Integrand size = 22, antiderivative size = 163 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=-\frac {\left (13 b^2 c^2-10 a b c d+a^2 d^2\right ) x}{4 c d^4}+\frac {b^2 x^3}{3 d^3}+\frac {(b c-a d)^2 x^5}{4 c d^2 \left (c+d x^2\right )^2}-\frac {(b c-a d) (9 b c-a d) x}{8 d^4 \left (c+d x^2\right )}+\frac {\left (35 b^2 c^2-30 a b c d+3 a^2 d^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 \sqrt {c} d^{9/2}} \]
-1/4*(a^2*d^2-10*a*b*c*d+13*b^2*c^2)*x/c/d^4+1/3*b^2*x^3/d^3+1/4*(-a*d+b*c )^2*x^5/c/d^2/(d*x^2+c)^2-1/8*(-a*d+b*c)*(-a*d+9*b*c)*x/d^4/(d*x^2+c)+1/8* (3*a^2*d^2-30*a*b*c*d+35*b^2*c^2)*arctan(x*d^(1/2)/c^(1/2))/d^(9/2)/c^(1/2 )
Time = 0.06 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.91 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=-\frac {b (3 b c-2 a d) x}{d^4}+\frac {b^2 x^3}{3 d^3}+\frac {c (b c-a d)^2 x}{4 d^4 \left (c+d x^2\right )^2}-\frac {\left (13 b^2 c^2-18 a b c d+5 a^2 d^2\right ) x}{8 d^4 \left (c+d x^2\right )}+\frac {\left (35 b^2 c^2-30 a b c d+3 a^2 d^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 \sqrt {c} d^{9/2}} \]
-((b*(3*b*c - 2*a*d)*x)/d^4) + (b^2*x^3)/(3*d^3) + (c*(b*c - a*d)^2*x)/(4* d^4*(c + d*x^2)^2) - ((13*b^2*c^2 - 18*a*b*c*d + 5*a^2*d^2)*x)/(8*d^4*(c + d*x^2)) + ((35*b^2*c^2 - 30*a*b*c*d + 3*a^2*d^2)*ArcTan[(Sqrt[d]*x)/Sqrt[ c]])/(8*Sqrt[c]*d^(9/2))
Time = 0.39 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {366, 25, 360, 25, 1467, 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.
| \(\displaystyle \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 366 |
| \(\displaystyle \frac {x^5 (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2}-\frac {\int -\frac {x^4 \left (4 a^2 d^2+4 b^2 c x^2 d-5 (b c-a d)^2\right )}{\left (d x^2+c\right )^2}dx}{4 c d^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {x^4 \left (4 a^2 d^2+4 b^2 c x^2 d-5 (b c-a d)^2\right )}{\left (d x^2+c\right )^2}dx}{4 c d^2}+\frac {x^5 (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2}\) |
| \(\Big \downarrow \) 360 |
| \(\displaystyle \frac {-\frac {\int -\frac {8 b^2 c d^3 x^4-2 d^2 (b c-a d) (9 b c-a d) x^2+c d (b c-a d) (9 b c-a d)}{d x^2+c}dx}{2 d^3}-\frac {c x (b c-a d) (9 b c-a d)}{2 d^2 \left (c+d x^2\right )}}{4 c d^2}+\frac {x^5 (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {8 b^2 c d^3 x^4-2 d^2 (b c-a d) (9 b c-a d) x^2+c d (b c-a d) (9 b c-a d)}{d x^2+c}dx}{2 d^3}-\frac {c x (b c-a d) (9 b c-a d)}{2 d^2 \left (c+d x^2\right )}}{4 c d^2}+\frac {x^5 (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2}\) |
| \(\Big \downarrow \) 1467 |
| \(\displaystyle \frac {\frac {\int \left (8 b^2 c d^2 x^2-2 d \left (13 b^2 c^2-10 a b d c+a^2 d^2\right )+\frac {35 b^2 d c^3-30 a b d^2 c^2+3 a^2 d^3 c}{d x^2+c}\right )dx}{2 d^3}-\frac {c x (b c-a d) (9 b c-a d)}{2 d^2 \left (c+d x^2\right )}}{4 c d^2}+\frac {x^5 (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\sqrt {c} \sqrt {d} \left (3 a^2 d^2-30 a b c d+35 b^2 c^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )-2 d x \left (a^2 d^2-10 a b c d+13 b^2 c^2\right )+\frac {8}{3} b^2 c d^2 x^3}{2 d^3}-\frac {c x (b c-a d) (9 b c-a d)}{2 d^2 \left (c+d x^2\right )}}{4 c d^2}+\frac {x^5 (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2}\) |
((b*c - a*d)^2*x^5)/(4*c*d^2*(c + d*x^2)^2) + (-1/2*(c*(b*c - a*d)*(9*b*c - a*d)*x)/(d^2*(c + d*x^2)) + (-2*d*(13*b^2*c^2 - 10*a*b*c*d + a^2*d^2)*x + (8*b^2*c*d^2*x^3)/3 + Sqrt[c]*Sqrt[d]*(35*b^2*c^2 - 30*a*b*c*d + 3*a^2*d ^2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(2*d^3))/(4*c*d^2)
3.2.89.3.1 Defintions of rubi rules used
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[(a + b*x^2)^(p + 1)*Expan dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & & (IntegerQ[p] || EqQ[m + 2*p + 1, 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]
Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Time = 2.63 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.84
| method | result | size |
| default | \(\frac {b \left (\frac {1}{3} b d \,x^{3}+2 a d x -3 b c x \right )}{d^{4}}+\frac {\frac {\left (-\frac {5}{8} a^{2} d^{3}+\frac {9}{4} a b c \,d^{2}-\frac {13}{8} b^{2} c^{2} d \right ) x^{3}-\frac {c \left (3 a^{2} d^{2}-14 a b c d +11 b^{2} c^{2}\right ) x}{8}}{\left (d \,x^{2}+c \right )^{2}}+\frac {\left (3 a^{2} d^{2}-30 a b c d +35 b^{2} c^{2}\right ) \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \sqrt {c d}}}{d^{4}}\) | \(137\) |
| risch | \(\frac {b^{2} x^{3}}{3 d^{3}}+\frac {2 b a x}{d^{3}}-\frac {3 b^{2} c x}{d^{4}}+\frac {\left (-\frac {5}{8} a^{2} d^{3}+\frac {9}{4} a b c \,d^{2}-\frac {13}{8} b^{2} c^{2} d \right ) x^{3}-\frac {c \left (3 a^{2} d^{2}-14 a b c d +11 b^{2} c^{2}\right ) x}{8}}{d^{4} \left (d \,x^{2}+c \right )^{2}}-\frac {3 \ln \left (d x +\sqrt {-c d}\right ) a^{2}}{16 d^{2} \sqrt {-c d}}+\frac {15 \ln \left (d x +\sqrt {-c d}\right ) a b c}{8 d^{3} \sqrt {-c d}}-\frac {35 \ln \left (d x +\sqrt {-c d}\right ) b^{2} c^{2}}{16 d^{4} \sqrt {-c d}}+\frac {3 \ln \left (-d x +\sqrt {-c d}\right ) a^{2}}{16 d^{2} \sqrt {-c d}}-\frac {15 \ln \left (-d x +\sqrt {-c d}\right ) a b c}{8 d^{3} \sqrt {-c d}}+\frac {35 \ln \left (-d x +\sqrt {-c d}\right ) b^{2} c^{2}}{16 d^{4} \sqrt {-c d}}\) | \(261\) |
b/d^4*(1/3*b*d*x^3+2*a*d*x-3*b*c*x)+1/d^4*(((-5/8*a^2*d^3+9/4*a*b*c*d^2-13 /8*b^2*c^2*d)*x^3-1/8*c*(3*a^2*d^2-14*a*b*c*d+11*b^2*c^2)*x)/(d*x^2+c)^2+1 /8*(3*a^2*d^2-30*a*b*c*d+35*b^2*c^2)/(c*d)^(1/2)*arctan(d*x/(c*d)^(1/2)))
Time = 0.26 (sec) , antiderivative size = 522, normalized size of antiderivative = 3.20 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\left [\frac {16 \, b^{2} c d^{4} x^{7} - 16 \, {\left (7 \, b^{2} c^{2} d^{3} - 6 \, a b c d^{4}\right )} x^{5} - 10 \, {\left (35 \, b^{2} c^{3} d^{2} - 30 \, a b c^{2} d^{3} + 3 \, a^{2} c d^{4}\right )} x^{3} - 3 \, {\left (35 \, b^{2} c^{4} - 30 \, a b c^{3} d + 3 \, a^{2} c^{2} d^{2} + {\left (35 \, b^{2} c^{2} d^{2} - 30 \, a b c d^{3} + 3 \, a^{2} d^{4}\right )} x^{4} + 2 \, {\left (35 \, b^{2} c^{3} d - 30 \, a b c^{2} d^{2} + 3 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {-c d} \log \left (\frac {d x^{2} - 2 \, \sqrt {-c d} x - c}{d x^{2} + c}\right ) - 6 \, {\left (35 \, b^{2} c^{4} d - 30 \, a b c^{3} d^{2} + 3 \, a^{2} c^{2} d^{3}\right )} x}{48 \, {\left (c d^{7} x^{4} + 2 \, c^{2} d^{6} x^{2} + c^{3} d^{5}\right )}}, \frac {8 \, b^{2} c d^{4} x^{7} - 8 \, {\left (7 \, b^{2} c^{2} d^{3} - 6 \, a b c d^{4}\right )} x^{5} - 5 \, {\left (35 \, b^{2} c^{3} d^{2} - 30 \, a b c^{2} d^{3} + 3 \, a^{2} c d^{4}\right )} x^{3} + 3 \, {\left (35 \, b^{2} c^{4} - 30 \, a b c^{3} d + 3 \, a^{2} c^{2} d^{2} + {\left (35 \, b^{2} c^{2} d^{2} - 30 \, a b c d^{3} + 3 \, a^{2} d^{4}\right )} x^{4} + 2 \, {\left (35 \, b^{2} c^{3} d - 30 \, a b c^{2} d^{2} + 3 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {c d} \arctan \left (\frac {\sqrt {c d} x}{c}\right ) - 3 \, {\left (35 \, b^{2} c^{4} d - 30 \, a b c^{3} d^{2} + 3 \, a^{2} c^{2} d^{3}\right )} x}{24 \, {\left (c d^{7} x^{4} + 2 \, c^{2} d^{6} x^{2} + c^{3} d^{5}\right )}}\right ] \]
[1/48*(16*b^2*c*d^4*x^7 - 16*(7*b^2*c^2*d^3 - 6*a*b*c*d^4)*x^5 - 10*(35*b^ 2*c^3*d^2 - 30*a*b*c^2*d^3 + 3*a^2*c*d^4)*x^3 - 3*(35*b^2*c^4 - 30*a*b*c^3 *d + 3*a^2*c^2*d^2 + (35*b^2*c^2*d^2 - 30*a*b*c*d^3 + 3*a^2*d^4)*x^4 + 2*( 35*b^2*c^3*d - 30*a*b*c^2*d^2 + 3*a^2*c*d^3)*x^2)*sqrt(-c*d)*log((d*x^2 - 2*sqrt(-c*d)*x - c)/(d*x^2 + c)) - 6*(35*b^2*c^4*d - 30*a*b*c^3*d^2 + 3*a^ 2*c^2*d^3)*x)/(c*d^7*x^4 + 2*c^2*d^6*x^2 + c^3*d^5), 1/24*(8*b^2*c*d^4*x^7 - 8*(7*b^2*c^2*d^3 - 6*a*b*c*d^4)*x^5 - 5*(35*b^2*c^3*d^2 - 30*a*b*c^2*d^ 3 + 3*a^2*c*d^4)*x^3 + 3*(35*b^2*c^4 - 30*a*b*c^3*d + 3*a^2*c^2*d^2 + (35* b^2*c^2*d^2 - 30*a*b*c*d^3 + 3*a^2*d^4)*x^4 + 2*(35*b^2*c^3*d - 30*a*b*c^2 *d^2 + 3*a^2*c*d^3)*x^2)*sqrt(c*d)*arctan(sqrt(c*d)*x/c) - 3*(35*b^2*c^4*d - 30*a*b*c^3*d^2 + 3*a^2*c^2*d^3)*x)/(c*d^7*x^4 + 2*c^2*d^6*x^2 + c^3*d^5 )]
Time = 1.09 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.47 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {b^{2} x^{3}}{3 d^{3}} + x \left (\frac {2 a b}{d^{3}} - \frac {3 b^{2} c}{d^{4}}\right ) - \frac {\sqrt {- \frac {1}{c d^{9}}} \cdot \left (3 a^{2} d^{2} - 30 a b c d + 35 b^{2} c^{2}\right ) \log {\left (- c d^{4} \sqrt {- \frac {1}{c d^{9}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{c d^{9}}} \cdot \left (3 a^{2} d^{2} - 30 a b c d + 35 b^{2} c^{2}\right ) \log {\left (c d^{4} \sqrt {- \frac {1}{c d^{9}}} + x \right )}}{16} + \frac {x^{3} \left (- 5 a^{2} d^{3} + 18 a b c d^{2} - 13 b^{2} c^{2} d\right ) + x \left (- 3 a^{2} c d^{2} + 14 a b c^{2} d - 11 b^{2} c^{3}\right )}{8 c^{2} d^{4} + 16 c d^{5} x^{2} + 8 d^{6} x^{4}} \]
b**2*x**3/(3*d**3) + x*(2*a*b/d**3 - 3*b**2*c/d**4) - sqrt(-1/(c*d**9))*(3 *a**2*d**2 - 30*a*b*c*d + 35*b**2*c**2)*log(-c*d**4*sqrt(-1/(c*d**9)) + x) /16 + sqrt(-1/(c*d**9))*(3*a**2*d**2 - 30*a*b*c*d + 35*b**2*c**2)*log(c*d* *4*sqrt(-1/(c*d**9)) + x)/16 + (x**3*(-5*a**2*d**3 + 18*a*b*c*d**2 - 13*b* *2*c**2*d) + x*(-3*a**2*c*d**2 + 14*a*b*c**2*d - 11*b**2*c**3))/(8*c**2*d* *4 + 16*c*d**5*x**2 + 8*d**6*x**4)
Time = 0.29 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.98 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=-\frac {{\left (13 \, b^{2} c^{2} d - 18 \, a b c d^{2} + 5 \, a^{2} d^{3}\right )} x^{3} + {\left (11 \, b^{2} c^{3} - 14 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x}{8 \, {\left (d^{6} x^{4} + 2 \, c d^{5} x^{2} + c^{2} d^{4}\right )}} + \frac {{\left (35 \, b^{2} c^{2} - 30 \, a b c d + 3 \, a^{2} d^{2}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, \sqrt {c d} d^{4}} + \frac {b^{2} d x^{3} - 3 \, {\left (3 \, b^{2} c - 2 \, a b d\right )} x}{3 \, d^{4}} \]
-1/8*((13*b^2*c^2*d - 18*a*b*c*d^2 + 5*a^2*d^3)*x^3 + (11*b^2*c^3 - 14*a*b *c^2*d + 3*a^2*c*d^2)*x)/(d^6*x^4 + 2*c*d^5*x^2 + c^2*d^4) + 1/8*(35*b^2*c ^2 - 30*a*b*c*d + 3*a^2*d^2)*arctan(d*x/sqrt(c*d))/(sqrt(c*d)*d^4) + 1/3*( b^2*d*x^3 - 3*(3*b^2*c - 2*a*b*d)*x)/d^4
Time = 0.29 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.94 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {{\left (35 \, b^{2} c^{2} - 30 \, a b c d + 3 \, a^{2} d^{2}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, \sqrt {c d} d^{4}} - \frac {13 \, b^{2} c^{2} d x^{3} - 18 \, a b c d^{2} x^{3} + 5 \, a^{2} d^{3} x^{3} + 11 \, b^{2} c^{3} x - 14 \, a b c^{2} d x + 3 \, a^{2} c d^{2} x}{8 \, {\left (d x^{2} + c\right )}^{2} d^{4}} + \frac {b^{2} d^{6} x^{3} - 9 \, b^{2} c d^{5} x + 6 \, a b d^{6} x}{3 \, d^{9}} \]
1/8*(35*b^2*c^2 - 30*a*b*c*d + 3*a^2*d^2)*arctan(d*x/sqrt(c*d))/(sqrt(c*d) *d^4) - 1/8*(13*b^2*c^2*d*x^3 - 18*a*b*c*d^2*x^3 + 5*a^2*d^3*x^3 + 11*b^2* c^3*x - 14*a*b*c^2*d*x + 3*a^2*c*d^2*x)/((d*x^2 + c)^2*d^4) + 1/3*(b^2*d^6 *x^3 - 9*b^2*c*d^5*x + 6*a*b*d^6*x)/d^9
Time = 5.11 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.98 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {b^2\,x^3}{3\,d^3}-\frac {\left (\frac {5\,a^2\,d^3}{8}-\frac {9\,a\,b\,c\,d^2}{4}+\frac {13\,b^2\,c^2\,d}{8}\right )\,x^3+\left (\frac {3\,a^2\,c\,d^2}{8}-\frac {7\,a\,b\,c^2\,d}{4}+\frac {11\,b^2\,c^3}{8}\right )\,x}{c^2\,d^4+2\,c\,d^5\,x^2+d^6\,x^4}-x\,\left (\frac {3\,b^2\,c}{d^4}-\frac {2\,a\,b}{d^3}\right )+\frac {\mathrm {atan}\left (\frac {\sqrt {d}\,x}{\sqrt {c}}\right )\,\left (3\,a^2\,d^2-30\,a\,b\,c\,d+35\,b^2\,c^2\right )}{8\,\sqrt {c}\,d^{9/2}} \]
(b^2*x^3)/(3*d^3) - (x^3*((5*a^2*d^3)/8 + (13*b^2*c^2*d)/8 - (9*a*b*c*d^2) /4) + x*((11*b^2*c^3)/8 + (3*a^2*c*d^2)/8 - (7*a*b*c^2*d)/4))/(c^2*d^4 + d ^6*x^4 + 2*c*d^5*x^2) - x*((3*b^2*c)/d^4 - (2*a*b)/d^3) + (atan((d^(1/2)*x )/c^(1/2))*(3*a^2*d^2 + 35*b^2*c^2 - 30*a*b*c*d))/(8*c^(1/2)*d^(9/2))