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

Optimal. Leaf size=58 \[ -\frac{3 \sqrt{x^4+5}}{10 x^2}-\frac{\sqrt{x^4+5}}{10 x^4}+\frac{\tanh ^{-1}\left (\frac{\sqrt{x^4+5}}{\sqrt{5}}\right )}{10 \sqrt{5}} \]

[Out]

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

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Rubi [A]  time = 0.0512227, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {1252, 835, 807, 266, 63, 207} \[ -\frac{3 \sqrt{x^4+5}}{10 x^2}-\frac{\sqrt{x^4+5}}{10 x^4}+\frac{\tanh ^{-1}\left (\frac{\sqrt{x^4+5}}{\sqrt{5}}\right )}{10 \sqrt{5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[5 + x^4]/(10*x^4) - (3*Sqrt[5 + x^4])/(10*x^2) + ArcTanh[Sqrt[5 + x^4]/Sqrt[5]]/(10*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 835

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

Rule 807

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

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{2+3 x^2}{x^5 \sqrt{5+x^4}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{2+3 x}{x^3 \sqrt{5+x^2}} \, dx,x,x^2\right )\\ &=-\frac{\sqrt{5+x^4}}{10 x^4}-\frac{1}{20} \operatorname{Subst}\left (\int \frac{-30+2 x}{x^2 \sqrt{5+x^2}} \, dx,x,x^2\right )\\ &=-\frac{\sqrt{5+x^4}}{10 x^4}-\frac{3 \sqrt{5+x^4}}{10 x^2}-\frac{1}{10} \operatorname{Subst}\left (\int \frac{1}{x \sqrt{5+x^2}} \, dx,x,x^2\right )\\ &=-\frac{\sqrt{5+x^4}}{10 x^4}-\frac{3 \sqrt{5+x^4}}{10 x^2}-\frac{1}{20} \operatorname{Subst}\left (\int \frac{1}{x \sqrt{5+x}} \, dx,x,x^4\right )\\ &=-\frac{\sqrt{5+x^4}}{10 x^4}-\frac{3 \sqrt{5+x^4}}{10 x^2}-\frac{1}{10} \operatorname{Subst}\left (\int \frac{1}{-5+x^2} \, dx,x,\sqrt{5+x^4}\right )\\ &=-\frac{\sqrt{5+x^4}}{10 x^4}-\frac{3 \sqrt{5+x^4}}{10 x^2}+\frac{\tanh ^{-1}\left (\frac{\sqrt{5+x^4}}{\sqrt{5}}\right )}{10 \sqrt{5}}\\ \end{align*}

Mathematica [A]  time = 0.0303463, size = 49, normalized size = 0.84 \[ \frac{\sqrt{5} x^4 \tanh ^{-1}\left (\sqrt{\frac{x^4}{5}+1}\right )-5 \left (3 x^2+1\right ) \sqrt{x^4+5}}{50 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*(1 + 3*x^2)*Sqrt[5 + x^4] + Sqrt[5]*x^4*ArcTanh[Sqrt[1 + x^4/5]])/(50*x^4)

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Maple [A]  time = 0.013, size = 43, normalized size = 0.7 \begin{align*} -{\frac{1}{10\,{x}^{4}}\sqrt{{x}^{4}+5}}+{\frac{\sqrt{5}}{50}{\it Artanh} \left ({\sqrt{5}{\frac{1}{\sqrt{{x}^{4}+5}}}} \right ) }-{\frac{3}{10\,{x}^{2}}\sqrt{{x}^{4}+5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10*(x^4+5)^(1/2)/x^4+1/50*5^(1/2)*arctanh(5^(1/2)/(x^4+5)^(1/2))-3/10*(x^4+5)^(1/2)/x^2

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Maxima [A]  time = 1.4286, size = 80, normalized size = 1.38 \begin{align*} -\frac{1}{100} \, \sqrt{5} \log \left (-\frac{\sqrt{5} - \sqrt{x^{4} + 5}}{\sqrt{5} + \sqrt{x^{4} + 5}}\right ) - \frac{3 \, \sqrt{x^{4} + 5}}{10 \, x^{2}} - \frac{\sqrt{x^{4} + 5}}{10 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/100*sqrt(5)*log(-(sqrt(5) - sqrt(x^4 + 5))/(sqrt(5) + sqrt(x^4 + 5))) - 3/10*sqrt(x^4 + 5)/x^2 - 1/10*sqrt(
x^4 + 5)/x^4

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Fricas [A]  time = 1.5432, size = 132, normalized size = 2.28 \begin{align*} \frac{\sqrt{5} x^{4} \log \left (\frac{\sqrt{5} + \sqrt{x^{4} + 5}}{x^{2}}\right ) - 15 \, x^{4} - 5 \, \sqrt{x^{4} + 5}{\left (3 \, x^{2} + 1\right )}}{50 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/50*(sqrt(5)*x^4*log((sqrt(5) + sqrt(x^4 + 5))/x^2) - 15*x^4 - 5*sqrt(x^4 + 5)*(3*x^2 + 1))/x^4

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Sympy [A]  time = 7.57052, size = 88, normalized size = 1.52 \begin{align*} \frac{\sqrt{5} \left (- \frac{\log{\left (\sqrt{\frac{x^{4}}{5} + 1} - 1 \right )}}{4} + \frac{\log{\left (\sqrt{\frac{x^{4}}{5} + 1} + 1 \right )}}{4} - \frac{1}{4 \left (\sqrt{\frac{x^{4}}{5} + 1} + 1\right )} - \frac{1}{4 \left (\sqrt{\frac{x^{4}}{5} + 1} - 1\right )}\right )}{25} - \frac{3 \sqrt{5} \sqrt{5 x^{4} + 25}}{50 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(5)*(-log(sqrt(x**4/5 + 1) - 1)/4 + log(sqrt(x**4/5 + 1) + 1)/4 - 1/(4*(sqrt(x**4/5 + 1) + 1)) - 1/(4*(sqr
t(x**4/5 + 1) - 1)))/25 - 3*sqrt(5)*sqrt(5*x**4 + 25)/(50*x**2)

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Giac [A]  time = 1.17223, size = 72, normalized size = 1.24 \begin{align*} -\frac{1}{10} \,{\left (\frac{1}{x^{2}} + 3\right )} \sqrt{\frac{5}{x^{4}} + 1} + \frac{1}{100} \, \sqrt{5} \log \left (\sqrt{5} + \sqrt{x^{4} + 5}\right ) - \frac{1}{100} \, \sqrt{5} \log \left (-\sqrt{5} + \sqrt{x^{4} + 5}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10*(1/x^2 + 3)*sqrt(5/x^4 + 1) + 1/100*sqrt(5)*log(sqrt(5) + sqrt(x^4 + 5)) - 1/100*sqrt(5)*log(-sqrt(5) +
sqrt(x^4 + 5))