Integrand size = 24, antiderivative size = 1243 \[ \int \frac {1}{x^3 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=-\frac {(5 b c-4 a d) \sqrt {c+d x^8}}{8 a^2 c (b c-a d) x^2}+\frac {\sqrt {d} (5 b c-4 a d) x^2 \sqrt {c+d x^8}}{8 a^2 c (b c-a d) \left (\sqrt {c}+\sqrt {d} x^4\right )}+\frac {b \sqrt {c+d x^8}}{8 a (b c-a d) x^2 \left (a+b x^8\right )}-\frac {b^{3/4} (5 b c-7 a d) \arctan \left (\frac {\sqrt {b c-a d} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {c+d x^8}}\right )}{32 (-a)^{9/4} (b c-a d)^{3/2}}-\frac {b^{3/4} (5 b c-7 a d) \arctan \left (\frac {\sqrt {-b c+a d} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {c+d x^8}}\right )}{32 (-a)^{9/4} (-b c+a d)^{3/2}}-\frac {\sqrt [4]{d} (5 b c-4 a d) \left (\sqrt {c}+\sqrt {d} x^4\right ) \sqrt {\frac {c+d x^8}{\left (\sqrt {c}+\sqrt {d} x^4\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{8 a^2 c^{3/4} (b c-a d) \sqrt {c+d x^8}}+\frac {\sqrt [4]{d} (5 b c-4 a d) \left (\sqrt {c}+\sqrt {d} x^4\right ) \sqrt {\frac {c+d x^8}{\left (\sqrt {c}+\sqrt {d} x^4\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right ),\frac {1}{2}\right )}{16 a^2 c^{3/4} (b c-a d) \sqrt {c+d x^8}}+\frac {b \left (\sqrt {c}-\frac {\sqrt {-a} \sqrt {d}}{\sqrt {b}}\right ) \sqrt [4]{d} (5 b c-7 a d) \left (\sqrt {c}+\sqrt {d} x^4\right ) \sqrt {\frac {c+d x^8}{\left (\sqrt {c}+\sqrt {d} x^4\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right ),\frac {1}{2}\right )}{32 a^2 \sqrt [4]{c} (b c-a d) (b c+a d) \sqrt {c+d x^8}}+\frac {b \left (\sqrt {c}+\frac {\sqrt {-a} \sqrt {d}}{\sqrt {b}}\right ) \sqrt [4]{d} (5 b c-7 a d) \left (\sqrt {c}+\sqrt {d} x^4\right ) \sqrt {\frac {c+d x^8}{\left (\sqrt {c}+\sqrt {d} x^4\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right ),\frac {1}{2}\right )}{32 a^2 \sqrt [4]{c} (b c-a d) (b c+a d) \sqrt {c+d x^8}}-\frac {\sqrt {b} \left (\sqrt {b} \sqrt {c}+\sqrt {-a} \sqrt {d}\right )^2 (5 b c-7 a d) \left (\sqrt {c}+\sqrt {d} x^4\right ) \sqrt {\frac {c+d x^8}{\left (\sqrt {c}+\sqrt {d} x^4\right )^2}} \operatorname {EllipticPi}\left (-\frac {\left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right )^2}{4 \sqrt {-a} \sqrt {b} \sqrt {c} \sqrt {d}},2 \arctan \left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right ),\frac {1}{2}\right )}{64 (-a)^{5/2} \sqrt [4]{c} \sqrt [4]{d} (b c-a d) (b c+a d) \sqrt {c+d x^8}}+\frac {\sqrt {b} \left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right )^2 (5 b c-7 a d) \left (\sqrt {c}+\sqrt {d} x^4\right ) \sqrt {\frac {c+d x^8}{\left (\sqrt {c}+\sqrt {d} x^4\right )^2}} \operatorname {EllipticPi}\left (\frac {\left (\sqrt {b} \sqrt {c}+\sqrt {-a} \sqrt {d}\right )^2}{4 \sqrt {-a} \sqrt {b} \sqrt {c} \sqrt {d}},2 \arctan \left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right ),\frac {1}{2}\right )}{64 (-a)^{5/2} \sqrt [4]{c} \sqrt [4]{d} (b c-a d) (b c+a d) \sqrt {c+d x^8}} \]
-1/32*b^(3/4)*(-7*a*d+5*b*c)*arctan(x^2*(-a*d+b*c)^(1/2)/(-a)^(1/4)/b^(1/4 )/(d*x^8+c)^(1/2))/(-a)^(9/4)/(-a*d+b*c)^(3/2)-1/32*b^(3/4)*(-7*a*d+5*b*c) *arctan(x^2*(a*d-b*c)^(1/2)/(-a)^(1/4)/b^(1/4)/(d*x^8+c)^(1/2))/(-a)^(9/4) /(a*d-b*c)^(3/2)-1/8*(-4*a*d+5*b*c)*(d*x^8+c)^(1/2)/a^2/c/(-a*d+b*c)/x^2+1 /8*b*(d*x^8+c)^(1/2)/a/(-a*d+b*c)/x^2/(b*x^8+a)+1/8*(-4*a*d+5*b*c)*x^2*d^( 1/2)*(d*x^8+c)^(1/2)/a^2/c/(-a*d+b*c)/(c^(1/2)+x^4*d^(1/2))-1/8*d^(1/4)*(- 4*a*d+5*b*c)*(cos(2*arctan(d^(1/4)*x^2/c^(1/4)))^2)^(1/2)/cos(2*arctan(d^( 1/4)*x^2/c^(1/4)))*EllipticE(sin(2*arctan(d^(1/4)*x^2/c^(1/4))),1/2*2^(1/2 ))*(c^(1/2)+x^4*d^(1/2))*((d*x^8+c)/(c^(1/2)+x^4*d^(1/2))^2)^(1/2)/a^2/c^( 3/4)/(-a*d+b*c)/(d*x^8+c)^(1/2)+1/16*d^(1/4)*(-4*a*d+5*b*c)*(cos(2*arctan( d^(1/4)*x^2/c^(1/4)))^2)^(1/2)/cos(2*arctan(d^(1/4)*x^2/c^(1/4)))*Elliptic F(sin(2*arctan(d^(1/4)*x^2/c^(1/4))),1/2*2^(1/2))*(c^(1/2)+x^4*d^(1/2))*(( d*x^8+c)/(c^(1/2)+x^4*d^(1/2))^2)^(1/2)/a^2/c^(3/4)/(-a*d+b*c)/(d*x^8+c)^( 1/2)+1/64*(-7*a*d+5*b*c)*(cos(2*arctan(d^(1/4)*x^2/c^(1/4)))^2)^(1/2)/cos( 2*arctan(d^(1/4)*x^2/c^(1/4)))*EllipticPi(sin(2*arctan(d^(1/4)*x^2/c^(1/4) )),1/4*(b^(1/2)*c^(1/2)+(-a)^(1/2)*d^(1/2))^2/(-a)^(1/2)/b^(1/2)/c^(1/2)/d ^(1/2),1/2*2^(1/2))*b^(1/2)*(c^(1/2)+x^4*d^(1/2))*(b^(1/2)*c^(1/2)-(-a)^(1 /2)*d^(1/2))^2*((d*x^8+c)/(c^(1/2)+x^4*d^(1/2))^2)^(1/2)/(-a)^(5/2)/c^(1/4 )/d^(1/4)/(-a*d+b*c)/(a*d+b*c)/(d*x^8+c)^(1/2)-1/64*(-7*a*d+5*b*c)*(cos(2* arctan(d^(1/4)*x^2/c^(1/4)))^2)^(1/2)/cos(2*arctan(d^(1/4)*x^2/c^(1/4))...
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 10.31 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.18 \[ \int \frac {1}{x^3 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\frac {21 a \left (c+d x^8\right ) \left (4 a^2 d-5 b^2 c x^8-4 a b \left (c-d x^8\right )\right )-7 \left (5 b^2 c^2-12 a b c d+4 a^2 d^2\right ) x^8 \left (a+b x^8\right ) \sqrt {1+\frac {d x^8}{c}} \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},-\frac {d x^8}{c},-\frac {b x^8}{a}\right )+3 b d (5 b c-4 a d) x^{16} \left (a+b x^8\right ) \sqrt {1+\frac {d x^8}{c}} \operatorname {AppellF1}\left (\frac {7}{4},\frac {1}{2},1,\frac {11}{4},-\frac {d x^8}{c},-\frac {b x^8}{a}\right )}{168 a^3 c (b c-a d) x^2 \left (a+b x^8\right ) \sqrt {c+d x^8}} \]
(21*a*(c + d*x^8)*(4*a^2*d - 5*b^2*c*x^8 - 4*a*b*(c - d*x^8)) - 7*(5*b^2*c ^2 - 12*a*b*c*d + 4*a^2*d^2)*x^8*(a + b*x^8)*Sqrt[1 + (d*x^8)/c]*AppellF1[ 3/4, 1/2, 1, 7/4, -((d*x^8)/c), -((b*x^8)/a)] + 3*b*d*(5*b*c - 4*a*d)*x^16 *(a + b*x^8)*Sqrt[1 + (d*x^8)/c]*AppellF1[7/4, 1/2, 1, 11/4, -((d*x^8)/c), -((b*x^8)/a)])/(168*a^3*c*(b*c - a*d)*x^2*(a + b*x^8)*Sqrt[c + d*x^8])
Time = 1.53 (sec) , antiderivative size = 1170, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {965, 972, 25, 1053, 1054, 2009}
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.
\(\displaystyle \int \frac {1}{x^3 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx\) |
\(\Big \downarrow \) 965 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^4 \left (b x^8+a\right )^2 \sqrt {d x^8+c}}dx^2\) |
\(\Big \downarrow \) 972 |
\(\displaystyle \frac {1}{2} \left (\frac {b \sqrt {c+d x^8}}{4 a x^2 \left (a+b x^8\right ) (b c-a d)}-\frac {\int -\frac {3 b d x^8+5 b c-4 a d}{x^4 \left (b x^8+a\right ) \sqrt {d x^8+c}}dx^2}{4 a (b c-a d)}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {3 b d x^8+5 b c-4 a d}{x^4 \left (b x^8+a\right ) \sqrt {d x^8+c}}dx^2}{4 a (b c-a d)}+\frac {b \sqrt {c+d x^8}}{4 a x^2 \left (a+b x^8\right ) (b c-a d)}\right )\) |
\(\Big \downarrow \) 1053 |
\(\displaystyle \frac {1}{2} \left (\frac {-\frac {\int \frac {x^4 \left ((b c-2 a d) (5 b c-2 a d)-b d (5 b c-4 a d) x^8\right )}{\left (b x^8+a\right ) \sqrt {d x^8+c}}dx^2}{a c}-\frac {\sqrt {c+d x^8} (5 b c-4 a d)}{a c x^2}}{4 a (b c-a d)}+\frac {b \sqrt {c+d x^8}}{4 a x^2 \left (a+b x^8\right ) (b c-a d)}\right )\) |
\(\Big \downarrow \) 1054 |
\(\displaystyle \frac {1}{2} \left (\frac {-\frac {\int \left (\frac {\left (5 b^2 c^2-7 a b c d\right ) x^4}{\left (b x^8+a\right ) \sqrt {d x^8+c}}-\frac {d (5 b c-4 a d) x^4}{\sqrt {d x^8+c}}\right )dx^2}{a c}-\frac {\sqrt {c+d x^8} (5 b c-4 a d)}{a c x^2}}{4 a (b c-a d)}+\frac {b \sqrt {c+d x^8}}{4 a x^2 \left (a+b x^8\right ) (b c-a d)}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {\sqrt {d x^8+c} b}{4 a (b c-a d) x^2 \left (b x^8+a\right )}+\frac {-\frac {\sqrt {d x^8+c} (5 b c-4 a d)}{a c x^2}-\frac {-\frac {\sqrt {b} c^{3/4} (5 b c-7 a d) \left (\sqrt {d} x^4+\sqrt {c}\right ) \sqrt {\frac {d x^8+c}{\left (\sqrt {d} x^4+\sqrt {c}\right )^2}} \operatorname {EllipticPi}\left (\frac {\left (\sqrt {b} \sqrt {c}+\sqrt {-a} \sqrt {d}\right )^2}{4 \sqrt {-a} \sqrt {b} \sqrt {c} \sqrt {d}},2 \arctan \left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right ),\frac {1}{2}\right ) \left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right )^2}{8 \sqrt {-a} \sqrt [4]{d} (b c+a d) \sqrt {d x^8+c}}+\frac {b^{3/4} c (5 b c-7 a d) \arctan \left (\frac {\sqrt {b c-a d} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {d x^8+c}}\right )}{4 \sqrt [4]{-a} \sqrt {b c-a d}}-\frac {b^{3/4} c (5 b c-7 a d) \text {arctanh}\left (\frac {\sqrt {b c-a d} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {d x^8+c}}\right )}{4 \sqrt [4]{-a} \sqrt {b c-a d}}+\frac {\sqrt [4]{c} \sqrt [4]{d} (5 b c-4 a d) \left (\sqrt {d} x^4+\sqrt {c}\right ) \sqrt {\frac {d x^8+c}{\left (\sqrt {d} x^4+\sqrt {c}\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{\sqrt {d x^8+c}}-\frac {\sqrt [4]{c} \sqrt [4]{d} (5 b c-4 a d) \left (\sqrt {d} x^4+\sqrt {c}\right ) \sqrt {\frac {d x^8+c}{\left (\sqrt {d} x^4+\sqrt {c}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right ),\frac {1}{2}\right )}{2 \sqrt {d x^8+c}}-\frac {b c^{3/4} \left (\sqrt {c}-\frac {\sqrt {-a} \sqrt {d}}{\sqrt {b}}\right ) \sqrt [4]{d} (5 b c-7 a d) \left (\sqrt {d} x^4+\sqrt {c}\right ) \sqrt {\frac {d x^8+c}{\left (\sqrt {d} x^4+\sqrt {c}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right ),\frac {1}{2}\right )}{4 (b c+a d) \sqrt {d x^8+c}}-\frac {b c^{3/4} \left (\sqrt {c}+\frac {\sqrt {-a} \sqrt {d}}{\sqrt {b}}\right ) \sqrt [4]{d} (5 b c-7 a d) \left (\sqrt {d} x^4+\sqrt {c}\right ) \sqrt {\frac {d x^8+c}{\left (\sqrt {d} x^4+\sqrt {c}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right ),\frac {1}{2}\right )}{4 (b c+a d) \sqrt {d x^8+c}}+\frac {\sqrt {b} c^{3/4} \left (\sqrt {b} \sqrt {c}+\sqrt {-a} \sqrt {d}\right )^2 (5 b c-7 a d) \left (\sqrt {d} x^4+\sqrt {c}\right ) \sqrt {\frac {d x^8+c}{\left (\sqrt {d} x^4+\sqrt {c}\right )^2}} \operatorname {EllipticPi}\left (-\frac {\sqrt {c} \left (\sqrt {b}-\frac {\sqrt {-a} \sqrt {d}}{\sqrt {c}}\right )^2}{4 \sqrt {-a} \sqrt {b} \sqrt {d}},2 \arctan \left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right ),\frac {1}{2}\right )}{8 \sqrt {-a} \sqrt [4]{d} (b c+a d) \sqrt {d x^8+c}}-\frac {\sqrt {d} (5 b c-4 a d) x^2 \sqrt {d x^8+c}}{\sqrt {d} x^4+\sqrt {c}}}{a c}}{4 a (b c-a d)}\right )\) |
((b*Sqrt[c + d*x^8])/(4*a*(b*c - a*d)*x^2*(a + b*x^8)) + (-(((5*b*c - 4*a* d)*Sqrt[c + d*x^8])/(a*c*x^2)) - (-((Sqrt[d]*(5*b*c - 4*a*d)*x^2*Sqrt[c + d*x^8])/(Sqrt[c] + Sqrt[d]*x^4)) + (b^(3/4)*c*(5*b*c - 7*a*d)*ArcTan[(Sqrt [b*c - a*d]*x^2)/((-a)^(1/4)*b^(1/4)*Sqrt[c + d*x^8])])/(4*(-a)^(1/4)*Sqrt [b*c - a*d]) - (b^(3/4)*c*(5*b*c - 7*a*d)*ArcTanh[(Sqrt[b*c - a*d]*x^2)/(( -a)^(1/4)*b^(1/4)*Sqrt[c + d*x^8])])/(4*(-a)^(1/4)*Sqrt[b*c - a*d]) + (c^( 1/4)*d^(1/4)*(5*b*c - 4*a*d)*(Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqr t[c] + Sqrt[d]*x^4)^2]*EllipticE[2*ArcTan[(d^(1/4)*x^2)/c^(1/4)], 1/2])/Sq rt[c + d*x^8] - (c^(1/4)*d^(1/4)*(5*b*c - 4*a*d)*(Sqrt[c] + Sqrt[d]*x^4)*S qrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticF[2*ArcTan[(d^(1/4)*x^2 )/c^(1/4)], 1/2])/(2*Sqrt[c + d*x^8]) - (b*c^(3/4)*(Sqrt[c] - (Sqrt[-a]*Sq rt[d])/Sqrt[b])*d^(1/4)*(5*b*c - 7*a*d)*(Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticF[2*ArcTan[(d^(1/4)*x^2)/c^(1/4) ], 1/2])/(4*(b*c + a*d)*Sqrt[c + d*x^8]) - (b*c^(3/4)*(Sqrt[c] + (Sqrt[-a] *Sqrt[d])/Sqrt[b])*d^(1/4)*(5*b*c - 7*a*d)*(Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticF[2*ArcTan[(d^(1/4)*x^2)/c^(1 /4)], 1/2])/(4*(b*c + a*d)*Sqrt[c + d*x^8]) - (Sqrt[b]*c^(3/4)*(Sqrt[b]*Sq rt[c] - Sqrt[-a]*Sqrt[d])^2*(5*b*c - 7*a*d)*(Sqrt[c] + Sqrt[d]*x^4)*Sqrt[( c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticPi[(Sqrt[b]*Sqrt[c] + Sqrt[- a]*Sqrt[d])^2/(4*Sqrt[-a]*Sqrt[b]*Sqrt[c]*Sqrt[d]), 2*ArcTan[(d^(1/4)*x...
3.10.23.3.1 Defintions of rubi rules used
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^q, x], x, x^k], x] /; k != 1] /; Free Q[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IntegerQ[m]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[(-b)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x ^n)^(q + 1)/(a*e*n*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*n*(b*c - a*d)*(p + 1)) Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*( b*c - a*d)*(p + 1) + d*b*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{ a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] & & IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_ ))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b *x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*c*g*(m + 1))), x] + Simp[1/(a*c*g^n*( m + 1)) Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2 ) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && IGtQ[n, 0] && LtQ[m, -1]
Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n _)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && IGtQ[n, 0]
\[\int \frac {1}{x^{3} \left (b \,x^{8}+a \right )^{2} \sqrt {d \,x^{8}+c}}d x\]
Timed out. \[ \int \frac {1}{x^3 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{x^3 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int \frac {1}{x^{3} \left (a + b x^{8}\right )^{2} \sqrt {c + d x^{8}}}\, dx \]
\[ \int \frac {1}{x^3 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int { \frac {1}{{\left (b x^{8} + a\right )}^{2} \sqrt {d x^{8} + c} x^{3}} \,d x } \]
\[ \int \frac {1}{x^3 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int { \frac {1}{{\left (b x^{8} + a\right )}^{2} \sqrt {d x^{8} + c} x^{3}} \,d x } \]
Timed out. \[ \int \frac {1}{x^3 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int \frac {1}{x^3\,{\left (b\,x^8+a\right )}^2\,\sqrt {d\,x^8+c}} \,d x \]