\(\int \frac {5 x+e^{\frac {10+x^3}{5 x}} (30 x^2-30 x^3-6 x^5+e^5 (-10+2 x^3))}{25 x^2} \, dx\) [7771]

Optimal result

Integrand size = 53, antiderivative size = 32 \[ \int \frac {5 x+e^{\frac {10+x^3}{5 x}} \left (30 x^2-30 x^3-6 x^5+e^5 \left (-10+2 x^3\right )\right )}{25 x^2} \, dx=\frac {1}{5} \left (e^{\frac {1}{5} \left (\frac {10}{x}+x^2\right )} \left (e^5-3 x^2\right )+\log (x)\right ) \]

[Out]

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

Rubi [F]

\[ \int \frac {5 x+e^{\frac {10+x^3}{5 x}} \left (30 x^2-30 x^3-6 x^5+e^5 \left (-10+2 x^3\right )\right )}{25 x^2} \, dx=\int \frac {5 x+e^{\frac {10+x^3}{5 x}} \left (30 x^2-30 x^3-6 x^5+e^5 \left (-10+2 x^3\right )\right )}{25 x^2} \, dx \]

[In]

Int[(5*x + E^((10 + x^3)/(5*x))*(30*x^2 - 30*x^3 - 6*x^5 + E^5*(-10 + 2*x^3)))/(25*x^2),x]

[Out]

Log[x]/5 + (6*Defer[Int][E^((10 + x^3)/(5*x)), x])/5 - (2*Defer[Int][E^((10 + 25*x + x^3)/(5*x))/x^2, x])/5 -
(2*(15 - E^5)*Defer[Int][E^((10 + x^3)/(5*x))*x, x])/25 - (6*Defer[Int][E^((10 + x^3)/(5*x))*x^3, x])/25

Rubi steps \begin{align*} \text {integral}& = \frac {1}{25} \int \frac {5 x+e^{\frac {10+x^3}{5 x}} \left (30 x^2-30 x^3-6 x^5+e^5 \left (-10+2 x^3\right )\right )}{x^2} \, dx \\ & = \frac {1}{25} \int \left (\frac {5}{x}+\frac {2 e^{\frac {10+x^3}{5 x}} \left (-5 e^5+15 x^2-15 \left (1-\frac {e^5}{15}\right ) x^3-3 x^5\right )}{x^2}\right ) \, dx \\ & = \frac {\log (x)}{5}+\frac {2}{25} \int \frac {e^{\frac {10+x^3}{5 x}} \left (-5 e^5+15 x^2-15 \left (1-\frac {e^5}{15}\right ) x^3-3 x^5\right )}{x^2} \, dx \\ & = \frac {\log (x)}{5}+\frac {2}{25} \int \left (15 e^{\frac {10+x^3}{5 x}}-\frac {5 e^{5+\frac {10+x^3}{5 x}}}{x^2}+e^{\frac {10+x^3}{5 x}} \left (-15+e^5\right ) x-3 e^{\frac {10+x^3}{5 x}} x^3\right ) \, dx \\ & = \frac {\log (x)}{5}-\frac {6}{25} \int e^{\frac {10+x^3}{5 x}} x^3 \, dx-\frac {2}{5} \int \frac {e^{5+\frac {10+x^3}{5 x}}}{x^2} \, dx+\frac {6}{5} \int e^{\frac {10+x^3}{5 x}} \, dx-\frac {1}{25} \left (2 \left (15-e^5\right )\right ) \int e^{\frac {10+x^3}{5 x}} x \, dx \\ & = \frac {\log (x)}{5}-\frac {6}{25} \int e^{\frac {10+x^3}{5 x}} x^3 \, dx-\frac {2}{5} \int \frac {e^{\frac {10+25 x+x^3}{5 x}}}{x^2} \, dx+\frac {6}{5} \int e^{\frac {10+x^3}{5 x}} \, dx-\frac {1}{25} \left (2 \left (15-e^5\right )\right ) \int e^{\frac {10+x^3}{5 x}} x \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {5 x+e^{\frac {10+x^3}{5 x}} \left (30 x^2-30 x^3-6 x^5+e^5 \left (-10+2 x^3\right )\right )}{25 x^2} \, dx=\frac {1}{5} \left (e^{\frac {2}{x}+\frac {x^2}{5}} \left (e^5-3 x^2\right )+\log (x)\right ) \]

[In]

Integrate[(5*x + E^((10 + x^3)/(5*x))*(30*x^2 - 30*x^3 - 6*x^5 + E^5*(-10 + 2*x^3)))/(25*x^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91

method	result	size
risch	\(\frac {\ln \left (x \right )}{5}+\frac {\left (-15 x^{2}+5 \,{\mathrm e}^{5}\right ) {\mathrm e}^{\frac {x^{3}+10}{5 x}}}{25}\)	\(29\)
parallelrisch	\(\frac {\ln \left (x \right )}{5}-\frac {3 \,{\mathrm e}^{\frac {x^{3}+10}{5 x}} x^{2}}{5}+\frac {{\mathrm e}^{5} {\mathrm e}^{\frac {x^{3}+10}{5 x}}}{5}\)	\(37\)
norman	\(\frac {-\frac {3 x^{3} {\mathrm e}^{\frac {x^{3}+10}{5 x}}}{5}+\frac {{\mathrm e}^{5} {\mathrm e}^{\frac {x^{3}+10}{5 x}} x}{5}}{x}+\frac {\ln \left (x \right )}{5}\)	\(43\)
parts	\(\frac {-\frac {3 x^{3} {\mathrm e}^{\frac {x^{3}+10}{5 x}}}{5}+\frac {{\mathrm e}^{5} {\mathrm e}^{\frac {x^{3}+10}{5 x}} x}{5}}{x}+\frac {\ln \left (x \right )}{5}\)	\(43\)

[In]

int(1/25*(((2*x^3-10)*exp(5)-6*x^5-30*x^3+30*x^2)*exp(1/5*(x^3+10)/x)+5*x)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/5*ln(x)+1/25*(-15*x^2+5*exp(5))*exp(1/5*(x^3+10)/x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {5 x+e^{\frac {10+x^3}{5 x}} \left (30 x^2-30 x^3-6 x^5+e^5 \left (-10+2 x^3\right )\right )}{25 x^2} \, dx=-\frac {1}{5} \, {\left (3 \, x^{2} - e^{5}\right )} e^{\left (\frac {x^{3} + 10}{5 \, x}\right )} + \frac {1}{5} \, \log \left (x\right ) \]

[In]

integrate(1/25*(((2*x^3-10)*exp(5)-6*x^5-30*x^3+30*x^2)*exp(1/5*(x^3+10)/x)+5*x)/x^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {5 x+e^{\frac {10+x^3}{5 x}} \left (30 x^2-30 x^3-6 x^5+e^5 \left (-10+2 x^3\right )\right )}{25 x^2} \, dx=\frac {\left (- 3 x^{2} + e^{5}\right ) e^{\frac {\frac {x^{3}}{5} + 2}{x}}}{5} + \frac {\log {\left (x \right )}}{5} \]

[In]

integrate(1/25*(((2*x**3-10)*exp(5)-6*x**5-30*x**3+30*x**2)*exp(1/5*(x**3+10)/x)+5*x)/x**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {5 x+e^{\frac {10+x^3}{5 x}} \left (30 x^2-30 x^3-6 x^5+e^5 \left (-10+2 x^3\right )\right )}{25 x^2} \, dx=-\frac {1}{5} \, {\left (3 \, x^{2} - e^{5}\right )} e^{\left (\frac {1}{5} \, x^{2} + \frac {2}{x}\right )} + \frac {1}{5} \, \log \left (x\right ) \]

[In]

integrate(1/25*(((2*x^3-10)*exp(5)-6*x^5-30*x^3+30*x^2)*exp(1/5*(x^3+10)/x)+5*x)/x^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {5 x+e^{\frac {10+x^3}{5 x}} \left (30 x^2-30 x^3-6 x^5+e^5 \left (-10+2 x^3\right )\right )}{25 x^2} \, dx=-\frac {3}{5} \, x^{2} e^{\left (\frac {x^{3} + 10}{5 \, x}\right )} + \frac {1}{5} \, e^{\left (\frac {x^{3} + 25 \, x + 10}{5 \, x}\right )} + \frac {1}{5} \, \log \left (x\right ) \]

[In]

integrate(1/25*(((2*x^3-10)*exp(5)-6*x^5-30*x^3+30*x^2)*exp(1/5*(x^3+10)/x)+5*x)/x^2,x, algorithm="giac")

[Out]

-3/5*x^2*e^(1/5*(x^3 + 10)/x) + 1/5*e^(1/5*(x^3 + 25*x + 10)/x) + 1/5*log(x)

Mupad [B] (verification not implemented)

Time = 12.49 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.19 \[ \int \frac {5 x+e^{\frac {10+x^3}{5 x}} \left (30 x^2-30 x^3-6 x^5+e^5 \left (-10+2 x^3\right )\right )}{25 x^2} \, dx=\frac {\ln \left (x\right )}{5}-\frac {3\,x^2\,{\left ({\mathrm {e}}^{x^2}\right )}^{1/5}\,{\mathrm {e}}^{2/x}}{5}+\frac {{\left ({\mathrm {e}}^{x^2}\right )}^{1/5}\,{\mathrm {e}}^5\,{\mathrm {e}}^{2/x}}{5} \]

[In]

int((x/5 + (exp((x^3/5 + 2)/x)*(exp(5)*(2*x^3 - 10) + 30*x^2 - 30*x^3 - 6*x^5))/25)/x^2,x)

[Out]

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