3.86.70 \(\int \frac {x+5 x^2+5 e^x x^2-\log (-e^x)}{5 x^2} \, dx\)

Optimal. Leaf size=24 \[ 5-e^9+e^x+x+\frac {\log \left (-e^x\right )}{5 x} \]

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Rubi [A]  time = 0.03, antiderivative size = 18, normalized size of antiderivative = 0.75, number of steps used = 10, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {12, 14, 2194, 43, 2168, 29} \begin {gather*} x+e^x+\frac {\log \left (-e^x\right )}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^x + x + Log[-E^x]/(5*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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2168

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(u^(m + 1)*v^
n)/(a*(m + 1)), x] - Dist[(b*n)/(a*(m + 1)), Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m
, n}, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0] &&  !(ILtQ[m + n, -2] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 18, normalized size = 0.75 \begin {gather*} e^x+x+\frac {\log \left (-e^x\right )}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^x + x + Log[-E^x]/(5*x)

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fricas [C]  time = 0.48, size = 19, normalized size = 0.79 \begin {gather*} \frac {i \, \pi + 5 \, x^{2} + 5 \, x e^{x}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(I*pi + 5*x^2 + 5*x*e^x)/x

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giac [A]  time = 0.17, size = 4, normalized size = 0.17 \begin {gather*} x + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + e^x

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maple [A]  time = 0.06, size = 15, normalized size = 0.62

method	result	size
default	\(x +\frac {\ln \left (-{\mathrm e}^{x}\right )}{5 x}+{\mathrm e}^{x}\)	\(15\)
norman	\(\frac {{\mathrm e}^{x} x +x \ln \left (-{\mathrm e}^{x}\right )+\frac {\ln \left (-{\mathrm e}^{x}\right )}{5}}{x}\)	\(24\)
risch	\(\frac {\ln \left ({\mathrm e}^{x}\right )}{5 x}+\frac {-2 i \pi \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2}+2 i \pi \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{3}+2 i \pi +10 x^{2}+10 \,{\mathrm e}^{x} x}{10 x}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x+1/5*ln(-exp(x))/x+exp(x)

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maxima [A]  time = 0.37, size = 14, normalized size = 0.58 \begin {gather*} x + \frac {\log \left (-e^{x}\right )}{5 \, x} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 1/5*log(-e^x)/x + e^x

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mupad [B]  time = 5.18, size = 11, normalized size = 0.46 \begin {gather*} x+{\mathrm {e}}^x+\frac {\pi \,1{}\mathrm {i}}{5\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x/5 - log(-exp(x))/5 + x^2*exp(x) + x^2)/x^2,x)

[Out]

x + exp(x) + (pi*1i)/(5*x)

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sympy [C]  time = 0.12, size = 10, normalized size = 0.42 \begin {gather*} x + e^{x} + \frac {i \pi }{5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-ln(-exp(x))+5*exp(x)*x**2+5*x**2+x)/x**2,x)

[Out]

x + exp(x) + I*pi/(5*x)

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