3.30.9 \(\int \frac {-25 x+10 x^2+e^{\frac {2}{5} (25+e^{25+x}+5 \log (5)+5 \log (x))} (10+2 e^{25+x} x)+e^{\frac {1}{5} (25+e^{25+x}+5 \log (5)+5 \log (x))} (-30+20 x+e^{25+x} (-6 x+2 x^2))}{5 x} \, dx\)

Optimal. Leaf size=25 \[ x+\left (3-x-5 e^{5+\frac {e^{25+x}}{5}} x\right )^2 \]

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Rubi [B]  time = 0.14, antiderivative size = 69, normalized size of antiderivative = 2.76, number of steps used = 5, number of rules used = 3, integrand size = 90, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {12, 14, 2288} \begin {gather*} 25 e^{\frac {2}{5} \left (e^{x+25}+25\right )} x^2-10 e^{\frac {1}{5} \left (e^{x+25}+25\right )-x-25} \left (3 e^{x+25} x-e^{x+25} x^2\right )+\frac {1}{4} (5-2 x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-25*x + 10*x^2 + E^((2*(25 + E^(25 + x) + 5*Log[5] + 5*Log[x]))/5)*(10 + 2*E^(25 + x)*x) + E^((25 + E^(25
+ x) + 5*Log[5] + 5*Log[x])/5)*(-30 + 20*x + E^(25 + x)*(-6*x + 2*x^2)))/(5*x),x]

[Out]

(5 - 2*x)^2/4 + 25*E^((2*(25 + E^(25 + x)))/5)*x^2 - 10*E^(-25 + (25 + E^(25 + x))/5 - x)*(3*E^(25 + x)*x - E^
(25 + x)*x^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {-25 x+10 x^2+e^{\frac {2}{5} \left (25+e^{25+x}+5 \log (5)+5 \log (x)\right )} \left (10+2 e^{25+x} x\right )+e^{\frac {1}{5} \left (25+e^{25+x}+5 \log (5)+5 \log (x)\right )} \left (-30+20 x+e^{25+x} \left (-6 x+2 x^2\right )\right )}{x} \, dx\\ &=\frac {1}{5} \int \left (5 (-5+2 x)+50 e^{\frac {2}{5} \left (25+e^{25+x}\right )} x \left (5+e^{25+x} x\right )+10 e^{\frac {1}{5} \left (25+e^{25+x}\right )} \left (-15+10 x-3 e^{25+x} x+e^{25+x} x^2\right )\right ) \, dx\\ &=\frac {1}{4} (5-2 x)^2+2 \int e^{\frac {1}{5} \left (25+e^{25+x}\right )} \left (-15+10 x-3 e^{25+x} x+e^{25+x} x^2\right ) \, dx+10 \int e^{\frac {2}{5} \left (25+e^{25+x}\right )} x \left (5+e^{25+x} x\right ) \, dx\\ &=\frac {1}{4} (5-2 x)^2+25 e^{\frac {2}{5} \left (25+e^{25+x}\right )} x^2-10 e^{-25+\frac {1}{5} \left (25+e^{25+x}\right )-x} \left (3 e^{25+x} x-e^{25+x} x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.09, size = 52, normalized size = 2.08 \begin {gather*} -5 x+x^2+25 e^{10+\frac {2 e^{25+x}}{5}} x^2+e^{\frac {e^{25+x}}{5}} \left (-30 e^5 x+10 e^5 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25*x + 10*x^2 + E^((2*(25 + E^(25 + x) + 5*Log[5] + 5*Log[x]))/5)*(10 + 2*E^(25 + x)*x) + E^((25 +
E^(25 + x) + 5*Log[5] + 5*Log[x])/5)*(-30 + 20*x + E^(25 + x)*(-6*x + 2*x^2)))/(5*x),x]

[Out]

-5*x + x^2 + 25*E^(10 + (2*E^(25 + x))/5)*x^2 + E^(E^(25 + x)/5)*(-30*E^5*x + 10*E^5*x^2)

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fricas [B]  time = 0.85, size = 42, normalized size = 1.68 \begin {gather*} x^{2} + 2 \, {\left (x - 3\right )} e^{\left (\frac {1}{5} \, e^{\left (x + 25\right )} + \log \relax (5) + \log \relax (x) + 5\right )} - 5 \, x + e^{\left (\frac {2}{5} \, e^{\left (x + 25\right )} + 2 \, \log \relax (5) + 2 \, \log \relax (x) + 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((2*x*exp(x+25)+10)*exp(log(x)+1/5*exp(x+25)+log(5)+5)^2+((2*x^2-6*x)*exp(x+25)+20*x-30)*exp(log
(x)+1/5*exp(x+25)+log(5)+5)+10*x^2-25*x)/x,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {10 \, x^{2} + 2 \, {\left (x e^{\left (x + 25\right )} + 5\right )} e^{\left (\frac {2}{5} \, e^{\left (x + 25\right )} + 2 \, \log \relax (5) + 2 \, \log \relax (x) + 10\right )} + 2 \, {\left ({\left (x^{2} - 3 \, x\right )} e^{\left (x + 25\right )} + 10 \, x - 15\right )} e^{\left (\frac {1}{5} \, e^{\left (x + 25\right )} + \log \relax (5) + \log \relax (x) + 5\right )} - 25 \, x}{5 \, x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((2*x*exp(x+25)+10)*exp(log(x)+1/5*exp(x+25)+log(5)+5)^2+((2*x^2-6*x)*exp(x+25)+20*x-30)*exp(log
(x)+1/5*exp(x+25)+log(5)+5)+10*x^2-25*x)/x,x, algorithm="giac")

[Out]

integrate(1/5*(10*x^2 + 2*(x*e^(x + 25) + 5)*e^(2/5*e^(x + 25) + 2*log(5) + 2*log(x) + 10) + 2*((x^2 - 3*x)*e^
(x + 25) + 10*x - 15)*e^(1/5*e^(x + 25) + log(5) + log(x) + 5) - 25*x)/x, x)

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maple [A]  time = 0.12, size = 38, normalized size = 1.52

method	result	size
risch	\(x^{2}+25 x^{2} {\mathrm e}^{10+\frac {2 \,{\mathrm e}^{x +25}}{5}}-5 x +\left (10 x -30\right ) x \,{\mathrm e}^{5+\frac {{\mathrm e}^{x +25}}{5}}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((2*x*exp(x+25)+10)*exp(ln(x)+1/5*exp(x+25)+ln(5)+5)^2+((2*x^2-6*x)*exp(x+25)+20*x-30)*exp(ln(x)+1/5*e
xp(x+25)+ln(5)+5)+10*x^2-25*x)/x,x,method=_RETURNVERBOSE)

[Out]

x^2+25*x^2*exp(10+2/5*exp(x+25))-5*x+(10*x-30)*x*exp(5+1/5*exp(x+25))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 25 \, x^{2} e^{\left (\frac {2}{5} \, e^{\left (x + 25\right )} + 10\right )} + x^{2} - 30 \, {\rm Ei}\left (\frac {1}{5} \, e^{\left (x + 25\right )}\right ) e^{5} + 10 \, {\left (x^{2} e^{5} - 3 \, x e^{5}\right )} e^{\left (\frac {1}{5} \, e^{\left (x + 25\right )}\right )} - 5 \, x + 30 \, \int e^{\left (\frac {1}{5} \, e^{\left (x + 25\right )} + 5\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((2*x*exp(x+25)+10)*exp(log(x)+1/5*exp(x+25)+log(5)+5)^2+((2*x^2-6*x)*exp(x+25)+20*x-30)*exp(log
(x)+1/5*exp(x+25)+log(5)+5)+10*x^2-25*x)/x,x, algorithm="maxima")

[Out]

25*x^2*e^(2/5*e^(x + 25) + 10) + x^2 - 30*Ei(1/5*e^(x + 25))*e^5 + 10*(x^2*e^5 - 3*x*e^5)*e^(1/5*e^(x + 25)) -
5*x + 30*integrate(e^(1/5*e^(x + 25) + 5), x)

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mupad [B]  time = 1.82, size = 40, normalized size = 1.60 \begin {gather*} x\,\left (x-30\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{25}\,{\mathrm {e}}^x}{5}+5}+10\,x\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{25}\,{\mathrm {e}}^x}{5}+5}+25\,x\,{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^{25}\,{\mathrm {e}}^x}{5}+10}-5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x - (exp((2*exp(x + 25))/5 + 2*log(5) + 2*log(x) + 10)*(2*x*exp(x + 25) + 10))/5 + (exp(exp(x + 25)/5
+ log(5) + log(x) + 5)*(exp(x + 25)*(6*x - 2*x^2) - 20*x + 30))/5 - 2*x^2)/x,x)

[Out]

x*(x - 30*exp((exp(25)*exp(x))/5 + 5) + 10*x*exp((exp(25)*exp(x))/5 + 5) + 25*x*exp((2*exp(25)*exp(x))/5 + 10)
- 5)

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sympy [B]  time = 0.21, size = 41, normalized size = 1.64 \begin {gather*} 25 x^{2} e^{\frac {2 e^{x + 25}}{5} + 10} + x^{2} - 5 x + \left (10 x^{2} - 30 x\right ) e^{\frac {e^{x + 25}}{5} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((2*x*exp(x+25)+10)*exp(ln(x)+1/5*exp(x+25)+ln(5)+5)**2+((2*x**2-6*x)*exp(x+25)+20*x-30)*exp(ln(
x)+1/5*exp(x+25)+ln(5)+5)+10*x**2-25*x)/x,x)

[Out]

25*x**2*exp(2*exp(x + 25)/5 + 10) + x**2 - 5*x + (10*x**2 - 30*x)*exp(exp(x + 25)/5 + 5)

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