Integrand size = 28, antiderivative size = 176 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^2} \, dx=\frac {(3 A b-5 a C) x}{2 b^3}+\frac {(2 b B-3 a D) x^2}{2 b^3}-\frac {(3 A b-5 a C) x^3}{6 a b^2}+\frac {D x^4}{4 b^2}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-(A b-a C) x\right )}{2 a b \left (a+b x^2\right )}-\frac {\sqrt {a} (3 A b-5 a C) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{7/2}}-\frac {a (2 b B-3 a D) \log \left (a+b x^2\right )}{2 b^4} \]
1/2*(3*A*b-5*C*a)*x/b^3+1/2*(2*B*b-3*D*a)*x^2/b^3-1/6*(3*A*b-5*C*a)*x^3/a/ b^2+1/4*D*x^4/b^2-1/2*x^4*(a*(B-a*D/b)-(A*b-C*a)*x)/a/b/(b*x^2+a)-1/2*a*(2 *B*b-3*D*a)*ln(b*x^2+a)/b^4-1/2*(3*A*b-5*C*a)*arctan(x*b^(1/2)/a^(1/2))*a^ (1/2)/b^(7/2)
Time = 0.08 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.79 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^2} \, dx=\frac {12 b (A b-2 a C) x+6 b (b B-2 a D) x^2+4 b^2 C x^3+3 b^2 D x^4+\frac {6 a \left (a^2 D+A b^2 x-a b (B+C x)\right )}{a+b x^2}+6 \sqrt {a} \sqrt {b} (-3 A b+5 a C) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )+6 a (-2 b B+3 a D) \log \left (a+b x^2\right )}{12 b^4} \]
(12*b*(A*b - 2*a*C)*x + 6*b*(b*B - 2*a*D)*x^2 + 4*b^2*C*x^3 + 3*b^2*D*x^4 + (6*a*(a^2*D + A*b^2*x - a*b*(B + C*x)))/(a + b*x^2) + 6*Sqrt[a]*Sqrt[b]* (-3*A*b + 5*a*C)*ArcTan[(Sqrt[b]*x)/Sqrt[a]] + 6*a*(-2*b*B + 3*a*D)*Log[a + b*x^2])/(12*b^4)
Time = 0.53 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2335, 25, 2333, 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.
| \(\displaystyle \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 2335 |
| \(\displaystyle -\frac {\int -\frac {x^3 \left (2 a D x^2-(3 A b-5 a C) x+4 a \left (B-\frac {a D}{b}\right )\right )}{b x^2+a}dx}{2 a b}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{2 a b \left (a+b x^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {x^3 \left (2 a D x^2-(3 A b-5 a C) x+4 a \left (B-\frac {a D}{b}\right )\right )}{b x^2+a}dx}{2 a b}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{2 a b \left (a+b x^2\right )}\) |
| \(\Big \downarrow \) 2333 |
| \(\displaystyle \frac {\int \left (\frac {2 a D x^3}{b}-\frac {(3 A b-5 a C) x^2}{b}+\frac {2 a (2 b B-3 a D) x}{b^2}+\frac {a (3 A b-5 a C)}{b^2}-\frac {(3 A b-5 a C) a^2+2 (2 b B-3 a D) x a^2}{b^2 \left (b x^2+a\right )}\right )dx}{2 a b}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{2 a b \left (a+b x^2\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {a^{3/2} (3 A b-5 a C) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{5/2}}-\frac {a^2 (2 b B-3 a D) \log \left (a+b x^2\right )}{b^3}+\frac {a x (3 A b-5 a C)}{b^2}-\frac {x^3 (3 A b-5 a C)}{3 b}+\frac {a x^2 (2 b B-3 a D)}{b^2}+\frac {a D x^4}{2 b}}{2 a b}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{2 a b \left (a+b x^2\right )}\) |
-1/2*(x^4*(a*(B - (a*D)/b) - (A*b - a*C)*x))/(a*b*(a + b*x^2)) + ((a*(3*A* b - 5*a*C)*x)/b^2 + (a*(2*b*B - 3*a*D)*x^2)/b^2 - ((3*A*b - 5*a*C)*x^3)/(3 *b) + (a*D*x^4)/(2*b) - (a^(3/2)*(3*A*b - 5*a*C)*ArcTan[(Sqrt[b]*x)/Sqrt[a ]])/b^(5/2) - (a^2*(2*b*B - 3*a*D)*Log[a + b*x^2])/b^3)/(2*a*b)
3.1.94.3.1 Defintions of rubi rules used
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
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 = 3.55 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.76
| method | result | size |
| default | \(\frac {\frac {1}{4} D b \,x^{4}+\frac {1}{3} b C \,x^{3}+\frac {1}{2} b B \,x^{2}-D a \,x^{2}+A b x -2 C a x}{b^{3}}-\frac {a \left (\frac {\left (-\frac {A b}{2}+\frac {C a}{2}\right ) x +\frac {a \left (B b -D a \right )}{2 b}}{b \,x^{2}+a}+\frac {\left (4 B b -6 D a \right ) \ln \left (b \,x^{2}+a \right )}{4 b}+\frac {\left (3 A b -5 C a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{3}}\) | \(134\) |
1/b^3*(1/4*D*b*x^4+1/3*b*C*x^3+1/2*b*B*x^2-D*a*x^2+A*b*x-2*C*a*x)-a/b^3*(( (-1/2*A*b+1/2*C*a)*x+1/2*a*(B*b-D*a)/b)/(b*x^2+a)+1/4*(4*B*b-6*D*a)/b*ln(b *x^2+a)+1/2*(3*A*b-5*C*a)/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))
Time = 0.29 (sec) , antiderivative size = 468, normalized size of antiderivative = 2.66 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^2} \, dx=\left [\frac {3 \, D b^{3} x^{6} + 4 \, C b^{3} x^{5} - 3 \, {\left (3 \, D a b^{2} - 2 \, B b^{3}\right )} x^{4} + 6 \, D a^{3} - 6 \, B a^{2} b - 4 \, {\left (5 \, C a b^{2} - 3 \, A b^{3}\right )} x^{3} - 6 \, {\left (2 \, D a^{2} b - B a b^{2}\right )} x^{2} - 3 \, {\left (5 \, C a^{2} b - 3 \, A a b^{2} + {\left (5 \, C a b^{2} - 3 \, A b^{3}\right )} x^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) - 6 \, {\left (5 \, C a^{2} b - 3 \, A a b^{2}\right )} x + 6 \, {\left (3 \, D a^{3} - 2 \, B a^{2} b + {\left (3 \, D a^{2} b - 2 \, B a b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right )}{12 \, {\left (b^{5} x^{2} + a b^{4}\right )}}, \frac {3 \, D b^{3} x^{6} + 4 \, C b^{3} x^{5} - 3 \, {\left (3 \, D a b^{2} - 2 \, B b^{3}\right )} x^{4} + 6 \, D a^{3} - 6 \, B a^{2} b - 4 \, {\left (5 \, C a b^{2} - 3 \, A b^{3}\right )} x^{3} - 6 \, {\left (2 \, D a^{2} b - B a b^{2}\right )} x^{2} + 6 \, {\left (5 \, C a^{2} b - 3 \, A a b^{2} + {\left (5 \, C a b^{2} - 3 \, A b^{3}\right )} x^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) - 6 \, {\left (5 \, C a^{2} b - 3 \, A a b^{2}\right )} x + 6 \, {\left (3 \, D a^{3} - 2 \, B a^{2} b + {\left (3 \, D a^{2} b - 2 \, B a b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right )}{12 \, {\left (b^{5} x^{2} + a b^{4}\right )}}\right ] \]
[1/12*(3*D*b^3*x^6 + 4*C*b^3*x^5 - 3*(3*D*a*b^2 - 2*B*b^3)*x^4 + 6*D*a^3 - 6*B*a^2*b - 4*(5*C*a*b^2 - 3*A*b^3)*x^3 - 6*(2*D*a^2*b - B*a*b^2)*x^2 - 3 *(5*C*a^2*b - 3*A*a*b^2 + (5*C*a*b^2 - 3*A*b^3)*x^2)*sqrt(-a/b)*log((b*x^2 - 2*b*x*sqrt(-a/b) - a)/(b*x^2 + a)) - 6*(5*C*a^2*b - 3*A*a*b^2)*x + 6*(3 *D*a^3 - 2*B*a^2*b + (3*D*a^2*b - 2*B*a*b^2)*x^2)*log(b*x^2 + a))/(b^5*x^2 + a*b^4), 1/12*(3*D*b^3*x^6 + 4*C*b^3*x^5 - 3*(3*D*a*b^2 - 2*B*b^3)*x^4 + 6*D*a^3 - 6*B*a^2*b - 4*(5*C*a*b^2 - 3*A*b^3)*x^3 - 6*(2*D*a^2*b - B*a*b^ 2)*x^2 + 6*(5*C*a^2*b - 3*A*a*b^2 + (5*C*a*b^2 - 3*A*b^3)*x^2)*sqrt(a/b)*a rctan(b*x*sqrt(a/b)/a) - 6*(5*C*a^2*b - 3*A*a*b^2)*x + 6*(3*D*a^3 - 2*B*a^ 2*b + (3*D*a^2*b - 2*B*a*b^2)*x^2)*log(b*x^2 + a))/(b^5*x^2 + a*b^4)]
Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (160) = 320\).
Time = 1.87 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.90 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^2} \, dx=\frac {C x^{3}}{3 b^{2}} + \frac {D x^{4}}{4 b^{2}} + x^{2} \left (\frac {B}{2 b^{2}} - \frac {D a}{b^{3}}\right ) + x \left (\frac {A}{b^{2}} - \frac {2 C a}{b^{3}}\right ) + \left (\frac {a \left (- 2 B b + 3 D a\right )}{2 b^{4}} - \frac {\sqrt {- a b^{9}} \left (- 3 A b + 5 C a\right )}{4 b^{8}}\right ) \log {\left (x + \frac {4 B a b - 6 D a^{2} + 4 b^{4} \left (\frac {a \left (- 2 B b + 3 D a\right )}{2 b^{4}} - \frac {\sqrt {- a b^{9}} \left (- 3 A b + 5 C a\right )}{4 b^{8}}\right )}{- 3 A b^{2} + 5 C a b} \right )} + \left (\frac {a \left (- 2 B b + 3 D a\right )}{2 b^{4}} + \frac {\sqrt {- a b^{9}} \left (- 3 A b + 5 C a\right )}{4 b^{8}}\right ) \log {\left (x + \frac {4 B a b - 6 D a^{2} + 4 b^{4} \left (\frac {a \left (- 2 B b + 3 D a\right )}{2 b^{4}} + \frac {\sqrt {- a b^{9}} \left (- 3 A b + 5 C a\right )}{4 b^{8}}\right )}{- 3 A b^{2} + 5 C a b} \right )} + \frac {- B a^{2} b + D a^{3} + x \left (A a b^{2} - C a^{2} b\right )}{2 a b^{4} + 2 b^{5} x^{2}} \]
C*x**3/(3*b**2) + D*x**4/(4*b**2) + x**2*(B/(2*b**2) - D*a/b**3) + x*(A/b* *2 - 2*C*a/b**3) + (a*(-2*B*b + 3*D*a)/(2*b**4) - sqrt(-a*b**9)*(-3*A*b + 5*C*a)/(4*b**8))*log(x + (4*B*a*b - 6*D*a**2 + 4*b**4*(a*(-2*B*b + 3*D*a)/ (2*b**4) - sqrt(-a*b**9)*(-3*A*b + 5*C*a)/(4*b**8)))/(-3*A*b**2 + 5*C*a*b) ) + (a*(-2*B*b + 3*D*a)/(2*b**4) + sqrt(-a*b**9)*(-3*A*b + 5*C*a)/(4*b**8) )*log(x + (4*B*a*b - 6*D*a**2 + 4*b**4*(a*(-2*B*b + 3*D*a)/(2*b**4) + sqrt (-a*b**9)*(-3*A*b + 5*C*a)/(4*b**8)))/(-3*A*b**2 + 5*C*a*b)) + (-B*a**2*b + D*a**3 + x*(A*a*b**2 - C*a**2*b))/(2*a*b**4 + 2*b**5*x**2)
Time = 0.28 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.85 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^2} \, dx=\frac {D a^{3} - B a^{2} b - {\left (C a^{2} b - A a b^{2}\right )} x}{2 \, {\left (b^{5} x^{2} + a b^{4}\right )}} + \frac {{\left (5 \, C a^{2} - 3 \, A a b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{3}} + \frac {3 \, D b x^{4} + 4 \, C b x^{3} - 6 \, {\left (2 \, D a - B b\right )} x^{2} - 12 \, {\left (2 \, C a - A b\right )} x}{12 \, b^{3}} + \frac {{\left (3 \, D a^{2} - 2 \, B a b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{4}} \]
1/2*(D*a^3 - B*a^2*b - (C*a^2*b - A*a*b^2)*x)/(b^5*x^2 + a*b^4) + 1/2*(5*C *a^2 - 3*A*a*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) + 1/12*(3*D*b*x^4 + 4*C*b*x^3 - 6*(2*D*a - B*b)*x^2 - 12*(2*C*a - A*b)*x)/b^3 + 1/2*(3*D*a^2 - 2*B*a*b)*log(b*x^2 + a)/b^4
Time = 0.29 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.90 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^2} \, dx=\frac {{\left (5 \, C a^{2} - 3 \, A a b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{3}} + \frac {{\left (3 \, D a^{2} - 2 \, B a b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{4}} + \frac {D a^{3} - B a^{2} b - {\left (C a^{2} b - A a b^{2}\right )} x}{2 \, {\left (b x^{2} + a\right )} b^{4}} + \frac {3 \, D b^{6} x^{4} + 4 \, C b^{6} x^{3} - 12 \, D a b^{5} x^{2} + 6 \, B b^{6} x^{2} - 24 \, C a b^{5} x + 12 \, A b^{6} x}{12 \, b^{8}} \]
1/2*(5*C*a^2 - 3*A*a*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) + 1/2*(3*D*a ^2 - 2*B*a*b)*log(b*x^2 + a)/b^4 + 1/2*(D*a^3 - B*a^2*b - (C*a^2*b - A*a*b ^2)*x)/((b*x^2 + a)*b^4) + 1/12*(3*D*b^6*x^4 + 4*C*b^6*x^3 - 12*D*a*b^5*x^ 2 + 6*B*b^6*x^2 - 24*C*a*b^5*x + 12*A*b^6*x)/b^8
Timed out. \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^2} \, dx=\int \frac {x^4\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (b\,x^2+a\right )}^2} \,d x \]