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

Optimal. Leaf size=56 \[ \frac {149}{16} \tanh ^{-1}\left (\frac {2 x^2+5}{2 \sqrt {x^4+5 x^2+3}}\right )-\frac {1}{8} \left (37-6 x^2\right ) \sqrt {x^4+5 x^2+3} \]

[Out]

149/16*arctanh(1/2*(2*x^2+5)/(x^4+5*x^2+3)^(1/2))-1/8*(-6*x^2+37)*(x^4+5*x^2+3)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1251, 779, 621, 206} \[ \frac {149}{16} \tanh ^{-1}\left (\frac {2 x^2+5}{2 \sqrt {x^4+5 x^2+3}}\right )-\frac {1}{8} \left (37-6 x^2\right ) \sqrt {x^4+5 x^2+3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((37 - 6*x^2)*Sqrt[3 + 5*x^2 + x^4])/8 + (149*ArcTanh[(5 + 2*x^2)/(2*Sqrt[3 + 5*x^2 + x^4])])/16

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 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 \frac {x^3 \left (2+3 x^2\right )}{\sqrt {3+5 x^2+x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x (2+3 x)}{\sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=-\frac {1}{8} \left (37-6 x^2\right ) \sqrt {3+5 x^2+x^4}+\frac {149}{16} \operatorname {Subst}\left (\int \frac {1}{\sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=-\frac {1}{8} \left (37-6 x^2\right ) \sqrt {3+5 x^2+x^4}+\frac {149}{8} \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 {1}{8} \left (37-6 x^2\right ) \sqrt {3+5 x^2+x^4}+\frac {149}{16} \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 = 56, normalized size = 1.00 \[ \frac {1}{16} \left (2 \sqrt {x^4+5 x^2+3} \left (6 x^2-37\right )+149 \tanh ^{-1}\left (\frac {2 x^2+5}{2 \sqrt {x^4+5 x^2+3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-37 + 6*x^2)*Sqrt[3 + 5*x^2 + x^4] + 149*ArcTanh[(5 + 2*x^2)/(2*Sqrt[3 + 5*x^2 + x^4])])/16

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fricas [A]  time = 0.54, size = 46, normalized size = 0.82 \[ \frac {1}{8} \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (6 \, x^{2} - 37\right )} - \frac {149}{16} \, \log \left (-2 \, x^{2} + 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} - 5\right ) \]

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]

1/8*sqrt(x^4 + 5*x^2 + 3)*(6*x^2 - 37) - 149/16*log(-2*x^2 + 2*sqrt(x^4 + 5*x^2 + 3) - 5)

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giac [A]  time = 0.35, size = 46, normalized size = 0.82 \[ \frac {1}{8} \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (6 \, x^{2} - 37\right )} - \frac {149}{16} \, \log \left (2 \, x^{2} - 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} + 5\right ) \]

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]

1/8*sqrt(x^4 + 5*x^2 + 3)*(6*x^2 - 37) - 149/16*log(2*x^2 - 2*sqrt(x^4 + 5*x^2 + 3) + 5)

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maple [A]  time = 0.01, size = 53, normalized size = 0.95 \[ \frac {3 \sqrt {x^{4}+5 x^{2}+3}\, x^{2}}{4}+\frac {149 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{16}-\frac {37 \sqrt {x^{4}+5 x^{2}+3}}{8} \]

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]

3/4*(x^4+5*x^2+3)^(1/2)*x^2-37/8*(x^4+5*x^2+3)^(1/2)+149/16*ln(x^2+5/2+(x^4+5*x^2+3)^(1/2))

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maxima [A]  time = 1.04, size = 56, normalized size = 1.00 \[ \frac {3}{4} \, \sqrt {x^{4} + 5 \, x^{2} + 3} x^{2} - \frac {37}{8} \, \sqrt {x^{4} + 5 \, x^{2} + 3} + \frac {149}{16} \, \log \left (2 \, x^{2} + 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} + 5\right ) \]

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]

3/4*sqrt(x^4 + 5*x^2 + 3)*x^2 - 37/8*sqrt(x^4 + 5*x^2 + 3) + 149/16*log(2*x^2 + 2*sqrt(x^4 + 5*x^2 + 3) + 5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^3\,\left (3\,x^2+2\right )}{\sqrt {x^4+5\,x^2+3}} \,d x \]

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]

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

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

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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