3.72.98 \(\int \frac {-64+16 x+8 x^2+e^{2 x^2} (2 x^2+4 x^4)+(32-12 x-8 x^2) \log (4)+(-4+2 x+2 x^2) \log ^2(4)+e^{x^2} (-8 x-8 x^2-32 x^3-8 x^4+(2 x+4 x^2+8 x^3+4 x^4) \log (4))+(32-16 x+(-16+12 x) \log (4)+(2-2 x) \log ^2(4)+e^{x^2} (8 x+16 x^3+(-2 x-4 x^3) \log (4))) \log (x)}{16 x-8 x \log (4)+x \log ^2(4)} \, dx\)

Optimal. Leaf size=30 \[ \left (2+x-\frac {2 x+e^{x^2} x}{4-\log (4)}-\log (x)\right )^2 \]

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Rubi [B]  time = 0.43, antiderivative size = 109, normalized size of antiderivative = 3.63, number of steps used = 12, number of rules used = 8, integrand size = 175, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.046, Rules used = {6, 12, 14, 2226, 2209, 2212, 6686, 2288} \begin {gather*} \frac {e^{2 x^2} x^2}{(4-\log (4))^2}-\frac {4 e^{x^2} \left (x^3 (1-\log (2))-x^2 (2-\log (2)) \log (x)+2 x^2 (2-\log (2))\right )}{x (4-\log (4))^2}+\frac {(2 x (1-\log (2))-2 (2-\log (2)) \log (x)+8-\log (16))^2}{(4-\log (4))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-64 + 16*x + 8*x^2 + E^(2*x^2)*(2*x^2 + 4*x^4) + (32 - 12*x - 8*x^2)*Log[4] + (-4 + 2*x + 2*x^2)*Log[4]^2
+ E^x^2*(-8*x - 8*x^2 - 32*x^3 - 8*x^4 + (2*x + 4*x^2 + 8*x^3 + 4*x^4)*Log[4]) + (32 - 16*x + (-16 + 12*x)*Lo
g[4] + (2 - 2*x)*Log[4]^2 + E^x^2*(8*x + 16*x^3 + (-2*x - 4*x^3)*Log[4]))*Log[x])/(16*x - 8*x*Log[4] + x*Log[4
]^2),x]

[Out]

(E^(2*x^2)*x^2)/(4 - Log[4])^2 + (8 + 2*x*(1 - Log[2]) - Log[16] - 2*(2 - Log[2])*Log[x])^2/(4 - Log[4])^2 - (
4*E^x^2*(x^3*(1 - Log[2]) + 2*x^2*(2 - Log[2]) - x^2*(2 - Log[2])*Log[x]))/(x*(4 - 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 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 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2212

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
- n + 1)*F^(a + b*(c + d*x)^n))/(b*d*n*Log[F]), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 2226

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, 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]

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} &=\int \frac {-64+16 x+8 x^2+e^{2 x^2} \left (2 x^2+4 x^4\right )+\left (32-12 x-8 x^2\right ) \log (4)+\left (-4+2 x+2 x^2\right ) \log ^2(4)+e^{x^2} \left (-8 x-8 x^2-32 x^3-8 x^4+\left (2 x+4 x^2+8 x^3+4 x^4\right ) \log (4)\right )+\left (32-16 x+(-16+12 x) \log (4)+(2-2 x) \log ^2(4)+e^{x^2} \left (8 x+16 x^3+\left (-2 x-4 x^3\right ) \log (4)\right )\right ) \log (x)}{x (16-8 \log (4))+x \log ^2(4)} \, dx\\ &=\int \frac {-64+16 x+8 x^2+e^{2 x^2} \left (2 x^2+4 x^4\right )+\left (32-12 x-8 x^2\right ) \log (4)+\left (-4+2 x+2 x^2\right ) \log ^2(4)+e^{x^2} \left (-8 x-8 x^2-32 x^3-8 x^4+\left (2 x+4 x^2+8 x^3+4 x^4\right ) \log (4)\right )+\left (32-16 x+(-16+12 x) \log (4)+(2-2 x) \log ^2(4)+e^{x^2} \left (8 x+16 x^3+\left (-2 x-4 x^3\right ) \log (4)\right )\right ) \log (x)}{x \left (16-8 \log (4)+\log ^2(4)\right )} \, dx\\ &=\frac {\int \frac {-64+16 x+8 x^2+e^{2 x^2} \left (2 x^2+4 x^4\right )+\left (32-12 x-8 x^2\right ) \log (4)+\left (-4+2 x+2 x^2\right ) \log ^2(4)+e^{x^2} \left (-8 x-8 x^2-32 x^3-8 x^4+\left (2 x+4 x^2+8 x^3+4 x^4\right ) \log (4)\right )+\left (32-16 x+(-16+12 x) \log (4)+(2-2 x) \log ^2(4)+e^{x^2} \left (8 x+16 x^3+\left (-2 x-4 x^3\right ) \log (4)\right )\right ) \log (x)}{x} \, dx}{16-8 \log (4)+\log ^2(4)}\\ &=\frac {\int \left (2 e^{2 x^2} x \left (1+2 x^2\right )+\frac {2 (4-x (2-\log (4))-\log (4)) \left (-2 x (1-\log (2))-8 \left (1-\frac {\log (2)}{2}\right )+4 \left (1-\frac {\log (2)}{2}\right ) \log (x)\right )}{x}+2 e^{x^2} \left (-4 x (1-\log (2))-4 x^3 (1-\log (2))-4 \left (1-\frac {\log (2)}{2}\right )-16 x^2 \left (1-\frac {\log (2)}{2}\right )+4 \left (1-\frac {\log (2)}{2}\right ) \log (x)+8 x^2 \left (1-\frac {\log (2)}{2}\right ) \log (x)\right )\right ) \, dx}{16-8 \log (4)+\log ^2(4)}\\ &=\frac {2 \int e^{2 x^2} x \left (1+2 x^2\right ) \, dx}{(4-\log (4))^2}+\frac {2 \int \frac {(4-x (2-\log (4))-\log (4)) \left (-2 x (1-\log (2))-8 \left (1-\frac {\log (2)}{2}\right )+4 \left (1-\frac {\log (2)}{2}\right ) \log (x)\right )}{x} \, dx}{(4-\log (4))^2}+\frac {2 \int e^{x^2} \left (-4 x (1-\log (2))-4 x^3 (1-\log (2))-4 \left (1-\frac {\log (2)}{2}\right )-16 x^2 \left (1-\frac {\log (2)}{2}\right )+4 \left (1-\frac {\log (2)}{2}\right ) \log (x)+8 x^2 \left (1-\frac {\log (2)}{2}\right ) \log (x)\right ) \, dx}{(4-\log (4))^2}\\ &=\frac {(8+2 x (1-\log (2))-\log (16)-2 (2-\log (2)) \log (x))^2}{(4-\log (4))^2}-\frac {4 e^{x^2} \left (x^3 (1-\log (2))+2 x^2 (2-\log (2))-x^2 (2-\log (2)) \log (x)\right )}{x (4-\log (4))^2}+\frac {2 \int \left (e^{2 x^2} x+2 e^{2 x^2} x^3\right ) \, dx}{(4-\log (4))^2}\\ &=\frac {(8+2 x (1-\log (2))-\log (16)-2 (2-\log (2)) \log (x))^2}{(4-\log (4))^2}-\frac {4 e^{x^2} \left (x^3 (1-\log (2))+2 x^2 (2-\log (2))-x^2 (2-\log (2)) \log (x)\right )}{x (4-\log (4))^2}+\frac {2 \int e^{2 x^2} x \, dx}{(4-\log (4))^2}+\frac {4 \int e^{2 x^2} x^3 \, dx}{(4-\log (4))^2}\\ &=\frac {e^{2 x^2}}{2 (4-\log (4))^2}+\frac {e^{2 x^2} x^2}{(4-\log (4))^2}+\frac {(8+2 x (1-\log (2))-\log (16)-2 (2-\log (2)) \log (x))^2}{(4-\log (4))^2}-\frac {4 e^{x^2} \left (x^3 (1-\log (2))+2 x^2 (2-\log (2))-x^2 (2-\log (2)) \log (x)\right )}{x (4-\log (4))^2}-\frac {2 \int e^{2 x^2} x \, dx}{(4-\log (4))^2}\\ &=\frac {e^{2 x^2} x^2}{(4-\log (4))^2}+\frac {(8+2 x (1-\log (2))-\log (16)-2 (2-\log (2)) \log (x))^2}{(4-\log (4))^2}-\frac {4 e^{x^2} \left (x^3 (1-\log (2))+2 x^2 (2-\log (2))-x^2 (2-\log (2)) \log (x)\right )}{x (4-\log (4))^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 32, normalized size = 1.07 \begin {gather*} \frac {\left (-8+x \left (-2+e^{x^2}+\log (4)\right )+\log (16)-(-4+\log (4)) \log (x)\right )^2}{(-4+\log (4))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-64 + 16*x + 8*x^2 + E^(2*x^2)*(2*x^2 + 4*x^4) + (32 - 12*x - 8*x^2)*Log[4] + (-4 + 2*x + 2*x^2)*Lo
g[4]^2 + E^x^2*(-8*x - 8*x^2 - 32*x^3 - 8*x^4 + (2*x + 4*x^2 + 8*x^3 + 4*x^4)*Log[4]) + (32 - 16*x + (-16 + 12
*x)*Log[4] + (2 - 2*x)*Log[4]^2 + E^x^2*(8*x + 16*x^3 + (-2*x - 4*x^3)*Log[4]))*Log[x])/(16*x - 8*x*Log[4] + x
*Log[4]^2),x]

[Out]

(-8 + x*(-2 + E^x^2 + Log[4]) + Log[16] - (-4 + Log[4])*Log[x])^2/(-4 + Log[4])^2

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fricas [B]  time = 0.55, size = 137, normalized size = 4.57 \begin {gather*} \frac {x^{2} e^{\left (2 \, x^{2}\right )} + 4 \, {\left (x^{2} + 4 \, x\right )} \log \relax (2)^{2} + 4 \, {\left (\log \relax (2)^{2} - 4 \, \log \relax (2) + 4\right )} \log \relax (x)^{2} + 4 \, x^{2} - 4 \, {\left (x^{2} - {\left (x^{2} + 2 \, x\right )} \log \relax (2) + 4 \, x\right )} e^{\left (x^{2}\right )} - 8 \, {\left (x^{2} + 6 \, x\right )} \log \relax (2) - 4 \, {\left (2 \, {\left (x + 2\right )} \log \relax (2)^{2} + {\left (x \log \relax (2) - 2 \, x\right )} e^{\left (x^{2}\right )} - 2 \, {\left (3 \, x + 8\right )} \log \relax (2) + 4 \, x + 16\right )} \log \relax (x) + 32 \, x}{4 \, {\left (\log \relax (2)^{2} - 4 \, \log \relax (2) + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*(-4*x^3-2*x)*log(2)+16*x^3+8*x)*exp(x^2)+4*(-2*x+2)*log(2)^2+2*(12*x-16)*log(2)-16*x+32)*log(x)
+(4*x^4+2*x^2)*exp(x^2)^2+(2*(4*x^4+8*x^3+4*x^2+2*x)*log(2)-8*x^4-32*x^3-8*x^2-8*x)*exp(x^2)+4*(2*x^2+2*x-4)*l
og(2)^2+2*(-8*x^2-12*x+32)*log(2)+8*x^2+16*x-64)/(4*x*log(2)^2-16*x*log(2)+16*x),x, algorithm="fricas")

[Out]

1/4*(x^2*e^(2*x^2) + 4*(x^2 + 4*x)*log(2)^2 + 4*(log(2)^2 - 4*log(2) + 4)*log(x)^2 + 4*x^2 - 4*(x^2 - (x^2 + 2
*x)*log(2) + 4*x)*e^(x^2) - 8*(x^2 + 6*x)*log(2) - 4*(2*(x + 2)*log(2)^2 + (x*log(2) - 2*x)*e^(x^2) - 2*(3*x +
8)*log(2) + 4*x + 16)*log(x) + 32*x)/(log(2)^2 - 4*log(2) + 4)

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giac [B]  time = 0.27, size = 180, normalized size = 6.00 \begin {gather*} \frac {4 \, x^{2} e^{\left (x^{2}\right )} \log \relax (2) + 4 \, x^{2} \log \relax (2)^{2} - 4 \, x e^{\left (x^{2}\right )} \log \relax (2) \log \relax (x) - 8 \, x \log \relax (2)^{2} \log \relax (x) + 4 \, \log \relax (2)^{2} \log \relax (x)^{2} + x^{2} e^{\left (2 \, x^{2}\right )} - 4 \, x^{2} e^{\left (x^{2}\right )} - 8 \, x^{2} \log \relax (2) + 8 \, x e^{\left (x^{2}\right )} \log \relax (2) + 16 \, x \log \relax (2)^{2} + 8 \, x e^{\left (x^{2}\right )} \log \relax (x) + 24 \, x \log \relax (2) \log \relax (x) - 16 \, \log \relax (2)^{2} \log \relax (x) - 16 \, \log \relax (2) \log \relax (x)^{2} + 4 \, x^{2} - 16 \, x e^{\left (x^{2}\right )} - 48 \, x \log \relax (2) - 16 \, x \log \relax (x) + 64 \, \log \relax (2) \log \relax (x) + 16 \, \log \relax (x)^{2} + 32 \, x - 64 \, \log \relax (x)}{4 \, {\left (\log \relax (2)^{2} - 4 \, \log \relax (2) + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*(-4*x^3-2*x)*log(2)+16*x^3+8*x)*exp(x^2)+4*(-2*x+2)*log(2)^2+2*(12*x-16)*log(2)-16*x+32)*log(x)
+(4*x^4+2*x^2)*exp(x^2)^2+(2*(4*x^4+8*x^3+4*x^2+2*x)*log(2)-8*x^4-32*x^3-8*x^2-8*x)*exp(x^2)+4*(2*x^2+2*x-4)*l
og(2)^2+2*(-8*x^2-12*x+32)*log(2)+8*x^2+16*x-64)/(4*x*log(2)^2-16*x*log(2)+16*x),x, algorithm="giac")

[Out]

1/4*(4*x^2*e^(x^2)*log(2) + 4*x^2*log(2)^2 - 4*x*e^(x^2)*log(2)*log(x) - 8*x*log(2)^2*log(x) + 4*log(2)^2*log(
x)^2 + x^2*e^(2*x^2) - 4*x^2*e^(x^2) - 8*x^2*log(2) + 8*x*e^(x^2)*log(2) + 16*x*log(2)^2 + 8*x*e^(x^2)*log(x)
+ 24*x*log(2)*log(x) - 16*log(2)^2*log(x) - 16*log(2)*log(x)^2 + 4*x^2 - 16*x*e^(x^2) - 48*x*log(2) - 16*x*log
(x) + 64*log(2)*log(x) + 16*log(x)^2 + 32*x - 64*log(x))/(log(2)^2 - 4*log(2) + 4)

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maple [B]  time = 0.12, size = 209, normalized size = 6.97

method	result	size
risch	\(\ln \relax (x )^{2}-\frac {x \left (2 \ln \relax (2)+{\mathrm e}^{x^{2}}-2\right ) \ln \relax (x )}{\ln \relax (2)-2}+\frac {\ln \relax (2)^{2} x^{2}}{\left (\ln \relax (2)-2\right )^{2}}+\frac {x^{2} {\mathrm e}^{x^{2}} \ln \relax (2)}{\left (\ln \relax (2)-2\right )^{2}}-\frac {4 \ln \relax (x ) \ln \relax (2)^{2}}{\left (\ln \relax (2)-2\right )^{2}}+\frac {4 x \ln \relax (2)^{2}}{\left (\ln \relax (2)-2\right )^{2}}+\frac {2 x \,{\mathrm e}^{x^{2}} \ln \relax (2)}{\left (\ln \relax (2)-2\right )^{2}}-\frac {2 x^{2} \ln \relax (2)}{\left (\ln \relax (2)-2\right )^{2}}-\frac {x^{2} {\mathrm e}^{x^{2}}}{\left (\ln \relax (2)-2\right )^{2}}+\frac {x^{2} {\mathrm e}^{2 x^{2}}}{4 \left (\ln \relax (2)-2\right )^{2}}+\frac {16 \ln \relax (x ) \ln \relax (2)}{\left (\ln \relax (2)-2\right )^{2}}-\frac {12 x \ln \relax (2)}{\left (\ln \relax (2)-2\right )^{2}}-\frac {4 x \,{\mathrm e}^{x^{2}}}{\left (\ln \relax (2)-2\right )^{2}}+\frac {x^{2}}{\left (\ln \relax (2)-2\right )^{2}}-\frac {16 \ln \relax (x )}{\left (\ln \relax (2)-2\right )^{2}}+\frac {8 x}{\left (\ln \relax (2)-2\right )^{2}}\)	\(209\)
default	\(-\frac {x \,{\mathrm e}^{x^{2}} \ln \relax (x ) \ln \relax (2)}{\left (\ln \relax (2)-2\right )^{2}}+\frac {x^{2} {\mathrm e}^{x^{2}} \ln \relax (2)}{\left (\ln \relax (2)-2\right )^{2}}+\frac {2 x \,{\mathrm e}^{x^{2}} \ln \relax (2)}{\left (\ln \relax (2)-2\right )^{2}}+\frac {2 x \,{\mathrm e}^{x^{2}} \ln \relax (x )}{\left (\ln \relax (2)-2\right )^{2}}-\frac {x^{2} {\mathrm e}^{x^{2}}}{\left (\ln \relax (2)-2\right )^{2}}-\frac {4 x \,{\mathrm e}^{x^{2}}}{\left (\ln \relax (2)-2\right )^{2}}+\frac {\ln \relax (2)^{2} x^{2}}{\ln \relax (2)^{2}-4 \ln \relax (2)+4}+\frac {2 x \ln \relax (2)^{2}}{\ln \relax (2)^{2}-4 \ln \relax (2)+4}-\frac {2 x^{2} \ln \relax (2)}{\ln \relax (2)^{2}-4 \ln \relax (2)+4}-\frac {6 x \ln \relax (2)}{\ln \relax (2)^{2}-4 \ln \relax (2)+4}+\frac {x^{2}}{\ln \relax (2)^{2}-4 \ln \relax (2)+4}+\frac {4 x}{\ln \relax (2)^{2}-4 \ln \relax (2)+4}-\frac {4 \ln \relax (2)^{2} \ln \relax (x )}{\ln \relax (2)^{2}-4 \ln \relax (2)+4}+\frac {16 \ln \relax (2) \ln \relax (x )}{\ln \relax (2)^{2}-4 \ln \relax (2)+4}-\frac {16 \ln \relax (x )}{\ln \relax (2)^{2}-4 \ln \relax (2)+4}+\frac {x^{2} {\mathrm e}^{2 x^{2}}}{4 \ln \relax (2)^{2}-16 \ln \relax (2)+16}-\frac {2 \ln \relax (2) \ln \relax (x ) x}{\ln \relax (2)-2}+\frac {2 x \ln \relax (2)}{\ln \relax (2)-2}+\frac {\ln \relax (x )^{2} \ln \relax (2)}{\ln \relax (2)-2}+\frac {2 x \ln \relax (x )}{\ln \relax (2)-2}-\frac {2 x}{\ln \relax (2)-2}-\frac {2 \ln \relax (x )^{2}}{\ln \relax (2)-2}\)	\(345\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*(-4*x^3-2*x)*ln(2)+16*x^3+8*x)*exp(x^2)+4*(-2*x+2)*ln(2)^2+2*(12*x-16)*ln(2)-16*x+32)*ln(x)+(4*x^4+2*
x^2)*exp(x^2)^2+(2*(4*x^4+8*x^3+4*x^2+2*x)*ln(2)-8*x^4-32*x^3-8*x^2-8*x)*exp(x^2)+4*(2*x^2+2*x-4)*ln(2)^2+2*(-
8*x^2-12*x+32)*ln(2)+8*x^2+16*x-64)/(4*x*ln(2)^2-16*x*ln(2)+16*x),x,method=_RETURNVERBOSE)

[Out]

ln(x)^2-x*(2*ln(2)+exp(x^2)-2)/(ln(2)-2)*ln(x)+1/(ln(2)-2)^2*ln(2)^2*x^2+1/(ln(2)-2)^2*x^2*exp(x^2)*ln(2)-4/(l
n(2)-2)^2*ln(x)*ln(2)^2+4/(ln(2)-2)^2*x*ln(2)^2+2/(ln(2)-2)^2*x*exp(x^2)*ln(2)-2/(ln(2)-2)^2*x^2*ln(2)-1/(ln(2
)-2)^2*x^2*exp(x^2)+1/4/(ln(2)-2)^2*x^2*exp(2*x^2)+16/(ln(2)-2)^2*ln(x)*ln(2)-12/(ln(2)-2)^2*x*ln(2)-4/(ln(2)-
2)^2*x*exp(x^2)+1/(ln(2)-2)^2*x^2-16/(ln(2)-2)^2*ln(x)+8/(ln(2)-2)^2*x

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maxima [C]  time = 0.50, size = 728, normalized size = 24.27 result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*(-4*x^3-2*x)*log(2)+16*x^3+8*x)*exp(x^2)+4*(-2*x+2)*log(2)^2+2*(12*x-16)*log(2)-16*x+32)*log(x)
+(4*x^4+2*x^2)*exp(x^2)^2+(2*(4*x^4+8*x^3+4*x^2+2*x)*log(2)-8*x^4-32*x^3-8*x^2-8*x)*exp(x^2)+4*(2*x^2+2*x-4)*l
og(2)^2+2*(-8*x^2-12*x+32)*log(2)+8*x^2+16*x-64)/(4*x*log(2)^2-16*x*log(2)+16*x),x, algorithm="maxima")

[Out]

x^2*log(2)^2/(log(2)^2 - 4*log(2) + 4) - (2*log(x*log(2)^2 - 4*x*log(2) + 4*x)*log(x)/(log(2)^2 - 4*log(2) + 4
) - log(x)^2/(log(2)^2 - 4*log(2) + 4))*log(2)^2 - 2*x*log(2)^2*log(x)/(log(2)^2 - 4*log(2) + 4) + 2*log(2)^2*
log(x*log(2)^2 - 4*x*log(2) + 4*x)*log(x)/(log(2)^2 - 4*log(2) + 4) - 2*x^2*log(2)/(log(2)^2 - 4*log(2) + 4) +
(2*x*e^(x^2)/(log(2)^2 - 4*log(2) + 4) + I*sqrt(pi)*erf(I*x)/(log(2)^2 - 4*log(2) + 4))*log(2) + 4*(2*log(x*l
og(2)^2 - 4*x*log(2) + 4*x)*log(x)/(log(2)^2 - 4*log(2) + 4) - log(x)^2/(log(2)^2 - 4*log(2) + 4))*log(2) + (x
^2 - 1)*e^(x^2)*log(2)/(log(2)^2 - 4*log(2) + 4) + 4*x*log(2)^2/(log(2)^2 - 4*log(2) + 4) - 4*log(2)^2*log(x*l
og(2)^2 - 4*x*log(2) + 4*x)/(log(2)^2 - 4*log(2) + 4) - x*e^(x^2)*log(x)/(log(2) - 2) + 6*x*log(2)*log(x)/(log
(2)^2 - 4*log(2) + 4) - 8*log(2)*log(x*log(2)^2 - 4*x*log(2) + 4*x)*log(x)/(log(2)^2 - 4*log(2) + 4) - 1/2*I*s
qrt(pi)*erf(I*x)*log(2)/(log(2)^2 - 4*log(2) + 4) + x^2/(log(2)^2 - 4*log(2) + 4) + 1/8*(2*x^2 - 1)*e^(2*x^2)/
(log(2)^2 - 4*log(2) + 4) - (x^2 - 1)*e^(x^2)/(log(2)^2 - 4*log(2) + 4) - 4*x*e^(x^2)/(log(2)^2 - 4*log(2) + 4
) - 12*x*log(2)/(log(2)^2 - 4*log(2) + 4) + e^(x^2)*log(2)/(log(2)^2 - 4*log(2) + 4) + 16*log(2)*log(x*log(2)^
2 - 4*x*log(2) + 4*x)/(log(2)^2 - 4*log(2) + 4) - 4*x*log(x)/(log(2)^2 - 4*log(2) + 4) + 4*log(x)^2/(log(2)^2
- 4*log(2) + 4) - I*sqrt(pi)*erf(I*x)/(log(2)^2 - 4*log(2) + 4) - 1/2*I*sqrt(pi)*erf(I*x)/(log(2) - 2) + 8*x/(
log(2)^2 - 4*log(2) + 4) + 1/8*e^(2*x^2)/(log(2)^2 - 4*log(2) + 4) - e^(x^2)/(log(2)^2 - 4*log(2) + 4) - 16*lo
g(x*log(2)^2 - 4*x*log(2) + 4*x)/(log(2)^2 - 4*log(2) + 4)

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mupad [B]  time = 5.74, size = 216, normalized size = 7.20 \begin {gather*} \frac {4\,x}{{\ln \relax (2)}^2-\ln \left (16\right )+4}+\frac {x^2\,{\mathrm {e}}^{x^2}\,\left (\ln \relax (2)-1\right )+x\,{\mathrm {e}}^{x^2}\,\left (\ln \relax (4)-4\right )}{{\ln \relax (2)}^2-\ln \left (16\right )+4}+\frac {x^2}{{\ln \relax (2)}^2-\ln \left (16\right )+4}-\frac {16\,\ln \relax (x)}{{\ln \relax (2)}^2-\ln \left (16\right )+4}+\frac {x^3\,\left (\ln \relax (4)-2\right )+x^2\,{\ln \relax (x)}^2\,\left (\ln \relax (2)-2\right )-x^3\,{\mathrm {e}}^{x^2}\,\ln \relax (x)-x^3\,\ln \relax (x)\,\left (\ln \relax (4)-2\right )}{x^2\,\left (\ln \relax (2)-2\right )}-\frac {2\,\ln \relax (2)\,\left (3\,x-8\,\ln \relax (x)+x^2\right )}{{\ln \relax (2)}^2-\ln \left (16\right )+4}+\frac {x^2\,{\mathrm {e}}^{2\,x^2}}{2\,\left (2\,{\ln \relax (2)}^2-\ln \left (256\right )+8\right )}+\frac {{\ln \relax (2)}^2\,\left (2\,x-4\,\ln \relax (x)+x^2\right )}{{\ln \relax (2)}^2-\ln \left (16\right )+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x - exp(x^2)*(8*x - 2*log(2)*(2*x + 4*x^2 + 8*x^3 + 4*x^4) + 8*x^2 + 32*x^3 + 8*x^4) + log(x)*(2*log(2
)*(12*x - 16) - 16*x + exp(x^2)*(8*x - 2*log(2)*(2*x + 4*x^3) + 16*x^3) - 4*log(2)^2*(2*x - 2) + 32) - 2*log(2
)*(12*x + 8*x^2 - 32) + 4*log(2)^2*(2*x + 2*x^2 - 4) + exp(2*x^2)*(2*x^2 + 4*x^4) + 8*x^2 - 64)/(16*x - 16*x*l
og(2) + 4*x*log(2)^2),x)

[Out]

(4*x)/(log(2)^2 - log(16) + 4) + (x^2*exp(x^2)*(log(2) - 1) + x*exp(x^2)*(log(4) - 4))/(log(2)^2 - log(16) + 4
) + x^2/(log(2)^2 - log(16) + 4) - (16*log(x))/(log(2)^2 - log(16) + 4) + (x^3*(log(4) - 2) + x^2*log(x)^2*(lo
g(2) - 2) - x^3*exp(x^2)*log(x) - x^3*log(x)*(log(4) - 2))/(x^2*(log(2) - 2)) - (2*log(2)*(3*x - 8*log(x) + x^
2))/(log(2)^2 - log(16) + 4) + (x^2*exp(2*x^2))/(2*(2*log(2)^2 - log(256) + 8)) + (log(2)^2*(2*x - 4*log(x) +
x^2))/(log(2)^2 - log(16) + 4)

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sympy [B]  time = 1.10, size = 240, normalized size = 8.00 \begin {gather*} \frac {\left (- 2 x \log {\relax (2 )} + 2 x\right ) \log {\relax (x )}}{-2 + \log {\relax (2 )}} + \frac {\left (- 4 x^{2} \log {\relax (2 )} + x^{2} \log {\relax (2 )}^{2} + 4 x^{2}\right ) e^{2 x^{2}} + \left (- 16 x^{2} - 20 x^{2} \log {\relax (2 )}^{2} + 4 x^{2} \log {\relax (2 )}^{3} + 32 x^{2} \log {\relax (2 )} - 48 x \log {\relax (2 )} \log {\relax (x )} - 4 x \log {\relax (2 )}^{3} \log {\relax (x )} + 24 x \log {\relax (2 )}^{2} \log {\relax (x )} + 32 x \log {\relax (x )} - 64 x - 48 x \log {\relax (2 )}^{2} + 8 x \log {\relax (2 )}^{3} + 96 x \log {\relax (2 )}\right ) e^{x^{2}}}{- 128 \log {\relax (2 )} - 32 \log {\relax (2 )}^{3} + 4 \log {\relax (2 )}^{4} + 96 \log {\relax (2 )}^{2} + 64} + \frac {x^{2} \left (- 2 \log {\relax (2 )} + \log {\relax (2 )}^{2} + 1\right ) + x \left (- 12 \log {\relax (2 )} + 4 \log {\relax (2 )}^{2} + 8\right ) - 4 \left (-2 + \log {\relax (2 )}\right )^{2} \log {\relax (x )}}{- 4 \log {\relax (2 )} + \log {\relax (2 )}^{2} + 4} + \log {\relax (x )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*(-4*x**3-2*x)*ln(2)+16*x**3+8*x)*exp(x**2)+4*(-2*x+2)*ln(2)**2+2*(12*x-16)*ln(2)-16*x+32)*ln(x)
+(4*x**4+2*x**2)*exp(x**2)**2+(2*(4*x**4+8*x**3+4*x**2+2*x)*ln(2)-8*x**4-32*x**3-8*x**2-8*x)*exp(x**2)+4*(2*x*
*2+2*x-4)*ln(2)**2+2*(-8*x**2-12*x+32)*ln(2)+8*x**2+16*x-64)/(4*x*ln(2)**2-16*x*ln(2)+16*x),x)

[Out]

(-2*x*log(2) + 2*x)*log(x)/(-2 + log(2)) + ((-4*x**2*log(2) + x**2*log(2)**2 + 4*x**2)*exp(2*x**2) + (-16*x**2
- 20*x**2*log(2)**2 + 4*x**2*log(2)**3 + 32*x**2*log(2) - 48*x*log(2)*log(x) - 4*x*log(2)**3*log(x) + 24*x*lo
g(2)**2*log(x) + 32*x*log(x) - 64*x - 48*x*log(2)**2 + 8*x*log(2)**3 + 96*x*log(2))*exp(x**2))/(-128*log(2) -
32*log(2)**3 + 4*log(2)**4 + 96*log(2)**2 + 64) + (x**2*(-2*log(2) + log(2)**2 + 1) + x*(-12*log(2) + 4*log(2)
**2 + 8) - 4*(-2 + log(2))**2*log(x))/(-4*log(2) + log(2)**2 + 4) + log(x)**2

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