3.58.48 \(\int \frac {-16 x+(5+8 x) \log (\frac {25+80 x+64 x^2}{64 \log ^2(12)})+e^{4 x^2} (40 x+64 x^2) \log ^2(\frac {25+80 x+64 x^2}{64 \log ^2(12)})}{(5+8 x) \log ^2(\frac {25+80 x+64 x^2}{64 \log ^2(12)})} \, dx\)

Optimal. Leaf size=25 \[ e^{4 x^2}+\frac {x}{\log \left (\frac {\left (\frac {5}{8}+x\right )^2}{\log ^2(12)}\right )} \]

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Rubi [B]  time = 0.44, antiderivative size = 57, normalized size of antiderivative = 2.28, number of steps used = 14, number of rules used = 10, integrand size = 96, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.104, Rules used = {6688, 2209, 2411, 2353, 2297, 2300, 2178, 2302, 30, 2389} \begin {gather*} e^{4 x^2}+\frac {8 x+5}{8 \log \left (\frac {(8 x+5)^2}{64 \log ^2(12)}\right )}-\frac {5}{8 \log \left (\frac {(8 x+5)^2}{64 \log ^2(12)}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16*x + (5 + 8*x)*Log[(25 + 80*x + 64*x^2)/(64*Log[12]^2)] + E^(4*x^2)*(40*x + 64*x^2)*Log[(25 + 80*x + 6
4*x^2)/(64*Log[12]^2)]^2)/((5 + 8*x)*Log[(25 + 80*x + 64*x^2)/(64*Log[12]^2)]^2),x]

[Out]

E^(4*x^2) - 5/(8*Log[(5 + 8*x)^2/(64*Log[12]^2)]) + (5 + 8*x)/(8*Log[(5 + 8*x)^2/(64*Log[12]^2)])

Rule 30

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

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

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 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2300

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[E^(x/n)*(a +
b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (8 e^{4 x^2} x-\frac {16 x}{(5+8 x) \log ^2\left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )}+\frac {1}{\log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )}\right ) \, dx\\ &=8 \int e^{4 x^2} x \, dx-16 \int \frac {x}{(5+8 x) \log ^2\left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )} \, dx+\int \frac {1}{\log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )} \, dx\\ &=e^{4 x^2}+\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{\log \left (\frac {x^2}{64 \log ^2(12)}\right )} \, dx,x,5+8 x\right )-2 \operatorname {Subst}\left (\int \frac {-\frac {5}{8}+\frac {x}{8}}{x \log ^2\left (\frac {x^2}{64 \log ^2(12)}\right )} \, dx,x,5+8 x\right )\\ &=e^{4 x^2}-2 \operatorname {Subst}\left (\int \left (\frac {1}{8 \log ^2\left (\frac {x^2}{64 \log ^2(12)}\right )}-\frac {5}{8 x \log ^2\left (\frac {x^2}{64 \log ^2(12)}\right )}\right ) \, dx,x,5+8 x\right )+\frac {((5+8 x) \log (12)) \operatorname {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )\right )}{2 \sqrt {(5+8 x)^2}}\\ &=e^{4 x^2}+\frac {(5+8 x) \text {Ei}\left (\frac {1}{2} \log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )\right ) \log (12)}{2 \sqrt {(5+8 x)^2}}-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{\log ^2\left (\frac {x^2}{64 \log ^2(12)}\right )} \, dx,x,5+8 x\right )+\frac {5}{4} \operatorname {Subst}\left (\int \frac {1}{x \log ^2\left (\frac {x^2}{64 \log ^2(12)}\right )} \, dx,x,5+8 x\right )\\ &=e^{4 x^2}+\frac {(5+8 x) \text {Ei}\left (\frac {1}{2} \log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )\right ) \log (12)}{2 \sqrt {(5+8 x)^2}}+\frac {5+8 x}{8 \log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )}-\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{\log \left (\frac {x^2}{64 \log ^2(12)}\right )} \, dx,x,5+8 x\right )+\frac {5}{8} \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )\right )\\ &=e^{4 x^2}+\frac {(5+8 x) \text {Ei}\left (\frac {1}{2} \log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )\right ) \log (12)}{2 \sqrt {(5+8 x)^2}}-\frac {5}{8 \log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )}+\frac {5+8 x}{8 \log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )}-\frac {((5+8 x) \log (12)) \operatorname {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )\right )}{2 \sqrt {(5+8 x)^2}}\\ &=e^{4 x^2}-\frac {5}{8 \log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )}+\frac {5+8 x}{8 \log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 28, normalized size = 1.12 \begin {gather*} e^{4 x^2}+\frac {x}{\log \left (\frac {(5+8 x)^2}{64 \log ^2(12)}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16*x + (5 + 8*x)*Log[(25 + 80*x + 64*x^2)/(64*Log[12]^2)] + E^(4*x^2)*(40*x + 64*x^2)*Log[(25 + 80
*x + 64*x^2)/(64*Log[12]^2)]^2)/((5 + 8*x)*Log[(25 + 80*x + 64*x^2)/(64*Log[12]^2)]^2),x]

[Out]

E^(4*x^2) + x/Log[(5 + 8*x)^2/(64*Log[12]^2)]

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fricas [A]  time = 0.60, size = 46, normalized size = 1.84 \begin {gather*} \frac {e^{\left (4 \, x^{2}\right )} \log \left (\frac {64 \, x^{2} + 80 \, x + 25}{64 \, \log \left (12\right )^{2}}\right ) + x}{\log \left (\frac {64 \, x^{2} + 80 \, x + 25}{64 \, \log \left (12\right )^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x^2+40*x)*exp(4*x^2)*log(1/64*(64*x^2+80*x+25)/log(12)^2)^2+(8*x+5)*log(1/64*(64*x^2+80*x+25)/l
og(12)^2)-16*x)/(8*x+5)/log(1/64*(64*x^2+80*x+25)/log(12)^2)^2,x, algorithm="fricas")

[Out]

(e^(4*x^2)*log(1/64*(64*x^2 + 80*x + 25)/log(12)^2) + x)/log(1/64*(64*x^2 + 80*x + 25)/log(12)^2)

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giac [B]  time = 0.38, size = 70, normalized size = 2.80 \begin {gather*} \frac {6 \, e^{\left (4 \, x^{2}\right )} \log \relax (2) - e^{\left (4 \, x^{2}\right )} \log \left (64 \, x^{2} + 80 \, x + 25\right ) + 2 \, e^{\left (4 \, x^{2}\right )} \log \left (\log \left (12\right )\right ) - x}{6 \, \log \relax (2) - \log \left (64 \, x^{2} + 80 \, x + 25\right ) + 2 \, \log \left (\log \left (12\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x^2+40*x)*exp(4*x^2)*log(1/64*(64*x^2+80*x+25)/log(12)^2)^2+(8*x+5)*log(1/64*(64*x^2+80*x+25)/l
og(12)^2)-16*x)/(8*x+5)/log(1/64*(64*x^2+80*x+25)/log(12)^2)^2,x, algorithm="giac")

[Out]

(6*e^(4*x^2)*log(2) - e^(4*x^2)*log(64*x^2 + 80*x + 25) + 2*e^(4*x^2)*log(log(12)) - x)/(6*log(2) - log(64*x^2
+ 80*x + 25) + 2*log(log(12)))

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maple [A]  time = 0.34, size = 38, normalized size = 1.52

method	result	size
default	\(-\frac {x}{6 \ln \relax (2)+2 \ln \left (\ln \relax (3)+2 \ln \relax (2)\right )-\ln \left (\left (8 x +5\right )^{2}\right )}+{\mathrm e}^{4 x^{2}}\)	\(38\)
norman	\(\frac {x +{\mathrm e}^{4 x^{2}} \ln \left (\frac {64 x^{2}+80 x +25}{64 \ln \left (12\right )^{2}}\right )}{\ln \left (\frac {64 x^{2}+80 x +25}{64 \ln \left (12\right )^{2}}\right )}\)	\(47\)
risch	\({\mathrm e}^{4 x^{2}}+\frac {2 i x}{\pi \mathrm {csgn}\left (i \left (\frac {5}{8}+x \right )\right )^{2} \mathrm {csgn}\left (i \left (\frac {5}{8}+x \right )^{2}\right )-2 \pi \,\mathrm {csgn}\left (i \left (\frac {5}{8}+x \right )\right ) \mathrm {csgn}\left (i \left (\frac {5}{8}+x \right )^{2}\right )^{2}+\pi \mathrm {csgn}\left (i \left (\frac {5}{8}+x \right )^{2}\right )^{3}-4 i \ln \left (\ln \relax (3)+2 \ln \relax (2)\right )+4 i \ln \left (\frac {5}{8}+x \right )}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((64*x^2+40*x)*exp(4*x^2)*ln(1/64*(64*x^2+80*x+25)/ln(12)^2)^2+(8*x+5)*ln(1/64*(64*x^2+80*x+25)/ln(12)^2)-
16*x)/(8*x+5)/ln(1/64*(64*x^2+80*x+25)/ln(12)^2)^2,x,method=_RETURNVERBOSE)

[Out]

-x/(6*ln(2)+2*ln(ln(3)+2*ln(2))-ln((8*x+5)^2))+exp(4*x^2)

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maxima [B]  time = 0.51, size = 64, normalized size = 2.56 \begin {gather*} \frac {2 \, {\left (3 \, \log \relax (2) + \log \left (\log \relax (3) + 2 \, \log \relax (2)\right )\right )} e^{\left (4 \, x^{2}\right )} - 2 \, e^{\left (4 \, x^{2}\right )} \log \left (8 \, x + 5\right ) - x}{2 \, {\left (3 \, \log \relax (2) - \log \left (8 \, x + 5\right ) + \log \left (\log \relax (3) + 2 \, \log \relax (2)\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x^2+40*x)*exp(4*x^2)*log(1/64*(64*x^2+80*x+25)/log(12)^2)^2+(8*x+5)*log(1/64*(64*x^2+80*x+25)/l
og(12)^2)-16*x)/(8*x+5)/log(1/64*(64*x^2+80*x+25)/log(12)^2)^2,x, algorithm="maxima")

[Out]

1/2*(2*(3*log(2) + log(log(3) + 2*log(2)))*e^(4*x^2) - 2*e^(4*x^2)*log(8*x + 5) - x)/(3*log(2) - log(8*x + 5)
+ log(log(3) + 2*log(2)))

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mupad [B]  time = 3.85, size = 25, normalized size = 1.00 \begin {gather*} {\mathrm {e}}^{4\,x^2}+\frac {x}{\ln \left (\frac {x^2+\frac {5\,x}{4}+\frac {25}{64}}{{\ln \left (12\right )}^2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(((5*x)/4 + x^2 + 25/64)/log(12)^2)*(8*x + 5) - 16*x + exp(4*x^2)*log(((5*x)/4 + x^2 + 25/64)/log(12)^
2)^2*(40*x + 64*x^2))/(log(((5*x)/4 + x^2 + 25/64)/log(12)^2)^2*(8*x + 5)),x)

[Out]

exp(4*x^2) + x/log(((5*x)/4 + x^2 + 25/64)/log(12)^2)

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sympy [A]  time = 0.38, size = 26, normalized size = 1.04 \begin {gather*} \frac {x}{\log {\left (\frac {x^{2} + \frac {5 x}{4} + \frac {25}{64}}{\log {\left (12 \right )}^{2}} \right )}} + e^{4 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x**2+40*x)*exp(4*x**2)*ln(1/64*(64*x**2+80*x+25)/ln(12)**2)**2+(8*x+5)*ln(1/64*(64*x**2+80*x+25
)/ln(12)**2)-16*x)/(8*x+5)/ln(1/64*(64*x**2+80*x+25)/ln(12)**2)**2,x)

[Out]

x/log((x**2 + 5*x/4 + 25/64)/log(12)**2) + exp(4*x**2)

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