Integrand size = 22, antiderivative size = 74 \[ \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=-\frac {c}{2 d (b c-a d) \left (c+d x^2\right )}-\frac {a \log \left (a+b x^2\right )}{2 (b c-a d)^2}+\frac {a \log \left (c+d x^2\right )}{2 (b c-a d)^2} \] Output:
-1/2*c/d/(-a*d+b*c)/(d*x^2+c)-1/2*a*ln(b*x^2+a)/(-a*d+b*c)^2+1/2*a*ln(d*x^ 2+c)/(-a*d+b*c)^2
Time = 0.02 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00 \[ \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=\frac {c}{2 d (-b c+a d) \left (c+d x^2\right )}-\frac {a \log \left (a+b x^2\right )}{2 (b c-a d)^2}+\frac {a \log \left (c+d x^2\right )}{2 (b c-a d)^2} \] Input:
Integrate[x^3/((a + b*x^2)*(c + d*x^2)^2),x]
Output:
c/(2*d*(-(b*c) + a*d)*(c + d*x^2)) - (a*Log[a + b*x^2])/(2*(b*c - a*d)^2) + (a*Log[c + d*x^2])/(2*(b*c - a*d)^2)
Time = 0.22 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.96, 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.
\(\displaystyle \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {x^2}{\left (b x^2+a\right ) \left (d x^2+c\right )^2}dx^2\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} \int \left (-\frac {a b}{(b c-a d)^2 \left (b x^2+a\right )}+\frac {a d}{(a d-b c)^2 \left (d x^2+c\right )}+\frac {c}{(b c-a d) \left (d x^2+c\right )^2}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {c}{d \left (c+d x^2\right ) (b c-a d)}-\frac {a \log \left (a+b x^2\right )}{(b c-a d)^2}+\frac {a \log \left (c+d x^2\right )}{(b c-a d)^2}\right )\) |
Input:
Int[x^3/((a + b*x^2)*(c + d*x^2)^2),x]
Output:
(-(c/(d*(b*c - a*d)*(c + d*x^2))) - (a*Log[a + b*x^2])/(b*c - a*d)^2 + (a* Log[c + d*x^2])/(b*c - a*d)^2)/2
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 = 0.56 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.92
method | result | size |
default | \(-\frac {a \ln \left (b \,x^{2}+a \right )}{2 \left (a d -b c \right )^{2}}+\frac {\frac {\left (a d -b c \right ) c}{d \left (x^{2} d +c \right )}+a \ln \left (x^{2} d +c \right )}{2 \left (a d -b c \right )^{2}}\) | \(68\) |
norman | \(-\frac {x^{2}}{2 \left (a d -b c \right ) \left (x^{2} d +c \right )}-\frac {a \ln \left (b \,x^{2}+a \right )}{2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {a \ln \left (x^{2} d +c \right )}{2 a^{2} d^{2}-4 a b c d +2 b^{2} c^{2}}\) | \(94\) |
risch | \(\frac {c}{2 \left (a d -b c \right ) d \left (x^{2} d +c \right )}-\frac {a \ln \left (-b \,x^{2}-a \right )}{2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {a \ln \left (x^{2} d +c \right )}{2 a^{2} d^{2}-4 a b c d +2 b^{2} c^{2}}\) | \(98\) |
parallelrisch | \(-\frac {\ln \left (b \,x^{2}+a \right ) x^{2} a \,d^{2}-\ln \left (x^{2} d +c \right ) x^{2} a \,d^{2}+\ln \left (b \,x^{2}+a \right ) a c d -\ln \left (x^{2} d +c \right ) a c d -a c d +b \,c^{2}}{2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (x^{2} d +c \right ) d}\) | \(107\) |
Input:
int(x^3/(b*x^2+a)/(d*x^2+c)^2,x,method=_RETURNVERBOSE)
Output:
-1/2*a/(a*d-b*c)^2*ln(b*x^2+a)+1/2/(a*d-b*c)^2*((a*d-b*c)*c/d/(d*x^2+c)+a* ln(d*x^2+c))
Time = 0.09 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.58 \[ \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=-\frac {b c^{2} - a c d + {\left (a d^{2} x^{2} + a c d\right )} \log \left (b x^{2} + a\right ) - {\left (a d^{2} x^{2} + a c d\right )} \log \left (d x^{2} + c\right )}{2 \, {\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3} + {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x^{2}\right )}} \] Input:
integrate(x^3/(b*x^2+a)/(d*x^2+c)^2,x, algorithm="fricas")
Output:
-1/2*(b*c^2 - a*c*d + (a*d^2*x^2 + a*c*d)*log(b*x^2 + a) - (a*d^2*x^2 + a* c*d)*log(d*x^2 + c))/(b^2*c^3*d - 2*a*b*c^2*d^2 + a^2*c*d^3 + (b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*x^2)
Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (58) = 116\).
Time = 0.91 (sec) , antiderivative size = 253, normalized size of antiderivative = 3.42 \[ \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=\frac {a \log {\left (x^{2} + \frac {- \frac {a^{4} d^{3}}{\left (a d - b c\right )^{2}} + \frac {3 a^{3} b c d^{2}}{\left (a d - b c\right )^{2}} - \frac {3 a^{2} b^{2} c^{2} d}{\left (a d - b c\right )^{2}} + a^{2} d + \frac {a b^{3} c^{3}}{\left (a d - b c\right )^{2}} + a b c}{2 a b d} \right )}}{2 \left (a d - b c\right )^{2}} - \frac {a \log {\left (x^{2} + \frac {\frac {a^{4} d^{3}}{\left (a d - b c\right )^{2}} - \frac {3 a^{3} b c d^{2}}{\left (a d - b c\right )^{2}} + \frac {3 a^{2} b^{2} c^{2} d}{\left (a d - b c\right )^{2}} + a^{2} d - \frac {a b^{3} c^{3}}{\left (a d - b c\right )^{2}} + a b c}{2 a b d} \right )}}{2 \left (a d - b c\right )^{2}} + \frac {c}{2 a c d^{2} - 2 b c^{2} d + x^{2} \cdot \left (2 a d^{3} - 2 b c d^{2}\right )} \] Input:
integrate(x**3/(b*x**2+a)/(d*x**2+c)**2,x)
Output:
a*log(x**2 + (-a**4*d**3/(a*d - b*c)**2 + 3*a**3*b*c*d**2/(a*d - b*c)**2 - 3*a**2*b**2*c**2*d/(a*d - b*c)**2 + a**2*d + a*b**3*c**3/(a*d - b*c)**2 + a*b*c)/(2*a*b*d))/(2*(a*d - b*c)**2) - a*log(x**2 + (a**4*d**3/(a*d - b*c )**2 - 3*a**3*b*c*d**2/(a*d - b*c)**2 + 3*a**2*b**2*c**2*d/(a*d - b*c)**2 + a**2*d - a*b**3*c**3/(a*d - b*c)**2 + a*b*c)/(2*a*b*d))/(2*(a*d - b*c)** 2) + c/(2*a*c*d**2 - 2*b*c**2*d + x**2*(2*a*d**3 - 2*b*c*d**2))
Time = 0.04 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.42 \[ \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=-\frac {a \log \left (b x^{2} + a\right )}{2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )}} + \frac {a \log \left (d x^{2} + c\right )}{2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )}} - \frac {c}{2 \, {\left (b c^{2} d - a c d^{2} + {\left (b c d^{2} - a d^{3}\right )} x^{2}\right )}} \] Input:
integrate(x^3/(b*x^2+a)/(d*x^2+c)^2,x, algorithm="maxima")
Output:
-1/2*a*log(b*x^2 + a)/(b^2*c^2 - 2*a*b*c*d + a^2*d^2) + 1/2*a*log(d*x^2 + c)/(b^2*c^2 - 2*a*b*c*d + a^2*d^2) - 1/2*c/(b*c^2*d - a*c*d^2 + (b*c*d^2 - a*d^3)*x^2)
Time = 0.13 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.23 \[ \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=-\frac {\frac {a d^{2} \log \left ({\left | b - \frac {b c}{d x^{2} + c} + \frac {a d}{d x^{2} + c} \right |}\right )}{b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}} + \frac {c d}{{\left (b c d - a d^{2}\right )} {\left (d x^{2} + c\right )}}}{2 \, d} \] Input:
integrate(x^3/(b*x^2+a)/(d*x^2+c)^2,x, algorithm="giac")
Output:
-1/2*(a*d^2*log(abs(b - b*c/(d*x^2 + c) + a*d/(d*x^2 + c)))/(b^2*c^2*d - 2 *a*b*c*d^2 + a^2*d^3) + c*d/((b*c*d - a*d^2)*(d*x^2 + c)))/d
Time = 0.63 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.34 \[ \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=-\frac {b\,c^2-c\,\left (a\,d-a\,d\,\mathrm {atan}\left (\frac {a\,d\,x^2\,1{}\mathrm {i}-b\,c\,x^2\,1{}\mathrm {i}}{2\,a\,c+a\,d\,x^2+b\,c\,x^2}\right )\,2{}\mathrm {i}\right )+a\,d^2\,x^2\,\mathrm {atan}\left (\frac {a\,d\,x^2\,1{}\mathrm {i}-b\,c\,x^2\,1{}\mathrm {i}}{2\,a\,c+a\,d\,x^2+b\,c\,x^2}\right )\,2{}\mathrm {i}}{2\,a^2\,c\,d^3+2\,a^2\,d^4\,x^2-4\,a\,b\,c^2\,d^2-4\,a\,b\,c\,d^3\,x^2+2\,b^2\,c^3\,d+2\,b^2\,c^2\,d^2\,x^2} \] Input:
int(x^3/((a + b*x^2)*(c + d*x^2)^2),x)
Output:
-(b*c^2 - c*(a*d - a*d*atan((a*d*x^2*1i - b*c*x^2*1i)/(2*a*c + a*d*x^2 + b *c*x^2))*2i) + a*d^2*x^2*atan((a*d*x^2*1i - b*c*x^2*1i)/(2*a*c + a*d*x^2 + b*c*x^2))*2i)/(2*a^2*c*d^3 + 2*b^2*c^3*d + 2*a^2*d^4*x^2 + 2*b^2*c^2*d^2* x^2 - 4*a*b*c^2*d^2 - 4*a*b*c*d^3*x^2)
Time = 0.19 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.74 \[ \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=\frac {-\mathrm {log}\left (b \,x^{2}+a \right ) a c -\mathrm {log}\left (b \,x^{2}+a \right ) a d \,x^{2}+\mathrm {log}\left (d \,x^{2}+c \right ) a c +\mathrm {log}\left (d \,x^{2}+c \right ) a d \,x^{2}-a d \,x^{2}+b c \,x^{2}}{2 a^{2} d^{3} x^{2}-4 a b c \,d^{2} x^{2}+2 b^{2} c^{2} d \,x^{2}+2 a^{2} c \,d^{2}-4 a b \,c^{2} d +2 b^{2} c^{3}} \] Input:
int(x^3/(b*x^2+a)/(d*x^2+c)^2,x)
Output:
( - log(a + b*x**2)*a*c - log(a + b*x**2)*a*d*x**2 + log(c + d*x**2)*a*c + log(c + d*x**2)*a*d*x**2 - a*d*x**2 + b*c*x**2)/(2*(a**2*c*d**2 + a**2*d* *3*x**2 - 2*a*b*c**2*d - 2*a*b*c*d**2*x**2 + b**2*c**3 + b**2*c**2*d*x**2) )