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

Optimal. Leaf size=63 \[ -\frac{\sqrt{x^4+5} \left (3 x^2+1\right )}{2 x^4}+\frac{3}{2} \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right )-\frac{\tanh ^{-1}\left (\frac{\sqrt{x^4+5}}{\sqrt{5}}\right )}{2 \sqrt{5}} \]

[Out]

-((1 + 3*x^2)*Sqrt[5 + x^4])/(2*x^4) + (3*ArcSinh[x^2/Sqrt[5]])/2 - ArcTanh[Sqrt[5 + x^4]/Sqrt[5]]/(2*Sqrt[5])

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Rubi [A]  time = 0.0548039, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {1252, 811, 844, 215, 266, 63, 207} \[ -\frac{\sqrt{x^4+5} \left (3 x^2+1\right )}{2 x^4}+\frac{3}{2} \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right )-\frac{\tanh ^{-1}\left (\frac{\sqrt{x^4+5}}{\sqrt{5}}\right )}{2 \sqrt{5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 + 3*x^2)*Sqrt[5 + x^4])/(2*x^4) + (3*ArcSinh[x^2/Sqrt[5]])/2 - ArcTanh[Sqrt[5 + x^4]/Sqrt[5]]/(2*Sqrt[5])

Rule 1252

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m
- 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]

Rule 811

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((d + e*x)^
(m + 1)*(a + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 + a*e^2) - 2*c*d^2*p*(e*f - d*g) - e*(g*(m + 1)*(c*d^2 + a*e
^2) + 2*c*d*p*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2 + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2
+ a*e^2)), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) - c*(2*c*d*(d*g*(2*p + 1
) - e*f*(m + 2*p + 2)) - 2*a*e^2*g*(m + 1))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2
, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

Rule 844

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

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 207

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

Rubi steps

\begin{align*} \int \frac{\left (2+3 x^2\right ) \sqrt{5+x^4}}{x^5} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(2+3 x) \sqrt{5+x^2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac{\left (1+3 x^2\right ) \sqrt{5+x^4}}{2 x^4}-\frac{1}{40} \operatorname{Subst}\left (\int \frac{-20-60 x}{x \sqrt{5+x^2}} \, dx,x,x^2\right )\\ &=-\frac{\left (1+3 x^2\right ) \sqrt{5+x^4}}{2 x^4}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x \sqrt{5+x^2}} \, dx,x,x^2\right )+\frac{3}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{5+x^2}} \, dx,x,x^2\right )\\ &=-\frac{\left (1+3 x^2\right ) \sqrt{5+x^4}}{2 x^4}+\frac{3}{2} \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right )+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{x \sqrt{5+x}} \, dx,x,x^4\right )\\ &=-\frac{\left (1+3 x^2\right ) \sqrt{5+x^4}}{2 x^4}+\frac{3}{2} \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{-5+x^2} \, dx,x,\sqrt{5+x^4}\right )\\ &=-\frac{\left (1+3 x^2\right ) \sqrt{5+x^4}}{2 x^4}+\frac{3}{2} \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right )-\frac{\tanh ^{-1}\left (\frac{\sqrt{5+x^4}}{\sqrt{5}}\right )}{2 \sqrt{5}}\\ \end{align*}

Mathematica [A]  time = 0.076863, size = 59, normalized size = 0.94 \[ \frac{1}{10} \left (-\frac{5 \sqrt{x^4+5} \left (3 x^2+1\right )}{x^4}+15 \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right )-\sqrt{5} \tanh ^{-1}\left (\sqrt{\frac{x^4}{5}+1}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-5*(1 + 3*x^2)*Sqrt[5 + x^4])/x^4 + 15*ArcSinh[x^2/Sqrt[5]] - Sqrt[5]*ArcTanh[Sqrt[1 + x^4/5]])/10

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Maple [A]  time = 0.011, size = 75, normalized size = 1.2 \begin{align*} -{\frac{3}{10\,{x}^{2}} \left ({x}^{4}+5 \right ) ^{{\frac{3}{2}}}}+{\frac{3\,{x}^{2}}{10}\sqrt{{x}^{4}+5}}+{\frac{3}{2}{\it Arcsinh} \left ({\frac{{x}^{2}\sqrt{5}}{5}} \right ) }-{\frac{1}{10\,{x}^{4}} \left ({x}^{4}+5 \right ) ^{{\frac{3}{2}}}}+{\frac{1}{10}\sqrt{{x}^{4}+5}}-{\frac{\sqrt{5}}{10}{\it Artanh} \left ({\sqrt{5}{\frac{1}{\sqrt{{x}^{4}+5}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/10/x^2*(x^4+5)^(3/2)+3/10*x^2*(x^4+5)^(1/2)+3/2*arcsinh(1/5*x^2*5^(1/2))-1/10/x^4*(x^4+5)^(3/2)+1/10*(x^4+5
)^(1/2)-1/10*5^(1/2)*arctanh(5^(1/2)/(x^4+5)^(1/2))

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Maxima [A]  time = 1.44143, size = 123, normalized size = 1.95 \begin{align*} \frac{1}{20} \, \sqrt{5} \log \left (-\frac{\sqrt{5} - \sqrt{x^{4} + 5}}{\sqrt{5} + \sqrt{x^{4} + 5}}\right ) - \frac{3 \, \sqrt{x^{4} + 5}}{2 \, x^{2}} - \frac{\sqrt{x^{4} + 5}}{2 \, x^{4}} + \frac{3}{4} \, \log \left (\frac{\sqrt{x^{4} + 5}}{x^{2}} + 1\right ) - \frac{3}{4} \, \log \left (\frac{\sqrt{x^{4} + 5}}{x^{2}} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2)*(x^4+5)^(1/2)/x^5,x, algorithm="maxima")

[Out]

1/20*sqrt(5)*log(-(sqrt(5) - sqrt(x^4 + 5))/(sqrt(5) + sqrt(x^4 + 5))) - 3/2*sqrt(x^4 + 5)/x^2 - 1/2*sqrt(x^4
+ 5)/x^4 + 3/4*log(sqrt(x^4 + 5)/x^2 + 1) - 3/4*log(sqrt(x^4 + 5)/x^2 - 1)

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Fricas [A]  time = 1.53979, size = 181, normalized size = 2.87 \begin{align*} \frac{\sqrt{5} x^{4} \log \left (-\frac{\sqrt{5} - \sqrt{x^{4} + 5}}{x^{2}}\right ) - 15 \, x^{4} \log \left (-x^{2} + \sqrt{x^{4} + 5}\right ) - 15 \, x^{4} - 5 \, \sqrt{x^{4} + 5}{\left (3 \, x^{2} + 1\right )}}{10 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2)*(x^4+5)^(1/2)/x^5,x, algorithm="fricas")

[Out]

1/10*(sqrt(5)*x^4*log(-(sqrt(5) - sqrt(x^4 + 5))/x^2) - 15*x^4*log(-x^2 + sqrt(x^4 + 5)) - 15*x^4 - 5*sqrt(x^4
+ 5)*(3*x^2 + 1))/x^4

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Sympy [A]  time = 5.12512, size = 76, normalized size = 1.21 \begin{align*} - \frac{3 x^{2}}{2 \sqrt{x^{4} + 5}} - \frac{\sqrt{5} \operatorname{asinh}{\left (\frac{\sqrt{5}}{x^{2}} \right )}}{10} + \frac{3 \operatorname{asinh}{\left (\frac{\sqrt{5} x^{2}}{5} \right )}}{2} - \frac{\sqrt{1 + \frac{5}{x^{4}}}}{2 x^{2}} - \frac{15}{2 x^{2} \sqrt{x^{4} + 5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*x**2/(2*sqrt(x**4 + 5)) - sqrt(5)*asinh(sqrt(5)/x**2)/10 + 3*asinh(sqrt(5)*x**2/5)/2 - sqrt(1 + 5/x**4)/(2*
x**2) - 15/(2*x**2*sqrt(x**4 + 5))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2)*(x^4+5)^(1/2)/x^5,x, algorithm="giac")

[Out]

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