\(\int \frac {5+10 x^3+5 e^x x^3+2 e^3 x^4-10 \log (x)}{5 x^3} \, dx\) [4478]

Optimal result

Integrand size = 34, antiderivative size = 24 \[ \int \frac {5+10 x^3+5 e^x x^3+2 e^3 x^4-10 \log (x)}{5 x^3} \, dx=1+e^x+x+x \left (1+\frac {e^3 x}{5}\right )+\frac {\log (x)}{x^2} \]

[Out]

x+x*(1+1/5*x*exp(3))+1+exp(x)+ln(x)/x^2

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 14, 2225, 2341} \[ \int \frac {5+10 x^3+5 e^x x^3+2 e^3 x^4-10 \log (x)}{5 x^3} \, dx=\frac {e^3 x^2}{5}+\frac {\log (x)}{x^2}+2 x+e^x \]

[In]

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

[Out]

E^x + 2*x + (E^3*x^2)/5 + Log[x]/x^2

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 2225

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 2341

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
]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \frac {5+10 x^3+5 e^x x^3+2 e^3 x^4-10 \log (x)}{x^3} \, dx \\ & = \frac {1}{5} \int \left (5 e^x+\frac {5+10 x^3+2 e^3 x^4-10 \log (x)}{x^3}\right ) \, dx \\ & = \frac {1}{5} \int \frac {5+10 x^3+2 e^3 x^4-10 \log (x)}{x^3} \, dx+\int e^x \, dx \\ & = e^x+\frac {1}{5} \int \left (\frac {5+10 x^3+2 e^3 x^4}{x^3}-\frac {10 \log (x)}{x^3}\right ) \, dx \\ & = e^x+\frac {1}{5} \int \frac {5+10 x^3+2 e^3 x^4}{x^3} \, dx-2 \int \frac {\log (x)}{x^3} \, dx \\ & = e^x+\frac {1}{2 x^2}+\frac {\log (x)}{x^2}+\frac {1}{5} \int \left (10+\frac {5}{x^3}+2 e^3 x\right ) \, dx \\ & = e^x+2 x+\frac {e^3 x^2}{5}+\frac {\log (x)}{x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {5+10 x^3+5 e^x x^3+2 e^3 x^4-10 \log (x)}{5 x^3} \, dx=e^x+2 x+\frac {e^3 x^2}{5}+\frac {\log (x)}{x^2} \]

[In]

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

[Out]

E^x + 2*x + (E^3*x^2)/5 + Log[x]/x^2

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83

method	result	size
default	\(2 x +\frac {x^{2} {\mathrm e}^{3}}{5}+\frac {\ln \left (x \right )}{x^{2}}+{\mathrm e}^{x}\)	\(20\)
risch	\(2 x +\frac {x^{2} {\mathrm e}^{3}}{5}+\frac {\ln \left (x \right )}{x^{2}}+{\mathrm e}^{x}\)	\(20\)
parts	\(2 x +\frac {x^{2} {\mathrm e}^{3}}{5}+\frac {\ln \left (x \right )}{x^{2}}+{\mathrm e}^{x}\)	\(20\)
norman	\(\frac {{\mathrm e}^{x} x^{2}+2 x^{3}+\frac {x^{4} {\mathrm e}^{3}}{5}+\ln \left (x \right )}{x^{2}}\)	\(26\)
parallelrisch	\(\frac {x^{4} {\mathrm e}^{3}+5 \,{\mathrm e}^{x} x^{2}+10 x^{3}+5 \ln \left (x \right )}{5 x^{2}}\)	\(29\)

[In]

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

[Out]

2*x+1/5*x^2*exp(3)+ln(x)/x^2+exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {5+10 x^3+5 e^x x^3+2 e^3 x^4-10 \log (x)}{5 x^3} \, dx=\frac {x^{4} e^{3} + 10 \, x^{3} + 5 \, x^{2} e^{x} + 5 \, \log \left (x\right )}{5 \, x^{2}} \]

[In]

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

[Out]

1/5*(x^4*e^3 + 10*x^3 + 5*x^2*e^x + 5*log(x))/x^2

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {5+10 x^3+5 e^x x^3+2 e^3 x^4-10 \log (x)}{5 x^3} \, dx=\frac {x^{2} e^{3}}{5} + 2 x + e^{x} + \frac {\log {\left (x \right )}}{x^{2}} \]

[In]

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

[Out]

x**2*exp(3)/5 + 2*x + exp(x) + log(x)/x**2

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {5+10 x^3+5 e^x x^3+2 e^3 x^4-10 \log (x)}{5 x^3} \, dx=\frac {1}{5} \, x^{2} e^{3} + 2 \, x + \frac {\log \left (x\right )}{x^{2}} + e^{x} \]

[In]

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

[Out]

1/5*x^2*e^3 + 2*x + log(x)/x^2 + e^x

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {5+10 x^3+5 e^x x^3+2 e^3 x^4-10 \log (x)}{5 x^3} \, dx=\frac {x^{4} e^{3} + 10 \, x^{3} + 5 \, x^{2} e^{x} + 5 \, \log \left (x\right )}{5 \, x^{2}} \]

[In]

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

[Out]

1/5*(x^4*e^3 + 10*x^3 + 5*x^2*e^x + 5*log(x))/x^2

Mupad [B] (verification not implemented)

Time = 10.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {5+10 x^3+5 e^x x^3+2 e^3 x^4-10 \log (x)}{5 x^3} \, dx=2\,x+{\mathrm {e}}^x+\frac {\ln \left (x\right )}{x^2}+\frac {x^2\,{\mathrm {e}}^3}{5} \]

[In]

int((x^3*exp(x) - 2*log(x) + (2*x^4*exp(3))/5 + 2*x^3 + 1)/x^3,x)

[Out]

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