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]

-1/48*(-18*x^2+59)*(x^4+5*x^2+3)^(3/2)-3367/256*arctanh(1/2*(2*x^2+5)/(x^4+5*x^2+3)^(1/2))+259/128*(2*x^2+5)*(
x^4+5*x^2+3)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.02, 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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fricas [A]  time = 0.71, size = 56, normalized size = 0.69 \[ \frac {1}{384} \, {\left (144 \, x^{6} + 248 \, x^{4} - 374 \, x^{2} + 2469\right )} \sqrt {x^{4} + 5 \, x^{2} + 3} + \frac {3367}{256} \, \log \left (-2 \, x^{2} + 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} - 5\right ) \]

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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giac [A]  time = 0.36, size = 88, normalized size = 1.09 \[ \frac {1}{128} \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (2 \, {\left (4 \, {\left (6 \, x^{2} + 5\right )} x^{2} - 89\right )} x^{2} + 1095\right )} + \frac {1}{24} \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (2 \, {\left (4 \, x^{2} + 5\right )} x^{2} - 51\right )} + \frac {3367}{256} \, \log \left (2 \, x^{2} - 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} + 5\right ) \]

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]

1/128*sqrt(x^4 + 5*x^2 + 3)*(2*(4*(6*x^2 + 5)*x^2 - 89)*x^2 + 1095) + 1/24*sqrt(x^4 + 5*x^2 + 3)*(2*(4*x^2 + 5
)*x^2 - 51) + 3367/256*log(2*x^2 - 2*sqrt(x^4 + 5*x^2 + 3) + 5)

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maple [A]  time = 0.02, size = 74, normalized size = 0.91 \[ \frac {3 \left (x^{4}+5 x^{2}+3\right )^{\frac {3}{2}} x^{2}}{8}-\frac {3367 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{256}-\frac {59 \left (x^{4}+5 x^{2}+3\right )^{\frac {3}{2}}}{48}+\frac {259 \left (2 x^{2}+5\right ) \sqrt {x^{4}+5 x^{2}+3}}{128} \]

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^4+5*x^2+3)^(3/2)*x^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(x^2+5/
2+(x^4+5*x^2+3)^(1/2))

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maxima [A]  time = 0.59, size = 87, normalized size = 1.07 \[ \frac {3}{8} \, {\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac {3}{2}} x^{2} + \frac {259}{64} \, \sqrt {x^{4} + 5 \, x^{2} + 3} x^{2} - \frac {59}{48} \, {\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac {3}{2}} + \frac {1295}{128} \, \sqrt {x^{4} + 5 \, x^{2} + 3} - \frac {3367}{256} \, \log \left (2 \, x^{2} + 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} + 5\right ) \]

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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mupad [B]  time = 0.43, size = 85, normalized size = 1.05 \[ \frac {3\,x^2\,{\left (x^4+5\,x^2+3\right )}^{3/2}}{8}-\frac {3367\,\ln \left (\sqrt {x^4+5\,x^2+3}+x^2+\frac {5}{2}\right )}{256}-\frac {9\,\left (\frac {x^2}{2}+\frac {5}{4}\right )\,\sqrt {x^4+5\,x^2+3}}{8}-\frac {59\,\sqrt {x^4+5\,x^2+3}\,\left (8\,x^4+10\,x^2-51\right )}{384} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(3*x^2 + 2)*(5*x^2 + x^4 + 3)^(1/2),x)

[Out]

(3*x^2*(5*x^2 + x^4 + 3)^(3/2))/8 - (3367*log((5*x^2 + x^4 + 3)^(1/2) + x^2 + 5/2))/256 - (9*(x^2/2 + 5/4)*(5*
x^2 + x^4 + 3)^(1/2))/8 - (59*(5*x^2 + x^4 + 3)^(1/2)*(10*x^2 + 8*x^4 - 51))/384

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \left (3 x^{2} + 2\right ) \sqrt {x^{4} + 5 x^{2} + 3}\, dx \]

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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