Integrand size = 20, antiderivative size = 82 \[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^7} \, dx=-\frac {\left (4-9 x^2\right ) \sqrt {5+x^4}}{4 x^2}-\frac {\left (4+9 x^2\right ) \left (5+x^4\right )^{3/2}}{12 x^6}+\text {arcsinh}\left (\frac {x^2}{\sqrt {5}}\right )-\frac {9}{4} \sqrt {5} \text {arctanh}\left (\frac {\sqrt {5+x^4}}{\sqrt {5}}\right ) \]
-1/12*(9*x^2+4)*(x^4+5)^(3/2)/x^6+arcsinh(1/5*x^2*5^(1/2))-9/4*arctanh(1/5 *(x^4+5)^(1/2)*5^(1/2))*5^(1/2)-1/4*(-9*x^2+4)*(x^4+5)^(1/2)/x^2
Time = 0.23 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.01 \[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^7} \, dx=\frac {\sqrt {5+x^4} \left (-20-45 x^2-16 x^4+18 x^6\right )}{12 x^6}+\frac {9}{2} \sqrt {5} \text {arctanh}\left (\frac {x^2-\sqrt {5+x^4}}{\sqrt {5}}\right )-\log \left (-x^2+\sqrt {5+x^4}\right ) \]
(Sqrt[5 + x^4]*(-20 - 45*x^2 - 16*x^4 + 18*x^6))/(12*x^6) + (9*Sqrt[5]*Arc Tanh[(x^2 - Sqrt[5 + x^4])/Sqrt[5]])/2 - Log[-x^2 + Sqrt[5 + x^4]]
Time = 0.23 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.09, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {1579, 537, 25, 536, 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.
\(\displaystyle \int \frac {\left (3 x^2+2\right ) \left (x^4+5\right )^{3/2}}{x^7} \, dx\) |
\(\Big \downarrow \) 1579 |
\(\displaystyle \frac {1}{2} \int \frac {\left (3 x^2+2\right ) \left (x^4+5\right )^{3/2}}{x^8}dx^2\) |
\(\Big \downarrow \) 537 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{2} \int -\frac {\left (9 x^2+4\right ) \sqrt {x^4+5}}{x^4}dx^2-\frac {\left (9 x^2+4\right ) \left (x^4+5\right )^{3/2}}{6 x^6}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {\left (9 x^2+4\right ) \sqrt {x^4+5}}{x^4}dx^2-\frac {\left (9 x^2+4\right ) \left (x^4+5\right )^{3/2}}{6 x^6}\right )\) |
\(\Big \downarrow \) 536 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\int \frac {4 x^2+45}{x^2 \sqrt {x^4+5}}dx^2-\frac {\left (4-9 x^2\right ) \sqrt {x^4+5}}{x^2}\right )-\frac {\left (9 x^2+4\right ) \left (x^4+5\right )^{3/2}}{6 x^6}\right )\) |
\(\Big \downarrow \) 538 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (4 \int \frac {1}{\sqrt {x^4+5}}dx^2+45 \int \frac {1}{x^2 \sqrt {x^4+5}}dx^2-\frac {\sqrt {x^4+5} \left (4-9 x^2\right )}{x^2}\right )-\frac {\left (9 x^2+4\right ) \left (x^4+5\right )^{3/2}}{6 x^6}\right )\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (45 \int \frac {1}{x^2 \sqrt {x^4+5}}dx^2+4 \text {arcsinh}\left (\frac {x^2}{\sqrt {5}}\right )-\frac {\sqrt {x^4+5} \left (4-9 x^2\right )}{x^2}\right )-\frac {\left (9 x^2+4\right ) \left (x^4+5\right )^{3/2}}{6 x^6}\right )\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {45}{2} \int \frac {1}{x^2 \sqrt {x^4+5}}dx^4+4 \text {arcsinh}\left (\frac {x^2}{\sqrt {5}}\right )-\frac {\sqrt {x^4+5} \left (4-9 x^2\right )}{x^2}\right )-\frac {\left (9 x^2+4\right ) \left (x^4+5\right )^{3/2}}{6 x^6}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (45 \int \frac {1}{\sqrt {x^4+5}-5}d\sqrt {x^4+5}+4 \text {arcsinh}\left (\frac {x^2}{\sqrt {5}}\right )-\frac {\sqrt {x^4+5} \left (4-9 x^2\right )}{x^2}\right )-\frac {\left (9 x^2+4\right ) \left (x^4+5\right )^{3/2}}{6 x^6}\right )\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (4 \text {arcsinh}\left (\frac {x^2}{\sqrt {5}}\right )-9 \sqrt {5} \text {arctanh}\left (\frac {\sqrt {x^4+5}}{\sqrt {5}}\right )-\frac {\sqrt {x^4+5} \left (4-9 x^2\right )}{x^2}\right )-\frac {\left (9 x^2+4\right ) \left (x^4+5\right )^{3/2}}{6 x^6}\right )\) |
(-1/6*((4 + 9*x^2)*(5 + x^4)^(3/2))/x^6 + (-(((4 - 9*x^2)*Sqrt[5 + x^4])/x ^2) + 4*ArcSinh[x^2/Sqrt[5]] - 9*Sqrt[5]*ArcTanh[Sqrt[5 + x^4]/Sqrt[5]])/2 )/2
3.1.26.3.1 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_)^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[(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 = 0.48 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.79
method | result | size |
pseudoelliptic | \(\frac {-9 \sqrt {5}\, \operatorname {arctanh}\left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right ) x^{6}+4 \,\operatorname {arcsinh}\left (\frac {x^{2} \sqrt {5}}{5}\right ) x^{6}+6 \left (x^{6}-\frac {8}{9} x^{4}-\frac {5}{2} x^{2}-\frac {10}{9}\right ) \sqrt {x^{4}+5}}{4 x^{6}}\) | \(65\) |
risch | \(-\frac {16 x^{8}+45 x^{6}+100 x^{4}+225 x^{2}+100}{12 x^{6} \sqrt {x^{4}+5}}+\operatorname {arcsinh}\left (\frac {x^{2} \sqrt {5}}{5}\right )+\frac {3 \sqrt {x^{4}+5}}{2}-\frac {9 \sqrt {5}\, \operatorname {arctanh}\left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right )}{4}\) | \(71\) |
default | \(\operatorname {arcsinh}\left (\frac {x^{2} \sqrt {5}}{5}\right )-\frac {4 \sqrt {x^{4}+5}}{3 x^{2}}-\frac {5 \sqrt {x^{4}+5}}{3 x^{6}}+\frac {3 \sqrt {x^{4}+5}}{2}-\frac {15 \sqrt {x^{4}+5}}{4 x^{4}}-\frac {9 \sqrt {5}\, \operatorname {arctanh}\left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right )}{4}\) | \(73\) |
elliptic | \(\operatorname {arcsinh}\left (\frac {x^{2} \sqrt {5}}{5}\right )-\frac {4 \sqrt {x^{4}+5}}{3 x^{2}}-\frac {5 \sqrt {x^{4}+5}}{3 x^{6}}+\frac {3 \sqrt {x^{4}+5}}{2}-\frac {15 \sqrt {x^{4}+5}}{4 x^{4}}-\frac {9 \sqrt {5}\, \operatorname {arctanh}\left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right )}{4}\) | \(73\) |
trager | \(\frac {\left (18 x^{6}-16 x^{4}-45 x^{2}-20\right ) \sqrt {x^{4}+5}}{12 x^{6}}-\ln \left (x^{2}-\sqrt {x^{4}+5}\right )+\frac {9 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) \ln \left (\frac {\sqrt {x^{4}+5}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right )}{x^{2}}\right )}{4}\) | \(76\) |
meijerg | \(\frac {-\frac {5 \sqrt {\pi }\, \sqrt {5}\, \left (\frac {4 x^{4}}{5}+1\right ) \sqrt {1+\frac {x^{4}}{5}}}{3 x^{6}}+\sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {x^{2} \sqrt {5}}{5}\right )}{\sqrt {\pi }}+\frac {9 \sqrt {5}\, \left (\frac {5 \sqrt {\pi }\, \left (-\frac {12 x^{4}}{5}+8\right )}{6 x^{4}}-\frac {5 \sqrt {\pi }\, \left (8-\frac {16 x^{4}}{5}\right ) \sqrt {1+\frac {x^{4}}{5}}}{6 x^{4}}-4 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {1+\frac {x^{4}}{5}}}{2}\right )+2 \left (1-2 \ln \left (2\right )+4 \ln \left (x \right )-\ln \left (5\right )\right ) \sqrt {\pi }-\frac {20 \sqrt {\pi }}{3 x^{4}}\right )}{16 \sqrt {\pi }}\) | \(143\) |
1/4*(-9*5^(1/2)*arctanh(5^(1/2)/(x^4+5)^(1/2))*x^6+4*arcsinh(1/5*x^2*5^(1/ 2))*x^6+6*(x^6-8/9*x^4-5/2*x^2-10/9)*(x^4+5)^(1/2))/x^6
Time = 0.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00 \[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^7} \, dx=\frac {27 \, \sqrt {5} x^{6} \log \left (-\frac {\sqrt {5} - \sqrt {x^{4} + 5}}{x^{2}}\right ) - 12 \, x^{6} \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) - 16 \, x^{6} + {\left (18 \, x^{6} - 16 \, x^{4} - 45 \, x^{2} - 20\right )} \sqrt {x^{4} + 5}}{12 \, x^{6}} \]
1/12*(27*sqrt(5)*x^6*log(-(sqrt(5) - sqrt(x^4 + 5))/x^2) - 12*x^6*log(-x^2 + sqrt(x^4 + 5)) - 16*x^6 + (18*x^6 - 16*x^4 - 45*x^2 - 20)*sqrt(x^4 + 5) )/x^6
Time = 5.79 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.80 \[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^7} \, dx=- \frac {x^{2}}{\sqrt {x^{4} + 5}} - \frac {\sqrt {1 + \frac {5}{x^{4}}}}{3} + \frac {3 \sqrt {x^{4} + 5}}{2} + \frac {3 \sqrt {5} \log {\left (x^{4} \right )}}{4} - \frac {3 \sqrt {5} \log {\left (\sqrt {\frac {x^{4}}{5} + 1} + 1 \right )}}{2} - \frac {3 \sqrt {5} \operatorname {asinh}{\left (\frac {\sqrt {5}}{x^{2}} \right )}}{4} + \operatorname {asinh}{\left (\frac {\sqrt {5} x^{2}}{5} \right )} - \frac {15 \sqrt {1 + \frac {5}{x^{4}}}}{4 x^{2}} - \frac {5}{x^{2} \sqrt {x^{4} + 5}} - \frac {5 \sqrt {1 + \frac {5}{x^{4}}}}{3 x^{4}} \]
-x**2/sqrt(x**4 + 5) - sqrt(1 + 5/x**4)/3 + 3*sqrt(x**4 + 5)/2 + 3*sqrt(5) *log(x**4)/4 - 3*sqrt(5)*log(sqrt(x**4/5 + 1) + 1)/2 - 3*sqrt(5)*asinh(sqr t(5)/x**2)/4 + asinh(sqrt(5)*x**2/5) - 15*sqrt(1 + 5/x**4)/(4*x**2) - 5/(x **2*sqrt(x**4 + 5)) - 5*sqrt(1 + 5/x**4)/(3*x**4)
Time = 0.29 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.37 \[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^7} \, dx=\frac {9}{8} \, \sqrt {5} \log \left (-\frac {\sqrt {5} - \sqrt {x^{4} + 5}}{\sqrt {5} + \sqrt {x^{4} + 5}}\right ) + \frac {3}{2} \, \sqrt {x^{4} + 5} - \frac {\sqrt {x^{4} + 5}}{x^{2}} - \frac {15 \, \sqrt {x^{4} + 5}}{4 \, x^{4}} - \frac {{\left (x^{4} + 5\right )}^{\frac {3}{2}}}{3 \, x^{6}} + \frac {1}{2} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} + 1\right ) - \frac {1}{2} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} - 1\right ) \]
9/8*sqrt(5)*log(-(sqrt(5) - sqrt(x^4 + 5))/(sqrt(5) + sqrt(x^4 + 5))) + 3/ 2*sqrt(x^4 + 5) - sqrt(x^4 + 5)/x^2 - 15/4*sqrt(x^4 + 5)/x^4 - 1/3*(x^4 + 5)^(3/2)/x^6 + 1/2*log(sqrt(x^4 + 5)/x^2 + 1) - 1/2*log(sqrt(x^4 + 5)/x^2 - 1)
Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (66) = 132\).
Time = 0.31 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.93 \[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^7} \, dx=\frac {9}{4} \, \sqrt {5} \log \left (-\frac {x^{2} + \sqrt {5} - \sqrt {x^{4} + 5}}{x^{2} - \sqrt {5} - \sqrt {x^{4} + 5}}\right ) + \frac {3}{2} \, \sqrt {x^{4} + 5} + \frac {5 \, {\left (9 \, {\left (x^{2} - \sqrt {x^{4} + 5}\right )}^{5} + 24 \, {\left (x^{2} - \sqrt {x^{4} + 5}\right )}^{4} - 120 \, {\left (x^{2} - \sqrt {x^{4} + 5}\right )}^{2} - 225 \, x^{2} + 225 \, \sqrt {x^{4} + 5} + 400\right )}}{6 \, {\left ({\left (x^{2} - \sqrt {x^{4} + 5}\right )}^{2} - 5\right )}^{3}} - \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) \]
9/4*sqrt(5)*log(-(x^2 + sqrt(5) - sqrt(x^4 + 5))/(x^2 - sqrt(5) - sqrt(x^4 + 5))) + 3/2*sqrt(x^4 + 5) + 5/6*(9*(x^2 - sqrt(x^4 + 5))^5 + 24*(x^2 - s qrt(x^4 + 5))^4 - 120*(x^2 - sqrt(x^4 + 5))^2 - 225*x^2 + 225*sqrt(x^4 + 5 ) + 400)/((x^2 - sqrt(x^4 + 5))^2 - 5)^3 - log(-x^2 + sqrt(x^4 + 5))
Time = 8.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00 \[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^7} \, dx=\mathrm {asinh}\left (\frac {\sqrt {5}\,x^2}{5}\right )+\frac {3\,\sqrt {x^4+5}}{2}+\sqrt {x^4+5}\,\left (\frac {2}{3\,x^2}-\frac {5}{3\,x^6}\right )-\frac {2\,\sqrt {x^4+5}}{x^2}-\frac {15\,\sqrt {x^4+5}}{4\,x^4}+\frac {\sqrt {5}\,\mathrm {atan}\left (\frac {\sqrt {5}\,\sqrt {x^4+5}\,1{}\mathrm {i}}{5}\right )\,9{}\mathrm {i}}{4} \]