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

Optimal. Leaf size=65 \[ \frac{3 x^2+2}{10 x^2 \sqrt{x^4+5}}-\frac{2 \sqrt{x^4+5}}{25 x^2}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{x^4+5}}{\sqrt{5}}\right )}{10 \sqrt{5}} \]

[Out]

(2 + 3*x^2)/(10*x^2*Sqrt[5 + x^4]) - (2*Sqrt[5 + x^4])/(25*x^2) - (3*ArcTanh[Sqrt[5 + x^4]/Sqrt[5]])/(10*Sqrt[
5])

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Rubi [A]  time = 0.055294, antiderivative size = 65, 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, 823, 807, 266, 63, 207} \[ \frac{3 x^2+2}{10 x^2 \sqrt{x^4+5}}-\frac{2 \sqrt{x^4+5}}{25 x^2}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{x^4+5}}{\sqrt{5}}\right )}{10 \sqrt{5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2 + 3*x^2)/(10*x^2*Sqrt[5 + x^4]) - (2*Sqrt[5 + x^4])/(25*x^2) - (3*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 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[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^3 \left (5+x^4\right )^{3/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{2+3 x}{x^2 \left (5+x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=\frac{2+3 x^2}{10 x^2 \sqrt{5+x^4}}-\frac{1}{50} \operatorname{Subst}\left (\int \frac{-20-15 x}{x^2 \sqrt{5+x^2}} \, dx,x,x^2\right )\\ &=\frac{2+3 x^2}{10 x^2 \sqrt{5+x^4}}-\frac{2 \sqrt{5+x^4}}{25 x^2}+\frac{3}{10} \operatorname{Subst}\left (\int \frac{1}{x \sqrt{5+x^2}} \, dx,x,x^2\right )\\ &=\frac{2+3 x^2}{10 x^2 \sqrt{5+x^4}}-\frac{2 \sqrt{5+x^4}}{25 x^2}+\frac{3}{20} \operatorname{Subst}\left (\int \frac{1}{x \sqrt{5+x}} \, dx,x,x^4\right )\\ &=\frac{2+3 x^2}{10 x^2 \sqrt{5+x^4}}-\frac{2 \sqrt{5+x^4}}{25 x^2}+\frac{3}{10} \operatorname{Subst}\left (\int \frac{1}{-5+x^2} \, dx,x,\sqrt{5+x^4}\right )\\ &=\frac{2+3 x^2}{10 x^2 \sqrt{5+x^4}}-\frac{2 \sqrt{5+x^4}}{25 x^2}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{5+x^4}}{\sqrt{5}}\right )}{10 \sqrt{5}}\\ \end{align*}

Mathematica [C]  time = 0.0284293, size = 45, normalized size = 0.69 \[ \frac{15 x^2 \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{x^4}{5}+1\right )-4 x^4-10}{50 x^2 \sqrt{x^4+5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-10 - 4*x^4 + 15*x^2*Hypergeometric2F1[-1/2, 1, 1/2, 1 + x^4/5])/(50*x^2*Sqrt[5 + x^4])

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Maple [A]  time = 0.013, size = 47, normalized size = 0.7 \begin{align*} -{\frac{2\,{x}^{4}+5}{25\,{x}^{2}}{\frac{1}{\sqrt{{x}^{4}+5}}}}+{\frac{3}{10}{\frac{1}{\sqrt{{x}^{4}+5}}}}-{\frac{3\,\sqrt{5}}{50}{\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^3/(x^4+5)^(3/2),x)

[Out]

-1/25/x^2*(2*x^4+5)/(x^4+5)^(1/2)+3/10/(x^4+5)^(1/2)-3/50*5^(1/2)*arctanh(5^(1/2)/(x^4+5)^(1/2))

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Maxima [A]  time = 1.42585, size = 92, normalized size = 1.42 \begin{align*} -\frac{x^{2}}{25 \, \sqrt{x^{4} + 5}} + \frac{3}{100} \, \sqrt{5} \log \left (-\frac{\sqrt{5} - \sqrt{x^{4} + 5}}{\sqrt{5} + \sqrt{x^{4} + 5}}\right ) + \frac{3}{10 \, \sqrt{x^{4} + 5}} - \frac{\sqrt{x^{4} + 5}}{25 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/25*x^2/sqrt(x^4 + 5) + 3/100*sqrt(5)*log(-(sqrt(5) - sqrt(x^4 + 5))/(sqrt(5) + sqrt(x^4 + 5))) + 3/10/sqrt(
x^4 + 5) - 1/25*sqrt(x^4 + 5)/x^2

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Fricas [A]  time = 1.53314, size = 186, normalized size = 2.86 \begin{align*} -\frac{4 \, x^{6} - 3 \, \sqrt{5}{\left (x^{6} + 5 \, x^{2}\right )} \log \left (-\frac{\sqrt{5} - \sqrt{x^{4} + 5}}{x^{2}}\right ) + 20 \, x^{2} +{\left (4 \, x^{4} - 15 \, x^{2} + 10\right )} \sqrt{x^{4} + 5}}{50 \,{\left (x^{6} + 5 \, x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/50*(4*x^6 - 3*sqrt(5)*(x^6 + 5*x^2)*log(-(sqrt(5) - sqrt(x^4 + 5))/x^2) + 20*x^2 + (4*x^4 - 15*x^2 + 10)*sq
rt(x^4 + 5))/(x^6 + 5*x^2)

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Sympy [B]  time = 10.2049, size = 228, normalized size = 3.51 \begin{align*} \frac{3 x^{4} \log{\left (x^{4} \right )}}{20 \sqrt{5} x^{4} + 100 \sqrt{5}} - \frac{6 x^{4} \log{\left (\sqrt{\frac{x^{4}}{5} + 1} + 1 \right )}}{20 \sqrt{5} x^{4} + 100 \sqrt{5}} - \frac{3 x^{4} \log{\left (5 \right )}}{20 \sqrt{5} x^{4} + 100 \sqrt{5}} + \frac{6 \sqrt{5} \sqrt{x^{4} + 5}}{20 \sqrt{5} x^{4} + 100 \sqrt{5}} + \frac{15 \log{\left (x^{4} \right )}}{20 \sqrt{5} x^{4} + 100 \sqrt{5}} - \frac{30 \log{\left (\sqrt{\frac{x^{4}}{5} + 1} + 1 \right )}}{20 \sqrt{5} x^{4} + 100 \sqrt{5}} - \frac{15 \log{\left (5 \right )}}{20 \sqrt{5} x^{4} + 100 \sqrt{5}} - \frac{2}{25 \sqrt{1 + \frac{5}{x^{4}}}} - \frac{1}{5 x^{4} \sqrt{1 + \frac{5}{x^{4}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*x**4*log(x**4)/(20*sqrt(5)*x**4 + 100*sqrt(5)) - 6*x**4*log(sqrt(x**4/5 + 1) + 1)/(20*sqrt(5)*x**4 + 100*sqr
t(5)) - 3*x**4*log(5)/(20*sqrt(5)*x**4 + 100*sqrt(5)) + 6*sqrt(5)*sqrt(x**4 + 5)/(20*sqrt(5)*x**4 + 100*sqrt(5
)) + 15*log(x**4)/(20*sqrt(5)*x**4 + 100*sqrt(5)) - 30*log(sqrt(x**4/5 + 1) + 1)/(20*sqrt(5)*x**4 + 100*sqrt(5
)) - 15*log(5)/(20*sqrt(5)*x**4 + 100*sqrt(5)) - 2/(25*sqrt(1 + 5/x**4)) - 1/(5*x**4*sqrt(1 + 5/x**4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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