Integrand size = 20, antiderivative size = 93 \[ \int \frac {A+B x^2}{x^7 \left (a+b x^2\right )} \, dx=-\frac {A}{6 a x^6}+\frac {A b-a B}{4 a^2 x^4}-\frac {b (A b-a B)}{2 a^3 x^2}-\frac {b^2 (A b-a B) \log (x)}{a^4}+\frac {b^2 (A b-a B) \log \left (a+b x^2\right )}{2 a^4} \] Output:
-1/6*A/a/x^6+1/4*(A*b-B*a)/a^2/x^4-1/2*b*(A*b-B*a)/a^3/x^2-b^2*(A*b-B*a)*l n(x)/a^4+1/2*b^2*(A*b-B*a)*ln(b*x^2+a)/a^4
Time = 0.03 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.03 \[ \int \frac {A+B x^2}{x^7 \left (a+b x^2\right )} \, dx=-\frac {A}{6 a x^6}+\frac {A b-a B}{4 a^2 x^4}+\frac {b (-A b+a B)}{2 a^3 x^2}+\frac {\left (-A b^3+a b^2 B\right ) \log (x)}{a^4}+\frac {\left (A b^3-a b^2 B\right ) \log \left (a+b x^2\right )}{2 a^4} \] Input:
Integrate[(A + B*x^2)/(x^7*(a + b*x^2)),x]
Output:
-1/6*A/(a*x^6) + (A*b - a*B)/(4*a^2*x^4) + (b*(-(A*b) + a*B))/(2*a^3*x^2) + ((-(A*b^3) + a*b^2*B)*Log[x])/a^4 + ((A*b^3 - a*b^2*B)*Log[a + b*x^2])/( 2*a^4)
Time = 0.24 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.01, 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.
\(\displaystyle \int \frac {A+B x^2}{x^7 \left (a+b x^2\right )} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {B x^2+A}{x^8 \left (b x^2+a\right )}dx^2\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} \int \left (-\frac {(a B-A b) b^3}{a^4 \left (b x^2+a\right )}+\frac {(a B-A b) b^2}{a^4 x^2}-\frac {(a B-A b) b}{a^3 x^4}+\frac {a B-A b}{a^2 x^6}+\frac {A}{a x^8}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {b^2 \log \left (x^2\right ) (A b-a B)}{a^4}+\frac {b^2 (A b-a B) \log \left (a+b x^2\right )}{a^4}-\frac {b (A b-a B)}{a^3 x^2}+\frac {A b-a B}{2 a^2 x^4}-\frac {A}{3 a x^6}\right )\) |
Input:
Int[(A + B*x^2)/(x^7*(a + b*x^2)),x]
Output:
(-1/3*A/(a*x^6) + (A*b - a*B)/(2*a^2*x^4) - (b*(A*b - a*B))/(a^3*x^2) - (b ^2*(A*b - a*B)*Log[x^2])/a^4 + (b^2*(A*b - a*B)*Log[a + b*x^2])/a^4)/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 = 0.48 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.92
method | result | size |
default | \(\frac {b^{2} \left (A b -B a \right ) \ln \left (b \,x^{2}+a \right )}{2 a^{4}}-\frac {A}{6 a \,x^{6}}-\frac {-A b +B a}{4 a^{2} x^{4}}-\frac {b \left (A b -B a \right )}{2 a^{3} x^{2}}-\frac {b^{2} \left (A b -B a \right ) \ln \left (x \right )}{a^{4}}\) | \(86\) |
norman | \(\frac {-\frac {A}{6 a}+\frac {\left (A b -B a \right ) x^{2}}{4 a^{2}}-\frac {\left (A b -B a \right ) b \,x^{4}}{2 a^{3}}}{x^{6}}-\frac {b^{2} \left (A b -B a \right ) \ln \left (x \right )}{a^{4}}+\frac {b^{2} \left (A b -B a \right ) \ln \left (b \,x^{2}+a \right )}{2 a^{4}}\) | \(88\) |
risch | \(\frac {-\frac {A}{6 a}+\frac {\left (A b -B a \right ) x^{2}}{4 a^{2}}-\frac {\left (A b -B a \right ) b \,x^{4}}{2 a^{3}}}{x^{6}}-\frac {b^{3} \ln \left (x \right ) A}{a^{4}}+\frac {b^{2} \ln \left (x \right ) B}{a^{3}}+\frac {b^{3} \ln \left (-b \,x^{2}-a \right ) A}{2 a^{4}}-\frac {b^{2} \ln \left (-b \,x^{2}-a \right ) B}{2 a^{3}}\) | \(107\) |
parallelrisch | \(-\frac {12 A \ln \left (x \right ) x^{6} b^{3}-6 A \ln \left (b \,x^{2}+a \right ) x^{6} b^{3}-12 B \ln \left (x \right ) x^{6} a \,b^{2}+6 B \ln \left (b \,x^{2}+a \right ) x^{6} a \,b^{2}+6 a A \,b^{2} x^{4}-6 B \,a^{2} b \,x^{4}-3 a^{2} A b \,x^{2}+3 B \,a^{3} x^{2}+2 a^{3} A}{12 a^{4} x^{6}}\) | \(113\) |
Input:
int((B*x^2+A)/x^7/(b*x^2+a),x,method=_RETURNVERBOSE)
Output:
1/2*b^2*(A*b-B*a)*ln(b*x^2+a)/a^4-1/6*A/a/x^6-1/4*(-A*b+B*a)/a^2/x^4-1/2*b *(A*b-B*a)/a^3/x^2-b^2*(A*b-B*a)*ln(x)/a^4
Time = 0.09 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.05 \[ \int \frac {A+B x^2}{x^7 \left (a+b x^2\right )} \, dx=-\frac {6 \, {\left (B a b^{2} - A b^{3}\right )} x^{6} \log \left (b x^{2} + a\right ) - 12 \, {\left (B a b^{2} - A b^{3}\right )} x^{6} \log \left (x\right ) - 6 \, {\left (B a^{2} b - A a b^{2}\right )} x^{4} + 2 \, A a^{3} + 3 \, {\left (B a^{3} - A a^{2} b\right )} x^{2}}{12 \, a^{4} x^{6}} \] Input:
integrate((B*x^2+A)/x^7/(b*x^2+a),x, algorithm="fricas")
Output:
-1/12*(6*(B*a*b^2 - A*b^3)*x^6*log(b*x^2 + a) - 12*(B*a*b^2 - A*b^3)*x^6*l og(x) - 6*(B*a^2*b - A*a*b^2)*x^4 + 2*A*a^3 + 3*(B*a^3 - A*a^2*b)*x^2)/(a^ 4*x^6)
Time = 0.48 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.95 \[ \int \frac {A+B x^2}{x^7 \left (a+b x^2\right )} \, dx=\frac {- 2 A a^{2} + x^{4} \left (- 6 A b^{2} + 6 B a b\right ) + x^{2} \cdot \left (3 A a b - 3 B a^{2}\right )}{12 a^{3} x^{6}} + \frac {b^{2} \left (- A b + B a\right ) \log {\left (x \right )}}{a^{4}} - \frac {b^{2} \left (- A b + B a\right ) \log {\left (\frac {a}{b} + x^{2} \right )}}{2 a^{4}} \] Input:
integrate((B*x**2+A)/x**7/(b*x**2+a),x)
Output:
(-2*A*a**2 + x**4*(-6*A*b**2 + 6*B*a*b) + x**2*(3*A*a*b - 3*B*a**2))/(12*a **3*x**6) + b**2*(-A*b + B*a)*log(x)/a**4 - b**2*(-A*b + B*a)*log(a/b + x* *2)/(2*a**4)
Time = 0.05 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.03 \[ \int \frac {A+B x^2}{x^7 \left (a+b x^2\right )} \, dx=-\frac {{\left (B a b^{2} - A b^{3}\right )} \log \left (b x^{2} + a\right )}{2 \, a^{4}} + \frac {{\left (B a b^{2} - A b^{3}\right )} \log \left (x^{2}\right )}{2 \, a^{4}} + \frac {6 \, {\left (B a b - A b^{2}\right )} x^{4} - 2 \, A a^{2} - 3 \, {\left (B a^{2} - A a b\right )} x^{2}}{12 \, a^{3} x^{6}} \] Input:
integrate((B*x^2+A)/x^7/(b*x^2+a),x, algorithm="maxima")
Output:
-1/2*(B*a*b^2 - A*b^3)*log(b*x^2 + a)/a^4 + 1/2*(B*a*b^2 - A*b^3)*log(x^2) /a^4 + 1/12*(6*(B*a*b - A*b^2)*x^4 - 2*A*a^2 - 3*(B*a^2 - A*a*b)*x^2)/(a^3 *x^6)
Time = 0.12 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.35 \[ \int \frac {A+B x^2}{x^7 \left (a+b x^2\right )} \, dx=\frac {{\left (B a b^{2} - A b^{3}\right )} \log \left (x^{2}\right )}{2 \, a^{4}} - \frac {{\left (B a b^{3} - A b^{4}\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{4} b} - \frac {11 \, B a b^{2} x^{6} - 11 \, A b^{3} x^{6} - 6 \, B a^{2} b x^{4} + 6 \, A a b^{2} x^{4} + 3 \, B a^{3} x^{2} - 3 \, A a^{2} b x^{2} + 2 \, A a^{3}}{12 \, a^{4} x^{6}} \] Input:
integrate((B*x^2+A)/x^7/(b*x^2+a),x, algorithm="giac")
Output:
1/2*(B*a*b^2 - A*b^3)*log(x^2)/a^4 - 1/2*(B*a*b^3 - A*b^4)*log(abs(b*x^2 + a))/(a^4*b) - 1/12*(11*B*a*b^2*x^6 - 11*A*b^3*x^6 - 6*B*a^2*b*x^4 + 6*A*a *b^2*x^4 + 3*B*a^3*x^2 - 3*A*a^2*b*x^2 + 2*A*a^3)/(a^4*x^6)
Time = 0.36 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.99 \[ \int \frac {A+B x^2}{x^7 \left (a+b x^2\right )} \, dx=\frac {\ln \left (b\,x^2+a\right )\,\left (A\,b^3-B\,a\,b^2\right )}{2\,a^4}-\frac {\frac {A}{6\,a}-\frac {x^2\,\left (A\,b-B\,a\right )}{4\,a^2}+\frac {b\,x^4\,\left (A\,b-B\,a\right )}{2\,a^3}}{x^6}-\frac {\ln \left (x\right )\,\left (A\,b^3-B\,a\,b^2\right )}{a^4} \] Input:
int((A + B*x^2)/(x^7*(a + b*x^2)),x)
Output:
(log(a + b*x^2)*(A*b^3 - B*a*b^2))/(2*a^4) - (A/(6*a) - (x^2*(A*b - B*a))/ (4*a^2) + (b*x^4*(A*b - B*a))/(2*a^3))/x^6 - (log(x)*(A*b^3 - B*a*b^2))/a^ 4
Time = 0.19 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.05 \[ \int \frac {A+B x^2}{x^7 \left (a+b x^2\right )} \, dx=-\frac {1}{6 x^{6}} \] Input:
int((B*x^2+A)/x^7/(b*x^2+a),x)
Output:
( - 1)/(6*x**6)