3.83.11 \(\int \frac {e^4 x^2+e^4 (8 e x+5 e^2 x^2)+e^6 x^2 \log (4)-e^6 x^2 \log (3 e^2)}{e^2 x^2+e^2 (8 e x+10 e^2 x^2)+e^4 (16+40 e x+25 e^2 x^2)+(2 e^4 x^2+e^4 (8 e x+10 e^2 x^2)) \log (4)+e^6 x^2 \log ^2(4)+(-2 e^4 x^2+e^4 (-8 e x-10 e^2 x^2)-2 e^6 x^2 \log (4)) \log (3 e^2)+e^6 x^2 \log ^2(3 e^2)} \, dx\)

Optimal. Leaf size=27 \[ \frac {x}{5+\frac {1}{e^2}+\frac {4}{e x}+\log (4)-\log \left (3 e^2\right )} \]

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Rubi [B]  time = 0.19, antiderivative size = 59, normalized size of antiderivative = 2.19, number of steps used = 11, number of rules used = 5, integrand size = 192, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6, 1984, 27, 12, 683} \begin {gather*} \frac {e^2 x}{1+e^2 \left (3+\log \left (\frac {4}{3}\right )\right )}+\frac {16 e^4}{\left (1+e^2 \left (3+\log \left (\frac {4}{3}\right )\right )\right )^2 \left (x \left (1+e^2 \left (3+\log \left (\frac {4}{3}\right )\right )\right )+4 e\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^4*x^2 + E^4*(8*E*x + 5*E^2*x^2) + E^6*x^2*Log[4] - E^6*x^2*Log[3*E^2])/(E^2*x^2 + E^2*(8*E*x + 10*E^2*x
^2) + E^4*(16 + 40*E*x + 25*E^2*x^2) + (2*E^4*x^2 + E^4*(8*E*x + 10*E^2*x^2))*Log[4] + E^6*x^2*Log[4]^2 + (-2*
E^4*x^2 + E^4*(-8*E*x - 10*E^2*x^2) - 2*E^6*x^2*Log[4])*Log[3*E^2] + E^6*x^2*Log[3*E^2]^2),x]

[Out]

(E^2*x)/(1 + E^2*(3 + Log[4/3])) + (16*E^4)/((1 + E^2*(3 + Log[4/3]))^2*(4*E + x*(1 + E^2*(3 + Log[4/3]))))

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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 683

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

Rule 1984

Int[(u_)^(p_.)*(v_)^(q_.), x_Symbol] :> Int[ExpandToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{p, q}, x] &&
QuadraticQ[{u, v}, x] &&  !QuadraticMatchQ[{u, v}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^4 \left (8 e x+5 e^2 x^2\right )+x^2 \left (e^4+e^6 \log (4)\right )-e^6 x^2 \log \left (3 e^2\right )}{e^2 x^2+e^2 \left (8 e x+10 e^2 x^2\right )+e^4 \left (16+40 e x+25 e^2 x^2\right )+\left (2 e^4 x^2+e^4 \left (8 e x+10 e^2 x^2\right )\right ) \log (4)+e^6 x^2 \log ^2(4)+\left (-2 e^4 x^2+e^4 \left (-8 e x-10 e^2 x^2\right )-2 e^6 x^2 \log (4)\right ) \log \left (3 e^2\right )+e^6 x^2 \log ^2\left (3 e^2\right )} \, dx\\ &=\int \frac {e^4 \left (8 e x+5 e^2 x^2\right )+x^2 \left (e^4+e^6 \log (4)-e^6 \log \left (3 e^2\right )\right )}{e^2 x^2+e^2 \left (8 e x+10 e^2 x^2\right )+e^4 \left (16+40 e x+25 e^2 x^2\right )+\left (2 e^4 x^2+e^4 \left (8 e x+10 e^2 x^2\right )\right ) \log (4)+e^6 x^2 \log ^2(4)+\left (-2 e^4 x^2+e^4 \left (-8 e x-10 e^2 x^2\right )-2 e^6 x^2 \log (4)\right ) \log \left (3 e^2\right )+e^6 x^2 \log ^2\left (3 e^2\right )} \, dx\\ &=\int \frac {e^4 \left (8 e x+5 e^2 x^2\right )+x^2 \left (e^4+e^6 \log (4)-e^6 \log \left (3 e^2\right )\right )}{e^2 \left (8 e x+10 e^2 x^2\right )+e^4 \left (16+40 e x+25 e^2 x^2\right )+\left (2 e^4 x^2+e^4 \left (8 e x+10 e^2 x^2\right )\right ) \log (4)+x^2 \left (e^2+e^6 \log ^2(4)\right )+\left (-2 e^4 x^2+e^4 \left (-8 e x-10 e^2 x^2\right )-2 e^6 x^2 \log (4)\right ) \log \left (3 e^2\right )+e^6 x^2 \log ^2\left (3 e^2\right )} \, dx\\ &=\int \frac {e^4 \left (8 e x+5 e^2 x^2\right )+x^2 \left (e^4+e^6 \log (4)-e^6 \log \left (3 e^2\right )\right )}{e^2 \left (8 e x+10 e^2 x^2\right )+e^4 \left (16+40 e x+25 e^2 x^2\right )+\left (2 e^4 x^2+e^4 \left (8 e x+10 e^2 x^2\right )\right ) \log (4)+\left (-2 e^4 x^2+e^4 \left (-8 e x-10 e^2 x^2\right )-2 e^6 x^2 \log (4)\right ) \log \left (3 e^2\right )+x^2 \left (e^2+e^6 \log ^2(4)+e^6 \log ^2\left (3 e^2\right )\right )} \, dx\\ &=\int \frac {8 e^5 x+e^4 x^2 \left (1+e^2 \left (3+\log \left (\frac {4}{3}\right )\right )\right )}{16 e^4+8 e^3 x \left (1+e^2 \left (3+\log \left (\frac {4}{3}\right )\right )\right )+e^2 x^2 \left (1+e^2 \left (3+\log \left (\frac {4}{3}\right )\right )\right )^2} \, dx\\ &=\int \frac {8 e^5 x+e^4 x^2 \left (1+e^2 \left (3+\log \left (\frac {4}{3}\right )\right )\right )}{e^2 \left (4 e+x+3 e^2 x+e^2 x \log \left (\frac {4}{3}\right )\right )^2} \, dx\\ &=\int \frac {8 e^5 x+e^4 x^2 \left (1+e^2 \left (3+\log \left (\frac {4}{3}\right )\right )\right )}{e^2 \left (4 e+\left (1+3 e^2\right ) x+e^2 x \log \left (\frac {4}{3}\right )\right )^2} \, dx\\ &=\int \frac {8 e^5 x+e^4 x^2 \left (1+e^2 \left (3+\log \left (\frac {4}{3}\right )\right )\right )}{e^2 \left (4 e+x \left (1+3 e^2+e^2 \log \left (\frac {4}{3}\right )\right )\right )^2} \, dx\\ &=\frac {\int \frac {8 e^5 x+e^4 x^2 \left (1+e^2 \left (3+\log \left (\frac {4}{3}\right )\right )\right )}{\left (4 e+x \left (1+3 e^2+e^2 \log \left (\frac {4}{3}\right )\right )\right )^2} \, dx}{e^2}\\ &=\frac {\int \left (\frac {e^4}{1+e^2 \left (3+\log \left (\frac {4}{3}\right )\right )}+\frac {16 e^6}{\left (-1-e^2 \left (3+\log \left (\frac {4}{3}\right )\right )\right ) \left (4 e+x \left (1+e^2 \left (3+\log \left (\frac {4}{3}\right )\right )\right )\right )^2}\right ) \, dx}{e^2}\\ &=\frac {e^2 x}{1+e^2 \left (3+\log \left (\frac {4}{3}\right )\right )}+\frac {16 e^4}{\left (1+e^2 \left (3+\log \left (\frac {4}{3}\right )\right )\right )^2 \left (4 e+x \left (1+e^2 \left (3+\log \left (\frac {4}{3}\right )\right )\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.04, size = 88, normalized size = 3.26 \begin {gather*} \frac {e^2 \left (8 e x+x^2+8 e^3 x \left (3+\log \left (\frac {4}{3}\right )\right )+e^4 x^2 \left (3+\log \left (\frac {4}{3}\right )\right )^2+2 e^2 \left (16+x^2 \left (3+\log \left (\frac {4}{3}\right )\right )\right )\right )}{\left (1+e^2 \left (3+\log \left (\frac {4}{3}\right )\right )\right )^2 \left (4 e+x+e^2 x \left (3+\log \left (\frac {4}{3}\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^4*x^2 + E^4*(8*E*x + 5*E^2*x^2) + E^6*x^2*Log[4] - E^6*x^2*Log[3*E^2])/(E^2*x^2 + E^2*(8*E*x + 10
*E^2*x^2) + E^4*(16 + 40*E*x + 25*E^2*x^2) + (2*E^4*x^2 + E^4*(8*E*x + 10*E^2*x^2))*Log[4] + E^6*x^2*Log[4]^2
+ (-2*E^4*x^2 + E^4*(-8*E*x - 10*E^2*x^2) - 2*E^6*x^2*Log[4])*Log[3*E^2] + E^6*x^2*Log[3*E^2]^2),x]

[Out]

(E^2*(8*E*x + x^2 + 8*E^3*x*(3 + Log[4/3]) + E^4*x^2*(3 + Log[4/3])^2 + 2*E^2*(16 + x^2*(3 + Log[4/3]))))/((1
+ E^2*(3 + Log[4/3]))^2*(4*E + x + E^2*x*(3 + Log[4/3])))

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fricas [B]  time = 1.52, size = 295, normalized size = 10.93 \begin {gather*} -\frac {x^{2} e^{6} \log \relax (3)^{2} + 4 \, x^{2} e^{6} \log \relax (2)^{2} + 9 \, x^{2} e^{6} + x^{2} e^{2} + 12 \, x e^{5} + 2 \, {\left (3 \, x^{2} + 8\right )} e^{4} + 4 \, x e^{3} - 2 \, {\left (2 \, x^{2} e^{6} \log \relax (2) + 3 \, x^{2} e^{6} + x^{2} e^{4} + 2 \, x e^{5}\right )} \log \relax (3) + 4 \, {\left (3 \, x^{2} e^{6} + x^{2} e^{4} + 2 \, x e^{5}\right )} \log \relax (2)}{x e^{6} \log \relax (3)^{3} - 8 \, x e^{6} \log \relax (2)^{3} - {\left (6 \, x e^{6} \log \relax (2) + 9 \, x e^{6} + 3 \, x e^{4} + 4 \, e^{5}\right )} \log \relax (3)^{2} - 4 \, {\left (9 \, x e^{6} + 3 \, x e^{4} + 4 \, e^{5}\right )} \log \relax (2)^{2} - 27 \, x e^{6} - 27 \, x e^{4} - 9 \, x e^{2} + {\left (12 \, x e^{6} \log \relax (2)^{2} + 27 \, x e^{6} + 18 \, x e^{4} + 3 \, x e^{2} + 4 \, {\left (9 \, x e^{6} + 3 \, x e^{4} + 4 \, e^{5}\right )} \log \relax (2) + 24 \, e^{5} + 8 \, e^{3}\right )} \log \relax (3) - 2 \, {\left (27 \, x e^{6} + 18 \, x e^{4} + 3 \, x e^{2} + 24 \, e^{5} + 8 \, e^{3}\right )} \log \relax (2) - x - 36 \, e^{5} - 24 \, e^{3} - 4 \, e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2*exp(1)^2*exp(2)^2*log(3*exp(2))+2*x^2*exp(1)^2*exp(2)^2*log(2)+(5*x^2*exp(1)^2+8*x*exp(1))*exp
(2)^2+x^2*exp(1)^2*exp(2))/(x^2*exp(1)^2*exp(2)^2*log(3*exp(2))^2+(-4*x^2*exp(1)^2*exp(2)^2*log(2)+(-10*x^2*ex
p(1)^2-8*x*exp(1))*exp(2)^2-2*x^2*exp(1)^2*exp(2))*log(3*exp(2))+4*x^2*exp(1)^2*exp(2)^2*log(2)^2+2*((10*x^2*e
xp(1)^2+8*x*exp(1))*exp(2)^2+2*x^2*exp(1)^2*exp(2))*log(2)+(25*x^2*exp(1)^2+40*x*exp(1)+16)*exp(2)^2+(10*x^2*e
xp(1)^2+8*x*exp(1))*exp(2)+x^2*exp(1)^2),x, algorithm="fricas")

[Out]

-(x^2*e^6*log(3)^2 + 4*x^2*e^6*log(2)^2 + 9*x^2*e^6 + x^2*e^2 + 12*x*e^5 + 2*(3*x^2 + 8)*e^4 + 4*x*e^3 - 2*(2*
x^2*e^6*log(2) + 3*x^2*e^6 + x^2*e^4 + 2*x*e^5)*log(3) + 4*(3*x^2*e^6 + x^2*e^4 + 2*x*e^5)*log(2))/(x*e^6*log(
3)^3 - 8*x*e^6*log(2)^3 - (6*x*e^6*log(2) + 9*x*e^6 + 3*x*e^4 + 4*e^5)*log(3)^2 - 4*(9*x*e^6 + 3*x*e^4 + 4*e^5
)*log(2)^2 - 27*x*e^6 - 27*x*e^4 - 9*x*e^2 + (12*x*e^6*log(2)^2 + 27*x*e^6 + 18*x*e^4 + 3*x*e^2 + 4*(9*x*e^6 +
3*x*e^4 + 4*e^5)*log(2) + 24*e^5 + 8*e^3)*log(3) - 2*(27*x*e^6 + 18*x*e^4 + 3*x*e^2 + 24*e^5 + 8*e^3)*log(2)
- x - 36*e^5 - 24*e^3 - 4*e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2*exp(1)^2*exp(2)^2*log(3*exp(2))+2*x^2*exp(1)^2*exp(2)^2*log(2)+(5*x^2*exp(1)^2+8*x*exp(1))*exp
(2)^2+x^2*exp(1)^2*exp(2))/(x^2*exp(1)^2*exp(2)^2*log(3*exp(2))^2+(-4*x^2*exp(1)^2*exp(2)^2*log(2)+(-10*x^2*ex
p(1)^2-8*x*exp(1))*exp(2)^2-2*x^2*exp(1)^2*exp(2))*log(3*exp(2))+4*x^2*exp(1)^2*exp(2)^2*log(2)^2+2*((10*x^2*e
xp(1)^2+8*x*exp(1))*exp(2)^2+2*x^2*exp(1)^2*exp(2))*log(2)+(25*x^2*exp(1)^2+40*x*exp(1)+16)*exp(2)^2+(10*x^2*e
xp(1)^2+8*x*exp(1))*exp(2)+x^2*exp(1)^2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: (-8*exp(6)*ln(2)*exp(2)*exp(1)-40*exp(6)
*ln(2)*exp(1)*exp(4)+4*exp(6)*ln(3*exp(2))*exp(2)*exp(1)+20*exp(6)*ln(3*exp(2))*exp(1)*exp(4)+40*ln(2)*exp(2)*
exp(1)*exp(4)^2+8*ln(

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maple [A]  time = 0.54, size = 45, normalized size = 1.67

method	result	size
norman	\(\frac {x^{2} {\mathrm e} \,{\mathrm e}^{2}}{2 \,{\mathrm e}^{2} {\mathrm e} \ln \relax (2) x -{\mathrm e}^{2} {\mathrm e} \ln \relax (3) x +3 x \,{\mathrm e} \,{\mathrm e}^{2}+x \,{\mathrm e}+4 \,{\mathrm e}^{2}}\)	\(45\)
gosper	\(\frac {x^{2} {\mathrm e}^{3}}{2 \,{\mathrm e}^{2} {\mathrm e} \ln \relax (2) x -{\mathrm e}^{2} {\mathrm e} \ln \left (3 \,{\mathrm e}^{2}\right ) x +5 x \,{\mathrm e} \,{\mathrm e}^{2}+x \,{\mathrm e}+4 \,{\mathrm e}^{2}}\)	\(46\)
risch	\(\frac {x \,{\mathrm e}^{2}}{2 \,{\mathrm e}^{2} \ln \relax (2)-{\mathrm e}^{2} \ln \relax (3)+3 \,{\mathrm e}^{2}+1}+\frac {8 \,{\mathrm e}^{4}}{\left (2 \,{\mathrm e}^{2} \ln \relax (2)-{\mathrm e}^{2} \ln \relax (3)+3 \,{\mathrm e}^{2}+1\right )^{2} \left (x \,{\mathrm e}^{2} \ln \relax (2)-\frac {x \,{\mathrm e}^{2} \ln \relax (3)}{2}+\frac {3 \,{\mathrm e}^{2} x}{2}+2 \,{\mathrm e}+\frac {x}{2}\right )}\)	\(78\)
meijerg	\(\frac {4 \left (2 \ln \relax (2) {\mathrm e}^{6}-\ln \left (3 \,{\mathrm e}^{2}\right ) {\mathrm e}^{6}+5 \,{\mathrm e}^{6}+{\mathrm e}^{4}\right ) {\mathrm e} \left (\frac {x \,{\mathrm e}^{-1} \left (2 \,{\mathrm e}^{2} \ln \relax (2)-{\mathrm e}^{2} \ln \left (3 \,{\mathrm e}^{2}\right )+5 \,{\mathrm e}^{2}+1\right ) \left (\frac {3 x \,{\mathrm e}^{-1} \left (2 \,{\mathrm e}^{2} \ln \relax (2)-{\mathrm e}^{2} \ln \left (3 \,{\mathrm e}^{2}\right )+5 \,{\mathrm e}^{2}+1\right )}{4}+6\right )}{12+3 x \,{\mathrm e}^{-1} \left (2 \,{\mathrm e}^{2} \ln \relax (2)-{\mathrm e}^{2} \ln \left (3 \,{\mathrm e}^{2}\right )+5 \,{\mathrm e}^{2}+1\right )}-2 \ln \left (1+\frac {x \,{\mathrm e}^{-1} \left (2 \,{\mathrm e}^{2} \ln \relax (2)-{\mathrm e}^{2} \ln \left (3 \,{\mathrm e}^{2}\right )+5 \,{\mathrm e}^{2}+1\right )}{4}\right )\right )}{\left (4 \ln \relax (2)^{2} {\mathrm e}^{6}-4 \ln \relax (2) \ln \left (3 \,{\mathrm e}^{2}\right ) {\mathrm e}^{6}+\ln \left (3 \,{\mathrm e}^{2}\right )^{2} {\mathrm e}^{6}+4 \,{\mathrm e}^{4} \ln \relax (2)-2 \,{\mathrm e}^{4} \ln \left (3 \,{\mathrm e}^{2}\right )+20 \ln \relax (2) {\mathrm e}^{6}-10 \ln \left (3 \,{\mathrm e}^{2}\right ) {\mathrm e}^{6}+{\mathrm e}^{2}+10 \,{\mathrm e}^{4}+25 \,{\mathrm e}^{6}\right ) \left (2 \,{\mathrm e}^{2} \ln \relax (2)-{\mathrm e}^{2} \ln \left (3 \,{\mathrm e}^{2}\right )+5 \,{\mathrm e}^{2}+1\right )}+\frac {8 \,{\mathrm e}^{5} \left (-\frac {x \,{\mathrm e}^{-1} \left (2 \,{\mathrm e}^{2} \ln \relax (2)-{\mathrm e}^{2} \ln \left (3 \,{\mathrm e}^{2}\right )+5 \,{\mathrm e}^{2}+1\right )}{4 \left (1+\frac {x \,{\mathrm e}^{-1} \left (2 \,{\mathrm e}^{2} \ln \relax (2)-{\mathrm e}^{2} \ln \left (3 \,{\mathrm e}^{2}\right )+5 \,{\mathrm e}^{2}+1\right )}{4}\right )}+\ln \left (1+\frac {x \,{\mathrm e}^{-1} \left (2 \,{\mathrm e}^{2} \ln \relax (2)-{\mathrm e}^{2} \ln \left (3 \,{\mathrm e}^{2}\right )+5 \,{\mathrm e}^{2}+1\right )}{4}\right )\right )}{4 \ln \relax (2)^{2} {\mathrm e}^{6}-4 \ln \relax (2) \ln \left (3 \,{\mathrm e}^{2}\right ) {\mathrm e}^{6}+\ln \left (3 \,{\mathrm e}^{2}\right )^{2} {\mathrm e}^{6}+4 \,{\mathrm e}^{4} \ln \relax (2)-2 \,{\mathrm e}^{4} \ln \left (3 \,{\mathrm e}^{2}\right )+20 \ln \relax (2) {\mathrm e}^{6}-10 \ln \left (3 \,{\mathrm e}^{2}\right ) {\mathrm e}^{6}+{\mathrm e}^{2}+10 \,{\mathrm e}^{4}+25 \,{\mathrm e}^{6}}\)	\(401\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^2*exp(1)^2*exp(2)^2*ln(3*exp(2))+2*x^2*exp(1)^2*exp(2)^2*ln(2)+(5*x^2*exp(1)^2+8*x*exp(1))*exp(2)^2+x^
2*exp(1)^2*exp(2))/(x^2*exp(1)^2*exp(2)^2*ln(3*exp(2))^2+(-4*x^2*exp(1)^2*exp(2)^2*ln(2)+(-10*x^2*exp(1)^2-8*x
*exp(1))*exp(2)^2-2*x^2*exp(1)^2*exp(2))*ln(3*exp(2))+4*x^2*exp(1)^2*exp(2)^2*ln(2)^2+2*((10*x^2*exp(1)^2+8*x*
exp(1))*exp(2)^2+2*x^2*exp(1)^2*exp(2))*ln(2)+(25*x^2*exp(1)^2+40*x*exp(1)+16)*exp(2)^2+(10*x^2*exp(1)^2+8*x*e
xp(1))*exp(2)+x^2*exp(1)^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.37, size = 219, normalized size = 8.11 \begin {gather*} \frac {x e^{2}}{2 \, e^{2} \log \relax (2) - e^{2} \log \left (3 \, e^{2}\right ) + 5 \, e^{2} + 1} + \frac {16 \, e^{4}}{16 \, e^{5} \log \relax (2)^{2} + 4 \, e^{5} \log \left (3 \, e^{2}\right )^{2} + {\left (8 \, e^{6} \log \relax (2)^{3} - e^{6} \log \left (3 \, e^{2}\right )^{3} + 12 \, {\left (5 \, e^{6} + e^{4}\right )} \log \relax (2)^{2} + 3 \, {\left (2 \, e^{6} \log \relax (2) + 5 \, e^{6} + e^{4}\right )} \log \left (3 \, e^{2}\right )^{2} + 6 \, {\left (25 \, e^{6} + 10 \, e^{4} + e^{2}\right )} \log \relax (2) - 3 \, {\left (4 \, e^{6} \log \relax (2)^{2} + 4 \, {\left (5 \, e^{6} + e^{4}\right )} \log \relax (2) + 25 \, e^{6} + 10 \, e^{4} + e^{2}\right )} \log \left (3 \, e^{2}\right ) + 125 \, e^{6} + 75 \, e^{4} + 15 \, e^{2} + 1\right )} x + 16 \, {\left (5 \, e^{5} + e^{3}\right )} \log \relax (2) - 8 \, {\left (2 \, e^{5} \log \relax (2) + 5 \, e^{5} + e^{3}\right )} \log \left (3 \, e^{2}\right ) + 100 \, e^{5} + 40 \, e^{3} + 4 \, e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2*exp(1)^2*exp(2)^2*log(3*exp(2))+2*x^2*exp(1)^2*exp(2)^2*log(2)+(5*x^2*exp(1)^2+8*x*exp(1))*exp
(2)^2+x^2*exp(1)^2*exp(2))/(x^2*exp(1)^2*exp(2)^2*log(3*exp(2))^2+(-4*x^2*exp(1)^2*exp(2)^2*log(2)+(-10*x^2*ex
p(1)^2-8*x*exp(1))*exp(2)^2-2*x^2*exp(1)^2*exp(2))*log(3*exp(2))+4*x^2*exp(1)^2*exp(2)^2*log(2)^2+2*((10*x^2*e
xp(1)^2+8*x*exp(1))*exp(2)^2+2*x^2*exp(1)^2*exp(2))*log(2)+(25*x^2*exp(1)^2+40*x*exp(1)+16)*exp(2)^2+(10*x^2*e
xp(1)^2+8*x*exp(1))*exp(2)+x^2*exp(1)^2),x, algorithm="maxima")

[Out]

x*e^2/(2*e^2*log(2) - e^2*log(3*e^2) + 5*e^2 + 1) + 16*e^4/(16*e^5*log(2)^2 + 4*e^5*log(3*e^2)^2 + (8*e^6*log(
2)^3 - e^6*log(3*e^2)^3 + 12*(5*e^6 + e^4)*log(2)^2 + 3*(2*e^6*log(2) + 5*e^6 + e^4)*log(3*e^2)^2 + 6*(25*e^6
+ 10*e^4 + e^2)*log(2) - 3*(4*e^6*log(2)^2 + 4*(5*e^6 + e^4)*log(2) + 25*e^6 + 10*e^4 + e^2)*log(3*e^2) + 125*
e^6 + 75*e^4 + 15*e^2 + 1)*x + 16*(5*e^5 + e^3)*log(2) - 8*(2*e^5*log(2) + 5*e^5 + e^3)*log(3*e^2) + 100*e^5 +
40*e^3 + 4*e)

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mupad [B]  time = 148.24, size = 1870, normalized size = 69.26 result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*exp(4) + exp(4)*(8*x*exp(1) + 5*x^2*exp(2)) + 2*x^2*exp(6)*log(2) - x^2*log(3*exp(2))*exp(6))/(x^2*ex
p(2) - log(3*exp(2))*(2*x^2*exp(4) + exp(4)*(8*x*exp(1) + 10*x^2*exp(2)) + 4*x^2*exp(6)*log(2)) + 2*log(2)*(2*
x^2*exp(4) + exp(4)*(8*x*exp(1) + 10*x^2*exp(2))) + exp(2)*(8*x*exp(1) + 10*x^2*exp(2)) + exp(4)*(40*x*exp(1)
+ 25*x^2*exp(2) + 16) + x^2*log(3*exp(2))^2*exp(6) + 4*x^2*exp(6)*log(2)^2),x)

[Out]

(log(16*exp(2) + 8*x*exp(1) + 24*x*exp(3) + 6*x^2*exp(2) + 9*x^2*exp(4) + x^2 + 4*x^2*exp(4)*log(2)^2 + x^2*ex
p(4)*log(3)^2 - 8*x*exp(3)*log(3/4) - 2*x^2*exp(2)*log(3/4)*(3*exp(2) + 1) - 4*x^2*exp(4)*log(2)*log(3))*(128*
exp(7)*log(2)^4 + 8*exp(7)*log(3)^4 + 8*exp(7)*log(3/4)^4 + 192*exp(7)*log(2)^2*log(3)^2 - 64*exp(7)*log(2)^2*
log(3/4)^2 - 16*exp(7)*log(3)^2*log(3/4)^2 - 64*exp(7)*log(2)*log(3)^3 - 256*exp(7)*log(2)^3*log(3) + 64*exp(7
)*log(2)*log(3)*log(3/4)^2))/(2*(48*exp(2)*log(2)^2 - 4*log(2)*log(3) + 12*exp(2)*log(3)^2 + 216*exp(4)*log(2)
^2 + 54*exp(4)*log(3)^2 + 32*exp(4)*log(2)^4 + 432*exp(6)*log(2)^2 - 12*exp(2)*log(3/4)^2 + 2*exp(4)*log(3)^4
+ 108*exp(6)*log(3)^2 + 4*exp(2)*log(3/4)^3 + 192*exp(6)*log(2)^4 + 324*exp(8)*log(2)^2 - 54*exp(4)*log(3/4)^2
+ 12*exp(6)*log(3)^4 + 81*exp(8)*log(3)^2 + 36*exp(4)*log(3/4)^3 + 288*exp(8)*log(2)^4 - 4*exp(4)*log(3/4)^4
- 108*exp(6)*log(3/4)^2 + 18*exp(8)*log(3)^4 + 108*exp(6)*log(3/4)^3 + 64*exp(8)*log(2)^6 - 24*exp(6)*log(3/4)
^4 + exp(8)*log(3)^6 - 81*exp(8)*log(3/4)^2 + 108*exp(8)*log(3/4)^3 - 36*exp(8)*log(3/4)^4 + 4*log(2)^2 + log(
3)^2 - log(3/4)^2 + 48*exp(4)*log(2)^2*log(3)^2 + 288*exp(6)*log(2)^2*log(3)^2 + 8*exp(4)*log(2)^2*log(3/4)^2
+ 432*exp(8)*log(2)^2*log(3)^2 + 2*exp(4)*log(3)^2*log(3/4)^2 + 48*exp(6)*log(2)^2*log(3/4)^2 + 60*exp(8)*log(
2)^2*log(3)^4 - 160*exp(8)*log(2)^3*log(3)^3 + 240*exp(8)*log(2)^4*log(3)^2 + 16*exp(6)*log(2)^2*log(3/4)^3 +
12*exp(6)*log(3)^2*log(3/4)^2 + 4*exp(6)*log(3)^2*log(3/4)^3 + 72*exp(8)*log(2)^2*log(3/4)^2 + 48*exp(8)*log(2
)^2*log(3/4)^3 + 18*exp(8)*log(3)^2*log(3/4)^2 + 12*exp(8)*log(3)^2*log(3/4)^3 - 16*exp(8)*log(2)^4*log(3/4)^2
- exp(8)*log(3)^4*log(3/4)^2 - 48*exp(2)*log(2)*log(3) - 216*exp(4)*log(2)*log(3) - 432*exp(6)*log(2)*log(3)
- 324*exp(8)*log(2)*log(3) - 16*exp(4)*log(2)*log(3)^3 - 64*exp(4)*log(2)^3*log(3) - 16*exp(2)*log(2)^2*log(3/
4) - 4*exp(2)*log(3)^2*log(3/4) - 96*exp(6)*log(2)*log(3)^3 - 384*exp(6)*log(2)^3*log(3) - 144*exp(4)*log(2)^2
*log(3/4) - 36*exp(4)*log(3)^2*log(3/4) - 144*exp(8)*log(2)*log(3)^3 - 576*exp(8)*log(2)^3*log(3) - 432*exp(6)
*log(2)^2*log(3/4) - 108*exp(6)*log(3)^2*log(3/4) - 12*exp(8)*log(2)*log(3)^5 - 192*exp(8)*log(2)^5*log(3) - 6
4*exp(6)*log(2)^4*log(3/4) - 432*exp(8)*log(2)^2*log(3/4) - 4*exp(6)*log(3)^4*log(3/4) - 108*exp(8)*log(3)^2*l
og(3/4) - 192*exp(8)*log(2)^4*log(3/4) - 12*exp(8)*log(3)^4*log(3/4) + 16*exp(2)*log(2)*log(3)*log(3/4) + 144*
exp(4)*log(2)*log(3)*log(3/4) + 432*exp(6)*log(2)*log(3)*log(3/4) + 432*exp(8)*log(2)*log(3)*log(3/4) - 24*exp
(8)*log(2)^2*log(3)^2*log(3/4)^2 - 8*exp(4)*log(2)*log(3)*log(3/4)^2 - 48*exp(6)*log(2)*log(3)*log(3/4)^2 - 16
*exp(6)*log(2)*log(3)*log(3/4)^3 + 32*exp(6)*log(2)*log(3)^3*log(3/4) + 128*exp(6)*log(2)^3*log(3)*log(3/4) -
72*exp(8)*log(2)*log(3)*log(3/4)^2 - 48*exp(8)*log(2)*log(3)*log(3/4)^3 + 96*exp(8)*log(2)*log(3)^3*log(3/4) +
384*exp(8)*log(2)^3*log(3)*log(3/4) - 96*exp(6)*log(2)^2*log(3)^2*log(3/4) - 288*exp(8)*log(2)^2*log(3)^2*log
(3/4) + 8*exp(8)*log(2)*log(3)^3*log(3/4)^2 + 32*exp(8)*log(2)^3*log(3)*log(3/4)^2)) + (x*(exp(2) + 3*exp(4) -
exp(4)*log(3/4)))/(6*exp(2) + 9*exp(4) + 4*exp(4)*log(2)^2 + exp(4)*log(3)^2 - 2*exp(2)*log(3/4)*(3*exp(2) +
1) - 4*exp(4)*log(2)*log(3) + 1) + (4*atan((exp(-3)*(4*exp(1)*(4*log(2)^2 - 4*log(2)*log(3) + log(3)^2 - log(3
/4)^2)^(1/2) + 12*exp(3)*(4*log(2)^2 - 4*log(2)*log(3) + log(3)^2 - log(3/4)^2)^(1/2) + x*(4*log(2)^2 - 4*log(
2)*log(3) + log(3)^2 - log(3/4)^2)^(1/2) - 4*exp(3)*log(3/4)*(4*log(2)^2 - 4*log(2)*log(3) + log(3)^2 - log(3/
4)^2)^(1/2) + 6*x*exp(2)*(4*log(2)^2 - 4*log(2)*log(3) + log(3)^2 - log(3/4)^2)^(1/2) + 9*x*exp(4)*(4*log(2)^2
- 4*log(2)*log(3) + log(3)^2 - log(3/4)^2)^(1/2) + 4*x*exp(4)*log(2)^2*(4*log(2)^2 - 4*log(2)*log(3) + log(3)
^2 - log(3/4)^2)^(1/2) + x*exp(4)*log(3)^2*(4*log(2)^2 - 4*log(2)*log(3) + log(3)^2 - log(3/4)^2)^(1/2) - 2*x*
exp(2)*log(3/4)*(4*log(2)^2 - 4*log(2)*log(3) + log(3)^2 - log(3/4)^2)^(1/2) - 6*x*exp(4)*log(3/4)*(4*log(2)^2
- 4*log(2)*log(3) + log(3)^2 - log(3/4)^2)^(1/2) - 4*x*exp(4)*log(2)*log(3)*(4*log(2)^2 - 4*log(2)*log(3) + l
og(3)^2 - log(3/4)^2)^(1/2)))/(4*(4*log(2)*log(3) - 4*log(2)^2 - log(3)^2 + log(3/4)^2)))*exp(1)*(3*exp(2) - e
xp(2)*log(3/4) + 1)*(6*exp(2) + 9*exp(4) - 2*exp(2)*log(3/4) - 6*exp(4)*log(3/4) + 12*exp(4)*log(2)^2 + 3*exp(
4)*log(3)^2 - 2*exp(4)*log(3/4)^2 - 12*exp(4)*log(2)*log(3) + 1))/((4*log(2)^2 - 4*log(2)*log(3) + log(3)^2 -
log(3/4)^2)^(1/2)*(12*exp(2) + 54*exp(4) + 108*exp(6) + 81*exp(8) - 4*exp(2)*log(3/4) - 36*exp(4)*log(3/4) - 1
08*exp(6)*log(3/4) - 108*exp(8)*log(3/4) + 8*exp(4)*log(2)^2 + 2*exp(4)*log(3)^2 + 48*exp(6)*log(2)^2 + 12*exp
(6)*log(3)^2 + 72*exp(8)*log(2)^2 + 4*exp(4)*log(3/4)^2 + 18*exp(8)*log(3)^2 + 16*exp(8)*log(2)^4 + 24*exp(6)*
log(3/4)^2 + exp(8)*log(3)^4 + 36*exp(8)*log(3/4)^2 + 24*exp(8)*log(2)^2*log(3)^2 - 8*exp(4)*log(2)*log(3) - 4
8*exp(6)*log(2)*log(3) - 72*exp(8)*log(2)*log(3) - 8*exp(8)*log(2)*log(3)^3 - 32*exp(8)*log(2)^3*log(3) - 16*e
xp(6)*log(2)^2*log(3/4) - 4*exp(6)*log(3)^2*log(3/4) - 48*exp(8)*log(2)^2*log(3/4) - 12*exp(8)*log(3)^2*log(3/
4) + 16*exp(6)*log(2)*log(3)*log(3/4) + 48*exp(8)*log(2)*log(3)*log(3/4) + 1))

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sympy [B]  time = 0.88, size = 292, normalized size = 10.81 \begin {gather*} \frac {x e^{2}}{- e^{2} \log {\relax (3 )} + 1 + 2 e^{2} \log {\relax (2 )} + 3 e^{2}} + \frac {16 e^{4}}{x \left (- 27 e^{6} \log {\relax (3 )} - 36 e^{6} \log {\relax (2 )} \log {\relax (3 )} - 12 e^{6} \log {\relax (2 )}^{2} \log {\relax (3 )} - 18 e^{4} \log {\relax (3 )} - e^{6} \log {\relax (3 )}^{3} - 12 e^{4} \log {\relax (2 )} \log {\relax (3 )} - 3 e^{2} \log {\relax (3 )} + 1 + 6 e^{2} \log {\relax (2 )} + 9 e^{2} + 3 e^{4} \log {\relax (3 )}^{2} + 12 e^{4} \log {\relax (2 )}^{2} + 8 e^{6} \log {\relax (2 )}^{3} + 36 e^{4} \log {\relax (2 )} + 27 e^{4} + 6 e^{6} \log {\relax (2 )} \log {\relax (3 )}^{2} + 9 e^{6} \log {\relax (3 )}^{2} + 36 e^{6} \log {\relax (2 )}^{2} + 27 e^{6} + 54 e^{6} \log {\relax (2 )}\right ) - 24 e^{5} \log {\relax (3 )} - 16 e^{5} \log {\relax (2 )} \log {\relax (3 )} - 8 e^{3} \log {\relax (3 )} + 4 e + 16 e^{3} \log {\relax (2 )} + 24 e^{3} + 4 e^{5} \log {\relax (3 )}^{2} + 16 e^{5} \log {\relax (2 )}^{2} + 48 e^{5} \log {\relax (2 )} + 36 e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**2*exp(1)**2*exp(2)**2*ln(3*exp(2))+2*x**2*exp(1)**2*exp(2)**2*ln(2)+(5*x**2*exp(1)**2+8*x*exp(1
))*exp(2)**2+x**2*exp(1)**2*exp(2))/(x**2*exp(1)**2*exp(2)**2*ln(3*exp(2))**2+(-4*x**2*exp(1)**2*exp(2)**2*ln(
2)+(-10*x**2*exp(1)**2-8*x*exp(1))*exp(2)**2-2*x**2*exp(1)**2*exp(2))*ln(3*exp(2))+4*x**2*exp(1)**2*exp(2)**2*
ln(2)**2+2*((10*x**2*exp(1)**2+8*x*exp(1))*exp(2)**2+2*x**2*exp(1)**2*exp(2))*ln(2)+(25*x**2*exp(1)**2+40*x*ex
p(1)+16)*exp(2)**2+(10*x**2*exp(1)**2+8*x*exp(1))*exp(2)+x**2*exp(1)**2),x)

[Out]

x*exp(2)/(-exp(2)*log(3) + 1 + 2*exp(2)*log(2) + 3*exp(2)) + 16*exp(4)/(x*(-27*exp(6)*log(3) - 36*exp(6)*log(2
)*log(3) - 12*exp(6)*log(2)**2*log(3) - 18*exp(4)*log(3) - exp(6)*log(3)**3 - 12*exp(4)*log(2)*log(3) - 3*exp(
2)*log(3) + 1 + 6*exp(2)*log(2) + 9*exp(2) + 3*exp(4)*log(3)**2 + 12*exp(4)*log(2)**2 + 8*exp(6)*log(2)**3 + 3
6*exp(4)*log(2) + 27*exp(4) + 6*exp(6)*log(2)*log(3)**2 + 9*exp(6)*log(3)**2 + 36*exp(6)*log(2)**2 + 27*exp(6)
+ 54*exp(6)*log(2)) - 24*exp(5)*log(3) - 16*exp(5)*log(2)*log(3) - 8*exp(3)*log(3) + 4*E + 16*exp(3)*log(2) +
24*exp(3) + 4*exp(5)*log(3)**2 + 16*exp(5)*log(2)**2 + 48*exp(5)*log(2) + 36*exp(5))

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