Integrand size = 20, antiderivative size = 58 \[ \int \frac {\left (2+3 x^2\right ) \sqrt {5+x^4}}{x} \, dx=\frac {1}{4} \left (4+3 x^2\right ) \sqrt {5+x^4}+\frac {15}{4} \text {arcsinh}\left (\frac {x^2}{\sqrt {5}}\right )-\sqrt {5} \text {arctanh}\left (\frac {\sqrt {5+x^4}}{\sqrt {5}}\right ) \]
15/4*arcsinh(1/5*x^2*5^(1/2))-arctanh(1/5*(x^4+5)^(1/2)*5^(1/2))*5^(1/2)+1 /4*(3*x^2+4)*(x^4+5)^(1/2)
Time = 0.13 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.19 \[ \int \frac {\left (2+3 x^2\right ) \sqrt {5+x^4}}{x} \, dx=\frac {1}{4} \left (\left (4+3 x^2\right ) \sqrt {5+x^4}+8 \sqrt {5} \text {arctanh}\left (\frac {x^2-\sqrt {5+x^4}}{\sqrt {5}}\right )-15 \log \left (-x^2+\sqrt {5+x^4}\right )\right ) \]
((4 + 3*x^2)*Sqrt[5 + x^4] + 8*Sqrt[5]*ArcTanh[(x^2 - Sqrt[5 + x^4])/Sqrt[ 5]] - 15*Log[-x^2 + Sqrt[5 + x^4]])/4
Time = 0.21 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1579, 535, 538, 222, 243, 73, 220}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (3 x^2+2\right ) \sqrt {x^4+5}}{x} \, dx\) |
\(\Big \downarrow \) 1579 |
\(\displaystyle \frac {1}{2} \int \frac {\left (3 x^2+2\right ) \sqrt {x^4+5}}{x^2}dx^2\) |
\(\Big \downarrow \) 535 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{2} \int \frac {3 x^2+4}{x^2 \sqrt {x^4+5}}dx^2+\frac {1}{2} \sqrt {x^4+5} \left (3 x^2+4\right )\right )\) |
\(\Big \downarrow \) 538 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{2} \left (3 \int \frac {1}{\sqrt {x^4+5}}dx^2+4 \int \frac {1}{x^2 \sqrt {x^4+5}}dx^2\right )+\frac {1}{2} \sqrt {x^4+5} \left (3 x^2+4\right )\right )\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{2} \left (4 \int \frac {1}{x^2 \sqrt {x^4+5}}dx^2+3 \text {arcsinh}\left (\frac {x^2}{\sqrt {5}}\right )\right )+\frac {1}{2} \sqrt {x^4+5} \left (3 x^2+4\right )\right )\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{2} \left (2 \int \frac {1}{x^2 \sqrt {x^4+5}}dx^4+3 \text {arcsinh}\left (\frac {x^2}{\sqrt {5}}\right )\right )+\frac {1}{2} \sqrt {x^4+5} \left (3 x^2+4\right )\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{2} \left (4 \int \frac {1}{\sqrt {x^4+5}-5}d\sqrt {x^4+5}+3 \text {arcsinh}\left (\frac {x^2}{\sqrt {5}}\right )\right )+\frac {1}{2} \sqrt {x^4+5} \left (3 x^2+4\right )\right )\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{2} \left (3 \text {arcsinh}\left (\frac {x^2}{\sqrt {5}}\right )-\frac {4 \text {arctanh}\left (\frac {\sqrt {x^4+5}}{\sqrt {5}}\right )}{\sqrt {5}}\right )+\frac {1}{2} \sqrt {x^4+5} \left (3 x^2+4\right )\right )\) |
(((4 + 3*x^2)*Sqrt[5 + x^4])/2 + (5*(3*ArcSinh[x^2/Sqrt[5]] - (4*ArcTanh[S qrt[5 + x^4]/Sqrt[5]])/Sqrt[5]))/2)/2
3.1.11.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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])
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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_))/(x_), x_Symbol] :> Sim p[(c*(2*p + 1) + 2*d*p*x)*((a + b*x^2)^p/(2*p*(2*p + 1))), x] + Simp[a/(2*p + 1) Int[(c*(2*p + 1) + 2*d*p*x)*((a + b*x^2)^(p - 1)/x), x], x] /; Free Q[{a, b, c, d}, x] && GtQ[p, 0] && IntegerQ[2*p]
Int[((c_) + (d_.)*(x_))/((x_)*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Simp [c Int[1/(x*Sqrt[a + b*x^2]), x], x] + Simp[d Int[1/Sqrt[a + b*x^2], x] , x] /; FreeQ[{a, b, c, d}, x]
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_S ymbol] :> Simp[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]
Time = 0.46 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.84
method | result | size |
default | \(\frac {3 x^{2} \sqrt {x^{4}+5}}{4}+\frac {15 \,\operatorname {arcsinh}\left (\frac {x^{2} \sqrt {5}}{5}\right )}{4}+\sqrt {x^{4}+5}-\sqrt {5}\, \operatorname {arctanh}\left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right )\) | \(49\) |
elliptic | \(\frac {3 x^{2} \sqrt {x^{4}+5}}{4}+\frac {15 \,\operatorname {arcsinh}\left (\frac {x^{2} \sqrt {5}}{5}\right )}{4}+\sqrt {x^{4}+5}-\sqrt {5}\, \operatorname {arctanh}\left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right )\) | \(49\) |
pseudoelliptic | \(\frac {3 x^{2} \sqrt {x^{4}+5}}{4}+\frac {15 \,\operatorname {arcsinh}\left (\frac {x^{2} \sqrt {5}}{5}\right )}{4}+\sqrt {x^{4}+5}-\sqrt {5}\, \operatorname {arctanh}\left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right )\) | \(49\) |
trager | \(\left (\frac {3 x^{2}}{4}+1\right ) \sqrt {x^{4}+5}-\frac {15 \ln \left (x^{2}-\sqrt {x^{4}+5}\right )}{4}+\operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) \ln \left (\frac {\sqrt {x^{4}+5}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right )}{x^{2}}\right )\) | \(61\) |
meijerg | \(-\frac {\sqrt {5}\, \left (4 \sqrt {\pi }-4 \sqrt {\pi }\, \sqrt {1+\frac {x^{4}}{5}}+4 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {1+\frac {x^{4}}{5}}}{2}\right )-2 \left (2-2 \ln \left (2\right )+4 \ln \left (x \right )-\ln \left (5\right )\right ) \sqrt {\pi }\right )}{4 \sqrt {\pi }}-\frac {15 \left (-\frac {2 \sqrt {\pi }\, x^{2} \sqrt {5}\, \sqrt {1+\frac {x^{4}}{5}}}{5}-2 \sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {x^{2} \sqrt {5}}{5}\right )\right )}{8 \sqrt {\pi }}\) | \(108\) |
3/4*x^2*(x^4+5)^(1/2)+15/4*arcsinh(1/5*x^2*5^(1/2))+(x^4+5)^(1/2)-5^(1/2)* arctanh(5^(1/2)/(x^4+5)^(1/2))
Time = 0.25 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.97 \[ \int \frac {\left (2+3 x^2\right ) \sqrt {5+x^4}}{x} \, dx=\frac {1}{4} \, \sqrt {x^{4} + 5} {\left (3 \, x^{2} + 4\right )} + \sqrt {5} \log \left (-\frac {\sqrt {5} - \sqrt {x^{4} + 5}}{x^{2}}\right ) - \frac {15}{4} \, \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) \]
1/4*sqrt(x^4 + 5)*(3*x^2 + 4) + sqrt(5)*log(-(sqrt(5) - sqrt(x^4 + 5))/x^2 ) - 15/4*log(-x^2 + sqrt(x^4 + 5))
Time = 4.78 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.29 \[ \int \frac {\left (2+3 x^2\right ) \sqrt {5+x^4}}{x} \, dx=\frac {3 x^{2} \sqrt {x^{4} + 5}}{4} + \sqrt {5} \left (\sqrt {\frac {x^{4}}{5} + 1} + \frac {\log {\left (\sqrt {\frac {x^{4}}{5} + 1} - 1 \right )}}{2} - \frac {\log {\left (\sqrt {\frac {x^{4}}{5} + 1} + 1 \right )}}{2}\right ) + \frac {15 \operatorname {asinh}{\left (\frac {\sqrt {5} x^{2}}{5} \right )}}{4} \]
3*x**2*sqrt(x**4 + 5)/4 + sqrt(5)*(sqrt(x**4/5 + 1) + log(sqrt(x**4/5 + 1) - 1)/2 - log(sqrt(x**4/5 + 1) + 1)/2) + 15*asinh(sqrt(5)*x**2/5)/4
Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (46) = 92\).
Time = 0.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.71 \[ \int \frac {\left (2+3 x^2\right ) \sqrt {5+x^4}}{x} \, dx=\frac {1}{2} \, \sqrt {5} \log \left (-\frac {\sqrt {5} - \sqrt {x^{4} + 5}}{\sqrt {5} + \sqrt {x^{4} + 5}}\right ) + \sqrt {x^{4} + 5} + \frac {15 \, \sqrt {x^{4} + 5}}{4 \, x^{2} {\left (\frac {x^{4} + 5}{x^{4}} - 1\right )}} + \frac {15}{8} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} + 1\right ) - \frac {15}{8} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} - 1\right ) \]
1/2*sqrt(5)*log(-(sqrt(5) - sqrt(x^4 + 5))/(sqrt(5) + sqrt(x^4 + 5))) + sq rt(x^4 + 5) + 15/4*sqrt(x^4 + 5)/(x^2*((x^4 + 5)/x^4 - 1)) + 15/8*log(sqrt (x^4 + 5)/x^2 + 1) - 15/8*log(sqrt(x^4 + 5)/x^2 - 1)
Time = 0.29 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.31 \[ \int \frac {\left (2+3 x^2\right ) \sqrt {5+x^4}}{x} \, dx=\frac {1}{4} \, \sqrt {x^{4} + 5} {\left (3 \, x^{2} + 4\right )} + \sqrt {5} \log \left (-\frac {x^{2} + \sqrt {5} - \sqrt {x^{4} + 5}}{x^{2} - \sqrt {5} - \sqrt {x^{4} + 5}}\right ) - \frac {15}{4} \, \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) \]
1/4*sqrt(x^4 + 5)*(3*x^2 + 4) + sqrt(5)*log(-(x^2 + sqrt(5) - sqrt(x^4 + 5 ))/(x^2 - sqrt(5) - sqrt(x^4 + 5))) - 15/4*log(-x^2 + sqrt(x^4 + 5))
Time = 7.61 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.78 \[ \int \frac {\left (2+3 x^2\right ) \sqrt {5+x^4}}{x} \, dx=\frac {15\,\mathrm {asinh}\left (\frac {\sqrt {5}\,x^2}{5}\right )}{4}-\sqrt {5}\,\mathrm {atanh}\left (\frac {\sqrt {5}\,\sqrt {x^4+5}}{5}\right )+\sqrt {x^4+5}\,\left (\frac {3\,x^2}{4}+1\right ) \]