Integrand size = 22, antiderivative size = 145 \[ \int \frac {x^4 \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx=\frac {(3 b c-7 a d) (b c-a d) x}{2 b^4}-\frac {(3 b c-7 a d) (b c-a d) x^3}{6 a b^3}+\frac {d^2 x^5}{5 b^2}+\frac {(b c-a d)^2 x^5}{2 a b^2 \left (a+b x^2\right )}-\frac {\sqrt {a} (3 b c-7 a d) (b c-a d) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{9/2}} \]
1/2*(-7*a*d+3*b*c)*(-a*d+b*c)*x/b^4-1/6*(-7*a*d+3*b*c)*(-a*d+b*c)*x^3/a/b^ 3+1/5*d^2*x^5/b^2+1/2*(-a*d+b*c)^2*x^5/a/b^2/(b*x^2+a)-1/2*(-7*a*d+3*b*c)* (-a*d+b*c)*arctan(x*b^(1/2)/a^(1/2))*a^(1/2)/b^(9/2)
Time = 0.06 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.95 \[ \int \frac {x^4 \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx=\frac {\left (b^2 c^2-4 a b c d+3 a^2 d^2\right ) x}{b^4}+\frac {2 d (b c-a d) x^3}{3 b^3}+\frac {d^2 x^5}{5 b^2}+\frac {a (b c-a d)^2 x}{2 b^4 \left (a+b x^2\right )}-\frac {\sqrt {a} \left (3 b^2 c^2-10 a b c d+7 a^2 d^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{9/2}} \]
((b^2*c^2 - 4*a*b*c*d + 3*a^2*d^2)*x)/b^4 + (2*d*(b*c - a*d)*x^3)/(3*b^3) + (d^2*x^5)/(5*b^2) + (a*(b*c - a*d)^2*x)/(2*b^4*(a + b*x^2)) - (Sqrt[a]*( 3*b^2*c^2 - 10*a*b*c*d + 7*a^2*d^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*b^(9/2 ))
Time = 0.27 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.80, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {366, 25, 363, 254, 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.
\(\displaystyle \int \frac {x^4 \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 366 |
\(\displaystyle \frac {x^5 (b c-a d)^2}{2 a b^2 \left (a+b x^2\right )}-\frac {\int -\frac {x^4 \left (2 b^2 c^2-5 (b c-a d)^2+2 a b d^2 x^2\right )}{b x^2+a}dx}{2 a b^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {x^4 \left (2 b^2 c^2-5 (b c-a d)^2+2 a b d^2 x^2\right )}{b x^2+a}dx}{2 a b^2}+\frac {x^5 (b c-a d)^2}{2 a b^2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 363 |
\(\displaystyle \frac {\frac {2}{5} a d^2 x^5-(3 b c-7 a d) (b c-a d) \int \frac {x^4}{b x^2+a}dx}{2 a b^2}+\frac {x^5 (b c-a d)^2}{2 a b^2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {\frac {2}{5} a d^2 x^5-(3 b c-7 a d) (b c-a d) \int \left (\frac {a^2}{b^2 \left (b x^2+a\right )}-\frac {a}{b^2}+\frac {x^2}{b}\right )dx}{2 a b^2}+\frac {x^5 (b c-a d)^2}{2 a b^2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {2}{5} a d^2 x^5-\left (\frac {a^{3/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{5/2}}-\frac {a x}{b^2}+\frac {x^3}{3 b}\right ) (3 b c-7 a d) (b c-a d)}{2 a b^2}+\frac {x^5 (b c-a d)^2}{2 a b^2 \left (a+b x^2\right )}\) |
((b*c - a*d)^2*x^5)/(2*a*b^2*(a + b*x^2)) + ((2*a*d^2*x^5)/5 - (3*b*c - 7* a*d)*(b*c - a*d)*(-((a*x)/b^2) + x^3/(3*b) + (a^(3/2)*ArcTan[(Sqrt[b]*x)/S qrt[a]])/b^(5/2)))/(2*a*b^2)
3.3.71.3.1 Defintions of rubi rules used
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x_Symbol] :> Simp[(-(b*c - a*d)^2)*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a* b^2*e*(p + 1))), x] + Simp[1/(2*a*b^2*(p + 1)) Int[(e*x)^m*(a + b*x^2)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + 2*b^2*c^2*(p + 1) + 2*a*b*d^2*(p + 1)*x^ 2, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && LtQ[p , -1]
Time = 2.65 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.97
method | result | size |
default | \(\frac {\frac {1}{5} b^{2} d^{2} x^{5}-\frac {2}{3} x^{3} a b \,d^{2}+\frac {2}{3} x^{3} b^{2} c d +3 a^{2} d^{2} x -4 a b c d x +b^{2} c^{2} x}{b^{4}}-\frac {a \left (\frac {\left (-\frac {1}{2} a^{2} d^{2}+a b c d -\frac {1}{2} b^{2} c^{2}\right ) x}{b \,x^{2}+a}+\frac {\left (7 a^{2} d^{2}-10 a b c d +3 b^{2} c^{2}\right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{4}}\) | \(141\) |
risch | \(\frac {d^{2} x^{5}}{5 b^{2}}-\frac {2 x^{3} a \,d^{2}}{3 b^{3}}+\frac {2 x^{3} c d}{3 b^{2}}+\frac {3 a^{2} d^{2} x}{b^{4}}-\frac {4 a c d x}{b^{3}}+\frac {c^{2} x}{b^{2}}+\frac {\left (\frac {1}{2} a^{3} d^{2}-a^{2} b c d +\frac {1}{2} b^{2} c^{2} a \right ) x}{b^{4} \left (b \,x^{2}+a \right )}+\frac {7 \sqrt {-a b}\, \ln \left (-\sqrt {-a b}\, x -a \right ) a^{2} d^{2}}{4 b^{5}}-\frac {5 \sqrt {-a b}\, \ln \left (-\sqrt {-a b}\, x -a \right ) a c d}{2 b^{4}}+\frac {3 \sqrt {-a b}\, \ln \left (-\sqrt {-a b}\, x -a \right ) c^{2}}{4 b^{3}}-\frac {7 \sqrt {-a b}\, \ln \left (\sqrt {-a b}\, x -a \right ) a^{2} d^{2}}{4 b^{5}}+\frac {5 \sqrt {-a b}\, \ln \left (\sqrt {-a b}\, x -a \right ) a c d}{2 b^{4}}-\frac {3 \sqrt {-a b}\, \ln \left (\sqrt {-a b}\, x -a \right ) c^{2}}{4 b^{3}}\) | \(275\) |
1/b^4*(1/5*b^2*d^2*x^5-2/3*x^3*a*b*d^2+2/3*x^3*b^2*c*d+3*a^2*d^2*x-4*a*b*c *d*x+b^2*c^2*x)-a/b^4*((-1/2*a^2*d^2+a*b*c*d-1/2*b^2*c^2)*x/(b*x^2+a)+1/2* (7*a^2*d^2-10*a*b*c*d+3*b^2*c^2)/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))
Time = 0.25 (sec) , antiderivative size = 400, normalized size of antiderivative = 2.76 \[ \int \frac {x^4 \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx=\left [\frac {12 \, b^{3} d^{2} x^{7} + 4 \, {\left (10 \, b^{3} c d - 7 \, a b^{2} d^{2}\right )} x^{5} + 20 \, {\left (3 \, b^{3} c^{2} - 10 \, a b^{2} c d + 7 \, a^{2} b d^{2}\right )} x^{3} + 15 \, {\left (3 \, a b^{2} c^{2} - 10 \, a^{2} b c d + 7 \, a^{3} d^{2} + {\left (3 \, b^{3} c^{2} - 10 \, a b^{2} c d + 7 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) + 30 \, {\left (3 \, a b^{2} c^{2} - 10 \, a^{2} b c d + 7 \, a^{3} d^{2}\right )} x}{60 \, {\left (b^{5} x^{2} + a b^{4}\right )}}, \frac {6 \, b^{3} d^{2} x^{7} + 2 \, {\left (10 \, b^{3} c d - 7 \, a b^{2} d^{2}\right )} x^{5} + 10 \, {\left (3 \, b^{3} c^{2} - 10 \, a b^{2} c d + 7 \, a^{2} b d^{2}\right )} x^{3} - 15 \, {\left (3 \, a b^{2} c^{2} - 10 \, a^{2} b c d + 7 \, a^{3} d^{2} + {\left (3 \, b^{3} c^{2} - 10 \, a b^{2} c d + 7 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) + 15 \, {\left (3 \, a b^{2} c^{2} - 10 \, a^{2} b c d + 7 \, a^{3} d^{2}\right )} x}{30 \, {\left (b^{5} x^{2} + a b^{4}\right )}}\right ] \]
[1/60*(12*b^3*d^2*x^7 + 4*(10*b^3*c*d - 7*a*b^2*d^2)*x^5 + 20*(3*b^3*c^2 - 10*a*b^2*c*d + 7*a^2*b*d^2)*x^3 + 15*(3*a*b^2*c^2 - 10*a^2*b*c*d + 7*a^3* d^2 + (3*b^3*c^2 - 10*a*b^2*c*d + 7*a^2*b*d^2)*x^2)*sqrt(-a/b)*log((b*x^2 - 2*b*x*sqrt(-a/b) - a)/(b*x^2 + a)) + 30*(3*a*b^2*c^2 - 10*a^2*b*c*d + 7* a^3*d^2)*x)/(b^5*x^2 + a*b^4), 1/30*(6*b^3*d^2*x^7 + 2*(10*b^3*c*d - 7*a*b ^2*d^2)*x^5 + 10*(3*b^3*c^2 - 10*a*b^2*c*d + 7*a^2*b*d^2)*x^3 - 15*(3*a*b^ 2*c^2 - 10*a^2*b*c*d + 7*a^3*d^2 + (3*b^3*c^2 - 10*a*b^2*c*d + 7*a^2*b*d^2 )*x^2)*sqrt(a/b)*arctan(b*x*sqrt(a/b)/a) + 15*(3*a*b^2*c^2 - 10*a^2*b*c*d + 7*a^3*d^2)*x)/(b^5*x^2 + a*b^4)]
Leaf count of result is larger than twice the leaf count of optimal. 286 vs. \(2 (133) = 266\).
Time = 0.57 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.97 \[ \int \frac {x^4 \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx=x^{3} \left (- \frac {2 a d^{2}}{3 b^{3}} + \frac {2 c d}{3 b^{2}}\right ) + x \left (\frac {3 a^{2} d^{2}}{b^{4}} - \frac {4 a c d}{b^{3}} + \frac {c^{2}}{b^{2}}\right ) + \frac {x \left (a^{3} d^{2} - 2 a^{2} b c d + a b^{2} c^{2}\right )}{2 a b^{4} + 2 b^{5} x^{2}} + \frac {\sqrt {- \frac {a}{b^{9}}} \left (a d - b c\right ) \left (7 a d - 3 b c\right ) \log {\left (- \frac {b^{4} \sqrt {- \frac {a}{b^{9}}} \left (a d - b c\right ) \left (7 a d - 3 b c\right )}{7 a^{2} d^{2} - 10 a b c d + 3 b^{2} c^{2}} + x \right )}}{4} - \frac {\sqrt {- \frac {a}{b^{9}}} \left (a d - b c\right ) \left (7 a d - 3 b c\right ) \log {\left (\frac {b^{4} \sqrt {- \frac {a}{b^{9}}} \left (a d - b c\right ) \left (7 a d - 3 b c\right )}{7 a^{2} d^{2} - 10 a b c d + 3 b^{2} c^{2}} + x \right )}}{4} + \frac {d^{2} x^{5}}{5 b^{2}} \]
x**3*(-2*a*d**2/(3*b**3) + 2*c*d/(3*b**2)) + x*(3*a**2*d**2/b**4 - 4*a*c*d /b**3 + c**2/b**2) + x*(a**3*d**2 - 2*a**2*b*c*d + a*b**2*c**2)/(2*a*b**4 + 2*b**5*x**2) + sqrt(-a/b**9)*(a*d - b*c)*(7*a*d - 3*b*c)*log(-b**4*sqrt( -a/b**9)*(a*d - b*c)*(7*a*d - 3*b*c)/(7*a**2*d**2 - 10*a*b*c*d + 3*b**2*c* *2) + x)/4 - sqrt(-a/b**9)*(a*d - b*c)*(7*a*d - 3*b*c)*log(b**4*sqrt(-a/b* *9)*(a*d - b*c)*(7*a*d - 3*b*c)/(7*a**2*d**2 - 10*a*b*c*d + 3*b**2*c**2) + x)/4 + d**2*x**5/(5*b**2)
Time = 0.29 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.03 \[ \int \frac {x^4 \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx=\frac {{\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} x}{2 \, {\left (b^{5} x^{2} + a b^{4}\right )}} - \frac {{\left (3 \, a b^{2} c^{2} - 10 \, a^{2} b c d + 7 \, a^{3} d^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{4}} + \frac {3 \, b^{2} d^{2} x^{5} + 10 \, {\left (b^{2} c d - a b d^{2}\right )} x^{3} + 15 \, {\left (b^{2} c^{2} - 4 \, a b c d + 3 \, a^{2} d^{2}\right )} x}{15 \, b^{4}} \]
1/2*(a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2)*x/(b^5*x^2 + a*b^4) - 1/2*(3*a*b^2 *c^2 - 10*a^2*b*c*d + 7*a^3*d^2)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^4) + 1 /15*(3*b^2*d^2*x^5 + 10*(b^2*c*d - a*b*d^2)*x^3 + 15*(b^2*c^2 - 4*a*b*c*d + 3*a^2*d^2)*x)/b^4
Time = 0.30 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.08 \[ \int \frac {x^4 \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx=-\frac {{\left (3 \, a b^{2} c^{2} - 10 \, a^{2} b c d + 7 \, a^{3} d^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{4}} + \frac {a b^{2} c^{2} x - 2 \, a^{2} b c d x + a^{3} d^{2} x}{2 \, {\left (b x^{2} + a\right )} b^{4}} + \frac {3 \, b^{8} d^{2} x^{5} + 10 \, b^{8} c d x^{3} - 10 \, a b^{7} d^{2} x^{3} + 15 \, b^{8} c^{2} x - 60 \, a b^{7} c d x + 45 \, a^{2} b^{6} d^{2} x}{15 \, b^{10}} \]
-1/2*(3*a*b^2*c^2 - 10*a^2*b*c*d + 7*a^3*d^2)*arctan(b*x/sqrt(a*b))/(sqrt( a*b)*b^4) + 1/2*(a*b^2*c^2*x - 2*a^2*b*c*d*x + a^3*d^2*x)/((b*x^2 + a)*b^4 ) + 1/15*(3*b^8*d^2*x^5 + 10*b^8*c*d*x^3 - 10*a*b^7*d^2*x^3 + 15*b^8*c^2*x - 60*a*b^7*c*d*x + 45*a^2*b^6*d^2*x)/b^10
Time = 4.90 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.38 \[ \int \frac {x^4 \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx=x\,\left (\frac {c^2}{b^2}+\frac {2\,a\,\left (\frac {2\,a\,d^2}{b^3}-\frac {2\,c\,d}{b^2}\right )}{b}-\frac {a^2\,d^2}{b^4}\right )-x^3\,\left (\frac {2\,a\,d^2}{3\,b^3}-\frac {2\,c\,d}{3\,b^2}\right )+\frac {d^2\,x^5}{5\,b^2}+\frac {x\,\left (\frac {a^3\,d^2}{2}-a^2\,b\,c\,d+\frac {a\,b^2\,c^2}{2}\right )}{b^5\,x^2+a\,b^4}-\frac {\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,x\,\left (a\,d-b\,c\right )\,\left (7\,a\,d-3\,b\,c\right )}{7\,a^3\,d^2-10\,a^2\,b\,c\,d+3\,a\,b^2\,c^2}\right )\,\left (a\,d-b\,c\right )\,\left (7\,a\,d-3\,b\,c\right )}{2\,b^{9/2}} \]
x*(c^2/b^2 + (2*a*((2*a*d^2)/b^3 - (2*c*d)/b^2))/b - (a^2*d^2)/b^4) - x^3* ((2*a*d^2)/(3*b^3) - (2*c*d)/(3*b^2)) + (d^2*x^5)/(5*b^2) + (x*((a^3*d^2)/ 2 + (a*b^2*c^2)/2 - a^2*b*c*d))/(a*b^4 + b^5*x^2) - (a^(1/2)*atan((a^(1/2) *b^(1/2)*x*(a*d - b*c)*(7*a*d - 3*b*c))/(7*a^3*d^2 + 3*a*b^2*c^2 - 10*a^2* b*c*d))*(a*d - b*c)*(7*a*d - 3*b*c))/(2*b^(9/2))