3.2.83 \(\int \frac {x^3 (2+3 x^2)}{\sqrt {3+5 x^2+x^4}} \, dx\) [183]

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

[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)

________________________________________________________________________________________

Rubi [A]
time = 0.03, 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 = {1265, 793, 635, 212} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*((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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

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 793

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 1265

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} \text {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} \text {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} \text {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*}

________________________________________________________________________________________

Mathematica [A]
time = 0.09, size = 54, normalized size = 0.96 \begin {gather*} \frac {1}{8} \left (-37+6 x^2\right ) \sqrt {3+5 x^2+x^4}-\frac {149}{16} \log \left (-5-2 x^2+2 \sqrt {3+5 x^2+x^4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(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*Log[-5 - 2*x^2 + 2*Sqrt[3 + 5*x^2 + x^4]])/16

________________________________________________________________________________________

Maple [A]
time = 0.10, size = 53, normalized size = 0.95

method	result	size
risch	\(\frac {\left (6 x^{2}-37\right ) \sqrt {x^{4}+5 x^{2}+3}}{8}+\frac {149 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{16}\)	\(43\)
trager	\(\left (\frac {3 x^{2}}{4}-\frac {37}{8}\right ) \sqrt {x^{4}+5 x^{2}+3}+\frac {149 \ln \left (2 x^{2}+5+2 \sqrt {x^{4}+5 x^{2}+3}\right )}{16}\)	\(46\)
default	\(\frac {3 x^{2} \sqrt {x^{4}+5 x^{2}+3}}{4}-\frac {37 \sqrt {x^{4}+5 x^{2}+3}}{8}+\frac {149 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{16}\)	\(53\)
elliptic	\(\frac {3 x^{2} \sqrt {x^{4}+5 x^{2}+3}}{4}-\frac {37 \sqrt {x^{4}+5 x^{2}+3}}{8}+\frac {149 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{16}\)	\(53\)

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,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.27, size = 56, normalized size = 1.00 \begin {gather*} \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 ) \end {gather*}

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)

________________________________________________________________________________________

Fricas [A]
time = 0.37, size = 46, normalized size = 0.82 \begin {gather*} \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 ) \end {gather*}

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

Giac [A]
time = 4.03, size = 46, normalized size = 0.82 \begin {gather*} \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 ) \end {gather*}

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________