Integrand size = 26, antiderivative size = 119 \[ \int \frac {x \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {x \left (a \left (B-\frac {a D}{b}\right )-(A b-a C) x\right )}{4 a b \left (a+b x^2\right )^2}-\frac {2 (A b+a C)-(b B-5 a D) x}{8 a b^2 \left (a+b x^2\right )}+\frac {(b B+3 a D) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{5/2}} \]
-1/4*x*(a*(B-a*D/b)-(A*b-C*a)*x)/a/b/(b*x^2+a)^2+1/8*(-2*A*b-2*C*a+(B*b-5* D*a)*x)/a/b^2/(b*x^2+a)+1/8*(B*b+3*D*a)*arctan(x*b^(1/2)/a^(1/2))/a^(3/2)/ b^(5/2)
Time = 0.07 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.83 \[ \int \frac {x \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=\frac {\frac {\sqrt {b} \left (b^2 B x^3-a^2 (2 C+3 D x)-a b \left (2 A+x \left (B+4 C x+5 D x^2\right )\right )\right )}{a \left (a+b x^2\right )^2}+\frac {(b B+3 a D) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2}}}{8 b^{5/2}} \]
((Sqrt[b]*(b^2*B*x^3 - a^2*(2*C + 3*D*x) - a*b*(2*A + x*(B + 4*C*x + 5*D*x ^2))))/(a*(a + b*x^2)^2) + ((b*B + 3*a*D)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/a^( 3/2))/(8*b^(5/2))
Time = 0.35 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2335, 25, 2345, 25, 27, 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.
\(\displaystyle \int \frac {x \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 2335 |
\(\displaystyle -\frac {\int -\frac {4 a D x^2+2 (A b+a C) x+\frac {a (b B-a D)}{b}}{\left (b x^2+a\right )^2}dx}{4 a b}-\frac {x \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {4 a D x^2+2 (A b+a C) x+\frac {a (b B-a D)}{b}}{\left (b x^2+a\right )^2}dx}{4 a b}-\frac {x \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle \frac {-\frac {\int -\frac {a (b B+3 a D)}{b \left (b x^2+a\right )}dx}{2 a}-\frac {2 (a C+A b)-x (b B-5 a D)}{2 b \left (a+b x^2\right )}}{4 a b}-\frac {x \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\int \frac {a (b B+3 a D)}{b \left (b x^2+a\right )}dx}{2 a}-\frac {2 (a C+A b)-x (b B-5 a D)}{2 b \left (a+b x^2\right )}}{4 a b}-\frac {x \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {(3 a D+b B) \int \frac {1}{b x^2+a}dx}{2 b}-\frac {2 (a C+A b)-x (b B-5 a D)}{2 b \left (a+b x^2\right )}}{4 a b}-\frac {x \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) (3 a D+b B)}{2 \sqrt {a} b^{3/2}}-\frac {2 (a C+A b)-x (b B-5 a D)}{2 b \left (a+b x^2\right )}}{4 a b}-\frac {x \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}\) |
-1/4*(x*(a*(B - (a*D)/b) - (A*b - a*C)*x))/(a*b*(a + b*x^2)^2) + (-1/2*(2* (A*b + a*C) - (b*B - 5*a*D)*x)/(b*(a + b*x^2)) + ((b*B + 3*a*D)*ArcTan[(Sq rt[b]*x)/Sqrt[a]])/(2*Sqrt[a]*b^(3/2)))/(4*a*b)
3.2.5.3.1 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[(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]
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 = 3.49 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.82
method | result | size |
default | \(\frac {\frac {\left (B b -5 D a \right ) x^{3}}{8 a b}-\frac {C \,x^{2}}{2 b}-\frac {\left (B b +3 D a \right ) x}{8 b^{2}}-\frac {A b +C a}{4 b^{2}}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\left (B b +3 D a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 b^{2} a \sqrt {a b}}\) | \(97\) |
(1/8*(B*b-5*D*a)/a/b*x^3-1/2*C*x^2/b-1/8*(B*b+3*D*a)/b^2*x-1/4*(A*b+C*a)/b ^2)/(b*x^2+a)^2+1/8*(B*b+3*D*a)/b^2/a/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))
Time = 0.29 (sec) , antiderivative size = 357, normalized size of antiderivative = 3.00 \[ \int \frac {x \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=\left [-\frac {8 \, C a^{2} b^{2} x^{2} + 4 \, C a^{3} b + 4 \, A a^{2} b^{2} + 2 \, {\left (5 \, D a^{2} b^{2} - B a b^{3}\right )} x^{3} + {\left ({\left (3 \, D a b^{2} + B b^{3}\right )} x^{4} + 3 \, D a^{3} + B a^{2} b + 2 \, {\left (3 \, D a^{2} b + B a b^{2}\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + 2 \, {\left (3 \, D a^{3} b + B a^{2} b^{2}\right )} x}{16 \, {\left (a^{2} b^{5} x^{4} + 2 \, a^{3} b^{4} x^{2} + a^{4} b^{3}\right )}}, -\frac {4 \, C a^{2} b^{2} x^{2} + 2 \, C a^{3} b + 2 \, A a^{2} b^{2} + {\left (5 \, D a^{2} b^{2} - B a b^{3}\right )} x^{3} - {\left ({\left (3 \, D a b^{2} + B b^{3}\right )} x^{4} + 3 \, D a^{3} + B a^{2} b + 2 \, {\left (3 \, D a^{2} b + B a b^{2}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + {\left (3 \, D a^{3} b + B a^{2} b^{2}\right )} x}{8 \, {\left (a^{2} b^{5} x^{4} + 2 \, a^{3} b^{4} x^{2} + a^{4} b^{3}\right )}}\right ] \]
[-1/16*(8*C*a^2*b^2*x^2 + 4*C*a^3*b + 4*A*a^2*b^2 + 2*(5*D*a^2*b^2 - B*a*b ^3)*x^3 + ((3*D*a*b^2 + B*b^3)*x^4 + 3*D*a^3 + B*a^2*b + 2*(3*D*a^2*b + B* a*b^2)*x^2)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) + 2*( 3*D*a^3*b + B*a^2*b^2)*x)/(a^2*b^5*x^4 + 2*a^3*b^4*x^2 + a^4*b^3), -1/8*(4 *C*a^2*b^2*x^2 + 2*C*a^3*b + 2*A*a^2*b^2 + (5*D*a^2*b^2 - B*a*b^3)*x^3 - ( (3*D*a*b^2 + B*b^3)*x^4 + 3*D*a^3 + B*a^2*b + 2*(3*D*a^2*b + B*a*b^2)*x^2) *sqrt(a*b)*arctan(sqrt(a*b)*x/a) + (3*D*a^3*b + B*a^2*b^2)*x)/(a^2*b^5*x^4 + 2*a^3*b^4*x^2 + a^4*b^3)]
Time = 7.84 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.50 \[ \int \frac {x \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=- \frac {\sqrt {- \frac {1}{a^{3} b^{5}}} \left (B b + 3 D a\right ) \log {\left (- a^{2} b^{2} \sqrt {- \frac {1}{a^{3} b^{5}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{a^{3} b^{5}}} \left (B b + 3 D a\right ) \log {\left (a^{2} b^{2} \sqrt {- \frac {1}{a^{3} b^{5}}} + x \right )}}{16} + \frac {- 2 A a b - 2 C a^{2} - 4 C a b x^{2} + x^{3} \left (B b^{2} - 5 D a b\right ) + x \left (- B a b - 3 D a^{2}\right )}{8 a^{3} b^{2} + 16 a^{2} b^{3} x^{2} + 8 a b^{4} x^{4}} \]
-sqrt(-1/(a**3*b**5))*(B*b + 3*D*a)*log(-a**2*b**2*sqrt(-1/(a**3*b**5)) + x)/16 + sqrt(-1/(a**3*b**5))*(B*b + 3*D*a)*log(a**2*b**2*sqrt(-1/(a**3*b** 5)) + x)/16 + (-2*A*a*b - 2*C*a**2 - 4*C*a*b*x**2 + x**3*(B*b**2 - 5*D*a*b ) + x*(-B*a*b - 3*D*a**2))/(8*a**3*b**2 + 16*a**2*b**3*x**2 + 8*a*b**4*x** 4)
Time = 0.28 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.93 \[ \int \frac {x \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {4 \, C a b x^{2} + {\left (5 \, D a b - B b^{2}\right )} x^{3} + 2 \, C a^{2} + 2 \, A a b + {\left (3 \, D a^{2} + B a b\right )} x}{8 \, {\left (a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\right )}} + \frac {{\left (3 \, D a + B b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a b^{2}} \]
-1/8*(4*C*a*b*x^2 + (5*D*a*b - B*b^2)*x^3 + 2*C*a^2 + 2*A*a*b + (3*D*a^2 + B*a*b)*x)/(a*b^4*x^4 + 2*a^2*b^3*x^2 + a^3*b^2) + 1/8*(3*D*a + B*b)*arcta n(b*x/sqrt(a*b))/(sqrt(a*b)*a*b^2)
Time = 0.28 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.82 \[ \int \frac {x \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=\frac {{\left (3 \, D a + B b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a b^{2}} - \frac {5 \, D a b x^{3} - B b^{2} x^{3} + 4 \, C a b x^{2} + 3 \, D a^{2} x + B a b x + 2 \, C a^{2} + 2 \, A a b}{8 \, {\left (b x^{2} + a\right )}^{2} a b^{2}} \]
1/8*(3*D*a + B*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a*b^2) - 1/8*(5*D*a*b*x ^3 - B*b^2*x^3 + 4*C*a*b*x^2 + 3*D*a^2*x + B*a*b*x + 2*C*a^2 + 2*A*a*b)/(( b*x^2 + a)^2*a*b^2)
Timed out. \[ \int \frac {x \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=\int \frac {x\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (b\,x^2+a\right )}^3} \,d x \]