3.11.73 \(\int \frac {e^x (3+90 x-15 x^3+5 x^5)+e^x (-12+3 x) \log (x)}{5 x^5} \, dx\)

Optimal. Leaf size=24 \[ e^x-\frac {3 e^x \left (2+x-\frac {\log (x)}{5 x}\right )}{x^3} \]

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Rubi [A]  time = 0.73, antiderivative size = 32, normalized size of antiderivative = 1.33, number of steps used = 54, number of rules used = 9, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.237, Rules used = {12, 14, 2194, 2177, 2178, 2554, 2199, 6483, 6475} \begin {gather*} \frac {3 e^x \log (x)}{5 x^4}-\frac {6 e^x}{x^3}-\frac {3 e^x}{x^2}+e^x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(3 + 90*x - 15*x^3 + 5*x^5) + E^x*(-12 + 3*x)*Log[x])/(5*x^5),x]

[Out]

E^x - (6*E^x)/x^3 - (3*E^x)/x^2 + (3*E^x*Log[x])/(5*x^4)

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 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 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

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^x \left (3+90 x-15 x^3+5 x^5\right )+e^x (-12+3 x) \log (x)}{x^5} \, dx\\ &=\frac {1}{5} \int \left (5 e^x+\frac {3 e^x}{x^5}+\frac {90 e^x}{x^4}-\frac {15 e^x}{x^2}-\frac {12 e^x \log (x)}{x^5}+\frac {3 e^x \log (x)}{x^4}\right ) \, dx\\ &=\frac {3}{5} \int \frac {e^x}{x^5} \, dx+\frac {3}{5} \int \frac {e^x \log (x)}{x^4} \, dx-\frac {12}{5} \int \frac {e^x \log (x)}{x^5} \, dx-3 \int \frac {e^x}{x^2} \, dx+18 \int \frac {e^x}{x^4} \, dx+\int e^x \, dx\\ &=e^x-\frac {3 e^x}{20 x^4}-\frac {6 e^x}{x^3}+\frac {3 e^x}{x}+\frac {3 e^x \log (x)}{5 x^4}+\frac {3}{20} \int \frac {e^x}{x^4} \, dx-\frac {3}{5} \int \frac {-e^x \left (2+x+x^2\right )+x^3 \text {Ei}(x)}{6 x^4} \, dx+\frac {12}{5} \int \frac {-e^x \left (6+2 x+x^2+x^3\right )+x^4 \text {Ei}(x)}{24 x^5} \, dx-3 \int \frac {e^x}{x} \, dx+6 \int \frac {e^x}{x^3} \, dx\\ &=e^x-\frac {3 e^x}{20 x^4}-\frac {121 e^x}{20 x^3}-\frac {3 e^x}{x^2}+\frac {3 e^x}{x}-3 \text {Ei}(x)+\frac {3 e^x \log (x)}{5 x^4}+\frac {1}{20} \int \frac {e^x}{x^3} \, dx-\frac {1}{10} \int \frac {-e^x \left (2+x+x^2\right )+x^3 \text {Ei}(x)}{x^4} \, dx+\frac {1}{10} \int \frac {-e^x \left (6+2 x+x^2+x^3\right )+x^4 \text {Ei}(x)}{x^5} \, dx+3 \int \frac {e^x}{x^2} \, dx\\ &=e^x-\frac {3 e^x}{20 x^4}-\frac {121 e^x}{20 x^3}-\frac {121 e^x}{40 x^2}-3 \text {Ei}(x)+\frac {3 e^x \log (x)}{5 x^4}+\frac {1}{40} \int \frac {e^x}{x^2} \, dx-\frac {1}{10} \int \left (-\frac {e^x \left (2+x+x^2\right )}{x^4}+\frac {\text {Ei}(x)}{x}\right ) \, dx+\frac {1}{10} \int \left (-\frac {e^x \left (6+2 x+x^2+x^3\right )}{x^5}+\frac {\text {Ei}(x)}{x}\right ) \, dx+3 \int \frac {e^x}{x} \, dx\\ &=e^x-\frac {3 e^x}{20 x^4}-\frac {121 e^x}{20 x^3}-\frac {121 e^x}{40 x^2}-\frac {e^x}{40 x}+\frac {3 e^x \log (x)}{5 x^4}+\frac {1}{40} \int \frac {e^x}{x} \, dx+\frac {1}{10} \int \frac {e^x \left (2+x+x^2\right )}{x^4} \, dx-\frac {1}{10} \int \frac {e^x \left (6+2 x+x^2+x^3\right )}{x^5} \, dx\\ &=e^x-\frac {3 e^x}{20 x^4}-\frac {121 e^x}{20 x^3}-\frac {121 e^x}{40 x^2}-\frac {e^x}{40 x}+\frac {\text {Ei}(x)}{40}+\frac {3 e^x \log (x)}{5 x^4}+\frac {1}{10} \int \left (\frac {2 e^x}{x^4}+\frac {e^x}{x^3}+\frac {e^x}{x^2}\right ) \, dx-\frac {1}{10} \int \left (\frac {6 e^x}{x^5}+\frac {2 e^x}{x^4}+\frac {e^x}{x^3}+\frac {e^x}{x^2}\right ) \, dx\\ &=e^x-\frac {3 e^x}{20 x^4}-\frac {121 e^x}{20 x^3}-\frac {121 e^x}{40 x^2}-\frac {e^x}{40 x}+\frac {\text {Ei}(x)}{40}+\frac {3 e^x \log (x)}{5 x^4}-\frac {3}{5} \int \frac {e^x}{x^5} \, dx\\ &=e^x-\frac {121 e^x}{20 x^3}-\frac {121 e^x}{40 x^2}-\frac {e^x}{40 x}+\frac {\text {Ei}(x)}{40}+\frac {3 e^x \log (x)}{5 x^4}-\frac {3}{20} \int \frac {e^x}{x^4} \, dx\\ &=e^x-\frac {6 e^x}{x^3}-\frac {121 e^x}{40 x^2}-\frac {e^x}{40 x}+\frac {\text {Ei}(x)}{40}+\frac {3 e^x \log (x)}{5 x^4}-\frac {1}{20} \int \frac {e^x}{x^3} \, dx\\ &=e^x-\frac {6 e^x}{x^3}-\frac {3 e^x}{x^2}-\frac {e^x}{40 x}+\frac {\text {Ei}(x)}{40}+\frac {3 e^x \log (x)}{5 x^4}-\frac {1}{40} \int \frac {e^x}{x^2} \, dx\\ &=e^x-\frac {6 e^x}{x^3}-\frac {3 e^x}{x^2}+\frac {\text {Ei}(x)}{40}+\frac {3 e^x \log (x)}{5 x^4}-\frac {1}{40} \int \frac {e^x}{x} \, dx\\ &=e^x-\frac {6 e^x}{x^3}-\frac {3 e^x}{x^2}+\frac {3 e^x \log (x)}{5 x^4}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 26, normalized size = 1.08 \begin {gather*} \frac {e^x \left (5 x \left (-6-3 x+x^3\right )+3 \log (x)\right )}{5 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(3 + 90*x - 15*x^3 + 5*x^5) + E^x*(-12 + 3*x)*Log[x])/(5*x^5),x]

[Out]

(E^x*(5*x*(-6 - 3*x + x^3) + 3*Log[x]))/(5*x^4)

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fricas [A]  time = 0.64, size = 28, normalized size = 1.17 \begin {gather*} \frac {5 \, {\left (x^{4} - 3 \, x^{2} - 6 \, x\right )} e^{x} + 3 \, e^{x} \log \relax (x)}{5 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((3*x-12)*exp(x)*log(x)+(5*x^5-15*x^3+90*x+3)*exp(x))/x^5,x, algorithm="fricas")

[Out]

1/5*(5*(x^4 - 3*x^2 - 6*x)*e^x + 3*e^x*log(x))/x^4

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giac [A]  time = 0.34, size = 31, normalized size = 1.29 \begin {gather*} \frac {5 \, x^{4} e^{x} - 15 \, x^{2} e^{x} - 30 \, x e^{x} + 3 \, e^{x} \log \relax (x)}{5 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((3*x-12)*exp(x)*log(x)+(5*x^5-15*x^3+90*x+3)*exp(x))/x^5,x, algorithm="giac")

[Out]

1/5*(5*x^4*e^x - 15*x^2*e^x - 30*x*e^x + 3*e^x*log(x))/x^4

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maple [A]  time = 0.04, size = 25, normalized size = 1.04

method	result	size
risch	\(\frac {3 \,{\mathrm e}^{x} \ln \relax (x )}{5 x^{4}}+\frac {\left (x^{3}-3 x -6\right ) {\mathrm e}^{x}}{x^{3}}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((3*x-12)*exp(x)*ln(x)+(5*x^5-15*x^3+90*x+3)*exp(x))/x^5,x,method=_RETURNVERBOSE)

[Out]

3/5/x^4*exp(x)*ln(x)+(x^3-3*x-6)/x^3*exp(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {3 \, e^{x} \log \relax (x)}{5 \, x^{4}} + e^{x} - 3 \, \Gamma \left (-1, -x\right ) + 18 \, \Gamma \left (-3, -x\right ) - \frac {3}{5} \, \Gamma \left (-4, -x\right ) - \frac {3}{5} \, \int \frac {e^{x}}{x^{5}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((3*x-12)*exp(x)*log(x)+(5*x^5-15*x^3+90*x+3)*exp(x))/x^5,x, algorithm="maxima")

[Out]

3/5*e^x*log(x)/x^4 + e^x - 3*gamma(-1, -x) + 18*gamma(-3, -x) - 3/5*gamma(-4, -x) - 3/5*integrate(e^x/x^5, x)

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mupad [B]  time = 1.10, size = 26, normalized size = 1.08 \begin {gather*} {\mathrm {e}}^x-\frac {3\,{\mathrm {e}}^x}{x^2}-\frac {6\,{\mathrm {e}}^x}{x^3}+\frac {3\,{\mathrm {e}}^x\,\ln \relax (x)}{5\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x)*(90*x - 15*x^3 + 5*x^5 + 3))/5 + (exp(x)*log(x)*(3*x - 12))/5)/x^5,x)

[Out]

exp(x) - (3*exp(x))/x^2 - (6*exp(x))/x^3 + (3*exp(x)*log(x))/(5*x^4)

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sympy [A]  time = 0.27, size = 26, normalized size = 1.08 \begin {gather*} \frac {\left (5 x^{4} - 15 x^{2} - 30 x + 3 \log {\relax (x )}\right ) e^{x}}{5 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((3*x-12)*exp(x)*ln(x)+(5*x**5-15*x**3+90*x+3)*exp(x))/x**5,x)

[Out]

(5*x**4 - 15*x**2 - 30*x + 3*log(x))*exp(x)/(5*x**4)

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