Integrand size = 22, antiderivative size = 92 \[ \int \frac {x^9}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {x^2}{2 b d}+\frac {a^{3/2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 b^{3/2} (b c-a d)}-\frac {c^{3/2} \arctan \left (\frac {\sqrt {d} x^2}{\sqrt {c}}\right )}{2 d^{3/2} (b c-a d)} \]
1/2*x^2/b/d+1/2*a^(3/2)*arctan(x^2*b^(1/2)/a^(1/2))/b^(3/2)/(-a*d+b*c)-1/2 *c^(3/2)*arctan(x^2*d^(1/2)/c^(1/2))/d^(3/2)/(-a*d+b*c)
Time = 0.15 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.89 \[ \int \frac {x^9}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {\left (-\frac {a}{b}+\frac {c}{d}\right ) x^2+\frac {a^{3/2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{b^{3/2}}-\frac {c^{3/2} \arctan \left (\frac {\sqrt {d} x^2}{\sqrt {c}}\right )}{d^{3/2}}}{2 b c-2 a d} \]
((-(a/b) + c/d)*x^2 + (a^(3/2)*ArcTan[(Sqrt[b]*x^2)/Sqrt[a]])/b^(3/2) - (c ^(3/2)*ArcTan[(Sqrt[d]*x^2)/Sqrt[c]])/d^(3/2))/(2*b*c - 2*a*d)
Time = 0.25 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {965, 381, 397, 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.
\(\displaystyle \int \frac {x^9}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx\) |
\(\Big \downarrow \) 965 |
\(\displaystyle \frac {1}{2} \int \frac {x^8}{\left (b x^4+a\right ) \left (d x^4+c\right )}dx^2\) |
\(\Big \downarrow \) 381 |
\(\displaystyle \frac {1}{2} \left (\frac {x^2}{b d}-\frac {\int \frac {(b c+a d) x^4+a c}{\left (b x^4+a\right ) \left (d x^4+c\right )}dx^2}{b d}\right )\) |
\(\Big \downarrow \) 397 |
\(\displaystyle \frac {1}{2} \left (\frac {x^2}{b d}-\frac {\frac {b c^2 \int \frac {1}{d x^4+c}dx^2}{b c-a d}-\frac {a^2 d \int \frac {1}{b x^4+a}dx^2}{b c-a d}}{b d}\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {1}{2} \left (\frac {x^2}{b d}-\frac {\frac {b c^{3/2} \arctan \left (\frac {\sqrt {d} x^2}{\sqrt {c}}\right )}{\sqrt {d} (b c-a d)}-\frac {a^{3/2} d \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{\sqrt {b} (b c-a d)}}{b d}\right )\) |
(x^2/(b*d) - (-((a^(3/2)*d*ArcTan[(Sqrt[b]*x^2)/Sqrt[a]])/(Sqrt[b]*(b*c - a*d))) + (b*c^(3/2)*ArcTan[(Sqrt[d]*x^2)/Sqrt[c]])/(Sqrt[d]*(b*c - a*d)))/ (b*d))/2
3.8.74.3.1 Defintions of rubi rules used
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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(b*d*(m + 2*(p + q) + 1))), x] - Simp[e^4/(b*d*(m + 2*(p + q) + 1)) Int[(e*x)^(m - 4)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 2*q - 1) + b*c*(m + 2*p - 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p, q }, x] && NeQ[b*c - a*d, 0] && GtQ[m, 3] && IntBinomialQ[a, b, c, d, e, m, 2 , p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, 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 = 4.70 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.88
method | result | size |
default | \(\frac {x^{2}}{2 b d}-\frac {a^{2} \arctan \left (\frac {b \,x^{2}}{\sqrt {a b}}\right )}{2 \left (a d -b c \right ) b \sqrt {a b}}+\frac {c^{2} \arctan \left (\frac {d \,x^{2}}{\sqrt {c d}}\right )}{2 \left (a d -b c \right ) d \sqrt {c d}}\) | \(81\) |
risch | \(\frac {x^{2}}{2 b d}+\frac {\sqrt {-a b}\, a \ln \left (\left (b^{3} c \,d^{3} a^{3}-b^{6} c^{4}\right ) x^{2}+\left (-a b \right )^{\frac {3}{2}} a^{3} d^{4}+\left (-a b \right )^{\frac {3}{2}} a^{2} b c \,d^{3}+a^{4} \sqrt {-a b}\, d^{4} b +b^{5} c^{4} \sqrt {-a b}\right )}{4 b^{2} \left (a d -b c \right )}-\frac {\sqrt {-a b}\, a \ln \left (\left (b^{3} c \,d^{3} a^{3}-b^{6} c^{4}\right ) x^{2}-\left (-a b \right )^{\frac {3}{2}} a^{3} d^{4}-\left (-a b \right )^{\frac {3}{2}} a^{2} b c \,d^{3}-a^{4} \sqrt {-a b}\, d^{4} b -b^{5} c^{4} \sqrt {-a b}\right )}{4 b^{2} \left (a d -b c \right )}+\frac {\sqrt {-c d}\, c \ln \left (\left (a^{4} d^{6}-a \,d^{3} c^{3} b^{3}\right ) x^{2}+\left (-c d \right )^{\frac {3}{2}} a \,b^{3} c^{2} d +\left (-c d \right )^{\frac {3}{2}} b^{4} c^{3}+a^{4} \sqrt {-c d}\, d^{5}+b^{4} c^{4} \sqrt {-c d}\, d \right )}{4 d^{2} \left (a d -b c \right )}-\frac {\sqrt {-c d}\, c \ln \left (\left (a^{4} d^{6}-a \,d^{3} c^{3} b^{3}\right ) x^{2}-\left (-c d \right )^{\frac {3}{2}} a \,b^{3} c^{2} d -\left (-c d \right )^{\frac {3}{2}} b^{4} c^{3}-a^{4} \sqrt {-c d}\, d^{5}-b^{4} c^{4} \sqrt {-c d}\, d \right )}{4 d^{2} \left (a d -b c \right )}\) | \(433\) |
1/2*x^2/b/d-1/2*a^2/(a*d-b*c)/b/(a*b)^(1/2)*arctan(b*x^2/(a*b)^(1/2))+1/2* c^2/(a*d-b*c)/d/(c*d)^(1/2)*arctan(d*x^2/(c*d)^(1/2))
Time = 0.47 (sec) , antiderivative size = 416, normalized size of antiderivative = 4.52 \[ \int \frac {x^9}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\left [-\frac {a d \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{4} - 2 \, b x^{2} \sqrt {-\frac {a}{b}} - a}{b x^{4} + a}\right ) + b c \sqrt {-\frac {c}{d}} \log \left (\frac {d x^{4} + 2 \, d x^{2} \sqrt {-\frac {c}{d}} - c}{d x^{4} + c}\right ) - 2 \, {\left (b c - a d\right )} x^{2}}{4 \, {\left (b^{2} c d - a b d^{2}\right )}}, \frac {2 \, a d \sqrt {\frac {a}{b}} \arctan \left (\frac {b x^{2} \sqrt {\frac {a}{b}}}{a}\right ) - b c \sqrt {-\frac {c}{d}} \log \left (\frac {d x^{4} + 2 \, d x^{2} \sqrt {-\frac {c}{d}} - c}{d x^{4} + c}\right ) + 2 \, {\left (b c - a d\right )} x^{2}}{4 \, {\left (b^{2} c d - a b d^{2}\right )}}, -\frac {2 \, b c \sqrt {\frac {c}{d}} \arctan \left (\frac {d x^{2} \sqrt {\frac {c}{d}}}{c}\right ) + a d \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{4} - 2 \, b x^{2} \sqrt {-\frac {a}{b}} - a}{b x^{4} + a}\right ) - 2 \, {\left (b c - a d\right )} x^{2}}{4 \, {\left (b^{2} c d - a b d^{2}\right )}}, \frac {a d \sqrt {\frac {a}{b}} \arctan \left (\frac {b x^{2} \sqrt {\frac {a}{b}}}{a}\right ) - b c \sqrt {\frac {c}{d}} \arctan \left (\frac {d x^{2} \sqrt {\frac {c}{d}}}{c}\right ) + {\left (b c - a d\right )} x^{2}}{2 \, {\left (b^{2} c d - a b d^{2}\right )}}\right ] \]
[-1/4*(a*d*sqrt(-a/b)*log((b*x^4 - 2*b*x^2*sqrt(-a/b) - a)/(b*x^4 + a)) + b*c*sqrt(-c/d)*log((d*x^4 + 2*d*x^2*sqrt(-c/d) - c)/(d*x^4 + c)) - 2*(b*c - a*d)*x^2)/(b^2*c*d - a*b*d^2), 1/4*(2*a*d*sqrt(a/b)*arctan(b*x^2*sqrt(a/ b)/a) - b*c*sqrt(-c/d)*log((d*x^4 + 2*d*x^2*sqrt(-c/d) - c)/(d*x^4 + c)) + 2*(b*c - a*d)*x^2)/(b^2*c*d - a*b*d^2), -1/4*(2*b*c*sqrt(c/d)*arctan(d*x^ 2*sqrt(c/d)/c) + a*d*sqrt(-a/b)*log((b*x^4 - 2*b*x^2*sqrt(-a/b) - a)/(b*x^ 4 + a)) - 2*(b*c - a*d)*x^2)/(b^2*c*d - a*b*d^2), 1/2*(a*d*sqrt(a/b)*arcta n(b*x^2*sqrt(a/b)/a) - b*c*sqrt(c/d)*arctan(d*x^2*sqrt(c/d)/c) + (b*c - a* d)*x^2)/(b^2*c*d - a*b*d^2)]
Timed out. \[ \int \frac {x^9}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\text {Timed out} \]
Time = 0.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.87 \[ \int \frac {x^9}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {a^{2} \arctan \left (\frac {b x^{2}}{\sqrt {a b}}\right )}{2 \, {\left (b^{2} c - a b d\right )} \sqrt {a b}} - \frac {c^{2} \arctan \left (\frac {d x^{2}}{\sqrt {c d}}\right )}{2 \, {\left (b c d - a d^{2}\right )} \sqrt {c d}} + \frac {x^{2}}{2 \, b d} \]
1/2*a^2*arctan(b*x^2/sqrt(a*b))/((b^2*c - a*b*d)*sqrt(a*b)) - 1/2*c^2*arct an(d*x^2/sqrt(c*d))/((b*c*d - a*d^2)*sqrt(c*d)) + 1/2*x^2/(b*d)
Time = 0.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.87 \[ \int \frac {x^9}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {a^{2} \arctan \left (\frac {b x^{2}}{\sqrt {a b}}\right )}{2 \, {\left (b^{2} c - a b d\right )} \sqrt {a b}} - \frac {c^{2} \arctan \left (\frac {d x^{2}}{\sqrt {c d}}\right )}{2 \, {\left (b c d - a d^{2}\right )} \sqrt {c d}} + \frac {x^{2}}{2 \, b d} \]
1/2*a^2*arctan(b*x^2/sqrt(a*b))/((b^2*c - a*b*d)*sqrt(a*b)) - 1/2*c^2*arct an(d*x^2/sqrt(c*d))/((b*c*d - a*d^2)*sqrt(c*d)) + 1/2*x^2/(b*d)
Time = 10.54 (sec) , antiderivative size = 518, normalized size of antiderivative = 5.63 \[ \int \frac {x^9}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {\ln \left (b^9\,c^6\,\sqrt {-a^3\,b^3}-a^3\,d^6\,{\left (-a^3\,b^3\right )}^{3/2}+a\,b^{11}\,c^6\,x^2+a^7\,b^5\,d^6\,x^2+2\,b^3\,c^3\,d^3\,{\left (-a^3\,b^3\right )}^{3/2}-2\,a^4\,b^8\,c^3\,d^3\,x^2\right )\,\sqrt {-a^3\,b^3}}{4\,b^4\,c-4\,a\,b^3\,d}-\frac {\ln \left (a^3\,d^6\,{\left (-a^3\,b^3\right )}^{3/2}-b^9\,c^6\,\sqrt {-a^3\,b^3}+a\,b^{11}\,c^6\,x^2+a^7\,b^5\,d^6\,x^2-2\,b^3\,c^3\,d^3\,{\left (-a^3\,b^3\right )}^{3/2}-2\,a^4\,b^8\,c^3\,d^3\,x^2\right )\,\sqrt {-a^3\,b^3}}{4\,\left (b^4\,c-a\,b^3\,d\right )}-\frac {\ln \left (b^6\,c^3\,{\left (-c^3\,d^3\right )}^{3/2}-a^6\,d^9\,\sqrt {-c^3\,d^3}+a^6\,c\,d^{11}\,x^2+b^6\,c^7\,d^5\,x^2-2\,a^3\,b^3\,d^3\,{\left (-c^3\,d^3\right )}^{3/2}-2\,a^3\,b^3\,c^4\,d^8\,x^2\right )\,\sqrt {-c^3\,d^3}}{4\,\left (a\,d^4-b\,c\,d^3\right )}+\frac {\ln \left (a^6\,d^9\,\sqrt {-c^3\,d^3}-b^6\,c^3\,{\left (-c^3\,d^3\right )}^{3/2}+a^6\,c\,d^{11}\,x^2+b^6\,c^7\,d^5\,x^2+2\,a^3\,b^3\,d^3\,{\left (-c^3\,d^3\right )}^{3/2}-2\,a^3\,b^3\,c^4\,d^8\,x^2\right )\,\sqrt {-c^3\,d^3}}{4\,a\,d^4-4\,b\,c\,d^3}+\frac {x^2}{2\,b\,d} \]
(log(b^9*c^6*(-a^3*b^3)^(1/2) - a^3*d^6*(-a^3*b^3)^(3/2) + a*b^11*c^6*x^2 + a^7*b^5*d^6*x^2 + 2*b^3*c^3*d^3*(-a^3*b^3)^(3/2) - 2*a^4*b^8*c^3*d^3*x^2 )*(-a^3*b^3)^(1/2))/(4*b^4*c - 4*a*b^3*d) - (log(a^3*d^6*(-a^3*b^3)^(3/2) - b^9*c^6*(-a^3*b^3)^(1/2) + a*b^11*c^6*x^2 + a^7*b^5*d^6*x^2 - 2*b^3*c^3* d^3*(-a^3*b^3)^(3/2) - 2*a^4*b^8*c^3*d^3*x^2)*(-a^3*b^3)^(1/2))/(4*(b^4*c - a*b^3*d)) - (log(b^6*c^3*(-c^3*d^3)^(3/2) - a^6*d^9*(-c^3*d^3)^(1/2) + a ^6*c*d^11*x^2 + b^6*c^7*d^5*x^2 - 2*a^3*b^3*d^3*(-c^3*d^3)^(3/2) - 2*a^3*b ^3*c^4*d^8*x^2)*(-c^3*d^3)^(1/2))/(4*(a*d^4 - b*c*d^3)) + (log(a^6*d^9*(-c ^3*d^3)^(1/2) - b^6*c^3*(-c^3*d^3)^(3/2) + a^6*c*d^11*x^2 + b^6*c^7*d^5*x^ 2 + 2*a^3*b^3*d^3*(-c^3*d^3)^(3/2) - 2*a^3*b^3*c^4*d^8*x^2)*(-c^3*d^3)^(1/ 2))/(4*a*d^4 - 4*b*c*d^3) + x^2/(2*b*d)