3.70.5 \(\int \frac {96+e^3 (-48-8 x)+16 x+4 e^{4+2 x^2} x^2+e^6 (6+x)+e^{2+x^2} (8+48 x^2+e^3 (-2-12 x^2))+(32-16 e^3+2 e^6+e^{2+x^2} (16 x^2-4 e^3 x^2)) \log (x)}{16 x-8 e^3 x+e^6 x} \, dx\)

Optimal. Leaf size=24 \[ x+\left (3-\frac {e^{2+x^2}}{-4+e^3}+\log (x)\right )^2 \]

________________________________________________________________________________________

Rubi [B]  time = 0.18, antiderivative size = 62, normalized size of antiderivative = 2.58, number of steps used = 12, number of rules used = 7, integrand size = 116, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {6, 12, 14, 2209, 43, 2301, 2288} \begin {gather*} \frac {e^{2 x^2+4}}{\left (4-e^3\right )^2}+\frac {2 e^{x^2+2} \left (3 x^2+x^2 \log (x)\right )}{\left (4-e^3\right ) x^2}+x+\log ^2(x)+6 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(96 + E^3*(-48 - 8*x) + 16*x + 4*E^(4 + 2*x^2)*x^2 + E^6*(6 + x) + E^(2 + x^2)*(8 + 48*x^2 + E^3*(-2 - 12*
x^2)) + (32 - 16*E^3 + 2*E^6 + E^(2 + x^2)*(16*x^2 - 4*E^3*x^2))*Log[x])/(16*x - 8*E^3*x + E^6*x),x]

[Out]

E^(4 + 2*x^2)/(4 - E^3)^2 + x + 6*Log[x] + Log[x]^2 + (2*E^(2 + x^2)*(3*x^2 + x^2*Log[x]))/((4 - E^3)*x^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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 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 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {96+e^3 (-48-8 x)+16 x+4 e^{4+2 x^2} x^2+e^6 (6+x)+e^{2+x^2} \left (8+48 x^2+e^3 \left (-2-12 x^2\right )\right )+\left (32-16 e^3+2 e^6+e^{2+x^2} \left (16 x^2-4 e^3 x^2\right )\right ) \log (x)}{e^6 x+\left (16-8 e^3\right ) x} \, dx\\ &=\int \frac {96+e^3 (-48-8 x)+16 x+4 e^{4+2 x^2} x^2+e^6 (6+x)+e^{2+x^2} \left (8+48 x^2+e^3 \left (-2-12 x^2\right )\right )+\left (32-16 e^3+2 e^6+e^{2+x^2} \left (16 x^2-4 e^3 x^2\right )\right ) \log (x)}{\left (16-8 e^3+e^6\right ) x} \, dx\\ &=\frac {\int \frac {96+e^3 (-48-8 x)+16 x+4 e^{4+2 x^2} x^2+e^6 (6+x)+e^{2+x^2} \left (8+48 x^2+e^3 \left (-2-12 x^2\right )\right )+\left (32-16 e^3+2 e^6+e^{2+x^2} \left (16 x^2-4 e^3 x^2\right )\right ) \log (x)}{x} \, dx}{16-8 e^3+e^6}\\ &=\frac {\int \left (4 e^{4+2 x^2} x+\frac {\left (-4+e^3\right )^2 (6+x+2 \log (x))}{x}-\frac {2 e^{2+x^2} \left (-4+e^3\right ) \left (1+6 x^2+2 x^2 \log (x)\right )}{x}\right ) \, dx}{16-8 e^3+e^6}\\ &=\frac {4 \int e^{4+2 x^2} x \, dx}{\left (4-e^3\right )^2}+\frac {2 \int \frac {e^{2+x^2} \left (1+6 x^2+2 x^2 \log (x)\right )}{x} \, dx}{4-e^3}+\int \frac {6+x+2 \log (x)}{x} \, dx\\ &=\frac {e^{4+2 x^2}}{\left (4-e^3\right )^2}+\frac {2 e^{2+x^2} \left (3 x^2+x^2 \log (x)\right )}{\left (4-e^3\right ) x^2}+\int \left (\frac {6+x}{x}+\frac {2 \log (x)}{x}\right ) \, dx\\ &=\frac {e^{4+2 x^2}}{\left (4-e^3\right )^2}+\frac {2 e^{2+x^2} \left (3 x^2+x^2 \log (x)\right )}{\left (4-e^3\right ) x^2}+2 \int \frac {\log (x)}{x} \, dx+\int \frac {6+x}{x} \, dx\\ &=\frac {e^{4+2 x^2}}{\left (4-e^3\right )^2}+\log ^2(x)+\frac {2 e^{2+x^2} \left (3 x^2+x^2 \log (x)\right )}{\left (4-e^3\right ) x^2}+\int \left (1+\frac {6}{x}\right ) \, dx\\ &=\frac {e^{4+2 x^2}}{\left (4-e^3\right )^2}+x+6 \log (x)+\log ^2(x)+\frac {2 e^{2+x^2} \left (3 x^2+x^2 \log (x)\right )}{\left (4-e^3\right ) x^2}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [B]  time = 0.08, size = 80, normalized size = 3.33 \begin {gather*} \frac {e^{4+2 x^2}-6 e^{2+x^2} \left (-4+e^3\right )+\left (-4+e^3\right )^2 x-2 e^{2+x^2} \left (-4+e^3\right ) \log (x)+6 \left (-4+e^3\right )^2 \log (x)+\left (-4+e^3\right )^2 \log ^2(x)}{\left (-4+e^3\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(96 + E^3*(-48 - 8*x) + 16*x + 4*E^(4 + 2*x^2)*x^2 + E^6*(6 + x) + E^(2 + x^2)*(8 + 48*x^2 + E^3*(-2
- 12*x^2)) + (32 - 16*E^3 + 2*E^6 + E^(2 + x^2)*(16*x^2 - 4*E^3*x^2))*Log[x])/(16*x - 8*E^3*x + E^6*x),x]

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.92, size = 82, normalized size = 3.42 \begin {gather*} \frac {{\left (e^{6} - 8 \, e^{3} + 16\right )} \log \relax (x)^{2} + x e^{6} - 8 \, x e^{3} - 6 \, {\left (e^{3} - 4\right )} e^{\left (x^{2} + 2\right )} - 2 \, {\left ({\left (e^{3} - 4\right )} e^{\left (x^{2} + 2\right )} - 3 \, e^{6} + 24 \, e^{3} - 48\right )} \log \relax (x) + 16 \, x + e^{\left (2 \, x^{2} + 4\right )}}{e^{6} - 8 \, e^{3} + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^2*exp(3)+16*x^2)*exp(x^2+2)+2*exp(3)^2-16*exp(3)+32)*log(x)+4*x^2*exp(x^2+2)^2+((-12*x^2-2)*
exp(3)+48*x^2+8)*exp(x^2+2)+(x+6)*exp(3)^2+(-8*x-48)*exp(3)+16*x+96)/(x*exp(3)^2-8*x*exp(3)+16*x),x, algorithm
="fricas")

[Out]

((e^6 - 8*e^3 + 16)*log(x)^2 + x*e^6 - 8*x*e^3 - 6*(e^3 - 4)*e^(x^2 + 2) - 2*((e^3 - 4)*e^(x^2 + 2) - 3*e^6 +
24*e^3 - 48)*log(x) + 16*x + e^(2*x^2 + 4))/(e^6 - 8*e^3 + 16)

________________________________________________________________________________________

giac [B]  time = 0.23, size = 105, normalized size = 4.38 \begin {gather*} \frac {e^{6} \log \relax (x)^{2} - 8 \, e^{3} \log \relax (x)^{2} + x e^{6} - 8 \, x e^{3} + 6 \, e^{6} \log \relax (x) - 48 \, e^{3} \log \relax (x) - 2 \, e^{\left (x^{2} + 5\right )} \log \relax (x) + 8 \, e^{\left (x^{2} + 2\right )} \log \relax (x) + 16 \, \log \relax (x)^{2} + 16 \, x + e^{\left (2 \, x^{2} + 4\right )} - 6 \, e^{\left (x^{2} + 5\right )} + 24 \, e^{\left (x^{2} + 2\right )} + 96 \, \log \relax (x)}{e^{6} - 8 \, e^{3} + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^2*exp(3)+16*x^2)*exp(x^2+2)+2*exp(3)^2-16*exp(3)+32)*log(x)+4*x^2*exp(x^2+2)^2+((-12*x^2-2)*
exp(3)+48*x^2+8)*exp(x^2+2)+(x+6)*exp(3)^2+(-8*x-48)*exp(3)+16*x+96)/(x*exp(3)^2-8*x*exp(3)+16*x),x, algorithm
="giac")

[Out]

(e^6*log(x)^2 - 8*e^3*log(x)^2 + x*e^6 - 8*x*e^3 + 6*e^6*log(x) - 48*e^3*log(x) - 2*e^(x^2 + 5)*log(x) + 8*e^(
x^2 + 2)*log(x) + 16*log(x)^2 + 16*x + e^(2*x^2 + 4) - 6*e^(x^2 + 5) + 24*e^(x^2 + 2) + 96*log(x))/(e^6 - 8*e^
3 + 16)

________________________________________________________________________________________

maple [B]  time = 0.12, size = 62, normalized size = 2.58

method	result	size
default	\(x +6 \ln \relax (x )-\frac {6 \,{\mathrm e}^{x^{2}+2}}{{\mathrm e}^{3}-4}-\frac {2 \,{\mathrm e}^{x^{2}+2} \ln \relax (x )}{{\mathrm e}^{3}-4}+\ln \relax (x )^{2}+\frac {{\mathrm e}^{2 x^{2}+4}}{{\mathrm e}^{6}-8 \,{\mathrm e}^{3}+16}\)	\(62\)
norman	\(\frac {\frac {{\mathrm e}^{2 x^{2}+4}}{{\mathrm e}^{3}-4}+\left ({\mathrm e}^{3}-4\right ) x +\left ({\mathrm e}^{3}-4\right ) \ln \relax (x )^{2}+\left (6 \,{\mathrm e}^{3}-24\right ) \ln \relax (x )-2 \,{\mathrm e}^{x^{2}+2} \ln \relax (x )-6 \,{\mathrm e}^{x^{2}+2}}{{\mathrm e}^{3}-4}\)	\(66\)
risch	\(\ln \relax (x )^{2}-\frac {2 \,{\mathrm e}^{x^{2}+2} \ln \relax (x )}{{\mathrm e}^{3}-4}-\frac {48 \ln \relax (x ) {\mathrm e}^{3}}{\left ({\mathrm e}^{3}-4\right )^{2}}+\frac {6 \,{\mathrm e}^{6} \ln \relax (x )}{\left ({\mathrm e}^{3}-4\right )^{2}}-\frac {8 x \,{\mathrm e}^{3}}{\left ({\mathrm e}^{3}-4\right )^{2}}+\frac {x \,{\mathrm e}^{6}}{\left ({\mathrm e}^{3}-4\right )^{2}}+\frac {96 \ln \relax (x )}{\left ({\mathrm e}^{3}-4\right )^{2}}+\frac {24 \,{\mathrm e}^{x^{2}+2}}{\left ({\mathrm e}^{3}-4\right )^{2}}-\frac {6 \,{\mathrm e}^{x^{2}+5}}{\left ({\mathrm e}^{3}-4\right )^{2}}+\frac {{\mathrm e}^{2 x^{2}+4}}{\left ({\mathrm e}^{3}-4\right )^{2}}+\frac {16 x}{\left ({\mathrm e}^{3}-4\right )^{2}}\)	\(129\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*x^2*exp(3)+16*x^2)*exp(x^2+2)+2*exp(3)^2-16*exp(3)+32)*ln(x)+4*x^2*exp(x^2+2)^2+((-12*x^2-2)*exp(3)+
48*x^2+8)*exp(x^2+2)+(x+6)*exp(3)^2+(-8*x-48)*exp(3)+16*x+96)/(x*exp(3)^2-8*x*exp(3)+16*x),x,method=_RETURNVER
BOSE)

[Out]

x+6*ln(x)-6*exp(x^2+2)/(exp(3)-4)-2/(exp(3)-4)*exp(x^2+2)*ln(x)+ln(x)^2+1/(exp(3)^2-8*exp(3)+16)*exp(x^2+2)^2

________________________________________________________________________________________

maxima [C]  time = 0.50, size = 457, normalized size = 19.04 \begin {gather*} -{\left (\frac {2 \, \log \left (x e^{6} - 8 \, x e^{3} + 16 \, x\right ) \log \relax (x)}{e^{6} - 8 \, e^{3} + 16} - \frac {\log \relax (x)^{2}}{e^{6} - 8 \, e^{3} + 16}\right )} e^{6} + 8 \, {\left (\frac {2 \, \log \left (x e^{6} - 8 \, x e^{3} + 16 \, x\right ) \log \relax (x)}{e^{6} - 8 \, e^{3} + 16} - \frac {\log \relax (x)^{2}}{e^{6} - 8 \, e^{3} + 16}\right )} e^{3} + \frac {2 \, e^{6} \log \left (x e^{6} - 8 \, x e^{3} + 16 \, x\right ) \log \relax (x)}{e^{6} - 8 \, e^{3} + 16} - \frac {16 \, e^{3} \log \left (x e^{6} - 8 \, x e^{3} + 16 \, x\right ) \log \relax (x)}{e^{6} - 8 \, e^{3} + 16} + \frac {x e^{6}}{e^{6} - 8 \, e^{3} + 16} + \frac {{\rm Ei}\left (x^{2}\right ) e^{5}}{e^{6} - 8 \, e^{3} + 16} - \frac {8 \, x e^{3}}{e^{6} - 8 \, e^{3} + 16} - \frac {4 \, {\rm Ei}\left (x^{2}\right ) e^{2}}{e^{6} - 8 \, e^{3} + 16} + \frac {6 \, e^{6} \log \left (x e^{6} - 8 \, x e^{3} + 16 \, x\right )}{e^{6} - 8 \, e^{3} + 16} - \frac {48 \, e^{3} \log \left (x e^{6} - 8 \, x e^{3} + 16 \, x\right )}{e^{6} - 8 \, e^{3} + 16} - \frac {2 \, e^{\left (x^{2} + 5\right )} \log \relax (x)}{e^{6} - 8 \, e^{3} + 16} + \frac {8 \, e^{\left (x^{2} + 2\right )} \log \relax (x)}{e^{6} - 8 \, e^{3} + 16} + \frac {16 \, \log \relax (x)^{2}}{e^{6} - 8 \, e^{3} + 16} + \frac {16 \, x}{e^{6} - 8 \, e^{3} + 16} - \frac {{\rm Ei}\left (x^{2}\right ) e^{5}}{{\left (e^{3} - 4\right )}^{2}} + \frac {4 \, {\rm Ei}\left (x^{2}\right ) e^{2}}{{\left (e^{3} - 4\right )}^{2}} + \frac {e^{\left (2 \, x^{2} + 4\right )}}{e^{6} - 8 \, e^{3} + 16} - \frac {6 \, e^{\left (x^{2} + 5\right )}}{e^{6} - 8 \, e^{3} + 16} + \frac {24 \, e^{\left (x^{2} + 2\right )}}{e^{6} - 8 \, e^{3} + 16} + \frac {96 \, \log \left (x e^{6} - 8 \, x e^{3} + 16 \, x\right )}{e^{6} - 8 \, e^{3} + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^2*exp(3)+16*x^2)*exp(x^2+2)+2*exp(3)^2-16*exp(3)+32)*log(x)+4*x^2*exp(x^2+2)^2+((-12*x^2-2)*
exp(3)+48*x^2+8)*exp(x^2+2)+(x+6)*exp(3)^2+(-8*x-48)*exp(3)+16*x+96)/(x*exp(3)^2-8*x*exp(3)+16*x),x, algorithm
="maxima")

[Out]

-(2*log(x*e^6 - 8*x*e^3 + 16*x)*log(x)/(e^6 - 8*e^3 + 16) - log(x)^2/(e^6 - 8*e^3 + 16))*e^6 + 8*(2*log(x*e^6
- 8*x*e^3 + 16*x)*log(x)/(e^6 - 8*e^3 + 16) - log(x)^2/(e^6 - 8*e^3 + 16))*e^3 + 2*e^6*log(x*e^6 - 8*x*e^3 + 1
6*x)*log(x)/(e^6 - 8*e^3 + 16) - 16*e^3*log(x*e^6 - 8*x*e^3 + 16*x)*log(x)/(e^6 - 8*e^3 + 16) + x*e^6/(e^6 - 8
*e^3 + 16) + Ei(x^2)*e^5/(e^6 - 8*e^3 + 16) - 8*x*e^3/(e^6 - 8*e^3 + 16) - 4*Ei(x^2)*e^2/(e^6 - 8*e^3 + 16) +
6*e^6*log(x*e^6 - 8*x*e^3 + 16*x)/(e^6 - 8*e^3 + 16) - 48*e^3*log(x*e^6 - 8*x*e^3 + 16*x)/(e^6 - 8*e^3 + 16) -
2*e^(x^2 + 5)*log(x)/(e^6 - 8*e^3 + 16) + 8*e^(x^2 + 2)*log(x)/(e^6 - 8*e^3 + 16) + 16*log(x)^2/(e^6 - 8*e^3
+ 16) + 16*x/(e^6 - 8*e^3 + 16) - Ei(x^2)*e^5/(e^3 - 4)^2 + 4*Ei(x^2)*e^2/(e^3 - 4)^2 + e^(2*x^2 + 4)/(e^6 - 8
*e^3 + 16) - 6*e^(x^2 + 5)/(e^6 - 8*e^3 + 16) + 24*e^(x^2 + 2)/(e^6 - 8*e^3 + 16) + 96*log(x*e^6 - 8*x*e^3 + 1
6*x)/(e^6 - 8*e^3 + 16)

________________________________________________________________________________________

mupad [B]  time = 4.48, size = 52, normalized size = 2.17 \begin {gather*} x+6\,\ln \relax (x)+{\ln \relax (x)}^2-{\mathrm {e}}^{x^2+2}\,\left (\frac {6}{{\mathrm {e}}^3-4}+\frac {2\,\ln \relax (x)}{{\mathrm {e}}^3-4}\right )+\frac {{\mathrm {e}}^{2\,x^2+4}}{{\left ({\mathrm {e}}^3-4\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x + 4*x^2*exp(2*x^2 + 4) - log(x)*(16*exp(3) - 2*exp(6) + exp(x^2 + 2)*(4*x^2*exp(3) - 16*x^2) - 32) +
exp(6)*(x + 6) + exp(x^2 + 2)*(48*x^2 - exp(3)*(12*x^2 + 2) + 8) - exp(3)*(8*x + 48) + 96)/(16*x - 8*x*exp(3)
+ x*exp(6)),x)

[Out]

x + 6*log(x) + log(x)^2 - exp(x^2 + 2)*(6/(exp(3) - 4) + (2*log(x))/(exp(3) - 4)) + exp(2*x^2 + 4)/(exp(3) - 4
)^2

________________________________________________________________________________________

sympy [B]  time = 0.59, size = 80, normalized size = 3.33 \begin {gather*} x + \frac {\left (- 2 e^{6} \log {\relax (x )} - 32 \log {\relax (x )} + 16 e^{3} \log {\relax (x )} - 6 e^{6} - 96 + 48 e^{3}\right ) e^{x^{2} + 2} + \left (-4 + e^{3}\right ) e^{2 x^{2} + 4}}{- 12 e^{6} - 64 + 48 e^{3} + e^{9}} + \log {\relax (x )}^{2} + 6 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x**2*exp(3)+16*x**2)*exp(x**2+2)+2*exp(3)**2-16*exp(3)+32)*ln(x)+4*x**2*exp(x**2+2)**2+((-12*x
**2-2)*exp(3)+48*x**2+8)*exp(x**2+2)+(x+6)*exp(3)**2+(-8*x-48)*exp(3)+16*x+96)/(x*exp(3)**2-8*x*exp(3)+16*x),x
)

[Out]

x + ((-2*exp(6)*log(x) - 32*log(x) + 16*exp(3)*log(x) - 6*exp(6) - 96 + 48*exp(3))*exp(x**2 + 2) + (-4 + exp(3
))*exp(2*x**2 + 4))/(-12*exp(6) - 64 + 48*exp(3) + exp(9)) + log(x)**2 + 6*log(x)

________________________________________________________________________________________