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

Optimal. Leaf size=46 \[ \frac {3 x^2+2}{10 \sqrt {x^4+5}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {x^4+5}}{\sqrt {5}}\right )}{5 \sqrt {5}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1252, 823, 12, 266, 63, 207} \[ \frac {3 x^2+2}{10 \sqrt {x^4+5}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {x^4+5}}{\sqrt {5}}\right )}{5 \sqrt {5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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])

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 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 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]

Rubi steps

\begin {align*} \int \frac {2+3 x^2}{x \left (5+x^4\right )^{3/2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {2+3 x}{x \left (5+x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=\frac {2+3 x^2}{10 \sqrt {5+x^4}}-\frac {1}{50} \operatorname {Subst}\left (\int -\frac {10}{x \sqrt {5+x^2}} \, dx,x,x^2\right )\\ &=\frac {2+3 x^2}{10 \sqrt {5+x^4}}+\frac {1}{5} \operatorname {Subst}\left (\int \frac {1}{x \sqrt {5+x^2}} \, dx,x,x^2\right )\\ &=\frac {2+3 x^2}{10 \sqrt {5+x^4}}+\frac {1}{10} \operatorname {Subst}\left (\int \frac {1}{x \sqrt {5+x}} \, dx,x,x^4\right )\\ &=\frac {2+3 x^2}{10 \sqrt {5+x^4}}+\frac {1}{5} \operatorname {Subst}\left (\int \frac {1}{-5+x^2} \, dx,x,\sqrt {5+x^4}\right )\\ &=\frac {2+3 x^2}{10 \sqrt {5+x^4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {5+x^4}}{\sqrt {5}}\right )}{5 \sqrt {5}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 46, normalized size = 1.00 \[ \frac {1}{50} \left (\frac {5 \left (3 x^2+2\right )}{\sqrt {x^4+5}}-2 \sqrt {5} \tanh ^{-1}\left (\frac {\sqrt {x^4+5}}{\sqrt {5}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.54, size = 61, normalized size = 1.33 \[ \frac {15 \, x^{4} + 2 \, \sqrt {5} {\left (x^{4} + 5\right )} \log \left (-\frac {\sqrt {5} - \sqrt {x^{4} + 5}}{x^{2}}\right ) + 5 \, \sqrt {x^{4} + 5} {\left (3 \, x^{2} + 2\right )} + 75}{50 \, {\left (x^{4} + 5\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/50*(15*x^4 + 2*sqrt(5)*(x^4 + 5)*log(-(sqrt(5) - sqrt(x^4 + 5))/x^2) + 5*sqrt(x^4 + 5)*(3*x^2 + 2) + 75)/(x^
4 + 5)

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giac [A]  time = 0.25, size = 61, normalized size = 1.33 \[ \frac {1}{25} \, \sqrt {5} \log \left (x^{2} + \sqrt {5} - \sqrt {x^{4} + 5}\right ) - \frac {1}{25} \, \sqrt {5} \log \left (-x^{2} + \sqrt {5} + \sqrt {x^{4} + 5}\right ) + \frac {3 \, x^{2} + 2}{10 \, \sqrt {x^{4} + 5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 40, normalized size = 0.87 \[ \frac {3 x^{2}}{10 \sqrt {x^{4}+5}}-\frac {\sqrt {5}\, \arctanh \left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right )}{25}+\frac {1}{5 \sqrt {x^{4}+5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.41, size = 56, normalized size = 1.22 \[ \frac {3 \, x^{2}}{10 \, \sqrt {x^{4} + 5}} + \frac {1}{50} \, \sqrt {5} \log \left (-\frac {\sqrt {5} - \sqrt {x^{4} + 5}}{\sqrt {5} + \sqrt {x^{4} + 5}}\right ) + \frac {1}{5 \, \sqrt {x^{4} + 5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.48, size = 40, normalized size = 0.87 \[ \frac {1}{5\,\sqrt {x^4+5}}-\frac {\sqrt {5}\,\mathrm {atanh}\left (\frac {\sqrt {5}\,\sqrt {x^4+5}}{5}\right )}{25}+\frac {3\,x^2}{10\,\sqrt {x^4+5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 19.62, size = 212, normalized size = 4.61 \[ \frac {2 x^{4} \log {\left (x^{4} \right )}}{20 \sqrt {5} x^{4} + 100 \sqrt {5}} - \frac {4 x^{4} \log {\left (\sqrt {\frac {x^{4}}{5} + 1} + 1 \right )}}{20 \sqrt {5} x^{4} + 100 \sqrt {5}} - \frac {2 x^{4} \log {\relax (5 )}}{20 \sqrt {5} x^{4} + 100 \sqrt {5}} + \frac {3 x^{2}}{10 \sqrt {x^{4} + 5}} + \frac {4 \sqrt {5} \sqrt {x^{4} + 5}}{20 \sqrt {5} x^{4} + 100 \sqrt {5}} + \frac {10 \log {\left (x^{4} \right )}}{20 \sqrt {5} x^{4} + 100 \sqrt {5}} - \frac {20 \log {\left (\sqrt {\frac {x^{4}}{5} + 1} + 1 \right )}}{20 \sqrt {5} x^{4} + 100 \sqrt {5}} - \frac {10 \log {\relax (5 )}}{20 \sqrt {5} x^{4} + 100 \sqrt {5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x**4*log(x**4)/(20*sqrt(5)*x**4 + 100*sqrt(5)) - 4*x**4*log(sqrt(x**4/5 + 1) + 1)/(20*sqrt(5)*x**4 + 100*sqr
t(5)) - 2*x**4*log(5)/(20*sqrt(5)*x**4 + 100*sqrt(5)) + 3*x**2/(10*sqrt(x**4 + 5)) + 4*sqrt(5)*sqrt(x**4 + 5)/
(20*sqrt(5)*x**4 + 100*sqrt(5)) + 10*log(x**4)/(20*sqrt(5)*x**4 + 100*sqrt(5)) - 20*log(sqrt(x**4/5 + 1) + 1)/
(20*sqrt(5)*x**4 + 100*sqrt(5)) - 10*log(5)/(20*sqrt(5)*x**4 + 100*sqrt(5))

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