3.1006 \(\int \frac {1}{\sqrt {5-2 x+x^2} (8+x^3)} \, dx\)

Optimal. Leaf size=84 \[ -\frac {\tan ^{-1}\left (\frac {1-x}{\sqrt {3} \sqrt {x^2-2 x+5}}\right )}{4 \sqrt {3}}-\frac {\tanh ^{-1}\left (\frac {7-3 x}{\sqrt {13} \sqrt {x^2-2 x+5}}\right )}{12 \sqrt {13}}+\frac {1}{12} \tanh ^{-1}\left (\sqrt {x^2-2 x+5}\right ) \]

[Out]

1/12*arctanh((x^2-2*x+5)^(1/2))-1/12*arctan(1/3*(1-x)*3^(1/2)/(x^2-2*x+5)^(1/2))*3^(1/2)-1/156*arctanh(1/13*(7
-3*x)*13^(1/2)/(x^2-2*x+5)^(1/2))*13^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2074, 724, 206, 1025, 982, 203, 1024, 207} \[ -\frac {\tan ^{-1}\left (\frac {1-x}{\sqrt {3} \sqrt {x^2-2 x+5}}\right )}{4 \sqrt {3}}-\frac {\tanh ^{-1}\left (\frac {7-3 x}{\sqrt {13} \sqrt {x^2-2 x+5}}\right )}{12 \sqrt {13}}+\frac {1}{12} \tanh ^{-1}\left (\sqrt {x^2-2 x+5}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[5 - 2*x + x^2]*(8 + x^3)),x]

[Out]

-ArcTan[(1 - x)/(Sqrt[3]*Sqrt[5 - 2*x + x^2])]/(4*Sqrt[3]) - ArcTanh[(7 - 3*x)/(Sqrt[13]*Sqrt[5 - 2*x + x^2])]
/(12*Sqrt[13]) + ArcTanh[Sqrt[5 - 2*x + x^2]]/12

Rule 203

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

Rule 206

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

Rule 207

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

Rule 724

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

Rule 982

Int[1/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[-2*e, Su
bst[Int[1/(e*(b*e - 4*a*f) - (b*d - a*e)*x^2), x], x, (e + 2*f*x)/Sqrt[d + e*x + f*x^2]], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] && EqQ[c*e - b*f, 0]

Rule 1024

Int[((g_) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol
] :> Dist[-2*g, Subst[Int[1/(b*d - a*e - b*x^2), x], x, Sqrt[d + e*x + f*x^2]], x] /; FreeQ[{a, b, c, d, e, f,
g, h}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] && EqQ[c*e - b*f, 0] && EqQ[h*e - 2*g*f, 0]

Rule 1025

Int[((g_.) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbo
l] :> -Dist[(h*e - 2*g*f)/(2*f), Int[1/((a + b*x + c*x^2)*Sqrt[d + e*x + f*x^2]), x], x] + Dist[h/(2*f), Int[(
e + 2*f*x)/((a + b*x + c*x^2)*Sqrt[d + e*x + f*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && NeQ[b^2
- 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] && EqQ[c*e - b*f, 0] && NeQ[h*e - 2*g*f, 0]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {5-2 x+x^2} \left (8+x^3\right )} \, dx &=\int \left (\frac {1}{12 (2+x) \sqrt {5-2 x+x^2}}+\frac {4-x}{12 \left (4-2 x+x^2\right ) \sqrt {5-2 x+x^2}}\right ) \, dx\\ &=\frac {1}{12} \int \frac {1}{(2+x) \sqrt {5-2 x+x^2}} \, dx+\frac {1}{12} \int \frac {4-x}{\left (4-2 x+x^2\right ) \sqrt {5-2 x+x^2}} \, dx\\ &=-\left (\frac {1}{24} \int \frac {-2+2 x}{\left (4-2 x+x^2\right ) \sqrt {5-2 x+x^2}} \, dx\right )-\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{52-x^2} \, dx,x,\frac {14-6 x}{\sqrt {5-2 x+x^2}}\right )+\frac {1}{4} \int \frac {1}{\left (4-2 x+x^2\right ) \sqrt {5-2 x+x^2}} \, dx\\ &=-\frac {\tanh ^{-1}\left (\frac {7-3 x}{\sqrt {13} \sqrt {5-2 x+x^2}}\right )}{12 \sqrt {13}}-\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{-2+2 x^2} \, dx,x,\sqrt {5-2 x+x^2}\right )+\operatorname {Subst}\left (\int \frac {1}{24+2 x^2} \, dx,x,\frac {-2+2 x}{\sqrt {5-2 x+x^2}}\right )\\ &=\frac {\tan ^{-1}\left (\frac {-2+2 x}{2 \sqrt {3} \sqrt {5-2 x+x^2}}\right )}{4 \sqrt {3}}-\frac {\tanh ^{-1}\left (\frac {7-3 x}{\sqrt {13} \sqrt {5-2 x+x^2}}\right )}{12 \sqrt {13}}+\frac {1}{12} \tanh ^{-1}\left (\sqrt {5-2 x+x^2}\right )\\ \end {align*}

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Mathematica [C]  time = 0.33, size = 160, normalized size = 1.90 \[ \frac {1}{312} \left (-2 \sqrt {13} \tanh ^{-1}\left (\frac {7-3 x}{\sqrt {13} \sqrt {x^2-2 x+5}}\right )-13 \left (\left (\sqrt {3}+i\right ) \tan ^{-1}\left (\frac {-2 \sqrt [3]{-1} x+4 x+5 i \sqrt {3}+1}{\sqrt {2-2 i \sqrt {3}} \sqrt {x^2-2 x+5}}\right )+\left (\sqrt {3}-i\right ) \tan ^{-1}\left (\frac {2 \left (2+(-1)^{2/3}\right ) x-5 i \sqrt {3}+1}{\sqrt {2+2 i \sqrt {3}} \sqrt {x^2-2 x+5}}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[5 - 2*x + x^2]*(8 + x^3)),x]

[Out]

(-13*((I + Sqrt[3])*ArcTan[(1 + (5*I)*Sqrt[3] + 4*x - 2*(-1)^(1/3)*x)/(Sqrt[2 - (2*I)*Sqrt[3]]*Sqrt[5 - 2*x +
x^2])] + (-I + Sqrt[3])*ArcTan[(1 - (5*I)*Sqrt[3] + 2*(2 + (-1)^(2/3))*x)/(Sqrt[2 + (2*I)*Sqrt[3]]*Sqrt[5 - 2*
x + x^2])]) - 2*Sqrt[13]*ArcTanh[(7 - 3*x)/(Sqrt[13]*Sqrt[5 - 2*x + x^2])])/312

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fricas [B]  time = 0.69, size = 154, normalized size = 1.83 \[ \frac {1}{12} \, \sqrt {3} \arctan \left (-\frac {1}{3} \, \sqrt {3} {\left (x - 2\right )} + \frac {1}{3} \, \sqrt {3} \sqrt {x^{2} - 2 \, x + 5}\right ) - \frac {1}{12} \, \sqrt {3} \arctan \left (-\frac {1}{3} \, \sqrt {3} x + \frac {1}{3} \, \sqrt {3} \sqrt {x^{2} - 2 \, x + 5}\right ) + \frac {1}{156} \, \sqrt {13} \log \left (-\frac {\sqrt {13} {\left (3 \, x - 7\right )} + \sqrt {x^{2} - 2 \, x + 5} {\left (3 \, \sqrt {13} + 13\right )} + 9 \, x - 21}{x + 2}\right ) + \frac {1}{24} \, \log \left (x^{2} - \sqrt {x^{2} - 2 \, x + 5} {\left (x - 2\right )} - 3 \, x + 6\right ) - \frac {1}{24} \, \log \left (x^{2} - \sqrt {x^{2} - 2 \, x + 5} x - x + 4\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3+8)/(x^2-2*x+5)^(1/2),x, algorithm="fricas")

[Out]

1/12*sqrt(3)*arctan(-1/3*sqrt(3)*(x - 2) + 1/3*sqrt(3)*sqrt(x^2 - 2*x + 5)) - 1/12*sqrt(3)*arctan(-1/3*sqrt(3)
*x + 1/3*sqrt(3)*sqrt(x^2 - 2*x + 5)) + 1/156*sqrt(13)*log(-(sqrt(13)*(3*x - 7) + sqrt(x^2 - 2*x + 5)*(3*sqrt(
13) + 13) + 9*x - 21)/(x + 2)) + 1/24*log(x^2 - sqrt(x^2 - 2*x + 5)*(x - 2) - 3*x + 6) - 1/24*log(x^2 - sqrt(x
^2 - 2*x + 5)*x - x + 4)

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giac [B]  time = 0.59, size = 164, normalized size = 1.95 \[ -\frac {1}{12} \, \sqrt {3} \arctan \left (-\frac {1}{3} \, \sqrt {3} {\left (x - \sqrt {x^{2} - 2 \, x + 5}\right )}\right ) + \frac {1}{12} \, \sqrt {3} \arctan \left (-\frac {1}{3} \, \sqrt {3} {\left (x - \sqrt {x^{2} - 2 \, x + 5} - 2\right )}\right ) + \frac {1}{156} \, \sqrt {13} \log \left (\frac {{\left | -2 \, x - 2 \, \sqrt {13} + 2 \, \sqrt {x^{2} - 2 \, x + 5} - 4 \right |}}{{\left | -2 \, x + 2 \, \sqrt {13} + 2 \, \sqrt {x^{2} - 2 \, x + 5} - 4 \right |}}\right ) + \frac {1}{24} \, \log \left ({\left (x - \sqrt {x^{2} - 2 \, x + 5}\right )}^{2} - 4 \, x + 4 \, \sqrt {x^{2} - 2 \, x + 5} + 7\right ) - \frac {1}{24} \, \log \left ({\left (x - \sqrt {x^{2} - 2 \, x + 5}\right )}^{2} + 3\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3+8)/(x^2-2*x+5)^(1/2),x, algorithm="giac")

[Out]

-1/12*sqrt(3)*arctan(-1/3*sqrt(3)*(x - sqrt(x^2 - 2*x + 5))) + 1/12*sqrt(3)*arctan(-1/3*sqrt(3)*(x - sqrt(x^2
- 2*x + 5) - 2)) + 1/156*sqrt(13)*log(abs(-2*x - 2*sqrt(13) + 2*sqrt(x^2 - 2*x + 5) - 4)/abs(-2*x + 2*sqrt(13)
+ 2*sqrt(x^2 - 2*x + 5) - 4)) + 1/24*log((x - sqrt(x^2 - 2*x + 5))^2 - 4*x + 4*sqrt(x^2 - 2*x + 5) + 7) - 1/2
4*log((x - sqrt(x^2 - 2*x + 5))^2 + 3)

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maple [A]  time = 0.03, size = 69, normalized size = 0.82 \[ \frac {\arctanh \left (\sqrt {x^{2}-2 x +5}\right )}{12}-\frac {\sqrt {13}\, \arctanh \left (\frac {\left (-6 x +14\right ) \sqrt {13}}{26 \sqrt {-6 x +\left (x +2\right )^{2}+1}}\right )}{156}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (2 x -2\right )}{6 \sqrt {x^{2}-2 x +5}}\right )}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3+8)/(x^2-2*x+5)^(1/2),x)

[Out]

-1/156*13^(1/2)*arctanh(1/26*(14-6*x)*13^(1/2)/((x+2)^2-6*x+1)^(1/2))+1/12*arctanh((x^2-2*x+5)^(1/2))+1/12*3^(
1/2)*arctan(1/6*3^(1/2)/(x^2-2*x+5)^(1/2)*(2*x-2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (x^{3} + 8\right )} \sqrt {x^{2} - 2 \, x + 5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3+8)/(x^2-2*x+5)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((x^3 + 8)*sqrt(x^2 - 2*x + 5)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\left (x^3+8\right )\,\sqrt {x^2-2\,x+5}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((x^3 + 8)*(x^2 - 2*x + 5)^(1/2)),x)

[Out]

int(1/((x^3 + 8)*(x^2 - 2*x + 5)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (x + 2\right ) \left (x^{2} - 2 x + 4\right ) \sqrt {x^{2} - 2 x + 5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x**3+8)/(x**2-2*x+5)**(1/2),x)

[Out]

Integral(1/((x + 2)*(x**2 - 2*x + 4)*sqrt(x**2 - 2*x + 5)), x)

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