Integrand size = 40, antiderivative size = 352 \[ \int \frac {\left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^4+a \left (q+p x^4\right )^2} \, dx=\frac {1}{4} \text {RootSum}\left [16 a p^2 q^2-32 a p^{3/2} q^{3/2} \text {$\#$1}^2+16 b \text {$\#$1}^4+24 a p q \text {$\#$1}^4-8 a \sqrt {p} \sqrt {q} \text {$\#$1}^6+a \text {$\#$1}^8\&,\frac {8 p^{3/2} q^{3/2} \log (x)-8 p^{3/2} q^{3/2} \log \left (\sqrt {q}+\sqrt {p} x^2+\sqrt {q+p x^4}-x \text {$\#$1}\right )-4 p q \log (x) \text {$\#$1}^2+4 p q \log \left (\sqrt {q}+\sqrt {p} x^2+\sqrt {q+p x^4}-x \text {$\#$1}\right ) \text {$\#$1}^2-2 \sqrt {p} \sqrt {q} \log (x) \text {$\#$1}^4+2 \sqrt {p} \sqrt {q} \log \left (\sqrt {q}+\sqrt {p} x^2+\sqrt {q+p x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4+\log (x) \text {$\#$1}^6-\log \left (\sqrt {q}+\sqrt {p} x^2+\sqrt {q+p x^4}-x \text {$\#$1}\right ) \text {$\#$1}^6}{8 a p^{3/2} q^{3/2} \text {$\#$1}-8 b \text {$\#$1}^3-12 a p q \text {$\#$1}^3+6 a \sqrt {p} \sqrt {q} \text {$\#$1}^5-a \text {$\#$1}^7}\&\right ] \]
Time = 1.33 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.37 \[ \int \frac {\left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^4+a \left (q+p x^4\right )^2} \, dx=-\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x \sqrt {q+p x^4}}{-\sqrt {b} x^2+\sqrt {a} \left (q+p x^4\right )}\right )+\text {arctanh}\left (\frac {\sqrt {b} x^2+\sqrt {a} \left (q+p x^4\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x \sqrt {q+p x^4}}\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{b}} \]
-1/2*(ArcTan[(Sqrt[2]*a^(1/4)*b^(1/4)*x*Sqrt[q + p*x^4])/(-(Sqrt[b]*x^2) + Sqrt[a]*(q + p*x^4))] + ArcTanh[(Sqrt[b]*x^2 + Sqrt[a]*(q + p*x^4))/(Sqrt [2]*a^(1/4)*b^(1/4)*x*Sqrt[q + p*x^4])])/(Sqrt[2]*a^(3/4)*b^(1/4))
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.
| \(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{a \left (p x^4+q\right )^2+b x^4} \, dx\) |
| \(\Big \downarrow \) 2091 |
| \(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\) |
3.30.49.3.1 Defintions of rubi rules used
Int[(Px_)*(u_)^(p_.)*(z_)^(q_.), x_Symbol] :> Int[Px*ExpandToSum[z, x]^q*Ex pandToSum[u, x]^p, x] /; FreeQ[{p, q}, x] && PolyQ[Px, x] && BinomialQ[z, x ] && TrinomialQ[u, x] && !(BinomialMatchQ[z, x] && TrinomialMatchQ[u, x])
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 0.60 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.48
| method | result | size |
| default | \(\frac {\left (\ln \left (\frac {\frac {p \,x^{4}+q}{2 x^{2}}-\frac {\left (\frac {b}{a}\right )^{\frac {1}{4}} \sqrt {p \,x^{4}+q}\, \sqrt {2}}{2 x}+\frac {\sqrt {\frac {b}{a}}}{2}}{\frac {p \,x^{4}+q}{2 x^{2}}+\frac {\left (\frac {b}{a}\right )^{\frac {1}{4}} \sqrt {p \,x^{4}+q}\, \sqrt {2}}{2 x}+\frac {\sqrt {\frac {b}{a}}}{2}}\right )+2 \arctan \left (\frac {\sqrt {p \,x^{4}+q}\, \sqrt {2}}{\left (\frac {b}{a}\right )^{\frac {1}{4}} x}+1\right )+2 \arctan \left (\frac {\sqrt {p \,x^{4}+q}\, \sqrt {2}}{\left (\frac {b}{a}\right )^{\frac {1}{4}} x}-1\right )\right ) \sqrt {2}}{8 a \left (\frac {b}{a}\right )^{\frac {1}{4}}}\) | \(169\) |
| elliptic | \(\frac {\left (\ln \left (\frac {\frac {p \,x^{4}+q}{2 x^{2}}-\frac {\left (\frac {b}{a}\right )^{\frac {1}{4}} \sqrt {p \,x^{4}+q}\, \sqrt {2}}{2 x}+\frac {\sqrt {\frac {b}{a}}}{2}}{\frac {p \,x^{4}+q}{2 x^{2}}+\frac {\left (\frac {b}{a}\right )^{\frac {1}{4}} \sqrt {p \,x^{4}+q}\, \sqrt {2}}{2 x}+\frac {\sqrt {\frac {b}{a}}}{2}}\right )+2 \arctan \left (\frac {\sqrt {p \,x^{4}+q}\, \sqrt {2}}{\left (\frac {b}{a}\right )^{\frac {1}{4}} x}+1\right )+2 \arctan \left (\frac {\sqrt {p \,x^{4}+q}\, \sqrt {2}}{\left (\frac {b}{a}\right )^{\frac {1}{4}} x}-1\right )\right ) \sqrt {2}}{8 a \left (\frac {b}{a}\right )^{\frac {1}{4}}}\) | \(169\) |
| pseudoelliptic | \(\frac {\sqrt {2}\, \left (\ln \left (\frac {p \,x^{4}-\left (\frac {b}{a}\right )^{\frac {1}{4}} \sqrt {p \,x^{4}+q}\, \sqrt {2}\, x +\sqrt {\frac {b}{a}}\, x^{2}+q}{p \,x^{4}+\left (\frac {b}{a}\right )^{\frac {1}{4}} \sqrt {p \,x^{4}+q}\, \sqrt {2}\, x +\sqrt {\frac {b}{a}}\, x^{2}+q}\right )+2 \arctan \left (\frac {\sqrt {p \,x^{4}+q}\, \sqrt {2}+\left (\frac {b}{a}\right )^{\frac {1}{4}} x}{\left (\frac {b}{a}\right )^{\frac {1}{4}} x}\right )+2 \arctan \left (\frac {\sqrt {p \,x^{4}+q}\, \sqrt {2}-\left (\frac {b}{a}\right )^{\frac {1}{4}} x}{\left (\frac {b}{a}\right )^{\frac {1}{4}} x}\right )\right )}{8 \left (\frac {b}{a}\right )^{\frac {1}{4}} a}\) | \(175\) |
1/8/a/(b/a)^(1/4)*(ln((1/2*(p*x^4+q)/x^2-1/2*(b/a)^(1/4)*(p*x^4+q)^(1/2)*2 ^(1/2)/x+1/2*(b/a)^(1/2))/(1/2*(p*x^4+q)/x^2+1/2*(b/a)^(1/4)*(p*x^4+q)^(1/ 2)*2^(1/2)/x+1/2*(b/a)^(1/2)))+2*arctan(1/(b/a)^(1/4)*(p*x^4+q)^(1/2)*2^(1 /2)/x+1)+2*arctan(1/(b/a)^(1/4)*(p*x^4+q)^(1/2)*2^(1/2)/x-1))*2^(1/2)
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 3.16 (sec) , antiderivative size = 607, normalized size of antiderivative = 1.72 \[ \int \frac {\left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^4+a \left (q+p x^4\right )^2} \, dx=\frac {1}{8} \, \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}} \log \left (\frac {2 \, {\left (a^{2} b p x^{6} + a^{2} b q x^{2}\right )} \left (-\frac {1}{a^{3} b}\right )^{\frac {3}{4}} + 2 \, {\left (p x^{5} - a b x^{3} \sqrt {-\frac {1}{a^{3} b}} + q x\right )} \sqrt {p x^{4} + q} - {\left (a p^{2} x^{8} + {\left (2 \, a p q - b\right )} x^{4} + a q^{2}\right )} \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}}}{2 \, {\left (a p^{2} x^{8} + {\left (2 \, a p q + b\right )} x^{4} + a q^{2}\right )}}\right ) - \frac {1}{8} \, \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}} \log \left (-\frac {2 \, {\left (a^{2} b p x^{6} + a^{2} b q x^{2}\right )} \left (-\frac {1}{a^{3} b}\right )^{\frac {3}{4}} - 2 \, {\left (p x^{5} - a b x^{3} \sqrt {-\frac {1}{a^{3} b}} + q x\right )} \sqrt {p x^{4} + q} - {\left (a p^{2} x^{8} + {\left (2 \, a p q - b\right )} x^{4} + a q^{2}\right )} \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}}}{2 \, {\left (a p^{2} x^{8} + {\left (2 \, a p q + b\right )} x^{4} + a q^{2}\right )}}\right ) + \frac {1}{8} i \, \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}} \log \left (-\frac {2 \, {\left (i \, a^{2} b p x^{6} + i \, a^{2} b q x^{2}\right )} \left (-\frac {1}{a^{3} b}\right )^{\frac {3}{4}} - 2 \, {\left (p x^{5} + a b x^{3} \sqrt {-\frac {1}{a^{3} b}} + q x\right )} \sqrt {p x^{4} + q} + {\left (i \, a p^{2} x^{8} + i \, {\left (2 \, a p q - b\right )} x^{4} + i \, a q^{2}\right )} \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}}}{2 \, {\left (a p^{2} x^{8} + {\left (2 \, a p q + b\right )} x^{4} + a q^{2}\right )}}\right ) - \frac {1}{8} i \, \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}} \log \left (-\frac {2 \, {\left (-i \, a^{2} b p x^{6} - i \, a^{2} b q x^{2}\right )} \left (-\frac {1}{a^{3} b}\right )^{\frac {3}{4}} - 2 \, {\left (p x^{5} + a b x^{3} \sqrt {-\frac {1}{a^{3} b}} + q x\right )} \sqrt {p x^{4} + q} + {\left (-i \, a p^{2} x^{8} - i \, {\left (2 \, a p q - b\right )} x^{4} - i \, a q^{2}\right )} \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}}}{2 \, {\left (a p^{2} x^{8} + {\left (2 \, a p q + b\right )} x^{4} + a q^{2}\right )}}\right ) \]
1/8*(-1/(a^3*b))^(1/4)*log(1/2*(2*(a^2*b*p*x^6 + a^2*b*q*x^2)*(-1/(a^3*b)) ^(3/4) + 2*(p*x^5 - a*b*x^3*sqrt(-1/(a^3*b)) + q*x)*sqrt(p*x^4 + q) - (a*p ^2*x^8 + (2*a*p*q - b)*x^4 + a*q^2)*(-1/(a^3*b))^(1/4))/(a*p^2*x^8 + (2*a* p*q + b)*x^4 + a*q^2)) - 1/8*(-1/(a^3*b))^(1/4)*log(-1/2*(2*(a^2*b*p*x^6 + a^2*b*q*x^2)*(-1/(a^3*b))^(3/4) - 2*(p*x^5 - a*b*x^3*sqrt(-1/(a^3*b)) + q *x)*sqrt(p*x^4 + q) - (a*p^2*x^8 + (2*a*p*q - b)*x^4 + a*q^2)*(-1/(a^3*b)) ^(1/4))/(a*p^2*x^8 + (2*a*p*q + b)*x^4 + a*q^2)) + 1/8*I*(-1/(a^3*b))^(1/4 )*log(-1/2*(2*(I*a^2*b*p*x^6 + I*a^2*b*q*x^2)*(-1/(a^3*b))^(3/4) - 2*(p*x^ 5 + a*b*x^3*sqrt(-1/(a^3*b)) + q*x)*sqrt(p*x^4 + q) + (I*a*p^2*x^8 + I*(2* a*p*q - b)*x^4 + I*a*q^2)*(-1/(a^3*b))^(1/4))/(a*p^2*x^8 + (2*a*p*q + b)*x ^4 + a*q^2)) - 1/8*I*(-1/(a^3*b))^(1/4)*log(-1/2*(2*(-I*a^2*b*p*x^6 - I*a^ 2*b*q*x^2)*(-1/(a^3*b))^(3/4) - 2*(p*x^5 + a*b*x^3*sqrt(-1/(a^3*b)) + q*x) *sqrt(p*x^4 + q) + (-I*a*p^2*x^8 - I*(2*a*p*q - b)*x^4 - I*a*q^2)*(-1/(a^3 *b))^(1/4))/(a*p^2*x^8 + (2*a*p*q + b)*x^4 + a*q^2))
Timed out. \[ \int \frac {\left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^4+a \left (q+p x^4\right )^2} \, dx=\text {Timed out} \]
Not integrable
Time = 0.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.11 \[ \int \frac {\left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^4+a \left (q+p x^4\right )^2} \, dx=\int { \frac {\sqrt {p x^{4} + q} {\left (p x^{4} - q\right )}}{b x^{4} + {\left (p x^{4} + q\right )}^{2} a} \,d x } \]
Timed out. \[ \int \frac {\left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^4+a \left (q+p x^4\right )^2} \, dx=\text {Timed out} \]
Not integrable
Time = 7.46 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.11 \[ \int \frac {\left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^4+a \left (q+p x^4\right )^2} \, dx=\int -\frac {\sqrt {p\,x^4+q}\,\left (q-p\,x^4\right )}{a\,{\left (p\,x^4+q\right )}^2+b\,x^4} \,d x \]