Integrand size = 28, antiderivative size = 185 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {3 (A b-5 a C) x}{8 a b^3}-\frac {(b B-3 a D) x^2}{2 a b^3}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-(A b-a C) x\right )}{4 a b \left (a+b x^2\right )^2}+\frac {x^3 (A b-5 a C+4 (b B-2 a D) x)}{8 a b^2 \left (a+b x^2\right )}+\frac {3 (A b-5 a C) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{7/2}}+\frac {(b B-3 a D) \log \left (a+b x^2\right )}{2 b^4} \]
-3/8*(A*b-5*C*a)*x/a/b^3-1/2*(B*b-3*D*a)*x^2/a/b^3-1/4*x^4*(a*(B-a*D/b)-(A *b-C*a)*x)/a/b/(b*x^2+a)^2+1/8*x^3*(A*b-5*C*a+4*(B*b-2*D*a)*x)/a/b^2/(b*x^ 2+a)+1/2*(B*b-3*D*a)*ln(b*x^2+a)/b^4+3/8*(A*b-5*C*a)*arctan(x*b^(1/2)/a^(1 /2))/b^(7/2)/a^(1/2)
Time = 0.07 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.75 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=\frac {8 b C x+4 b D x^2+\frac {8 a b B-12 a^2 D-5 A b^2 x+9 a b C x}{a+b x^2}+\frac {2 a \left (a^2 D+A b^2 x-a b (B+C x)\right )}{\left (a+b x^2\right )^2}+\frac {3 \sqrt {b} (A b-5 a C) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a}}+4 (b B-3 a D) \log \left (a+b x^2\right )}{8 b^4} \]
(8*b*C*x + 4*b*D*x^2 + (8*a*b*B - 12*a^2*D - 5*A*b^2*x + 9*a*b*C*x)/(a + b *x^2) + (2*a*(a^2*D + A*b^2*x - a*b*(B + C*x)))/(a + b*x^2)^2 + (3*Sqrt[b] *(A*b - 5*a*C)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/Sqrt[a] + 4*(b*B - 3*a*D)*Log[ a + b*x^2])/(8*b^4)
Time = 0.58 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2335, 25, 2335, 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.
| \(\displaystyle \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 2335 |
| \(\displaystyle -\frac {\int -\frac {x^3 \left (4 a D x^2-(A b-5 a C) x+4 a \left (B-\frac {a D}{b}\right )\right )}{\left (b x^2+a\right )^2}dx}{4 a b}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {x^3 \left (4 a D x^2-(A b-5 a C) x+4 a \left (B-\frac {a D}{b}\right )\right )}{\left (b x^2+a\right )^2}dx}{4 a b}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}\) |
| \(\Big \downarrow \) 2335 |
| \(\displaystyle \frac {\frac {x^3 (4 x (b B-2 a D)-5 a C+A b)}{2 b \left (a+b x^2\right )}-\frac {\int \frac {a x^2 (3 (A b-5 a C)+8 (b B-3 a D) x)}{b x^2+a}dx}{2 a b}}{4 a b}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {x^3 (4 x (b B-2 a D)-5 a C+A b)}{2 b \left (a+b x^2\right )}-\frac {\int \frac {x^2 (3 (A b-5 a C)+8 (b B-3 a D) x)}{b x^2+a}dx}{2 b}}{4 a b}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}\) |
| \(\Big \downarrow \) 523 |
| \(\displaystyle \frac {\frac {x^3 (4 x (b B-2 a D)-5 a C+A b)}{2 b \left (a+b x^2\right )}-\frac {\int \left (3 \left (A-\frac {5 a C}{b}\right )+\frac {8 (b B-3 a D) x}{b}-\frac {3 a (A b-5 a C)+8 a (b B-3 a D) x}{b \left (b x^2+a\right )}\right )dx}{2 b}}{4 a b}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {x^3 (4 x (b B-2 a D)-5 a C+A b)}{2 b \left (a+b x^2\right )}-\frac {-\frac {3 \sqrt {a} (A b-5 a C) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{3/2}}+3 x \left (A-\frac {5 a C}{b}\right )-\frac {4 a (b B-3 a D) \log \left (a+b x^2\right )}{b^2}+\frac {4 x^2 (b B-3 a D)}{b}}{2 b}}{4 a b}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}\) |
-1/4*(x^4*(a*(B - (a*D)/b) - (A*b - a*C)*x))/(a*b*(a + b*x^2)^2) + ((x^3*( A*b - 5*a*C + 4*(b*B - 2*a*D)*x))/(2*b*(a + b*x^2)) - (3*(A - (5*a*C)/b)*x + (4*(b*B - 3*a*D)*x^2)/b - (3*Sqrt[a]*(A*b - 5*a*C)*ArcTan[(Sqrt[b]*x)/S qrt[a]])/b^(3/2) - (4*a*(b*B - 3*a*D)*Log[a + b*x^2])/b^2)/(2*b))/(4*a*b)
3.2.2.3.1 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 = 3.46 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.76
| method | result | size |
| default | \(\frac {\frac {1}{2} D x^{2}+C x}{b^{3}}+\frac {\frac {\left (-\frac {5}{8} b^{2} A +\frac {9}{8} C a b \right ) x^{3}+\left (a b B -\frac {3}{2} D a^{2}\right ) x^{2}-\frac {a \left (3 A b -7 C a \right ) x}{8}+\frac {a^{2} \left (3 B b -5 D a \right )}{4 b}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\left (8 B b -24 D a \right ) \ln \left (b \,x^{2}+a \right )}{16 b}+\frac {\left (3 A b -15 C a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}}{b^{3}}\) | \(140\) |
1/b^3*(1/2*D*x^2+C*x)+1/b^3*(((-5/8*b^2*A+9/8*C*a*b)*x^3+(a*b*B-3/2*D*a^2) *x^2-1/8*a*(3*A*b-7*C*a)*x+1/4*a^2*(3*B*b-5*D*a)/b)/(b*x^2+a)^2+1/16*(8*B* b-24*D*a)/b*ln(b*x^2+a)+1/8*(3*A*b-15*C*a)/(a*b)^(1/2)*arctan(b*x/(a*b)^(1 /2)))
Time = 0.28 (sec) , antiderivative size = 574, normalized size of antiderivative = 3.10 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=\left [\frac {8 \, D a b^{3} x^{6} + 16 \, C a b^{3} x^{5} + 16 \, D a^{2} b^{2} x^{4} - 20 \, D a^{4} + 12 \, B a^{3} b + 10 \, {\left (5 \, C a^{2} b^{2} - A a b^{3}\right )} x^{3} - 16 \, {\left (D a^{3} b - B a^{2} b^{2}\right )} x^{2} + 3 \, {\left ({\left (5 \, C a b^{2} - A b^{3}\right )} x^{4} + 5 \, C a^{3} - A a^{2} b + 2 \, {\left (5 \, C a^{2} b - A a b^{2}\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + 6 \, {\left (5 \, C a^{3} b - A a^{2} b^{2}\right )} x - 8 \, {\left (3 \, D a^{4} - B a^{3} b + {\left (3 \, D a^{2} b^{2} - B a b^{3}\right )} x^{4} + 2 \, {\left (3 \, D a^{3} b - B a^{2} b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right )}{16 \, {\left (a b^{6} x^{4} + 2 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}}, \frac {4 \, D a b^{3} x^{6} + 8 \, C a b^{3} x^{5} + 8 \, D a^{2} b^{2} x^{4} - 10 \, D a^{4} + 6 \, B a^{3} b + 5 \, {\left (5 \, C a^{2} b^{2} - A a b^{3}\right )} x^{3} - 8 \, {\left (D a^{3} b - B a^{2} b^{2}\right )} x^{2} - 3 \, {\left ({\left (5 \, C a b^{2} - A b^{3}\right )} x^{4} + 5 \, C a^{3} - A a^{2} b + 2 \, {\left (5 \, C a^{2} b - A a b^{2}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + 3 \, {\left (5 \, C a^{3} b - A a^{2} b^{2}\right )} x - 4 \, {\left (3 \, D a^{4} - B a^{3} b + {\left (3 \, D a^{2} b^{2} - B a b^{3}\right )} x^{4} + 2 \, {\left (3 \, D a^{3} b - B a^{2} b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right )}{8 \, {\left (a b^{6} x^{4} + 2 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}}\right ] \]
[1/16*(8*D*a*b^3*x^6 + 16*C*a*b^3*x^5 + 16*D*a^2*b^2*x^4 - 20*D*a^4 + 12*B *a^3*b + 10*(5*C*a^2*b^2 - A*a*b^3)*x^3 - 16*(D*a^3*b - B*a^2*b^2)*x^2 + 3 *((5*C*a*b^2 - A*b^3)*x^4 + 5*C*a^3 - A*a^2*b + 2*(5*C*a^2*b - A*a*b^2)*x^ 2)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) + 6*(5*C*a^3*b - A*a^2*b^2)*x - 8*(3*D*a^4 - B*a^3*b + (3*D*a^2*b^2 - B*a*b^3)*x^4 + 2*( 3*D*a^3*b - B*a^2*b^2)*x^2)*log(b*x^2 + a))/(a*b^6*x^4 + 2*a^2*b^5*x^2 + a ^3*b^4), 1/8*(4*D*a*b^3*x^6 + 8*C*a*b^3*x^5 + 8*D*a^2*b^2*x^4 - 10*D*a^4 + 6*B*a^3*b + 5*(5*C*a^2*b^2 - A*a*b^3)*x^3 - 8*(D*a^3*b - B*a^2*b^2)*x^2 - 3*((5*C*a*b^2 - A*b^3)*x^4 + 5*C*a^3 - A*a^2*b + 2*(5*C*a^2*b - A*a*b^2)* x^2)*sqrt(a*b)*arctan(sqrt(a*b)*x/a) + 3*(5*C*a^3*b - A*a^2*b^2)*x - 4*(3* D*a^4 - B*a^3*b + (3*D*a^2*b^2 - B*a*b^3)*x^4 + 2*(3*D*a^3*b - B*a^2*b^2)* x^2)*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. 357 vs. \(2 (172) = 344\).
Time = 98.08 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.93 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=\frac {C x}{b^{3}} + \frac {D x^{2}}{2 b^{3}} + \left (- \frac {- B b + 3 D a}{2 b^{4}} - \frac {3 \sqrt {- a b^{9}} \left (- A b + 5 C a\right )}{16 a b^{8}}\right ) \log {\left (x + \frac {8 B a b - 24 D a^{2} - 16 a b^{4} \left (- \frac {- B b + 3 D a}{2 b^{4}} - \frac {3 \sqrt {- a b^{9}} \left (- A b + 5 C a\right )}{16 a b^{8}}\right )}{- 3 A b^{2} + 15 C a b} \right )} + \left (- \frac {- B b + 3 D a}{2 b^{4}} + \frac {3 \sqrt {- a b^{9}} \left (- A b + 5 C a\right )}{16 a b^{8}}\right ) \log {\left (x + \frac {8 B a b - 24 D a^{2} - 16 a b^{4} \left (- \frac {- B b + 3 D a}{2 b^{4}} + \frac {3 \sqrt {- a b^{9}} \left (- A b + 5 C a\right )}{16 a b^{8}}\right )}{- 3 A b^{2} + 15 C a b} \right )} + \frac {6 B a^{2} b - 10 D a^{3} + x^{3} \left (- 5 A b^{3} + 9 C a b^{2}\right ) + x^{2} \cdot \left (8 B a b^{2} - 12 D a^{2} b\right ) + x \left (- 3 A a b^{2} + 7 C a^{2} b\right )}{8 a^{2} b^{4} + 16 a b^{5} x^{2} + 8 b^{6} x^{4}} \]
C*x/b**3 + D*x**2/(2*b**3) + (-(-B*b + 3*D*a)/(2*b**4) - 3*sqrt(-a*b**9)*( -A*b + 5*C*a)/(16*a*b**8))*log(x + (8*B*a*b - 24*D*a**2 - 16*a*b**4*(-(-B* b + 3*D*a)/(2*b**4) - 3*sqrt(-a*b**9)*(-A*b + 5*C*a)/(16*a*b**8)))/(-3*A*b **2 + 15*C*a*b)) + (-(-B*b + 3*D*a)/(2*b**4) + 3*sqrt(-a*b**9)*(-A*b + 5*C *a)/(16*a*b**8))*log(x + (8*B*a*b - 24*D*a**2 - 16*a*b**4*(-(-B*b + 3*D*a) /(2*b**4) + 3*sqrt(-a*b**9)*(-A*b + 5*C*a)/(16*a*b**8)))/(-3*A*b**2 + 15*C *a*b)) + (6*B*a**2*b - 10*D*a**3 + x**3*(-5*A*b**3 + 9*C*a*b**2) + x**2*(8 *B*a*b**2 - 12*D*a**2*b) + x*(-3*A*a*b**2 + 7*C*a**2*b))/(8*a**2*b**4 + 16 *a*b**5*x**2 + 8*b**6*x**4)
Time = 0.29 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.89 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {10 \, D a^{3} - 6 \, B a^{2} b - {\left (9 \, C a b^{2} - 5 \, A b^{3}\right )} x^{3} + 4 \, {\left (3 \, D a^{2} b - 2 \, B a b^{2}\right )} x^{2} - {\left (7 \, C a^{2} b - 3 \, A a b^{2}\right )} x}{8 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}} - \frac {3 \, {\left (5 \, C a - A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3}} + \frac {D x^{2} + 2 \, C x}{2 \, b^{3}} - \frac {{\left (3 \, D a - B b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{4}} \]
-1/8*(10*D*a^3 - 6*B*a^2*b - (9*C*a*b^2 - 5*A*b^3)*x^3 + 4*(3*D*a^2*b - 2* B*a*b^2)*x^2 - (7*C*a^2*b - 3*A*a*b^2)*x)/(b^6*x^4 + 2*a*b^5*x^2 + a^2*b^4 ) - 3/8*(5*C*a - A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) + 1/2*(D*x^2 + 2*C*x)/b^3 - 1/2*(3*D*a - B*b)*log(b*x^2 + a)/b^4
Time = 0.29 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.85 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {3 \, {\left (5 \, C a - A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3}} - \frac {{\left (3 \, D a - B b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{4}} + \frac {D b^{3} x^{2} + 2 \, C b^{3} x}{2 \, b^{6}} - \frac {10 \, D a^{3} - 6 \, B a^{2} b - {\left (9 \, C a b^{2} - 5 \, A b^{3}\right )} x^{3} + 4 \, {\left (3 \, D a^{2} b - 2 \, B a b^{2}\right )} x^{2} - {\left (7 \, C a^{2} b - 3 \, A a b^{2}\right )} x}{8 \, {\left (b x^{2} + a\right )}^{2} b^{4}} \]
-3/8*(5*C*a - A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) - 1/2*(3*D*a - B* b)*log(b*x^2 + a)/b^4 + 1/2*(D*b^3*x^2 + 2*C*b^3*x)/b^6 - 1/8*(10*D*a^3 - 6*B*a^2*b - (9*C*a*b^2 - 5*A*b^3)*x^3 + 4*(3*D*a^2*b - 2*B*a*b^2)*x^2 - (7 *C*a^2*b - 3*A*a*b^2)*x)/((b*x^2 + a)^2*b^4)
Time = 6.17 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.25 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=\frac {\frac {7\,C\,a^2\,x}{8}+\frac {9\,C\,b\,a\,x^3}{8}}{a^2\,b^3+2\,a\,b^4\,x^2+b^5\,x^4}-\frac {\frac {5\,A\,x^3}{8\,b}+\frac {3\,A\,a\,x}{8\,b^2}}{a^2+2\,a\,b\,x^2+b^2\,x^4}+\frac {\frac {3\,B\,a^2}{4\,b^3}+\frac {B\,a\,x^2}{b^2}}{a^2+2\,a\,b\,x^2+b^2\,x^4}-\frac {D\,\left (3\,a\,\ln \left (b\,x^2+a\right )-b\,x^2+\frac {3\,a^2}{b\,x^2+a}-\frac {a^3}{2\,{\left (b\,x^2+a\right )}^2}\right )}{2\,b^4}+\frac {B\,\ln \left (b\,x^2+a\right )}{2\,b^3}+\frac {C\,x}{b^3}+\frac {3\,A\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,\sqrt {a}\,b^{5/2}}-\frac {15\,C\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,b^{7/2}} \]
((7*C*a^2*x)/8 + (9*C*a*b*x^3)/8)/(a^2*b^3 + b^5*x^4 + 2*a*b^4*x^2) - ((5* A*x^3)/(8*b) + (3*A*a*x)/(8*b^2))/(a^2 + b^2*x^4 + 2*a*b*x^2) + ((3*B*a^2) /(4*b^3) + (B*a*x^2)/b^2)/(a^2 + b^2*x^4 + 2*a*b*x^2) - (D*(3*a*log(a + b* x^2) - b*x^2 + (3*a^2)/(a + b*x^2) - a^3/(2*(a + b*x^2)^2)))/(2*b^4) + (B* log(a + b*x^2))/(2*b^3) + (C*x)/b^3 + (3*A*atan((b^(1/2)*x)/a^(1/2)))/(8*a ^(1/2)*b^(5/2)) - (15*C*a^(1/2)*atan((b^(1/2)*x)/a^(1/2)))/(8*b^(7/2))