3.68.44 \(\int \frac {48 x^2+e^3 (80 x-24 x^2)+e^6 (25-20 x+3 x^2)+(e^3 (40-16 x)+64 x) \log (4)+16 \log ^2(4)+e^{9+e^{x^2}} (-24 x^2+e^3 (-20 x+6 x^2)+(-32 x+e^3 (-10+4 x)) \log (4)-8 \log ^2(4)+e^{x^2} (-16 x^4+e^3 (-20 x^3+4 x^4)+(-32 x^3+e^3 (-20 x^2+4 x^3)) \log (4)-16 x^2 \log ^2(4)))+e^{18+2 e^{x^2}} (3 x^2+4 x \log (4)+\log ^2(4)+e^{x^2} (4 x^4+8 x^3 \log (4)+4 x^2 \log ^2(4)))}{e^6} \, dx\)

Optimal. Leaf size=26 \[ x \left (-5+x+\frac {\left (-4+e^{9+e^{x^2}}\right ) (x+\log (4))}{e^3}\right )^2 \]

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Rubi [B]  time = 0.26, antiderivative size = 179, normalized size of antiderivative = 6.88, number of steps used = 6, number of rules used = 2, integrand size = 225, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used = {12, 2288} \begin {gather*} -\frac {8 x^3}{e^3}+\frac {16 x^3}{e^6}+x^3+\frac {40 x^2}{e^3}-10 x^2+\frac {e^{2 e^{x^2}+12} \left (x^4+2 x^3 \log (4)+x^2 \log ^2(4)\right )}{x}-\frac {2 e^{e^{x^2}+3} \left (4 x^4+4 x^2 \log ^2(4)+e^3 \left (5 x^3-x^4\right )+\left (8 x^3+e^3 \left (5 x^2-x^3\right )\right ) \log (4)\right )}{x}+25 x+\frac {16 x \log ^2(4)}{e^6}+\frac {2 \left (e^3 (5-2 x)+8 x\right )^2 \log (4)}{e^6 \left (4-e^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(48*x^2 + E^3*(80*x - 24*x^2) + E^6*(25 - 20*x + 3*x^2) + (E^3*(40 - 16*x) + 64*x)*Log[4] + 16*Log[4]^2 +
E^(9 + E^x^2)*(-24*x^2 + E^3*(-20*x + 6*x^2) + (-32*x + E^3*(-10 + 4*x))*Log[4] - 8*Log[4]^2 + E^x^2*(-16*x^4
+ E^3*(-20*x^3 + 4*x^4) + (-32*x^3 + E^3*(-20*x^2 + 4*x^3))*Log[4] - 16*x^2*Log[4]^2)) + E^(18 + 2*E^x^2)*(3*x
^2 + 4*x*Log[4] + Log[4]^2 + E^x^2*(4*x^4 + 8*x^3*Log[4] + 4*x^2*Log[4]^2)))/E^6,x]

[Out]

25*x - 10*x^2 + (40*x^2)/E^3 + x^3 + (16*x^3)/E^6 - (8*x^3)/E^3 + (2*(E^3*(5 - 2*x) + 8*x)^2*Log[4])/(E^6*(4 -
E^3)) + (16*x*Log[4]^2)/E^6 + (E^(12 + 2*E^x^2)*(x^4 + 2*x^3*Log[4] + x^2*Log[4]^2))/x - (2*E^(3 + E^x^2)*(4*
x^4 + E^3*(5*x^3 - x^4) + (8*x^3 + E^3*(5*x^2 - x^3))*Log[4] + 4*x^2*Log[4]^2))/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 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 {\int \left (48 x^2+e^3 \left (80 x-24 x^2\right )+e^6 \left (25-20 x+3 x^2\right )+\left (e^3 (40-16 x)+64 x\right ) \log (4)+16 \log ^2(4)+e^{9+e^{x^2}} \left (-24 x^2+e^3 \left (-20 x+6 x^2\right )+\left (-32 x+e^3 (-10+4 x)\right ) \log (4)-8 \log ^2(4)+e^{x^2} \left (-16 x^4+e^3 \left (-20 x^3+4 x^4\right )+\left (-32 x^3+e^3 \left (-20 x^2+4 x^3\right )\right ) \log (4)-16 x^2 \log ^2(4)\right )\right )+e^{18+2 e^{x^2}} \left (3 x^2+4 x \log (4)+\log ^2(4)+e^{x^2} \left (4 x^4+8 x^3 \log (4)+4 x^2 \log ^2(4)\right )\right )\right ) \, dx}{e^6}\\ &=\frac {16 x^3}{e^6}+\frac {2 \left (e^3 (5-2 x)+8 x\right )^2 \log (4)}{e^6 \left (4-e^3\right )}+\frac {16 x \log ^2(4)}{e^6}+\frac {\int e^{9+e^{x^2}} \left (-24 x^2+e^3 \left (-20 x+6 x^2\right )+\left (-32 x+e^3 (-10+4 x)\right ) \log (4)-8 \log ^2(4)+e^{x^2} \left (-16 x^4+e^3 \left (-20 x^3+4 x^4\right )+\left (-32 x^3+e^3 \left (-20 x^2+4 x^3\right )\right ) \log (4)-16 x^2 \log ^2(4)\right )\right ) \, dx}{e^6}+\frac {\int e^{18+2 e^{x^2}} \left (3 x^2+4 x \log (4)+\log ^2(4)+e^{x^2} \left (4 x^4+8 x^3 \log (4)+4 x^2 \log ^2(4)\right )\right ) \, dx}{e^6}+\frac {\int \left (80 x-24 x^2\right ) \, dx}{e^3}+\int \left (25-20 x+3 x^2\right ) \, dx\\ &=25 x-10 x^2+\frac {40 x^2}{e^3}+x^3+\frac {16 x^3}{e^6}-\frac {8 x^3}{e^3}+\frac {2 \left (e^3 (5-2 x)+8 x\right )^2 \log (4)}{e^6 \left (4-e^3\right )}+\frac {16 x \log ^2(4)}{e^6}+\frac {e^{12+2 e^{x^2}} \left (x^4+2 x^3 \log (4)+x^2 \log ^2(4)\right )}{x}-\frac {2 e^{3+e^{x^2}} \left (4 x^4+e^3 \left (5 x^3-x^4\right )+\left (8 x^3+e^3 \left (5 x^2-x^3\right )\right ) \log (4)+4 x^2 \log ^2(4)\right )}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 35, normalized size = 1.35 \begin {gather*} \frac {x \left (e^3 (-5+x)-4 (x+\log (4))+e^{9+e^{x^2}} (x+\log (4))\right )^2}{e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(48*x^2 + E^3*(80*x - 24*x^2) + E^6*(25 - 20*x + 3*x^2) + (E^3*(40 - 16*x) + 64*x)*Log[4] + 16*Log[4
]^2 + E^(9 + E^x^2)*(-24*x^2 + E^3*(-20*x + 6*x^2) + (-32*x + E^3*(-10 + 4*x))*Log[4] - 8*Log[4]^2 + E^x^2*(-1
6*x^4 + E^3*(-20*x^3 + 4*x^4) + (-32*x^3 + E^3*(-20*x^2 + 4*x^3))*Log[4] - 16*x^2*Log[4]^2)) + E^(18 + 2*E^x^2
)*(3*x^2 + 4*x*Log[4] + Log[4]^2 + E^x^2*(4*x^4 + 8*x^3*Log[4] + 4*x^2*Log[4]^2)))/E^6,x]

[Out]

(x*(E^3*(-5 + x) - 4*(x + Log[4]) + E^(9 + E^x^2)*(x + Log[4]))^2)/E^6

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fricas [B]  time = 0.56, size = 149, normalized size = 5.73 \begin {gather*} {\left (16 \, x^{3} + 64 \, x \log \relax (2)^{2} + {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} e^{6} - 8 \, {\left (x^{3} - 5 \, x^{2}\right )} e^{3} + {\left (x^{3} + 4 \, x^{2} \log \relax (2) + 4 \, x \log \relax (2)^{2}\right )} e^{\left (2 \, e^{\left (x^{2}\right )} + 18\right )} - 2 \, {\left (4 \, x^{3} + 16 \, x \log \relax (2)^{2} - {\left (x^{3} - 5 \, x^{2}\right )} e^{3} + 2 \, {\left (8 \, x^{2} - {\left (x^{2} - 5 \, x\right )} e^{3}\right )} \log \relax (2)\right )} e^{\left (e^{\left (x^{2}\right )} + 9\right )} + 16 \, {\left (4 \, x^{2} - {\left (x^{2} - 5 \, x\right )} e^{3}\right )} \log \relax (2)\right )} e^{\left (-6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^2*log(2)^2+16*x^3*log(2)+4*x^4)*exp(x^2)+4*log(2)^2+8*x*log(2)+3*x^2)*exp(exp(x^2)+9)^2+((-6
4*x^2*log(2)^2+2*((4*x^3-20*x^2)*exp(3)-32*x^3)*log(2)+(4*x^4-20*x^3)*exp(3)-16*x^4)*exp(x^2)-32*log(2)^2+2*((
4*x-10)*exp(3)-32*x)*log(2)+(6*x^2-20*x)*exp(3)-24*x^2)*exp(exp(x^2)+9)+64*log(2)^2+2*((-16*x+40)*exp(3)+64*x)
*log(2)+(3*x^2-20*x+25)*exp(3)^2+(-24*x^2+80*x)*exp(3)+48*x^2)/exp(3)^2,x, algorithm="fricas")

[Out]

(16*x^3 + 64*x*log(2)^2 + (x^3 - 10*x^2 + 25*x)*e^6 - 8*(x^3 - 5*x^2)*e^3 + (x^3 + 4*x^2*log(2) + 4*x*log(2)^2
)*e^(2*e^(x^2) + 18) - 2*(4*x^3 + 16*x*log(2)^2 - (x^3 - 5*x^2)*e^3 + 2*(8*x^2 - (x^2 - 5*x)*e^3)*log(2))*e^(e
^(x^2) + 9) + 16*(4*x^2 - (x^2 - 5*x)*e^3)*log(2))*e^(-6)

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giac [B]  time = 0.24, size = 229, normalized size = 8.81 \begin {gather*} {\left (x^{3} e^{\left (2 \, e^{\left (x^{2}\right )} + 18\right )} + 4 \, x^{2} e^{\left (2 \, e^{\left (x^{2}\right )} + 18\right )} \log \relax (2) + 4 \, x e^{\left (2 \, e^{\left (x^{2}\right )} + 18\right )} \log \relax (2)^{2} + 16 \, x^{3} + 64 \, x \log \relax (2)^{2} + {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} e^{6} - 8 \, {\left (x^{3} - 5 \, x^{2}\right )} e^{3} + 2 \, {\left (x^{3} e^{\left (x^{2} + e^{\left (x^{2}\right )} + 12\right )} - 4 \, x^{3} e^{\left (x^{2} + e^{\left (x^{2}\right )} + 9\right )} + 2 \, x^{2} e^{\left (x^{2} + e^{\left (x^{2}\right )} + 12\right )} \log \relax (2) - 16 \, x^{2} e^{\left (x^{2} + e^{\left (x^{2}\right )} + 9\right )} \log \relax (2) - 16 \, x e^{\left (x^{2} + e^{\left (x^{2}\right )} + 9\right )} \log \relax (2)^{2} - 5 \, x^{2} e^{\left (x^{2} + e^{\left (x^{2}\right )} + 12\right )} - 10 \, x e^{\left (x^{2} + e^{\left (x^{2}\right )} + 12\right )} \log \relax (2)\right )} e^{\left (-x^{2}\right )} + 16 \, {\left (4 \, x^{2} - {\left (x^{2} - 5 \, x\right )} e^{3}\right )} \log \relax (2)\right )} e^{\left (-6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^2*log(2)^2+16*x^3*log(2)+4*x^4)*exp(x^2)+4*log(2)^2+8*x*log(2)+3*x^2)*exp(exp(x^2)+9)^2+((-6
4*x^2*log(2)^2+2*((4*x^3-20*x^2)*exp(3)-32*x^3)*log(2)+(4*x^4-20*x^3)*exp(3)-16*x^4)*exp(x^2)-32*log(2)^2+2*((
4*x-10)*exp(3)-32*x)*log(2)+(6*x^2-20*x)*exp(3)-24*x^2)*exp(exp(x^2)+9)+64*log(2)^2+2*((-16*x+40)*exp(3)+64*x)
*log(2)+(3*x^2-20*x+25)*exp(3)^2+(-24*x^2+80*x)*exp(3)+48*x^2)/exp(3)^2,x, algorithm="giac")

[Out]

(x^3*e^(2*e^(x^2) + 18) + 4*x^2*e^(2*e^(x^2) + 18)*log(2) + 4*x*e^(2*e^(x^2) + 18)*log(2)^2 + 16*x^3 + 64*x*lo
g(2)^2 + (x^3 - 10*x^2 + 25*x)*e^6 - 8*(x^3 - 5*x^2)*e^3 + 2*(x^3*e^(x^2 + e^(x^2) + 12) - 4*x^3*e^(x^2 + e^(x
^2) + 9) + 2*x^2*e^(x^2 + e^(x^2) + 12)*log(2) - 16*x^2*e^(x^2 + e^(x^2) + 9)*log(2) - 16*x*e^(x^2 + e^(x^2) +
9)*log(2)^2 - 5*x^2*e^(x^2 + e^(x^2) + 12) - 10*x*e^(x^2 + e^(x^2) + 12)*log(2))*e^(-x^2) + 16*(4*x^2 - (x^2
- 5*x)*e^3)*log(2))*e^(-6)

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maple [B]  time = 0.20, size = 145, normalized size = 5.58

method	result	size
risch	\(x^{3}-10 x^{2}-16 \ln \relax (2) {\mathrm e}^{-3} x^{2}-8 \,{\mathrm e}^{-3} x^{3}+25 x +80 \,{\mathrm e}^{-3} \ln \relax (2) x +40 x^{2} {\mathrm e}^{-3}+64 \,{\mathrm e}^{-6} x \ln \relax (2)^{2}+64 \,{\mathrm e}^{-6} x^{2} \ln \relax (2)+16 \,{\mathrm e}^{-6} x^{3}+\left (4 \ln \relax (2)^{2}+4 x \ln \relax (2)+x^{2}\right ) x \,{\mathrm e}^{12+2 \,{\mathrm e}^{x^{2}}}+2 \left (2 \,{\mathrm e}^{3} \ln \relax (2) x +x^{2} {\mathrm e}^{3}-10 \,{\mathrm e}^{3} \ln \relax (2)-5 x \,{\mathrm e}^{3}-16 \ln \relax (2)^{2}-16 x \ln \relax (2)-4 x^{2}\right ) x \,{\mathrm e}^{{\mathrm e}^{x^{2}}+3}\)	\(145\)
default	\({\mathrm e}^{-6} \left (x^{3} {\mathrm e}^{2 \,{\mathrm e}^{x^{2}}+18}+4 x \ln \relax (2)^{2} {\mathrm e}^{2 \,{\mathrm e}^{x^{2}}+18}+4 x^{2} \ln \relax (2) {\mathrm e}^{2 \,{\mathrm e}^{x^{2}}+18}+\left (-20 \,{\mathrm e}^{3} \ln \relax (2)-32 \ln \relax (2)^{2}\right ) x \,{\mathrm e}^{{\mathrm e}^{x^{2}}+9}+\left (2 \,{\mathrm e}^{3}-8\right ) x^{3} {\mathrm e}^{{\mathrm e}^{x^{2}}+9}+\left (4 \,{\mathrm e}^{3} \ln \relax (2)-10 \,{\mathrm e}^{3}-32 \ln \relax (2)\right ) x^{2} {\mathrm e}^{{\mathrm e}^{x^{2}}+9}+{\mathrm e}^{3} \left (-8 x^{3}+40 x^{2}\right )+{\mathrm e}^{6} \left (x^{3}-10 x^{2}+25 x \right )+16 x^{3}+64 x \ln \relax (2)^{2}-16 \ln \relax (2) x^{2} {\mathrm e}^{3}+80 \,{\mathrm e}^{3} \ln \relax (2) x +64 x^{2} \ln \relax (2)\right )\)	\(183\)
norman	\(\left (\left ({\mathrm e}^{6}-8 \,{\mathrm e}^{3}+16\right ) {\mathrm e}^{-3} x^{3}+\left (25 \,{\mathrm e}^{6}+64 \ln \relax (2)^{2}+80 \,{\mathrm e}^{3} \ln \relax (2)\right ) {\mathrm e}^{-3} x +{\mathrm e}^{-3} x^{3} {\mathrm e}^{2 \,{\mathrm e}^{x^{2}}+18}-2 \left (5 \,{\mathrm e}^{6}+8 \,{\mathrm e}^{3} \ln \relax (2)-20 \,{\mathrm e}^{3}-32 \ln \relax (2)\right ) {\mathrm e}^{-3} x^{2}+2 \left ({\mathrm e}^{3}-4\right ) {\mathrm e}^{-3} x^{3} {\mathrm e}^{{\mathrm e}^{x^{2}}+9}+2 \left (2 \,{\mathrm e}^{3} \ln \relax (2)-5 \,{\mathrm e}^{3}-16 \ln \relax (2)\right ) {\mathrm e}^{-3} x^{2} {\mathrm e}^{{\mathrm e}^{x^{2}}+9}+4 \,{\mathrm e}^{-3} \ln \relax (2)^{2} x \,{\mathrm e}^{2 \,{\mathrm e}^{x^{2}}+18}+4 \ln \relax (2) {\mathrm e}^{-3} x^{2} {\mathrm e}^{2 \,{\mathrm e}^{x^{2}}+18}-4 \ln \relax (2) \left (5 \,{\mathrm e}^{3}+8 \ln \relax (2)\right ) {\mathrm e}^{-3} x \,{\mathrm e}^{{\mathrm e}^{x^{2}}+9}\right ) {\mathrm e}^{-3}\)	\(213\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((16*x^2*ln(2)^2+16*x^3*ln(2)+4*x^4)*exp(x^2)+4*ln(2)^2+8*x*ln(2)+3*x^2)*exp(exp(x^2)+9)^2+((-64*x^2*ln(2
)^2+2*((4*x^3-20*x^2)*exp(3)-32*x^3)*ln(2)+(4*x^4-20*x^3)*exp(3)-16*x^4)*exp(x^2)-32*ln(2)^2+2*((4*x-10)*exp(3
)-32*x)*ln(2)+(6*x^2-20*x)*exp(3)-24*x^2)*exp(exp(x^2)+9)+64*ln(2)^2+2*((-16*x+40)*exp(3)+64*x)*ln(2)+(3*x^2-2
0*x+25)*exp(3)^2+(-24*x^2+80*x)*exp(3)+48*x^2)/exp(3)^2,x,method=_RETURNVERBOSE)

[Out]

x^3-10*x^2-16*ln(2)*exp(-3)*x^2-8*exp(-3)*x^3+25*x+80*exp(-3)*ln(2)*x+40*x^2*exp(-3)+64*exp(-6)*x*ln(2)^2+64*e
xp(-6)*x^2*ln(2)+16*exp(-6)*x^3+(4*ln(2)^2+4*x*ln(2)+x^2)*x*exp(12+2*exp(x^2))+2*(2*exp(3)*ln(2)*x+x^2*exp(3)-
10*exp(3)*ln(2)-5*x*exp(3)-16*ln(2)^2-16*x*ln(2)-4*x^2)*x*exp(exp(x^2)+3)

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maxima [B]  time = 0.46, size = 155, normalized size = 5.96 \begin {gather*} {\left (16 \, x^{3} + 64 \, x \log \relax (2)^{2} + {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} e^{6} - 8 \, {\left (x^{3} - 5 \, x^{2}\right )} e^{3} + {\left (x^{3} e^{18} + 4 \, x^{2} e^{18} \log \relax (2) + 4 \, x e^{18} \log \relax (2)^{2}\right )} e^{\left (2 \, e^{\left (x^{2}\right )}\right )} + 2 \, {\left (x^{3} {\left (e^{12} - 4 \, e^{9}\right )} + {\left ({\left (2 \, \log \relax (2) - 5\right )} e^{12} - 16 \, e^{9} \log \relax (2)\right )} x^{2} - 2 \, {\left (8 \, e^{9} \log \relax (2)^{2} + 5 \, e^{12} \log \relax (2)\right )} x\right )} e^{\left (e^{\left (x^{2}\right )}\right )} + 16 \, {\left (4 \, x^{2} - {\left (x^{2} - 5 \, x\right )} e^{3}\right )} \log \relax (2)\right )} e^{\left (-6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^2*log(2)^2+16*x^3*log(2)+4*x^4)*exp(x^2)+4*log(2)^2+8*x*log(2)+3*x^2)*exp(exp(x^2)+9)^2+((-6
4*x^2*log(2)^2+2*((4*x^3-20*x^2)*exp(3)-32*x^3)*log(2)+(4*x^4-20*x^3)*exp(3)-16*x^4)*exp(x^2)-32*log(2)^2+2*((
4*x-10)*exp(3)-32*x)*log(2)+(6*x^2-20*x)*exp(3)-24*x^2)*exp(exp(x^2)+9)+64*log(2)^2+2*((-16*x+40)*exp(3)+64*x)
*log(2)+(3*x^2-20*x+25)*exp(3)^2+(-24*x^2+80*x)*exp(3)+48*x^2)/exp(3)^2,x, algorithm="maxima")

[Out]

(16*x^3 + 64*x*log(2)^2 + (x^3 - 10*x^2 + 25*x)*e^6 - 8*(x^3 - 5*x^2)*e^3 + (x^3*e^18 + 4*x^2*e^18*log(2) + 4*
x*e^18*log(2)^2)*e^(2*e^(x^2)) + 2*(x^3*(e^12 - 4*e^9) + ((2*log(2) - 5)*e^12 - 16*e^9*log(2))*x^2 - 2*(8*e^9*
log(2)^2 + 5*e^12*log(2))*x)*e^(e^(x^2)) + 16*(4*x^2 - (x^2 - 5*x)*e^3)*log(2))*e^(-6)

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mupad [B]  time = 0.37, size = 44, normalized size = 1.69 \begin {gather*} x\,{\mathrm {e}}^{-6}\,{\left (4\,x+5\,{\mathrm {e}}^3+8\,\ln \relax (2)-x\,{\mathrm {e}}^3-2\,{\mathrm {e}}^9\,{\mathrm {e}}^{{\mathrm {e}}^{x^2}}\,\ln \relax (2)-x\,{\mathrm {e}}^9\,{\mathrm {e}}^{{\mathrm {e}}^{x^2}}\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-6)*(exp(6)*(3*x^2 - 20*x + 25) - exp(exp(x^2) + 9)*(exp(3)*(20*x - 6*x^2) + exp(x^2)*(64*x^2*log(2)^2
+ exp(3)*(20*x^3 - 4*x^4) + 16*x^4 + 2*log(2)*(exp(3)*(20*x^2 - 4*x^3) + 32*x^3)) + 32*log(2)^2 + 24*x^2 + 2*
log(2)*(32*x - exp(3)*(4*x - 10))) + exp(3)*(80*x - 24*x^2) + exp(2*exp(x^2) + 18)*(8*x*log(2) + 4*log(2)^2 +
3*x^2 + exp(x^2)*(16*x^2*log(2)^2 + 16*x^3*log(2) + 4*x^4)) + 64*log(2)^2 + 48*x^2 + 2*log(2)*(64*x - exp(3)*(
16*x - 40))),x)

[Out]

x*exp(-6)*(4*x + 5*exp(3) + 8*log(2) - x*exp(3) - 2*exp(9)*exp(exp(x^2))*log(2) - x*exp(9)*exp(exp(x^2)))^2

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sympy [B]  time = 5.97, size = 196, normalized size = 7.54 \begin {gather*} \frac {x^{3} \left (- 8 e^{3} + 16 + e^{6}\right )}{e^{6}} + \frac {x^{2} \left (- 10 e^{6} - 16 e^{3} \log {\relax (2 )} + 64 \log {\relax (2 )} + 40 e^{3}\right )}{e^{6}} + \frac {x \left (64 \log {\relax (2 )}^{2} + 80 e^{3} \log {\relax (2 )} + 25 e^{6}\right )}{e^{6}} + \frac {\left (x^{3} e^{6} + 4 x^{2} e^{6} \log {\relax (2 )} + 4 x e^{6} \log {\relax (2 )}^{2}\right ) e^{2 e^{x^{2}} + 18} + \left (- 8 x^{3} e^{6} + 2 x^{3} e^{9} - 10 x^{2} e^{9} - 32 x^{2} e^{6} \log {\relax (2 )} + 4 x^{2} e^{9} \log {\relax (2 )} - 20 x e^{9} \log {\relax (2 )} - 32 x e^{6} \log {\relax (2 )}^{2}\right ) e^{e^{x^{2}} + 9}}{e^{12}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x**2*ln(2)**2+16*x**3*ln(2)+4*x**4)*exp(x**2)+4*ln(2)**2+8*x*ln(2)+3*x**2)*exp(exp(x**2)+9)**2
+((-64*x**2*ln(2)**2+2*((4*x**3-20*x**2)*exp(3)-32*x**3)*ln(2)+(4*x**4-20*x**3)*exp(3)-16*x**4)*exp(x**2)-32*l
n(2)**2+2*((4*x-10)*exp(3)-32*x)*ln(2)+(6*x**2-20*x)*exp(3)-24*x**2)*exp(exp(x**2)+9)+64*ln(2)**2+2*((-16*x+40
)*exp(3)+64*x)*ln(2)+(3*x**2-20*x+25)*exp(3)**2+(-24*x**2+80*x)*exp(3)+48*x**2)/exp(3)**2,x)

[Out]

x**3*(-8*exp(3) + 16 + exp(6))*exp(-6) + x**2*(-10*exp(6) - 16*exp(3)*log(2) + 64*log(2) + 40*exp(3))*exp(-6)
+ x*(64*log(2)**2 + 80*exp(3)*log(2) + 25*exp(6))*exp(-6) + ((x**3*exp(6) + 4*x**2*exp(6)*log(2) + 4*x*exp(6)*
log(2)**2)*exp(2*exp(x**2) + 18) + (-8*x**3*exp(6) + 2*x**3*exp(9) - 10*x**2*exp(9) - 32*x**2*exp(6)*log(2) +
4*x**2*exp(9)*log(2) - 20*x*exp(9)*log(2) - 32*x*exp(6)*log(2)**2)*exp(exp(x**2) + 9))*exp(-12)

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