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]

-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

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Rubi [A]  time = 0.122149, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, 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 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]

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 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 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 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 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 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 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])

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*}

Mathematica [C]  time = 0.309522, size = 160, normalized size = 1.9 \[ \frac{1}{312} \left (-2 \sqrt{13} \tanh ^{-1}\left (\frac{7-3 x}{\sqrt{13} \sqrt{x^2-2 x+5}}\right )-13 \left (\left (\sqrt{3}+i\right ) \tan ^{-1}\left (\frac{-2 \sqrt [3]{-1} x+4 x+5 i \sqrt{3}+1}{\sqrt{2-2 i \sqrt{3}} \sqrt{x^2-2 x+5}}\right )+\left (\sqrt{3}-i\right ) \tan ^{-1}\left (\frac{2 \left (2+(-1)^{2/3}\right ) x-5 i \sqrt{3}+1}{\sqrt{2+2 i \sqrt{3}} \sqrt{x^2-2 x+5}}\right )\right )\right ) \]

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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Maple [A]  time = 0.018, size = 69, normalized size = 0.8 \begin{align*} -{\frac{\sqrt{13}}{156}{\it Artanh} \left ({\frac{ \left ( 14-6\,x \right ) \sqrt{13}}{26}{\frac{1}{\sqrt{ \left ( 2+x \right ) ^{2}-6\,x+1}}}} \right ) }+{\frac{1}{12}{\it Artanh} \left ( \sqrt{{x}^{2}-2\,x+5} \right ) }+{\frac{\sqrt{3}}{12}\arctan \left ({\frac{\sqrt{3} \left ( 2\,x-2 \right ) }{6}{\frac{1}{\sqrt{{x}^{2}-2\,x+5}}}} \right ) } \end{align*}

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)/((2+x)^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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x^{3} + 8\right )} \sqrt{x^{2} - 2 \, x + 5}}\,{d x} \end{align*}

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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Fricas [B]  time = 1.75505, size = 481, normalized size = 5.73 \begin{align*} \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 ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (x + 2\right ) \left (x^{2} - 2 x + 4\right ) \sqrt{x^{2} - 2 x + 5}}\, dx \end{align*}

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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Giac [B]  time = 1.34712, size = 221, normalized size = 2.63 \begin{align*} -\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 ) \end{align*}

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)