3.1.14 \(\int \frac {d+e x+f x^2+g x^3+h x^4+i x^5}{4-5 x^2+x^4} \, dx\) [14]

Optimal. Leaf size=76 \[ h x+\frac {i x^2}{2}-\frac {1}{6} (d+4 f+16 h) \tanh ^{-1}\left (\frac {x}{2}\right )+\frac {1}{3} (d+f+h) \tanh ^{-1}(x)-\frac {1}{6} (e+g+i) \log \left (1-x^2\right )+\frac {1}{6} (e+4 g+16 i) \log \left (4-x^2\right ) \]

[Out]

h*x+1/2*i*x^2-1/6*(d+4*f+16*h)*arctanh(1/2*x)+1/3*(d+f+h)*arctanh(x)-1/6*(e+g+i)*ln(-x^2+1)+1/6*(e+4*g+16*i)*l
n(-x^2+4)

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Rubi [A]
time = 0.12, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {1687, 1690, 1180, 213, 1677, 1671, 646, 31} \begin {gather*} -\frac {1}{6} \tanh ^{-1}\left (\frac {x}{2}\right ) (d+4 f+16 h)+\frac {1}{3} \tanh ^{-1}(x) (d+f+h)-\frac {1}{6} \log \left (1-x^2\right ) (e+g+i)+\frac {1}{6} \log \left (4-x^2\right ) (e+4 g+16 i)+h x+\frac {i x^2}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x + f*x^2 + g*x^3 + h*x^4 + i*x^5)/(4 - 5*x^2 + x^4),x]

[Out]

h*x + (i*x^2)/2 - ((d + 4*f + 16*h)*ArcTanh[x/2])/6 + ((d + f + h)*ArcTanh[x])/3 - ((e + g + i)*Log[1 - x^2])/
6 + ((e + 4*g + 16*i)*Log[4 - x^2])/6

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 646

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), 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 1180

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rule 1671

Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 1677

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)
*SubstFor[x^2, Pq, x]*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x^2] && Inte
gerQ[(m - 1)/2]

Rule 1687

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], k}, Int[Sum[Coeff[
Pq, x, 2*k]*x^(2*k), {k, 0, q/2}]*(a + b*x^2 + c*x^4)^p, x] + Int[x*Sum[Coeff[Pq, x, 2*k + 1]*x^(2*k), {k, 0,
(q - 1)/2}]*(a + b*x^2 + c*x^4)^p, x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] &&  !PolyQ[Pq, x^2]

Rule 1690

Int[(Pq_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^2 + c*x^4), x], x
] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x^2] && Expon[Pq, x^2] > 1

Rubi steps

\begin {align*} \int \frac {d+e x+f x^2+g x^3+h x^4+14 x^5}{4-5 x^2+x^4} \, dx &=\int \frac {x \left (e+g x^2+14 x^4\right )}{4-5 x^2+x^4} \, dx+\int \frac {d+f x^2+h x^4}{4-5 x^2+x^4} \, dx\\ &=\frac {1}{2} \text {Subst}\left (\int \frac {e+g x+14 x^2}{4-5 x+x^2} \, dx,x,x^2\right )+\int \left (h+\frac {d-4 h+(f+5 h) x^2}{4-5 x^2+x^4}\right ) \, dx\\ &=h x+\frac {1}{2} \text {Subst}\left (\int \left (14-\frac {56-e-(70+g) x}{4-5 x+x^2}\right ) \, dx,x,x^2\right )+\int \frac {d-4 h+(f+5 h) x^2}{4-5 x^2+x^4} \, dx\\ &=h x+7 x^2-\frac {1}{2} \text {Subst}\left (\int \frac {56-e-(70+g) x}{4-5 x+x^2} \, dx,x,x^2\right )-\frac {1}{3} (d+f+h) \int \frac {1}{-1+x^2} \, dx+\frac {1}{3} (d+4 f+16 h) \int \frac {1}{-4+x^2} \, dx\\ &=h x+7 x^2-\frac {1}{6} (d+4 f+16 h) \tanh ^{-1}\left (\frac {x}{2}\right )+\frac {1}{3} (d+f+h) \tanh ^{-1}(x)-\frac {1}{6} (-224-e-4 g) \text {Subst}\left (\int \frac {1}{-4+x} \, dx,x,x^2\right )-\frac {1}{6} (14+e+g) \text {Subst}\left (\int \frac {1}{-1+x} \, dx,x,x^2\right )\\ &=h x+7 x^2-\frac {1}{6} (d+4 f+16 h) \tanh ^{-1}\left (\frac {x}{2}\right )+\frac {1}{3} (d+f+h) \tanh ^{-1}(x)-\frac {1}{6} (14+e+g) \log \left (1-x^2\right )+\frac {1}{6} (224+e+4 g) \log \left (4-x^2\right )\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 98, normalized size = 1.29 \begin {gather*} \frac {1}{12} \left (12 h x+6 i x^2-2 (d+e+f+g+h+i) \log (1-x)+(d+2 e+4 (f+2 g+4 h+8 i)) \log (2-x)+2 (d-e+f-g+h-i) \log (1+x)-(d-2 (e-2 f+4 g-8 h+16 i)) \log (2+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x + f*x^2 + g*x^3 + h*x^4 + i*x^5)/(4 - 5*x^2 + x^4),x]

[Out]

(12*h*x + 6*i*x^2 - 2*(d + e + f + g + h + i)*Log[1 - x] + (d + 2*e + 4*(f + 2*g + 4*h + 8*i))*Log[2 - x] + 2*
(d - e + f - g + h - i)*Log[1 + x] - (d - 2*(e - 2*f + 4*g - 8*h + 16*i))*Log[2 + x])/12

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Maple [A]
time = 0.05, size = 107, normalized size = 1.41

method	result	size
default	\(\frac {i \,x^{2}}{2}+h x +\left (-\frac {d}{12}+\frac {e}{6}-\frac {f}{3}+\frac {2 g}{3}-\frac {4 h}{3}+\frac {8 i}{3}\right ) \ln \left (x +2\right )+\left (\frac {d}{12}+\frac {e}{6}+\frac {f}{3}+\frac {2 g}{3}+\frac {4 h}{3}+\frac {8 i}{3}\right ) \ln \left (x -2\right )+\left (-\frac {d}{6}-\frac {e}{6}-\frac {f}{6}-\frac {g}{6}-\frac {h}{6}-\frac {i}{6}\right ) \ln \left (-1+x \right )+\left (\frac {d}{6}-\frac {e}{6}+\frac {f}{6}-\frac {g}{6}+\frac {h}{6}-\frac {i}{6}\right ) \ln \left (1+x \right )\)	\(107\)
norman	\(\frac {i \,x^{2}}{2}+h x +\left (-\frac {d}{12}+\frac {e}{6}-\frac {f}{3}+\frac {2 g}{3}-\frac {4 h}{3}+\frac {8 i}{3}\right ) \ln \left (x +2\right )+\left (\frac {d}{12}+\frac {e}{6}+\frac {f}{3}+\frac {2 g}{3}+\frac {4 h}{3}+\frac {8 i}{3}\right ) \ln \left (x -2\right )+\left (-\frac {d}{6}-\frac {e}{6}-\frac {f}{6}-\frac {g}{6}-\frac {h}{6}-\frac {i}{6}\right ) \ln \left (-1+x \right )+\left (\frac {d}{6}-\frac {e}{6}+\frac {f}{6}-\frac {g}{6}+\frac {h}{6}-\frac {i}{6}\right ) \ln \left (1+x \right )\)	\(107\)
risch	\(h x +\frac {2 \ln \left (x +2\right ) g}{3}+\frac {2 \ln \left (2-x \right ) g}{3}-\frac {\ln \left (1+x \right ) g}{6}-\frac {\ln \left (1-x \right ) g}{6}-\frac {\ln \left (1-x \right ) f}{6}+\frac {\ln \left (2-x \right ) f}{3}-\frac {\ln \left (x +2\right ) f}{3}+\frac {\ln \left (1+x \right ) f}{6}+\frac {\ln \left (2-x \right ) e}{6}+\frac {\ln \left (2-x \right ) d}{12}-\frac {\ln \left (x +2\right ) d}{12}+\frac {\ln \left (x +2\right ) e}{6}+\frac {\ln \left (1+x \right ) d}{6}-\frac {\ln \left (1+x \right ) e}{6}-\frac {\ln \left (1-x \right ) d}{6}-\frac {\ln \left (1-x \right ) e}{6}-\frac {\ln \left (1-x \right ) h}{6}+\frac {\ln \left (1+x \right ) h}{6}+\frac {4 \ln \left (2-x \right ) h}{3}-\frac {4 \ln \left (x +2\right ) h}{3}+\frac {8 \ln \left (2-x \right ) i}{3}+\frac {i \,x^{2}}{2}-\frac {\ln \left (1+x \right ) i}{6}-\frac {\ln \left (1-x \right ) i}{6}+\frac {8 \ln \left (x +2\right ) i}{3}\)	\(203\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((i*x^5+h*x^4+g*x^3+f*x^2+e*x+d)/(x^4-5*x^2+4),x,method=_RETURNVERBOSE)

[Out]

1/2*i*x^2+h*x+(-1/12*d+1/6*e-1/3*f+2/3*g-4/3*h+8/3*i)*ln(x+2)+(1/12*d+1/6*e+1/3*f+2/3*g+4/3*h+8/3*i)*ln(x-2)+(
-1/6*d-1/6*e-1/6*f-1/6*g-1/6*h-1/6*i)*ln(-1+x)+(1/6*d-1/6*e+1/6*f-1/6*g+1/6*h-1/6*i)*ln(1+x)

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Maxima [A]
time = 0.28, size = 85, normalized size = 1.12 \begin {gather*} h x + \frac {1}{2} i \, x^{2} - \frac {1}{12} \, {\left (d + 4 \, f - 8 \, g + 16 \, h - 2 \, e - 32 i\right )} \log \left (x + 2\right ) + \frac {1}{6} \, {\left (d + f - g + h - e - i\right )} \log \left (x + 1\right ) - \frac {1}{6} \, {\left (d + f + g + h + e + i\right )} \log \left (x - 1\right ) + \frac {1}{12} \, {\left (d + 4 \, f + 8 \, g + 16 \, h + 2 \, e + 32 i\right )} \log \left (x - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x^5+h*x^4+g*x^3+f*x^2+e*x+d)/(x^4-5*x^2+4),x, algorithm="maxima")

[Out]

h*x + 1/2*I*x^2 - 1/12*(d + 4*f - 8*g + 16*h - 2*e - 32*I)*log(x + 2) + 1/6*(d + f - g + h - e - I)*log(x + 1)
- 1/6*(d + f + g + h + e + I)*log(x - 1) + 1/12*(d + 4*f + 8*g + 16*h + 2*e + 32*I)*log(x - 2)

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Fricas [A]
time = 6.52, size = 88, normalized size = 1.16 \begin {gather*} \frac {1}{2} \, i x^{2} + h x - \frac {1}{12} \, {\left (d - 2 \, e + 4 \, f - 8 \, g + 16 \, h - 32 \, i\right )} \log \left (x + 2\right ) + \frac {1}{6} \, {\left (d - e + f - g + h - i\right )} \log \left (x + 1\right ) - \frac {1}{6} \, {\left (d + e + f + g + h + i\right )} \log \left (x - 1\right ) + \frac {1}{12} \, {\left (d + 2 \, e + 4 \, f + 8 \, g + 16 \, h + 32 \, i\right )} \log \left (x - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x^5+h*x^4+g*x^3+f*x^2+e*x+d)/(x^4-5*x^2+4),x, algorithm="fricas")

[Out]

1/2*i*x^2 + h*x - 1/12*(d - 2*e + 4*f - 8*g + 16*h - 32*i)*log(x + 2) + 1/6*(d - e + f - g + h - i)*log(x + 1)
- 1/6*(d + e + f + g + h + i)*log(x - 1) + 1/12*(d + 2*e + 4*f + 8*g + 16*h + 32*i)*log(x - 2)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x**5+h*x**4+g*x**3+f*x**2+e*x+d)/(x**4-5*x**2+4),x)

[Out]

Timed out

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Giac [A]
time = 4.50, size = 89, normalized size = 1.17 \begin {gather*} h x + \frac {1}{2} i \, x^{2} - \frac {1}{12} \, {\left (d + 4 \, f - 8 \, g + 16 \, h - 2 \, e - 32 i\right )} \log \left ({\left | x + 2 \right |}\right ) + \frac {1}{6} \, {\left (d + f - g + h - e - i\right )} \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{6} \, {\left (d + f + g + h + e + i\right )} \log \left ({\left | x - 1 \right |}\right ) + \frac {1}{12} \, {\left (d + 4 \, f + 8 \, g + 16 \, h + 2 \, e + 32 i\right )} \log \left ({\left | x - 2 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x^5+h*x^4+g*x^3+f*x^2+e*x+d)/(x^4-5*x^2+4),x, algorithm="giac")

[Out]

h*x + 1/2*I*x^2 - 1/12*(d + 4*f - 8*g + 16*h - 2*e - 32*I)*log(abs(x + 2)) + 1/6*(d + f - g + h - e - I)*log(a
bs(x + 1)) - 1/6*(d + f + g + h + e + I)*log(abs(x - 1)) + 1/12*(d + 4*f + 8*g + 16*h + 2*e + 32*I)*log(abs(x
- 2))

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Mupad [B]
time = 1.19, size = 108, normalized size = 1.42 \begin {gather*} h\,x+\frac {i\,x^2}{2}-\ln \left (x-1\right )\,\left (\frac {d}{6}+\frac {e}{6}+\frac {f}{6}+\frac {g}{6}+\frac {h}{6}+\frac {i}{6}\right )+\ln \left (x+1\right )\,\left (\frac {d}{6}-\frac {e}{6}+\frac {f}{6}-\frac {g}{6}+\frac {h}{6}-\frac {i}{6}\right )+\ln \left (x-2\right )\,\left (\frac {d}{12}+\frac {e}{6}+\frac {f}{3}+\frac {2\,g}{3}+\frac {4\,h}{3}+\frac {8\,i}{3}\right )-\ln \left (x+2\right )\,\left (\frac {d}{12}-\frac {e}{6}+\frac {f}{3}-\frac {2\,g}{3}+\frac {4\,h}{3}-\frac {8\,i}{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x + f*x^2 + g*x^3 + h*x^4 + i*x^5)/(x^4 - 5*x^2 + 4),x)

[Out]

h*x + (i*x^2)/2 - log(x - 1)*(d/6 + e/6 + f/6 + g/6 + h/6 + i/6) + log(x + 1)*(d/6 - e/6 + f/6 - g/6 + h/6 - i
/6) + log(x - 2)*(d/12 + e/6 + f/3 + (2*g)/3 + (4*h)/3 + (8*i)/3) - log(x + 2)*(d/12 - e/6 + f/3 - (2*g)/3 + (
4*h)/3 - (8*i)/3)

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