Integrand size = 24, antiderivative size = 149 \[ \int \frac {1}{x^5 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=-\frac {(3 b c-2 a d) \sqrt {c+d x^8}}{8 a^2 c (b c-a d) x^4}+\frac {b \sqrt {c+d x^8}}{8 a (b c-a d) x^4 \left (a+b x^8\right )}-\frac {b (3 b c-4 a d) \arctan \left (\frac {\sqrt {b c-a d} x^4}{\sqrt {a} \sqrt {c+d x^8}}\right )}{8 a^{5/2} (b c-a d)^{3/2}} \]
-1/8*b*(-4*a*d+3*b*c)*arctan(x^4*(-a*d+b*c)^(1/2)/a^(1/2)/(d*x^8+c)^(1/2)) /a^(5/2)/(-a*d+b*c)^(3/2)-1/8*(-2*a*d+3*b*c)*(d*x^8+c)^(1/2)/a^2/c/(-a*d+b *c)/x^4+1/8*b*(d*x^8+c)^(1/2)/a/(-a*d+b*c)/x^4/(b*x^8+a)
Time = 2.13 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.05 \[ \int \frac {1}{x^5 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\frac {\sqrt {c+d x^8} \left (2 a b c-2 a^2 d+3 b^2 c x^8-2 a b d x^8\right )}{8 a^2 c (-b c+a d) x^4 \left (a+b x^8\right )}-\frac {b (3 b c-4 a d) \arctan \left (\frac {a \sqrt {d}+b \sqrt {d} x^8+b x^4 \sqrt {c+d x^8}}{\sqrt {a} \sqrt {b c-a d}}\right )}{8 a^{5/2} (b c-a d)^{3/2}} \]
(Sqrt[c + d*x^8]*(2*a*b*c - 2*a^2*d + 3*b^2*c*x^8 - 2*a*b*d*x^8))/(8*a^2*c *(-(b*c) + a*d)*x^4*(a + b*x^8)) - (b*(3*b*c - 4*a*d)*ArcTan[(a*Sqrt[d] + b*Sqrt[d]*x^8 + b*x^4*Sqrt[c + d*x^8])/(Sqrt[a]*Sqrt[b*c - a*d])])/(8*a^(5 /2)*(b*c - a*d)^(3/2))
Time = 0.33 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {965, 374, 25, 445, 27, 291, 218}
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.
\(\displaystyle \int \frac {1}{x^5 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx\) |
\(\Big \downarrow \) 965 |
\(\displaystyle \frac {1}{4} \int \frac {1}{x^8 \left (b x^8+a\right )^2 \sqrt {d x^8+c}}dx^4\) |
\(\Big \downarrow \) 374 |
\(\displaystyle \frac {1}{4} \left (\frac {b \sqrt {c+d x^8}}{2 a x^4 \left (a+b x^8\right ) (b c-a d)}-\frac {\int -\frac {2 b d x^8+3 b c-2 a d}{x^8 \left (b x^8+a\right ) \sqrt {d x^8+c}}dx^4}{2 a (b c-a d)}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {2 b d x^8+3 b c-2 a d}{x^8 \left (b x^8+a\right ) \sqrt {d x^8+c}}dx^4}{2 a (b c-a d)}+\frac {b \sqrt {c+d x^8}}{2 a x^4 \left (a+b x^8\right ) (b c-a d)}\right )\) |
\(\Big \downarrow \) 445 |
\(\displaystyle \frac {1}{4} \left (\frac {-\frac {\int \frac {b c (3 b c-4 a d)}{\left (b x^8+a\right ) \sqrt {d x^8+c}}dx^4}{a c}-\frac {\sqrt {c+d x^8} (3 b c-2 a d)}{a c x^4}}{2 a (b c-a d)}+\frac {b \sqrt {c+d x^8}}{2 a x^4 \left (a+b x^8\right ) (b c-a d)}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \left (\frac {-\frac {b (3 b c-4 a d) \int \frac {1}{\left (b x^8+a\right ) \sqrt {d x^8+c}}dx^4}{a}-\frac {\sqrt {c+d x^8} (3 b c-2 a d)}{a c x^4}}{2 a (b c-a d)}+\frac {b \sqrt {c+d x^8}}{2 a x^4 \left (a+b x^8\right ) (b c-a d)}\right )\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {1}{4} \left (\frac {-\frac {b (3 b c-4 a d) \int \frac {1}{a-(a d-b c) x^8}d\frac {x^4}{\sqrt {d x^8+c}}}{a}-\frac {\sqrt {c+d x^8} (3 b c-2 a d)}{a c x^4}}{2 a (b c-a d)}+\frac {b \sqrt {c+d x^8}}{2 a x^4 \left (a+b x^8\right ) (b c-a d)}\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {1}{4} \left (\frac {-\frac {b (3 b c-4 a d) \arctan \left (\frac {x^4 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^8}}\right )}{a^{3/2} \sqrt {b c-a d}}-\frac {\sqrt {c+d x^8} (3 b c-2 a d)}{a c x^4}}{2 a (b c-a d)}+\frac {b \sqrt {c+d x^8}}{2 a x^4 \left (a+b x^8\right ) (b c-a d)}\right )\) |
((b*Sqrt[c + d*x^8])/(2*a*(b*c - a*d)*x^4*(a + b*x^8)) + (-(((3*b*c - 2*a* d)*Sqrt[c + d*x^8])/(a*c*x^4)) - (b*(3*b*c - 4*a*d)*ArcTan[(Sqrt[b*c - a*d ]*x^4)/(Sqrt[a]*Sqrt[c + d*x^8])])/(a^(3/2)*Sqrt[b*c - a*d]))/(2*a*(b*c - a*d)))/4
3.10.16.3.1 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[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-b)*(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*e*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(e*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[b*c*(m + 1) + 2*(b*c - a*d)*(p + 1) + d*b*(m + 2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_ .)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*c*g*(m + 1))), x] + Simp[1/(a*c*g^2*(m + 1)) Int[(g*x)^(m + 2)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + 2 + 1) - e*2*(b*c*p + a*d*q) - b*e*d*(m + 2*(p + q + 2) + 1)*x^ 2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && LtQ[m, -1]
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]
Time = 31.25 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.75
method | result | size |
pseudoelliptic | \(\frac {-\frac {\sqrt {d \,x^{8}+c}}{x^{4}}+\frac {b c \left (\frac {b \sqrt {d \,x^{8}+c}\, x^{4}}{b \,x^{8}+a}-\frac {\left (4 a d -3 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{8}+c}\, a}{x^{4} \sqrt {\left (a d -b c \right ) a}}\right )}{\sqrt {\left (a d -b c \right ) a}}\right )}{2 a d -2 b c}}{4 a^{2} c}\) | \(112\) |
1/4/a^2*(-(d*x^8+c)^(1/2)/x^4+1/2*b*c/(a*d-b*c)*(b*(d*x^8+c)^(1/2)*x^4/(b* x^8+a)-(4*a*d-3*b*c)/((a*d-b*c)*a)^(1/2)*arctanh((d*x^8+c)^(1/2)/x^4*a/((a *d-b*c)*a)^(1/2))))/c
Leaf count of result is larger than twice the leaf count of optimal. 286 vs. \(2 (129) = 258\).
Time = 0.55 (sec) , antiderivative size = 612, normalized size of antiderivative = 4.11 \[ \int \frac {1}{x^5 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\left [-\frac {{\left ({\left (3 \, b^{3} c^{2} - 4 \, a b^{2} c d\right )} x^{12} + {\left (3 \, a b^{2} c^{2} - 4 \, a^{2} b c d\right )} x^{4}\right )} \sqrt {-a b c + a^{2} d} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{16} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{8} + a^{2} c^{2} + 4 \, {\left ({\left (b c - 2 \, a d\right )} x^{12} - a c x^{4}\right )} \sqrt {d x^{8} + c} \sqrt {-a b c + a^{2} d}}{b^{2} x^{16} + 2 \, a b x^{8} + a^{2}}\right ) + 4 \, {\left ({\left (3 \, a b^{3} c^{2} - 5 \, a^{2} b^{2} c d + 2 \, a^{3} b d^{2}\right )} x^{8} + 2 \, a^{2} b^{2} c^{2} - 4 \, a^{3} b c d + 2 \, a^{4} d^{2}\right )} \sqrt {d x^{8} + c}}{32 \, {\left ({\left (a^{3} b^{3} c^{3} - 2 \, a^{4} b^{2} c^{2} d + a^{5} b c d^{2}\right )} x^{12} + {\left (a^{4} b^{2} c^{3} - 2 \, a^{5} b c^{2} d + a^{6} c d^{2}\right )} x^{4}\right )}}, -\frac {{\left ({\left (3 \, b^{3} c^{2} - 4 \, a b^{2} c d\right )} x^{12} + {\left (3 \, a b^{2} c^{2} - 4 \, a^{2} b c d\right )} x^{4}\right )} \sqrt {a b c - a^{2} d} \arctan \left (\frac {{\left ({\left (b c - 2 \, a d\right )} x^{8} - a c\right )} \sqrt {d x^{8} + c} \sqrt {a b c - a^{2} d}}{2 \, {\left ({\left (a b c d - a^{2} d^{2}\right )} x^{12} + {\left (a b c^{2} - a^{2} c d\right )} x^{4}\right )}}\right ) + 2 \, {\left ({\left (3 \, a b^{3} c^{2} - 5 \, a^{2} b^{2} c d + 2 \, a^{3} b d^{2}\right )} x^{8} + 2 \, a^{2} b^{2} c^{2} - 4 \, a^{3} b c d + 2 \, a^{4} d^{2}\right )} \sqrt {d x^{8} + c}}{16 \, {\left ({\left (a^{3} b^{3} c^{3} - 2 \, a^{4} b^{2} c^{2} d + a^{5} b c d^{2}\right )} x^{12} + {\left (a^{4} b^{2} c^{3} - 2 \, a^{5} b c^{2} d + a^{6} c d^{2}\right )} x^{4}\right )}}\right ] \]
[-1/32*(((3*b^3*c^2 - 4*a*b^2*c*d)*x^12 + (3*a*b^2*c^2 - 4*a^2*b*c*d)*x^4) *sqrt(-a*b*c + a^2*d)*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^16 - 2*(3*a *b*c^2 - 4*a^2*c*d)*x^8 + a^2*c^2 + 4*((b*c - 2*a*d)*x^12 - a*c*x^4)*sqrt( d*x^8 + c)*sqrt(-a*b*c + a^2*d))/(b^2*x^16 + 2*a*b*x^8 + a^2)) + 4*((3*a*b ^3*c^2 - 5*a^2*b^2*c*d + 2*a^3*b*d^2)*x^8 + 2*a^2*b^2*c^2 - 4*a^3*b*c*d + 2*a^4*d^2)*sqrt(d*x^8 + c))/((a^3*b^3*c^3 - 2*a^4*b^2*c^2*d + a^5*b*c*d^2) *x^12 + (a^4*b^2*c^3 - 2*a^5*b*c^2*d + a^6*c*d^2)*x^4), -1/16*(((3*b^3*c^2 - 4*a*b^2*c*d)*x^12 + (3*a*b^2*c^2 - 4*a^2*b*c*d)*x^4)*sqrt(a*b*c - a^2*d )*arctan(1/2*((b*c - 2*a*d)*x^8 - a*c)*sqrt(d*x^8 + c)*sqrt(a*b*c - a^2*d) /((a*b*c*d - a^2*d^2)*x^12 + (a*b*c^2 - a^2*c*d)*x^4)) + 2*((3*a*b^3*c^2 - 5*a^2*b^2*c*d + 2*a^3*b*d^2)*x^8 + 2*a^2*b^2*c^2 - 4*a^3*b*c*d + 2*a^4*d^ 2)*sqrt(d*x^8 + c))/((a^3*b^3*c^3 - 2*a^4*b^2*c^2*d + a^5*b*c*d^2)*x^12 + (a^4*b^2*c^3 - 2*a^5*b*c^2*d + a^6*c*d^2)*x^4)]
\[ \int \frac {1}{x^5 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int \frac {1}{x^{5} \left (a + b x^{8}\right )^{2} \sqrt {c + d x^{8}}}\, dx \]
\[ \int \frac {1}{x^5 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int { \frac {1}{{\left (b x^{8} + a\right )}^{2} \sqrt {d x^{8} + c} x^{5}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 418 vs. \(2 (129) = 258\).
Time = 0.93 (sec) , antiderivative size = 418, normalized size of antiderivative = 2.81 \[ \int \frac {1}{x^5 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\frac {1}{8} \, d^{\frac {5}{2}} {\left (\frac {{\left (3 \, b^{2} c - 4 \, a b d\right )} \arctan \left (\frac {{\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{{\left (a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {a b c d - a^{2} d^{2}}} + \frac {2 \, {\left (3 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{4} b^{2} c - 4 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{4} a b d - 6 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} b^{2} c^{2} + 14 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} a b c d - 8 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} a^{2} d^{2} + 3 \, b^{2} c^{3} - 2 \, a b c^{2} d\right )}}{{\left ({\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{6} b - 3 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{4} b c + 4 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{4} a d + 3 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} b c^{2} - 4 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} a c d - b c^{3}\right )} {\left (a^{2} b c d^{2} - a^{3} d^{3}\right )}}\right )} \]
1/8*d^(5/2)*((3*b^2*c - 4*a*b*d)*arctan(1/2*((sqrt(d)*x^4 - sqrt(d*x^8 + c ))^2*b - b*c + 2*a*d)/sqrt(a*b*c*d - a^2*d^2))/((a^2*b*c*d^2 - a^3*d^3)*sq rt(a*b*c*d - a^2*d^2)) + 2*(3*(sqrt(d)*x^4 - sqrt(d*x^8 + c))^4*b^2*c - 4* (sqrt(d)*x^4 - sqrt(d*x^8 + c))^4*a*b*d - 6*(sqrt(d)*x^4 - sqrt(d*x^8 + c) )^2*b^2*c^2 + 14*(sqrt(d)*x^4 - sqrt(d*x^8 + c))^2*a*b*c*d - 8*(sqrt(d)*x^ 4 - sqrt(d*x^8 + c))^2*a^2*d^2 + 3*b^2*c^3 - 2*a*b*c^2*d)/(((sqrt(d)*x^4 - sqrt(d*x^8 + c))^6*b - 3*(sqrt(d)*x^4 - sqrt(d*x^8 + c))^4*b*c + 4*(sqrt( d)*x^4 - sqrt(d*x^8 + c))^4*a*d + 3*(sqrt(d)*x^4 - sqrt(d*x^8 + c))^2*b*c^ 2 - 4*(sqrt(d)*x^4 - sqrt(d*x^8 + c))^2*a*c*d - b*c^3)*(a^2*b*c*d^2 - a^3* d^3)))
Timed out. \[ \int \frac {1}{x^5 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int \frac {1}{x^5\,{\left (b\,x^8+a\right )}^2\,\sqrt {d\,x^8+c}} \,d x \]