Integrand size = 28, antiderivative size = 155 \[ \int \frac {x^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {3 (b B-5 a D) x}{8 a b^3}-\frac {x^3 \left (a \left (B-\frac {a D}{b}\right )-(A b-a C) x\right )}{4 a b \left (a+b x^2\right )^2}-\frac {x^2 (4 a C-(3 b B-7 a D) x)}{8 a b^2 \left (a+b x^2\right )}+\frac {3 (b B-5 a D) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{7/2}}+\frac {C \log \left (a+b x^2\right )}{2 b^3} \]
-3/8*(B*b-5*D*a)*x/a/b^3-1/4*x^3*(a*(B-a*D/b)-(A*b-C*a)*x)/a/b/(b*x^2+a)^2 -1/8*x^2*(4*C*a-(3*B*b-7*D*a)*x)/a/b^2/(b*x^2+a)+1/2*C*ln(b*x^2+a)/b^3+3/8 *(B*b-5*D*a)*arctan(x*b^(1/2)/a^(1/2))/b^(7/2)/a^(1/2)
Time = 0.05 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.81 \[ \int \frac {x^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=\frac {D x}{b^3}+\frac {-4 A b+8 a C-5 b B x+9 a D x}{8 b^3 \left (a+b x^2\right )}+\frac {a (A b+b B x-a (C+D x))}{4 b^3 \left (a+b x^2\right )^2}+\frac {3 (b B-5 a D) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{7/2}}+\frac {C \log \left (a+b x^2\right )}{2 b^3} \]
(D*x)/b^3 + (-4*A*b + 8*a*C - 5*b*B*x + 9*a*D*x)/(8*b^3*(a + b*x^2)) + (a* (A*b + b*B*x - a*(C + D*x)))/(4*b^3*(a + b*x^2)^2) + (3*(b*B - 5*a*D)*ArcT an[(Sqrt[b]*x)/Sqrt[a]])/(8*Sqrt[a]*b^(7/2)) + (C*Log[a + b*x^2])/(2*b^3)
Time = 0.50 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2335, 25, 2335, 25, 27, 523, 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.
| \(\displaystyle \int \frac {x^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 2335 |
| \(\displaystyle -\frac {\int -\frac {x^2 \left (4 a D x^2+4 a C x+3 a \left (B-\frac {a D}{b}\right )\right )}{\left (b x^2+a\right )^2}dx}{4 a b}-\frac {x^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {x^2 \left (4 a D x^2+4 a C x+3 a \left (B-\frac {a D}{b}\right )\right )}{\left (b x^2+a\right )^2}dx}{4 a b}-\frac {x^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}\) |
| \(\Big \downarrow \) 2335 |
| \(\displaystyle \frac {-\frac {\int -\frac {a x (8 a C-3 (b B-5 a D) x)}{b x^2+a}dx}{2 a b}-\frac {x^2 (4 a C-x (3 b B-7 a D))}{2 b \left (a+b x^2\right )}}{4 a b}-\frac {x^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {a x (8 a C-3 (b B-5 a D) x)}{b x^2+a}dx}{2 a b}-\frac {x^2 (4 a C-x (3 b B-7 a D))}{2 b \left (a+b x^2\right )}}{4 a b}-\frac {x^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {x (8 a C-3 (b B-5 a D) x)}{b x^2+a}dx}{2 b}-\frac {x^2 (4 a C-x (3 b B-7 a D))}{2 b \left (a+b x^2\right )}}{4 a b}-\frac {x^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}\) |
| \(\Big \downarrow \) 523 |
| \(\displaystyle \frac {\frac {\int \left (\frac {3 a (b B-5 a D)+8 a b C x}{b \left (b x^2+a\right )}-3 \left (B-\frac {5 a D}{b}\right )\right )dx}{2 b}-\frac {x^2 (4 a C-x (3 b B-7 a D))}{2 b \left (a+b x^2\right )}}{4 a b}-\frac {x^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\frac {3 \sqrt {a} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) (b B-5 a D)}{b^{3/2}}-3 x \left (B-\frac {5 a D}{b}\right )+\frac {4 a C \log \left (a+b x^2\right )}{b}}{2 b}-\frac {x^2 (4 a C-x (3 b B-7 a D))}{2 b \left (a+b x^2\right )}}{4 a b}-\frac {x^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}\) |
-1/4*(x^3*(a*(B - (a*D)/b) - (A*b - a*C)*x))/(a*b*(a + b*x^2)^2) + (-1/2*( x^2*(4*a*C - (3*b*B - 7*a*D)*x))/(b*(a + b*x^2)) + (-3*(B - (5*a*D)/b)*x + (3*Sqrt[a]*(b*B - 5*a*D)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/b^(3/2) + (4*a*C*Lo g[a + b*x^2])/b)/(2*b))/(4*a*b)
3.2.3.3.1 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[((x_)^(m_.)*((c_) + (d_.)*(x_)))/((a_) + (b_.)*(x_)^2), x_Symbol] :> In t[ExpandIntegrand[x^m*((c + d*x)/(a + b*x^2)), x], x] /; FreeQ[{a, b, c, d} , x] && IntegerQ[m]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ {Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq , a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(c*x)^m*(a + b*x^2)^(p + 1)*((a*g - b*f*x)/(2*a*b*(p + 1))), x] + Simp[c/(2*a*b*(p + 1)) Int[(c*x)^(m - 1)*(a + b*x^2)^(p + 1)*ExpandToSu m[2*a*b*(p + 1)*x*Q - a*g*m + b*f*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && GtQ[m, 0]
Time = 3.47 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.74
| method | result | size |
| default | \(\frac {D x}{b^{3}}+\frac {\frac {\left (-\frac {5}{8} B \,b^{2}+\frac {9}{8} D a b \right ) x^{3}+\left (-\frac {1}{2} b^{2} A +C a b \right ) x^{2}-\frac {a \left (3 B b -7 D a \right ) x}{8}-\frac {a b A}{4}+\frac {3 C \,a^{2}}{4}}{\left (b \,x^{2}+a \right )^{2}}+\frac {C \ln \left (b \,x^{2}+a \right )}{2}+\frac {\left (3 B b -15 D a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}}{b^{3}}\) | \(115\) |
D/b^3*x+1/b^3*(((-5/8*B*b^2+9/8*D*a*b)*x^3+(-1/2*b^2*A+C*a*b)*x^2-1/8*a*(3 *B*b-7*D*a)*x-1/4*a*b*A+3/4*C*a^2)/(b*x^2+a)^2+1/2*C*ln(b*x^2+a)+1/8*(3*B* b-15*D*a)/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))
Time = 0.28 (sec) , antiderivative size = 480, normalized size of antiderivative = 3.10 \[ \int \frac {x^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=\left [\frac {16 \, D a b^{3} x^{5} + 12 \, C a^{3} b - 4 \, A a^{2} b^{2} + 10 \, {\left (5 \, D a^{2} b^{2} - B a b^{3}\right )} x^{3} + 8 \, {\left (2 \, C a^{2} b^{2} - A a b^{3}\right )} x^{2} - 3 \, {\left ({\left (5 \, D a b^{2} - B b^{3}\right )} x^{4} + 5 \, D a^{3} - B a^{2} b + 2 \, {\left (5 \, D a^{2} b - B a b^{2}\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + 6 \, {\left (5 \, D a^{3} b - B a^{2} b^{2}\right )} x + 8 \, {\left (C a b^{3} x^{4} + 2 \, C a^{2} b^{2} x^{2} + C a^{3} b\right )} \log \left (b x^{2} + a\right )}{16 \, {\left (a b^{6} x^{4} + 2 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}}, \frac {8 \, D a b^{3} x^{5} + 6 \, C a^{3} b - 2 \, A a^{2} b^{2} + 5 \, {\left (5 \, D a^{2} b^{2} - B a b^{3}\right )} x^{3} + 4 \, {\left (2 \, C a^{2} b^{2} - A a b^{3}\right )} x^{2} - 3 \, {\left ({\left (5 \, D a b^{2} - B b^{3}\right )} x^{4} + 5 \, D a^{3} - B a^{2} b + 2 \, {\left (5 \, D a^{2} b - B a b^{2}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + 3 \, {\left (5 \, D a^{3} b - B a^{2} b^{2}\right )} x + 4 \, {\left (C a b^{3} x^{4} + 2 \, C a^{2} b^{2} x^{2} + C a^{3} b\right )} \log \left (b x^{2} + a\right )}{8 \, {\left (a b^{6} x^{4} + 2 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}}\right ] \]
[1/16*(16*D*a*b^3*x^5 + 12*C*a^3*b - 4*A*a^2*b^2 + 10*(5*D*a^2*b^2 - B*a*b ^3)*x^3 + 8*(2*C*a^2*b^2 - A*a*b^3)*x^2 - 3*((5*D*a*b^2 - B*b^3)*x^4 + 5*D *a^3 - B*a^2*b + 2*(5*D*a^2*b - B*a*b^2)*x^2)*sqrt(-a*b)*log((b*x^2 + 2*sq rt(-a*b)*x - a)/(b*x^2 + a)) + 6*(5*D*a^3*b - B*a^2*b^2)*x + 8*(C*a*b^3*x^ 4 + 2*C*a^2*b^2*x^2 + C*a^3*b)*log(b*x^2 + a))/(a*b^6*x^4 + 2*a^2*b^5*x^2 + a^3*b^4), 1/8*(8*D*a*b^3*x^5 + 6*C*a^3*b - 2*A*a^2*b^2 + 5*(5*D*a^2*b^2 - B*a*b^3)*x^3 + 4*(2*C*a^2*b^2 - A*a*b^3)*x^2 - 3*((5*D*a*b^2 - B*b^3)*x^ 4 + 5*D*a^3 - B*a^2*b + 2*(5*D*a^2*b - B*a*b^2)*x^2)*sqrt(a*b)*arctan(sqrt (a*b)*x/a) + 3*(5*D*a^3*b - B*a^2*b^2)*x + 4*(C*a*b^3*x^4 + 2*C*a^2*b^2*x^ 2 + C*a^3*b)*log(b*x^2 + a))/(a*b^6*x^4 + 2*a^2*b^5*x^2 + a^3*b^4)]
Leaf count of result is larger than twice the leaf count of optimal. 282 vs. \(2 (139) = 278\).
Time = 87.26 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.82 \[ \int \frac {x^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=\frac {D x}{b^{3}} + \left (\frac {C}{2 b^{3}} - \frac {3 \sqrt {- a b^{7}} \left (- B b + 5 D a\right )}{16 a b^{7}}\right ) \log {\left (x + \frac {8 C a - 16 a b^{3} \left (\frac {C}{2 b^{3}} - \frac {3 \sqrt {- a b^{7}} \left (- B b + 5 D a\right )}{16 a b^{7}}\right )}{- 3 B b + 15 D a} \right )} + \left (\frac {C}{2 b^{3}} + \frac {3 \sqrt {- a b^{7}} \left (- B b + 5 D a\right )}{16 a b^{7}}\right ) \log {\left (x + \frac {8 C a - 16 a b^{3} \left (\frac {C}{2 b^{3}} + \frac {3 \sqrt {- a b^{7}} \left (- B b + 5 D a\right )}{16 a b^{7}}\right )}{- 3 B b + 15 D a} \right )} + \frac {- 2 A a b + 6 C a^{2} + x^{3} \left (- 5 B b^{2} + 9 D a b\right ) + x^{2} \left (- 4 A b^{2} + 8 C a b\right ) + x \left (- 3 B a b + 7 D a^{2}\right )}{8 a^{2} b^{3} + 16 a b^{4} x^{2} + 8 b^{5} x^{4}} \]
D*x/b**3 + (C/(2*b**3) - 3*sqrt(-a*b**7)*(-B*b + 5*D*a)/(16*a*b**7))*log(x + (8*C*a - 16*a*b**3*(C/(2*b**3) - 3*sqrt(-a*b**7)*(-B*b + 5*D*a)/(16*a*b **7)))/(-3*B*b + 15*D*a)) + (C/(2*b**3) + 3*sqrt(-a*b**7)*(-B*b + 5*D*a)/( 16*a*b**7))*log(x + (8*C*a - 16*a*b**3*(C/(2*b**3) + 3*sqrt(-a*b**7)*(-B*b + 5*D*a)/(16*a*b**7)))/(-3*B*b + 15*D*a)) + (-2*A*a*b + 6*C*a**2 + x**3*( -5*B*b**2 + 9*D*a*b) + x**2*(-4*A*b**2 + 8*C*a*b) + x*(-3*B*a*b + 7*D*a**2 ))/(8*a**2*b**3 + 16*a*b**4*x**2 + 8*b**5*x**4)
Time = 0.29 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.88 \[ \int \frac {x^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=\frac {{\left (9 \, D a b - 5 \, B b^{2}\right )} x^{3} + 6 \, C a^{2} - 2 \, A a b + 4 \, {\left (2 \, C a b - A b^{2}\right )} x^{2} + {\left (7 \, D a^{2} - 3 \, B a b\right )} x}{8 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}} + \frac {D x}{b^{3}} + \frac {C \log \left (b x^{2} + a\right )}{2 \, b^{3}} - \frac {3 \, {\left (5 \, D a - B b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3}} \]
1/8*((9*D*a*b - 5*B*b^2)*x^3 + 6*C*a^2 - 2*A*a*b + 4*(2*C*a*b - A*b^2)*x^2 + (7*D*a^2 - 3*B*a*b)*x)/(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3) + D*x/b^3 + 1/ 2*C*log(b*x^2 + a)/b^3 - 3/8*(5*D*a - B*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b )*b^3)
Time = 0.30 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.79 \[ \int \frac {x^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=\frac {D x}{b^{3}} + \frac {C \log \left (b x^{2} + a\right )}{2 \, b^{3}} - \frac {3 \, {\left (5 \, D a - B b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3}} + \frac {{\left (9 \, D a b - 5 \, B b^{2}\right )} x^{3} + 6 \, C a^{2} - 2 \, A a b + 4 \, {\left (2 \, C a b - A b^{2}\right )} x^{2} + {\left (7 \, D a^{2} - 3 \, B a b\right )} x}{8 \, {\left (b x^{2} + a\right )}^{2} b^{3}} \]
D*x/b^3 + 1/2*C*log(b*x^2 + a)/b^3 - 3/8*(5*D*a - B*b)*arctan(b*x/sqrt(a*b ))/(sqrt(a*b)*b^3) + 1/8*((9*D*a*b - 5*B*b^2)*x^3 + 6*C*a^2 - 2*A*a*b + 4* (2*C*a*b - A*b^2)*x^2 + (7*D*a^2 - 3*B*a*b)*x)/((b*x^2 + a)^2*b^3)
Timed out. \[ \int \frac {x^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=\int \frac {x^3\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (b\,x^2+a\right )}^3} \,d x \]