3.6.56 \(\int \frac {(600+630 x+30 x^2) \log (\frac {20+x}{4}) \log ^2(\log (\frac {20+x}{4}))+e^{\frac {2 (e^5 x^2-3 x^3 \log (\log (\frac {20+x}{4})))}{15 \log (\log (\frac {20+x}{4}))}} (-2 e^5 x^2+e^5 (80 x+4 x^2) \log (\frac {20+x}{4}) \log (\log (\frac {20+x}{4}))+(-360 x^2-18 x^3) \log (\frac {20+x}{4}) \log ^2(\log (\frac {20+x}{4})))+e^{\frac {e^5 x^2-3 x^3 \log (\log (\frac {20+x}{4}))}{15 \log (\log (\frac {20+x}{4}))}} (e^5 (-2 x^2-2 x^3)+e^5 (80 x+84 x^2+4 x^3) \log (\frac {20+x}{4}) \log (\log (\frac {20+x}{4}))+(600+30 x-360 x^2-378 x^3-18 x^4) \log (\frac {20+x}{4}) \log ^2(\log (\frac {20+x}{4})))}{(300+15 x) \log (\frac {20+x}{4}) \log ^2(\log (\frac {20+x}{4}))} \, dx\)

Optimal. Leaf size=36 \[ \left (1+e^{\frac {1}{5} x^2 \left (-x+\frac {e^5}{3 \log \left (\log \left (5+\frac {x}{4}\right )\right )}\right )}+x\right )^2 \]

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Rubi [B]  time = 16.32, antiderivative size = 314, normalized size of antiderivative = 8.72, number of steps used = 6, number of rules used = 5, integrand size = 301, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {6741, 6742, 6688, 6706, 2288} \begin {gather*} e^{-\frac {2}{15} x^2 \left (3 x-\frac {e^5}{\log \left (\log \left (\frac {x}{4}+5\right )\right )}\right )}+\frac {2 e^{-\frac {1}{15} x^2 \left (3 x-\frac {e^5}{\log \left (\log \left (\frac {x}{4}+5\right )\right )}\right )} \left (9 x^4 \log \left (\frac {x}{4}+5\right ) \log ^2\left (\log \left (\frac {x}{4}+5\right )\right )+e^5 x^3+189 x^3 \log \left (\frac {x}{4}+5\right ) \log ^2\left (\log \left (\frac {x}{4}+5\right )\right )-2 e^5 x^3 \log \left (\frac {x}{4}+5\right ) \log \left (\log \left (\frac {x}{4}+5\right )\right )+e^5 x^2+180 x^2 \log \left (\frac {x}{4}+5\right ) \log ^2\left (\log \left (\frac {x}{4}+5\right )\right )-42 e^5 x^2 \log \left (\frac {x}{4}+5\right ) \log \left (\log \left (\frac {x}{4}+5\right )\right )-40 e^5 x \log \left (\frac {x}{4}+5\right ) \log \left (\log \left (\frac {x}{4}+5\right )\right )\right )}{(x+20) \log \left (\frac {x}{4}+5\right ) \left (x^2 \left (\frac {e^5}{(x+20) \log ^2\left (\log \left (\frac {x}{4}+5\right )\right ) \log \left (\frac {x}{4}+5\right )}+3\right )+2 x \left (3 x-\frac {e^5}{\log \left (\log \left (\frac {x}{4}+5\right )\right )}\right )\right ) \log ^2\left (\log \left (\frac {x}{4}+5\right )\right )}+(x+1)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((600 + 630*x + 30*x^2)*Log[(20 + x)/4]*Log[Log[(20 + x)/4]]^2 + E^((2*(E^5*x^2 - 3*x^3*Log[Log[(20 + x)/4
]]))/(15*Log[Log[(20 + x)/4]]))*(-2*E^5*x^2 + E^5*(80*x + 4*x^2)*Log[(20 + x)/4]*Log[Log[(20 + x)/4]] + (-360*
x^2 - 18*x^3)*Log[(20 + x)/4]*Log[Log[(20 + x)/4]]^2) + E^((E^5*x^2 - 3*x^3*Log[Log[(20 + x)/4]])/(15*Log[Log[
(20 + x)/4]]))*(E^5*(-2*x^2 - 2*x^3) + E^5*(80*x + 84*x^2 + 4*x^3)*Log[(20 + x)/4]*Log[Log[(20 + x)/4]] + (600
+ 30*x - 360*x^2 - 378*x^3 - 18*x^4)*Log[(20 + x)/4]*Log[Log[(20 + x)/4]]^2))/((300 + 15*x)*Log[(20 + x)/4]*L
og[Log[(20 + x)/4]]^2),x]

[Out]

E^((-2*x^2*(3*x - E^5/Log[Log[5 + x/4]]))/15) + (1 + x)^2 + (2*(E^5*x^2 + E^5*x^3 - 40*E^5*x*Log[5 + x/4]*Log[
Log[5 + x/4]] - 42*E^5*x^2*Log[5 + x/4]*Log[Log[5 + x/4]] - 2*E^5*x^3*Log[5 + x/4]*Log[Log[5 + x/4]] + 180*x^2
*Log[5 + x/4]*Log[Log[5 + x/4]]^2 + 189*x^3*Log[5 + x/4]*Log[Log[5 + x/4]]^2 + 9*x^4*Log[5 + x/4]*Log[Log[5 +
x/4]]^2))/(E^((x^2*(3*x - E^5/Log[Log[5 + x/4]]))/15)*(20 + x)*Log[5 + x/4]*(x^2*(3 + E^5/((20 + x)*Log[5 + x/
4]*Log[Log[5 + x/4]]^2)) + 2*x*(3*x - E^5/Log[Log[5 + x/4]]))*Log[Log[5 + x/4]]^2)

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 6688

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

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rule 6741

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

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 {\left (600+630 x+30 x^2\right ) \log \left (\frac {20+x}{4}\right ) \log ^2\left (\log \left (\frac {20+x}{4}\right )\right )+\exp \left (\frac {2 \left (e^5 x^2-3 x^3 \log \left (\log \left (\frac {20+x}{4}\right )\right )\right )}{15 \log \left (\log \left (\frac {20+x}{4}\right )\right )}\right ) \left (-2 e^5 x^2+e^5 \left (80 x+4 x^2\right ) \log \left (\frac {20+x}{4}\right ) \log \left (\log \left (\frac {20+x}{4}\right )\right )+\left (-360 x^2-18 x^3\right ) \log \left (\frac {20+x}{4}\right ) \log ^2\left (\log \left (\frac {20+x}{4}\right )\right )\right )+\exp \left (\frac {e^5 x^2-3 x^3 \log \left (\log \left (\frac {20+x}{4}\right )\right )}{15 \log \left (\log \left (\frac {20+x}{4}\right )\right )}\right ) \left (e^5 \left (-2 x^2-2 x^3\right )+e^5 \left (80 x+84 x^2+4 x^3\right ) \log \left (\frac {20+x}{4}\right ) \log \left (\log \left (\frac {20+x}{4}\right )\right )+\left (600+30 x-360 x^2-378 x^3-18 x^4\right ) \log \left (\frac {20+x}{4}\right ) \log ^2\left (\log \left (\frac {20+x}{4}\right )\right )\right )}{(300+15 x) \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )} \, dx\\ &=\int \left (2 (1+x)-\frac {2 e^{-\frac {2}{15} x^2 \left (3 x-\frac {e^5}{\log \left (\log \left (5+\frac {x}{4}\right )\right )}\right )} x \left (e^5 x-40 e^5 \log \left (5+\frac {x}{4}\right ) \log \left (\log \left (5+\frac {x}{4}\right )\right )-2 e^5 x \log \left (5+\frac {x}{4}\right ) \log \left (\log \left (5+\frac {x}{4}\right )\right )+180 x \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )+9 x^2 \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )\right )}{15 (20+x) \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )}-\frac {2 e^{-\frac {1}{15} x^2 \left (3 x-\frac {e^5}{\log \left (\log \left (5+\frac {x}{4}\right )\right )}\right )} \left (e^5 x^2+e^5 x^3-40 e^5 x \log \left (5+\frac {x}{4}\right ) \log \left (\log \left (5+\frac {x}{4}\right )\right )-42 e^5 x^2 \log \left (5+\frac {x}{4}\right ) \log \left (\log \left (5+\frac {x}{4}\right )\right )-2 e^5 x^3 \log \left (5+\frac {x}{4}\right ) \log \left (\log \left (5+\frac {x}{4}\right )\right )-300 \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )-15 x \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )+180 x^2 \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )+189 x^3 \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )+9 x^4 \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )\right )}{15 (20+x) \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )}\right ) \, dx\\ &=(1+x)^2-\frac {2}{15} \int \frac {e^{-\frac {2}{15} x^2 \left (3 x-\frac {e^5}{\log \left (\log \left (5+\frac {x}{4}\right )\right )}\right )} x \left (e^5 x-40 e^5 \log \left (5+\frac {x}{4}\right ) \log \left (\log \left (5+\frac {x}{4}\right )\right )-2 e^5 x \log \left (5+\frac {x}{4}\right ) \log \left (\log \left (5+\frac {x}{4}\right )\right )+180 x \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )+9 x^2 \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )\right )}{(20+x) \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )} \, dx-\frac {2}{15} \int \frac {e^{-\frac {1}{15} x^2 \left (3 x-\frac {e^5}{\log \left (\log \left (5+\frac {x}{4}\right )\right )}\right )} \left (e^5 x^2+e^5 x^3-40 e^5 x \log \left (5+\frac {x}{4}\right ) \log \left (\log \left (5+\frac {x}{4}\right )\right )-42 e^5 x^2 \log \left (5+\frac {x}{4}\right ) \log \left (\log \left (5+\frac {x}{4}\right )\right )-2 e^5 x^3 \log \left (5+\frac {x}{4}\right ) \log \left (\log \left (5+\frac {x}{4}\right )\right )-300 \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )-15 x \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )+180 x^2 \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )+189 x^3 \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )+9 x^4 \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )\right )}{(20+x) \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )} \, dx\\ &=(1+x)^2+\frac {2 e^{-\frac {1}{15} x^2 \left (3 x-\frac {e^5}{\log \left (\log \left (5+\frac {x}{4}\right )\right )}\right )} \left (e^5 x^2+e^5 x^3-40 e^5 x \log \left (5+\frac {x}{4}\right ) \log \left (\log \left (5+\frac {x}{4}\right )\right )-42 e^5 x^2 \log \left (5+\frac {x}{4}\right ) \log \left (\log \left (5+\frac {x}{4}\right )\right )-2 e^5 x^3 \log \left (5+\frac {x}{4}\right ) \log \left (\log \left (5+\frac {x}{4}\right )\right )+180 x^2 \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )+189 x^3 \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )+9 x^4 \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )\right )}{(20+x) \log \left (5+\frac {x}{4}\right ) \left (x^2 \left (3+\frac {e^5}{(20+x) \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )}\right )+2 x \left (3 x-\frac {e^5}{\log \left (\log \left (5+\frac {x}{4}\right )\right )}\right )\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )}-\frac {2}{15} \int \frac {e^{-\frac {2}{15} x^2 \left (3 x-\frac {e^5}{\log \left (\log \left (5+\frac {x}{4}\right )\right )}\right )} x \left (e^5 x+(20+x) \log \left (5+\frac {x}{4}\right ) \log \left (\log \left (5+\frac {x}{4}\right )\right ) \left (-2 e^5+9 x \log \left (\log \left (5+\frac {x}{4}\right )\right )\right )\right )}{(20+x) \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )} \, dx\\ &=e^{-\frac {2}{15} x^2 \left (3 x-\frac {e^5}{\log \left (\log \left (5+\frac {x}{4}\right )\right )}\right )}+(1+x)^2+\frac {2 e^{-\frac {1}{15} x^2 \left (3 x-\frac {e^5}{\log \left (\log \left (5+\frac {x}{4}\right )\right )}\right )} \left (e^5 x^2+e^5 x^3-40 e^5 x \log \left (5+\frac {x}{4}\right ) \log \left (\log \left (5+\frac {x}{4}\right )\right )-42 e^5 x^2 \log \left (5+\frac {x}{4}\right ) \log \left (\log \left (5+\frac {x}{4}\right )\right )-2 e^5 x^3 \log \left (5+\frac {x}{4}\right ) \log \left (\log \left (5+\frac {x}{4}\right )\right )+180 x^2 \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )+189 x^3 \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )+9 x^4 \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )\right )}{(20+x) \log \left (5+\frac {x}{4}\right ) \left (x^2 \left (3+\frac {e^5}{(20+x) \log \left (5+\frac {x}{4}\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )}\right )+2 x \left (3 x-\frac {e^5}{\log \left (\log \left (5+\frac {x}{4}\right )\right )}\right )\right ) \log ^2\left (\log \left (5+\frac {x}{4}\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.37, size = 81, normalized size = 2.25 \begin {gather*} e^{-\frac {2 x^3}{5}} \left (e^{\frac {2 e^5 x^2}{15 \log \left (\log \left (5+\frac {x}{4}\right )\right )}}+2 e^{\frac {1}{15} x^2 \left (3 x+\frac {e^5}{\log \left (\log \left (5+\frac {x}{4}\right )\right )}\right )} (1+x)+e^{\frac {2 x^3}{5}} x (2+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((600 + 630*x + 30*x^2)*Log[(20 + x)/4]*Log[Log[(20 + x)/4]]^2 + E^((2*(E^5*x^2 - 3*x^3*Log[Log[(20
+ x)/4]]))/(15*Log[Log[(20 + x)/4]]))*(-2*E^5*x^2 + E^5*(80*x + 4*x^2)*Log[(20 + x)/4]*Log[Log[(20 + x)/4]] +
(-360*x^2 - 18*x^3)*Log[(20 + x)/4]*Log[Log[(20 + x)/4]]^2) + E^((E^5*x^2 - 3*x^3*Log[Log[(20 + x)/4]])/(15*Lo
g[Log[(20 + x)/4]]))*(E^5*(-2*x^2 - 2*x^3) + E^5*(80*x + 84*x^2 + 4*x^3)*Log[(20 + x)/4]*Log[Log[(20 + x)/4]]
+ (600 + 30*x - 360*x^2 - 378*x^3 - 18*x^4)*Log[(20 + x)/4]*Log[Log[(20 + x)/4]]^2))/((300 + 15*x)*Log[(20 + x
)/4]*Log[Log[(20 + x)/4]]^2),x]

[Out]

(E^((2*E^5*x^2)/(15*Log[Log[5 + x/4]])) + 2*E^((x^2*(3*x + E^5/Log[Log[5 + x/4]]))/15)*(1 + x) + E^((2*x^3)/5)
*x*(2 + x))/E^((2*x^3)/5)

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fricas [B]  time = 0.73, size = 76, normalized size = 2.11 \begin {gather*} x^{2} + 2 \, {\left (x + 1\right )} e^{\left (-\frac {3 \, x^{3} \log \left (\log \left (\frac {1}{4} \, x + 5\right )\right ) - x^{2} e^{5}}{15 \, \log \left (\log \left (\frac {1}{4} \, x + 5\right )\right )}\right )} + 2 \, x + e^{\left (-\frac {2 \, {\left (3 \, x^{3} \log \left (\log \left (\frac {1}{4} \, x + 5\right )\right ) - x^{2} e^{5}\right )}}{15 \, \log \left (\log \left (\frac {1}{4} \, x + 5\right )\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-18*x^3-360*x^2)*log(5+1/4*x)*log(log(5+1/4*x))^2+(4*x^2+80*x)*exp(5)*log(5+1/4*x)*log(log(5+1/4*
x))-2*x^2*exp(5))*exp(1/15*(-3*x^3*log(log(5+1/4*x))+x^2*exp(5))/log(log(5+1/4*x)))^2+((-18*x^4-378*x^3-360*x^
2+30*x+600)*log(5+1/4*x)*log(log(5+1/4*x))^2+(4*x^3+84*x^2+80*x)*exp(5)*log(5+1/4*x)*log(log(5+1/4*x))+(-2*x^3
-2*x^2)*exp(5))*exp(1/15*(-3*x^3*log(log(5+1/4*x))+x^2*exp(5))/log(log(5+1/4*x)))+(30*x^2+630*x+600)*log(5+1/4
*x)*log(log(5+1/4*x))^2)/(15*x+300)/log(5+1/4*x)/log(log(5+1/4*x))^2,x, algorithm="fricas")

[Out]

x^2 + 2*(x + 1)*e^(-1/15*(3*x^3*log(log(1/4*x + 5)) - x^2*e^5)/log(log(1/4*x + 5))) + 2*x + e^(-2/15*(3*x^3*lo
g(log(1/4*x + 5)) - x^2*e^5)/log(log(1/4*x + 5)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-18*x^3-360*x^2)*log(5+1/4*x)*log(log(5+1/4*x))^2+(4*x^2+80*x)*exp(5)*log(5+1/4*x)*log(log(5+1/4*
x))-2*x^2*exp(5))*exp(1/15*(-3*x^3*log(log(5+1/4*x))+x^2*exp(5))/log(log(5+1/4*x)))^2+((-18*x^4-378*x^3-360*x^
2+30*x+600)*log(5+1/4*x)*log(log(5+1/4*x))^2+(4*x^3+84*x^2+80*x)*exp(5)*log(5+1/4*x)*log(log(5+1/4*x))+(-2*x^3
-2*x^2)*exp(5))*exp(1/15*(-3*x^3*log(log(5+1/4*x))+x^2*exp(5))/log(log(5+1/4*x)))+(30*x^2+630*x+600)*log(5+1/4
*x)*log(log(5+1/4*x))^2)/(15*x+300)/log(5+1/4*x)/log(log(5+1/4*x))^2,x, algorithm="giac")

[Out]

undef

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maple [B]  time = 0.39, size = 70, normalized size = 1.94

method	result	size
risch	\(x^{2}+{\mathrm e}^{\frac {2 x^{2} \left (-3 \ln \left (\ln \left (5+\frac {x}{4}\right )\right ) x +{\mathrm e}^{5}\right )}{15 \ln \left (\ln \left (5+\frac {x}{4}\right )\right )}}+2 x +\left (2 x +2\right ) {\mathrm e}^{\frac {x^{2} \left (-3 \ln \left (\ln \left (5+\frac {x}{4}\right )\right ) x +{\mathrm e}^{5}\right )}{15 \ln \left (\ln \left (5+\frac {x}{4}\right )\right )}}\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-18*x^3-360*x^2)*ln(5+1/4*x)*ln(ln(5+1/4*x))^2+(4*x^2+80*x)*exp(5)*ln(5+1/4*x)*ln(ln(5+1/4*x))-2*x^2*ex
p(5))*exp(1/15*(-3*x^3*ln(ln(5+1/4*x))+x^2*exp(5))/ln(ln(5+1/4*x)))^2+((-18*x^4-378*x^3-360*x^2+30*x+600)*ln(5
+1/4*x)*ln(ln(5+1/4*x))^2+(4*x^3+84*x^2+80*x)*exp(5)*ln(5+1/4*x)*ln(ln(5+1/4*x))+(-2*x^3-2*x^2)*exp(5))*exp(1/
15*(-3*x^3*ln(ln(5+1/4*x))+x^2*exp(5))/ln(ln(5+1/4*x)))+(30*x^2+630*x+600)*ln(5+1/4*x)*ln(ln(5+1/4*x))^2)/(15*
x+300)/ln(5+1/4*x)/ln(ln(5+1/4*x))^2,x,method=_RETURNVERBOSE)

[Out]

x^2+exp(2/15*x^2*(-3*ln(ln(5+1/4*x))*x+exp(5))/ln(ln(5+1/4*x)))+2*x+(2*x+2)*exp(1/15*x^2*(-3*ln(ln(5+1/4*x))*x
+exp(5))/ln(ln(5+1/4*x)))

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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((((-18*x^3-360*x^2)*log(5+1/4*x)*log(log(5+1/4*x))^2+(4*x^2+80*x)*exp(5)*log(5+1/4*x)*log(log(5+1/4*
x))-2*x^2*exp(5))*exp(1/15*(-3*x^3*log(log(5+1/4*x))+x^2*exp(5))/log(log(5+1/4*x)))^2+((-18*x^4-378*x^3-360*x^
2+30*x+600)*log(5+1/4*x)*log(log(5+1/4*x))^2+(4*x^3+84*x^2+80*x)*exp(5)*log(5+1/4*x)*log(log(5+1/4*x))+(-2*x^3
-2*x^2)*exp(5))*exp(1/15*(-3*x^3*log(log(5+1/4*x))+x^2*exp(5))/log(log(5+1/4*x)))+(30*x^2+630*x+600)*log(5+1/4
*x)*log(log(5+1/4*x))^2)/(15*x+300)/log(5+1/4*x)/log(log(5+1/4*x))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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mupad [B]  time = 1.25, size = 81, normalized size = 2.25 \begin {gather*} 2\,x+2\,{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^5}{15\,\ln \left (\ln \left (\frac {x}{4}+5\right )\right )}-\frac {x^3}{5}}+{\mathrm {e}}^{\frac {2\,x^2\,{\mathrm {e}}^5}{15\,\ln \left (\ln \left (\frac {x}{4}+5\right )\right )}-\frac {2\,x^3}{5}}+x^2+2\,x\,{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^5}{15\,\ln \left (\ln \left (\frac {x}{4}+5\right )\right )}-\frac {x^3}{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-((x^3*log(log(x/4 + 5)))/5 - (x^2*exp(5))/15)/log(log(x/4 + 5)))*(exp(5)*(2*x^2 + 2*x^3) + log(log(
x/4 + 5))^2*log(x/4 + 5)*(360*x^2 - 30*x + 378*x^3 + 18*x^4 - 600) - log(log(x/4 + 5))*exp(5)*log(x/4 + 5)*(80
*x + 84*x^2 + 4*x^3)) + exp(-(2*((x^3*log(log(x/4 + 5)))/5 - (x^2*exp(5))/15))/log(log(x/4 + 5)))*(2*x^2*exp(5
) + log(log(x/4 + 5))^2*log(x/4 + 5)*(360*x^2 + 18*x^3) - log(log(x/4 + 5))*exp(5)*log(x/4 + 5)*(80*x + 4*x^2)
) - log(log(x/4 + 5))^2*log(x/4 + 5)*(630*x + 30*x^2 + 600))/(log(log(x/4 + 5))^2*log(x/4 + 5)*(15*x + 300)),x
)

[Out]

2*x + 2*exp((x^2*exp(5))/(15*log(log(x/4 + 5))) - x^3/5) + exp((2*x^2*exp(5))/(15*log(log(x/4 + 5))) - (2*x^3)
/5) + x^2 + 2*x*exp((x^2*exp(5))/(15*log(log(x/4 + 5))) - x^3/5)

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sympy [B]  time = 4.27, size = 78, normalized size = 2.17 \begin {gather*} x^{2} + 2 x + \left (2 x + 2\right ) e^{\frac {- \frac {x^{3} \log {\left (\log {\left (\frac {x}{4} + 5 \right )} \right )}}{5} + \frac {x^{2} e^{5}}{15}}{\log {\left (\log {\left (\frac {x}{4} + 5 \right )} \right )}}} + e^{\frac {2 \left (- \frac {x^{3} \log {\left (\log {\left (\frac {x}{4} + 5 \right )} \right )}}{5} + \frac {x^{2} e^{5}}{15}\right )}{\log {\left (\log {\left (\frac {x}{4} + 5 \right )} \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-18*x**3-360*x**2)*ln(5+1/4*x)*ln(ln(5+1/4*x))**2+(4*x**2+80*x)*exp(5)*ln(5+1/4*x)*ln(ln(5+1/4*x)
)-2*x**2*exp(5))*exp(1/15*(-3*x**3*ln(ln(5+1/4*x))+x**2*exp(5))/ln(ln(5+1/4*x)))**2+((-18*x**4-378*x**3-360*x*
*2+30*x+600)*ln(5+1/4*x)*ln(ln(5+1/4*x))**2+(4*x**3+84*x**2+80*x)*exp(5)*ln(5+1/4*x)*ln(ln(5+1/4*x))+(-2*x**3-
2*x**2)*exp(5))*exp(1/15*(-3*x**3*ln(ln(5+1/4*x))+x**2*exp(5))/ln(ln(5+1/4*x)))+(30*x**2+630*x+600)*ln(5+1/4*x
)*ln(ln(5+1/4*x))**2)/(15*x+300)/ln(5+1/4*x)/ln(ln(5+1/4*x))**2,x)

[Out]

x**2 + 2*x + (2*x + 2)*exp((-x**3*log(log(x/4 + 5))/5 + x**2*exp(5)/15)/log(log(x/4 + 5))) + exp(2*(-x**3*log(
log(x/4 + 5))/5 + x**2*exp(5)/15)/log(log(x/4 + 5)))

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