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

Optimal. Leaf size=32 \[ \frac {1}{5} \left (e^{\frac {1}{5} \left (\frac {10}{x}+x^2\right )} \left (e^5-3 x^2\right )+\log (x)\right ) \]

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Rubi [F]  time = 0.40, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 36, normalized size = 1.12 \begin {gather*} \frac {1}{25} \left (e^{\frac {2}{x}+\frac {x^2}{5}} \left (5 e^5-15 x^2\right )+5 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[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)*(5*E^5 - 15*x^2) + 5*Log[x])/25

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fricas [A]  time = 0.68, size = 28, normalized size = 0.88 \begin {gather*} -\frac {1}{5} \, {\left (3 \, x^{2} - e^{5}\right )} e^{\left (\frac {x^{3} + 10}{5 \, x}\right )} + \frac {1}{5} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.16, size = 37, normalized size = 1.16 \begin {gather*} -\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 \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.16, size = 29, normalized size = 0.91

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.40, size = 29, normalized size = 0.91 \begin {gather*} -\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 \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 4.97, size = 38, normalized size = 1.19 \begin {gather*} \frac {\ln \relax (x)}{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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.23, size = 24, normalized size = 0.75 \begin {gather*} \frac {\left (- 3 x^{2} + e^{5}\right ) e^{\frac {\frac {x^{3}}{5} + 2}{x}}}{5} + \frac {\log {\relax (x )}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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