Integrand size = 25, antiderivative size = 104 \[ \int \frac {2+3 x^2}{x^7 \sqrt {3+5 x^2+x^4}} \, dx=-\frac {\sqrt {3+5 x^2+x^4}}{9 x^6}-\frac {\sqrt {3+5 x^2+x^4}}{54 x^4}+\frac {13 \sqrt {3+5 x^2+x^4}}{108 x^2}-\frac {61 \text {arctanh}\left (\frac {6+5 x^2}{2 \sqrt {3} \sqrt {3+5 x^2+x^4}}\right )}{216 \sqrt {3}} \]
-61/648*arctanh(1/6*(5*x^2+6)*3^(1/2)/(x^4+5*x^2+3)^(1/2))*3^(1/2)-1/9*(x^ 4+5*x^2+3)^(1/2)/x^6-1/54*(x^4+5*x^2+3)^(1/2)/x^4+13/108*(x^4+5*x^2+3)^(1/ 2)/x^2
Time = 0.24 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.72 \[ \int \frac {2+3 x^2}{x^7 \sqrt {3+5 x^2+x^4}} \, dx=\frac {\sqrt {3+5 x^2+x^4} \left (-12-2 x^2+13 x^4\right )}{108 x^6}+\frac {61 \text {arctanh}\left (\frac {x^2}{\sqrt {3}}-\frac {\sqrt {3+5 x^2+x^4}}{\sqrt {3}}\right )}{108 \sqrt {3}} \]
(Sqrt[3 + 5*x^2 + x^4]*(-12 - 2*x^2 + 13*x^4))/(108*x^6) + (61*ArcTanh[x^2 /Sqrt[3] - Sqrt[3 + 5*x^2 + x^4]/Sqrt[3]])/(108*Sqrt[3])
Time = 0.27 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {1578, 1237, 27, 1237, 27, 1228, 1154, 219}
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.
\(\displaystyle \int \frac {3 x^2+2}{x^7 \sqrt {x^4+5 x^2+3}} \, dx\) |
\(\Big \downarrow \) 1578 |
\(\displaystyle \frac {1}{2} \int \frac {3 x^2+2}{x^8 \sqrt {x^4+5 x^2+3}}dx^2\) |
\(\Big \downarrow \) 1237 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{9} \int -\frac {2 \left (1-2 x^2\right )}{x^6 \sqrt {x^4+5 x^2+3}}dx^2-\frac {2 \sqrt {x^4+5 x^2+3}}{9 x^6}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (\frac {2}{9} \int \frac {1-2 x^2}{x^6 \sqrt {x^4+5 x^2+3}}dx^2-\frac {2 \sqrt {x^4+5 x^2+3}}{9 x^6}\right )\) |
\(\Big \downarrow \) 1237 |
\(\displaystyle \frac {1}{2} \left (\frac {2}{9} \left (-\frac {1}{6} \int \frac {2 x^2+39}{2 x^4 \sqrt {x^4+5 x^2+3}}dx^2-\frac {\sqrt {x^4+5 x^2+3}}{6 x^4}\right )-\frac {2 \sqrt {x^4+5 x^2+3}}{9 x^6}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (\frac {2}{9} \left (-\frac {1}{12} \int \frac {2 x^2+39}{x^4 \sqrt {x^4+5 x^2+3}}dx^2-\frac {\sqrt {x^4+5 x^2+3}}{6 x^4}\right )-\frac {2 \sqrt {x^4+5 x^2+3}}{9 x^6}\right )\) |
\(\Big \downarrow \) 1228 |
\(\displaystyle \frac {1}{2} \left (\frac {2}{9} \left (\frac {1}{12} \left (\frac {61}{2} \int \frac {1}{x^2 \sqrt {x^4+5 x^2+3}}dx^2+\frac {13 \sqrt {x^4+5 x^2+3}}{x^2}\right )-\frac {\sqrt {x^4+5 x^2+3}}{6 x^4}\right )-\frac {2 \sqrt {x^4+5 x^2+3}}{9 x^6}\right )\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \frac {1}{2} \left (\frac {2}{9} \left (\frac {1}{12} \left (\frac {13 \sqrt {x^4+5 x^2+3}}{x^2}-61 \int \frac {1}{12-x^4}d\frac {5 x^2+6}{\sqrt {x^4+5 x^2+3}}\right )-\frac {\sqrt {x^4+5 x^2+3}}{6 x^4}\right )-\frac {2 \sqrt {x^4+5 x^2+3}}{9 x^6}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {2}{9} \left (\frac {1}{12} \left (\frac {13 \sqrt {x^4+5 x^2+3}}{x^2}-\frac {61 \text {arctanh}\left (\frac {5 x^2+6}{2 \sqrt {3} \sqrt {x^4+5 x^2+3}}\right )}{2 \sqrt {3}}\right )-\frac {\sqrt {x^4+5 x^2+3}}{6 x^4}\right )-\frac {2 \sqrt {x^4+5 x^2+3}}{9 x^6}\right )\) |
((-2*Sqrt[3 + 5*x^2 + x^4])/(9*x^6) + (2*(-1/6*Sqrt[3 + 5*x^2 + x^4]/x^4 + ((13*Sqrt[3 + 5*x^2 + x^4])/x^2 - (61*ArcTanh[(6 + 5*x^2)/(2*Sqrt[3]*Sqrt [3 + 5*x^2 + x^4])])/(2*Sqrt[3]))/12))/9)/2
3.2.88.3.1 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_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-2 Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c , d, e}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Simp[(b*(e *f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^ (m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x ] && EqQ[Simplify[m + 2*p + 3], 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*f - d*g)*(d + e*x)^(m + 1)*((a + b* x + c*x^2)^(p + 1)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[1/((m + 1) *(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[ (c*d*f - f*b*e + a*e*g)*(m + 1) + b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && LtQ[m, -1 ] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ )^4)^(p_.), x_Symbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && Int egerQ[(m - 1)/2]
Time = 0.38 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.63
method | result | size |
pseudoelliptic | \(\frac {-61 \,\operatorname {arctanh}\left (\frac {\left (5 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}+5 x^{2}+3}}\right ) \sqrt {3}\, x^{6}+6 \sqrt {x^{4}+5 x^{2}+3}\, \left (13 x^{4}-2 x^{2}-12\right )}{648 x^{6}}\) | \(66\) |
risch | \(\frac {13 x^{8}+63 x^{6}+17 x^{4}-66 x^{2}-36}{108 x^{6} \sqrt {x^{4}+5 x^{2}+3}}-\frac {61 \,\operatorname {arctanh}\left (\frac {\left (5 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}+5 x^{2}+3}}\right ) \sqrt {3}}{648}\) | \(71\) |
trager | \(\frac {\left (13 x^{4}-2 x^{2}-12\right ) \sqrt {x^{4}+5 x^{2}+3}}{108 x^{6}}+\frac {61 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (-\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x^{2}+6 \sqrt {x^{4}+5 x^{2}+3}-6 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right )}{x^{2}}\right )}{648}\) | \(79\) |
default | \(-\frac {61 \,\operatorname {arctanh}\left (\frac {\left (5 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}+5 x^{2}+3}}\right ) \sqrt {3}}{648}-\frac {\sqrt {x^{4}+5 x^{2}+3}}{9 x^{6}}-\frac {\sqrt {x^{4}+5 x^{2}+3}}{54 x^{4}}+\frac {13 \sqrt {x^{4}+5 x^{2}+3}}{108 x^{2}}\) | \(83\) |
elliptic | \(-\frac {61 \,\operatorname {arctanh}\left (\frac {\left (5 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}+5 x^{2}+3}}\right ) \sqrt {3}}{648}-\frac {\sqrt {x^{4}+5 x^{2}+3}}{9 x^{6}}-\frac {\sqrt {x^{4}+5 x^{2}+3}}{54 x^{4}}+\frac {13 \sqrt {x^{4}+5 x^{2}+3}}{108 x^{2}}\) | \(83\) |
1/648*(-61*arctanh(1/6*(5*x^2+6)*3^(1/2)/(x^4+5*x^2+3)^(1/2))*3^(1/2)*x^6+ 6*(x^4+5*x^2+3)^(1/2)*(13*x^4-2*x^2-12))/x^6
Time = 0.26 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.87 \[ \int \frac {2+3 x^2}{x^7 \sqrt {3+5 x^2+x^4}} \, dx=\frac {61 \, \sqrt {3} x^{6} \log \left (\frac {25 \, x^{2} - 2 \, \sqrt {3} {\left (5 \, x^{2} + 6\right )} - 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (5 \, \sqrt {3} - 6\right )} + 30}{x^{2}}\right ) + 78 \, x^{6} + 6 \, {\left (13 \, x^{4} - 2 \, x^{2} - 12\right )} \sqrt {x^{4} + 5 \, x^{2} + 3}}{648 \, x^{6}} \]
1/648*(61*sqrt(3)*x^6*log((25*x^2 - 2*sqrt(3)*(5*x^2 + 6) - 2*sqrt(x^4 + 5 *x^2 + 3)*(5*sqrt(3) - 6) + 30)/x^2) + 78*x^6 + 6*(13*x^4 - 2*x^2 - 12)*sq rt(x^4 + 5*x^2 + 3))/x^6
\[ \int \frac {2+3 x^2}{x^7 \sqrt {3+5 x^2+x^4}} \, dx=\int \frac {3 x^{2} + 2}{x^{7} \sqrt {x^{4} + 5 x^{2} + 3}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.82 \[ \int \frac {2+3 x^2}{x^7 \sqrt {3+5 x^2+x^4}} \, dx=-\frac {61}{648} \, \sqrt {3} \log \left (\frac {2 \, \sqrt {3} \sqrt {x^{4} + 5 \, x^{2} + 3}}{x^{2}} + \frac {6}{x^{2}} + 5\right ) + \frac {13 \, \sqrt {x^{4} + 5 \, x^{2} + 3}}{108 \, x^{2}} - \frac {\sqrt {x^{4} + 5 \, x^{2} + 3}}{54 \, x^{4}} - \frac {\sqrt {x^{4} + 5 \, x^{2} + 3}}{9 \, x^{6}} \]
-61/648*sqrt(3)*log(2*sqrt(3)*sqrt(x^4 + 5*x^2 + 3)/x^2 + 6/x^2 + 5) + 13/ 108*sqrt(x^4 + 5*x^2 + 3)/x^2 - 1/54*sqrt(x^4 + 5*x^2 + 3)/x^4 - 1/9*sqrt( x^4 + 5*x^2 + 3)/x^6
Leaf count of result is larger than twice the leaf count of optimal. 167 vs. \(2 (82) = 164\).
Time = 0.30 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.61 \[ \int \frac {2+3 x^2}{x^7 \sqrt {3+5 x^2+x^4}} \, dx=\frac {61}{648} \, \sqrt {3} \log \left (\frac {x^{2} + \sqrt {3} - \sqrt {x^{4} + 5 \, x^{2} + 3}}{x^{2} - \sqrt {3} - \sqrt {x^{4} + 5 \, x^{2} + 3}}\right ) - \frac {61 \, {\left (x^{2} - \sqrt {x^{4} + 5 \, x^{2} + 3}\right )}^{5} - 920 \, {\left (x^{2} - \sqrt {x^{4} + 5 \, x^{2} + 3}\right )}^{3} - 2052 \, {\left (x^{2} - \sqrt {x^{4} + 5 \, x^{2} + 3}\right )}^{2} - 1449 \, x^{2} + 1449 \, \sqrt {x^{4} + 5 \, x^{2} + 3} - 108}{108 \, {\left ({\left (x^{2} - \sqrt {x^{4} + 5 \, x^{2} + 3}\right )}^{2} - 3\right )}^{3}} \]
61/648*sqrt(3)*log((x^2 + sqrt(3) - sqrt(x^4 + 5*x^2 + 3))/(x^2 - sqrt(3) - sqrt(x^4 + 5*x^2 + 3))) - 1/108*(61*(x^2 - sqrt(x^4 + 5*x^2 + 3))^5 - 92 0*(x^2 - sqrt(x^4 + 5*x^2 + 3))^3 - 2052*(x^2 - sqrt(x^4 + 5*x^2 + 3))^2 - 1449*x^2 + 1449*sqrt(x^4 + 5*x^2 + 3) - 108)/((x^2 - sqrt(x^4 + 5*x^2 + 3 ))^2 - 3)^3
Timed out. \[ \int \frac {2+3 x^2}{x^7 \sqrt {3+5 x^2+x^4}} \, dx=\int \frac {3\,x^2+2}{x^7\,\sqrt {x^4+5\,x^2+3}} \,d x \]