Integrand size = 20, antiderivative size = 86 \[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^5} \, dx=-\frac {3 \left (15-2 x^2\right ) \sqrt {5+x^4}}{4 x^2}-\frac {\left (2-3 x^2\right ) \left (5+x^4\right )^{3/2}}{4 x^4}+\frac {45}{4} \text {arcsinh}\left (\frac {x^2}{\sqrt {5}}\right )-\frac {3}{2} \sqrt {5} \text {arctanh}\left (\frac {\sqrt {5+x^4}}{\sqrt {5}}\right ) \]
-1/4*(-3*x^2+2)*(x^4+5)^(3/2)/x^4+45/4*arcsinh(1/5*x^2*5^(1/2))-3/2*arctan h(1/5*(x^4+5)^(1/2)*5^(1/2))*5^(1/2)-3/4*(-2*x^2+15)*(x^4+5)^(1/2)/x^2
Time = 0.21 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.97 \[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^5} \, dx=3 \sqrt {5} \text {arctanh}\left (\frac {x^2-\sqrt {5+x^4}}{\sqrt {5}}\right )+\frac {1}{4} \left (\frac {\sqrt {5+x^4} \left (-10-30 x^2+4 x^4+3 x^6\right )}{x^4}-45 \log \left (-x^2+\sqrt {5+x^4}\right )\right ) \]
3*Sqrt[5]*ArcTanh[(x^2 - Sqrt[5 + x^4])/Sqrt[5]] + ((Sqrt[5 + x^4]*(-10 - 30*x^2 + 4*x^4 + 3*x^6))/x^4 - 45*Log[-x^2 + Sqrt[5 + x^4]])/4
Time = 0.23 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {1579, 537, 27, 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.
\(\displaystyle \int \frac {\left (3 x^2+2\right ) \left (x^4+5\right )^{3/2}}{x^5} \, dx\) |
\(\Big \downarrow \) 1579 |
\(\displaystyle \frac {1}{2} \int \frac {\left (3 x^2+2\right ) \left (x^4+5\right )^{3/2}}{x^6}dx^2\) |
\(\Big \downarrow \) 537 |
\(\displaystyle \frac {1}{2} \left (-\frac {3}{2} \int -\frac {2 \left (3 x^2+1\right ) \sqrt {x^4+5}}{x^2}dx^2-\frac {\left (3 x^2+1\right ) \left (x^4+5\right )^{3/2}}{x^4}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (3 \int \frac {\left (3 x^2+1\right ) \sqrt {x^4+5}}{x^2}dx^2-\frac {\left (3 x^2+1\right ) \left (x^4+5\right )^{3/2}}{x^4}\right )\) |
\(\Big \downarrow \) 535 |
\(\displaystyle \frac {1}{2} \left (3 \left (\frac {5}{2} \int \frac {3 x^2+2}{x^2 \sqrt {x^4+5}}dx^2+\frac {1}{2} \sqrt {x^4+5} \left (3 x^2+2\right )\right )-\frac {\left (3 x^2+1\right ) \left (x^4+5\right )^{3/2}}{x^4}\right )\) |
\(\Big \downarrow \) 538 |
\(\displaystyle \frac {1}{2} \left (3 \left (\frac {5}{2} \left (3 \int \frac {1}{\sqrt {x^4+5}}dx^2+2 \int \frac {1}{x^2 \sqrt {x^4+5}}dx^2\right )+\frac {1}{2} \sqrt {x^4+5} \left (3 x^2+2\right )\right )-\frac {\left (3 x^2+1\right ) \left (x^4+5\right )^{3/2}}{x^4}\right )\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{2} \left (3 \left (\frac {5}{2} \left (2 \int \frac {1}{x^2 \sqrt {x^4+5}}dx^2+3 \text {arcsinh}\left (\frac {x^2}{\sqrt {5}}\right )\right )+\frac {1}{2} \sqrt {x^4+5} \left (3 x^2+2\right )\right )-\frac {\left (3 x^2+1\right ) \left (x^4+5\right )^{3/2}}{x^4}\right )\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \left (3 \left (\frac {5}{2} \left (\int \frac {1}{x^2 \sqrt {x^4+5}}dx^4+3 \text {arcsinh}\left (\frac {x^2}{\sqrt {5}}\right )\right )+\frac {1}{2} \sqrt {x^4+5} \left (3 x^2+2\right )\right )-\frac {\left (3 x^2+1\right ) \left (x^4+5\right )^{3/2}}{x^4}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (3 \left (\frac {5}{2} \left (2 \int \frac {1}{\sqrt {x^4+5}-5}d\sqrt {x^4+5}+3 \text {arcsinh}\left (\frac {x^2}{\sqrt {5}}\right )\right )+\frac {1}{2} \sqrt {x^4+5} \left (3 x^2+2\right )\right )-\frac {\left (3 x^2+1\right ) \left (x^4+5\right )^{3/2}}{x^4}\right )\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {1}{2} \left (3 \left (\frac {5}{2} \left (3 \text {arcsinh}\left (\frac {x^2}{\sqrt {5}}\right )-\frac {2 \text {arctanh}\left (\frac {\sqrt {x^4+5}}{\sqrt {5}}\right )}{\sqrt {5}}\right )+\frac {1}{2} \sqrt {x^4+5} \left (3 x^2+2\right )\right )-\frac {\left (3 x^2+1\right ) \left (x^4+5\right )^{3/2}}{x^4}\right )\) |
(-(((1 + 3*x^2)*(5 + x^4)^(3/2))/x^4) + 3*(((2 + 3*x^2)*Sqrt[5 + x^4])/2 + (5*(3*ArcSinh[x^2/Sqrt[5]] - (2*ArcTanh[Sqrt[5 + x^4]/Sqrt[5]])/Sqrt[5])) /2))/2
3.1.25.3.1 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[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*(c*(m + 2) + d*(m + 1)*x)*((a + b*x^2)^p/((m + 1)*(m + 2))), x] - Simp[2*b*(p/((m + 1)*(m + 2))) Int[x^(m + 2)*(c*(m + 2) + d*(m + 1) *x)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, -2] && GtQ[p, 0] && !ILtQ[m + 2*p + 3, 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 = 0.49 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.76
method | result | size |
pseudoelliptic | \(\frac {-6 \sqrt {5}\, \operatorname {arctanh}\left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right ) x^{4}+45 \,\operatorname {arcsinh}\left (\frac {x^{2} \sqrt {5}}{5}\right ) x^{4}+3 \sqrt {x^{4}+5}\, \left (x^{6}+\frac {4}{3} x^{4}-10 x^{2}-\frac {10}{3}\right )}{4 x^{4}}\) | \(65\) |
default | \(\frac {45 \,\operatorname {arcsinh}\left (\frac {x^{2} \sqrt {5}}{5}\right )}{4}+\frac {3 x^{2} \sqrt {x^{4}+5}}{4}-\frac {15 \sqrt {x^{4}+5}}{2 x^{2}}+\sqrt {x^{4}+5}-\frac {5 \sqrt {x^{4}+5}}{2 x^{4}}-\frac {3 \sqrt {5}\, \operatorname {arctanh}\left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right )}{2}\) | \(73\) |
elliptic | \(\frac {45 \,\operatorname {arcsinh}\left (\frac {x^{2} \sqrt {5}}{5}\right )}{4}+\frac {3 x^{2} \sqrt {x^{4}+5}}{4}-\frac {15 \sqrt {x^{4}+5}}{2 x^{2}}+\sqrt {x^{4}+5}-\frac {5 \sqrt {x^{4}+5}}{2 x^{4}}-\frac {3 \sqrt {5}\, \operatorname {arctanh}\left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right )}{2}\) | \(73\) |
trager | \(\frac {\left (3 x^{6}+4 x^{4}-30 x^{2}-10\right ) \sqrt {x^{4}+5}}{4 x^{4}}+\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) \ln \left (\frac {\sqrt {x^{4}+5}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right )}{x^{2}}\right )}{2}-\frac {45 \ln \left (x^{2}-\sqrt {x^{4}+5}\right )}{4}\) | \(76\) |
risch | \(-\frac {5 \left (3 x^{6}+x^{4}+15 x^{2}+5\right )}{2 x^{4} \sqrt {x^{4}+5}}+\frac {45 \,\operatorname {arcsinh}\left (\frac {x^{2} \sqrt {5}}{5}\right )}{4}+\frac {3 x^{2} \sqrt {x^{4}+5}}{4}+\sqrt {x^{4}+5}-\frac {3 \sqrt {5}\, \operatorname {arctanh}\left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right )}{2}\) | \(76\) |
meijerg | \(\frac {3 \sqrt {5}\, \left (\frac {5 \sqrt {\pi }\, \left (-\frac {12 x^{4}}{5}+8\right )}{6 x^{4}}-\frac {5 \sqrt {\pi }\, \left (8-\frac {16 x^{4}}{5}\right ) \sqrt {1+\frac {x^{4}}{5}}}{6 x^{4}}-4 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {1+\frac {x^{4}}{5}}}{2}\right )+2 \left (1-2 \ln \left (2\right )+4 \ln \left (x \right )-\ln \left (5\right )\right ) \sqrt {\pi }-\frac {20 \sqrt {\pi }}{3 x^{4}}\right )}{8 \sqrt {\pi }}+\frac {-\frac {15 \sqrt {\pi }\, \sqrt {5}\, \left (-\frac {x^{4}}{10}+1\right ) \sqrt {1+\frac {x^{4}}{5}}}{2 x^{2}}+\frac {45 \sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {x^{2} \sqrt {5}}{5}\right )}{4}}{\sqrt {\pi }}\) | \(143\) |
1/4*(-6*5^(1/2)*arctanh(5^(1/2)/(x^4+5)^(1/2))*x^4+45*arcsinh(1/5*x^2*5^(1 /2))*x^4+3*(x^4+5)^(1/2)*(x^6+4/3*x^4-10*x^2-10/3))/x^4
Time = 0.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.95 \[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^5} \, dx=\frac {6 \, \sqrt {5} x^{4} \log \left (-\frac {\sqrt {5} - \sqrt {x^{4} + 5}}{x^{2}}\right ) - 45 \, x^{4} \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) - 30 \, x^{4} + {\left (3 \, x^{6} + 4 \, x^{4} - 30 \, x^{2} - 10\right )} \sqrt {x^{4} + 5}}{4 \, x^{4}} \]
1/4*(6*sqrt(5)*x^4*log(-(sqrt(5) - sqrt(x^4 + 5))/x^2) - 45*x^4*log(-x^2 + sqrt(x^4 + 5)) - 30*x^4 + (3*x^6 + 4*x^4 - 30*x^2 - 10)*sqrt(x^4 + 5))/x^ 4
Time = 5.51 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.55 \[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^5} \, dx=\frac {3 x^{6}}{4 \sqrt {x^{4} + 5}} - \frac {15 x^{2}}{4 \sqrt {x^{4} + 5}} + \sqrt {x^{4} + 5} + \frac {\sqrt {5} \log {\left (x^{4} \right )}}{2} - \sqrt {5} \log {\left (\sqrt {\frac {x^{4}}{5} + 1} + 1 \right )} - \frac {\sqrt {5} \operatorname {asinh}{\left (\frac {\sqrt {5}}{x^{2}} \right )}}{2} + \frac {45 \operatorname {asinh}{\left (\frac {\sqrt {5} x^{2}}{5} \right )}}{4} - \frac {5 \sqrt {1 + \frac {5}{x^{4}}}}{2 x^{2}} - \frac {75}{2 x^{2} \sqrt {x^{4} + 5}} \]
3*x**6/(4*sqrt(x**4 + 5)) - 15*x**2/(4*sqrt(x**4 + 5)) + sqrt(x**4 + 5) + sqrt(5)*log(x**4)/2 - sqrt(5)*log(sqrt(x**4/5 + 1) + 1) - sqrt(5)*asinh(sq rt(5)/x**2)/2 + 45*asinh(sqrt(5)*x**2/5)/4 - 5*sqrt(1 + 5/x**4)/(2*x**2) - 75/(2*x**2*sqrt(x**4 + 5))
Time = 0.29 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.43 \[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^5} \, dx=\frac {3}{4} \, \sqrt {5} \log \left (-\frac {\sqrt {5} - \sqrt {x^{4} + 5}}{\sqrt {5} + \sqrt {x^{4} + 5}}\right ) + \sqrt {x^{4} + 5} - \frac {15 \, \sqrt {x^{4} + 5}}{2 \, x^{2}} + \frac {15 \, \sqrt {x^{4} + 5}}{4 \, x^{2} {\left (\frac {x^{4} + 5}{x^{4}} - 1\right )}} - \frac {5 \, \sqrt {x^{4} + 5}}{2 \, x^{4}} + \frac {45}{8} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} + 1\right ) - \frac {45}{8} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} - 1\right ) \]
3/4*sqrt(5)*log(-(sqrt(5) - sqrt(x^4 + 5))/(sqrt(5) + sqrt(x^4 + 5))) + sq rt(x^4 + 5) - 15/2*sqrt(x^4 + 5)/x^2 + 15/4*sqrt(x^4 + 5)/(x^2*((x^4 + 5)/ x^4 - 1)) - 5/2*sqrt(x^4 + 5)/x^4 + 45/8*log(sqrt(x^4 + 5)/x^2 + 1) - 45/8 *log(sqrt(x^4 + 5)/x^2 - 1)
Leaf count of result is larger than twice the leaf count of optimal. 146 vs. \(2 (68) = 136\).
Time = 0.28 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.70 \[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^5} \, dx=\frac {1}{4} \, \sqrt {x^{4} + 5} {\left (3 \, x^{2} + 4\right )} + \frac {3}{2} \, \sqrt {5} \log \left (-\frac {x^{2} + \sqrt {5} - \sqrt {x^{4} + 5}}{x^{2} - \sqrt {5} - \sqrt {x^{4} + 5}}\right ) + \frac {5 \, {\left ({\left (x^{2} - \sqrt {x^{4} + 5}\right )}^{3} + 15 \, {\left (x^{2} - \sqrt {x^{4} + 5}\right )}^{2} + 5 \, x^{2} - 5 \, \sqrt {x^{4} + 5} - 75\right )}}{{\left ({\left (x^{2} - \sqrt {x^{4} + 5}\right )}^{2} - 5\right )}^{2}} - \frac {45}{4} \, \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) \]
1/4*sqrt(x^4 + 5)*(3*x^2 + 4) + 3/2*sqrt(5)*log(-(x^2 + sqrt(5) - sqrt(x^4 + 5))/(x^2 - sqrt(5) - sqrt(x^4 + 5))) + 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 - 45/4*log(-x^2 + sqrt(x^4 + 5))
Time = 7.93 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.83 \[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^5} \, dx=\frac {45\,\mathrm {asinh}\left (\frac {\sqrt {5}\,x^2}{5}\right )}{4}+\sqrt {x^4+5}\,\left (\frac {3\,x^2}{4}+1\right )-\frac {15\,\sqrt {x^4+5}}{2\,x^2}-\frac {5\,\sqrt {x^4+5}}{2\,x^4}+\frac {\sqrt {5}\,\mathrm {atan}\left (\frac {\sqrt {5}\,\sqrt {x^4+5}\,1{}\mathrm {i}}{5}\right )\,3{}\mathrm {i}}{2} \]