Integrand size = 24, antiderivative size = 93 \[ \int \frac {x^{11}}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=-\frac {x^4 \sqrt {c+d x^8}}{8 (b c-a d) \left (a+b x^8\right )}+\frac {c \arctan \left (\frac {\sqrt {b c-a d} x^4}{\sqrt {a} \sqrt {c+d x^8}}\right )}{8 \sqrt {a} (b c-a d)^{3/2}} \]
1/8*c*arctan(x^4*(-a*d+b*c)^(1/2)/a^(1/2)/(d*x^8+c)^(1/2))/(-a*d+b*c)^(3/2 )/a^(1/2)-1/8*x^4*(d*x^8+c)^(1/2)/(-a*d+b*c)/(b*x^8+a)
Time = 2.03 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.20 \[ \int \frac {x^{11}}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\frac {1}{8} \left (-\frac {x^4 \sqrt {c+d x^8}}{(b c-a d) \left (a+b x^8\right )}+\frac {c \arctan \left (\frac {a \sqrt {d}+b x^4 \left (\sqrt {d} x^4+\sqrt {c+d x^8}\right )}{\sqrt {a} \sqrt {b c-a d}}\right )}{\sqrt {a} (b c-a d)^{3/2}}\right ) \]
(-((x^4*Sqrt[c + d*x^8])/((b*c - a*d)*(a + b*x^8))) + (c*ArcTan[(a*Sqrt[d] + b*x^4*(Sqrt[d]*x^4 + Sqrt[c + d*x^8]))/(Sqrt[a]*Sqrt[b*c - a*d])])/(Sqr t[a]*(b*c - a*d)^(3/2)))/8
Time = 0.24 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {965, 373, 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.
\(\displaystyle \int \frac {x^{11}}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx\) |
\(\Big \downarrow \) 965 |
\(\displaystyle \frac {1}{4} \int \frac {x^8}{\left (b x^8+a\right )^2 \sqrt {d x^8+c}}dx^4\) |
\(\Big \downarrow \) 373 |
\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {c}{\left (b x^8+a\right ) \sqrt {d x^8+c}}dx^4}{2 (b c-a d)}-\frac {x^4 \sqrt {c+d x^8}}{2 \left (a+b x^8\right ) (b c-a d)}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \left (\frac {c \int \frac {1}{\left (b x^8+a\right ) \sqrt {d x^8+c}}dx^4}{2 (b c-a d)}-\frac {x^4 \sqrt {c+d x^8}}{2 \left (a+b x^8\right ) (b c-a d)}\right )\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {1}{4} \left (\frac {c \int \frac {1}{a-(a d-b c) x^8}d\frac {x^4}{\sqrt {d x^8+c}}}{2 (b c-a d)}-\frac {x^4 \sqrt {c+d x^8}}{2 \left (a+b x^8\right ) (b c-a d)}\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {1}{4} \left (\frac {c \arctan \left (\frac {x^4 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^8}}\right )}{2 \sqrt {a} (b c-a d)^{3/2}}-\frac {x^4 \sqrt {c+d x^8}}{2 \left (a+b x^8\right ) (b c-a d)}\right )\) |
(-1/2*(x^4*Sqrt[c + d*x^8])/((b*c - a*d)*(a + b*x^8)) + (c*ArcTan[(Sqrt[b* c - a*d]*x^4)/(Sqrt[a]*Sqrt[c + d*x^8])])/(2*Sqrt[a]*(b*c - a*d)^(3/2)))/4
3.10.14.3.1 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[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1)) Int[(e *x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
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 = 20.28 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.87
method | result | size |
pseudoelliptic | \(-\frac {c \left (-\frac {\sqrt {d \,x^{8}+c}\, x^{4}}{c \left (b \,x^{8}+a \right )}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{8}+c}\, a}{x^{4} \sqrt {\left (a d -b c \right ) a}}\right )}{\sqrt {\left (a d -b c \right ) a}}\right )}{8 \left (a d -b c \right )}\) | \(81\) |
-1/8*c/(a*d-b*c)*(-(d*x^8+c)^(1/2)*x^4/c/(b*x^8+a)+1/((a*d-b*c)*a)^(1/2)*a rctanh((d*x^8+c)^(1/2)/x^4*a/((a*d-b*c)*a)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 193 vs. \(2 (77) = 154\).
Time = 0.35 (sec) , antiderivative size = 426, normalized size of antiderivative = 4.58 \[ \int \frac {x^{11}}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\left [-\frac {4 \, \sqrt {d x^{8} + c} {\left (a b c - a^{2} d\right )} x^{4} - {\left (b c x^{8} + a c\right )} \sqrt {-a b c + a^{2} d} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{16} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{8} + a^{2} c^{2} + 4 \, {\left ({\left (b c - 2 \, a d\right )} x^{12} - a c x^{4}\right )} \sqrt {d x^{8} + c} \sqrt {-a b c + a^{2} d}}{b^{2} x^{16} + 2 \, a b x^{8} + a^{2}}\right )}{32 \, {\left ({\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x^{8} + a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )}}, -\frac {2 \, \sqrt {d x^{8} + c} {\left (a b c - a^{2} d\right )} x^{4} - {\left (b c x^{8} + a c\right )} \sqrt {a b c - a^{2} d} \arctan \left (\frac {{\left ({\left (b c - 2 \, a d\right )} x^{8} - a c\right )} \sqrt {d x^{8} + c} \sqrt {a b c - a^{2} d}}{2 \, {\left ({\left (a b c d - a^{2} d^{2}\right )} x^{12} + {\left (a b c^{2} - a^{2} c d\right )} x^{4}\right )}}\right )}{16 \, {\left ({\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x^{8} + a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )}}\right ] \]
[-1/32*(4*sqrt(d*x^8 + c)*(a*b*c - a^2*d)*x^4 - (b*c*x^8 + a*c)*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)))/((a*b^3*c^2 - 2*a^2*b ^2*c*d + a^3*b*d^2)*x^8 + a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2), -1/16*(2*s qrt(d*x^8 + c)*(a*b*c - a^2*d)*x^4 - (b*c*x^8 + a*c)*sqrt(a*b*c - a^2*d)*a rctan(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)))/((a*b^3*c^2 - 2*a^2*b ^2*c*d + a^3*b*d^2)*x^8 + a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2)]
Timed out. \[ \int \frac {x^{11}}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\text {Timed out} \]
\[ \int \frac {x^{11}}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int { \frac {x^{11}}{{\left (b x^{8} + a\right )}^{2} \sqrt {d x^{8} + c}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 244 vs. \(2 (77) = 154\).
Time = 0.99 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.62 \[ \int \frac {x^{11}}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\frac {c \sqrt {d} \arctan \left (-\frac {{\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{8 \, \sqrt {a b c d - a^{2} d^{2}} {\left (b c - a d\right )}} + \frac {{\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} b c \sqrt {d} - 2 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} a d^{\frac {3}{2}} - b c^{2} \sqrt {d}}{4 \, {\left ({\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} a d + b c^{2}\right )} {\left (b^{2} c - a b d\right )}} \]
1/8*c*sqrt(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))/(sqrt(a*b*c*d - a^2*d^2)*(b*c - a*d)) + 1/4*( (sqrt(d)*x^4 - sqrt(d*x^8 + c))^2*b*c*sqrt(d) - 2*(sqrt(d)*x^4 - sqrt(d*x^ 8 + c))^2*a*d^(3/2) - b*c^2*sqrt(d))/(((sqrt(d)*x^4 - sqrt(d*x^8 + c))^4*b - 2*(sqrt(d)*x^4 - sqrt(d*x^8 + c))^2*b*c + 4*(sqrt(d)*x^4 - sqrt(d*x^8 + c))^2*a*d + b*c^2)*(b^2*c - a*b*d))
Timed out. \[ \int \frac {x^{11}}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int \frac {x^{11}}{{\left (b\,x^8+a\right )}^2\,\sqrt {d\,x^8+c}} \,d x \]