3.36.80 \(\int \frac {220 x^3+(27500 x-33000 e^4 x+9900 e^8 x+1090 x^2) \log (3)+(-1250+1500 e^4-450 e^8-50 x) \log ^2(3)+(2 x^3+(250 x-300 e^4 x+90 e^8 x+10 x^2) \log (3)) \log (25-30 e^4+9 e^8+x)}{625 x^3-750 e^4 x^3+225 e^8 x^3+25 x^4} \, dx\)

Optimal. Leaf size=26 \[ \left (-22+\frac {\log (3)}{x}-\frac {1}{5} \log \left (\left (5-3 e^4\right )^2+x\right )\right )^2 \]

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Rubi [B]  time = 1.21, antiderivative size = 121, normalized size of antiderivative = 4.65, number of steps used = 17, number of rules used = 13, integrand size = 125, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.104, Rules used = {6, 1593, 6688, 12, 6742, 1612, 2418, 2390, 2301, 2395, 36, 29, 31} \begin {gather*} \frac {\log ^2(3)}{x^2}+\frac {1}{25} \log ^2\left (x+\left (5-3 e^4\right )^2\right )+\frac {2 \left (550-660 e^4+198 e^8+\log (3)\right ) \log \left (x+\left (5-3 e^4\right )^2\right )}{5 \left (5-3 e^4\right )^2}-\frac {2 \log (3) \log \left (x+\left (5-3 e^4\right )^2\right )}{5 x}-\frac {2 \log (3) \log \left (x+\left (5-3 e^4\right )^2\right )}{5 \left (5-3 e^4\right )^2}-\frac {44 \log (3)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(220*x^3 + (27500*x - 33000*E^4*x + 9900*E^8*x + 1090*x^2)*Log[3] + (-1250 + 1500*E^4 - 450*E^8 - 50*x)*Lo
g[3]^2 + (2*x^3 + (250*x - 300*E^4*x + 90*E^8*x + 10*x^2)*Log[3])*Log[25 - 30*E^4 + 9*E^8 + x])/(625*x^3 - 750
*E^4*x^3 + 225*E^8*x^3 + 25*x^4),x]

[Out]

(-44*Log[3])/x + Log[3]^2/x^2 - (2*Log[3]*Log[(5 - 3*E^4)^2 + x])/(5*(5 - 3*E^4)^2) - (2*Log[3]*Log[(5 - 3*E^4
)^2 + x])/(5*x) + (2*(550 - 660*E^4 + 198*E^8 + Log[3])*Log[(5 - 3*E^4)^2 + x])/(5*(5 - 3*E^4)^2) + Log[(5 - 3
*E^4)^2 + x]^2/25

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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1612

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[E
xpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && Poly
Q[Px, x] && IntegersQ[m, n]

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]

Rule 2390

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

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 2418

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

Rule 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {220 x^3+\left (27500 x-33000 e^4 x+9900 e^8 x+1090 x^2\right ) \log (3)+\left (-1250+1500 e^4-450 e^8-50 x\right ) \log ^2(3)+\left (2 x^3+\left (250 x-300 e^4 x+90 e^8 x+10 x^2\right ) \log (3)\right ) \log \left (25-30 e^4+9 e^8+x\right )}{225 e^8 x^3+\left (625-750 e^4\right ) x^3+25 x^4} \, dx\\ &=\int \frac {220 x^3+\left (27500 x-33000 e^4 x+9900 e^8 x+1090 x^2\right ) \log (3)+\left (-1250+1500 e^4-450 e^8-50 x\right ) \log ^2(3)+\left (2 x^3+\left (250 x-300 e^4 x+90 e^8 x+10 x^2\right ) \log (3)\right ) \log \left (25-30 e^4+9 e^8+x\right )}{\left (625-750 e^4+225 e^8\right ) x^3+25 x^4} \, dx\\ &=\int \frac {220 x^3+\left (27500 x-33000 e^4 x+9900 e^8 x+1090 x^2\right ) \log (3)+\left (-1250+1500 e^4-450 e^8-50 x\right ) \log ^2(3)+\left (2 x^3+\left (250 x-300 e^4 x+90 e^8 x+10 x^2\right ) \log (3)\right ) \log \left (25-30 e^4+9 e^8+x\right )}{x^3 \left (625-750 e^4+225 e^8+25 x\right )} \, dx\\ &=\int \frac {2 \left (x^2+5 \left (5-3 e^4\right )^2 \log (3)+5 x \log (3)\right ) \left (110 x-5 \log (3)+x \log \left (25-30 e^4+9 e^8+x\right )\right )}{25 x^3 \left (25-30 e^4+9 e^8+x\right )} \, dx\\ &=\frac {2}{25} \int \frac {\left (x^2+5 \left (5-3 e^4\right )^2 \log (3)+5 x \log (3)\right ) \left (110 x-5 \log (3)+x \log \left (25-30 e^4+9 e^8+x\right )\right )}{x^3 \left (25-30 e^4+9 e^8+x\right )} \, dx\\ &=\frac {2}{25} \int \left (\frac {5 (22 x-\log (3)) \left (x^2+5 \left (5-3 e^4\right )^2 \log (3)+5 x \log (3)\right )}{x^3 \left (25-30 e^4+9 e^8+x\right )}+\frac {\left (x^2+5 \left (5-3 e^4\right )^2 \log (3)+5 x \log (3)\right ) \log \left (25-30 e^4+9 e^8+x\right )}{x^2 \left (25-30 e^4+9 e^8+x\right )}\right ) \, dx\\ &=\frac {2}{25} \int \frac {\left (x^2+5 \left (5-3 e^4\right )^2 \log (3)+5 x \log (3)\right ) \log \left (25-30 e^4+9 e^8+x\right )}{x^2 \left (25-30 e^4+9 e^8+x\right )} \, dx+\frac {2}{5} \int \frac {(22 x-\log (3)) \left (x^2+5 \left (5-3 e^4\right )^2 \log (3)+5 x \log (3)\right )}{x^3 \left (25-30 e^4+9 e^8+x\right )} \, dx\\ &=\frac {2}{25} \int \left (\frac {\log \left (25-30 e^4+9 e^8+x\right )}{25-30 e^4+9 e^8+x}+\frac {5 \log (3) \log \left (25-30 e^4+9 e^8+x\right )}{x^2}\right ) \, dx+\frac {2}{5} \int \left (\frac {110 \log (3)}{x^2}-\frac {\log (3)}{\left (-5+3 e^4\right )^2 x}-\frac {5 \log ^2(3)}{x^3}+\frac {550-660 e^4+198 e^8+\log (3)}{\left (-5+3 e^4\right )^2 \left (25-30 e^4+9 e^8+x\right )}\right ) \, dx\\ &=-\frac {44 \log (3)}{x}+\frac {\log ^2(3)}{x^2}-\frac {2 \log (3) \log (x)}{5 \left (5-3 e^4\right )^2}+\frac {2 \left (550-660 e^4+198 e^8+\log (3)\right ) \log \left (\left (5-3 e^4\right )^2+x\right )}{5 \left (5-3 e^4\right )^2}+\frac {2}{25} \int \frac {\log \left (25-30 e^4+9 e^8+x\right )}{25-30 e^4+9 e^8+x} \, dx+\frac {1}{5} (2 \log (3)) \int \frac {\log \left (25-30 e^4+9 e^8+x\right )}{x^2} \, dx\\ &=-\frac {44 \log (3)}{x}+\frac {\log ^2(3)}{x^2}-\frac {2 \log (3) \log (x)}{5 \left (5-3 e^4\right )^2}-\frac {2 \log (3) \log \left (\left (5-3 e^4\right )^2+x\right )}{5 x}+\frac {2 \left (550-660 e^4+198 e^8+\log (3)\right ) \log \left (\left (5-3 e^4\right )^2+x\right )}{5 \left (5-3 e^4\right )^2}+\frac {2}{25} \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,25-30 e^4+9 e^8+x\right )+\frac {1}{5} (2 \log (3)) \int \frac {1}{x \left (25-30 e^4+9 e^8+x\right )} \, dx\\ &=-\frac {44 \log (3)}{x}+\frac {\log ^2(3)}{x^2}-\frac {2 \log (3) \log (x)}{5 \left (5-3 e^4\right )^2}-\frac {2 \log (3) \log \left (\left (5-3 e^4\right )^2+x\right )}{5 x}+\frac {2 \left (550-660 e^4+198 e^8+\log (3)\right ) \log \left (\left (5-3 e^4\right )^2+x\right )}{5 \left (5-3 e^4\right )^2}+\frac {1}{25} \log ^2\left (\left (5-3 e^4\right )^2+x\right )+\frac {(2 \log (3)) \int \frac {1}{x} \, dx}{5 \left (5-3 e^4\right )^2}-\frac {(2 \log (3)) \int \frac {1}{25-30 e^4+9 e^8+x} \, dx}{5 \left (5-3 e^4\right )^2}\\ &=-\frac {44 \log (3)}{x}+\frac {\log ^2(3)}{x^2}-\frac {2 \log (3) \log \left (\left (5-3 e^4\right )^2+x\right )}{5 \left (5-3 e^4\right )^2}-\frac {2 \log (3) \log \left (\left (5-3 e^4\right )^2+x\right )}{5 x}+\frac {2 \left (550-660 e^4+198 e^8+\log (3)\right ) \log \left (\left (5-3 e^4\right )^2+x\right )}{5 \left (5-3 e^4\right )^2}+\frac {1}{25} \log ^2\left (\left (5-3 e^4\right )^2+x\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.20, size = 116, normalized size = 4.46 \begin {gather*} \frac {2}{25} \left (\frac {25 \log ^2(3)}{2 x^2}-\frac {5 \log (3) \log \left (\left (5-3 e^4\right )^2+x\right )}{\left (5-3 e^4\right )^2}-\frac {5 \log (3) \left (110+\log \left (\left (5-3 e^4\right )^2+x\right )\right )}{x}+\frac {\left (5 \left (550-660 e^4+198 e^8+\log (3)\right )+\left (5-3 e^4\right )^2 \log \left (\left (5-3 e^4\right )^2+x\right )\right )^2}{2 \left (5-3 e^4\right )^4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(220*x^3 + (27500*x - 33000*E^4*x + 9900*E^8*x + 1090*x^2)*Log[3] + (-1250 + 1500*E^4 - 450*E^8 - 50
*x)*Log[3]^2 + (2*x^3 + (250*x - 300*E^4*x + 90*E^8*x + 10*x^2)*Log[3])*Log[25 - 30*E^4 + 9*E^8 + x])/(625*x^3
- 750*E^4*x^3 + 225*E^8*x^3 + 25*x^4),x]

[Out]

(2*((25*Log[3]^2)/(2*x^2) - (5*Log[3]*Log[(5 - 3*E^4)^2 + x])/(5 - 3*E^4)^2 - (5*Log[3]*(110 + Log[(5 - 3*E^4)
^2 + x]))/x + (5*(550 - 660*E^4 + 198*E^8 + Log[3]) + (5 - 3*E^4)^2*Log[(5 - 3*E^4)^2 + x])^2/(2*(5 - 3*E^4)^4
)))/25

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fricas [B]  time = 0.58, size = 60, normalized size = 2.31 \begin {gather*} \frac {x^{2} \log \left (x + 9 \, e^{8} - 30 \, e^{4} + 25\right )^{2} - 1100 \, x \log \relax (3) + 25 \, \log \relax (3)^{2} + 10 \, {\left (22 \, x^{2} - x \log \relax (3)\right )} \log \left (x + 9 \, e^{8} - 30 \, e^{4} + 25\right )}{25 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((90*x*exp(4)^2-300*x*exp(4)+10*x^2+250*x)*log(3)+2*x^3)*log(9*exp(4)^2-30*exp(4)+x+25)+(-450*exp(4
)^2+1500*exp(4)-50*x-1250)*log(3)^2+(9900*x*exp(4)^2-33000*x*exp(4)+1090*x^2+27500*x)*log(3)+220*x^3)/(225*x^3
*exp(4)^2-750*x^3*exp(4)+25*x^4+625*x^3),x, algorithm="fricas")

[Out]

1/25*(x^2*log(x + 9*e^8 - 30*e^4 + 25)^2 - 1100*x*log(3) + 25*log(3)^2 + 10*(22*x^2 - x*log(3))*log(x + 9*e^8
- 30*e^4 + 25))/x^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left (110 \, x^{3} - 25 \, {\left (x + 9 \, e^{8} - 30 \, e^{4} + 25\right )} \log \relax (3)^{2} + 5 \, {\left (109 \, x^{2} + 990 \, x e^{8} - 3300 \, x e^{4} + 2750 \, x\right )} \log \relax (3) + {\left (x^{3} + 5 \, {\left (x^{2} + 9 \, x e^{8} - 30 \, x e^{4} + 25 \, x\right )} \log \relax (3)\right )} \log \left (x + 9 \, e^{8} - 30 \, e^{4} + 25\right )\right )}}{25 \, {\left (x^{4} + 9 \, x^{3} e^{8} - 30 \, x^{3} e^{4} + 25 \, x^{3}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((90*x*exp(4)^2-300*x*exp(4)+10*x^2+250*x)*log(3)+2*x^3)*log(9*exp(4)^2-30*exp(4)+x+25)+(-450*exp(4
)^2+1500*exp(4)-50*x-1250)*log(3)^2+(9900*x*exp(4)^2-33000*x*exp(4)+1090*x^2+27500*x)*log(3)+220*x^3)/(225*x^3
*exp(4)^2-750*x^3*exp(4)+25*x^4+625*x^3),x, algorithm="giac")

[Out]

integrate(2/25*(110*x^3 - 25*(x + 9*e^8 - 30*e^4 + 25)*log(3)^2 + 5*(109*x^2 + 990*x*e^8 - 3300*x*e^4 + 2750*x
)*log(3) + (x^3 + 5*(x^2 + 9*x*e^8 - 30*x*e^4 + 25*x)*log(3))*log(x + 9*e^8 - 30*e^4 + 25))/(x^4 + 9*x^3*e^8 -
30*x^3*e^4 + 25*x^3), x)

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maple [B]  time = 0.18, size = 71, normalized size = 2.73

method	result	size
risch	\(\frac {\ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )^{2}}{25}-\frac {2 \ln \relax (3) \ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )}{5 x}+\frac {44 x^{2} \ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )+5 \ln \relax (3)^{2}-220 x \ln \relax (3)}{5 x^{2}}\)	\(71\)
norman	\(\frac {\ln \relax (3)^{2}+\frac {44 x^{2} \ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )}{5}-44 x \ln \relax (3)+\frac {x^{2} \ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )^{2}}{25}-\frac {2 \ln \relax (3) \ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right ) x}{5}}{x^{2}}\)	\(74\)
derivativedivides	\(\frac {2 \ln \relax (3) \ln \left (-x \right )}{5 \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25\right )}-\frac {2 \ln \relax (3) \ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right ) \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )}{5 \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25\right ) x}-\frac {44 \ln \relax (3)}{x}-\frac {2 \ln \relax (3) \ln \relax (x )}{5 \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25\right )}+\frac {2 \ln \relax (3) \ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )}{5 \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25\right )}+\frac {44 \ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )}{5}+\frac {\ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )^{2}}{25}+\frac {\ln \relax (3)^{2}}{x^{2}}\)	\(211\)
default	\(\frac {2 \ln \relax (3) \ln \left (-x \right )}{5 \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25\right )}-\frac {2 \ln \relax (3) \ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right ) \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )}{5 \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25\right ) x}-\frac {44 \ln \relax (3)}{x}-\frac {2 \ln \relax (3) \ln \relax (x )}{5 \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25\right )}+\frac {2 \ln \relax (3) \ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )}{5 \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25\right )}+\frac {44 \ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )}{5}+\frac {\ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )^{2}}{25}+\frac {\ln \relax (3)^{2}}{x^{2}}\)	\(211\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((90*x*exp(4)^2-300*x*exp(4)+10*x^2+250*x)*ln(3)+2*x^3)*ln(9*exp(4)^2-30*exp(4)+x+25)+(-450*exp(4)^2+1500
*exp(4)-50*x-1250)*ln(3)^2+(9900*x*exp(4)^2-33000*x*exp(4)+1090*x^2+27500*x)*ln(3)+220*x^3)/(225*x^3*exp(4)^2-
750*x^3*exp(4)+25*x^4+625*x^3),x,method=_RETURNVERBOSE)

[Out]

1/25*ln(9*exp(8)-30*exp(4)+x+25)^2-2/5*ln(3)/x*ln(9*exp(8)-30*exp(4)+x+25)+1/5*(44*x^2*ln(9*exp(8)-30*exp(4)+x
+25)+5*ln(3)^2-220*x*ln(3))/x^2

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maxima [B]  time = 0.62, size = 841, normalized size = 32.35 result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((90*x*exp(4)^2-300*x*exp(4)+10*x^2+250*x)*log(3)+2*x^3)*log(9*exp(4)^2-30*exp(4)+x+25)+(-450*exp(4
)^2+1500*exp(4)-50*x-1250)*log(3)^2+(9900*x*exp(4)^2-33000*x*exp(4)+1090*x^2+27500*x)*log(3)+220*x^3)/(225*x^3
*exp(4)^2-750*x^3*exp(4)+25*x^4+625*x^3),x, algorithm="maxima")

[Out]

9*(2*log(x + 9*e^8 - 30*e^4 + 25)/(729*e^24 - 7290*e^20 + 30375*e^16 - 67500*e^12 + 84375*e^8 - 56250*e^4 + 15
625) - 2*log(x)/(729*e^24 - 7290*e^20 + 30375*e^16 - 67500*e^12 + 84375*e^8 - 56250*e^4 + 15625) - (2*x - 9*e^
8 + 30*e^4 - 25)/(x^2*(81*e^16 - 540*e^12 + 1350*e^8 - 1500*e^4 + 625)))*e^8*log(3)^2 - 30*(2*log(x + 9*e^8 -
30*e^4 + 25)/(729*e^24 - 7290*e^20 + 30375*e^16 - 67500*e^12 + 84375*e^8 - 56250*e^4 + 15625) - 2*log(x)/(729*
e^24 - 7290*e^20 + 30375*e^16 - 67500*e^12 + 84375*e^8 - 56250*e^4 + 15625) - (2*x - 9*e^8 + 30*e^4 - 25)/(x^2
*(81*e^16 - 540*e^12 + 1350*e^8 - 1500*e^4 + 625)))*e^4*log(3)^2 + 396*(log(x + 9*e^8 - 30*e^4 + 25)/(81*e^16
- 540*e^12 + 1350*e^8 - 1500*e^4 + 625) - log(x)/(81*e^16 - 540*e^12 + 1350*e^8 - 1500*e^4 + 625) - 1/(x*(9*e^
8 - 30*e^4 + 25)))*e^8*log(3) - 1320*(log(x + 9*e^8 - 30*e^4 + 25)/(81*e^16 - 540*e^12 + 1350*e^8 - 1500*e^4 +
625) - log(x)/(81*e^16 - 540*e^12 + 1350*e^8 - 1500*e^4 + 625) - 1/(x*(9*e^8 - 30*e^4 + 25)))*e^4*log(3) + 25
*(2*log(x + 9*e^8 - 30*e^4 + 25)/(729*e^24 - 7290*e^20 + 30375*e^16 - 67500*e^12 + 84375*e^8 - 56250*e^4 + 156
25) - 2*log(x)/(729*e^24 - 7290*e^20 + 30375*e^16 - 67500*e^12 + 84375*e^8 - 56250*e^4 + 15625) - (2*x - 9*e^8
+ 30*e^4 - 25)/(x^2*(81*e^16 - 540*e^12 + 1350*e^8 - 1500*e^4 + 625)))*log(3)^2 - 2*(log(x + 9*e^8 - 30*e^4 +
25)/(81*e^16 - 540*e^12 + 1350*e^8 - 1500*e^4 + 625) - log(x)/(81*e^16 - 540*e^12 + 1350*e^8 - 1500*e^4 + 625
) - 1/(x*(9*e^8 - 30*e^4 + 25)))*log(3)^2 + 1100*(log(x + 9*e^8 - 30*e^4 + 25)/(81*e^16 - 540*e^12 + 1350*e^8
- 1500*e^4 + 625) - log(x)/(81*e^16 - 540*e^12 + 1350*e^8 - 1500*e^4 + 625) - 1/(x*(9*e^8 - 30*e^4 + 25)))*log
(3) - 218/5*(log(x + 9*e^8 - 30*e^4 + 25)/(9*e^8 - 30*e^4 + 25) - log(x)/(9*e^8 - 30*e^4 + 25))*log(3) + 2/5*l
og(3)*log(x)/(9*e^8 - 30*e^4 + 25) + 1/25*(x*(9*e^8 - 30*e^4 + 25)*log(x + 9*e^8 - 30*e^4 + 25)^2 - 10*(x*log(
3) + 9*e^8*log(3) - 30*e^4*log(3) + 25*log(3))*log(x + 9*e^8 - 30*e^4 + 25))/(x*(9*e^8 - 30*e^4 + 25)) + 44/5*
log(x + 9*e^8 - 30*e^4 + 25)

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mupad [B]  time = 3.09, size = 88, normalized size = 3.38 \begin {gather*} \frac {44\,\ln \left (x+{\left (3\,{\mathrm {e}}^4-5\right )}^2\right )}{5}-\frac {220\,x\,\ln \relax (3)-5\,{\ln \relax (3)}^2}{5\,x^2}+\frac {{\ln \left (x-30\,{\mathrm {e}}^4+9\,{\mathrm {e}}^8+25\right )}^2}{25}-\frac {\ln \left (x-30\,{\mathrm {e}}^4+9\,{\mathrm {e}}^8+25\right )\,\left (\frac {12\,{\mathrm {e}}^4}{5}-\frac {18\,{\mathrm {e}}^8}{25}+\frac {2\,\ln \relax (3)}{5}+\frac {2\,{\left (3\,{\mathrm {e}}^4-5\right )}^2}{25}-2\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x - 30*exp(4) + 9*exp(8) + 25)*(log(3)*(250*x - 300*x*exp(4) + 90*x*exp(8) + 10*x^2) + 2*x^3) + log(3
)*(27500*x - 33000*x*exp(4) + 9900*x*exp(8) + 1090*x^2) + 220*x^3 - log(3)^2*(50*x - 1500*exp(4) + 450*exp(8)
+ 1250))/(225*x^3*exp(8) - 750*x^3*exp(4) + 625*x^3 + 25*x^4),x)

[Out]

(44*log(x + (3*exp(4) - 5)^2))/5 - (220*x*log(3) - 5*log(3)^2)/(5*x^2) + log(x - 30*exp(4) + 9*exp(8) + 25)^2/
25 - (log(x - 30*exp(4) + 9*exp(8) + 25)*((12*exp(4))/5 - (18*exp(8))/25 + (2*log(3))/5 + (2*(3*exp(4) - 5)^2)
/25 - 2))/x

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sympy [B]  time = 2.99, size = 75, normalized size = 2.88 \begin {gather*} \frac {\log {\left (x - 30 e^{4} + 25 + 9 e^{8} \right )}^{2}}{25} + \frac {44 \log {\left (x - 30 e^{4} + 25 + 9 e^{8} \right )}}{5} - \frac {2 \log {\relax (3 )} \log {\left (x - 30 e^{4} + 25 + 9 e^{8} \right )}}{5 x} + \frac {- 44 x \log {\relax (3 )} + \log {\relax (3 )}^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((90*x*exp(4)**2-300*x*exp(4)+10*x**2+250*x)*ln(3)+2*x**3)*ln(9*exp(4)**2-30*exp(4)+x+25)+(-450*exp
(4)**2+1500*exp(4)-50*x-1250)*ln(3)**2+(9900*x*exp(4)**2-33000*x*exp(4)+1090*x**2+27500*x)*ln(3)+220*x**3)/(22
5*x**3*exp(4)**2-750*x**3*exp(4)+25*x**4+625*x**3),x)

[Out]

log(x - 30*exp(4) + 25 + 9*exp(8))**2/25 + 44*log(x - 30*exp(4) + 25 + 9*exp(8))/5 - 2*log(3)*log(x - 30*exp(4
) + 25 + 9*exp(8))/(5*x) + (-44*x*log(3) + log(3)**2)/x**2

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