Integrand size = 24, antiderivative size = 92 \[ \int \frac {A+B x^2}{x^5 \left (b x^2+c x^4\right )} \, dx=-\frac {A}{6 b x^6}-\frac {b B-A c}{4 b^2 x^4}+\frac {c (b B-A c)}{2 b^3 x^2}+\frac {c^2 (b B-A c) \log (x)}{b^4}-\frac {c^2 (b B-A c) \log \left (b+c x^2\right )}{2 b^4} \] Output:
-1/6*A/b/x^6-1/4*(-A*c+B*b)/b^2/x^4+1/2*c*(-A*c+B*b)/b^3/x^2+c^2*(-A*c+B*b )*ln(x)/b^4-1/2*c^2*(-A*c+B*b)*ln(c*x^2+b)/b^4
Time = 0.03 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.04 \[ \int \frac {A+B x^2}{x^5 \left (b x^2+c x^4\right )} \, dx=-\frac {A}{6 b x^6}+\frac {-b B+A c}{4 b^2 x^4}+\frac {c (b B-A c)}{2 b^3 x^2}+\frac {\left (b B c^2-A c^3\right ) \log (x)}{b^4}+\frac {\left (-b B c^2+A c^3\right ) \log \left (b+c x^2\right )}{2 b^4} \] Input:
Integrate[(A + B*x^2)/(x^5*(b*x^2 + c*x^4)),x]
Output:
-1/6*A/(b*x^6) + (-(b*B) + A*c)/(4*b^2*x^4) + (c*(b*B - A*c))/(2*b^3*x^2) + ((b*B*c^2 - A*c^3)*Log[x])/b^4 + ((-(b*B*c^2) + A*c^3)*Log[b + c*x^2])/( 2*b^4)
Time = 0.43 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {9, 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.
\(\displaystyle \int \frac {A+B x^2}{x^5 \left (b x^2+c x^4\right )} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {A+B x^2}{x^7 \left (b+c x^2\right )}dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {B x^2+A}{x^8 \left (c x^2+b\right )}dx^2\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} \int \left (-\frac {(b B-A c) c^3}{b^4 \left (c x^2+b\right )}+\frac {(b B-A c) c^2}{b^4 x^2}-\frac {(b B-A c) c}{b^3 x^4}+\frac {b B-A c}{b^2 x^6}+\frac {A}{b x^8}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {c^2 \log \left (x^2\right ) (b B-A c)}{b^4}-\frac {c^2 (b B-A c) \log \left (b+c x^2\right )}{b^4}+\frac {c (b B-A c)}{b^3 x^2}-\frac {b B-A c}{2 b^2 x^4}-\frac {A}{3 b x^6}\right )\) |
Input:
Int[(A + B*x^2)/(x^5*(b*x^2 + c*x^4)),x]
Output:
(-1/3*A/(b*x^6) - (b*B - A*c)/(2*b^2*x^4) + (c*(b*B - A*c))/(b^3*x^2) + (c ^2*(b*B - A*c)*Log[x^2])/b^4 - (c^2*(b*B - A*c)*Log[b + c*x^2])/b^4)/2
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
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 = 0.40 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.93
method | result | size |
default | \(-\frac {A}{6 b \,x^{6}}-\frac {-A c +B b}{4 b^{2} x^{4}}-\frac {\left (A c -B b \right ) c}{2 b^{3} x^{2}}-\frac {\left (A c -B b \right ) c^{2} \ln \left (x \right )}{b^{4}}+\frac {c^{2} \left (A c -B b \right ) \ln \left (c \,x^{2}+b \right )}{2 b^{4}}\) | \(86\) |
norman | \(\frac {-\frac {A}{6 b}+\frac {\left (A c -B b \right ) x^{2}}{4 b^{2}}-\frac {\left (A c -B b \right ) c \,x^{4}}{2 b^{3}}}{x^{6}}-\frac {\left (A c -B b \right ) c^{2} \ln \left (x \right )}{b^{4}}+\frac {c^{2} \left (A c -B b \right ) \ln \left (c \,x^{2}+b \right )}{2 b^{4}}\) | \(88\) |
risch | \(\frac {-\frac {A}{6 b}+\frac {\left (A c -B b \right ) x^{2}}{4 b^{2}}-\frac {\left (A c -B b \right ) c \,x^{4}}{2 b^{3}}}{x^{6}}-\frac {c^{3} \ln \left (x \right ) A}{b^{4}}+\frac {c^{2} \ln \left (x \right ) B}{b^{3}}+\frac {c^{3} \ln \left (-c \,x^{2}-b \right ) A}{2 b^{4}}-\frac {c^{2} \ln \left (-c \,x^{2}-b \right ) B}{2 b^{3}}\) | \(107\) |
parallelrisch | \(-\frac {12 A \ln \left (x \right ) x^{6} c^{3}-6 A \ln \left (c \,x^{2}+b \right ) x^{6} c^{3}-12 B \ln \left (x \right ) x^{6} b \,c^{2}+6 B \ln \left (c \,x^{2}+b \right ) x^{6} b \,c^{2}+6 A b \,c^{2} x^{4}-6 x^{4} B \,b^{2} c -3 A \,b^{2} c \,x^{2}+3 x^{2} B \,b^{3}+2 A \,b^{3}}{12 b^{4} x^{6}}\) | \(113\) |
Input:
int((B*x^2+A)/x^5/(c*x^4+b*x^2),x,method=_RETURNVERBOSE)
Output:
-1/6*A/b/x^6-1/4*(-A*c+B*b)/b^2/x^4-1/2*(A*c-B*b)/b^3*c/x^2-(A*c-B*b)/b^4* c^2*ln(x)+1/2*c^2*(A*c-B*b)/b^4*ln(c*x^2+b)
Time = 0.07 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.07 \[ \int \frac {A+B x^2}{x^5 \left (b x^2+c x^4\right )} \, dx=-\frac {6 \, {\left (B b c^{2} - A c^{3}\right )} x^{6} \log \left (c x^{2} + b\right ) - 12 \, {\left (B b c^{2} - A c^{3}\right )} x^{6} \log \left (x\right ) - 6 \, {\left (B b^{2} c - A b c^{2}\right )} x^{4} + 2 \, A b^{3} + 3 \, {\left (B b^{3} - A b^{2} c\right )} x^{2}}{12 \, b^{4} x^{6}} \] Input:
integrate((B*x^2+A)/x^5/(c*x^4+b*x^2),x, algorithm="fricas")
Output:
-1/12*(6*(B*b*c^2 - A*c^3)*x^6*log(c*x^2 + b) - 12*(B*b*c^2 - A*c^3)*x^6*l og(x) - 6*(B*b^2*c - A*b*c^2)*x^4 + 2*A*b^3 + 3*(B*b^3 - A*b^2*c)*x^2)/(b^ 4*x^6)
Time = 0.48 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.96 \[ \int \frac {A+B x^2}{x^5 \left (b x^2+c x^4\right )} \, dx=\frac {- 2 A b^{2} + x^{4} \left (- 6 A c^{2} + 6 B b c\right ) + x^{2} \cdot \left (3 A b c - 3 B b^{2}\right )}{12 b^{3} x^{6}} + \frac {c^{2} \left (- A c + B b\right ) \log {\left (x \right )}}{b^{4}} - \frac {c^{2} \left (- A c + B b\right ) \log {\left (\frac {b}{c} + x^{2} \right )}}{2 b^{4}} \] Input:
integrate((B*x**2+A)/x**5/(c*x**4+b*x**2),x)
Output:
(-2*A*b**2 + x**4*(-6*A*c**2 + 6*B*b*c) + x**2*(3*A*b*c - 3*B*b**2))/(12*b **3*x**6) + c**2*(-A*c + B*b)*log(x)/b**4 - c**2*(-A*c + B*b)*log(b/c + x* *2)/(2*b**4)
Time = 0.03 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.04 \[ \int \frac {A+B x^2}{x^5 \left (b x^2+c x^4\right )} \, dx=-\frac {{\left (B b c^{2} - A c^{3}\right )} \log \left (c x^{2} + b\right )}{2 \, b^{4}} + \frac {{\left (B b c^{2} - A c^{3}\right )} \log \left (x^{2}\right )}{2 \, b^{4}} + \frac {6 \, {\left (B b c - A c^{2}\right )} x^{4} - 2 \, A b^{2} - 3 \, {\left (B b^{2} - A b c\right )} x^{2}}{12 \, b^{3} x^{6}} \] Input:
integrate((B*x^2+A)/x^5/(c*x^4+b*x^2),x, algorithm="maxima")
Output:
-1/2*(B*b*c^2 - A*c^3)*log(c*x^2 + b)/b^4 + 1/2*(B*b*c^2 - A*c^3)*log(x^2) /b^4 + 1/12*(6*(B*b*c - A*c^2)*x^4 - 2*A*b^2 - 3*(B*b^2 - A*b*c)*x^2)/(b^3 *x^6)
Time = 0.19 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.37 \[ \int \frac {A+B x^2}{x^5 \left (b x^2+c x^4\right )} \, dx=\frac {{\left (B b c^{2} - A c^{3}\right )} \log \left (x^{2}\right )}{2 \, b^{4}} - \frac {{\left (B b c^{3} - A c^{4}\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, b^{4} c} - \frac {11 \, B b c^{2} x^{6} - 11 \, A c^{3} x^{6} - 6 \, B b^{2} c x^{4} + 6 \, A b c^{2} x^{4} + 3 \, B b^{3} x^{2} - 3 \, A b^{2} c x^{2} + 2 \, A b^{3}}{12 \, b^{4} x^{6}} \] Input:
integrate((B*x^2+A)/x^5/(c*x^4+b*x^2),x, algorithm="giac")
Output:
1/2*(B*b*c^2 - A*c^3)*log(x^2)/b^4 - 1/2*(B*b*c^3 - A*c^4)*log(abs(c*x^2 + b))/(b^4*c) - 1/12*(11*B*b*c^2*x^6 - 11*A*c^3*x^6 - 6*B*b^2*c*x^4 + 6*A*b *c^2*x^4 + 3*B*b^3*x^2 - 3*A*b^2*c*x^2 + 2*A*b^3)/(b^4*x^6)
Time = 8.77 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x^2}{x^5 \left (b x^2+c x^4\right )} \, dx=\frac {\ln \left (c\,x^2+b\right )\,\left (A\,c^3-B\,b\,c^2\right )}{2\,b^4}-\frac {\frac {A}{6\,b}-\frac {x^2\,\left (A\,c-B\,b\right )}{4\,b^2}+\frac {c\,x^4\,\left (A\,c-B\,b\right )}{2\,b^3}}{x^6}-\frac {\ln \left (x\right )\,\left (A\,c^3-B\,b\,c^2\right )}{b^4} \] Input:
int((A + B*x^2)/(x^5*(b*x^2 + c*x^4)),x)
Output:
(log(b + c*x^2)*(A*c^3 - B*b*c^2))/(2*b^4) - (A/(6*b) - (x^2*(A*c - B*b))/ (4*b^2) + (c*x^4*(A*c - B*b))/(2*b^3))/x^6 - (log(x)*(A*c^3 - B*b*c^2))/b^ 4
Time = 0.19 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.22 \[ \int \frac {A+B x^2}{x^5 \left (b x^2+c x^4\right )} \, dx=\frac {6 \,\mathrm {log}\left (c \,x^{2}+b \right ) a \,c^{3} x^{6}-6 \,\mathrm {log}\left (c \,x^{2}+b \right ) b^{2} c^{2} x^{6}-12 \,\mathrm {log}\left (x \right ) a \,c^{3} x^{6}+12 \,\mathrm {log}\left (x \right ) b^{2} c^{2} x^{6}-2 a \,b^{3}+3 a \,b^{2} c \,x^{2}-6 a b \,c^{2} x^{4}-3 b^{4} x^{2}+6 b^{3} c \,x^{4}}{12 b^{4} x^{6}} \] Input:
int((B*x^2+A)/x^5/(c*x^4+b*x^2),x)
Output:
(6*log(b + c*x**2)*a*c**3*x**6 - 6*log(b + c*x**2)*b**2*c**2*x**6 - 12*log (x)*a*c**3*x**6 + 12*log(x)*b**2*c**2*x**6 - 2*a*b**3 + 3*a*b**2*c*x**2 - 6*a*b*c**2*x**4 - 3*b**4*x**2 + 6*b**3*c*x**4)/(12*b**4*x**6)