3.70.87 \(\int \frac {-15+e^{\frac {1}{5} (e^{40+x} x+e^{40} (5+10 x))} (-30 e^{40}+e^{40+x} (-3-3 x))}{(5 e^{\frac {1}{5} (e^{40+x} x+e^{40} (5+10 x))}+5 x) \log ^2(e^{\frac {1}{5} (e^{40+x} x+e^{40} (5+10 x))}+x)} \, dx\)

Optimal. Leaf size=28 \[ 1+\frac {3}{\log \left (e^{e^{40} \left (1+\left (2+\frac {e^x}{5}\right ) x\right )}+x\right )} \]

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Rubi [A]  time = 0.63, antiderivative size = 31, normalized size of antiderivative = 1.11, number of steps used = 1, number of rules used = 1, integrand size = 103, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.010, Rules used = {6686} \begin {gather*} \frac {3}{\log \left (x+e^{\frac {1}{5} \left (e^{x+40} x+5 e^{40} (2 x+1)\right )}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-15 + E^((E^(40 + x)*x + E^40*(5 + 10*x))/5)*(-30*E^40 + E^(40 + x)*(-3 - 3*x)))/((5*E^((E^(40 + x)*x + E
^40*(5 + 10*x))/5) + 5*x)*Log[E^((E^(40 + x)*x + E^40*(5 + 10*x))/5) + x]^2),x]

[Out]

3/Log[E^((E^(40 + x)*x + 5*E^40*(1 + 2*x))/5) + x]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {3}{\log \left (e^{\frac {1}{5} \left (e^{40+x} x+5 e^{40} (1+2 x)\right )}+x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 29, normalized size = 1.04 \begin {gather*} \frac {3}{\log \left (e^{e^{40}+2 e^{40} x+\frac {1}{5} e^{40+x} x}+x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15 + E^((E^(40 + x)*x + E^40*(5 + 10*x))/5)*(-30*E^40 + E^(40 + x)*(-3 - 3*x)))/((5*E^((E^(40 + x)
*x + E^40*(5 + 10*x))/5) + 5*x)*Log[E^((E^(40 + x)*x + E^40*(5 + 10*x))/5) + x]^2),x]

[Out]

3/Log[E^(E^40 + 2*E^40*x + (E^(40 + x)*x)/5) + x]

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fricas [A]  time = 0.67, size = 24, normalized size = 0.86 \begin {gather*} \frac {3}{\log \left (x + e^{\left ({\left (2 \, x + 1\right )} e^{40} + \frac {1}{5} \, x e^{\left (x + 40\right )}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x-3)*exp(20)^2*exp(x)-30*exp(20)^2)*exp(1/5*x*exp(20)^2*exp(x)+1/5*(10*x+5)*exp(20)^2)-15)/(5*
exp(1/5*x*exp(20)^2*exp(x)+1/5*(10*x+5)*exp(20)^2)+5*x)/log(exp(1/5*x*exp(20)^2*exp(x)+1/5*(10*x+5)*exp(20)^2)
+x)^2,x, algorithm="fricas")

[Out]

3/log(x + e^((2*x + 1)*e^40 + 1/5*x*e^(x + 40)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {3 \, {\left ({\left ({\left (x + 1\right )} e^{\left (x + 40\right )} + 10 \, e^{40}\right )} e^{\left ({\left (2 \, x + 1\right )} e^{40} + \frac {1}{5} \, x e^{\left (x + 40\right )}\right )} + 5\right )}}{5 \, {\left (x + e^{\left ({\left (2 \, x + 1\right )} e^{40} + \frac {1}{5} \, x e^{\left (x + 40\right )}\right )}\right )} \log \left (x + e^{\left ({\left (2 \, x + 1\right )} e^{40} + \frac {1}{5} \, x e^{\left (x + 40\right )}\right )}\right )^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x-3)*exp(20)^2*exp(x)-30*exp(20)^2)*exp(1/5*x*exp(20)^2*exp(x)+1/5*(10*x+5)*exp(20)^2)-15)/(5*
exp(1/5*x*exp(20)^2*exp(x)+1/5*(10*x+5)*exp(20)^2)+5*x)/log(exp(1/5*x*exp(20)^2*exp(x)+1/5*(10*x+5)*exp(20)^2)
+x)^2,x, algorithm="giac")

[Out]

integrate(-3/5*(((x + 1)*e^(x + 40) + 10*e^40)*e^((2*x + 1)*e^40 + 1/5*x*e^(x + 40)) + 5)/((x + e^((2*x + 1)*e
^40 + 1/5*x*e^(x + 40)))*log(x + e^((2*x + 1)*e^40 + 1/5*x*e^(x + 40)))^2), x)

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maple [A]  time = 0.08, size = 24, normalized size = 0.86

method	result	size
risch	\(\frac {3}{\ln \left ({\mathrm e}^{\frac {x \,{\mathrm e}^{40+x}}{5}+2 \,{\mathrm e}^{40} x +{\mathrm e}^{40}}+x \right )}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-3*x-3)*exp(20)^2*exp(x)-30*exp(20)^2)*exp(1/5*x*exp(20)^2*exp(x)+1/5*(10*x+5)*exp(20)^2)-15)/(5*exp(1/
5*x*exp(20)^2*exp(x)+1/5*(10*x+5)*exp(20)^2)+5*x)/ln(exp(1/5*x*exp(20)^2*exp(x)+1/5*(10*x+5)*exp(20)^2)+x)^2,x
,method=_RETURNVERBOSE)

[Out]

3/ln(exp(1/5*x*exp(40+x)+2*exp(40)*x+exp(40))+x)

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maxima [A]  time = 0.45, size = 23, normalized size = 0.82 \begin {gather*} \frac {3}{\log \left (x + e^{\left (2 \, x e^{40} + \frac {1}{5} \, x e^{\left (x + 40\right )} + e^{40}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x-3)*exp(20)^2*exp(x)-30*exp(20)^2)*exp(1/5*x*exp(20)^2*exp(x)+1/5*(10*x+5)*exp(20)^2)-15)/(5*
exp(1/5*x*exp(20)^2*exp(x)+1/5*(10*x+5)*exp(20)^2)+5*x)/log(exp(1/5*x*exp(20)^2*exp(x)+1/5*(10*x+5)*exp(20)^2)
+x)^2,x, algorithm="maxima")

[Out]

3/log(x + e^(2*x*e^40 + 1/5*x*e^(x + 40) + e^40))

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mupad [B]  time = 4.23, size = 25, normalized size = 0.89 \begin {gather*} \frac {3}{\ln \left (x+{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{40}\,{\mathrm {e}}^x}{5}}\,{\mathrm {e}}^{2\,x\,{\mathrm {e}}^{40}}\,{\mathrm {e}}^{{\mathrm {e}}^{40}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((exp(40)*(10*x + 5))/5 + (x*exp(40)*exp(x))/5)*(30*exp(40) + exp(40)*exp(x)*(3*x + 3)) + 15)/(log(x
+ exp((exp(40)*(10*x + 5))/5 + (x*exp(40)*exp(x))/5))^2*(5*x + 5*exp((exp(40)*(10*x + 5))/5 + (x*exp(40)*exp(x
))/5))),x)

[Out]

3/log(x + exp((x*exp(40)*exp(x))/5)*exp(2*x*exp(40))*exp(exp(40)))

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sympy [A]  time = 0.85, size = 24, normalized size = 0.86 \begin {gather*} \frac {3}{\log {\left (x + e^{\frac {x e^{40} e^{x}}{5} + \left (2 x + 1\right ) e^{40}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x-3)*exp(20)**2*exp(x)-30*exp(20)**2)*exp(1/5*x*exp(20)**2*exp(x)+1/5*(10*x+5)*exp(20)**2)-15)
/(5*exp(1/5*x*exp(20)**2*exp(x)+1/5*(10*x+5)*exp(20)**2)+5*x)/ln(exp(1/5*x*exp(20)**2*exp(x)+1/5*(10*x+5)*exp(
20)**2)+x)**2,x)

[Out]

3/log(x + exp(x*exp(40)*exp(x)/5 + (2*x + 1)*exp(40)))

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