3.2.45 \(\int \frac {(2+3 x^2) \sqrt {3+5 x^2+x^4}}{x} \, dx\) [145]

Optimal. Leaf size=94 \[ \frac {1}{8} \left (23+6 x^2\right ) \sqrt {3+5 x^2+x^4}+\frac {1}{16} \tanh ^{-1}\left (\frac {5+2 x^2}{2 \sqrt {3+5 x^2+x^4}}\right )-\sqrt {3} \tanh ^{-1}\left (\frac {6+5 x^2}{2 \sqrt {3} \sqrt {3+5 x^2+x^4}}\right ) \]

[Out]

1/16*arctanh(1/2*(2*x^2+5)/(x^4+5*x^2+3)^(1/2))-arctanh(1/6*(5*x^2+6)*3^(1/2)/(x^4+5*x^2+3)^(1/2))*3^(1/2)+1/8
*(6*x^2+23)*(x^4+5*x^2+3)^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1265, 828, 857, 635, 212, 738} \begin {gather*} \frac {1}{8} \sqrt {x^4+5 x^2+3} \left (6 x^2+23\right )+\frac {1}{16} \tanh ^{-1}\left (\frac {2 x^2+5}{2 \sqrt {x^4+5 x^2+3}}\right )-\sqrt {3} \tanh ^{-1}\left (\frac {5 x^2+6}{2 \sqrt {3} \sqrt {x^4+5 x^2+3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((23 + 6*x^2)*Sqrt[3 + 5*x^2 + x^4])/8 + ArcTanh[(5 + 2*x^2)/(2*Sqrt[3 + 5*x^2 + x^4])]/16 - Sqrt[3]*ArcTanh[(
6 + 5*x^2)/(2*Sqrt[3]*Sqrt[3 + 5*x^2 + x^4])]

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 738

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 828

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

Rule 857

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 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 {\left (2+3 x^2\right ) \sqrt {3+5 x^2+x^4}}{x} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {(2+3 x) \sqrt {3+5 x+x^2}}{x} \, dx,x,x^2\right )\\ &=\frac {1}{8} \left (23+6 x^2\right ) \sqrt {3+5 x^2+x^4}-\frac {1}{8} \text {Subst}\left (\int \frac {-24-\frac {x}{2}}{x \sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=\frac {1}{8} \left (23+6 x^2\right ) \sqrt {3+5 x^2+x^4}+\frac {1}{16} \text {Subst}\left (\int \frac {1}{\sqrt {3+5 x+x^2}} \, dx,x,x^2\right )+3 \text {Subst}\left (\int \frac {1}{x \sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=\frac {1}{8} \left (23+6 x^2\right ) \sqrt {3+5 x^2+x^4}+\frac {1}{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 )-6 \text {Subst}\left (\int \frac {1}{12-x^2} \, dx,x,\frac {6+5 x^2}{\sqrt {3+5 x^2+x^4}}\right )\\ &=\frac {1}{8} \left (23+6 x^2\right ) \sqrt {3+5 x^2+x^4}+\frac {1}{16} \tanh ^{-1}\left (\frac {5+2 x^2}{2 \sqrt {3+5 x^2+x^4}}\right )-\sqrt {3} \tanh ^{-1}\left (\frac {6+5 x^2}{2 \sqrt {3} \sqrt {3+5 x^2+x^4}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 88, normalized size = 0.94 \begin {gather*} \frac {1}{8} \left (23+6 x^2\right ) \sqrt {3+5 x^2+x^4}+2 \sqrt {3} \tanh ^{-1}\left (\frac {x^2-\sqrt {3+5 x^2+x^4}}{\sqrt {3}}\right )-\frac {1}{16} \log \left (-5-2 x^2+2 \sqrt {3+5 x^2+x^4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((23 + 6*x^2)*Sqrt[3 + 5*x^2 + x^4])/8 + 2*Sqrt[3]*ArcTanh[(x^2 - Sqrt[3 + 5*x^2 + x^4])/Sqrt[3]] - Log[-5 - 2
*x^2 + 2*Sqrt[3 + 5*x^2 + x^4]]/16

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Maple [A]
time = 0.30, size = 85, normalized size = 0.90

method	result	size
elliptic	\(\frac {3 x^{2} \sqrt {x^{4}+5 x^{2}+3}}{4}+\frac {23 \sqrt {x^{4}+5 x^{2}+3}}{8}+\frac {\ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{16}-\arctanh \left (\frac {\left (5 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}+5 x^{2}+3}}\right ) \sqrt {3}\)	\(83\)
default	\(\frac {3 \left (2 x^{2}+5\right ) \sqrt {x^{4}+5 x^{2}+3}}{8}+\frac {\ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{16}+\sqrt {x^{4}+5 x^{2}+3}-\arctanh \left (\frac {\left (5 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}+5 x^{2}+3}}\right ) \sqrt {3}\)	\(85\)
trager	\(\left (\frac {3 x^{2}}{4}+\frac {23}{8}\right ) \sqrt {x^{4}+5 x^{2}+3}+\frac {\ln \left (-2 x^{2}-2 \sqrt {x^{4}+5 x^{2}+3}-5\right )}{16}-\RootOf \left (\textit {\_Z}^{2}-3\right ) \ln \left (-\frac {5 \RootOf \left (\textit {\_Z}^{2}-3\right ) x^{2}+6 \RootOf \left (\textit {\_Z}^{2}-3\right )+6 \sqrt {x^{4}+5 x^{2}+3}}{x^{2}}\right )\)	\(94\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 89, normalized size = 0.95 \begin {gather*} \frac {3}{4} \, \sqrt {x^{4} + 5 \, x^{2} + 3} x^{2} - \sqrt {3} \log \left (\frac {2 \, \sqrt {3} \sqrt {x^{4} + 5 \, x^{2} + 3}}{x^{2}} + \frac {6}{x^{2}} + 5\right ) + \frac {23}{8} \, \sqrt {x^{4} + 5 \, x^{2} + 3} + \frac {1}{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((3*x^2+2)*(x^4+5*x^2+3)^(1/2)/x,x, algorithm="maxima")

[Out]

3/4*sqrt(x^4 + 5*x^2 + 3)*x^2 - sqrt(3)*log(2*sqrt(3)*sqrt(x^4 + 5*x^2 + 3)/x^2 + 6/x^2 + 5) + 23/8*sqrt(x^4 +
5*x^2 + 3) + 1/16*log(2*x^2 + 2*sqrt(x^4 + 5*x^2 + 3) + 5)

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Fricas [A]
time = 0.38, size = 95, normalized size = 1.01 \begin {gather*} \frac {1}{8} \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (6 \, x^{2} + 23\right )} + \sqrt {3} \log \left (\frac {25 \, x^{2} - 2 \, \sqrt {3} {\left (5 \, x^{2} + 6\right )} - 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (5 \, \sqrt {3} - 6\right )} + 30}{x^{2}}\right ) - \frac {1}{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((3*x^2+2)*(x^4+5*x^2+3)^(1/2)/x,x, algorithm="fricas")

[Out]

1/8*sqrt(x^4 + 5*x^2 + 3)*(6*x^2 + 23) + sqrt(3)*log((25*x^2 - 2*sqrt(3)*(5*x^2 + 6) - 2*sqrt(x^4 + 5*x^2 + 3)
*(5*sqrt(3) - 6) + 30)/x^2) - 1/16*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 {\left (3 x^{2} + 2\right ) \sqrt {x^{4} + 5 x^{2} + 3}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 4.34, size = 98, normalized size = 1.04 \begin {gather*} \frac {1}{8} \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (6 \, x^{2} + 23\right )} + \sqrt {3} \log \left (\frac {x^{2} + \sqrt {3} - \sqrt {x^{4} + 5 \, x^{2} + 3}}{x^{2} - \sqrt {3} - \sqrt {x^{4} + 5 \, x^{2} + 3}}\right ) - \frac {1}{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((3*x^2+2)*(x^4+5*x^2+3)^(1/2)/x,x, algorithm="giac")

[Out]

1/8*sqrt(x^4 + 5*x^2 + 3)*(6*x^2 + 23) + sqrt(3)*log((x^2 + sqrt(3) - sqrt(x^4 + 5*x^2 + 3))/(x^2 - sqrt(3) -
sqrt(x^4 + 5*x^2 + 3))) - 1/16*log(2*x^2 - 2*sqrt(x^4 + 5*x^2 + 3) + 5)

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Mupad [B]
time = 0.43, size = 86, normalized size = 0.91 \begin {gather*} \frac {\ln \left (\sqrt {x^4+5\,x^2+3}+x^2+\frac {5}{2}\right )}{16}-\sqrt {3}\,\ln \left (\frac {3}{x^2}+\frac {\sqrt {3}\,\sqrt {x^4+5\,x^2+3}}{x^2}+\frac {5}{2}\right )+\frac {3\,\left (\frac {x^2}{2}+\frac {5}{4}\right )\,\sqrt {x^4+5\,x^2+3}}{2}+\sqrt {x^4+5\,x^2+3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((5*x^2 + x^4 + 3)^(1/2) + x^2 + 5/2)/16 - 3^(1/2)*log(3/x^2 + (3^(1/2)*(5*x^2 + x^4 + 3)^(1/2))/x^2 + 5/2)
+ (3*(x^2/2 + 5/4)*(5*x^2 + x^4 + 3)^(1/2))/2 + (5*x^2 + x^4 + 3)^(1/2)

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