)*log(x)+(5*x^3+26*x)*exp((2*x+4)/x))/x^2,x, algorithm="maxima
")

[Out]

-26/5*Ei(4/x)*e^2 + 16*e^2*gamma(-2, -4/x) + 1/5*integrate(2*(5*x^3*e^2 - 10*x^2*e^2 - 52*e^2)*e^(4/x)*log(x)/
x^2, x)

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mupad [B]  time = 3.49, size = 19, normalized size = 0.83 \begin {gather*} \frac {{\mathrm {e}}^{\frac {4}{x}+2}\,\ln \relax (x)\,\left (5\,x^2+26\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp((2*x + 4)/x)*(26*x + 5*x^3))/5 - (exp((2*x + 4)/x)*log(x)*(20*x^2 - 10*x^3 + 104))/5)/x^2,x)

[Out]

(exp(4/x + 2)*log(x)*(5*x^2 + 26))/5

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sympy [A]  time = 0.36, size = 22, normalized size = 0.96 \begin {gather*} \frac {\left (5 x^{2} \log {\relax (x )} + 26 \log {\relax (x )}\right ) e^{\frac {2 x + 4}{x}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((10*x**3-20*x**2-104)*exp((2*x+4)/x)*ln(x)+(5*x**3+26*x)*exp((2*x+4)/x))/x**2,x)

[Out]

(5*x**2*log(x) + 26*log(x))*exp((2*x + 4)/x)/5

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