3.44.22 \(\int \frac {e^{2-x} (e^{-5+x} x-3 \log (3))^4 (8 e^{-5+x} x-24 \log (3)+(e^{-5+x} (4 x+3 x^2)+3 x \log (3)) \log (\frac {2}{x^2}))}{81 (e^{-5+x} x^2-3 x \log (3)) \log ^5(\frac {2}{x^2})} \, dx\)

Optimal. Leaf size=33 \[ \frac {e^{2-x} \left (\frac {1}{3} e^{-5+x} x-\log (3)\right )^4}{\log ^4\left (\frac {2}{x^2}\right )} \]

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Rubi [B]  time = 4.36, antiderivative size = 111, normalized size of antiderivative = 3.36, number of steps used = 20, number of rules used = 7, integrand size = 91, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 6742, 2360, 2297, 2300, 2178, 2288} \begin {gather*} \frac {e^{2-x} \log ^4(3)}{\log ^4\left (\frac {2}{x^2}\right )}-\frac {4 x \log ^3(3)}{3 e^3 \log ^4\left (\frac {2}{x^2}\right )}+\frac {2 e^{x-8} x^2 \log ^2(3)}{3 \log ^4\left (\frac {2}{x^2}\right )}+\frac {e^{3 x-18} x^4}{81 \log ^4\left (\frac {2}{x^2}\right )}-\frac {4 e^{2 x-13} x^3 \log (3)}{27 \log ^4\left (\frac {2}{x^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(2 - x)*(E^(-5 + x)*x - 3*Log[3])^4*(8*E^(-5 + x)*x - 24*Log[3] + (E^(-5 + x)*(4*x + 3*x^2) + 3*x*Log[3
])*Log[2/x^2]))/(81*(E^(-5 + x)*x^2 - 3*x*Log[3])*Log[2/x^2]^5),x]

[Out]

(E^(-18 + 3*x)*x^4)/(81*Log[2/x^2]^4) - (4*E^(-13 + 2*x)*x^3*Log[3])/(27*Log[2/x^2]^4) + (2*E^(-8 + x)*x^2*Log
[3]^2)/(3*Log[2/x^2]^4) - (4*x*Log[3]^3)/(3*E^3*Log[2/x^2]^4) + (E^(2 - x)*Log[3]^4)/Log[2/x^2]^4

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 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 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 2360

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Log[(c_.)*(x_)^(n_.)]*(e_.) + (d_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*Log[c*x^n])^p*(d + e*Log[c*x^n])^q, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && IntegerQ[p
] && IntegerQ[q]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{81} \int \frac {e^{2-x} \left (e^{-5+x} x-3 \log (3)\right )^4 \left (8 e^{-5+x} x-24 \log (3)+\left (e^{-5+x} \left (4 x+3 x^2\right )+3 x \log (3)\right ) \log \left (\frac {2}{x^2}\right )\right )}{\left (e^{-5+x} x^2-3 x \log (3)\right ) \log ^5\left (\frac {2}{x^2}\right )} \, dx\\ &=\frac {1}{81} \int \left (-\frac {108 \log ^3(3) \left (8+\log \left (\frac {2}{x^2}\right )\right )}{e^3 \log ^5\left (\frac {2}{x^2}\right )}-\frac {81 e^{2-x} \log ^4(3) \left (-8+x \log \left (\frac {2}{x^2}\right )\right )}{x \log ^5\left (\frac {2}{x^2}\right )}+\frac {54 e^{-8+x} x \log ^2(3) \left (8+2 \log \left (\frac {2}{x^2}\right )+x \log \left (\frac {2}{x^2}\right )\right )}{\log ^5\left (\frac {2}{x^2}\right )}-\frac {12 e^{-13+2 x} x^2 \log (3) \left (8+3 \log \left (\frac {2}{x^2}\right )+2 x \log \left (\frac {2}{x^2}\right )\right )}{\log ^5\left (\frac {2}{x^2}\right )}+\frac {e^{-18+3 x} x^3 \left (8+4 \log \left (\frac {2}{x^2}\right )+3 x \log \left (\frac {2}{x^2}\right )\right )}{\log ^5\left (\frac {2}{x^2}\right )}\right ) \, dx\\ &=\frac {1}{81} \int \frac {e^{-18+3 x} x^3 \left (8+4 \log \left (\frac {2}{x^2}\right )+3 x \log \left (\frac {2}{x^2}\right )\right )}{\log ^5\left (\frac {2}{x^2}\right )} \, dx-\frac {1}{27} (4 \log (3)) \int \frac {e^{-13+2 x} x^2 \left (8+3 \log \left (\frac {2}{x^2}\right )+2 x \log \left (\frac {2}{x^2}\right )\right )}{\log ^5\left (\frac {2}{x^2}\right )} \, dx+\frac {1}{3} \left (2 \log ^2(3)\right ) \int \frac {e^{-8+x} x \left (8+2 \log \left (\frac {2}{x^2}\right )+x \log \left (\frac {2}{x^2}\right )\right )}{\log ^5\left (\frac {2}{x^2}\right )} \, dx-\frac {\left (4 \log ^3(3)\right ) \int \frac {8+\log \left (\frac {2}{x^2}\right )}{\log ^5\left (\frac {2}{x^2}\right )} \, dx}{3 e^3}-\log ^4(3) \int \frac {e^{2-x} \left (-8+x \log \left (\frac {2}{x^2}\right )\right )}{x \log ^5\left (\frac {2}{x^2}\right )} \, dx\\ &=\frac {e^{-18+3 x} x^4}{81 \log ^4\left (\frac {2}{x^2}\right )}-\frac {4 e^{-13+2 x} x^3 \log (3)}{27 \log ^4\left (\frac {2}{x^2}\right )}+\frac {2 e^{-8+x} x^2 \log ^2(3)}{3 \log ^4\left (\frac {2}{x^2}\right )}+\frac {e^{2-x} \log ^4(3)}{\log ^4\left (\frac {2}{x^2}\right )}-\frac {\left (4 \log ^3(3)\right ) \int \left (\frac {8}{\log ^5\left (\frac {2}{x^2}\right )}+\frac {1}{\log ^4\left (\frac {2}{x^2}\right )}\right ) \, dx}{3 e^3}\\ &=\frac {e^{-18+3 x} x^4}{81 \log ^4\left (\frac {2}{x^2}\right )}-\frac {4 e^{-13+2 x} x^3 \log (3)}{27 \log ^4\left (\frac {2}{x^2}\right )}+\frac {2 e^{-8+x} x^2 \log ^2(3)}{3 \log ^4\left (\frac {2}{x^2}\right )}+\frac {e^{2-x} \log ^4(3)}{\log ^4\left (\frac {2}{x^2}\right )}-\frac {\left (4 \log ^3(3)\right ) \int \frac {1}{\log ^4\left (\frac {2}{x^2}\right )} \, dx}{3 e^3}-\frac {\left (32 \log ^3(3)\right ) \int \frac {1}{\log ^5\left (\frac {2}{x^2}\right )} \, dx}{3 e^3}\\ &=\frac {e^{-18+3 x} x^4}{81 \log ^4\left (\frac {2}{x^2}\right )}-\frac {4 e^{-13+2 x} x^3 \log (3)}{27 \log ^4\left (\frac {2}{x^2}\right )}+\frac {2 e^{-8+x} x^2 \log ^2(3)}{3 \log ^4\left (\frac {2}{x^2}\right )}-\frac {4 x \log ^3(3)}{3 e^3 \log ^4\left (\frac {2}{x^2}\right )}+\frac {e^{2-x} \log ^4(3)}{\log ^4\left (\frac {2}{x^2}\right )}-\frac {2 x \log ^3(3)}{9 e^3 \log ^3\left (\frac {2}{x^2}\right )}+\frac {\left (2 \log ^3(3)\right ) \int \frac {1}{\log ^3\left (\frac {2}{x^2}\right )} \, dx}{9 e^3}+\frac {\left (4 \log ^3(3)\right ) \int \frac {1}{\log ^4\left (\frac {2}{x^2}\right )} \, dx}{3 e^3}\\ &=\frac {e^{-18+3 x} x^4}{81 \log ^4\left (\frac {2}{x^2}\right )}-\frac {4 e^{-13+2 x} x^3 \log (3)}{27 \log ^4\left (\frac {2}{x^2}\right )}+\frac {2 e^{-8+x} x^2 \log ^2(3)}{3 \log ^4\left (\frac {2}{x^2}\right )}-\frac {4 x \log ^3(3)}{3 e^3 \log ^4\left (\frac {2}{x^2}\right )}+\frac {e^{2-x} \log ^4(3)}{\log ^4\left (\frac {2}{x^2}\right )}+\frac {x \log ^3(3)}{18 e^3 \log ^2\left (\frac {2}{x^2}\right )}-\frac {\log ^3(3) \int \frac {1}{\log ^2\left (\frac {2}{x^2}\right )} \, dx}{18 e^3}-\frac {\left (2 \log ^3(3)\right ) \int \frac {1}{\log ^3\left (\frac {2}{x^2}\right )} \, dx}{9 e^3}\\ &=\frac {e^{-18+3 x} x^4}{81 \log ^4\left (\frac {2}{x^2}\right )}-\frac {4 e^{-13+2 x} x^3 \log (3)}{27 \log ^4\left (\frac {2}{x^2}\right )}+\frac {2 e^{-8+x} x^2 \log ^2(3)}{3 \log ^4\left (\frac {2}{x^2}\right )}-\frac {4 x \log ^3(3)}{3 e^3 \log ^4\left (\frac {2}{x^2}\right )}+\frac {e^{2-x} \log ^4(3)}{\log ^4\left (\frac {2}{x^2}\right )}-\frac {x \log ^3(3)}{36 e^3 \log \left (\frac {2}{x^2}\right )}+\frac {\log ^3(3) \int \frac {1}{\log \left (\frac {2}{x^2}\right )} \, dx}{36 e^3}+\frac {\log ^3(3) \int \frac {1}{\log ^2\left (\frac {2}{x^2}\right )} \, dx}{18 e^3}\\ &=\frac {e^{-18+3 x} x^4}{81 \log ^4\left (\frac {2}{x^2}\right )}-\frac {4 e^{-13+2 x} x^3 \log (3)}{27 \log ^4\left (\frac {2}{x^2}\right )}+\frac {2 e^{-8+x} x^2 \log ^2(3)}{3 \log ^4\left (\frac {2}{x^2}\right )}-\frac {4 x \log ^3(3)}{3 e^3 \log ^4\left (\frac {2}{x^2}\right )}+\frac {e^{2-x} \log ^4(3)}{\log ^4\left (\frac {2}{x^2}\right )}-\frac {\log ^3(3) \int \frac {1}{\log \left (\frac {2}{x^2}\right )} \, dx}{36 e^3}-\frac {\left (\sqrt {\frac {1}{x^2}} x \log ^3(3)\right ) \operatorname {Subst}\left (\int \frac {e^{-x/2}}{x} \, dx,x,\log \left (\frac {2}{x^2}\right )\right )}{36 \sqrt {2} e^3}\\ &=-\frac {\sqrt {\frac {1}{x^2}} x \text {Ei}\left (-\frac {1}{2} \log \left (\frac {2}{x^2}\right )\right ) \log ^3(3)}{36 \sqrt {2} e^3}+\frac {e^{-18+3 x} x^4}{81 \log ^4\left (\frac {2}{x^2}\right )}-\frac {4 e^{-13+2 x} x^3 \log (3)}{27 \log ^4\left (\frac {2}{x^2}\right )}+\frac {2 e^{-8+x} x^2 \log ^2(3)}{3 \log ^4\left (\frac {2}{x^2}\right )}-\frac {4 x \log ^3(3)}{3 e^3 \log ^4\left (\frac {2}{x^2}\right )}+\frac {e^{2-x} \log ^4(3)}{\log ^4\left (\frac {2}{x^2}\right )}+\frac {\left (\sqrt {\frac {1}{x^2}} x \log ^3(3)\right ) \operatorname {Subst}\left (\int \frac {e^{-x/2}}{x} \, dx,x,\log \left (\frac {2}{x^2}\right )\right )}{36 \sqrt {2} e^3}\\ &=\frac {e^{-18+3 x} x^4}{81 \log ^4\left (\frac {2}{x^2}\right )}-\frac {4 e^{-13+2 x} x^3 \log (3)}{27 \log ^4\left (\frac {2}{x^2}\right )}+\frac {2 e^{-8+x} x^2 \log ^2(3)}{3 \log ^4\left (\frac {2}{x^2}\right )}-\frac {4 x \log ^3(3)}{3 e^3 \log ^4\left (\frac {2}{x^2}\right )}+\frac {e^{2-x} \log ^4(3)}{\log ^4\left (\frac {2}{x^2}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.95, size = 34, normalized size = 1.03 \begin {gather*} \frac {e^{-18-x} \left (e^x x-3 e^5 \log (3)\right )^4}{81 \log ^4\left (\frac {2}{x^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2 - x)*(E^(-5 + x)*x - 3*Log[3])^4*(8*E^(-5 + x)*x - 24*Log[3] + (E^(-5 + x)*(4*x + 3*x^2) + 3*x
*Log[3])*Log[2/x^2]))/(81*(E^(-5 + x)*x^2 - 3*x*Log[3])*Log[2/x^2]^5),x]

[Out]

(E^(-18 - x)*(E^x*x - 3*E^5*Log[3])^4)/(81*Log[2/x^2]^4)

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fricas [B]  time = 0.68, size = 72, normalized size = 2.18 \begin {gather*} \frac {{\left (x^{4} e^{\left (4 \, x - 20\right )} - 12 \, x^{3} e^{\left (3 \, x - 15\right )} \log \relax (3) + 54 \, x^{2} e^{\left (2 \, x - 10\right )} \log \relax (3)^{2} - 108 \, x e^{\left (x - 5\right )} \log \relax (3)^{3} + 81 \, \log \relax (3)^{4}\right )} e^{\left (-x + 2\right )}}{81 \, \log \left (\frac {2}{x^{2}}\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^2+4*x)*exp(x-5)+3*x*log(3))*log(2/x^2)+8*x*exp(x-5)-24*log(3))*exp(4*log(1/3*(x*exp(x-5)-3*lo
g(3))/log(2/x^2))+2-x)/(x^2*exp(x-5)-3*x*log(3))/log(2/x^2),x, algorithm="fricas")

[Out]

1/81*(x^4*e^(4*x - 20) - 12*x^3*e^(3*x - 15)*log(3) + 54*x^2*e^(2*x - 10)*log(3)^2 - 108*x*e^(x - 5)*log(3)^3
+ 81*log(3)^4)*e^(-x + 2)/log(2/x^2)^4

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giac [B]  time = 2.65, size = 256, normalized size = 7.76 \begin {gather*} \frac {x^{4} e^{\left (4 \, x\right )} - 12 \, x^{3} e^{\left (3 \, x + 5\right )} \log \relax (3) + 54 \, x^{2} e^{\left (2 \, x + 10\right )} \log \relax (3)^{2} - 108 \, x e^{\left (x + 15\right )} \log \relax (3)^{3} + 81 \, e^{20} \log \relax (3)^{4}}{81 \, {\left (e^{\left (x + 18\right )} \log \relax (2)^{4} - 8 \, e^{\left (x + 18\right )} \log \relax (2)^{3} \log \relax (x) + 24 \, e^{\left (x + 18\right )} \log \relax (2)^{2} \log \relax (x)^{2} - 32 \, e^{\left (x + 18\right )} \log \relax (2) \log \relax (x)^{3} + 16 \, e^{\left (x + 18\right )} \log \relax (x)^{4} - 8 \, e^{\left (x + 18\right )} \log \relax (2)^{3} \log \left (\mathrm {sgn}\relax (x)\right ) + 48 \, e^{\left (x + 18\right )} \log \relax (2)^{2} \log \relax (x) \log \left (\mathrm {sgn}\relax (x)\right ) - 96 \, e^{\left (x + 18\right )} \log \relax (2) \log \relax (x)^{2} \log \left (\mathrm {sgn}\relax (x)\right ) + 64 \, e^{\left (x + 18\right )} \log \relax (x)^{3} \log \left (\mathrm {sgn}\relax (x)\right ) + 24 \, e^{\left (x + 18\right )} \log \relax (2)^{2} \log \left (\mathrm {sgn}\relax (x)\right )^{2} - 96 \, e^{\left (x + 18\right )} \log \relax (2) \log \relax (x) \log \left (\mathrm {sgn}\relax (x)\right )^{2} + 96 \, e^{\left (x + 18\right )} \log \relax (x)^{2} \log \left (\mathrm {sgn}\relax (x)\right )^{2} - 32 \, e^{\left (x + 18\right )} \log \relax (2) \log \left (\mathrm {sgn}\relax (x)\right )^{3} + 64 \, e^{\left (x + 18\right )} \log \relax (x) \log \left (\mathrm {sgn}\relax (x)\right )^{3} + 16 \, e^{\left (x + 18\right )} \log \left (\mathrm {sgn}\relax (x)\right )^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^2+4*x)*exp(x-5)+3*x*log(3))*log(2/x^2)+8*x*exp(x-5)-24*log(3))*exp(4*log(1/3*(x*exp(x-5)-3*lo
g(3))/log(2/x^2))+2-x)/(x^2*exp(x-5)-3*x*log(3))/log(2/x^2),x, algorithm="giac")

[Out]

1/81*(x^4*e^(4*x) - 12*x^3*e^(3*x + 5)*log(3) + 54*x^2*e^(2*x + 10)*log(3)^2 - 108*x*e^(x + 15)*log(3)^3 + 81*
e^20*log(3)^4)/(e^(x + 18)*log(2)^4 - 8*e^(x + 18)*log(2)^3*log(x) + 24*e^(x + 18)*log(2)^2*log(x)^2 - 32*e^(x
+ 18)*log(2)*log(x)^3 + 16*e^(x + 18)*log(x)^4 - 8*e^(x + 18)*log(2)^3*log(sgn(x)) + 48*e^(x + 18)*log(2)^2*l
og(x)*log(sgn(x)) - 96*e^(x + 18)*log(2)*log(x)^2*log(sgn(x)) + 64*e^(x + 18)*log(x)^3*log(sgn(x)) + 24*e^(x +
18)*log(2)^2*log(sgn(x))^2 - 96*e^(x + 18)*log(2)*log(x)*log(sgn(x))^2 + 96*e^(x + 18)*log(x)^2*log(sgn(x))^2
- 32*e^(x + 18)*log(2)*log(sgn(x))^3 + 64*e^(x + 18)*log(x)*log(sgn(x))^3 + 16*e^(x + 18)*log(sgn(x))^4)

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (\left (3 x^{2}+4 x \right ) {\mathrm e}^{x -5}+3 x \ln \relax (3)\right ) \ln \left (\frac {2}{x^{2}}\right )+8 x \,{\mathrm e}^{x -5}-24 \ln \relax (3)\right ) {\mathrm e}^{4 \ln \left (\frac {x \,{\mathrm e}^{x -5}-3 \ln \relax (3)}{3 \ln \left (\frac {2}{x^{2}}\right )}\right )+2-x}}{\left (x^{2} {\mathrm e}^{x -5}-3 x \ln \relax (3)\right ) \ln \left (\frac {2}{x^{2}}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((3*x^2+4*x)*exp(x-5)+3*x*ln(3))*ln(2/x^2)+8*x*exp(x-5)-24*ln(3))*exp(4*ln(1/3*(x*exp(x-5)-3*ln(3))/ln(2/
x^2))+2-x)/(x^2*exp(x-5)-3*x*ln(3))/ln(2/x^2),x)

[Out]

int((((3*x^2+4*x)*exp(x-5)+3*x*ln(3))*ln(2/x^2)+8*x*exp(x-5)-24*ln(3))*exp(4*ln(1/3*(x*exp(x-5)-3*ln(3))/ln(2/
x^2))+2-x)/(x^2*exp(x-5)-3*x*ln(3))/ln(2/x^2),x)

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maxima [B]  time = 0.53, size = 108, normalized size = 3.27 \begin {gather*} \frac {x^{4} e^{\left (3 \, x\right )} - 12 \, x^{3} e^{\left (2 \, x + 5\right )} \log \relax (3) + 54 \, x^{2} e^{\left (x + 10\right )} \log \relax (3)^{2} - 108 \, x e^{15} \log \relax (3)^{3} + 81 \, e^{\left (-x + 20\right )} \log \relax (3)^{4}}{81 \, {\left (e^{18} \log \relax (2)^{4} - 8 \, e^{18} \log \relax (2)^{3} \log \relax (x) + 24 \, e^{18} \log \relax (2)^{2} \log \relax (x)^{2} - 32 \, e^{18} \log \relax (2) \log \relax (x)^{3} + 16 \, e^{18} \log \relax (x)^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^2+4*x)*exp(x-5)+3*x*log(3))*log(2/x^2)+8*x*exp(x-5)-24*log(3))*exp(4*log(1/3*(x*exp(x-5)-3*lo
g(3))/log(2/x^2))+2-x)/(x^2*exp(x-5)-3*x*log(3))/log(2/x^2),x, algorithm="maxima")

[Out]

1/81*(x^4*e^(3*x) - 12*x^3*e^(2*x + 5)*log(3) + 54*x^2*e^(x + 10)*log(3)^2 - 108*x*e^15*log(3)^3 + 81*e^(-x +
20)*log(3)^4)/(e^18*log(2)^4 - 8*e^18*log(2)^3*log(x) + 24*e^18*log(2)^2*log(x)^2 - 32*e^18*log(2)*log(x)^3 +
16*e^18*log(x)^4)

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mupad [B]  time = 4.42, size = 1729, normalized size = 52.39 result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(4*log(-(log(3) - (x*exp(x - 5))/3)/log(2/x^2)) - x + 2)*(8*x*exp(x - 5) - 24*log(3) + log(2/x^2)*(ex
p(x - 5)*(4*x + 3*x^2) + 3*x*log(3))))/(log(2/x^2)*(3*x*log(3) - x^2*exp(x - 5))),x)

[Out]

exp(x - 5)*((x^2*exp(-3)*log(3)^2)/36 + (65*x^3*exp(-3)*log(3)^2)/576 + (55*x^4*exp(-3)*log(3)^2)/576 + (7*x^5
*exp(-3)*log(3)^2)/288 + (x^6*exp(-3)*log(3)^2)/576) - ((x*exp(2 - x)*(4*exp(x - 5)*log(27)^3 - 432*exp(x - 5)
*log(3)^3 - 324*log(3)^4 + log(27)^4 + 12*x^3*exp(4*x - 20) + 9*x^4*exp(4*x - 20) + 432*x*exp(2*x - 10)*log(3)
^2 - 144*x^2*exp(3*x - 15)*log(3) - 96*x^3*exp(3*x - 15)*log(3) - 12*x*exp(2*x - 10)*log(27)^2 + 12*x^2*exp(3*
x - 15)*log(27) + 8*x^3*exp(3*x - 15)*log(27) + 216*x^2*exp(2*x - 10)*log(3)^2 - 6*x^2*exp(2*x - 10)*log(27)^2
))/1944 + (x*exp(2 - x)*log(2/x^2)*(x*log(27)^4 - 4*exp(x - 5)*log(27)^3 - log(27)^4 + 16*x^3*exp(4*x - 20) +
27*x^4*exp(4*x - 20) + 9*x^5*exp(4*x - 20) + 24*x*exp(2*x - 10)*log(27)^2 - 36*x^2*exp(3*x - 15)*log(27) - 56*
x^3*exp(3*x - 15)*log(27) - 16*x^4*exp(3*x - 15)*log(27) + 30*x^2*exp(2*x - 10)*log(27)^2 + 6*x^3*exp(2*x - 10
)*log(27)^2))/3888)/log(2/x^2)^3 + (exp(2 - x)*(log(3)^4 + (x^4*exp(4*x - 20))/81 - (4*x^3*exp(3*x - 15)*log(3
))/27 + (2*x^2*exp(2*x - 10)*log(3)^2)/3 - (4*x*exp(x - 5)*log(3)^3)/3) - (x*exp(2 - x)*log(2/x^2)*(log(27) -
x*exp(x - 5))^3*(4*exp(x - 5) + log(27) + 3*x*exp(x - 5)))/648)/log(2/x^2)^4 - ((x*exp(2 - x)*(3*x^2*log(27)^4
- 324*x^2*log(3)^4 - 432*exp(x - 5)*log(3)^3 + 12*exp(x - 5)*log(27)^3 + 972*x*log(3)^4 - 9*x*log(27)^4 - 324
*log(3)^4 + 3*log(27)^4 + 64*x^3*exp(4*x - 20) + 183*x^4*exp(4*x - 20) + 135*x^5*exp(4*x - 20) + 27*x^6*exp(4*
x - 20) + 1728*x*exp(2*x - 10)*log(3)^2 - 1296*x^2*exp(3*x - 15)*log(3) - 3552*x^3*exp(3*x - 15)*log(3) - 2304
*x^4*exp(3*x - 15)*log(3) - 384*x^5*exp(3*x - 15)*log(3) - 144*x*exp(2*x - 10)*log(27)^2 + 324*x^2*exp(3*x - 1
5)*log(27) + 888*x^3*exp(3*x - 15)*log(27) + 576*x^4*exp(3*x - 15)*log(27) + 96*x^5*exp(3*x - 15)*log(27) + 41
04*x^2*exp(2*x - 10)*log(3)^2 + 1944*x^3*exp(2*x - 10)*log(3)^2 + 216*x^4*exp(2*x - 10)*log(3)^2 - 342*x^2*exp
(2*x - 10)*log(27)^2 - 162*x^3*exp(2*x - 10)*log(27)^2 - 18*x^4*exp(2*x - 10)*log(27)^2))/15552 + (x*exp(2 - x
)*log(2/x^2)*(x^3*log(27)^4 - 6*x^2*log(27)^4 - 4*exp(x - 5)*log(27)^3 + 7*x*log(27)^4 - log(27)^4 + 256*x^3*e
xp(4*x - 20) + 1107*x^4*exp(4*x - 20) + 1359*x^5*exp(4*x - 20) + 594*x^6*exp(4*x - 20) + 81*x^7*exp(4*x - 20)
+ 96*x*exp(2*x - 10)*log(27)^2 - 324*x^2*exp(3*x - 15)*log(27) - 1400*x^3*exp(3*x - 15)*log(27) - 1552*x^4*exp
(3*x - 15)*log(27) - 576*x^5*exp(3*x - 15)*log(27) - 64*x^6*exp(3*x - 15)*log(27) + 390*x^2*exp(2*x - 10)*log(
27)^2 + 330*x^3*exp(2*x - 10)*log(27)^2 + 84*x^4*exp(2*x - 10)*log(27)^2 + 6*x^5*exp(2*x - 10)*log(27)^2))/311
04)/log(2/x^2) + exp(3*x - 15)*((2*x^4*exp(-3))/243 + (41*x^5*exp(-3))/1152 + (151*x^6*exp(-3))/3456 + (11*x^7
*exp(-3))/576 + (x^8*exp(-3))/384) - exp(2*x - 10)*((x^3*exp(-3)*log(3))/32 + (175*x^4*exp(-3)*log(3))/1296 +
(97*x^5*exp(-3)*log(3))/648 + (x^6*exp(-3)*log(3))/18 + (x^7*exp(-3)*log(3))/162) - exp(5 - x)*((x*exp(-3)*log
(3)^4)/384 - (7*x^2*exp(-3)*log(3)^4)/384 + (x^3*exp(-3)*log(3)^4)/64 - (x^4*exp(-3)*log(3)^4)/384) + ((x*exp(
2 - x)*(4*exp(x - 5)*log(27)^3 - 216*exp(x - 5)*log(3)^3 + 162*x*log(3)^4 - x*log(27)^4 - 162*log(3)^4 + log(2
7)^4 + 16*x^3*exp(4*x - 20) + 27*x^4*exp(4*x - 20) + 9*x^5*exp(4*x - 20) + 432*x*exp(2*x - 10)*log(3)^2 - 216*
x^2*exp(3*x - 15)*log(3) - 336*x^3*exp(3*x - 15)*log(3) - 96*x^4*exp(3*x - 15)*log(3) - 24*x*exp(2*x - 10)*log
(27)^2 + 36*x^2*exp(3*x - 15)*log(27) + 56*x^3*exp(3*x - 15)*log(27) + 16*x^4*exp(3*x - 15)*log(27) + 540*x^2*
exp(2*x - 10)*log(3)^2 + 108*x^3*exp(2*x - 10)*log(3)^2 - 30*x^2*exp(2*x - 10)*log(27)^2 - 6*x^3*exp(2*x - 10)
*log(27)^2))/3888 + (x*exp(2 - x)*log(2/x^2)*(3*x*log(27)^4 - 4*exp(x - 5)*log(27)^3 - x^2*log(27)^4 - log(27)
^4 + 64*x^3*exp(4*x - 20) + 183*x^4*exp(4*x - 20) + 135*x^5*exp(4*x - 20) + 27*x^6*exp(4*x - 20) + 48*x*exp(2*
x - 10)*log(27)^2 - 108*x^2*exp(3*x - 15)*log(27) - 296*x^3*exp(3*x - 15)*log(27) - 192*x^4*exp(3*x - 15)*log(
27) - 32*x^5*exp(3*x - 15)*log(27) + 114*x^2*exp(2*x - 10)*log(27)^2 + 54*x^3*exp(2*x - 10)*log(27)^2 + 6*x^4*
exp(2*x - 10)*log(27)^2))/15552)/log(2/x^2)^2 - (x*exp(-3)*log(3)^3)/288

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sympy [B]  time = 0.76, size = 136, normalized size = 4.12 \begin {gather*} - \frac {4 x \log {\relax (3 )}^{3}}{3 e^{3} \log {\left (\frac {2}{x^{2}} \right )}^{4}} + \frac {81 x^{4} e^{15} e^{3 x - 6} \log {\left (\frac {2}{x^{2}} \right )}^{12} - 972 x^{3} e^{18} e^{2 x - 4} \log {\relax (3 )} \log {\left (\frac {2}{x^{2}} \right )}^{12} + 4374 x^{2} e^{21} e^{x - 2} \log {\relax (3 )}^{2} \log {\left (\frac {2}{x^{2}} \right )}^{12} + 6561 e^{27} e^{2 - x} \log {\relax (3 )}^{4} \log {\left (\frac {2}{x^{2}} \right )}^{12}}{6561 e^{27} \log {\left (\frac {2}{x^{2}} \right )}^{16}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x**2+4*x)*exp(x-5)+3*x*ln(3))*ln(2/x**2)+8*x*exp(x-5)-24*ln(3))*exp(4*ln(1/3*(x*exp(x-5)-3*ln(3
))/ln(2/x**2))+2-x)/(x**2*exp(x-5)-3*x*ln(3))/ln(2/x**2),x)

[Out]

-4*x*exp(-3)*log(3)**3/(3*log(2/x**2)**4) + (81*x**4*exp(15)*exp(3*x - 6)*log(2/x**2)**12 - 972*x**3*exp(18)*e
xp(2*x - 4)*log(3)*log(2/x**2)**12 + 4374*x**2*exp(21)*exp(x - 2)*log(3)**2*log(2/x**2)**12 + 6561*exp(27)*exp
(2 - x)*log(3)**4*log(2/x**2)**12)*exp(-27)/(6561*log(2/x**2)**16)

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