3.1.38 \(\int \frac {2+3 x^2}{x^5 \sqrt {5+x^4}} \, dx\) [38]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 58, 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 = {1266, 849, 821, 272, 65, 213} \begin {gather*} -\frac {\sqrt {x^4+5}}{10 x^4}+\frac {\tanh ^{-1}\left (\frac {\sqrt {x^4+5}}{\sqrt {5}}\right )}{10 \sqrt {5}}-\frac {3 \sqrt {x^4+5}}{10 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

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 213

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

Rule 272

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 821

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 849

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 1266

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^5 \sqrt {5+x^4}} \, dx &=\frac {1}{2} \text {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} \text {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} \text {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} \text {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} \text {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*}

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Mathematica [A]
time = 0.14, size = 55, normalized size = 0.95 \begin {gather*} \frac {1}{50} \left (-\frac {5 \left (1+3 x^2\right ) \sqrt {5+x^4}}{x^4}-2 \sqrt {5} \tanh ^{-1}\left (\frac {x^2-\sqrt {5+x^4}}{\sqrt {5}}\right )\right ) \end {gather*}

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

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Maple [A]
time = 0.20, size = 43, normalized size = 0.74

method	result	size
default	\(-\frac {\sqrt {x^{4}+5}}{10 x^{4}}+\frac {\sqrt {5}\, \arctanh \left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right )}{50}-\frac {3 \sqrt {x^{4}+5}}{10 x^{2}}\)	\(43\)
elliptic	\(-\frac {\sqrt {x^{4}+5}}{10 x^{4}}+\frac {\sqrt {5}\, \arctanh \left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right )}{50}-\frac {3 \sqrt {x^{4}+5}}{10 x^{2}}\)	\(43\)
risch	\(-\frac {3 x^{6}+x^{4}+15 x^{2}+5}{10 x^{4} \sqrt {x^{4}+5}}+\frac {\sqrt {5}\, \arctanh \left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right )}{50}\)	\(46\)
trager	\(-\frac {\left (3 x^{2}+1\right ) \sqrt {x^{4}+5}}{10 x^{4}}-\frac {\RootOf \left (\textit {\_Z}^{2}-5\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{2}-5\right )-\sqrt {x^{4}+5}}{x^{2}}\right )}{50}\)	\(51\)
meijerg	\(\frac {\sqrt {5}\, \left (\frac {5 \sqrt {\pi }\, \left (\frac {4 x^{4}}{5}+8\right )}{8 x^{4}}-\frac {5 \sqrt {\pi }\, \sqrt {1+\frac {x^{4}}{5}}}{x^{4}}+\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {1+\frac {x^{4}}{5}}}{2}\right )-\frac {\left (1-2 \ln \left (2\right )+4 \ln \left (x \right )-\ln \left (5\right )\right ) \sqrt {\pi }}{2}-\frac {5 \sqrt {\pi }}{x^{4}}\right )}{50 \sqrt {\pi }}-\frac {3 \sqrt {5}\, \sqrt {1+\frac {x^{4}}{5}}}{10 x^{2}}\)	\(105\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.50, size = 59, normalized size = 1.02 \begin {gather*} -\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 {gather*}

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 = 0.34, size = 50, normalized size = 0.86 \begin {gather*} \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 {gather*}

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 = 5.66, size = 88, normalized size = 1.52 \begin {gather*} \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 {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (43) = 86\).
time = 5.76, size = 114, normalized size = 1.97 \begin {gather*} -\frac {1}{50} \, \sqrt {5} \log \left (-\frac {x^{2} + \sqrt {5} - \sqrt {x^{4} + 5}}{x^{2} - \sqrt {5} - \sqrt {x^{4} + 5}}\right ) + \frac {{\left (x^{2} - \sqrt {x^{4} + 5}\right )}^{3} + 15 \, {\left (x^{2} - \sqrt {x^{4} + 5}\right )}^{2} + 5 \, x^{2} - 5 \, \sqrt {x^{4} + 5} - 75}{5 \, {\left ({\left (x^{2} - \sqrt {x^{4} + 5}\right )}^{2} - 5\right )}^{2}} \end {gather*}

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]

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

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Mupad [B]
time = 0.69, size = 43, normalized size = 0.74 \begin {gather*} \frac {\sqrt {5}\,\mathrm {atanh}\left (\frac {\sqrt {5}\,\sqrt {x^4+5}}{5}\right )}{50}-\frac {3\,\sqrt {x^4+5}}{10\,x^2}-\frac {\sqrt {x^4+5}}{10\,x^4} \end {gather*}

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]

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

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