3.103.88 \(\int \frac {8 x^3-10 e x^5+8 x^3 \log (9)+(4 x^3-4 e x^5+4 x^3 \log (9)) \log (-1+e x^2-\log (9))}{1-e x^2+\log (9)} \, dx\)

Optimal. Leaf size=18 \[ x^4 \left (2+\log \left (-1+e x^2-\log (9)\right )\right ) \]

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Rubi [A]  time = 0.25, antiderivative size = 22, normalized size of antiderivative = 1.22, number of steps used = 10, number of rules used = 7, integrand size = 64, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.109, Rules used = {6, 6725, 446, 77, 2454, 2395, 43} \begin {gather*} 2 x^4+x^4 \log \left (e x^2-1-\log (9)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8*x^3 - 10*E*x^5 + 8*x^3*Log[9] + (4*x^3 - 4*E*x^5 + 4*x^3*Log[9])*Log[-1 + E*x^2 - Log[9]])/(1 - E*x^2 +
Log[9]),x]

[Out]

2*x^4 + x^4*Log[-1 + E*x^2 - Log[9]]

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

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 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-10 e x^5+x^3 (8+8 \log (9))+\left (4 x^3-4 e x^5+4 x^3 \log (9)\right ) \log \left (-1+e x^2-\log (9)\right )}{1-e x^2+\log (9)} \, dx\\ &=\int \left (\frac {2 x^3 \left (-4+5 e x^2-4 \log (9)\right )}{-1+e x^2-\log (9)}+4 x^3 \log \left (-1+e x^2-\log (9)\right )\right ) \, dx\\ &=2 \int \frac {x^3 \left (-4+5 e x^2-4 \log (9)\right )}{-1+e x^2-\log (9)} \, dx+4 \int x^3 \log \left (-1+e x^2-\log (9)\right ) \, dx\\ &=2 \operatorname {Subst}\left (\int x \log (-1+e x-\log (9)) \, dx,x,x^2\right )+\operatorname {Subst}\left (\int \frac {x (-4+5 e x-4 \log (9))}{-1+e x-\log (9)} \, dx,x,x^2\right )\\ &=x^4 \log \left (-1+e x^2-\log (9)\right )-e \operatorname {Subst}\left (\int \frac {x^2}{-1+e x-\log (9)} \, dx,x,x^2\right )+\operatorname {Subst}\left (\int \left (5 x+\frac {1+\log (9)}{e}+\frac {(1+\log (9))^2}{e (-1+e x-\log (9))}\right ) \, dx,x,x^2\right )\\ &=\frac {5 x^4}{2}+\frac {x^2 (1+\log (9))}{e}+x^4 \log \left (-1+e x^2-\log (9)\right )+\frac {(1+\log (9))^2 \log \left (1-e x^2+\log (9)\right )}{e^2}-e \operatorname {Subst}\left (\int \left (\frac {x}{e}+\frac {1+\log (9)}{e^2}+\frac {(1+\log (9))^2}{e^2 (-1+e x-\log (9))}\right ) \, dx,x,x^2\right )\\ &=2 x^4+x^4 \log \left (-1+e x^2-\log (9)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 25, normalized size = 1.39 \begin {gather*} 2 \left (x^4+\frac {1}{2} x^4 \log \left (-1+e x^2-\log (9)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8*x^3 - 10*E*x^5 + 8*x^3*Log[9] + (4*x^3 - 4*E*x^5 + 4*x^3*Log[9])*Log[-1 + E*x^2 - Log[9]])/(1 - E
*x^2 + Log[9]),x]

[Out]

2*(x^4 + (x^4*Log[-1 + E*x^2 - Log[9]])/2)

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fricas [A]  time = 0.61, size = 23, normalized size = 1.28 \begin {gather*} x^{4} \log \left (x^{2} e - 2 \, \log \relax (3) - 1\right ) + 2 \, x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^4*log(x^2*e - 2*log(3) - 1) + 2*x^4

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giac [B]  time = 0.17, size = 181, normalized size = 10.06 \begin {gather*} {\left (4 \, x^{2} e + {\left (x^{2} e - 2 \, \log \relax (3) - 1\right )}^{2} \log \left (x^{2} e - 2 \, \log \relax (3) - 1\right ) + 4 \, {\left (x^{2} e - 2 \, \log \relax (3) - 1\right )} \log \relax (3) \log \left (x^{2} e - 2 \, \log \relax (3) - 1\right ) + 4 \, \log \relax (3)^{2} \log \left (x^{2} e - 2 \, \log \relax (3) - 1\right ) + 2 \, {\left (x^{2} e - 2 \, \log \relax (3) - 1\right )}^{2} + 8 \, {\left (x^{2} e - 2 \, \log \relax (3) - 1\right )} \log \relax (3) + 2 \, {\left (x^{2} e - 2 \, \log \relax (3) - 1\right )} \log \left (x^{2} e - 2 \, \log \relax (3) - 1\right ) + 4 \, \log \relax (3) \log \left (x^{2} e - 2 \, \log \relax (3) - 1\right ) - 8 \, \log \relax (3) + \log \left (x^{2} e - 2 \, \log \relax (3) - 1\right ) - 4\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*x^2*e + (x^2*e - 2*log(3) - 1)^2*log(x^2*e - 2*log(3) - 1) + 4*(x^2*e - 2*log(3) - 1)*log(3)*log(x^2*e - 2*
log(3) - 1) + 4*log(3)^2*log(x^2*e - 2*log(3) - 1) + 2*(x^2*e - 2*log(3) - 1)^2 + 8*(x^2*e - 2*log(3) - 1)*log
(3) + 2*(x^2*e - 2*log(3) - 1)*log(x^2*e - 2*log(3) - 1) + 4*log(3)*log(x^2*e - 2*log(3) - 1) - 8*log(3) + log
(x^2*e - 2*log(3) - 1) - 4)*e^(-2)

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maple [A]  time = 0.14, size = 24, normalized size = 1.33

method	result	size
norman	\(x^{4} \ln \left (-2 \ln \relax (3)+x^{2} {\mathrm e}-1\right )+2 x^{4}\)	\(24\)
risch	\(x^{4} \ln \left (-2 \ln \relax (3)+x^{2} {\mathrm e}-1\right )+2 x^{4}\)	\(24\)
default	\(x^{4} \ln \left (-2 \ln \relax (3)+x^{2} {\mathrm e}-1\right )+2 x^{4}-4 \,{\mathrm e}^{-2} \ln \left (-2 \ln \relax (3)+x^{2} {\mathrm e}-1\right ) \ln \relax (3)^{2}-4 \,{\mathrm e}^{-2} \ln \left (-2 \ln \relax (3)+x^{2} {\mathrm e}-1\right ) \ln \relax (3)-{\mathrm e}^{-2} \ln \left (-2 \ln \relax (3)+x^{2} {\mathrm e}-1\right )+4 \left ({\mathrm e}^{-1}\right )^{2} \ln \left (-2 \ln \relax (3)+x^{2} {\mathrm e}-1\right ) \ln \relax (3)^{2}+4 \left ({\mathrm e}^{-1}\right )^{2} \ln \left (-2 \ln \relax (3)+x^{2} {\mathrm e}-1\right ) \ln \relax (3)+\left ({\mathrm e}^{-1}\right )^{2} \ln \left (-2 \ln \relax (3)+x^{2} {\mathrm e}-1\right )\)	\(149\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.45, size = 467, normalized size = 25.94 \begin {gather*} -4 \, x^{2} e^{\left (-1\right )} + 2 \, {\left (2 \, x^{2} e + {\left (2 \, \log \relax (3) + 1\right )} \log \left (x^{2} e - 2 \, \log \relax (3) - 1\right )^{2} + 2 \, {\left (2 \, \log \relax (3) + 1\right )} \log \left (x^{2} e - 2 \, \log \relax (3) - 1\right )\right )} e^{\left (-2\right )} \log \relax (3) + {\left (2 \, {\left (4 \, \log \relax (3)^{2} + 4 \, \log \relax (3) + 1\right )} e^{\left (-3\right )} \log \left (x^{2} e - 2 \, \log \relax (3) - 1\right ) + {\left (x^{4} e + 2 \, x^{2} {\left (2 \, \log \relax (3) + 1\right )}\right )} e^{\left (-2\right )}\right )} e \log \left (x^{2} e - 2 \, \log \relax (3) - 1\right ) - 4 \, {\left (2 \, \log \relax (3) + 1\right )} e^{\left (-2\right )} \log \left (x^{2} e - 2 \, \log \relax (3) - 1\right ) - 4 \, {\left (x^{2} e^{\left (-1\right )} + {\left (2 \, \log \relax (3) + 1\right )} e^{\left (-2\right )} \log \left (x^{2} e - 2 \, \log \relax (3) - 1\right )\right )} \log \relax (3) \log \left (x^{2} e - 2 \, \log \relax (3) - 1\right ) + \frac {5}{2} \, {\left (2 \, {\left (4 \, \log \relax (3)^{2} + 4 \, \log \relax (3) + 1\right )} e^{\left (-3\right )} \log \left (x^{2} e - 2 \, \log \relax (3) - 1\right ) + {\left (x^{4} e + 2 \, x^{2} {\left (2 \, \log \relax (3) + 1\right )}\right )} e^{\left (-2\right )}\right )} e - \frac {1}{2} \, {\left (x^{4} e^{2} + 6 \, x^{2} {\left (2 \, \log \relax (3) + 1\right )} e + 2 \, {\left (4 \, \log \relax (3)^{2} + 4 \, \log \relax (3) + 1\right )} \log \left (x^{2} e - 2 \, \log \relax (3) - 1\right )^{2} + 6 \, {\left (4 \, \log \relax (3)^{2} + 4 \, \log \relax (3) + 1\right )} \log \left (x^{2} e - 2 \, \log \relax (3) - 1\right )\right )} e^{\left (-2\right )} + {\left (2 \, x^{2} e + {\left (2 \, \log \relax (3) + 1\right )} \log \left (x^{2} e - 2 \, \log \relax (3) - 1\right )^{2} + 2 \, {\left (2 \, \log \relax (3) + 1\right )} \log \left (x^{2} e - 2 \, \log \relax (3) - 1\right )\right )} e^{\left (-2\right )} - 8 \, {\left (x^{2} e^{\left (-1\right )} + {\left (2 \, \log \relax (3) + 1\right )} e^{\left (-2\right )} \log \left (x^{2} e - 2 \, \log \relax (3) - 1\right )\right )} \log \relax (3) - 2 \, {\left (x^{2} e^{\left (-1\right )} + {\left (2 \, \log \relax (3) + 1\right )} e^{\left (-2\right )} \log \left (x^{2} e - 2 \, \log \relax (3) - 1\right )\right )} \log \left (x^{2} e - 2 \, \log \relax (3) - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*x^2*e^(-1) + 2*(2*x^2*e + (2*log(3) + 1)*log(x^2*e - 2*log(3) - 1)^2 + 2*(2*log(3) + 1)*log(x^2*e - 2*log(3
) - 1))*e^(-2)*log(3) + (2*(4*log(3)^2 + 4*log(3) + 1)*e^(-3)*log(x^2*e - 2*log(3) - 1) + (x^4*e + 2*x^2*(2*lo
g(3) + 1))*e^(-2))*e*log(x^2*e - 2*log(3) - 1) - 4*(2*log(3) + 1)*e^(-2)*log(x^2*e - 2*log(3) - 1) - 4*(x^2*e^
(-1) + (2*log(3) + 1)*e^(-2)*log(x^2*e - 2*log(3) - 1))*log(3)*log(x^2*e - 2*log(3) - 1) + 5/2*(2*(4*log(3)^2
+ 4*log(3) + 1)*e^(-3)*log(x^2*e - 2*log(3) - 1) + (x^4*e + 2*x^2*(2*log(3) + 1))*e^(-2))*e - 1/2*(x^4*e^2 + 6
*x^2*(2*log(3) + 1)*e + 2*(4*log(3)^2 + 4*log(3) + 1)*log(x^2*e - 2*log(3) - 1)^2 + 6*(4*log(3)^2 + 4*log(3) +
1)*log(x^2*e - 2*log(3) - 1))*e^(-2) + (2*x^2*e + (2*log(3) + 1)*log(x^2*e - 2*log(3) - 1)^2 + 2*(2*log(3) +
1)*log(x^2*e - 2*log(3) - 1))*e^(-2) - 8*(x^2*e^(-1) + (2*log(3) + 1)*e^(-2)*log(x^2*e - 2*log(3) - 1))*log(3)
- 2*(x^2*e^(-1) + (2*log(3) + 1)*e^(-2)*log(x^2*e - 2*log(3) - 1))*log(x^2*e - 2*log(3) - 1)

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mupad [B]  time = 20.72, size = 156, normalized size = 8.67 \begin {gather*} x^4\,\ln \left (\mathrm {e}\,x^2-2\,\ln \relax (3)-1\right )-4\,x^2\,{\mathrm {e}}^{-1}+2\,x^4-8\,x^2\,{\mathrm {e}}^{-1}\,\ln \relax (3)-\ln \left (\mathrm {e}\,x^2-\ln \relax (9)-1\right )\,{\mathrm {e}}^{-2}\,\left (4\,\ln \relax (9)+4\right )+\ln \left (\mathrm {e}\,x^2-\ln \relax (9)-1\right )\,{\mathrm {e}}^{-2}\,\left (10\,\ln \relax (9)+5\,{\ln \relax (9)}^2+5\right )-\ln \left (\mathrm {e}\,x^2-\ln \relax (9)-1\right )\,{\mathrm {e}}^{-2}\,\left (8\,\ln \relax (3)+8\,\ln \relax (3)\,\ln \relax (9)\right )+4\,x^2\,{\mathrm {e}}^{-1}\,\left (\ln \relax (9)+1\right )-\ln \left (\mathrm {e}\,x^2-\ln \relax (9)-1\right )\,{\mathrm {e}}^{-2}\,\left (2\,\ln \relax (9)+{\ln \relax (9)}^2+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^4*log(x^2*exp(1) - 2*log(3) - 1) - 4*x^2*exp(-1) + 2*x^4 - 8*x^2*exp(-1)*log(3) - log(x^2*exp(1) - log(9) -
1)*exp(-2)*(4*log(9) + 4) + log(x^2*exp(1) - log(9) - 1)*exp(-2)*(10*log(9) + 5*log(9)^2 + 5) - log(x^2*exp(1)
- log(9) - 1)*exp(-2)*(8*log(3) + 8*log(3)*log(9)) + 4*x^2*exp(-1)*(log(9) + 1) - log(x^2*exp(1) - log(9) - 1
)*exp(-2)*(2*log(9) + log(9)^2 + 1)

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sympy [A]  time = 0.21, size = 22, normalized size = 1.22 \begin {gather*} x^{4} \log {\left (e x^{2} - 2 \log {\relax (3 )} - 1 \right )} + 2 x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**4*log(E*x**2 - 2*log(3) - 1) + 2*x**4

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