Integrand size = 20, antiderivative size = 109 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {(A b-3 a B) x^2}{2 b^4}+\frac {B x^4}{4 b^3}+\frac {a^3 (A b-a B)}{4 b^5 \left (a+b x^2\right )^2}-\frac {a^2 (3 A b-4 a B)}{2 b^5 \left (a+b x^2\right )}-\frac {3 a (A b-2 a B) \log \left (a+b x^2\right )}{2 b^5} \] Output:
1/2*(A*b-3*B*a)*x^2/b^4+1/4*B*x^4/b^3+1/4*a^3*(A*b-B*a)/b^5/(b*x^2+a)^2-1/ 2*a^2*(3*A*b-4*B*a)/b^5/(b*x^2+a)-3/2*a*(A*b-2*B*a)*ln(b*x^2+a)/b^5
Time = 0.04 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.86 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {2 b (A b-3 a B) x^2+b^2 B x^4+\frac {a^3 (A b-a B)}{\left (a+b x^2\right )^2}+\frac {2 a^2 (-3 A b+4 a B)}{a+b x^2}+6 a (-A b+2 a B) \log \left (a+b x^2\right )}{4 b^5} \] Input:
Integrate[(x^7*(A + B*x^2))/(a + b*x^2)^3,x]
Output:
(2*b*(A*b - 3*a*B)*x^2 + b^2*B*x^4 + (a^3*(A*b - a*B))/(a + b*x^2)^2 + (2* a^2*(-3*A*b + 4*a*B))/(a + b*x^2) + 6*a*(-(A*b) + 2*a*B)*Log[a + b*x^2])/( 4*b^5)
Time = 0.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.97, 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.
\(\displaystyle \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {x^6 \left (B x^2+A\right )}{\left (b x^2+a\right )^3}dx^2\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} \int \left (\frac {(a B-A b) a^3}{b^4 \left (b x^2+a\right )^3}-\frac {(4 a B-3 A b) a^2}{b^4 \left (b x^2+a\right )^2}+\frac {3 (2 a B-A b) a}{b^4 \left (b x^2+a\right )}+\frac {B x^2}{b^3}+\frac {A b-3 a B}{b^4}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {a^3 (A b-a B)}{2 b^5 \left (a+b x^2\right )^2}-\frac {a^2 (3 A b-4 a B)}{b^5 \left (a+b x^2\right )}-\frac {3 a (A b-2 a B) \log \left (a+b x^2\right )}{b^5}+\frac {x^2 (A b-3 a B)}{b^4}+\frac {B x^4}{2 b^3}\right )\) |
Input:
Int[(x^7*(A + B*x^2))/(a + b*x^2)^3,x]
Output:
(((A*b - 3*a*B)*x^2)/b^4 + (B*x^4)/(2*b^3) + (a^3*(A*b - a*B))/(2*b^5*(a + b*x^2)^2) - (a^2*(3*A*b - 4*a*B))/(b^5*(a + b*x^2)) - (3*a*(A*b - 2*a*B)* Log[a + b*x^2])/b^5)/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 = 0.44 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.93
method | result | size |
norman | \(\frac {\frac {B \,x^{8}}{4 b}-\frac {a^{2} \left (9 a b A -18 a^{2} B \right )}{4 b^{5}}+\frac {\left (A b -2 B a \right ) x^{6}}{2 b^{2}}-\frac {a \left (3 a b A -6 a^{2} B \right ) x^{2}}{b^{4}}}{\left (b \,x^{2}+a \right )^{2}}-\frac {3 a \left (A b -2 B a \right ) \ln \left (b \,x^{2}+a \right )}{2 b^{5}}\) | \(101\) |
default | \(\frac {\left (b B \,x^{2}+A b -3 B a \right )^{2}}{4 b^{5} B}-\frac {a \left (-\frac {a^{2} \left (A b -B a \right )}{2 b \left (b \,x^{2}+a \right )^{2}}+\frac {a \left (3 A b -4 B a \right )}{b \left (b \,x^{2}+a \right )}+\frac {\left (3 A b -6 B a \right ) \ln \left (b \,x^{2}+a \right )}{b}\right )}{2 b^{4}}\) | \(102\) |
risch | \(\frac {B \,x^{4}}{4 b^{3}}+\frac {A \,x^{2}}{2 b^{3}}-\frac {3 B a \,x^{2}}{2 b^{4}}+\frac {A^{2}}{4 b^{3} B}-\frac {3 A a}{2 b^{4}}+\frac {9 B \,a^{2}}{4 b^{5}}+\frac {\left (-\frac {3}{2} a^{2} b A +2 a^{3} B \right ) x^{2}-\frac {a^{3} \left (5 A b -7 B a \right )}{4 b}}{b^{4} \left (b \,x^{2}+a \right )^{2}}-\frac {3 a \ln \left (b \,x^{2}+a \right ) A}{2 b^{4}}+\frac {3 a^{2} \ln \left (b \,x^{2}+a \right ) B}{b^{5}}\) | \(138\) |
parallelrisch | \(-\frac {-B \,x^{8} b^{4}-2 A \,x^{6} b^{4}+4 B \,x^{6} a \,b^{3}+6 A \ln \left (b \,x^{2}+a \right ) x^{4} a \,b^{3}-12 B \ln \left (b \,x^{2}+a \right ) x^{4} a^{2} b^{2}+12 A \ln \left (b \,x^{2}+a \right ) x^{2} a^{2} b^{2}-24 B \ln \left (b \,x^{2}+a \right ) x^{2} a^{3} b +12 A \,a^{2} b^{2} x^{2}-24 B \,a^{3} b \,x^{2}+6 A \ln \left (b \,x^{2}+a \right ) a^{3} b -12 B \ln \left (b \,x^{2}+a \right ) a^{4}+9 A \,a^{3} b -18 B \,a^{4}}{4 b^{5} \left (b \,x^{2}+a \right )^{2}}\) | \(184\) |
Input:
int(x^7*(B*x^2+A)/(b*x^2+a)^3,x,method=_RETURNVERBOSE)
Output:
(1/4*B*x^8/b-1/4*a^2*(9*A*a*b-18*B*a^2)/b^5+1/2*(A*b-2*B*a)/b^2*x^6-a*(3*A *a*b-6*B*a^2)/b^4*x^2)/(b*x^2+a)^2-3/2*a*(A*b-2*B*a)*ln(b*x^2+a)/b^5
Time = 0.08 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.64 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {B b^{4} x^{8} - 2 \, {\left (2 \, B a b^{3} - A b^{4}\right )} x^{6} + 7 \, B a^{4} - 5 \, A a^{3} b - {\left (11 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} x^{4} + 2 \, {\left (B a^{3} b - 2 \, A a^{2} b^{2}\right )} x^{2} + 6 \, {\left (2 \, B a^{4} - A a^{3} b + {\left (2 \, B a^{2} b^{2} - A a b^{3}\right )} x^{4} + 2 \, {\left (2 \, B a^{3} b - A a^{2} b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right )}{4 \, {\left (b^{7} x^{4} + 2 \, a b^{6} x^{2} + a^{2} b^{5}\right )}} \] Input:
integrate(x^7*(B*x^2+A)/(b*x^2+a)^3,x, algorithm="fricas")
Output:
1/4*(B*b^4*x^8 - 2*(2*B*a*b^3 - A*b^4)*x^6 + 7*B*a^4 - 5*A*a^3*b - (11*B*a ^2*b^2 - 4*A*a*b^3)*x^4 + 2*(B*a^3*b - 2*A*a^2*b^2)*x^2 + 6*(2*B*a^4 - A*a ^3*b + (2*B*a^2*b^2 - A*a*b^3)*x^4 + 2*(2*B*a^3*b - A*a^2*b^2)*x^2)*log(b* x^2 + a))/(b^7*x^4 + 2*a*b^6*x^2 + a^2*b^5)
Time = 0.75 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.09 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {B x^{4}}{4 b^{3}} + \frac {3 a \left (- A b + 2 B a\right ) \log {\left (a + b x^{2} \right )}}{2 b^{5}} + x^{2} \left (\frac {A}{2 b^{3}} - \frac {3 B a}{2 b^{4}}\right ) + \frac {- 5 A a^{3} b + 7 B a^{4} + x^{2} \left (- 6 A a^{2} b^{2} + 8 B a^{3} b\right )}{4 a^{2} b^{5} + 8 a b^{6} x^{2} + 4 b^{7} x^{4}} \] Input:
integrate(x**7*(B*x**2+A)/(b*x**2+a)**3,x)
Output:
B*x**4/(4*b**3) + 3*a*(-A*b + 2*B*a)*log(a + b*x**2)/(2*b**5) + x**2*(A/(2 *b**3) - 3*B*a/(2*b**4)) + (-5*A*a**3*b + 7*B*a**4 + x**2*(-6*A*a**2*b**2 + 8*B*a**3*b))/(4*a**2*b**5 + 8*a*b**6*x**2 + 4*b**7*x**4)
Time = 0.05 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.06 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {7 \, B a^{4} - 5 \, A a^{3} b + 2 \, {\left (4 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{2}}{4 \, {\left (b^{7} x^{4} + 2 \, a b^{6} x^{2} + a^{2} b^{5}\right )}} + \frac {B b x^{4} - 2 \, {\left (3 \, B a - A b\right )} x^{2}}{4 \, b^{4}} + \frac {3 \, {\left (2 \, B a^{2} - A a b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{5}} \] Input:
integrate(x^7*(B*x^2+A)/(b*x^2+a)^3,x, algorithm="maxima")
Output:
1/4*(7*B*a^4 - 5*A*a^3*b + 2*(4*B*a^3*b - 3*A*a^2*b^2)*x^2)/(b^7*x^4 + 2*a *b^6*x^2 + a^2*b^5) + 1/4*(B*b*x^4 - 2*(3*B*a - A*b)*x^2)/b^4 + 3/2*(2*B*a ^2 - A*a*b)*log(b*x^2 + a)/b^5
Time = 0.13 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.21 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {3 \, {\left (2 \, B a^{2} - A a b\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{5}} + \frac {B b^{3} x^{4} - 6 \, B a b^{2} x^{2} + 2 \, A b^{3} x^{2}}{4 \, b^{6}} - \frac {18 \, B a^{2} b^{2} x^{4} - 9 \, A a b^{3} x^{4} + 28 \, B a^{3} b x^{2} - 12 \, A a^{2} b^{2} x^{2} + 11 \, B a^{4} - 4 \, A a^{3} b}{4 \, {\left (b x^{2} + a\right )}^{2} b^{5}} \] Input:
integrate(x^7*(B*x^2+A)/(b*x^2+a)^3,x, algorithm="giac")
Output:
3/2*(2*B*a^2 - A*a*b)*log(abs(b*x^2 + a))/b^5 + 1/4*(B*b^3*x^4 - 6*B*a*b^2 *x^2 + 2*A*b^3*x^2)/b^6 - 1/4*(18*B*a^2*b^2*x^4 - 9*A*a*b^3*x^4 + 28*B*a^3 *b*x^2 - 12*A*a^2*b^2*x^2 + 11*B*a^4 - 4*A*a^3*b)/((b*x^2 + a)^2*b^5)
Time = 0.08 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.08 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {\frac {7\,B\,a^4-5\,A\,a^3\,b}{4\,b}+x^2\,\left (2\,B\,a^3-\frac {3\,A\,a^2\,b}{2}\right )}{a^2\,b^4+2\,a\,b^5\,x^2+b^6\,x^4}+x^2\,\left (\frac {A}{2\,b^3}-\frac {3\,B\,a}{2\,b^4}\right )+\frac {\ln \left (b\,x^2+a\right )\,\left (6\,B\,a^2-3\,A\,a\,b\right )}{2\,b^5}+\frac {B\,x^4}{4\,b^3} \] Input:
int((x^7*(A + B*x^2))/(a + b*x^2)^3,x)
Output:
((7*B*a^4 - 5*A*a^3*b)/(4*b) + x^2*(2*B*a^3 - (3*A*a^2*b)/2))/(a^2*b^4 + b ^6*x^4 + 2*a*b^5*x^2) + x^2*(A/(2*b^3) - (3*B*a)/(2*b^4)) + (log(a + b*x^2 )*(6*B*a^2 - 3*A*a*b))/(2*b^5) + (B*x^4)/(4*b^3)
Time = 0.21 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.64 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {6 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{3}+6 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} b \,x^{2}-6 a^{2} b \,x^{2}-3 a \,b^{2} x^{4}+b^{3} x^{6}}{4 b^{4} \left (b \,x^{2}+a \right )} \] Input:
int(x^7*(B*x^2+A)/(b*x^2+a)^3,x)
Output:
(6*log(a + b*x**2)*a**3 + 6*log(a + b*x**2)*a**2*b*x**2 - 6*a**2*b*x**2 - 3*a*b**2*x**4 + b**3*x**6)/(4*b**4*(a + b*x**2))