\(\int \frac {1+2 e^3-x+x^2+2 x^3+3 x^5+(x-x^3) \log (5)-2 \log (x)}{x^3} \, dx\) [7013]

Optimal result

Integrand size = 41, antiderivative size = 29 \[ \int \frac {1+2 e^3-x+x^2+2 x^3+3 x^5+\left (x-x^3\right ) \log (5)-2 \log (x)}{x^3} \, dx=\left (\frac {1}{x}+x\right ) \left (1-\frac {e^3}{x}+x^2-\log (5)+\frac {\log (x)}{x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.76, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {14, 2341} \[ \int \frac {1+2 e^3-x+x^2+2 x^3+3 x^5+\left (x-x^3\right ) \log (5)-2 \log (x)}{x^3} \, dx=x^3-\frac {1+2 e^3}{2 x^2}+\frac {1}{2 x^2}+\frac {\log (x)}{x^2}+x (2-\log (5))+\log (x)+\frac {1-\log (5)}{x} \]

[In]

Int[(1 + 2*E^3 - x + x^2 + 2*x^3 + 3*x^5 + (x - x^3)*Log[5] - 2*Log[x])/x^3,x]

[Out]

1/(2*x^2) - (1 + 2*E^3)/(2*x^2) + x^3 + (1 - Log[5])/x + x*(2 - Log[5]) + Log[x] + Log[x]/x^2

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 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}& = \int \left (\frac {1+2 e^3+x^2+3 x^5-x (1-\log (5))+2 x^3 \left (1-\frac {\log (5)}{2}\right )}{x^3}-\frac {2 \log (x)}{x^3}\right ) \, dx \\ & = -\left (2 \int \frac {\log (x)}{x^3} \, dx\right )+\int \frac {1+2 e^3+x^2+3 x^5-x (1-\log (5))+2 x^3 \left (1-\frac {\log (5)}{2}\right )}{x^3} \, dx \\ & = \frac {1}{2 x^2}+\frac {\log (x)}{x^2}+\int \left (\frac {1+2 e^3}{x^3}+\frac {1}{x}+3 x^2+2 \left (1-\frac {\log (5)}{2}\right )+\frac {-1+\log (5)}{x^2}\right ) \, dx \\ & = \frac {1}{2 x^2}-\frac {1+2 e^3}{2 x^2}+x^3+\frac {1-\log (5)}{x}+x (2-\log (5))+\log (x)+\frac {\log (x)}{x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int \frac {1+2 e^3-x+x^2+2 x^3+3 x^5+\left (x-x^3\right ) \log (5)-2 \log (x)}{x^3} \, dx=-\frac {e^3}{x^2}+\frac {1}{x}+2 x+x^3-\frac {\log (5)}{x}-x \log (5)+\log (x)+\frac {\log (x)}{x^2} \]

[In]

Integrate[(1 + 2*E^3 - x + x^2 + 2*x^3 + 3*x^5 + (x - x^3)*Log[5] - 2*Log[x])/x^3,x]

[Out]

-(E^3/x^2) + x^(-1) + 2*x + x^3 - Log[5]/x - x*Log[5] + Log[x] + Log[x]/x^2

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34

method	result	size
norman	\(\frac {x^{5}+\left (2-\ln \left (5\right )\right ) x^{3}+\left (-\ln \left (5\right )+1\right ) x +x^{2} \ln \left (x \right )-{\mathrm e}^{3}+\ln \left (x \right )}{x^{2}}\)	\(39\)
default	\(-\frac {{\mathrm e}^{3-\ln \left (x \right )}}{x}+x^{3}-x \ln \left (5\right )+2 x +\ln \left (x \right )+\frac {-\ln \left (5\right )+1}{x}+\frac {\ln \left (x \right )}{x^{2}}\)	\(43\)
parts	\(-\frac {{\mathrm e}^{3-\ln \left (x \right )}}{x}+x^{3}-x \ln \left (5\right )+2 x +\ln \left (x \right )+\frac {-\ln \left (5\right )+1}{x}+\frac {\ln \left (x \right )}{x^{2}}\)	\(43\)
risch	\(\frac {\ln \left (x \right )}{x^{2}}+\frac {x^{5}-x^{3} \ln \left (5\right )+x^{2} \ln \left (x \right )+2 x^{3}-x \ln \left (5\right )-{\mathrm e}^{3}+x}{x^{2}}\)	\(44\)
parallelrisch	\(-\frac {-x^{5}+x^{3} \ln \left (5\right )-2 x^{3}-x^{2} \ln \left (x \right )+x \ln \left (5\right )+x \,{\mathrm e}^{3-\ln \left (x \right )}-x -\ln \left (x \right )}{x^{2}}\)	\(50\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {1+2 e^3-x+x^2+2 x^3+3 x^5+\left (x-x^3\right ) \log (5)-2 \log (x)}{x^3} \, dx=\frac {x^{5} + 2 \, x^{3} - {\left (x^{3} + x\right )} \log \left (5\right ) + {\left (x^{2} + 1\right )} \log \left (x\right ) + x - e^{3}}{x^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {1+2 e^3-x+x^2+2 x^3+3 x^5+\left (x-x^3\right ) \log (5)-2 \log (x)}{x^3} \, dx=x^{3} + x \left (2 - \log {\left (5 \right )}\right ) + \log {\left (x \right )} + \frac {x \left (1 - \log {\left (5 \right )}\right ) - e^{3}}{x^{2}} + \frac {\log {\left (x \right )}}{x^{2}} \]

[In]

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

[Out]

x**3 + x*(2 - log(5)) + log(x) + (x*(1 - log(5)) - exp(3))/x**2 + log(x)/x**2

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {1+2 e^3-x+x^2+2 x^3+3 x^5+\left (x-x^3\right ) \log (5)-2 \log (x)}{x^3} \, dx=x^{3} - x \log \left (5\right ) + 2 \, x - \frac {\log \left (5\right )}{x} + \frac {1}{x} - \frac {e^{3}}{x^{2}} + \frac {\log \left (x\right )}{x^{2}} + \log \left (x\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int \frac {1+2 e^3-x+x^2+2 x^3+3 x^5+\left (x-x^3\right ) \log (5)-2 \log (x)}{x^3} \, dx=\frac {x^{5} - x^{3} \log \left (5\right ) + 2 \, x^{3} + x^{2} \log \left (x\right ) - x \log \left (5\right ) + x - e^{3} + \log \left (x\right )}{x^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 17.17 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {1+2 e^3-x+x^2+2 x^3+3 x^5+\left (x-x^3\right ) \log (5)-2 \log (x)}{x^3} \, dx=\ln \left (x\right )-\frac {x^2\,\left (\ln \left (5\right )-1\right )+x\,\left ({\mathrm {e}}^3-\ln \left (x\right )\right )}{x^3}-x\,\left (\ln \left (5\right )-2\right )+x^3 \]

[In]

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

[Out]

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