Integrand size = 20, antiderivative size = 76 \[ \int \frac {c+d x^2}{x^3 \left (a+b x^2\right )^2} \, dx=-\frac {c}{2 a^2 x^2}-\frac {b c-a d}{2 a^2 \left (a+b x^2\right )}-\frac {(2 b c-a d) \log (x)}{a^3}+\frac {(2 b c-a d) \log \left (a+b x^2\right )}{2 a^3} \] Output:
-1/2*c/a^2/x^2-1/2*(-a*d+b*c)/a^2/(b*x^2+a)-(-a*d+2*b*c)*ln(x)/a^3+1/2*(-a *d+2*b*c)*ln(b*x^2+a)/a^3
Time = 0.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.84 \[ \int \frac {c+d x^2}{x^3 \left (a+b x^2\right )^2} \, dx=\frac {-\frac {a c}{x^2}+\frac {a (-b c+a d)}{a+b x^2}+2 (-2 b c+a d) \log (x)+(2 b c-a d) \log \left (a+b x^2\right )}{2 a^3} \] Input:
Integrate[(c + d*x^2)/(x^3*(a + b*x^2)^2),x]
Output:
(-((a*c)/x^2) + (a*(-(b*c) + a*d))/(a + b*x^2) + 2*(-2*b*c + a*d)*Log[x] + (2*b*c - a*d)*Log[a + b*x^2])/(2*a^3)
Time = 0.24 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, 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.
\(\displaystyle \int \frac {c+d x^2}{x^3 \left (a+b x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {d x^2+c}{x^4 \left (b x^2+a\right )^2}dx^2\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} \int \left (\frac {c}{a^2 x^4}-\frac {b (a d-2 b c)}{a^3 \left (b x^2+a\right )}+\frac {a d-2 b c}{a^3 x^2}-\frac {b (a d-b c)}{a^2 \left (b x^2+a\right )^2}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {\log \left (x^2\right ) (2 b c-a d)}{a^3}+\frac {(2 b c-a d) \log \left (a+b x^2\right )}{a^3}-\frac {b c-a d}{a^2 \left (a+b x^2\right )}-\frac {c}{a^2 x^2}\right )\) |
Input:
Int[(c + d*x^2)/(x^3*(a + b*x^2)^2),x]
Output:
(-(c/(a^2*x^2)) - (b*c - a*d)/(a^2*(a + b*x^2)) - ((2*b*c - a*d)*Log[x^2]) /a^3 + ((2*b*c - a*d)*Log[a + b*x^2])/a^3)/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 = 0.42 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.99
method | result | size |
default | \(-\frac {b \left (-\frac {\left (a d -b c \right ) a}{b \left (b \,x^{2}+a \right )}+\frac {\left (a d -2 b c \right ) \ln \left (b \,x^{2}+a \right )}{b}\right )}{2 a^{3}}-\frac {c}{2 a^{2} x^{2}}+\frac {\left (a d -2 b c \right ) \ln \left (x \right )}{a^{3}}\) | \(75\) |
norman | \(\frac {-\frac {c}{2 a}+\frac {b \left (-a d +2 b c \right ) x^{4}}{2 a^{3}}}{x^{2} \left (b \,x^{2}+a \right )}+\frac {\left (a d -2 b c \right ) \ln \left (x \right )}{a^{3}}-\frac {\left (a d -2 b c \right ) \ln \left (b \,x^{2}+a \right )}{2 a^{3}}\) | \(75\) |
risch | \(\frac {\frac {\left (a d -2 b c \right ) x^{2}}{2 a^{2}}-\frac {c}{2 a}}{x^{2} \left (b \,x^{2}+a \right )}+\frac {\ln \left (x \right ) d}{a^{2}}-\frac {2 b c \ln \left (x \right )}{a^{3}}-\frac {\ln \left (b \,x^{2}+a \right ) d}{2 a^{2}}+\frac {b c \ln \left (b \,x^{2}+a \right )}{a^{3}}\) | \(82\) |
parallelrisch | \(\frac {2 \ln \left (x \right ) x^{4} a b d -4 \ln \left (x \right ) x^{4} b^{2} c -\ln \left (b \,x^{2}+a \right ) x^{4} a b d +2 \ln \left (b \,x^{2}+a \right ) x^{4} b^{2} c -a b d \,x^{4}+2 b^{2} c \,x^{4}+2 \ln \left (x \right ) x^{2} a^{2} d -4 \ln \left (x \right ) x^{2} a b c -\ln \left (b \,x^{2}+a \right ) x^{2} a^{2} d +2 \ln \left (b \,x^{2}+a \right ) x^{2} a b c -a^{2} c}{2 a^{3} x^{2} \left (b \,x^{2}+a \right )}\) | \(150\) |
Input:
int((d*x^2+c)/x^3/(b*x^2+a)^2,x,method=_RETURNVERBOSE)
Output:
-1/2/a^3*b*(-(a*d-b*c)*a/b/(b*x^2+a)+(a*d-2*b*c)/b*ln(b*x^2+a))-1/2*c/a^2/ x^2+(a*d-2*b*c)/a^3*ln(x)
Time = 0.09 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.61 \[ \int \frac {c+d x^2}{x^3 \left (a+b x^2\right )^2} \, dx=-\frac {a^{2} c + {\left (2 \, a b c - a^{2} d\right )} x^{2} - {\left ({\left (2 \, b^{2} c - a b d\right )} x^{4} + {\left (2 \, a b c - a^{2} d\right )} x^{2}\right )} \log \left (b x^{2} + a\right ) + 2 \, {\left ({\left (2 \, b^{2} c - a b d\right )} x^{4} + {\left (2 \, a b c - a^{2} d\right )} x^{2}\right )} \log \left (x\right )}{2 \, {\left (a^{3} b x^{4} + a^{4} x^{2}\right )}} \] Input:
integrate((d*x^2+c)/x^3/(b*x^2+a)^2,x, algorithm="fricas")
Output:
-1/2*(a^2*c + (2*a*b*c - a^2*d)*x^2 - ((2*b^2*c - a*b*d)*x^4 + (2*a*b*c - a^2*d)*x^2)*log(b*x^2 + a) + 2*((2*b^2*c - a*b*d)*x^4 + (2*a*b*c - a^2*d)* x^2)*log(x))/(a^3*b*x^4 + a^4*x^2)
Time = 0.45 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.92 \[ \int \frac {c+d x^2}{x^3 \left (a+b x^2\right )^2} \, dx=\frac {- a c + x^{2} \left (a d - 2 b c\right )}{2 a^{3} x^{2} + 2 a^{2} b x^{4}} + \frac {\left (a d - 2 b c\right ) \log {\left (x \right )}}{a^{3}} - \frac {\left (a d - 2 b c\right ) \log {\left (\frac {a}{b} + x^{2} \right )}}{2 a^{3}} \] Input:
integrate((d*x**2+c)/x**3/(b*x**2+a)**2,x)
Output:
(-a*c + x**2*(a*d - 2*b*c))/(2*a**3*x**2 + 2*a**2*b*x**4) + (a*d - 2*b*c)* log(x)/a**3 - (a*d - 2*b*c)*log(a/b + x**2)/(2*a**3)
Time = 0.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.03 \[ \int \frac {c+d x^2}{x^3 \left (a+b x^2\right )^2} \, dx=-\frac {{\left (2 \, b c - a d\right )} x^{2} + a c}{2 \, {\left (a^{2} b x^{4} + a^{3} x^{2}\right )}} + \frac {{\left (2 \, b c - a d\right )} \log \left (b x^{2} + a\right )}{2 \, a^{3}} - \frac {{\left (2 \, b c - a d\right )} \log \left (x^{2}\right )}{2 \, a^{3}} \] Input:
integrate((d*x^2+c)/x^3/(b*x^2+a)^2,x, algorithm="maxima")
Output:
-1/2*((2*b*c - a*d)*x^2 + a*c)/(a^2*b*x^4 + a^3*x^2) + 1/2*(2*b*c - a*d)*l og(b*x^2 + a)/a^3 - 1/2*(2*b*c - a*d)*log(x^2)/a^3
Time = 0.11 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.11 \[ \int \frac {c+d x^2}{x^3 \left (a+b x^2\right )^2} \, dx=-\frac {{\left (2 \, b c - a d\right )} \log \left (x^{2}\right )}{2 \, a^{3}} - \frac {2 \, b c x^{2} - a d x^{2} + a c}{2 \, {\left (b x^{4} + a x^{2}\right )} a^{2}} + \frac {{\left (2 \, b^{2} c - a b d\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{3} b} \] Input:
integrate((d*x^2+c)/x^3/(b*x^2+a)^2,x, algorithm="giac")
Output:
-1/2*(2*b*c - a*d)*log(x^2)/a^3 - 1/2*(2*b*c*x^2 - a*d*x^2 + a*c)/((b*x^4 + a*x^2)*a^2) + 1/2*(2*b^2*c - a*b*d)*log(abs(b*x^2 + a))/(a^3*b)
Time = 0.11 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.97 \[ \int \frac {c+d x^2}{x^3 \left (a+b x^2\right )^2} \, dx=\frac {\ln \left (x\right )\,\left (a\,d-2\,b\,c\right )}{a^3}-\frac {\ln \left (b\,x^2+a\right )\,\left (a\,d-2\,b\,c\right )}{2\,a^3}-\frac {\frac {c}{2\,a}-\frac {x^2\,\left (a\,d-2\,b\,c\right )}{2\,a^2}}{b\,x^4+a\,x^2} \] Input:
int((c + d*x^2)/(x^3*(a + b*x^2)^2),x)
Output:
(log(x)*(a*d - 2*b*c))/a^3 - (log(a + b*x^2)*(a*d - 2*b*c))/(2*a^3) - (c/( 2*a) - (x^2*(a*d - 2*b*c))/(2*a^2))/(a*x^2 + b*x^4)
Time = 0.20 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.96 \[ \int \frac {c+d x^2}{x^3 \left (a+b x^2\right )^2} \, dx=\frac {-\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} d \,x^{2}+2 \,\mathrm {log}\left (b \,x^{2}+a \right ) a b c \,x^{2}-\mathrm {log}\left (b \,x^{2}+a \right ) a b d \,x^{4}+2 \,\mathrm {log}\left (b \,x^{2}+a \right ) b^{2} c \,x^{4}+2 \,\mathrm {log}\left (x \right ) a^{2} d \,x^{2}-4 \,\mathrm {log}\left (x \right ) a b c \,x^{2}+2 \,\mathrm {log}\left (x \right ) a b d \,x^{4}-4 \,\mathrm {log}\left (x \right ) b^{2} c \,x^{4}-a^{2} c -a b d \,x^{4}+2 b^{2} c \,x^{4}}{2 a^{3} x^{2} \left (b \,x^{2}+a \right )} \] Input:
int((d*x^2+c)/x^3/(b*x^2+a)^2,x)
Output:
( - log(a + b*x**2)*a**2*d*x**2 + 2*log(a + b*x**2)*a*b*c*x**2 - log(a + b *x**2)*a*b*d*x**4 + 2*log(a + b*x**2)*b**2*c*x**4 + 2*log(x)*a**2*d*x**2 - 4*log(x)*a*b*c*x**2 + 2*log(x)*a*b*d*x**4 - 4*log(x)*b**2*c*x**4 - a**2*c - a*b*d*x**4 + 2*b**2*c*x**4)/(2*a**3*x**2*(a + b*x**2))