3.194 \(\int \frac{x^5 (2+3 x^2)}{(3+5 x^2+x^4)^{3/2}} \, dx\)

Optimal. Leaf size=77 \[ -\frac{\left (47 x^2+33\right ) x^2}{13 \sqrt{x^4+5 x^2+3}}+\frac{133}{26} \sqrt{x^4+5 x^2+3}-\frac{41}{4} \tanh ^{-1}\left (\frac{2 x^2+5}{2 \sqrt{x^4+5 x^2+3}}\right ) \]

[Out]

-(x^2*(33 + 47*x^2))/(13*Sqrt[3 + 5*x^2 + x^4]) + (133*Sqrt[3 + 5*x^2 + x^4])/26 - (41*ArcTanh[(5 + 2*x^2)/(2*
Sqrt[3 + 5*x^2 + x^4])])/4

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Rubi [A]  time = 0.0576965, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1251, 818, 640, 621, 206} \[ -\frac{\left (47 x^2+33\right ) x^2}{13 \sqrt{x^4+5 x^2+3}}+\frac{133}{26} \sqrt{x^4+5 x^2+3}-\frac{41}{4} \tanh ^{-1}\left (\frac{2 x^2+5}{2 \sqrt{x^4+5 x^2+3}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(2 + 3*x^2))/(3 + 5*x^2 + x^4)^(3/2),x]

[Out]

-(x^2*(33 + 47*x^2))/(13*Sqrt[3 + 5*x^2 + x^4]) + (133*Sqrt[3 + 5*x^2 + x^4])/26 - (41*ArcTanh[(5 + 2*x^2)/(2*
Sqrt[3 + 5*x^2 + x^4])])/4

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 818

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g
- c*(b*e*f + b*d*g + 2*a*e*g))*x))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[1/(c*(p + 1)*(b^2 - 4*a*c)), Int[(d +
e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Simp[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a
*e*(e*f*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*(m + p + 1) + 2*c^2*d*f*(m +
2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2*p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] &&
RationalQ[a, b, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 640

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

Rule 621

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

Rubi steps

\begin{align*} \int \frac{x^5 \left (2+3 x^2\right )}{\left (3+5 x^2+x^4\right )^{3/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2 (2+3 x)}{\left (3+5 x+x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=-\frac{x^2 \left (33+47 x^2\right )}{13 \sqrt{3+5 x^2+x^4}}+\frac{1}{13} \operatorname{Subst}\left (\int \frac{33+\frac{133 x}{2}}{\sqrt{3+5 x+x^2}} \, dx,x,x^2\right )\\ &=-\frac{x^2 \left (33+47 x^2\right )}{13 \sqrt{3+5 x^2+x^4}}+\frac{133}{26} \sqrt{3+5 x^2+x^4}-\frac{41}{4} \operatorname{Subst}\left (\int \frac{1}{\sqrt{3+5 x+x^2}} \, dx,x,x^2\right )\\ &=-\frac{x^2 \left (33+47 x^2\right )}{13 \sqrt{3+5 x^2+x^4}}+\frac{133}{26} \sqrt{3+5 x^2+x^4}-\frac{41}{2} \operatorname{Subst}\left (\int \frac{1}{4-x^2} \, dx,x,\frac{5+2 x^2}{\sqrt{3+5 x^2+x^4}}\right )\\ &=-\frac{x^2 \left (33+47 x^2\right )}{13 \sqrt{3+5 x^2+x^4}}+\frac{133}{26} \sqrt{3+5 x^2+x^4}-\frac{41}{4} \tanh ^{-1}\left (\frac{5+2 x^2}{2 \sqrt{3+5 x^2+x^4}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0243875, size = 72, normalized size = 0.94 \[ \frac{78 x^4+1198 x^2-533 \sqrt{x^4+5 x^2+3} \tanh ^{-1}\left (\frac{2 x^2+5}{2 \sqrt{x^4+5 x^2+3}}\right )+798}{52 \sqrt{x^4+5 x^2+3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(2 + 3*x^2))/(3 + 5*x^2 + x^4)^(3/2),x]

[Out]

(798 + 1198*x^2 + 78*x^4 - 533*Sqrt[3 + 5*x^2 + x^4]*ArcTanh[(5 + 2*x^2)/(2*Sqrt[3 + 5*x^2 + x^4])])/(52*Sqrt[
3 + 5*x^2 + x^4])

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Maple [A]  time = 0.019, size = 91, normalized size = 1.2 \begin{align*}{\frac{3\,{x}^{4}}{2}{\frac{1}{\sqrt{{x}^{4}+5\,{x}^{2}+3}}}}+{\frac{41\,{x}^{2}}{4}{\frac{1}{\sqrt{{x}^{4}+5\,{x}^{2}+3}}}}-{\frac{133}{8}{\frac{1}{\sqrt{{x}^{4}+5\,{x}^{2}+3}}}}+{\frac{1330\,{x}^{2}+3325}{104}{\frac{1}{\sqrt{{x}^{4}+5\,{x}^{2}+3}}}}-{\frac{41}{4}\ln \left ({\frac{5}{2}}+{x}^{2}+\sqrt{{x}^{4}+5\,{x}^{2}+3} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(3*x^2+2)/(x^4+5*x^2+3)^(3/2),x)

[Out]

3/2*x^4/(x^4+5*x^2+3)^(1/2)+41/4*x^2/(x^4+5*x^2+3)^(1/2)-133/8/(x^4+5*x^2+3)^(1/2)+665/104*(2*x^2+5)/(x^4+5*x^
2+3)^(1/2)-41/4*ln(5/2+x^2+(x^4+5*x^2+3)^(1/2))

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Maxima [A]  time = 0.965985, size = 99, normalized size = 1.29 \begin{align*} \frac{3 \, x^{4}}{2 \, \sqrt{x^{4} + 5 \, x^{2} + 3}} + \frac{599 \, x^{2}}{26 \, \sqrt{x^{4} + 5 \, x^{2} + 3}} + \frac{399}{26 \, \sqrt{x^{4} + 5 \, x^{2} + 3}} - \frac{41}{4} \, \log \left (2 \, x^{2} + 2 \, \sqrt{x^{4} + 5 \, x^{2} + 3} + 5\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(3*x^2+2)/(x^4+5*x^2+3)^(3/2),x, algorithm="maxima")

[Out]

3/2*x^4/sqrt(x^4 + 5*x^2 + 3) + 599/26*x^2/sqrt(x^4 + 5*x^2 + 3) + 399/26/sqrt(x^4 + 5*x^2 + 3) - 41/4*log(2*x
^2 + 2*sqrt(x^4 + 5*x^2 + 3) + 5)

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Fricas [A]  time = 1.45239, size = 232, normalized size = 3.01 \begin{align*} \frac{1811 \, x^{4} + 9055 \, x^{2} + 1066 \,{\left (x^{4} + 5 \, x^{2} + 3\right )} \log \left (-2 \, x^{2} + 2 \, \sqrt{x^{4} + 5 \, x^{2} + 3} - 5\right ) + 4 \,{\left (39 \, x^{4} + 599 \, x^{2} + 399\right )} \sqrt{x^{4} + 5 \, x^{2} + 3} + 5433}{104 \,{\left (x^{4} + 5 \, x^{2} + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(3*x^2+2)/(x^4+5*x^2+3)^(3/2),x, algorithm="fricas")

[Out]

1/104*(1811*x^4 + 9055*x^2 + 1066*(x^4 + 5*x^2 + 3)*log(-2*x^2 + 2*sqrt(x^4 + 5*x^2 + 3) - 5) + 4*(39*x^4 + 59
9*x^2 + 399)*sqrt(x^4 + 5*x^2 + 3) + 5433)/(x^4 + 5*x^2 + 3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5} \left (3 x^{2} + 2\right )}{\left (x^{4} + 5 x^{2} + 3\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(3*x**2+2)/(x**4+5*x**2+3)**(3/2),x)

[Out]

Integral(x**5*(3*x**2 + 2)/(x**4 + 5*x**2 + 3)**(3/2), x)

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Giac [A]  time = 1.12356, size = 70, normalized size = 0.91 \begin{align*} \frac{{\left (39 \, x^{2} + 599\right )} x^{2} + 399}{26 \, \sqrt{x^{4} + 5 \, x^{2} + 3}} + \frac{41}{4} \, \log \left (2 \, x^{2} - 2 \, \sqrt{x^{4} + 5 \, x^{2} + 3} + 5\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(3*x^2+2)/(x^4+5*x^2+3)^(3/2),x, algorithm="giac")

[Out]

1/26*((39*x^2 + 599)*x^2 + 399)/sqrt(x^4 + 5*x^2 + 3) + 41/4*log(2*x^2 - 2*sqrt(x^4 + 5*x^2 + 3) + 5)