3.148 \(\int \frac{(2+3 x^2) \sqrt{3+5 x^2+x^4}}{x^7} \, dx\)

Optimal. Leaf size=90 \[ -\frac{\left (x^4+5 x^2+3\right )^{3/2}}{9 x^6}-\frac{\left (5 x^2+6\right ) \sqrt{x^4+5 x^2+3}}{18 x^4}+\frac{13 \tanh ^{-1}\left (\frac{5 x^2+6}{2 \sqrt{3} \sqrt{x^4+5 x^2+3}}\right )}{36 \sqrt{3}} \]

[Out]

-((6 + 5*x^2)*Sqrt[3 + 5*x^2 + x^4])/(18*x^4) - (3 + 5*x^2 + x^4)^(3/2)/(9*x^6) + (13*ArcTanh[(6 + 5*x^2)/(2*S
qrt[3]*Sqrt[3 + 5*x^2 + x^4])])/(36*Sqrt[3])

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Rubi [A]  time = 0.0681416, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1251, 806, 720, 724, 206} \[ -\frac{\left (x^4+5 x^2+3\right )^{3/2}}{9 x^6}-\frac{\left (5 x^2+6\right ) \sqrt{x^4+5 x^2+3}}{18 x^4}+\frac{13 \tanh ^{-1}\left (\frac{5 x^2+6}{2 \sqrt{3} \sqrt{x^4+5 x^2+3}}\right )}{36 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[((2 + 3*x^2)*Sqrt[3 + 5*x^2 + x^4])/x^7,x]

[Out]

-((6 + 5*x^2)*Sqrt[3 + 5*x^2 + x^4])/(18*x^4) - (3 + 5*x^2 + x^4)^(3/2)/(9*x^6) + (13*ArcTanh[(6 + 5*x^2)/(2*S
qrt[3]*Sqrt[3 + 5*x^2 + x^4])])/(36*Sqrt[3])

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 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(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] && EqQ[Sim
plify[m + 2*p + 3], 0]

Rule 720

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

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])

Rubi steps

\begin{align*} \int \frac{\left (2+3 x^2\right ) \sqrt{3+5 x^2+x^4}}{x^7} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(2+3 x) \sqrt{3+5 x+x^2}}{x^4} \, dx,x,x^2\right )\\ &=-\frac{\left (3+5 x^2+x^4\right )^{3/2}}{9 x^6}+\frac{2}{3} \operatorname{Subst}\left (\int \frac{\sqrt{3+5 x+x^2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac{\left (6+5 x^2\right ) \sqrt{3+5 x^2+x^4}}{18 x^4}-\frac{\left (3+5 x^2+x^4\right )^{3/2}}{9 x^6}-\frac{13}{36} \operatorname{Subst}\left (\int \frac{1}{x \sqrt{3+5 x+x^2}} \, dx,x,x^2\right )\\ &=-\frac{\left (6+5 x^2\right ) \sqrt{3+5 x^2+x^4}}{18 x^4}-\frac{\left (3+5 x^2+x^4\right )^{3/2}}{9 x^6}+\frac{13}{18} \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{\left (6+5 x^2\right ) \sqrt{3+5 x^2+x^4}}{18 x^4}-\frac{\left (3+5 x^2+x^4\right )^{3/2}}{9 x^6}+\frac{13 \tanh ^{-1}\left (\frac{6+5 x^2}{2 \sqrt{3} \sqrt{3+5 x^2+x^4}}\right )}{36 \sqrt{3}}\\ \end{align*}

Mathematica [A]  time = 0.0262554, size = 74, normalized size = 0.82 \[ \frac{1}{108} \left (13 \sqrt{3} \tanh ^{-1}\left (\frac{5 x^2+6}{2 \sqrt{3} \sqrt{x^4+5 x^2+3}}\right )-\frac{6 \sqrt{x^4+5 x^2+3} \left (7 x^4+16 x^2+6\right )}{x^6}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((2 + 3*x^2)*Sqrt[3 + 5*x^2 + x^4])/x^7,x]

[Out]

((-6*Sqrt[3 + 5*x^2 + x^4]*(6 + 16*x^2 + 7*x^4))/x^6 + 13*Sqrt[3]*ArcTanh[(6 + 5*x^2)/(2*Sqrt[3]*Sqrt[3 + 5*x^
2 + x^4])])/108

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Maple [A]  time = 0.013, size = 118, normalized size = 1.3 \begin{align*} -{\frac{1}{9\,{x}^{4}} \left ({x}^{4}+5\,{x}^{2}+3 \right ) ^{{\frac{3}{2}}}}+{\frac{5}{54\,{x}^{2}} \left ({x}^{4}+5\,{x}^{2}+3 \right ) ^{{\frac{3}{2}}}}-{\frac{13}{108}\sqrt{{x}^{4}+5\,{x}^{2}+3}}+{\frac{13\,\sqrt{3}}{108}{\it Artanh} \left ({\frac{ \left ( 5\,{x}^{2}+6 \right ) \sqrt{3}}{6}{\frac{1}{\sqrt{{x}^{4}+5\,{x}^{2}+3}}}} \right ) }-{\frac{10\,{x}^{2}+25}{108}\sqrt{{x}^{4}+5\,{x}^{2}+3}}-{\frac{1}{9\,{x}^{6}} \left ({x}^{4}+5\,{x}^{2}+3 \right ) ^{{\frac{3}{2}}}} \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^7,x)

[Out]

-1/9/x^4*(x^4+5*x^2+3)^(3/2)+5/54/x^2*(x^4+5*x^2+3)^(3/2)-13/108*(x^4+5*x^2+3)^(1/2)+13/108*arctanh(1/6*(5*x^2
+6)*3^(1/2)/(x^4+5*x^2+3)^(1/2))*3^(1/2)-5/108*(2*x^2+5)*(x^4+5*x^2+3)^(1/2)-1/9*(x^4+5*x^2+3)^(3/2)/x^6

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Maxima [A]  time = 1.43131, size = 134, normalized size = 1.49 \begin{align*} \frac{13}{108} \, \sqrt{3} \log \left (\frac{2 \, \sqrt{3} \sqrt{x^{4} + 5 \, x^{2} + 3}}{x^{2}} + \frac{6}{x^{2}} + 5\right ) + \frac{1}{9} \, \sqrt{x^{4} + 5 \, x^{2} + 3} + \frac{5 \, \sqrt{x^{4} + 5 \, x^{2} + 3}}{18 \, x^{2}} - \frac{{\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac{3}{2}}}{9 \, x^{4}} - \frac{{\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac{3}{2}}}{9 \, x^{6}} \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^7,x, algorithm="maxima")

[Out]

13/108*sqrt(3)*log(2*sqrt(3)*sqrt(x^4 + 5*x^2 + 3)/x^2 + 6/x^2 + 5) + 1/9*sqrt(x^4 + 5*x^2 + 3) + 5/18*sqrt(x^
4 + 5*x^2 + 3)/x^2 - 1/9*(x^4 + 5*x^2 + 3)^(3/2)/x^4 - 1/9*(x^4 + 5*x^2 + 3)^(3/2)/x^6

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Fricas [A]  time = 1.391, size = 234, normalized size = 2.6 \begin{align*} \frac{13 \, \sqrt{3} x^{6} \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 ) - 42 \, x^{6} - 6 \,{\left (7 \, x^{4} + 16 \, x^{2} + 6\right )} \sqrt{x^{4} + 5 \, x^{2} + 3}}{108 \, x^{6}} \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^7,x, algorithm="fricas")

[Out]

1/108*(13*sqrt(3)*x^6*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)
- 42*x^6 - 6*(7*x^4 + 16*x^2 + 6)*sqrt(x^4 + 5*x^2 + 3))/x^6

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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^{7}}\, 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**7,x)

[Out]

Integral((3*x**2 + 2)*sqrt(x**4 + 5*x**2 + 3)/x**7, 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^{7}}\,{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^7,x, algorithm="giac")

[Out]

integrate(sqrt(x^4 + 5*x^2 + 3)*(3*x^2 + 2)/x^7, x)