3.143 \(\int x^3 (2+3 x^2) \sqrt{3+5 x^2+x^4} \, dx\)

Optimal. Leaf size=81 \[ -\frac{1}{48} \left (59-18 x^2\right ) \left (x^4+5 x^2+3\right )^{3/2}+\frac{259}{128} \left (2 x^2+5\right ) \sqrt{x^4+5 x^2+3}-\frac{3367}{256} \tanh ^{-1}\left (\frac{2 x^2+5}{2 \sqrt{x^4+5 x^2+3}}\right ) \]

[Out]

(259*(5 + 2*x^2)*Sqrt[3 + 5*x^2 + x^4])/128 - ((59 - 18*x^2)*(3 + 5*x^2 + x^4)^(3/2))/48 - (3367*ArcTanh[(5 +
2*x^2)/(2*Sqrt[3 + 5*x^2 + x^4])])/256

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Rubi [A]  time = 0.0562794, antiderivative size = 81, 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, 779, 612, 621, 206} \[ -\frac{1}{48} \left (59-18 x^2\right ) \left (x^4+5 x^2+3\right )^{3/2}+\frac{259}{128} \left (2 x^2+5\right ) \sqrt{x^4+5 x^2+3}-\frac{3367}{256} \tanh ^{-1}\left (\frac{2 x^2+5}{2 \sqrt{x^4+5 x^2+3}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(259*(5 + 2*x^2)*Sqrt[3 + 5*x^2 + x^4])/128 - ((59 - 18*x^2)*(3 + 5*x^2 + x^4)^(3/2))/48 - (3367*ArcTanh[(5 +
2*x^2)/(2*Sqrt[3 + 5*x^2 + x^4])])/256

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 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), 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))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

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 x^3 \left (2+3 x^2\right ) \sqrt{3+5 x^2+x^4} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x (2+3 x) \sqrt{3+5 x+x^2} \, dx,x,x^2\right )\\ &=-\frac{1}{48} \left (59-18 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}+\frac{259}{32} \operatorname{Subst}\left (\int \sqrt{3+5 x+x^2} \, dx,x,x^2\right )\\ &=\frac{259}{128} \left (5+2 x^2\right ) \sqrt{3+5 x^2+x^4}-\frac{1}{48} \left (59-18 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}-\frac{3367}{256} \operatorname{Subst}\left (\int \frac{1}{\sqrt{3+5 x+x^2}} \, dx,x,x^2\right )\\ &=\frac{259}{128} \left (5+2 x^2\right ) \sqrt{3+5 x^2+x^4}-\frac{1}{48} \left (59-18 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}-\frac{3367}{128} \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{259}{128} \left (5+2 x^2\right ) \sqrt{3+5 x^2+x^4}-\frac{1}{48} \left (59-18 x^2\right ) \left (3+5 x^2+x^4\right )^{3/2}-\frac{3367}{256} \tanh ^{-1}\left (\frac{5+2 x^2}{2 \sqrt{3+5 x^2+x^4}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0207215, size = 66, normalized size = 0.81 \[ \frac{1}{768} \left (2 \sqrt{x^4+5 x^2+3} \left (144 x^6+248 x^4-374 x^2+2469\right )-10101 \tanh ^{-1}\left (\frac{2 x^2+5}{2 \sqrt{x^4+5 x^2+3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[3 + 5*x^2 + x^4]*(2469 - 374*x^2 + 248*x^4 + 144*x^6) - 10101*ArcTanh[(5 + 2*x^2)/(2*Sqrt[3 + 5*x^2 +
x^4])])/768

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Maple [A]  time = 0.014, size = 74, normalized size = 0.9 \begin{align*}{\frac{3\,{x}^{2}}{8} \left ({x}^{4}+5\,{x}^{2}+3 \right ) ^{{\frac{3}{2}}}}-{\frac{59}{48} \left ({x}^{4}+5\,{x}^{2}+3 \right ) ^{{\frac{3}{2}}}}+{\frac{518\,{x}^{2}+1295}{128}\sqrt{{x}^{4}+5\,{x}^{2}+3}}-{\frac{3367}{256}\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^3*(3*x^2+2)*(x^4+5*x^2+3)^(1/2),x)

[Out]

3/8*x^2*(x^4+5*x^2+3)^(3/2)-59/48*(x^4+5*x^2+3)^(3/2)+259/128*(2*x^2+5)*(x^4+5*x^2+3)^(1/2)-3367/256*ln(5/2+x^
2+(x^4+5*x^2+3)^(1/2))

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Maxima [A]  time = 0.977678, size = 117, normalized size = 1.44 \begin{align*} \frac{3}{8} \,{\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac{3}{2}} x^{2} + \frac{259}{64} \, \sqrt{x^{4} + 5 \, x^{2} + 3} x^{2} - \frac{59}{48} \,{\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac{3}{2}} + \frac{1295}{128} \, \sqrt{x^{4} + 5 \, x^{2} + 3} - \frac{3367}{256} \, \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)^(1/2),x, algorithm="maxima")

[Out]

3/8*(x^4 + 5*x^2 + 3)^(3/2)*x^2 + 259/64*sqrt(x^4 + 5*x^2 + 3)*x^2 - 59/48*(x^4 + 5*x^2 + 3)^(3/2) + 1295/128*
sqrt(x^4 + 5*x^2 + 3) - 3367/256*log(2*x^2 + 2*sqrt(x^4 + 5*x^2 + 3) + 5)

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Fricas [A]  time = 1.63903, size = 161, normalized size = 1.99 \begin{align*} \frac{1}{384} \,{\left (144 \, x^{6} + 248 \, x^{4} - 374 \, x^{2} + 2469\right )} \sqrt{x^{4} + 5 \, x^{2} + 3} + \frac{3367}{256} \, \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)^(1/2),x, algorithm="fricas")

[Out]

1/384*(144*x^6 + 248*x^4 - 374*x^2 + 2469)*sqrt(x^4 + 5*x^2 + 3) + 3367/256*log(-2*x^2 + 2*sqrt(x^4 + 5*x^2 +
3) - 5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \left (3 x^{2} + 2\right ) \sqrt{x^{4} + 5 x^{2} + 3}\, 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)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.08371, size = 81, normalized size = 1. \begin{align*} \frac{1}{384} \, \sqrt{x^{4} + 5 \, x^{2} + 3}{\left (2 \,{\left (4 \,{\left (18 \, x^{2} + 31\right )} x^{2} - 187\right )} x^{2} + 2469\right )} + \frac{3367}{256} \, \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)^(1/2),x, algorithm="giac")

[Out]

1/384*sqrt(x^4 + 5*x^2 + 3)*(2*(4*(18*x^2 + 31)*x^2 - 187)*x^2 + 2469) + 3367/256*log(2*x^2 - 2*sqrt(x^4 + 5*x
^2 + 3) + 5)