Integrand size = 20, antiderivative size = 81 \[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^3} \, dx=\frac {3}{2} \left (5+x^2\right ) \sqrt {5+x^4}-\frac {\left (2-x^2\right ) \left (5+x^4\right )^{3/2}}{2 x^2}+\frac {15}{2} \text {arcsinh}\left (\frac {x^2}{\sqrt {5}}\right )-\frac {15}{2} \sqrt {5} \text {arctanh}\left (\frac {\sqrt {5+x^4}}{\sqrt {5}}\right ) \]
-1/2*(-x^2+2)*(x^4+5)^(3/2)/x^2+15/2*arcsinh(1/5*x^2*5^(1/2))-15/2*arctanh (1/5*(x^4+5)^(1/2)*5^(1/2))*5^(1/2)+3/2*(x^2+5)*(x^4+5)^(1/2)
Time = 0.16 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.96 \[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^3} \, dx=\frac {1}{2} \left (\frac {\sqrt {5+x^4} \left (-10+20 x^2+x^4+x^6\right )}{x^2}+30 \sqrt {5} \text {arctanh}\left (\frac {x^2-\sqrt {5+x^4}}{\sqrt {5}}\right )-15 \log \left (-x^2+\sqrt {5+x^4}\right )\right ) \]
((Sqrt[5 + x^4]*(-10 + 20*x^2 + x^4 + x^6))/x^2 + 30*Sqrt[5]*ArcTanh[(x^2 - Sqrt[5 + x^4])/Sqrt[5]] - 15*Log[-x^2 + Sqrt[5 + x^4]])/2
Time = 0.22 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.96, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {1579, 536, 535, 27, 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.
\(\displaystyle \int \frac {\left (3 x^2+2\right ) \left (x^4+5\right )^{3/2}}{x^3} \, dx\) |
\(\Big \downarrow \) 1579 |
\(\displaystyle \frac {1}{2} \int \frac {\left (3 x^2+2\right ) \left (x^4+5\right )^{3/2}}{x^4}dx^2\) |
\(\Big \downarrow \) 536 |
\(\displaystyle \frac {1}{2} \left (\int \frac {\left (6 x^2+15\right ) \sqrt {x^4+5}}{x^2}dx^2-\frac {\left (2-x^2\right ) \left (x^4+5\right )^{3/2}}{x^2}\right )\) |
\(\Big \downarrow \) 535 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{2} \int \frac {6 \left (x^2+5\right )}{x^2 \sqrt {x^4+5}}dx^2-\frac {\left (2-x^2\right ) \left (x^4+5\right )^{3/2}}{x^2}+3 \left (x^2+5\right ) \sqrt {x^4+5}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (15 \int \frac {x^2+5}{x^2 \sqrt {x^4+5}}dx^2-\frac {\left (2-x^2\right ) \left (x^4+5\right )^{3/2}}{x^2}+3 \left (x^2+5\right ) \sqrt {x^4+5}\right )\) |
\(\Big \downarrow \) 538 |
\(\displaystyle \frac {1}{2} \left (15 \left (\int \frac {1}{\sqrt {x^4+5}}dx^2+5 \int \frac {1}{x^2 \sqrt {x^4+5}}dx^2\right )-\frac {\left (2-x^2\right ) \left (x^4+5\right )^{3/2}}{x^2}+3 \left (x^2+5\right ) \sqrt {x^4+5}\right )\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{2} \left (15 \left (5 \int \frac {1}{x^2 \sqrt {x^4+5}}dx^2+\text {arcsinh}\left (\frac {x^2}{\sqrt {5}}\right )\right )-\frac {\left (2-x^2\right ) \left (x^4+5\right )^{3/2}}{x^2}+3 \left (x^2+5\right ) \sqrt {x^4+5}\right )\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \left (15 \left (\frac {5}{2} \int \frac {1}{x^2 \sqrt {x^4+5}}dx^4+\text {arcsinh}\left (\frac {x^2}{\sqrt {5}}\right )\right )-\frac {\left (2-x^2\right ) \left (x^4+5\right )^{3/2}}{x^2}+3 \left (x^2+5\right ) \sqrt {x^4+5}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (15 \left (5 \int \frac {1}{\sqrt {x^4+5}-5}d\sqrt {x^4+5}+\text {arcsinh}\left (\frac {x^2}{\sqrt {5}}\right )\right )-\frac {\left (2-x^2\right ) \left (x^4+5\right )^{3/2}}{x^2}+3 \left (x^2+5\right ) \sqrt {x^4+5}\right )\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {1}{2} \left (15 \left (\text {arcsinh}\left (\frac {x^2}{\sqrt {5}}\right )-\sqrt {5} \text {arctanh}\left (\frac {\sqrt {x^4+5}}{\sqrt {5}}\right )\right )-\frac {\left (2-x^2\right ) \left (x^4+5\right )^{3/2}}{x^2}+3 \left (x^2+5\right ) \sqrt {x^4+5}\right )\) |
(3*(5 + x^2)*Sqrt[5 + x^4] - ((2 - x^2)*(5 + x^4)^(3/2))/x^2 + 15*(ArcSinh [x^2/Sqrt[5]] - Sqrt[5]*ArcTanh[Sqrt[5 + x^4]/Sqrt[5]]))/2
3.1.24.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_))*((a_) + (b_.)*(x_)^2)^(p_))/(x_)^2, x_Symbol] :> S imp[(-(2*c*p - d*x))*((a + b*x^2)^p/(2*p*x)), x] + Int[(a*d + 2*b*c*p*x)*(( a + b*x^2)^(p - 1)/x), x] /; FreeQ[{a, b, c, d}, x] && GtQ[p, 0] && Integer Q[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 = 0.50 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.77
method | result | size |
pseudoelliptic | \(\frac {-15 \sqrt {5}\, \operatorname {arctanh}\left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right ) x^{2}+15 \,\operatorname {arcsinh}\left (\frac {x^{2} \sqrt {5}}{5}\right ) x^{2}+\left (x^{6}+x^{4}+20 x^{2}-10\right ) \sqrt {x^{4}+5}}{2 x^{2}}\) | \(62\) |
trager | \(\frac {\left (x^{6}+x^{4}+20 x^{2}-10\right ) \sqrt {x^{4}+5}}{2 x^{2}}+\frac {15 \ln \left (-x^{2}-\sqrt {x^{4}+5}\right )}{2}+\frac {15 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) \ln \left (\frac {\sqrt {x^{4}+5}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right )}{x^{2}}\right )}{2}\) | \(74\) |
default | \(\frac {x^{4} \sqrt {x^{4}+5}}{2}+10 \sqrt {x^{4}+5}-\frac {15 \sqrt {5}\, \operatorname {arctanh}\left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right )}{2}+\frac {15 \,\operatorname {arcsinh}\left (\frac {x^{2} \sqrt {5}}{5}\right )}{2}+\frac {x^{2} \sqrt {x^{4}+5}}{2}-\frac {5 \sqrt {x^{4}+5}}{x^{2}}\) | \(75\) |
elliptic | \(\frac {x^{4} \sqrt {x^{4}+5}}{2}+10 \sqrt {x^{4}+5}-\frac {15 \sqrt {5}\, \operatorname {arctanh}\left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right )}{2}+\frac {15 \,\operatorname {arcsinh}\left (\frac {x^{2} \sqrt {5}}{5}\right )}{2}+\frac {x^{2} \sqrt {x^{4}+5}}{2}-\frac {5 \sqrt {x^{4}+5}}{x^{2}}\) | \(75\) |
risch | \(-\frac {5 \sqrt {x^{4}+5}}{x^{2}}+\frac {15 \,\operatorname {arcsinh}\left (\frac {x^{2} \sqrt {5}}{5}\right )}{2}+\frac {\sqrt {x^{4}+5}\, \left (x^{4}-10\right )}{2}+\frac {x^{2} \sqrt {x^{4}+5}}{2}+15 \sqrt {x^{4}+5}-\frac {15 \sqrt {5}\, \operatorname {arctanh}\left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right )}{2}\) | \(77\) |
meijerg | \(\frac {-\frac {5 \sqrt {\pi }\, \sqrt {5}\, \left (-\frac {x^{4}}{10}+1\right ) \sqrt {1+\frac {x^{4}}{5}}}{x^{2}}+\frac {15 \sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {x^{2} \sqrt {5}}{5}\right )}{2}}{\sqrt {\pi }}+\frac {45 \sqrt {5}\, \left (-\frac {32 \sqrt {\pi }}{9}+\frac {2 \sqrt {\pi }\, \left (\frac {4 x^{4}}{5}+16\right ) \sqrt {1+\frac {x^{4}}{5}}}{9}-\frac {8 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {1+\frac {x^{4}}{5}}}{2}\right )}{3}+\frac {4 \left (\frac {8}{3}-2 \ln \left (2\right )+4 \ln \left (x \right )-\ln \left (5\right )\right ) \sqrt {\pi }}{3}\right )}{16 \sqrt {\pi }}\) | \(122\) |
1/2*(-15*5^(1/2)*arctanh(5^(1/2)/(x^4+5)^(1/2))*x^2+15*arcsinh(1/5*x^2*5^( 1/2))*x^2+(x^6+x^4+20*x^2-10)*(x^4+5)^(1/2))/x^2
Time = 0.25 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.96 \[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^3} \, dx=\frac {15 \, \sqrt {5} x^{2} \log \left (-\frac {\sqrt {5} - \sqrt {x^{4} + 5}}{x^{2}}\right ) - 15 \, x^{2} \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) - 10 \, x^{2} + {\left (x^{6} + x^{4} + 20 \, x^{2} - 10\right )} \sqrt {x^{4} + 5}}{2 \, x^{2}} \]
1/2*(15*sqrt(5)*x^2*log(-(sqrt(5) - sqrt(x^4 + 5))/x^2) - 15*x^2*log(-x^2 + sqrt(x^4 + 5)) - 10*x^2 + (x^6 + x^4 + 20*x^2 - 10)*sqrt(x^4 + 5))/x^2
Time = 4.35 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.41 \[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^3} \, dx=\frac {x^{6}}{2 \sqrt {x^{4} + 5}} - \frac {5 x^{2}}{2 \sqrt {x^{4} + 5}} + \frac {\left (x^{4} + 5\right )^{\frac {3}{2}}}{2} + \frac {15 \sqrt {x^{4} + 5}}{2} + \frac {15 \sqrt {5} \log {\left (x^{4} \right )}}{4} - \frac {15 \sqrt {5} \log {\left (\sqrt {\frac {x^{4}}{5} + 1} + 1 \right )}}{2} + \frac {15 \operatorname {asinh}{\left (\frac {\sqrt {5} x^{2}}{5} \right )}}{2} - \frac {25}{x^{2} \sqrt {x^{4} + 5}} \]
x**6/(2*sqrt(x**4 + 5)) - 5*x**2/(2*sqrt(x**4 + 5)) + (x**4 + 5)**(3/2)/2 + 15*sqrt(x**4 + 5)/2 + 15*sqrt(5)*log(x**4)/4 - 15*sqrt(5)*log(sqrt(x**4/ 5 + 1) + 1)/2 + 15*asinh(sqrt(5)*x**2/5)/2 - 25/(x**2*sqrt(x**4 + 5))
Time = 0.28 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.51 \[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^3} \, dx=\frac {1}{2} \, {\left (x^{4} + 5\right )}^{\frac {3}{2}} + \frac {15}{4} \, \sqrt {5} \log \left (-\frac {\sqrt {5} - \sqrt {x^{4} + 5}}{\sqrt {5} + \sqrt {x^{4} + 5}}\right ) + \frac {15}{2} \, \sqrt {x^{4} + 5} - \frac {5 \, \sqrt {x^{4} + 5}}{x^{2}} + \frac {5 \, \sqrt {x^{4} + 5}}{2 \, x^{2} {\left (\frac {x^{4} + 5}{x^{4}} - 1\right )}} + \frac {15}{4} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} + 1\right ) - \frac {15}{4} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} - 1\right ) \]
1/2*(x^4 + 5)^(3/2) + 15/4*sqrt(5)*log(-(sqrt(5) - sqrt(x^4 + 5))/(sqrt(5) + sqrt(x^4 + 5))) + 15/2*sqrt(x^4 + 5) - 5*sqrt(x^4 + 5)/x^2 + 5/2*sqrt(x ^4 + 5)/(x^2*((x^4 + 5)/x^4 - 1)) + 15/4*log(sqrt(x^4 + 5)/x^2 + 1) - 15/4 *log(sqrt(x^4 + 5)/x^2 - 1)
Time = 0.29 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.26 \[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^3} \, dx=\frac {1}{2} \, \sqrt {x^{4} + 5} {\left ({\left (x^{2} + 1\right )} x^{2} + 20\right )} + \frac {15}{2} \, \sqrt {5} \log \left (-\frac {x^{2} + \sqrt {5} - \sqrt {x^{4} + 5}}{x^{2} - \sqrt {5} - \sqrt {x^{4} + 5}}\right ) + \frac {50}{{\left (x^{2} - \sqrt {x^{4} + 5}\right )}^{2} - 5} - \frac {15}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) \]
1/2*sqrt(x^4 + 5)*((x^2 + 1)*x^2 + 20) + 15/2*sqrt(5)*log(-(x^2 + sqrt(5) - sqrt(x^4 + 5))/(x^2 - sqrt(5) - sqrt(x^4 + 5))) + 50/((x^2 - sqrt(x^4 + 5))^2 - 5) - 15/2*log(-x^2 + sqrt(x^4 + 5))
Time = 8.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.79 \[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^3} \, dx=\frac {15\,\mathrm {asinh}\left (\frac {\sqrt {5}\,x^2}{5}\right )}{2}+\sqrt {x^4+5}\,\left (\frac {x^4}{2}+\frac {x^2}{2}+10\right )-\frac {5\,\sqrt {x^4+5}}{x^2}+\frac {\sqrt {5}\,\mathrm {atan}\left (\frac {\sqrt {5}\,\sqrt {x^4+5}\,1{}\mathrm {i}}{5}\right )\,15{}\mathrm {i}}{2} \]