3.51.76 \(\int \frac {4-16 x+4 x^2+e^{\frac {1}{4} (5 x^2-x^3)} (10 x^2-23 x^3+16 x^4-3 x^5)}{4 x-8 x^2+4 x^3} \, dx\)

Optimal. Leaf size=29 \[ 2+e^{\frac {1}{4} (5-x) x^2}+\frac {2}{1-\frac {1}{x}}+\log (x) \]

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Rubi [A]  time = 0.69, antiderivative size = 26, normalized size of antiderivative = 0.90, number of steps used = 8, number of rules used = 6, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1594, 27, 12, 6742, 6706, 893} \begin {gather*} e^{\frac {1}{4} (5-x) x^2}-\frac {2}{1-x}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 - 16*x + 4*x^2 + E^((5*x^2 - x^3)/4)*(10*x^2 - 23*x^3 + 16*x^4 - 3*x^5))/(4*x - 8*x^2 + 4*x^3),x]

[Out]

E^(((5 - x)*x^2)/4) - 2/(1 - x) + 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 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 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

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]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4-16 x+4 x^2+e^{\frac {1}{4} \left (5 x^2-x^3\right )} \left (10 x^2-23 x^3+16 x^4-3 x^5\right )}{x \left (4-8 x+4 x^2\right )} \, dx\\ &=\int \frac {4-16 x+4 x^2+e^{\frac {1}{4} \left (5 x^2-x^3\right )} \left (10 x^2-23 x^3+16 x^4-3 x^5\right )}{4 (-1+x)^2 x} \, dx\\ &=\frac {1}{4} \int \frac {4-16 x+4 x^2+e^{\frac {1}{4} \left (5 x^2-x^3\right )} \left (10 x^2-23 x^3+16 x^4-3 x^5\right )}{(-1+x)^2 x} \, dx\\ &=\frac {1}{4} \int \left (-e^{-\frac {1}{4} (-5+x) x^2} x (-10+3 x)+\frac {4 \left (1-4 x+x^2\right )}{(-1+x)^2 x}\right ) \, dx\\ &=-\left (\frac {1}{4} \int e^{-\frac {1}{4} (-5+x) x^2} x (-10+3 x) \, dx\right )+\int \frac {1-4 x+x^2}{(-1+x)^2 x} \, dx\\ &=e^{\frac {1}{4} (5-x) x^2}+\int \left (-\frac {2}{(-1+x)^2}+\frac {1}{x}\right ) \, dx\\ &=e^{\frac {1}{4} (5-x) x^2}-\frac {2}{1-x}+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 27, normalized size = 0.93 \begin {gather*} e^{\frac {5 x^2}{4}-\frac {x^3}{4}}+\frac {2}{-1+x}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 - 16*x + 4*x^2 + E^((5*x^2 - x^3)/4)*(10*x^2 - 23*x^3 + 16*x^4 - 3*x^5))/(4*x - 8*x^2 + 4*x^3),x]

[Out]

E^((5*x^2)/4 - x^3/4) + 2/(-1 + x) + Log[x]

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fricas [A]  time = 0.59, size = 30, normalized size = 1.03 \begin {gather*} \frac {{\left (x - 1\right )} e^{\left (-\frac {1}{4} \, x^{3} + \frac {5}{4} \, x^{2}\right )} + {\left (x - 1\right )} \log \relax (x) + 2}{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^5+16*x^4-23*x^3+10*x^2)*exp(-1/4*x^3+5/4*x^2)+4*x^2-16*x+4)/(4*x^3-8*x^2+4*x),x, algorithm="f
ricas")

[Out]

((x - 1)*e^(-1/4*x^3 + 5/4*x^2) + (x - 1)*log(x) + 2)/(x - 1)

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giac [A]  time = 0.24, size = 44, normalized size = 1.52 \begin {gather*} \frac {x e^{\left (-\frac {1}{4} \, x^{3} + \frac {5}{4} \, x^{2}\right )} + x \log \relax (x) - e^{\left (-\frac {1}{4} \, x^{3} + \frac {5}{4} \, x^{2}\right )} - \log \relax (x) + 2}{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^5+16*x^4-23*x^3+10*x^2)*exp(-1/4*x^3+5/4*x^2)+4*x^2-16*x+4)/(4*x^3-8*x^2+4*x),x, algorithm="g
iac")

[Out]

(x*e^(-1/4*x^3 + 5/4*x^2) + x*log(x) - e^(-1/4*x^3 + 5/4*x^2) - log(x) + 2)/(x - 1)

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maple [A]  time = 0.09, size = 20, normalized size = 0.69

method	result	size
risch	\(\frac {2}{x -1}+\ln \relax (x )+{\mathrm e}^{-\frac {x^{2} \left (x -5\right )}{4}}\)	\(20\)
norman	\(\frac {{\mathrm e}^{-\frac {1}{4} x^{3}+\frac {5}{4} x^{2}} x -{\mathrm e}^{-\frac {1}{4} x^{3}+\frac {5}{4} x^{2}}+2}{x -1}+\ln \relax (x )\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*x^5+16*x^4-23*x^3+10*x^2)*exp(-1/4*x^3+5/4*x^2)+4*x^2-16*x+4)/(4*x^3-8*x^2+4*x),x,method=_RETURNVERBO
SE)

[Out]

2/(x-1)+ln(x)+exp(-1/4*x^2*(x-5))

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maxima [A]  time = 0.40, size = 22, normalized size = 0.76 \begin {gather*} \frac {2}{x - 1} + e^{\left (-\frac {1}{4} \, x^{3} + \frac {5}{4} \, x^{2}\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^5+16*x^4-23*x^3+10*x^2)*exp(-1/4*x^3+5/4*x^2)+4*x^2-16*x+4)/(4*x^3-8*x^2+4*x),x, algorithm="m
axima")

[Out]

2/(x - 1) + e^(-1/4*x^3 + 5/4*x^2) + log(x)

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mupad [B]  time = 3.49, size = 22, normalized size = 0.76 \begin {gather*} {\mathrm {e}}^{\frac {5\,x^2}{4}-\frac {x^3}{4}}+\ln \relax (x)+\frac {2}{x-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((5*x^2)/4 - x^3/4)*(10*x^2 - 23*x^3 + 16*x^4 - 3*x^5) - 16*x + 4*x^2 + 4)/(4*x - 8*x^2 + 4*x^3),x)

[Out]

exp((5*x^2)/4 - x^3/4) + log(x) + 2/(x - 1)

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sympy [A]  time = 0.18, size = 20, normalized size = 0.69 \begin {gather*} e^{- \frac {x^{3}}{4} + \frac {5 x^{2}}{4}} + \log {\relax (x )} + \frac {2}{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x**5+16*x**4-23*x**3+10*x**2)*exp(-1/4*x**3+5/4*x**2)+4*x**2-16*x+4)/(4*x**3-8*x**2+4*x),x)

[Out]

exp(-x**3/4 + 5*x**2/4) + log(x) + 2/(x - 1)

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