Integrand size = 20, antiderivative size = 94 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {B x}{b^3}+\frac {a (A b-a B) x}{4 b^3 \left (a+b x^2\right )^2}-\frac {(5 A b-9 a B) x}{8 b^3 \left (a+b x^2\right )}+\frac {3 (A b-5 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{7/2}} \]
B*x/b^3+1/4*a*(A*b-B*a)*x/b^3/(b*x^2+a)^2-1/8*(5*A*b-9*B*a)*x/b^3/(b*x^2+a )+3/8*(A*b-5*B*a)*arctan(x*b^(1/2)/a^(1/2))/b^(7/2)/a^(1/2)
Time = 0.05 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.97 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {x \left (15 a^2 B+b^2 x^2 \left (-5 A+8 B x^2\right )+a \left (-3 A b+25 b B x^2\right )\right )}{8 b^3 \left (a+b x^2\right )^2}+\frac {3 (A b-5 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{7/2}} \]
(x*(15*a^2*B + b^2*x^2*(-5*A + 8*B*x^2) + a*(-3*A*b + 25*b*B*x^2)))/(8*b^3 *(a + b*x^2)^2) + (3*(A*b - 5*a*B)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*Sqrt[a] *b^(7/2))
Time = 0.25 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {360, 1471, 27, 299, 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.
\(\displaystyle \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 360 |
\(\displaystyle \frac {a x (A b-a B)}{4 b^3 \left (a+b x^2\right )^2}-\frac {\int \frac {-4 b^2 B x^4-4 b (A b-a B) x^2+a (A b-a B)}{\left (b x^2+a\right )^2}dx}{4 b^3}\) |
\(\Big \downarrow \) 1471 |
\(\displaystyle \frac {a x (A b-a B)}{4 b^3 \left (a+b x^2\right )^2}-\frac {\frac {x (5 A b-9 a B)}{2 \left (a+b x^2\right )}-\frac {\int \frac {a \left (8 b B x^2+3 A b-7 a B\right )}{b x^2+a}dx}{2 a}}{4 b^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a x (A b-a B)}{4 b^3 \left (a+b x^2\right )^2}-\frac {\frac {x (5 A b-9 a B)}{2 \left (a+b x^2\right )}-\frac {1}{2} \int \frac {8 b B x^2+3 A b-7 a B}{b x^2+a}dx}{4 b^3}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {a x (A b-a B)}{4 b^3 \left (a+b x^2\right )^2}-\frac {\frac {1}{2} \left (-3 (A b-5 a B) \int \frac {1}{b x^2+a}dx-8 B x\right )+\frac {x (5 A b-9 a B)}{2 \left (a+b x^2\right )}}{4 b^3}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {a x (A b-a B)}{4 b^3 \left (a+b x^2\right )^2}-\frac {\frac {1}{2} \left (-\frac {3 (A b-5 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}}-8 B x\right )+\frac {x (5 A b-9 a B)}{2 \left (a+b x^2\right )}}{4 b^3}\) |
(a*(A*b - a*B)*x)/(4*b^3*(a + b*x^2)^2) - (((5*A*b - 9*a*B)*x)/(2*(a + b*x ^2)) + (-8*B*x - (3*(A*b - 5*a*B)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(Sqrt[a]*Sq rt[b]))/2)/(4*b^3)
3.2.1.3.1 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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[(a + b*x^2)^(p + 1)*Expan dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & & (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2 , x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], x , 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q + 1)) Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^ 2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]
Time = 2.54 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.82
method | result | size |
default | \(\frac {B x}{b^{3}}+\frac {\frac {\left (-\frac {5}{8} b^{2} A +\frac {9}{8} a b B \right ) x^{3}-\frac {a \left (3 A b -7 B a \right ) x}{8}}{\left (b \,x^{2}+a \right )^{2}}+\frac {3 \left (A b -5 B a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}}{b^{3}}\) | \(77\) |
risch | \(\frac {B x}{b^{3}}+\frac {\left (-\frac {5}{8} b^{2} A +\frac {9}{8} a b B \right ) x^{3}-\frac {a \left (3 A b -7 B a \right ) x}{8}}{b^{3} \left (b \,x^{2}+a \right )^{2}}-\frac {3 \ln \left (b x +\sqrt {-a b}\right ) A}{16 b^{2} \sqrt {-a b}}+\frac {15 \ln \left (b x +\sqrt {-a b}\right ) B a}{16 b^{3} \sqrt {-a b}}+\frac {3 \ln \left (-b x +\sqrt {-a b}\right ) A}{16 b^{2} \sqrt {-a b}}-\frac {15 \ln \left (-b x +\sqrt {-a b}\right ) B a}{16 b^{3} \sqrt {-a b}}\) | \(147\) |
B*x/b^3+1/b^3*(((-5/8*b^2*A+9/8*a*b*B)*x^3-1/8*a*(3*A*b-7*B*a)*x)/(b*x^2+a )^2+3/8*(A*b-5*B*a)/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))
Time = 0.32 (sec) , antiderivative size = 328, normalized size of antiderivative = 3.49 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\left [\frac {16 \, B a b^{3} x^{5} + 10 \, {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x^{3} + 3 \, {\left ({\left (5 \, B a b^{2} - A b^{3}\right )} x^{4} + 5 \, B a^{3} - A a^{2} b + 2 \, {\left (5 \, B a^{2} b - A a b^{2}\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + 6 \, {\left (5 \, B a^{3} b - A a^{2} b^{2}\right )} x}{16 \, {\left (a b^{6} x^{4} + 2 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}}, \frac {8 \, B a b^{3} x^{5} + 5 \, {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x^{3} - 3 \, {\left ({\left (5 \, B a b^{2} - A b^{3}\right )} x^{4} + 5 \, B a^{3} - A a^{2} b + 2 \, {\left (5 \, B a^{2} b - A a b^{2}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + 3 \, {\left (5 \, B a^{3} b - A a^{2} b^{2}\right )} x}{8 \, {\left (a b^{6} x^{4} + 2 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}}\right ] \]
[1/16*(16*B*a*b^3*x^5 + 10*(5*B*a^2*b^2 - A*a*b^3)*x^3 + 3*((5*B*a*b^2 - A *b^3)*x^4 + 5*B*a^3 - A*a^2*b + 2*(5*B*a^2*b - A*a*b^2)*x^2)*sqrt(-a*b)*lo g((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) + 6*(5*B*a^3*b - A*a^2*b^2)*x) /(a*b^6*x^4 + 2*a^2*b^5*x^2 + a^3*b^4), 1/8*(8*B*a*b^3*x^5 + 5*(5*B*a^2*b^ 2 - A*a*b^3)*x^3 - 3*((5*B*a*b^2 - A*b^3)*x^4 + 5*B*a^3 - A*a^2*b + 2*(5*B *a^2*b - A*a*b^2)*x^2)*sqrt(a*b)*arctan(sqrt(a*b)*x/a) + 3*(5*B*a^3*b - A* a^2*b^2)*x)/(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. 194 vs. \(2 (92) = 184\).
Time = 0.58 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.06 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {B x}{b^{3}} + \frac {3 \sqrt {- \frac {1}{a b^{7}}} \left (- A b + 5 B a\right ) \log {\left (- \frac {3 a b^{3} \sqrt {- \frac {1}{a b^{7}}} \left (- A b + 5 B a\right )}{- 3 A b + 15 B a} + x \right )}}{16} - \frac {3 \sqrt {- \frac {1}{a b^{7}}} \left (- A b + 5 B a\right ) \log {\left (\frac {3 a b^{3} \sqrt {- \frac {1}{a b^{7}}} \left (- A b + 5 B a\right )}{- 3 A b + 15 B a} + x \right )}}{16} + \frac {x^{3} \left (- 5 A b^{2} + 9 B a b\right ) + x \left (- 3 A a b + 7 B a^{2}\right )}{8 a^{2} b^{3} + 16 a b^{4} x^{2} + 8 b^{5} x^{4}} \]
B*x/b**3 + 3*sqrt(-1/(a*b**7))*(-A*b + 5*B*a)*log(-3*a*b**3*sqrt(-1/(a*b** 7))*(-A*b + 5*B*a)/(-3*A*b + 15*B*a) + x)/16 - 3*sqrt(-1/(a*b**7))*(-A*b + 5*B*a)*log(3*a*b**3*sqrt(-1/(a*b**7))*(-A*b + 5*B*a)/(-3*A*b + 15*B*a) + x)/16 + (x**3*(-5*A*b**2 + 9*B*a*b) + x*(-3*A*a*b + 7*B*a**2))/(8*a**2*b** 3 + 16*a*b**4*x**2 + 8*b**5*x**4)
Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {{\left (9 \, B a b - 5 \, A b^{2}\right )} x^{3} + {\left (7 \, B a^{2} - 3 \, A a b\right )} x}{8 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}} + \frac {B x}{b^{3}} - \frac {3 \, {\left (5 \, B a - A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3}} \]
1/8*((9*B*a*b - 5*A*b^2)*x^3 + (7*B*a^2 - 3*A*a*b)*x)/(b^5*x^4 + 2*a*b^4*x ^2 + a^2*b^3) + B*x/b^3 - 3/8*(5*B*a - A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a* b)*b^3)
Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.85 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {B x}{b^{3}} - \frac {3 \, {\left (5 \, B a - A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3}} + \frac {9 \, B a b x^{3} - 5 \, A b^{2} x^{3} + 7 \, B a^{2} x - 3 \, A a b x}{8 \, {\left (b x^{2} + a\right )}^{2} b^{3}} \]
B*x/b^3 - 3/8*(5*B*a - A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) + 1/8*(9 *B*a*b*x^3 - 5*A*b^2*x^3 + 7*B*a^2*x - 3*A*a*b*x)/((b*x^2 + a)^2*b^3)
Time = 5.00 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.98 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {B\,x}{b^3}-\frac {x^3\,\left (\frac {5\,A\,b^2}{8}-\frac {9\,B\,a\,b}{8}\right )-x\,\left (\frac {7\,B\,a^2}{8}-\frac {3\,A\,a\,b}{8}\right )}{a^2\,b^3+2\,a\,b^4\,x^2+b^5\,x^4}+\frac {3\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (A\,b-5\,B\,a\right )}{8\,\sqrt {a}\,b^{7/2}} \]