3.15.93 \(\int \frac {(30 x^2+60 x^3) \log (e^4+x+x^2)+(30 e^4 x+30 x^2+30 x^3) \log ^2(e^4+x+x^2)}{e^4+x+x^2} \, dx\)
Optimal. Leaf size=16 \[ 15 x^2 \log ^2\left (e^4+x+x^2\right ) \]
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Rubi [A] time = 0.24, antiderivative size = 16, normalized size of antiderivative = 1.00,
number of steps used = 3, number of rules used = 10, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used
= {6688, 12, 14, 800, 634, 618, 204, 628, 2525, 6687} \begin {gather*} 15 x^2 \log ^2\left (x^2+x+e^4\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
Int[((30*x^2 + 60*x^3)*Log[E^4 + x + x^2] + (30*E^4*x + 30*x^2 + 30*x^3)*Log[E^4 + x + x^2]^2)/(E^4 + x + x^2)
,x]
[Out]
15*x^2*Log[E^4 + x + x^2]^2
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 14
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&& !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Rule 204
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])
Rule 618
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]
Rule 628
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Rule 634
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && !NiceSqrtQ[b^2 - 4*a*c]
Rule 800
Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]
Rule 2525
Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m
+ 1)*(a + b*Log[c*RFx^p])^n)/(e*(m + 1)), x] - Dist[(b*n*p)/(e*(m + 1)), Int[SimplifyIntegrand[((d + e*x)^(m +
1)*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]
Rule 6687
Int[(u_)*(y_)^(m_.)*(z_)^(n_.), x_Symbol] :> With[{q = DerivativeDivides[y*z, u*z^(n - m), x]}, Simp[(q*y^(m +
1)*z^(m + 1))/(m + 1), x] /; !FalseQ[q]] /; FreeQ[{m, n}, x] && NeQ[m, -1]
Rule 6688
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int 30 x \log \left (e^4+x+x^2\right ) \left (\frac {x (1+2 x)}{e^4+x+x^2}+\log \left (e^4+x+x^2\right )\right ) \, dx\\ &=30 \int x \log \left (e^4+x+x^2\right ) \left (\frac {x (1+2 x)}{e^4+x+x^2}+\log \left (e^4+x+x^2\right )\right ) \, dx\\ &=15 x^2 \log ^2\left (e^4+x+x^2\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.24, size = 16, normalized size = 1.00 \begin {gather*} 15 x^2 \log ^2\left (e^4+x+x^2\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[((30*x^2 + 60*x^3)*Log[E^4 + x + x^2] + (30*E^4*x + 30*x^2 + 30*x^3)*Log[E^4 + x + x^2]^2)/(E^4 + x
+ x^2),x]
[Out]
15*x^2*Log[E^4 + x + x^2]^2
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fricas [A] time = 0.99, size = 15, normalized size = 0.94 \begin {gather*} 15 \, x^{2} \log \left (x^{2} + x + e^{4}\right )^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(((30*x*exp(4)+30*x^3+30*x^2)*log(exp(4)+x^2+x)^2+(60*x^3+30*x^2)*log(exp(4)+x^2+x))/(exp(4)+x^2+x),x
, algorithm="fricas")
[Out]
15*x^2*log(x^2 + x + e^4)^2
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giac [A] time = 0.65, size = 15, normalized size = 0.94 \begin {gather*} 15 \, x^{2} \log \left (x^{2} + x + e^{4}\right )^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(((30*x*exp(4)+30*x^3+30*x^2)*log(exp(4)+x^2+x)^2+(60*x^3+30*x^2)*log(exp(4)+x^2+x))/(exp(4)+x^2+x),x
, algorithm="giac")
[Out]
15*x^2*log(x^2 + x + e^4)^2
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maple [A] time = 0.61, size = 16, normalized size = 1.00
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\(15 \ln \left ({\mathrm e}^{4}+x^{2}+x \right )^{2} x^{2}\) |
\(16\) |
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\(15 \ln \left ({\mathrm e}^{4}+x^{2}+x \right )^{2} x^{2}\) |
\(16\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((30*x*exp(4)+30*x^3+30*x^2)*ln(exp(4)+x^2+x)^2+(60*x^3+30*x^2)*ln(exp(4)+x^2+x))/(exp(4)+x^2+x),x,method=
_RETURNVERBOSE)
[Out]
15*ln(exp(4)+x^2+x)^2*x^2
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maxima [A] time = 0.85, size = 15, normalized size = 0.94 \begin {gather*} 15 \, x^{2} \log \left (x^{2} + x + e^{4}\right )^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(((30*x*exp(4)+30*x^3+30*x^2)*log(exp(4)+x^2+x)^2+(60*x^3+30*x^2)*log(exp(4)+x^2+x))/(exp(4)+x^2+x),x
, algorithm="maxima")
[Out]
15*x^2*log(x^2 + x + e^4)^2
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mupad [B] time = 1.19, size = 15, normalized size = 0.94 \begin {gather*} 15\,x^2\,{\ln \left (x^2+x+{\mathrm {e}}^4\right )}^2 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((log(x + exp(4) + x^2)^2*(30*x*exp(4) + 30*x^2 + 30*x^3) + log(x + exp(4) + x^2)*(30*x^2 + 60*x^3))/(x + e
xp(4) + x^2),x)
[Out]
15*x^2*log(x + exp(4) + x^2)^2
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sympy [A] time = 0.14, size = 15, normalized size = 0.94 \begin {gather*} 15 x^{2} \log {\left (x^{2} + x + e^{4} \right )}^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(((30*x*exp(4)+30*x**3+30*x**2)*ln(exp(4)+x**2+x)**2+(60*x**3+30*x**2)*ln(exp(4)+x**2+x))/(exp(4)+x**
2+x),x)
[Out]
15*x**2*log(x**2 + x + exp(4))**2
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