3.9.52 \(\int \frac {e^{5/4} (-360+72 x) \log ^3(x)+e^{5/4} (-90+36 x) \log ^4(x)+(180-72 x) \log ^5(x)}{-e^{25/4}+10 e^5 \log (x)-40 e^{15/4} \log ^2(x)+80 e^{5/2} \log ^3(x)-80 e^{5/4} \log ^4(x)+32 \log ^5(x)} \, dx\)

Optimal. Leaf size=23 \[ \frac {18 (5-x) x}{\left (2-\frac {e^{5/4}}{\log (x)}\right )^4} \]

________________________________________________________________________________________

Rubi [C]  time = 2.10, antiderivative size = 909, normalized size of antiderivative = 39.52, number of steps used = 74, number of rules used = 9, integrand size = 98, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.092, Rules used = {6688, 12, 6742, 2320, 2330, 2299, 2178, 2309, 2297} \begin {gather*} -\frac {9}{32} (5-2 x)^2-\frac {15}{256} e^{\frac {15}{4}+\frac {e^{5/4}}{2}} \left (24-e^{5/4}\right ) \text {Ei}\left (\frac {1}{2} \left (2 \log (x)-e^{5/4}\right )\right )-\frac {45}{32} e^{\frac {1}{2} \left (5+e^{5/4}\right )} \left (6-e^{5/4}\right ) \text {Ei}\left (\frac {1}{2} \left (2 \log (x)-e^{5/4}\right )\right )-\frac {45}{16} e^{\frac {5}{4}+\frac {e^{5/4}}{2}} \left (4-3 e^{5/4}\right ) \text {Ei}\left (\frac {1}{2} \left (2 \log (x)-e^{5/4}\right )\right )-\frac {15}{256} e^{5+\frac {e^{5/4}}{2}} \text {Ei}\left (\frac {1}{2} \left (2 \log (x)-e^{5/4}\right )\right )+\frac {45}{4} e^{\frac {5}{4}+\frac {e^{5/4}}{2}} \text {Ei}\left (\frac {1}{2} \left (2 \log (x)-e^{5/4}\right )\right )+\frac {3}{16} e^{\frac {15}{4}+e^{5/4}} \left (12-e^{5/4}\right ) \text {Ei}\left (2 \log (x)-e^{5/4}\right )+\frac {9}{4} e^{\frac {5}{2}+e^{5/4}} \left (3-e^{5/4}\right ) \text {Ei}\left (2 \log (x)-e^{5/4}\right )+\frac {9}{4} e^{\frac {5}{4}+e^{5/4}} \left (2-3 e^{5/4}\right ) \text {Ei}\left (2 \log (x)-e^{5/4}\right )+\frac {3}{16} e^{5+e^{5/4}} \text {Ei}\left (2 \log (x)-e^{5/4}\right )-\frac {9}{2} e^{\frac {5}{4}+e^{5/4}} \text {Ei}\left (2 \log (x)-e^{5/4}\right )-\frac {3 e^5 (5-x) x}{16 \left (e^{5/4}-2 \log (x)\right )}+\frac {45 e^{15/4} \left (24-e^{5/4}\right ) x}{128 \left (e^{5/4}-2 \log (x)\right )}+\frac {45 e^{5/2} \left (6-e^{5/4}\right ) x}{16 \left (e^{5/4}-2 \log (x)\right )}+\frac {105 e^5 x}{128 \left (e^{5/4}-2 \log (x)\right )}-\frac {9 e^{5/4} x \left (5 \left (4-3 e^{5/4}\right )-2 \left (2-3 e^{5/4}\right ) x\right )}{8 \left (e^{5/4}-2 \log (x)\right )}-\frac {9 e^{5/2} x \left (5 \left (6-e^{5/4}\right )-2 \left (3-e^{5/4}\right ) x\right )}{8 \left (e^{5/4}-2 \log (x)\right )}-\frac {3 e^{15/4} x \left (5 \left (24-e^{5/4}\right )-2 \left (12-e^{5/4}\right ) x\right )}{32 \left (e^{5/4}-2 \log (x)\right )}+\frac {3 e^5 (5-x) x}{16 \left (e^{5/4}-2 \log (x)\right )^2}-\frac {15 e^{15/4} \left (24-e^{5/4}\right ) x}{64 \left (e^{5/4}-2 \log (x)\right )^2}-\frac {45 e^5 x}{64 \left (e^{5/4}-2 \log (x)\right )^2}+\frac {9 e^{5/2} x \left (5 \left (6-e^{5/4}\right )-2 \left (3-e^{5/4}\right ) x\right )}{8 \left (e^{5/4}-2 \log (x)\right )^2}+\frac {3 e^{15/4} x \left (5 \left (24-e^{5/4}\right )-2 \left (12-e^{5/4}\right ) x\right )}{32 \left (e^{5/4}-2 \log (x)\right )^2}-\frac {3 e^5 (5-x) x}{8 \left (e^{5/4}-2 \log (x)\right )^3}+\frac {15 e^5 x}{16 \left (e^{5/4}-2 \log (x)\right )^3}-\frac {3 e^{15/4} x \left (5 \left (24-e^{5/4}\right )-2 \left (12-e^{5/4}\right ) x\right )}{16 \left (e^{5/4}-2 \log (x)\right )^3}+\frac {9 e^5 (5-x) x}{8 \left (e^{5/4}-2 \log (x)\right )^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(5/4)*(-360 + 72*x)*Log[x]^3 + E^(5/4)*(-90 + 36*x)*Log[x]^4 + (180 - 72*x)*Log[x]^5)/(-E^(25/4) + 10*E
^5*Log[x] - 40*E^(15/4)*Log[x]^2 + 80*E^(5/2)*Log[x]^3 - 80*E^(5/4)*Log[x]^4 + 32*Log[x]^5),x]

[Out]

(-9*(5 - 2*x)^2)/32 + (45*E^(5/4 + E^(5/4)/2)*ExpIntegralEi[(-E^(5/4) + 2*Log[x])/2])/4 - (15*E^(5 + E^(5/4)/2
)*ExpIntegralEi[(-E^(5/4) + 2*Log[x])/2])/256 - (45*E^(5/4 + E^(5/4)/2)*(4 - 3*E^(5/4))*ExpIntegralEi[(-E^(5/4
) + 2*Log[x])/2])/16 - (45*E^((5 + E^(5/4))/2)*(6 - E^(5/4))*ExpIntegralEi[(-E^(5/4) + 2*Log[x])/2])/32 - (15*
E^(15/4 + E^(5/4)/2)*(24 - E^(5/4))*ExpIntegralEi[(-E^(5/4) + 2*Log[x])/2])/256 - (9*E^(5/4 + E^(5/4))*ExpInte
gralEi[-E^(5/4) + 2*Log[x]])/2 + (3*E^(5 + E^(5/4))*ExpIntegralEi[-E^(5/4) + 2*Log[x]])/16 + (9*E^(5/4 + E^(5/
4))*(2 - 3*E^(5/4))*ExpIntegralEi[-E^(5/4) + 2*Log[x]])/4 + (9*E^(5/2 + E^(5/4))*(3 - E^(5/4))*ExpIntegralEi[-
E^(5/4) + 2*Log[x]])/4 + (3*E^(15/4 + E^(5/4))*(12 - E^(5/4))*ExpIntegralEi[-E^(5/4) + 2*Log[x]])/16 + (9*E^5*
(5 - x)*x)/(8*(E^(5/4) - 2*Log[x])^4) + (15*E^5*x)/(16*(E^(5/4) - 2*Log[x])^3) - (3*E^5*(5 - x)*x)/(8*(E^(5/4)
- 2*Log[x])^3) - (3*E^(15/4)*x*(5*(24 - E^(5/4)) - 2*(12 - E^(5/4))*x))/(16*(E^(5/4) - 2*Log[x])^3) - (45*E^5
*x)/(64*(E^(5/4) - 2*Log[x])^2) - (15*E^(15/4)*(24 - E^(5/4))*x)/(64*(E^(5/4) - 2*Log[x])^2) + (3*E^5*(5 - x)*
x)/(16*(E^(5/4) - 2*Log[x])^2) + (9*E^(5/2)*x*(5*(6 - E^(5/4)) - 2*(3 - E^(5/4))*x))/(8*(E^(5/4) - 2*Log[x])^2
) + (3*E^(15/4)*x*(5*(24 - E^(5/4)) - 2*(12 - E^(5/4))*x))/(32*(E^(5/4) - 2*Log[x])^2) + (105*E^5*x)/(128*(E^(
5/4) - 2*Log[x])) + (45*E^(5/2)*(6 - E^(5/4))*x)/(16*(E^(5/4) - 2*Log[x])) + (45*E^(15/4)*(24 - E^(5/4))*x)/(1
28*(E^(5/4) - 2*Log[x])) - (3*E^5*(5 - x)*x)/(16*(E^(5/4) - 2*Log[x])) - (9*E^(5/4)*x*(5*(4 - 3*E^(5/4)) - 2*(
2 - 3*E^(5/4))*x))/(8*(E^(5/4) - 2*Log[x])) - (9*E^(5/2)*x*(5*(6 - E^(5/4)) - 2*(3 - E^(5/4))*x))/(8*(E^(5/4)
- 2*Log[x])) - (3*E^(15/4)*x*(5*(24 - E^(5/4)) - 2*(12 - E^(5/4))*x))/(32*(E^(5/4) - 2*Log[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 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 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 2299

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

Rule 2309

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

Rule 2320

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

Rule 2330

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

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 {18 \log ^3(x) \left (-4 e^{5/4} (-5+x)-e^{5/4} (-5+2 x) \log (x)-(10-4 x) \log ^2(x)\right )}{\left (e^{5/4}-2 \log (x)\right )^5} \, dx\\ &=18 \int \frac {\log ^3(x) \left (-4 e^{5/4} (-5+x)-e^{5/4} (-5+2 x) \log (x)-(10-4 x) \log ^2(x)\right )}{\left (e^{5/4}-2 \log (x)\right )^5} \, dx\\ &=18 \int \left (\frac {1}{16} (5-2 x)-\frac {e^5 (-5+x)}{2 \left (e^{5/4}-2 \log (x)\right )^5}+\frac {-5 e^{15/4} \left (24-e^{5/4}\right )+2 e^{15/4} \left (12-e^{5/4}\right ) x}{16 \left (e^{5/4}-2 \log (x)\right )^4}+\frac {e^{5/2} \left (5 \left (6-e^{5/4}\right )-2 \left (3-e^{5/4}\right ) x\right )}{4 \left (e^{5/4}-2 \log (x)\right )^3}+\frac {-5 e^{5/4} \left (4-3 e^{5/4}\right )+2 e^{5/4} \left (2-3 e^{5/4}\right ) x}{8 \left (e^{5/4}-2 \log (x)\right )^2}+\frac {e^{5/4} (-5+2 x)}{4 \left (e^{5/4}-2 \log (x)\right )}\right ) \, dx\\ &=-\frac {9}{32} (5-2 x)^2+\frac {9}{8} \int \frac {-5 e^{15/4} \left (24-e^{5/4}\right )+2 e^{15/4} \left (12-e^{5/4}\right ) x}{\left (e^{5/4}-2 \log (x)\right )^4} \, dx+\frac {9}{4} \int \frac {-5 e^{5/4} \left (4-3 e^{5/4}\right )+2 e^{5/4} \left (2-3 e^{5/4}\right ) x}{\left (e^{5/4}-2 \log (x)\right )^2} \, dx+\frac {1}{2} \left (9 e^{5/4}\right ) \int \frac {-5+2 x}{e^{5/4}-2 \log (x)} \, dx+\frac {1}{2} \left (9 e^{5/2}\right ) \int \frac {5 \left (6-e^{5/4}\right )-2 \left (3-e^{5/4}\right ) x}{\left (e^{5/4}-2 \log (x)\right )^3} \, dx-\left (9 e^5\right ) \int \frac {-5+x}{\left (e^{5/4}-2 \log (x)\right )^5} \, dx\\ &=-\frac {9}{32} (5-2 x)^2+\frac {9 e^5 (5-x) x}{8 \left (e^{5/4}-2 \log (x)\right )^4}-\frac {3 e^{15/4} x \left (5 \left (24-e^{5/4}\right )-2 \left (12-e^{5/4}\right ) x\right )}{16 \left (e^{5/4}-2 \log (x)\right )^3}+\frac {9 e^{5/2} x \left (5 \left (6-e^{5/4}\right )-2 \left (3-e^{5/4}\right ) x\right )}{8 \left (e^{5/4}-2 \log (x)\right )^2}-\frac {9 e^{5/4} x \left (5 \left (4-3 e^{5/4}\right )-2 \left (2-3 e^{5/4}\right ) x\right )}{8 \left (e^{5/4}-2 \log (x)\right )}-\frac {3}{8} \int \frac {-5 e^{15/4} \left (24-e^{5/4}\right )+2 e^{15/4} \left (12-e^{5/4}\right ) x}{\left (e^{5/4}-2 \log (x)\right )^3} \, dx-\frac {9}{4} \int \frac {-5 e^{5/4} \left (4-3 e^{5/4}\right )+2 e^{5/4} \left (2-3 e^{5/4}\right ) x}{e^{5/4}-2 \log (x)} \, dx+\frac {1}{2} \left (9 e^{5/4}\right ) \int \left (-\frac {5}{e^{5/4}-2 \log (x)}+\frac {2 x}{e^{5/4}-2 \log (x)}\right ) \, dx-\frac {1}{4} \left (9 e^{5/2}\right ) \int \frac {5 \left (6-e^{5/4}\right )-2 \left (3-e^{5/4}\right ) x}{\left (e^{5/4}-2 \log (x)\right )^2} \, dx+\frac {1}{4} \left (9 e^5\right ) \int \frac {-5+x}{\left (e^{5/4}-2 \log (x)\right )^4} \, dx+\frac {1}{8} \left (45 e^5\right ) \int \frac {1}{\left (e^{5/4}-2 \log (x)\right )^4} \, dx-\frac {1}{8} \left (45 e^{5/4} \left (4-3 e^{5/4}\right )\right ) \int \frac {1}{e^{5/4}-2 \log (x)} \, dx+\frac {1}{8} \left (45 e^{5/2} \left (6-e^{5/4}\right )\right ) \int \frac {1}{\left (e^{5/4}-2 \log (x)\right )^2} \, dx-\frac {1}{16} \left (15 e^{15/4} \left (24-e^{5/4}\right )\right ) \int \frac {1}{\left (e^{5/4}-2 \log (x)\right )^3} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.42, size = 22, normalized size = 0.96 \begin {gather*} -\frac {18 (-5+x) x \log ^4(x)}{\left (e^{5/4}-2 \log (x)\right )^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(5/4)*(-360 + 72*x)*Log[x]^3 + E^(5/4)*(-90 + 36*x)*Log[x]^4 + (180 - 72*x)*Log[x]^5)/(-E^(25/4)
+ 10*E^5*Log[x] - 40*E^(15/4)*Log[x]^2 + 80*E^(5/2)*Log[x]^3 - 80*E^(5/4)*Log[x]^4 + 32*Log[x]^5),x]

[Out]

(-18*(-5 + x)*x*Log[x]^4)/(E^(5/4) - 2*Log[x])^4

________________________________________________________________________________________

fricas [B]  time = 0.95, size = 48, normalized size = 2.09 \begin {gather*} \frac {18 \, {\left (x^{2} - 5 \, x\right )} \log \relax (x)^{4}}{32 \, e^{\frac {5}{4}} \log \relax (x)^{3} - 16 \, \log \relax (x)^{4} - 24 \, e^{\frac {5}{2}} \log \relax (x)^{2} + 8 \, e^{\frac {15}{4}} \log \relax (x) - e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-72*x+180)*log(x)^5+(36*x-90)*exp(5/4)*log(x)^4+(72*x-360)*exp(5/4)*log(x)^3)/(32*log(x)^5-80*exp(
5/4)*log(x)^4+80*exp(5/4)^2*log(x)^3-40*exp(5/4)^3*log(x)^2+10*exp(5/4)^4*log(x)-exp(5/4)^5),x, algorithm="fri
cas")

[Out]

18*(x^2 - 5*x)*log(x)^4/(32*e^(5/4)*log(x)^3 - 16*log(x)^4 - 24*e^(5/2)*log(x)^2 + 8*e^(15/4)*log(x) - e^5)

________________________________________________________________________________________

giac [B]  time = 2.98, size = 87, normalized size = 3.78 \begin {gather*} \frac {18 \, x^{2} \log \relax (x)^{4}}{32 \, e^{\frac {5}{4}} \log \relax (x)^{3} - 16 \, \log \relax (x)^{4} - 24 \, e^{\frac {5}{2}} \log \relax (x)^{2} + 8 \, e^{\frac {15}{4}} \log \relax (x) - e^{5}} - \frac {90 \, x \log \relax (x)^{4}}{32 \, e^{\frac {5}{4}} \log \relax (x)^{3} - 16 \, \log \relax (x)^{4} - 24 \, e^{\frac {5}{2}} \log \relax (x)^{2} + 8 \, e^{\frac {15}{4}} \log \relax (x) - e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-72*x+180)*log(x)^5+(36*x-90)*exp(5/4)*log(x)^4+(72*x-360)*exp(5/4)*log(x)^3)/(32*log(x)^5-80*exp(
5/4)*log(x)^4+80*exp(5/4)^2*log(x)^3-40*exp(5/4)^3*log(x)^2+10*exp(5/4)^4*log(x)-exp(5/4)^5),x, algorithm="gia
c")

[Out]

18*x^2*log(x)^4/(32*e^(5/4)*log(x)^3 - 16*log(x)^4 - 24*e^(5/2)*log(x)^2 + 8*e^(15/4)*log(x) - e^5) - 90*x*log
(x)^4/(32*e^(5/4)*log(x)^3 - 16*log(x)^4 - 24*e^(5/2)*log(x)^2 + 8*e^(15/4)*log(x) - e^5)

________________________________________________________________________________________

maple [A]  time = 0.12, size = 28, normalized size = 1.22

method	result	size
norman	\(\frac {90 x \ln \relax (x )^{4}-18 x^{2} \ln \relax (x )^{4}}{\left ({\mathrm e}^{\frac {5}{4}}-2 \ln \relax (x )\right )^{4}}\)	\(28\)
risch	\(-\frac {9 x^{2}}{8}+\frac {45 x}{8}+\frac {9 \,{\mathrm e}^{\frac {5}{4}} x \left (x \,{\mathrm e}^{\frac {15}{4}}-8 \ln \relax (x ) {\mathrm e}^{\frac {5}{2}} x +24 \,{\mathrm e}^{\frac {5}{4}} \ln \relax (x )^{2} x -32 x \ln \relax (x )^{3}-5 \,{\mathrm e}^{\frac {15}{4}}+40 \ln \relax (x ) {\mathrm e}^{\frac {5}{2}}-120 \ln \relax (x )^{2} {\mathrm e}^{\frac {5}{4}}+160 \ln \relax (x )^{3}\right )}{8 \left ({\mathrm e}^{\frac {5}{4}}-2 \ln \relax (x )\right )^{4}}\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-72*x+180)*ln(x)^5+(36*x-90)*exp(5/4)*ln(x)^4+(72*x-360)*exp(5/4)*ln(x)^3)/(32*ln(x)^5-80*exp(5/4)*ln(x)
^4+80*exp(5/4)^2*ln(x)^3-40*exp(5/4)^3*ln(x)^2+10*exp(5/4)^4*ln(x)-exp(5/4)^5),x,method=_RETURNVERBOSE)

[Out]

(90*x*ln(x)^4-18*x^2*ln(x)^4)/(exp(5/4)-2*ln(x))^4

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-72*x+180)*log(x)^5+(36*x-90)*exp(5/4)*log(x)^4+(72*x-360)*exp(5/4)*log(x)^3)/(32*log(x)^5-80*exp(
5/4)*log(x)^4+80*exp(5/4)^2*log(x)^3-40*exp(5/4)^3*log(x)^2+10*exp(5/4)^4*log(x)-exp(5/4)^5),x, algorithm="max
ima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

________________________________________________________________________________________

mupad [B]  time = 1.23, size = 738, normalized size = 32.09 result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(5/4)*log(x)^4*(36*x - 90) - log(x)^5*(72*x - 180) + exp(5/4)*log(x)^3*(72*x - 360))/(exp(25/4) - 80*
exp(5/2)*log(x)^3 + 80*exp(5/4)*log(x)^4 + 40*exp(15/4)*log(x)^2 - 32*log(x)^5 - 10*exp(5)*log(x)),x)

[Out]

((3*x*(100*exp(5) - 400*exp(15/4) - 5*exp(25/4) - 80*x*exp(5) + 160*x*exp(15/4) + 8*x*exp(25/4)))/2048 + (3*x*
log(x)*(25*exp(5) + 1200*exp(5/2) + 960*exp(5/4) - 400*exp(15/4) - 40*x*exp(5) - 480*x*exp(5/2) - 192*x*exp(5/
4) + 320*x*exp(15/4)))/1024 - (3*x*log(x)^4*(5*exp(5/4) - 8*x*exp(5/4)))/32 + (3*x*log(x)^3*(25*exp(5/2) - 140
*exp(5/4) - 40*x*exp(5/2) + 112*x*exp(5/4)))/128 + (3*x*log(x)^2*(300*exp(5/2) - 240*exp(5/4) - 25*exp(15/4) -
240*x*exp(5/2) + 96*x*exp(5/4) + 40*x*exp(15/4)))/256)/(exp(5/2)/4 + log(x)^2 - exp(5/4)*log(x)) + x*((45*exp
(5/4))/4 - (135*exp(5/2))/32 - (15*exp(5))/1024 + (15*exp(15/4))/32 + 45/8) - ((3*x*(150*exp(5) + 2400*exp(5/2
) + 1920*exp(5/4) - 1200*exp(15/4) - 5*exp(25/4) - 240*x*exp(5) - 960*x*exp(5/2) - 384*x*exp(5/4) + 960*x*exp(
15/4) + 16*x*exp(25/4)))/2048 + (3*x*log(x)*(25*exp(5) + 3600*exp(5/2) - 960*exp(5/4) - 600*exp(15/4) - 80*x*e
xp(5) - 2880*x*exp(5/2) + 384*x*exp(5/4) + 960*x*exp(15/4)))/1024 - (3*x*log(x)^4*(5*exp(5/4) - 16*x*exp(5/4))
)/32 + (3*x*log(x)^3*(25*exp(5/2) - 220*exp(5/4) - 80*x*exp(5/2) + 352*x*exp(5/4)))/128 + (3*x*log(x)^2*(450*e
xp(5/2) - 1080*exp(5/4) - 25*exp(15/4) - 720*x*exp(5/2) + 864*x*exp(5/4) + 80*x*exp(15/4)))/256)/(exp(5/4)/2 -
log(x)) - log(x)^2*((45*x*exp(5/4)*(exp(5/4) - 16))/128 - (9*x^2*exp(5/4)*(exp(5/4) - 8))/8) - ((3*x*(50*exp(
5) - 5*exp(25/4) - 20*x*exp(5) + 4*x*exp(25/4)))/1024 + (15*x*log(x)*(5*exp(5) - 40*exp(15/4) - 4*x*exp(5) + 1
6*x*exp(15/4)))/512 - (3*x*log(x)^4*(5*exp(5/4) - 4*x*exp(5/4)))/16 + (3*x*log(x)^3*(25*exp(5/2) - 60*exp(5/4)
- 20*x*exp(5/2) + 24*x*exp(5/4)))/64 + (3*x*log(x)^2*(150*exp(5/2) + 120*exp(5/4) - 25*exp(15/4) - 60*x*exp(5
/2) - 24*x*exp(5/4) + 20*x*exp(15/4)))/128)/(exp(15/4)/8 + (3*exp(5/4)*log(x)^2)/2 - log(x)^3 - (3*exp(5/2)*lo
g(x))/4) - x^2*((9*exp(5/4))/2 - (27*exp(5/2))/8 - (3*exp(5))/64 + (3*exp(15/4))/4 + 9/8) + ((9*x*exp(25/4)*(2
*x - 5))/1024 - (9*x*log(x)^4*(5*exp(5/4) - 2*x*exp(5/4)))/16 + (9*x*log(x)^3*(25*exp(5/2) + 20*exp(5/4) - 10*
x*exp(5/2) - 4*x*exp(5/4)))/64 - (45*x*exp(5)*log(x)*(2*x - 5))/512 + (45*x*exp(15/4)*log(x)^2*(2*x - 5))/128)
/(exp(5)/16 + (3*exp(5/2)*log(x)^2)/2 - 2*exp(5/4)*log(x)^3 + log(x)^4 - (exp(15/4)*log(x))/2) + log(x)*((15*x
*exp(5/4)*(exp(5/4) - 12)^2)/128 - (3*x^2*exp(5/4)*(exp(5/4) - 6)^2)/8) + log(x)^3*((15*x*exp(5/4))/32 - (3*x^
2*exp(5/4))/2)

________________________________________________________________________________________

sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: AttributeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-72*x+180)*ln(x)**5+(36*x-90)*exp(5/4)*ln(x)**4+(72*x-360)*exp(5/4)*ln(x)**3)/(32*ln(x)**5-80*exp(
5/4)*ln(x)**4+80*exp(5/4)**2*ln(x)**3-40*exp(5/4)**3*ln(x)**2+10*exp(5/4)**4*ln(x)-exp(5/4)**5),x)

[Out]

Exception raised: AttributeError

________________________________________________________________________________________