Integrand size = 25, antiderivative size = 116 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^3} \, dx=\frac {-a \left (B-\frac {a D}{b}\right )+(A b-a C) x}{4 a b \left (a+b x^2\right )^2}-\frac {4 a^2 D-b (3 A b+a C) x}{8 a^2 b^2 \left (a+b x^2\right )}+\frac {(3 A b+a C) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} b^{3/2}} \]
1/4*(-a*(B-a*D/b)+(A*b-C*a)*x)/a/b/(b*x^2+a)^2+1/8*(-4*a^2*D+b*(3*A*b+C*a) *x)/a^2/b^2/(b*x^2+a)+1/8*(3*A*b+C*a)*arctan(x*b^(1/2)/a^(1/2))/a^(5/2)/b^ (3/2)
Time = 0.07 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.90 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^3} \, dx=\frac {\frac {\sqrt {a} \left (-2 a^3 D+3 A b^3 x^3+a b^2 x \left (5 A+C x^2\right )-a^2 b (2 B+x (C+4 D x))\right )}{\left (a+b x^2\right )^2}+\sqrt {b} (3 A b+a C) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} b^2} \]
((Sqrt[a]*(-2*a^3*D + 3*A*b^3*x^3 + a*b^2*x*(5*A + C*x^2) - a^2*b*(2*B + x *(C + 4*D*x))))/(a + b*x^2)^2 + Sqrt[b]*(3*A*b + a*C)*ArcTan[(Sqrt[b]*x)/S qrt[a]])/(8*a^(5/2)*b^2)
Time = 0.27 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2345, 25, 27, 454, 218}
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.
\(\displaystyle \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle -\frac {\int -\frac {b \left (3 A+\frac {a C}{b}\right )+4 a D x}{b \left (b x^2+a\right )^2}dx}{4 a}-\frac {a \left (B-\frac {a D}{b}\right )-x (A b-a C)}{4 a b \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {3 A b+a C+4 a D x}{b \left (b x^2+a\right )^2}dx}{4 a}-\frac {a \left (B-\frac {a D}{b}\right )-x (A b-a C)}{4 a b \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {3 A b+a C+4 a D x}{\left (b x^2+a\right )^2}dx}{4 a b}-\frac {a \left (B-\frac {a D}{b}\right )-x (A b-a C)}{4 a b \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 454 |
\(\displaystyle \frac {\frac {(a C+3 A b) \int \frac {1}{b x^2+a}dx}{2 a}-\frac {4 a^2 D-b x (a C+3 A b)}{2 a b \left (a+b x^2\right )}}{4 a b}-\frac {a \left (B-\frac {a D}{b}\right )-x (A b-a C)}{4 a b \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {(a C+3 A b) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b}}-\frac {4 a^2 D-b x (a C+3 A b)}{2 a b \left (a+b x^2\right )}}{4 a b}-\frac {a \left (B-\frac {a D}{b}\right )-x (A b-a C)}{4 a b \left (a+b x^2\right )^2}\) |
-1/4*(a*(B - (a*D)/b) - (A*b - a*C)*x)/(a*b*(a + b*x^2)^2) + (-1/2*(4*a^2* D - b*(3*A*b + a*C)*x)/(a*b*(a + b*x^2)) + ((3*A*b + a*C)*ArcTan[(Sqrt[b]* x)/Sqrt[a]])/(2*a^(3/2)*Sqrt[b]))/(4*a*b)
3.2.6.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*d - b*c*x)/(2*a*b*(p + 1)))*(a + b*x^2)^(p + 1), x] + Simp[c*((2*p + 3)/(2*a *(p + 1))) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d}, x] && L tQ[p, -1] && NeQ[p, -3/2]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[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[(a*g - b *f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
Time = 3.43 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.84
method | result | size |
default | \(\frac {\frac {\left (3 A b +C a \right ) x^{3}}{8 a^{2}}-\frac {D x^{2}}{2 b}+\frac {\left (5 A b -C a \right ) x}{8 a b}-\frac {B b +D a}{4 b^{2}}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\left (3 A b +C a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 a^{2} b \sqrt {a b}}\) | \(98\) |
(1/8*(3*A*b+C*a)/a^2*x^3-1/2*D*x^2/b+1/8*(5*A*b-C*a)/a/b*x-1/4*(B*b+D*a)/b ^2)/(b*x^2+a)^2+1/8*(3*A*b+C*a)/a^2/b/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))
Time = 0.28 (sec) , antiderivative size = 346, normalized size of antiderivative = 2.98 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^3} \, dx=\left [-\frac {8 \, D a^{3} b x^{2} + 4 \, D a^{4} + 4 \, B a^{3} b - 2 \, {\left (C a^{2} b^{2} + 3 \, A a b^{3}\right )} x^{3} + {\left ({\left (C a b^{2} + 3 \, A b^{3}\right )} x^{4} + C a^{3} + 3 \, A a^{2} b + 2 \, {\left (C a^{2} b + 3 \, A a b^{2}\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + 2 \, {\left (C a^{3} b - 5 \, A a^{2} b^{2}\right )} x}{16 \, {\left (a^{3} b^{4} x^{4} + 2 \, a^{4} b^{3} x^{2} + a^{5} b^{2}\right )}}, -\frac {4 \, D a^{3} b x^{2} + 2 \, D a^{4} + 2 \, B a^{3} b - {\left (C a^{2} b^{2} + 3 \, A a b^{3}\right )} x^{3} - {\left ({\left (C a b^{2} + 3 \, A b^{3}\right )} x^{4} + C a^{3} + 3 \, A a^{2} b + 2 \, {\left (C a^{2} b + 3 \, A a b^{2}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + {\left (C a^{3} b - 5 \, A a^{2} b^{2}\right )} x}{8 \, {\left (a^{3} b^{4} x^{4} + 2 \, a^{4} b^{3} x^{2} + a^{5} b^{2}\right )}}\right ] \]
[-1/16*(8*D*a^3*b*x^2 + 4*D*a^4 + 4*B*a^3*b - 2*(C*a^2*b^2 + 3*A*a*b^3)*x^ 3 + ((C*a*b^2 + 3*A*b^3)*x^4 + C*a^3 + 3*A*a^2*b + 2*(C*a^2*b + 3*A*a*b^2) *x^2)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) + 2*(C*a^3* b - 5*A*a^2*b^2)*x)/(a^3*b^4*x^4 + 2*a^4*b^3*x^2 + a^5*b^2), -1/8*(4*D*a^3 *b*x^2 + 2*D*a^4 + 2*B*a^3*b - (C*a^2*b^2 + 3*A*a*b^3)*x^3 - ((C*a*b^2 + 3 *A*b^3)*x^4 + C*a^3 + 3*A*a^2*b + 2*(C*a^2*b + 3*A*a*b^2)*x^2)*sqrt(a*b)*a rctan(sqrt(a*b)*x/a) + (C*a^3*b - 5*A*a^2*b^2)*x)/(a^3*b^4*x^4 + 2*a^4*b^3 *x^2 + a^5*b^2)]
Time = 3.41 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.59 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^3} \, dx=- \frac {\sqrt {- \frac {1}{a^{5} b^{3}}} \cdot \left (3 A b + C a\right ) \log {\left (- a^{3} b \sqrt {- \frac {1}{a^{5} b^{3}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{a^{5} b^{3}}} \cdot \left (3 A b + C a\right ) \log {\left (a^{3} b \sqrt {- \frac {1}{a^{5} b^{3}}} + x \right )}}{16} + \frac {- 2 B a^{2} b - 2 D a^{3} - 4 D a^{2} b x^{2} + x^{3} \cdot \left (3 A b^{3} + C a b^{2}\right ) + x \left (5 A a b^{2} - C a^{2} b\right )}{8 a^{4} b^{2} + 16 a^{3} b^{3} x^{2} + 8 a^{2} b^{4} x^{4}} \]
-sqrt(-1/(a**5*b**3))*(3*A*b + C*a)*log(-a**3*b*sqrt(-1/(a**5*b**3)) + x)/ 16 + sqrt(-1/(a**5*b**3))*(3*A*b + C*a)*log(a**3*b*sqrt(-1/(a**5*b**3)) + x)/16 + (-2*B*a**2*b - 2*D*a**3 - 4*D*a**2*b*x**2 + x**3*(3*A*b**3 + C*a*b **2) + x*(5*A*a*b**2 - C*a**2*b))/(8*a**4*b**2 + 16*a**3*b**3*x**2 + 8*a** 2*b**4*x**4)
Time = 0.28 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.05 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^3} \, dx=-\frac {4 \, D a^{2} b x^{2} + 2 \, D a^{3} + 2 \, B a^{2} b - {\left (C a b^{2} + 3 \, A b^{3}\right )} x^{3} + {\left (C a^{2} b - 5 \, A a b^{2}\right )} x}{8 \, {\left (a^{2} b^{4} x^{4} + 2 \, a^{3} b^{3} x^{2} + a^{4} b^{2}\right )}} + \frac {{\left (C a + 3 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} b} \]
-1/8*(4*D*a^2*b*x^2 + 2*D*a^3 + 2*B*a^2*b - (C*a*b^2 + 3*A*b^3)*x^3 + (C*a ^2*b - 5*A*a*b^2)*x)/(a^2*b^4*x^4 + 2*a^3*b^3*x^2 + a^4*b^2) + 1/8*(C*a + 3*A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^2*b)
Time = 0.29 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.91 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^3} \, dx=\frac {{\left (C a + 3 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} b} + \frac {C a b^{2} x^{3} + 3 \, A b^{3} x^{3} - 4 \, D a^{2} b x^{2} - C a^{2} b x + 5 \, A a b^{2} x - 2 \, D a^{3} - 2 \, B a^{2} b}{8 \, {\left (b x^{2} + a\right )}^{2} a^{2} b^{2}} \]
1/8*(C*a + 3*A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^2*b) + 1/8*(C*a*b^2*x ^3 + 3*A*b^3*x^3 - 4*D*a^2*b*x^2 - C*a^2*b*x + 5*A*a*b^2*x - 2*D*a^3 - 2*B *a^2*b)/((b*x^2 + a)^2*a^2*b^2)
Time = 5.96 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.41 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^3} \, dx=\frac {\frac {C\,x^3}{8\,a}-\frac {C\,x}{8\,b}}{a^2+2\,a\,b\,x^2+b^2\,x^4}+\frac {\frac {5\,A\,x}{8\,a}+\frac {3\,A\,b\,x^3}{8\,a^2}}{a^2+2\,a\,b\,x^2+b^2\,x^4}-\frac {B}{4\,b\,\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}-\frac {\left (2\,b\,x^2+a\right )\,D}{4\,b^2\,{\left (b\,x^2+a\right )}^2}+\frac {3\,A\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,a^{5/2}\,\sqrt {b}}+\frac {C\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,a^{3/2}\,b^{3/2}} \]
((C*x^3)/(8*a) - (C*x)/(8*b))/(a^2 + b^2*x^4 + 2*a*b*x^2) + ((5*A*x)/(8*a) + (3*A*b*x^3)/(8*a^2))/(a^2 + b^2*x^4 + 2*a*b*x^2) - B/(4*b*(a^2 + b^2*x^ 4 + 2*a*b*x^2)) - ((a + 2*b*x^2)*D)/(4*b^2*(a + b*x^2)^2) + (3*A*atan((b^( 1/2)*x)/a^(1/2)))/(8*a^(5/2)*b^(1/2)) + (C*atan((b^(1/2)*x)/a^(1/2)))/(8*a ^(3/2)*b^(3/2))