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

Optimal. Leaf size=56 \[ \frac{3}{2} \tanh ^{-1}\left (\frac{2 x^2+5}{2 \sqrt{x^4+5 x^2+3}}\right )-\frac{47 x^2+33}{13 \sqrt{x^4+5 x^2+3}} \]

[Out]

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

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Rubi [A]  time = 0.0434781, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {1251, 777, 621, 206} \[ \frac{3}{2} \tanh ^{-1}\left (\frac{2 x^2+5}{2 \sqrt{x^4+5 x^2+3}}\right )-\frac{47 x^2+33}{13 \sqrt{x^4+5 x^2+3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 777

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((2
*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (b^2*e*g - b*c*(e*f + d*g) + 2*c*(c*d*f - a*e*g))*x)*(a + b*x + c*x^2)^
(p + 1))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p
+ 3))/(c*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && N
eQ[b^2 - 4*a*c, 0] && LtQ[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^3 \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+3 x)}{\left (3+5 x+x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=-\frac{33+47 x^2}{13 \sqrt{3+5 x^2+x^4}}+\frac{3}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{3+5 x+x^2}} \, dx,x,x^2\right )\\ &=-\frac{33+47 x^2}{13 \sqrt{3+5 x^2+x^4}}+3 \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{33+47 x^2}{13 \sqrt{3+5 x^2+x^4}}+\frac{3}{2} \tanh ^{-1}\left (\frac{5+2 x^2}{2 \sqrt{3+5 x^2+x^4}}\right )\\ \end{align*}

Mathematica [A]  time = 0.124872, size = 54, normalized size = 0.96 \[ \frac{3}{2} \log \left (2 x^2+2 \sqrt{x^4+5 x^2+3}+5\right )-\frac{47 x^2+33}{13 \sqrt{x^4+5 x^2+3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(33 + 47*x^2)/(13*Sqrt[3 + 5*x^2 + x^4]) + (3*Log[5 + 2*x^2 + 2*Sqrt[3 + 5*x^2 + x^4]])/2

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Maple [B]  time = 0.011, size = 95, normalized size = 1.7 \begin{align*} -{\frac{3\,{x}^{2}}{2}{\frac{1}{\sqrt{{x}^{4}+5\,{x}^{2}+3}}}}+{\frac{15}{4}{\frac{1}{\sqrt{{x}^{4}+5\,{x}^{2}+3}}}}-{\frac{150\,{x}^{2}+375}{52}{\frac{1}{\sqrt{{x}^{4}+5\,{x}^{2}+3}}}}+{\frac{3}{2}\ln \left ({\frac{5}{2}}+{x}^{2}+\sqrt{{x}^{4}+5\,{x}^{2}+3} \right ) }+{\frac{10\,{x}^{2}+12}{13}{\frac{1}{\sqrt{{x}^{4}+5\,{x}^{2}+3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*x^2/(x^4+5*x^2+3)^(1/2)+15/4/(x^4+5*x^2+3)^(1/2)-75/52*(2*x^2+5)/(x^4+5*x^2+3)^(1/2)+3/2*ln(5/2+x^2+(x^4+
5*x^2+3)^(1/2))+2/13/(x^4+5*x^2+3)^(1/2)*(5*x^2+6)

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Maxima [A]  time = 0.959788, size = 76, normalized size = 1.36 \begin{align*} -\frac{47 \, x^{2}}{13 \, \sqrt{x^{4} + 5 \, x^{2} + 3}} - \frac{33}{13 \, \sqrt{x^{4} + 5 \, x^{2} + 3}} + \frac{3}{2} \, \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^3*(3*x^2+2)/(x^4+5*x^2+3)^(3/2),x, algorithm="maxima")

[Out]

-47/13*x^2/sqrt(x^4 + 5*x^2 + 3) - 33/13/sqrt(x^4 + 5*x^2 + 3) + 3/2*log(2*x^2 + 2*sqrt(x^4 + 5*x^2 + 3) + 5)

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Fricas [A]  time = 1.23437, size = 209, normalized size = 3.73 \begin{align*} -\frac{94 \, x^{4} + 470 \, x^{2} + 39 \,{\left (x^{4} + 5 \, x^{2} + 3\right )} \log \left (-2 \, x^{2} + 2 \, \sqrt{x^{4} + 5 \, x^{2} + 3} - 5\right ) + 2 \, \sqrt{x^{4} + 5 \, x^{2} + 3}{\left (47 \, x^{2} + 33\right )} + 282}{26 \,{\left (x^{4} + 5 \, x^{2} + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/26*(94*x^4 + 470*x^2 + 39*(x^4 + 5*x^2 + 3)*log(-2*x^2 + 2*sqrt(x^4 + 5*x^2 + 3) - 5) + 2*sqrt(x^4 + 5*x^2
+ 3)*(47*x^2 + 33) + 282)/(x^4 + 5*x^2 + 3)

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

[Out]

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

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Giac [A]  time = 1.14948, size = 62, normalized size = 1.11 \begin{align*} -\frac{47 \, x^{2} + 33}{13 \, \sqrt{x^{4} + 5 \, x^{2} + 3}} - \frac{3}{2} \, \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^3*(3*x^2+2)/(x^4+5*x^2+3)^(3/2),x, algorithm="giac")

[Out]

-1/13*(47*x^2 + 33)/sqrt(x^4 + 5*x^2 + 3) - 3/2*log(2*x^2 - 2*sqrt(x^4 + 5*x^2 + 3) + 5)