Integrand size = 22, antiderivative size = 88 \[ \int \frac {x^3 \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx=\frac {d (b c-a d) x^2}{b^3}+\frac {d^2 x^4}{4 b^2}+\frac {a (b c-a d)^2}{2 b^4 \left (a+b x^2\right )}+\frac {(b c-3 a d) (b c-a d) \log \left (a+b x^2\right )}{2 b^4} \]
d*(-a*d+b*c)*x^2/b^3+1/4*d^2*x^4/b^2+1/2*a*(-a*d+b*c)^2/b^4/(b*x^2+a)+1/2* (-3*a*d+b*c)*(-a*d+b*c)*ln(b*x^2+a)/b^4
Time = 0.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.99 \[ \int \frac {x^3 \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx=\frac {4 b d (b c-a d) x^2+b^2 d^2 x^4+\frac {2 a (b c-a d)^2}{a+b x^2}+2 \left (b^2 c^2-4 a b c d+3 a^2 d^2\right ) \log \left (a+b x^2\right )}{4 b^4} \]
(4*b*d*(b*c - a*d)*x^2 + b^2*d^2*x^4 + (2*a*(b*c - a*d)^2)/(a + b*x^2) + 2 *(b^2*c^2 - 4*a*b*c*d + 3*a^2*d^2)*Log[a + b*x^2])/(4*b^4)
Time = 0.26 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {354, 86, 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.
\(\displaystyle \int \frac {x^3 \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {x^2 \left (d x^2+c\right )^2}{\left (b x^2+a\right )^2}dx^2\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} \int \left (-\frac {a (a d-b c)^2}{b^3 \left (b x^2+a\right )^2}+\frac {d^2 x^2}{b^2}+\frac {2 d (b c-a d)}{b^3}+\frac {(b c-3 a d) (b c-a d)}{b^3 \left (b x^2+a\right )}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {a (b c-a d)^2}{b^4 \left (a+b x^2\right )}+\frac {(b c-3 a d) (b c-a d) \log \left (a+b x^2\right )}{b^4}+\frac {2 d x^2 (b c-a d)}{b^3}+\frac {d^2 x^4}{2 b^2}\right )\) |
((2*d*(b*c - a*d)*x^2)/b^3 + (d^2*x^4)/(2*b^2) + (a*(b*c - a*d)^2)/(b^4*(a + b*x^2)) + ((b*c - 3*a*d)*(b*c - a*d)*Log[a + b*x^2])/b^4)/2
3.3.72.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Time = 2.74 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.93
method | result | size |
default | \(\frac {\left (-b d \,x^{2}+2 a d -2 b c \right )^{2}}{4 b^{4}}+\frac {\left (a d -b c \right ) \left (\frac {\left (3 a d -b c \right ) \ln \left (b \,x^{2}+a \right )}{b}+\frac {\left (a d -b c \right ) a}{b \left (b \,x^{2}+a \right )}\right )}{2 b^{3}}\) | \(82\) |
norman | \(\frac {\frac {d^{2} x^{6}}{4 b}-\frac {d \left (3 a d -4 b c \right ) x^{4}}{4 b^{2}}-\frac {\left (3 a^{3} d^{2}-4 a^{2} b c d +b^{2} c^{2} a \right ) x^{2}}{2 a \,b^{3}}}{b \,x^{2}+a}+\frac {\left (3 a^{2} d^{2}-4 a b c d +b^{2} c^{2}\right ) \ln \left (b \,x^{2}+a \right )}{2 b^{4}}\) | \(113\) |
risch | \(\frac {d^{2} x^{4}}{4 b^{2}}-\frac {x^{2} a \,d^{2}}{b^{3}}+\frac {x^{2} c d}{b^{2}}+\frac {a^{2} d^{2}}{b^{4}}-\frac {2 a c d}{b^{3}}+\frac {c^{2}}{b^{2}}+\frac {a^{3} d^{2}}{2 b^{4} \left (b \,x^{2}+a \right )}-\frac {a^{2} c d}{b^{3} \left (b \,x^{2}+a \right )}+\frac {a \,c^{2}}{2 b^{2} \left (b \,x^{2}+a \right )}+\frac {3 \ln \left (b \,x^{2}+a \right ) a^{2} d^{2}}{2 b^{4}}-\frac {2 \ln \left (b \,x^{2}+a \right ) a c d}{b^{3}}+\frac {\ln \left (b \,x^{2}+a \right ) c^{2}}{2 b^{2}}\) | \(167\) |
parallelrisch | \(\frac {b^{3} d^{2} x^{6}-3 x^{4} a \,b^{2} d^{2}+4 x^{4} b^{3} c d +6 \ln \left (b \,x^{2}+a \right ) x^{2} a^{2} b \,d^{2}-8 \ln \left (b \,x^{2}+a \right ) x^{2} a \,b^{2} c d +2 \ln \left (b \,x^{2}+a \right ) x^{2} b^{3} c^{2}+6 \ln \left (b \,x^{2}+a \right ) a^{3} d^{2}-8 \ln \left (b \,x^{2}+a \right ) a^{2} b c d +2 \ln \left (b \,x^{2}+a \right ) a \,b^{2} c^{2}+6 a^{3} d^{2}-8 a^{2} b c d +2 b^{2} c^{2} a}{4 b^{4} \left (b \,x^{2}+a \right )}\) | \(180\) |
1/4*(-b*d*x^2+2*a*d-2*b*c)^2/b^4+1/2/b^3*(a*d-b*c)*((3*a*d-b*c)/b*ln(b*x^2 +a)+(a*d-b*c)*a/b/(b*x^2+a))
Time = 0.23 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.82 \[ \int \frac {x^3 \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx=\frac {b^{3} d^{2} x^{6} + 2 \, a b^{2} c^{2} - 4 \, a^{2} b c d + 2 \, a^{3} d^{2} + {\left (4 \, b^{3} c d - 3 \, a b^{2} d^{2}\right )} x^{4} + 4 \, {\left (a b^{2} c d - a^{2} b d^{2}\right )} x^{2} + 2 \, {\left (a b^{2} c^{2} - 4 \, a^{2} b c d + 3 \, a^{3} d^{2} + {\left (b^{3} c^{2} - 4 \, a b^{2} c d + 3 \, a^{2} b d^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right )}{4 \, {\left (b^{5} x^{2} + a b^{4}\right )}} \]
1/4*(b^3*d^2*x^6 + 2*a*b^2*c^2 - 4*a^2*b*c*d + 2*a^3*d^2 + (4*b^3*c*d - 3* a*b^2*d^2)*x^4 + 4*(a*b^2*c*d - a^2*b*d^2)*x^2 + 2*(a*b^2*c^2 - 4*a^2*b*c* d + 3*a^3*d^2 + (b^3*c^2 - 4*a*b^2*c*d + 3*a^2*b*d^2)*x^2)*log(b*x^2 + a)) /(b^5*x^2 + a*b^4)
Time = 0.52 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.12 \[ \int \frac {x^3 \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx=x^{2} \left (- \frac {a d^{2}}{b^{3}} + \frac {c d}{b^{2}}\right ) + \frac {a^{3} d^{2} - 2 a^{2} b c d + a b^{2} c^{2}}{2 a b^{4} + 2 b^{5} x^{2}} + \frac {d^{2} x^{4}}{4 b^{2}} + \frac {\left (a d - b c\right ) \left (3 a d - b c\right ) \log {\left (a + b x^{2} \right )}}{2 b^{4}} \]
x**2*(-a*d**2/b**3 + c*d/b**2) + (a**3*d**2 - 2*a**2*b*c*d + a*b**2*c**2)/ (2*a*b**4 + 2*b**5*x**2) + d**2*x**4/(4*b**2) + (a*d - b*c)*(3*a*d - b*c)* log(a + b*x**2)/(2*b**4)
Time = 0.23 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.22 \[ \int \frac {x^3 \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx=\frac {a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}}{2 \, {\left (b^{5} x^{2} + a b^{4}\right )}} + \frac {b d^{2} x^{4} + 4 \, {\left (b c d - a d^{2}\right )} x^{2}}{4 \, b^{3}} + \frac {{\left (b^{2} c^{2} - 4 \, a b c d + 3 \, a^{2} d^{2}\right )} \log \left (b x^{2} + a\right )}{2 \, b^{4}} \]
1/2*(a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2)/(b^5*x^2 + a*b^4) + 1/4*(b*d^2*x^4 + 4*(b*c*d - a*d^2)*x^2)/b^3 + 1/2*(b^2*c^2 - 4*a*b*c*d + 3*a^2*d^2)*log( b*x^2 + a)/b^4
Time = 0.30 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.85 \[ \int \frac {x^3 \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx=\frac {\frac {{\left (b x^{2} + a\right )}^{2} {\left (d^{2} + \frac {2 \, {\left (2 \, b^{2} c d - 3 \, a b d^{2}\right )}}{{\left (b x^{2} + a\right )} b}\right )}}{b^{3}} - \frac {2 \, {\left (b^{2} c^{2} - 4 \, a b c d + 3 \, a^{2} d^{2}\right )} \log \left (\frac {{\left | b x^{2} + a \right |}}{{\left (b x^{2} + a\right )}^{2} {\left | b \right |}}\right )}{b^{3}} + \frac {2 \, {\left (\frac {a b^{4} c^{2}}{b x^{2} + a} - \frac {2 \, a^{2} b^{3} c d}{b x^{2} + a} + \frac {a^{3} b^{2} d^{2}}{b x^{2} + a}\right )}}{b^{5}}}{4 \, b} \]
1/4*((b*x^2 + a)^2*(d^2 + 2*(2*b^2*c*d - 3*a*b*d^2)/((b*x^2 + a)*b))/b^3 - 2*(b^2*c^2 - 4*a*b*c*d + 3*a^2*d^2)*log(abs(b*x^2 + a)/((b*x^2 + a)^2*abs (b)))/b^3 + 2*(a*b^4*c^2/(b*x^2 + a) - 2*a^2*b^3*c*d/(b*x^2 + a) + a^3*b^2 *d^2/(b*x^2 + a))/b^5)/b
Time = 0.08 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.27 \[ \int \frac {x^3 \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx=\frac {a^3\,d^2-2\,a^2\,b\,c\,d+a\,b^2\,c^2}{2\,b\,\left (b^4\,x^2+a\,b^3\right )}-x^2\,\left (\frac {a\,d^2}{b^3}-\frac {c\,d}{b^2}\right )+\frac {d^2\,x^4}{4\,b^2}+\frac {\ln \left (b\,x^2+a\right )\,\left (3\,a^2\,d^2-4\,a\,b\,c\,d+b^2\,c^2\right )}{2\,b^4} \]