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

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

[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])]

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Rubi [A]  time = 0.08329, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {1251, 814, 843, 621, 206, 724} \[ \frac{1}{8} \sqrt{x^4+5 x^2+3} \left (6 x^2+23\right )+\frac{1}{16} \tanh ^{-1}\left (\frac{2 x^2+5}{2 \sqrt{x^4+5 x^2+3}}\right )-\sqrt{3} \tanh ^{-1}\left (\frac{5 x^2+6}{2 \sqrt{3} \sqrt{x^4+5 x^2+3}}\right ) \]

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 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]

Rule 814

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 843

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 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 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 724

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]

Rubi steps

\begin{align*} \int \frac{\left (2+3 x^2\right ) \sqrt{3+5 x^2+x^4}}{x} \, dx &=\frac{1}{2} \operatorname{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} \operatorname{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} \operatorname{Subst}\left (\int \frac{1}{\sqrt{3+5 x+x^2}} \, dx,x,x^2\right )+3 \operatorname{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} \operatorname{Subst}\left (\int \frac{1}{4-x^2} \, dx,x,\frac{5+2 x^2}{\sqrt{3+5 x^2+x^4}}\right )-6 \operatorname{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*}

Mathematica [A]  time = 0.0377268, size = 92, normalized size = 0.98 \[ \frac{1}{16} \left (2 \sqrt{x^4+5 x^2+3} \left (6 x^2+23\right )+\tanh ^{-1}\left (\frac{2 x^2+5}{2 \sqrt{x^4+5 x^2+3}}\right )-16 \sqrt{3} \tanh ^{-1}\left (\frac{5 x^2+6}{2 \sqrt{3} \sqrt{x^4+5 x^2+3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.011, size = 85, normalized size = 0.9 \begin{align*}{\frac{6\,{x}^{2}+15}{8}\sqrt{{x}^{4}+5\,{x}^{2}+3}}+{\frac{1}{16}\ln \left ({\frac{5}{2}}+{x}^{2}+\sqrt{{x}^{4}+5\,{x}^{2}+3} \right ) }+\sqrt{{x}^{4}+5\,{x}^{2}+3}-{\it Artanh} \left ({\frac{ \left ( 5\,{x}^{2}+6 \right ) \sqrt{3}}{6}{\frac{1}{\sqrt{{x}^{4}+5\,{x}^{2}+3}}}} \right ) \sqrt{3} \end{align*}

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)

[Out]

3/8*(2*x^2+5)*(x^4+5*x^2+3)^(1/2)+1/16*ln(5/2+x^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 = 1.43885, size = 120, normalized size = 1.28 \begin{align*} \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{align*}

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 = 1.56655, size = 254, normalized size = 2.7 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (3 x^{2} + 2\right ) \sqrt{x^{4} + 5 x^{2} + 3}}{x}\, dx \end{align*}

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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{4} + 5 \, x^{2} + 3}{\left (3 \, x^{2} + 2\right )}}{x}\,{d x} \end{align*}

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]

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