3.101.67 \(\int (75 x^2+300 e^3 x^2+300 e^6 x^2+e^{4 x} (1+4 x)+(150 x^2+300 e^3 x^2) \log (4)+75 x^2 \log ^2(4)+e^{2 x} (-20 x-20 x^2+e^3 (-40 x-40 x^2)+(-20 x-20 x^2) \log (4))) \, dx\)

Optimal. Leaf size=26 \[ x \left (-e^{2 x}+5 \left (x+2 e^3 x+x \log (4)\right )\right )^2 \]

________________________________________________________________________________________

Rubi [B]  time = 0.16, antiderivative size = 89, normalized size of antiderivative = 3.42, number of steps used = 23, number of rules used = 6, integrand size = 99, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {6, 2176, 2194, 12, 30, 2196} \begin {gather*} 25 x^3 \left (1+4 e^3+4 e^6+\log ^2(4)\right )+50 \left (1+2 e^3\right ) x^3 \log (4)-10 e^{2 x} x^2-10 e^{2 x+3} x^2 \left (2+\frac {\log (4)}{e^3}\right )-\frac {e^{4 x}}{4}+\frac {1}{4} e^{4 x} (4 x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[75*x^2 + 300*E^3*x^2 + 300*E^6*x^2 + E^(4*x)*(1 + 4*x) + (150*x^2 + 300*E^3*x^2)*Log[4] + 75*x^2*Log[4]^2
+ E^(2*x)*(-20*x - 20*x^2 + E^3*(-40*x - 40*x^2) + (-20*x - 20*x^2)*Log[4]),x]

[Out]

-1/4*E^(4*x) - 10*E^(2*x)*x^2 + (E^(4*x)*(1 + 4*x))/4 + 50*(1 + 2*E^3)*x^3*Log[4] - 10*E^(3 + 2*x)*x^2*(2 + Lo
g[4]/E^3) + 25*x^3*(1 + 4*E^3 + 4*E^6 + Log[4]^2)

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 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (300 e^6 x^2+\left (75+300 e^3\right ) x^2+e^{4 x} (1+4 x)+\left (150 x^2+300 e^3 x^2\right ) \log (4)+75 x^2 \log ^2(4)+e^{2 x} \left (-20 x-20 x^2+e^3 \left (-40 x-40 x^2\right )+\left (-20 x-20 x^2\right ) \log (4)\right )\right ) \, dx\\ &=\int \left (\left (75+300 e^3+300 e^6\right ) x^2+e^{4 x} (1+4 x)+\left (150 x^2+300 e^3 x^2\right ) \log (4)+75 x^2 \log ^2(4)+e^{2 x} \left (-20 x-20 x^2+e^3 \left (-40 x-40 x^2\right )+\left (-20 x-20 x^2\right ) \log (4)\right )\right ) \, dx\\ &=\int \left (e^{4 x} (1+4 x)+\left (150 x^2+300 e^3 x^2\right ) \log (4)+e^{2 x} \left (-20 x-20 x^2+e^3 \left (-40 x-40 x^2\right )+\left (-20 x-20 x^2\right ) \log (4)\right )+x^2 \left (75+300 e^3+300 e^6+75 \log ^2(4)\right )\right ) \, dx\\ &=25 x^3 \left (1+4 e^3+4 e^6+\log ^2(4)\right )+\log (4) \int \left (150 x^2+300 e^3 x^2\right ) \, dx+\int e^{4 x} (1+4 x) \, dx+\int e^{2 x} \left (-20 x-20 x^2+e^3 \left (-40 x-40 x^2\right )+\left (-20 x-20 x^2\right ) \log (4)\right ) \, dx\\ &=\frac {1}{4} e^{4 x} (1+4 x)+25 x^3 \left (1+4 e^3+4 e^6+\log ^2(4)\right )+\log (4) \int \left (150+300 e^3\right ) x^2 \, dx-\int e^{4 x} \, dx+\int \left (-20 e^{2 x} x-20 e^{2 x} x^2-40 e^{3+2 x} x (1+x) \left (1+\frac {\log (4)}{2 e^3}\right )\right ) \, dx\\ &=-\frac {e^{4 x}}{4}+\frac {1}{4} e^{4 x} (1+4 x)+25 x^3 \left (1+4 e^3+4 e^6+\log ^2(4)\right )-20 \int e^{2 x} x \, dx-20 \int e^{2 x} x^2 \, dx+\left (150 \left (1+2 e^3\right ) \log (4)\right ) \int x^2 \, dx-\left (20 \left (2+\frac {\log (4)}{e^3}\right )\right ) \int e^{3+2 x} x (1+x) \, dx\\ &=-\frac {e^{4 x}}{4}-10 e^{2 x} x-10 e^{2 x} x^2+\frac {1}{4} e^{4 x} (1+4 x)+50 \left (1+2 e^3\right ) x^3 \log (4)+25 x^3 \left (1+4 e^3+4 e^6+\log ^2(4)\right )+10 \int e^{2 x} \, dx+20 \int e^{2 x} x \, dx-\left (20 \left (2+\frac {\log (4)}{e^3}\right )\right ) \int \left (e^{3+2 x} x+e^{3+2 x} x^2\right ) \, dx\\ &=5 e^{2 x}-\frac {e^{4 x}}{4}-10 e^{2 x} x^2+\frac {1}{4} e^{4 x} (1+4 x)+50 \left (1+2 e^3\right ) x^3 \log (4)+25 x^3 \left (1+4 e^3+4 e^6+\log ^2(4)\right )-10 \int e^{2 x} \, dx-\left (20 \left (2+\frac {\log (4)}{e^3}\right )\right ) \int e^{3+2 x} x \, dx-\left (20 \left (2+\frac {\log (4)}{e^3}\right )\right ) \int e^{3+2 x} x^2 \, dx\\ &=-\frac {e^{4 x}}{4}-10 e^{2 x} x^2+\frac {1}{4} e^{4 x} (1+4 x)+50 \left (1+2 e^3\right ) x^3 \log (4)-10 e^{3+2 x} x \left (2+\frac {\log (4)}{e^3}\right )-10 e^{3+2 x} x^2 \left (2+\frac {\log (4)}{e^3}\right )+25 x^3 \left (1+4 e^3+4 e^6+\log ^2(4)\right )+\left (10 \left (2+\frac {\log (4)}{e^3}\right )\right ) \int e^{3+2 x} \, dx+\left (20 \left (2+\frac {\log (4)}{e^3}\right )\right ) \int e^{3+2 x} x \, dx\\ &=-\frac {e^{4 x}}{4}-10 e^{2 x} x^2+\frac {1}{4} e^{4 x} (1+4 x)+50 \left (1+2 e^3\right ) x^3 \log (4)+5 e^{3+2 x} \left (2+\frac {\log (4)}{e^3}\right )-10 e^{3+2 x} x^2 \left (2+\frac {\log (4)}{e^3}\right )+25 x^3 \left (1+4 e^3+4 e^6+\log ^2(4)\right )-\left (10 \left (2+\frac {\log (4)}{e^3}\right )\right ) \int e^{3+2 x} \, dx\\ &=-\frac {e^{4 x}}{4}-10 e^{2 x} x^2+\frac {1}{4} e^{4 x} (1+4 x)+50 \left (1+2 e^3\right ) x^3 \log (4)-10 e^{3+2 x} x^2 \left (2+\frac {\log (4)}{e^3}\right )+25 x^3 \left (1+4 e^3+4 e^6+\log ^2(4)\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 23, normalized size = 0.88 \begin {gather*} x \left (e^{2 x}-10 e^3 x-5 x (1+\log (4))\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[75*x^2 + 300*E^3*x^2 + 300*E^6*x^2 + E^(4*x)*(1 + 4*x) + (150*x^2 + 300*E^3*x^2)*Log[4] + 75*x^2*Log
[4]^2 + E^(2*x)*(-20*x - 20*x^2 + E^3*(-40*x - 40*x^2) + (-20*x - 20*x^2)*Log[4]),x]

[Out]

x*(E^(2*x) - 10*E^3*x - 5*x*(1 + Log[4]))^2

________________________________________________________________________________________

fricas [B]  time = 0.75, size = 74, normalized size = 2.85 \begin {gather*} 100 \, x^{3} \log \relax (2)^{2} + 100 \, x^{3} e^{6} + 100 \, x^{3} e^{3} + 25 \, x^{3} + x e^{\left (4 \, x\right )} - 10 \, {\left (2 \, x^{2} e^{3} + 2 \, x^{2} \log \relax (2) + x^{2}\right )} e^{\left (2 \, x\right )} + 100 \, {\left (2 \, x^{3} e^{3} + x^{3}\right )} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x+1)*exp(2*x)^2+(2*(-20*x^2-20*x)*log(2)+(-40*x^2-40*x)*exp(3)-20*x^2-20*x)*exp(2*x)+300*x^2*log(
2)^2+2*(300*x^2*exp(3)+150*x^2)*log(2)+300*x^2*exp(3)^2+300*x^2*exp(3)+75*x^2,x, algorithm="fricas")

[Out]

100*x^3*log(2)^2 + 100*x^3*e^6 + 100*x^3*e^3 + 25*x^3 + x*e^(4*x) - 10*(2*x^2*e^3 + 2*x^2*log(2) + x^2)*e^(2*x
) + 100*(2*x^3*e^3 + x^3)*log(2)

________________________________________________________________________________________

giac [B]  time = 0.16, size = 78, normalized size = 3.00 \begin {gather*} 100 \, x^{3} \log \relax (2)^{2} + 100 \, x^{3} e^{6} + 100 \, x^{3} e^{3} + 25 \, x^{3} - 20 \, x^{2} e^{\left (2 \, x + 3\right )} + x e^{\left (4 \, x\right )} - 10 \, {\left (2 \, x^{2} \log \relax (2) + x^{2}\right )} e^{\left (2 \, x\right )} + 100 \, {\left (2 \, x^{3} e^{3} + x^{3}\right )} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x+1)*exp(2*x)^2+(2*(-20*x^2-20*x)*log(2)+(-40*x^2-40*x)*exp(3)-20*x^2-20*x)*exp(2*x)+300*x^2*log(
2)^2+2*(300*x^2*exp(3)+150*x^2)*log(2)+300*x^2*exp(3)^2+300*x^2*exp(3)+75*x^2,x, algorithm="giac")

[Out]

100*x^3*log(2)^2 + 100*x^3*e^6 + 100*x^3*e^3 + 25*x^3 - 20*x^2*e^(2*x + 3) + x*e^(4*x) - 10*(2*x^2*log(2) + x^
2)*e^(2*x) + 100*(2*x^3*e^3 + x^3)*log(2)

________________________________________________________________________________________

maple [B]  time = 0.07, size = 60, normalized size = 2.31

method	result	size
norman	\(x \,{\mathrm e}^{4 x}+\left (100 \,{\mathrm e}^{6}+200 \,{\mathrm e}^{3} \ln \relax (2)+100 \ln \relax (2)^{2}+100 \,{\mathrm e}^{3}+100 \ln \relax (2)+25\right ) x^{3}+\left (-20 \,{\mathrm e}^{3}-20 \ln \relax (2)-10\right ) x^{2} {\mathrm e}^{2 x}\)	\(60\)
risch	\(x \,{\mathrm e}^{4 x}-10 \left (2 \,{\mathrm e}^{3}+2 \ln \relax (2)+1\right ) x^{2} {\mathrm e}^{2 x}+100 x^{3} \ln \relax (2)^{2}+200 \ln \relax (2) x^{3} {\mathrm e}^{3}+100 x^{3} \ln \relax (2)+100 x^{3} {\mathrm e}^{6}+100 x^{3} {\mathrm e}^{3}+25 x^{3}\)	\(71\)
derivativedivides	\(200 \ln \relax (2) x^{3} {\mathrm e}^{3}+100 x^{3} \ln \relax (2)+25 x^{3}+100 x^{3} {\mathrm e}^{3}+100 x^{3} {\mathrm e}^{6}+100 x^{3} \ln \relax (2)^{2}+x \,{\mathrm e}^{4 x}-10 \,{\mathrm e}^{2 x} x^{2}-20 x^{2} {\mathrm e}^{3} {\mathrm e}^{2 x}-20 \,{\mathrm e}^{2 x} \ln \relax (2) x^{2}\)	\(87\)
default	\(200 \ln \relax (2) x^{3} {\mathrm e}^{3}+100 x^{3} \ln \relax (2)+25 x^{3}+100 x^{3} {\mathrm e}^{3}+100 x^{3} {\mathrm e}^{6}+100 x^{3} \ln \relax (2)^{2}+x \,{\mathrm e}^{4 x}-10 \,{\mathrm e}^{2 x} x^{2}-20 x^{2} {\mathrm e}^{3} {\mathrm e}^{2 x}-20 \,{\mathrm e}^{2 x} \ln \relax (2) x^{2}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x+1)*exp(2*x)^2+(2*(-20*x^2-20*x)*ln(2)+(-40*x^2-40*x)*exp(3)-20*x^2-20*x)*exp(2*x)+300*x^2*ln(2)^2+2*(
300*x^2*exp(3)+150*x^2)*ln(2)+300*x^2*exp(3)^2+300*x^2*exp(3)+75*x^2,x,method=_RETURNVERBOSE)

[Out]

x*exp(2*x)^2+(100*exp(3)^2+200*exp(3)*ln(2)+100*ln(2)^2+100*exp(3)+100*ln(2)+25)*x^3+(-20*exp(3)-20*ln(2)-10)*
x^2*exp(2*x)

________________________________________________________________________________________

maxima [B]  time = 0.44, size = 69, normalized size = 2.65 \begin {gather*} 100 \, x^{3} \log \relax (2)^{2} + 100 \, x^{3} e^{6} + 100 \, x^{3} e^{3} - 10 \, x^{2} {\left (2 \, e^{3} + 2 \, \log \relax (2) + 1\right )} e^{\left (2 \, x\right )} + 25 \, x^{3} + x e^{\left (4 \, x\right )} + 100 \, {\left (2 \, x^{3} e^{3} + x^{3}\right )} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x+1)*exp(2*x)^2+(2*(-20*x^2-20*x)*log(2)+(-40*x^2-40*x)*exp(3)-20*x^2-20*x)*exp(2*x)+300*x^2*log(
2)^2+2*(300*x^2*exp(3)+150*x^2)*log(2)+300*x^2*exp(3)^2+300*x^2*exp(3)+75*x^2,x, algorithm="maxima")

[Out]

100*x^3*log(2)^2 + 100*x^3*e^6 + 100*x^3*e^3 - 10*x^2*(2*e^3 + 2*log(2) + 1)*e^(2*x) + 25*x^3 + x*e^(4*x) + 10
0*(2*x^3*e^3 + x^3)*log(2)

________________________________________________________________________________________

mupad [B]  time = 0.19, size = 24, normalized size = 0.92 \begin {gather*} x\,{\left (5\,x-{\mathrm {e}}^{2\,x}+10\,x\,{\mathrm {e}}^3+10\,x\,\ln \relax (2)\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(300*x^2*log(2)^2 - exp(2*x)*(20*x + exp(3)*(40*x + 40*x^2) + 2*log(2)*(20*x + 20*x^2) + 20*x^2) + 300*x^2*
exp(3) + 300*x^2*exp(6) + exp(4*x)*(4*x + 1) + 75*x^2 + 2*log(2)*(300*x^2*exp(3) + 150*x^2),x)

[Out]

x*(5*x - exp(2*x) + 10*x*exp(3) + 10*x*log(2))^2

________________________________________________________________________________________

sympy [B]  time = 0.18, size = 70, normalized size = 2.69 \begin {gather*} x^{3} \left (25 + 100 \log {\relax (2 )}^{2} + 100 \log {\relax (2 )} + 100 e^{3} + 200 e^{3} \log {\relax (2 )} + 100 e^{6}\right ) + x e^{4 x} + \left (- 20 x^{2} e^{3} - 20 x^{2} \log {\relax (2 )} - 10 x^{2}\right ) e^{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x+1)*exp(2*x)**2+(2*(-20*x**2-20*x)*ln(2)+(-40*x**2-40*x)*exp(3)-20*x**2-20*x)*exp(2*x)+300*x**2*
ln(2)**2+2*(300*x**2*exp(3)+150*x**2)*ln(2)+300*x**2*exp(3)**2+300*x**2*exp(3)+75*x**2,x)

[Out]

x**3*(25 + 100*log(2)**2 + 100*log(2) + 100*exp(3) + 200*exp(3)*log(2) + 100*exp(6)) + x*exp(4*x) + (-20*x**2*
exp(3) - 20*x**2*log(2) - 10*x**2)*exp(2*x)

________________________________________________________________________________________