Integrand size = 42, antiderivative size = 82 \[ \int \frac {x^5 \left (-4 b+5 a x^2\right )}{\sqrt [4]{-b+a x^2} \left (b-b x^8+a x^{10}\right )} \, dx=\frac {1}{2} a \text {RootSum}\left [a^4 b+b^4 \text {$\#$1}^4+4 b^3 \text {$\#$1}^8+6 b^2 \text {$\#$1}^{12}+4 b \text {$\#$1}^{16}+\text {$\#$1}^{20}\&,\frac {\log \left (\sqrt [4]{-b+a x^2}-\text {$\#$1}\right )}{b \text {$\#$1}+\text {$\#$1}^5}\&\right ] \]
Time = 0.11 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00 \[ \int \frac {x^5 \left (-4 b+5 a x^2\right )}{\sqrt [4]{-b+a x^2} \left (b-b x^8+a x^{10}\right )} \, dx=\frac {1}{2} a \text {RootSum}\left [a^4 b+b^4 \text {$\#$1}^4+4 b^3 \text {$\#$1}^8+6 b^2 \text {$\#$1}^{12}+4 b \text {$\#$1}^{16}+\text {$\#$1}^{20}\&,\frac {\log \left (\sqrt [4]{-b+a x^2}-\text {$\#$1}\right )}{b \text {$\#$1}+\text {$\#$1}^5}\&\right ] \]
(a*RootSum[a^4*b + b^4*#1^4 + 4*b^3*#1^8 + 6*b^2*#1^12 + 4*b*#1^16 + #1^20 & , Log[(-b + a*x^2)^(1/4) - #1]/(b*#1 + #1^5) & ])/2
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.
| \(\displaystyle \int \frac {x^5 \left (5 a x^2-4 b\right )}{\sqrt [4]{a x^2-b} \left (a x^{10}-b x^8+b\right )} \, dx\) |
| \(\Big \downarrow \) 7283 |
| \(\displaystyle \frac {1}{2} \int -\frac {x^4 \left (4 b-5 a x^2\right )}{\sqrt [4]{a x^2-b} \left (a x^{10}-b x^8+b\right )}dx^2\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{2} \int \frac {x^4 \left (4 b-5 a x^2\right )}{\sqrt [4]{a x^2-b} \left (a x^{10}-b x^8+b\right )}dx^2\) |
| \(\Big \downarrow \) 7267 |
| \(\displaystyle 2 a \int \frac {x^4 \left (x^8+b\right )^2 \left (5 x^8+b\right )}{\left (x^8+b\right )^4 x^8+a^4 b}d\sqrt [4]{a x^2-b}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 2 a \int \left (\frac {5 x^{28}}{x^{40}+4 b x^{32}+6 b^2 x^{24}+4 b^3 x^{16}+b^4 x^8+a^4 b}+\frac {11 b x^{20}}{x^{40}+4 b x^{32}+6 b^2 x^{24}+4 b^3 x^{16}+b^4 x^8+a^4 b}+\frac {7 b^2 x^{12}}{x^{40}+4 b x^{32}+6 b^2 x^{24}+4 b^3 x^{16}+b^4 x^8+a^4 b}+\frac {b^3 x^4}{x^{40}+4 b x^{32}+6 b^2 x^{24}+4 b^3 x^{16}+b^4 x^8+a^4 b}\right )d\sqrt [4]{a x^2-b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 a \left (b^3 \int \frac {x^4}{\left (x^8+b\right )^4 x^8+a^4 b}d\sqrt [4]{a x^2-b}+7 b^2 \int \frac {x^{12}}{\left (x^8+b\right )^4 x^8+a^4 b}d\sqrt [4]{a x^2-b}+5 \int \frac {x^{28}}{\left (x^8+b\right )^4 x^8+a^4 b}d\sqrt [4]{a x^2-b}+11 b \int \frac {x^{20}}{\left (x^8+b\right )^4 x^8+a^4 b}d\sqrt [4]{a x^2-b}\right )\) |
3.12.8.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si mp[lst[[2]]*lst[[4]] Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x ] /; !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
Int[(u_)*(x_)^(m_.), x_Symbol] :> With[{lst = PowerVariableExpn[u, m + 1, x ]}, Simp[1/lst[[2]] Subst[Int[NormalizeIntegrand[Simplify[lst[[1]]/x], x] , x], x, (lst[[3]]*x)^lst[[2]]], x] /; !FalseQ[lst] && NeQ[lst[[2]], m + 1 ]] /; IntegerQ[m] && NeQ[m, -1] && NonsumQ[u] && (GtQ[m, 0] || !AlgebraicF unctionQ[u, x])
Time = 1.06 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.89
| method | result | size |
| pseudoelliptic | \(\frac {a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{20}+4 b \,\textit {\_Z}^{16}+6 b^{2} \textit {\_Z}^{12}+4 b^{3} \textit {\_Z}^{8}+b^{4} \textit {\_Z}^{4}+a^{4} b \right )}{\sum }\frac {\ln \left (\left (a \,x^{2}-b \right )^{\frac {1}{4}}-\textit {\_R} \right )}{\textit {\_R} \left (\textit {\_R}^{4}+b \right )}\right )}{2}\) | \(73\) |
1/2*a*sum(ln((a*x^2-b)^(1/4)-_R)/_R/(_R^4+b),_R=RootOf(_Z^20+4*_Z^16*b+6*_ Z^12*b^2+4*_Z^8*b^3+_Z^4*b^4+a^4*b))
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 0.32 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.76 \[ \int \frac {x^5 \left (-4 b+5 a x^2\right )}{\sqrt [4]{-b+a x^2} \left (b-b x^8+a x^{10}\right )} \, dx=\frac {1}{2} \, \left (-\frac {1}{b}\right )^{\frac {1}{4}} \log \left ({\left (a x^{2} - b\right )}^{\frac {1}{4}} x^{2} + b \left (-\frac {1}{b}\right )^{\frac {3}{4}}\right ) - \frac {1}{2} i \, \left (-\frac {1}{b}\right )^{\frac {1}{4}} \log \left ({\left (a x^{2} - b\right )}^{\frac {1}{4}} x^{2} + i \, b \left (-\frac {1}{b}\right )^{\frac {3}{4}}\right ) + \frac {1}{2} i \, \left (-\frac {1}{b}\right )^{\frac {1}{4}} \log \left ({\left (a x^{2} - b\right )}^{\frac {1}{4}} x^{2} - i \, b \left (-\frac {1}{b}\right )^{\frac {3}{4}}\right ) - \frac {1}{2} \, \left (-\frac {1}{b}\right )^{\frac {1}{4}} \log \left ({\left (a x^{2} - b\right )}^{\frac {1}{4}} x^{2} - b \left (-\frac {1}{b}\right )^{\frac {3}{4}}\right ) \]
1/2*(-1/b)^(1/4)*log((a*x^2 - b)^(1/4)*x^2 + b*(-1/b)^(3/4)) - 1/2*I*(-1/b )^(1/4)*log((a*x^2 - b)^(1/4)*x^2 + I*b*(-1/b)^(3/4)) + 1/2*I*(-1/b)^(1/4) *log((a*x^2 - b)^(1/4)*x^2 - I*b*(-1/b)^(3/4)) - 1/2*(-1/b)^(1/4)*log((a*x ^2 - b)^(1/4)*x^2 - b*(-1/b)^(3/4))
Not integrable
Time = 101.68 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.44 \[ \int \frac {x^5 \left (-4 b+5 a x^2\right )}{\sqrt [4]{-b+a x^2} \left (b-b x^8+a x^{10}\right )} \, dx=\int \frac {x^{5} \cdot \left (5 a x^{2} - 4 b\right )}{\sqrt [4]{a x^{2} - b} \left (a x^{10} - b x^{8} + b\right )}\, dx \]
Not integrable
Time = 0.24 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.51 \[ \int \frac {x^5 \left (-4 b+5 a x^2\right )}{\sqrt [4]{-b+a x^2} \left (b-b x^8+a x^{10}\right )} \, dx=\int { \frac {{\left (5 \, a x^{2} - 4 \, b\right )} x^{5}}{{\left (a x^{10} - b x^{8} + b\right )} {\left (a x^{2} - b\right )}^{\frac {1}{4}}} \,d x } \]
Timed out. \[ \int \frac {x^5 \left (-4 b+5 a x^2\right )}{\sqrt [4]{-b+a x^2} \left (b-b x^8+a x^{10}\right )} \, dx=\text {Timed out} \]
Time = 14.50 (sec) , antiderivative size = 3940, normalized size of antiderivative = 48.05 \[ \int \frac {x^5 \left (-4 b+5 a x^2\right )}{\sqrt [4]{-b+a x^2} \left (b-b x^8+a x^{10}\right )} \, dx=\text {Too large to display} \]
symsum(log(root(3840000000000*a^8*b^8*f^20 - 31250000000000*a^12*b^4*f^20 - 209715200000*a^4*b^12*f^20 + 4294967296*b^16*f^20 + 95367431640625*a^16* f^20 - 29296875000000*a^12*b^3*f^16 + 15200000000000*a^8*b^7*f^16 - 190054 4000000*a^4*b^11*f^16 + 69793218560*b^15*f^16 + 1626112000000*a^4*b^10*f^1 2 + 2575000000000*a^8*b^6*f^12 + 17280532480*b^14*f^12 - 58112000000*a^4*b ^9*f^8 + 1614807040*b^13*f^8 + 67174400*b^12*f^4 + 1048576*b^11, f, k)*(ro ot(3840000000000*a^8*b^8*f^20 - 31250000000000*a^12*b^4*f^20 - 20971520000 0*a^4*b^12*f^20 + 4294967296*b^16*f^20 + 95367431640625*a^16*f^20 - 292968 75000000*a^12*b^3*f^16 + 15200000000000*a^8*b^7*f^16 - 1900544000000*a^4*b ^11*f^16 + 69793218560*b^15*f^16 + 1626112000000*a^4*b^10*f^12 + 257500000 0000*a^8*b^6*f^12 + 17280532480*b^14*f^12 - 58112000000*a^4*b^9*f^8 + 1614 807040*b^13*f^8 + 67174400*b^12*f^4 + 1048576*b^11, f, k)^3*(root(38400000 00000*a^8*b^8*f^20 - 31250000000000*a^12*b^4*f^20 - 209715200000*a^4*b^12* f^20 + 4294967296*b^16*f^20 + 95367431640625*a^16*f^20 - 29296875000000*a^ 12*b^3*f^16 + 15200000000000*a^8*b^7*f^16 - 1900544000000*a^4*b^11*f^16 + 69793218560*b^15*f^16 + 1626112000000*a^4*b^10*f^12 + 2575000000000*a^8*b^ 6*f^12 + 17280532480*b^14*f^12 - 58112000000*a^4*b^9*f^8 + 1614807040*b^13 *f^8 + 67174400*b^12*f^4 + 1048576*b^11, f, k)*(root(3840000000000*a^8*b^8 *f^20 - 31250000000000*a^12*b^4*f^20 - 209715200000*a^4*b^12*f^20 + 429496 7296*b^16*f^20 + 95367431640625*a^16*f^20 - 29296875000000*a^12*b^3*f^1...