3.2.82 \(\int \frac {x^5 (2+3 x^2)}{\sqrt {3+5 x^2+x^4}} \, dx\) [182]

Optimal. Leaf size=77 \[ \frac {1}{2} x^4 \sqrt {3+5 x^2+x^4}+\frac {3}{16} \left (89-14 x^2\right ) \sqrt {3+5 x^2+x^4}-\frac {1083}{32} \tanh ^{-1}\left (\frac {5+2 x^2}{2 \sqrt {3+5 x^2+x^4}}\right ) \]

[Out]

-1083/32*arctanh(1/2*(2*x^2+5)/(x^4+5*x^2+3)^(1/2))+1/2*x^4*(x^4+5*x^2+3)^(1/2)+3/16*(-14*x^2+89)*(x^4+5*x^2+3
)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 77, 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 = {1265, 846, 793, 635, 212} \begin {gather*} \frac {1}{2} \sqrt {x^4+5 x^2+3} x^4+\frac {3}{16} \left (89-14 x^2\right ) \sqrt {x^4+5 x^2+3}-\frac {1083}{32} \tanh ^{-1}\left (\frac {2 x^2+5}{2 \sqrt {x^4+5 x^2+3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^4*Sqrt[3 + 5*x^2 + x^4])/2 + (3*(89 - 14*x^2)*Sqrt[3 + 5*x^2 + x^4])/16 - (1083*ArcTanh[(5 + 2*x^2)/(2*Sqrt
[3 + 5*x^2 + x^4])])/32

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 846

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m
- 1)*(a + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m*(c*e*f + c*d*g - b*e*g) + e*(p
+ 1)*(2*c*f - b*g))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
&&  !(IGtQ[m, 0] && EqQ[f, 0])

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^5 \left (2+3 x^2\right )}{\sqrt {3+5 x^2+x^4}} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {x^2 (2+3 x)}{\sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=\frac {1}{2} x^4 \sqrt {3+5 x^2+x^4}+\frac {1}{6} \text {Subst}\left (\int \frac {\left (-18-\frac {63 x}{2}\right ) x}{\sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=\frac {1}{2} x^4 \sqrt {3+5 x^2+x^4}+\frac {3}{16} \left (89-14 x^2\right ) \sqrt {3+5 x^2+x^4}-\frac {1083}{32} \text {Subst}\left (\int \frac {1}{\sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=\frac {1}{2} x^4 \sqrt {3+5 x^2+x^4}+\frac {3}{16} \left (89-14 x^2\right ) \sqrt {3+5 x^2+x^4}-\frac {1083}{16} \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}{2} x^4 \sqrt {3+5 x^2+x^4}+\frac {3}{16} \left (89-14 x^2\right ) \sqrt {3+5 x^2+x^4}-\frac {1083}{32} \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 = 0.10, size = 59, normalized size = 0.77 \begin {gather*} \frac {1}{16} \sqrt {3+5 x^2+x^4} \left (267-42 x^2+8 x^4\right )+\frac {1083}{32} \log \left (-5-2 x^2+2 \sqrt {3+5 x^2+x^4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[3 + 5*x^2 + x^4]*(267 - 42*x^2 + 8*x^4))/16 + (1083*Log[-5 - 2*x^2 + 2*Sqrt[3 + 5*x^2 + x^4]])/32

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Maple [A]
time = 0.10, size = 70, normalized size = 0.91

method	result	size
risch	\(\frac {\left (8 x^{4}-42 x^{2}+267\right ) \sqrt {x^{4}+5 x^{2}+3}}{16}-\frac {1083 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{32}\)	\(48\)
trager	\(\left (\frac {1}{2} x^{4}-\frac {21}{8} x^{2}+\frac {267}{16}\right ) \sqrt {x^{4}+5 x^{2}+3}-\frac {1083 \ln \left (2 x^{2}+5+2 \sqrt {x^{4}+5 x^{2}+3}\right )}{32}\)	\(51\)
default	\(\frac {x^{4} \sqrt {x^{4}+5 x^{2}+3}}{2}-\frac {21 x^{2} \sqrt {x^{4}+5 x^{2}+3}}{8}+\frac {267 \sqrt {x^{4}+5 x^{2}+3}}{16}-\frac {1083 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{32}\)	\(70\)
elliptic	\(\frac {x^{4} \sqrt {x^{4}+5 x^{2}+3}}{2}-\frac {21 x^{2} \sqrt {x^{4}+5 x^{2}+3}}{8}+\frac {267 \sqrt {x^{4}+5 x^{2}+3}}{16}-\frac {1083 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{32}\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^4*(x^4+5*x^2+3)^(1/2)-21/8*x^2*(x^4+5*x^2+3)^(1/2)+267/16*(x^4+5*x^2+3)^(1/2)-1083/32*ln(x^2+5/2+(x^4+5*
x^2+3)^(1/2))

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Maxima [A]
time = 0.28, size = 73, normalized size = 0.95 \begin {gather*} \frac {1}{2} \, \sqrt {x^{4} + 5 \, x^{2} + 3} x^{4} - \frac {21}{8} \, \sqrt {x^{4} + 5 \, x^{2} + 3} x^{2} + \frac {267}{16} \, \sqrt {x^{4} + 5 \, x^{2} + 3} - \frac {1083}{32} \, \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^5*(3*x^2+2)/(x^4+5*x^2+3)^(1/2),x, algorithm="maxima")

[Out]

1/2*sqrt(x^4 + 5*x^2 + 3)*x^4 - 21/8*sqrt(x^4 + 5*x^2 + 3)*x^2 + 267/16*sqrt(x^4 + 5*x^2 + 3) - 1083/32*log(2*
x^2 + 2*sqrt(x^4 + 5*x^2 + 3) + 5)

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Fricas [A]
time = 0.36, size = 51, normalized size = 0.66 \begin {gather*} \frac {1}{16} \, {\left (8 \, x^{4} - 42 \, x^{2} + 267\right )} \sqrt {x^{4} + 5 \, x^{2} + 3} + \frac {1083}{32} \, \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^5*(3*x^2+2)/(x^4+5*x^2+3)^(1/2),x, algorithm="fricas")

[Out]

1/16*(8*x^4 - 42*x^2 + 267)*sqrt(x^4 + 5*x^2 + 3) + 1083/32*log(-2*x^2 + 2*sqrt(x^4 + 5*x^2 + 3) - 5)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{5} \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**5*(3*x**2+2)/(x**4+5*x**2+3)**(1/2),x)

[Out]

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

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Giac [A]
time = 4.15, size = 53, normalized size = 0.69 \begin {gather*} \frac {1}{16} \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (2 \, {\left (4 \, x^{2} - 21\right )} x^{2} + 267\right )} + \frac {1083}{32} \, \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^5*(3*x^2+2)/(x^4+5*x^2+3)^(1/2),x, algorithm="giac")

[Out]

1/16*sqrt(x^4 + 5*x^2 + 3)*(2*(4*x^2 - 21)*x^2 + 267) + 1083/32*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 = -0.01 \begin {gather*} \int \frac {x^5\,\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^5*(3*x^2 + 2))/(5*x^2 + x^4 + 3)^(1/2),x)

[Out]

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

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