3.49.3 \(\int \frac {-30 x^2+e^{\frac {25+10 x+x^2}{x}} (-125+5 x^2)+6 x^2 \log (2)}{-75 x^2+5 e^{\frac {25+10 x+x^2}{x}} x^2-30 x^3+(15 x^2+6 x^3) \log (2)} \, dx\)

Optimal. Leaf size=29 \[ \log (2)+\log \left (x+5 \left (3+x+\frac {e^{\frac {(5+x)^2}{x}}}{-5+\log (2)}\right )\right ) \]

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Rubi [A]  time = 0.44, antiderivative size = 32, normalized size of antiderivative = 1.10, number of steps used = 3, number of rules used = 3, integrand size = 82, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {6, 6688, 6684} \begin {gather*} \log \left (5 e^{\frac {(x+5)^2}{x}}-(x (30-\log (64)))-15 (5-\log (2))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-30*x^2 + E^((25 + 10*x + x^2)/x)*(-125 + 5*x^2) + 6*x^2*Log[2])/(-75*x^2 + 5*E^((25 + 10*x + x^2)/x)*x^2
- 30*x^3 + (15*x^2 + 6*x^3)*Log[2]),x]

[Out]

Log[5*E^((5 + x)^2/x) - 15*(5 - Log[2]) - x*(30 - Log[64])]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {25+10 x+x^2}{x}} \left (-125+5 x^2\right )+x^2 (-30+6 \log (2))}{-75 x^2+5 e^{\frac {25+10 x+x^2}{x}} x^2-30 x^3+\left (15 x^2+6 x^3\right ) \log (2)} \, dx\\ &=\int \frac {5 e^{\frac {(5+x)^2}{x}} \left (-25+x^2\right )+x^2 (-30+\log (64))}{x^2 \left (5 e^{\frac {(5+x)^2}{x}}+15 (-5+\log (2))+x (-30+\log (64))\right )} \, dx\\ &=\log \left (5 e^{\frac {(5+x)^2}{x}}-15 (5-\log (2))-x (30-\log (64))\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.47, size = 27, normalized size = 0.93 \begin {gather*} \log \left (75-5 e^{10+\frac {25}{x}+x}+30 x-15 \log (2)-x \log (64)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-30*x^2 + E^((25 + 10*x + x^2)/x)*(-125 + 5*x^2) + 6*x^2*Log[2])/(-75*x^2 + 5*E^((25 + 10*x + x^2)/
x)*x^2 - 30*x^3 + (15*x^2 + 6*x^3)*Log[2]),x]

[Out]

Log[75 - 5*E^(10 + 25/x + x) + 30*x - 15*Log[2] - x*Log[64]]

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fricas [A]  time = 0.84, size = 30, normalized size = 1.03 \begin {gather*} \log \left (3 \, {\left (2 \, x + 5\right )} \log \relax (2) - 30 \, x + 5 \, e^{\left (\frac {x^{2} + 10 \, x + 25}{x}\right )} - 75\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2-125)*exp((x^2+10*x+25)/x)+6*x^2*log(2)-30*x^2)/(5*x^2*exp((x^2+10*x+25)/x)+(6*x^3+15*x^2)*lo
g(2)-30*x^3-75*x^2),x, algorithm="fricas")

[Out]

log(3*(2*x + 5)*log(2) - 30*x + 5*e^((x^2 + 10*x + 25)/x) - 75)

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giac [A]  time = 0.18, size = 30, normalized size = 1.03 \begin {gather*} \log \left (6 \, x \log \relax (2) - 30 \, x + 5 \, e^{\left (\frac {x^{2} + 10 \, x + 25}{x}\right )} + 15 \, \log \relax (2) - 75\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2-125)*exp((x^2+10*x+25)/x)+6*x^2*log(2)-30*x^2)/(5*x^2*exp((x^2+10*x+25)/x)+(6*x^3+15*x^2)*lo
g(2)-30*x^3-75*x^2),x, algorithm="giac")

[Out]

log(6*x*log(2) - 30*x + 5*e^((x^2 + 10*x + 25)/x) + 15*log(2) - 75)

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maple [A]  time = 0.16, size = 31, normalized size = 1.07

method	result	size
norman	\(\ln \left (6 x \ln \relax (2)+5 \,{\mathrm e}^{\frac {x^{2}+10 x +25}{x}}+15 \ln \relax (2)-30 x -75\right )\)	\(31\)
risch	\(x +\frac {25}{x}-\frac {x^{2}+10 x +25}{x}+\ln \left (\frac {6 x \ln \relax (2)}{5}+3 \ln \relax (2)-6 x +{\mathrm e}^{\frac {\left (5+x \right )^{2}}{x}}-15\right )\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x^2-125)*exp((x^2+10*x+25)/x)+6*x^2*ln(2)-30*x^2)/(5*x^2*exp((x^2+10*x+25)/x)+(6*x^3+15*x^2)*ln(2)-30*
x^3-75*x^2),x,method=_RETURNVERBOSE)

[Out]

ln(6*x*ln(2)+5*exp((x^2+10*x+25)/x)+15*ln(2)-30*x-75)

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maxima [A]  time = 0.48, size = 35, normalized size = 1.21 \begin {gather*} x + \log \left (\frac {1}{5} \, {\left (6 \, x {\left (\log \relax (2) - 5\right )} + 5 \, e^{\left (x + \frac {25}{x} + 10\right )} + 15 \, \log \relax (2) - 75\right )} e^{\left (-x - 10\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2-125)*exp((x^2+10*x+25)/x)+6*x^2*log(2)-30*x^2)/(5*x^2*exp((x^2+10*x+25)/x)+(6*x^3+15*x^2)*lo
g(2)-30*x^3-75*x^2),x, algorithm="maxima")

[Out]

x + log(1/5*(6*x*(log(2) - 5) + 5*e^(x + 25/x + 10) + 15*log(2) - 75)*e^(-x - 10))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{\frac {x^2+10\,x+25}{x}}\,\left (5\,x^2-125\right )+6\,x^2\,\ln \relax (2)-30\,x^2}{\ln \relax (2)\,\left (6\,x^3+15\,x^2\right )-75\,x^2-30\,x^3+5\,x^2\,{\mathrm {e}}^{\frac {x^2+10\,x+25}{x}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((10*x + x^2 + 25)/x)*(5*x^2 - 125) + 6*x^2*log(2) - 30*x^2)/(log(2)*(15*x^2 + 6*x^3) - 75*x^2 - 30*x^
3 + 5*x^2*exp((10*x + x^2 + 25)/x)),x)

[Out]

int((exp((10*x + x^2 + 25)/x)*(5*x^2 - 125) + 6*x^2*log(2) - 30*x^2)/(log(2)*(15*x^2 + 6*x^3) - 75*x^2 - 30*x^
3 + 5*x^2*exp((10*x + x^2 + 25)/x)), x)

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sympy [A]  time = 0.29, size = 31, normalized size = 1.07 \begin {gather*} \log {\left (- 6 x + \frac {6 x \log {\relax (2 )}}{5} + e^{\frac {x^{2} + 10 x + 25}{x}} - 15 + 3 \log {\relax (2 )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x**2-125)*exp((x**2+10*x+25)/x)+6*x**2*ln(2)-30*x**2)/(5*x**2*exp((x**2+10*x+25)/x)+(6*x**3+15*x
**2)*ln(2)-30*x**3-75*x**2),x)

[Out]

log(-6*x + 6*x*log(2)/5 + exp((x**2 + 10*x + 25)/x) - 15 + 3*log(2))

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