3.47.4 \(\int \frac {-27+27 x+72 x^2+e^{\frac {1}{9} (100 x^2+40 x^3+44 x^4+8 x^5+4 x^6)} (200 x^2+120 x^3+176 x^4+40 x^5+24 x^6)}{9 x} \, dx\)

Optimal. Leaf size=31 \[ e^{\frac {4}{9} x^2 \left (5+x+x^2\right )^2}+4 x^2+3 (x-\log (x)) \]

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Rubi [A]  time = 0.19, antiderivative size = 30, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 3, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {12, 14, 6706} \begin {gather*} 4 x^2+e^{\frac {4}{9} x^2 \left (x^2+x+5\right )^2}+3 x-3 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-27 + 27*x + 72*x^2 + E^((100*x^2 + 40*x^3 + 44*x^4 + 8*x^5 + 4*x^6)/9)*(200*x^2 + 120*x^3 + 176*x^4 + 40
*x^5 + 24*x^6))/(9*x),x]

[Out]

E^((4*x^2*(5 + x + x^2)^2)/9) + 3*x + 4*x^2 - 3*Log[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 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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{9} \int \frac {-27+27 x+72 x^2+\exp \left (\frac {1}{9} \left (100 x^2+40 x^3+44 x^4+8 x^5+4 x^6\right )\right ) \left (200 x^2+120 x^3+176 x^4+40 x^5+24 x^6\right )}{x} \, dx\\ &=\frac {1}{9} \int \left (8 e^{\frac {4}{9} x^2 \left (5+x+x^2\right )^2} x \left (5+x+x^2\right ) \left (5+2 x+3 x^2\right )+\frac {9 \left (-3+3 x+8 x^2\right )}{x}\right ) \, dx\\ &=\frac {8}{9} \int e^{\frac {4}{9} x^2 \left (5+x+x^2\right )^2} x \left (5+x+x^2\right ) \left (5+2 x+3 x^2\right ) \, dx+\int \frac {-3+3 x+8 x^2}{x} \, dx\\ &=e^{\frac {4}{9} x^2 \left (5+x+x^2\right )^2}+\int \left (3-\frac {3}{x}+8 x\right ) \, dx\\ &=e^{\frac {4}{9} x^2 \left (5+x+x^2\right )^2}+3 x+4 x^2-3 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 29, normalized size = 0.94 \begin {gather*} e^{\frac {4}{9} x^2 \left (5+x+x^2\right )^2}+x (3+4 x)-3 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-27 + 27*x + 72*x^2 + E^((100*x^2 + 40*x^3 + 44*x^4 + 8*x^5 + 4*x^6)/9)*(200*x^2 + 120*x^3 + 176*x^
4 + 40*x^5 + 24*x^6))/(9*x),x]

[Out]

E^((4*x^2*(5 + x + x^2)^2)/9) + x*(3 + 4*x) - 3*Log[x]

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fricas [A]  time = 0.61, size = 40, normalized size = 1.29 \begin {gather*} 4 \, x^{2} + 3 \, x + e^{\left (\frac {4}{9} \, x^{6} + \frac {8}{9} \, x^{5} + \frac {44}{9} \, x^{4} + \frac {40}{9} \, x^{3} + \frac {100}{9} \, x^{2}\right )} - 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((24*x^6+40*x^5+176*x^4+120*x^3+200*x^2)*exp(4/9*x^6+8/9*x^5+44/9*x^4+40/9*x^3+100/9*x^2)+72*x^2
+27*x-27)/x,x, algorithm="fricas")

[Out]

4*x^2 + 3*x + e^(4/9*x^6 + 8/9*x^5 + 44/9*x^4 + 40/9*x^3 + 100/9*x^2) - 3*log(x)

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giac [A]  time = 0.24, size = 40, normalized size = 1.29 \begin {gather*} 4 \, x^{2} + 3 \, x + e^{\left (\frac {4}{9} \, x^{6} + \frac {8}{9} \, x^{5} + \frac {44}{9} \, x^{4} + \frac {40}{9} \, x^{3} + \frac {100}{9} \, x^{2}\right )} - 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((24*x^6+40*x^5+176*x^4+120*x^3+200*x^2)*exp(4/9*x^6+8/9*x^5+44/9*x^4+40/9*x^3+100/9*x^2)+72*x^2
+27*x-27)/x,x, algorithm="giac")

[Out]

4*x^2 + 3*x + e^(4/9*x^6 + 8/9*x^5 + 44/9*x^4 + 40/9*x^3 + 100/9*x^2) - 3*log(x)

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maple [A]  time = 0.08, size = 28, normalized size = 0.90

method	result	size
risch	\(4 x^{2}+3 x -3 \ln \relax (x )+{\mathrm e}^{\frac {4 x^{2} \left (x^{2}+x +5\right )^{2}}{9}}\)	\(28\)
norman	\(3 x +4 x^{2}+{\mathrm e}^{\frac {4}{9} x^{6}+\frac {8}{9} x^{5}+\frac {44}{9} x^{4}+\frac {40}{9} x^{3}+\frac {100}{9} x^{2}}-3 \ln \relax (x )\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/9*((24*x^6+40*x^5+176*x^4+120*x^3+200*x^2)*exp(4/9*x^6+8/9*x^5+44/9*x^4+40/9*x^3+100/9*x^2)+72*x^2+27*x-
27)/x,x,method=_RETURNVERBOSE)

[Out]

4*x^2+3*x-3*ln(x)+exp(4/9*x^2*(x^2+x+5)^2)

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maxima [A]  time = 0.46, size = 40, normalized size = 1.29 \begin {gather*} 4 \, x^{2} + 3 \, x + e^{\left (\frac {4}{9} \, x^{6} + \frac {8}{9} \, x^{5} + \frac {44}{9} \, x^{4} + \frac {40}{9} \, x^{3} + \frac {100}{9} \, x^{2}\right )} - 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((24*x^6+40*x^5+176*x^4+120*x^3+200*x^2)*exp(4/9*x^6+8/9*x^5+44/9*x^4+40/9*x^3+100/9*x^2)+72*x^2
+27*x-27)/x,x, algorithm="maxima")

[Out]

4*x^2 + 3*x + e^(4/9*x^6 + 8/9*x^5 + 44/9*x^4 + 40/9*x^3 + 100/9*x^2) - 3*log(x)

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mupad [B]  time = 0.21, size = 40, normalized size = 1.29 \begin {gather*} 3\,x+{\mathrm {e}}^{\frac {4\,x^6}{9}+\frac {8\,x^5}{9}+\frac {44\,x^4}{9}+\frac {40\,x^3}{9}+\frac {100\,x^2}{9}}-3\,\ln \relax (x)+4\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x + (exp((100*x^2)/9 + (40*x^3)/9 + (44*x^4)/9 + (8*x^5)/9 + (4*x^6)/9)*(200*x^2 + 120*x^3 + 176*x^4 +
40*x^5 + 24*x^6))/9 + 8*x^2 - 3)/x,x)

[Out]

3*x + exp((100*x^2)/9 + (40*x^3)/9 + (44*x^4)/9 + (8*x^5)/9 + (4*x^6)/9) - 3*log(x) + 4*x^2

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sympy [A]  time = 0.17, size = 48, normalized size = 1.55 \begin {gather*} 4 x^{2} + 3 x + e^{\frac {4 x^{6}}{9} + \frac {8 x^{5}}{9} + \frac {44 x^{4}}{9} + \frac {40 x^{3}}{9} + \frac {100 x^{2}}{9}} - 3 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((24*x**6+40*x**5+176*x**4+120*x**3+200*x**2)*exp(4/9*x**6+8/9*x**5+44/9*x**4+40/9*x**3+100/9*x*
*2)+72*x**2+27*x-27)/x,x)

[Out]

4*x**2 + 3*x + exp(4*x**6/9 + 8*x**5/9 + 44*x**4/9 + 40*x**3/9 + 100*x**2/9) - 3*log(x)

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