Integrand size = 20, antiderivative size = 107 \[ \int \frac {x^{11} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {(A b-3 a B) x^3}{3 b^4}+\frac {B x^6}{6 b^3}+\frac {a^3 (A b-a B)}{6 b^5 \left (a+b x^3\right )^2}-\frac {a^2 (3 A b-4 a B)}{3 b^5 \left (a+b x^3\right )}-\frac {a (A b-2 a B) \log \left (a+b x^3\right )}{b^5} \] Output:
1/3*(A*b-3*B*a)*x^3/b^4+1/6*B*x^6/b^3+1/6*a^3*(A*b-B*a)/b^5/(b*x^3+a)^2-1/ 3*a^2*(3*A*b-4*B*a)/b^5/(b*x^3+a)-a*(A*b-2*B*a)*ln(b*x^3+a)/b^5
Time = 0.07 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.88 \[ \int \frac {x^{11} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {2 b (A b-3 a B) x^3+b^2 B x^6+\frac {a^3 (A b-a B)}{\left (a+b x^3\right )^2}+\frac {2 a^2 (-3 A b+4 a B)}{a+b x^3}+6 a (-A b+2 a B) \log \left (a+b x^3\right )}{6 b^5} \] Input:
Integrate[(x^11*(A + B*x^3))/(a + b*x^3)^3,x]
Output:
(2*b*(A*b - 3*a*B)*x^3 + b^2*B*x^6 + (a^3*(A*b - a*B))/(a + b*x^3)^2 + (2* a^2*(-3*A*b + 4*a*B))/(a + b*x^3) + 6*a*(-(A*b) + 2*a*B)*Log[a + b*x^3])/( 6*b^5)
Time = 0.49 (sec) , antiderivative size = 106, 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 = {948, 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.
\(\displaystyle \int \frac {x^{11} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {x^9 \left (B x^3+A\right )}{\left (b x^3+a\right )^3}dx^3\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{3} \int \left (\frac {(a B-A b) a^3}{b^4 \left (b x^3+a\right )^3}-\frac {(4 a B-3 A b) a^2}{b^4 \left (b x^3+a\right )^2}+\frac {3 (2 a B-A b) a}{b^4 \left (b x^3+a\right )}+\frac {B x^3}{b^3}+\frac {A b-3 a B}{b^4}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {a^3 (A b-a B)}{2 b^5 \left (a+b x^3\right )^2}-\frac {a^2 (3 A b-4 a B)}{b^5 \left (a+b x^3\right )}-\frac {3 a (A b-2 a B) \log \left (a+b x^3\right )}{b^5}+\frac {x^3 (A b-3 a B)}{b^4}+\frac {B x^6}{2 b^3}\right )\) |
Input:
Int[(x^11*(A + B*x^3))/(a + b*x^3)^3,x]
Output:
(((A*b - 3*a*B)*x^3)/b^4 + (B*x^6)/(2*b^3) + (a^3*(A*b - a*B))/(2*b^5*(a + b*x^3)^2) - (a^2*(3*A*b - 4*a*B))/(b^5*(a + b*x^3)) - (3*a*(A*b - 2*a*B)* Log[a + b*x^3])/b^5)/3
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_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Time = 0.86 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.93
method | result | size |
norman | \(\frac {\frac {B \,x^{12}}{6 b}-\frac {a^{2} \left (3 a b A -6 a^{2} B \right )}{2 b^{5}}+\frac {\left (A b -2 B a \right ) x^{9}}{3 b^{2}}-\frac {2 a \left (a b A -2 a^{2} B \right ) x^{3}}{b^{4}}}{\left (b \,x^{3}+a \right )^{2}}-\frac {a \left (A b -2 B a \right ) \ln \left (b \,x^{3}+a \right )}{b^{5}}\) | \(100\) |
default | \(\frac {\left (b B \,x^{3}+A b -3 B a \right )^{2}}{6 b^{5} B}-\frac {a \left (-\frac {a^{2} \left (A b -B a \right )}{2 b \left (b \,x^{3}+a \right )^{2}}+\frac {a \left (3 A b -4 B a \right )}{b \left (b \,x^{3}+a \right )}+\frac {\left (3 A b -6 B a \right ) \ln \left (b \,x^{3}+a \right )}{b}\right )}{3 b^{4}}\) | \(102\) |
risch | \(\frac {B \,x^{6}}{6 b^{3}}+\frac {A \,x^{3}}{3 b^{3}}-\frac {B a \,x^{3}}{b^{4}}+\frac {A^{2}}{6 b^{3} B}-\frac {A a}{b^{4}}+\frac {3 B \,a^{2}}{2 b^{5}}+\frac {\left (-a^{2} b A +\frac {4}{3} a^{3} B \right ) x^{3}-\frac {a^{3} \left (5 A b -7 B a \right )}{6 b}}{b^{4} \left (b \,x^{3}+a \right )^{2}}-\frac {a \ln \left (b \,x^{3}+a \right ) A}{b^{4}}+\frac {2 a^{2} \ln \left (b \,x^{3}+a \right ) B}{b^{5}}\) | \(138\) |
parallelrisch | \(-\frac {-B \,x^{12} b^{4}-2 A \,x^{9} b^{4}+4 B \,x^{9} a \,b^{3}+6 A \ln \left (b \,x^{3}+a \right ) x^{6} a \,b^{3}-12 B \ln \left (b \,x^{3}+a \right ) x^{6} a^{2} b^{2}+12 A \ln \left (b \,x^{3}+a \right ) x^{3} a^{2} b^{2}-24 B \ln \left (b \,x^{3}+a \right ) x^{3} a^{3} b +12 A \,a^{2} b^{2} x^{3}-24 B \,a^{3} b \,x^{3}+6 A \ln \left (b \,x^{3}+a \right ) a^{3} b -12 B \ln \left (b \,x^{3}+a \right ) a^{4}+9 A \,a^{3} b -18 B \,a^{4}}{6 b^{5} \left (b \,x^{3}+a \right )^{2}}\) | \(184\) |
Input:
int(x^11*(B*x^3+A)/(b*x^3+a)^3,x,method=_RETURNVERBOSE)
Output:
(1/6*B/b*x^12-1/2*a^2*(3*A*a*b-6*B*a^2)/b^5+1/3*(A*b-2*B*a)/b^2*x^9-2*a*(A *a*b-2*B*a^2)/b^4*x^3)/(b*x^3+a)^2-a*(A*b-2*B*a)*ln(b*x^3+a)/b^5
Time = 0.11 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.67 \[ \int \frac {x^{11} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {B b^{4} x^{12} - 2 \, {\left (2 \, B a b^{3} - A b^{4}\right )} x^{9} - {\left (11 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} x^{6} + 7 \, B a^{4} - 5 \, A a^{3} b + 2 \, {\left (B a^{3} b - 2 \, A a^{2} b^{2}\right )} x^{3} + 6 \, {\left ({\left (2 \, B a^{2} b^{2} - A a b^{3}\right )} x^{6} + 2 \, B a^{4} - A a^{3} b + 2 \, {\left (2 \, B a^{3} b - A a^{2} b^{2}\right )} x^{3}\right )} \log \left (b x^{3} + a\right )}{6 \, {\left (b^{7} x^{6} + 2 \, a b^{6} x^{3} + a^{2} b^{5}\right )}} \] Input:
integrate(x^11*(B*x^3+A)/(b*x^3+a)^3,x, algorithm="fricas")
Output:
1/6*(B*b^4*x^12 - 2*(2*B*a*b^3 - A*b^4)*x^9 - (11*B*a^2*b^2 - 4*A*a*b^3)*x ^6 + 7*B*a^4 - 5*A*a^3*b + 2*(B*a^3*b - 2*A*a^2*b^2)*x^3 + 6*((2*B*a^2*b^2 - A*a*b^3)*x^6 + 2*B*a^4 - A*a^3*b + 2*(2*B*a^3*b - A*a^2*b^2)*x^3)*log(b *x^3 + a))/(b^7*x^6 + 2*a*b^6*x^3 + a^2*b^5)
Time = 1.38 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.05 \[ \int \frac {x^{11} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {B x^{6}}{6 b^{3}} + \frac {a \left (- A b + 2 B a\right ) \log {\left (a + b x^{3} \right )}}{b^{5}} + x^{3} \left (\frac {A}{3 b^{3}} - \frac {B a}{b^{4}}\right ) + \frac {- 5 A a^{3} b + 7 B a^{4} + x^{3} \left (- 6 A a^{2} b^{2} + 8 B a^{3} b\right )}{6 a^{2} b^{5} + 12 a b^{6} x^{3} + 6 b^{7} x^{6}} \] Input:
integrate(x**11*(B*x**3+A)/(b*x**3+a)**3,x)
Output:
B*x**6/(6*b**3) + a*(-A*b + 2*B*a)*log(a + b*x**3)/b**5 + x**3*(A/(3*b**3) - B*a/b**4) + (-5*A*a**3*b + 7*B*a**4 + x**3*(-6*A*a**2*b**2 + 8*B*a**3*b ))/(6*a**2*b**5 + 12*a*b**6*x**3 + 6*b**7*x**6)
Time = 0.03 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.07 \[ \int \frac {x^{11} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {7 \, B a^{4} - 5 \, A a^{3} b + 2 \, {\left (4 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{3}}{6 \, {\left (b^{7} x^{6} + 2 \, a b^{6} x^{3} + a^{2} b^{5}\right )}} + \frac {B b x^{6} - 2 \, {\left (3 \, B a - A b\right )} x^{3}}{6 \, b^{4}} + \frac {{\left (2 \, B a^{2} - A a b\right )} \log \left (b x^{3} + a\right )}{b^{5}} \] Input:
integrate(x^11*(B*x^3+A)/(b*x^3+a)^3,x, algorithm="maxima")
Output:
1/6*(7*B*a^4 - 5*A*a^3*b + 2*(4*B*a^3*b - 3*A*a^2*b^2)*x^3)/(b^7*x^6 + 2*a *b^6*x^3 + a^2*b^5) + 1/6*(B*b*x^6 - 2*(3*B*a - A*b)*x^3)/b^4 + (2*B*a^2 - A*a*b)*log(b*x^3 + a)/b^5
Time = 0.13 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.22 \[ \int \frac {x^{11} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {{\left (2 \, B a^{2} - A a b\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{b^{5}} + \frac {B b^{3} x^{6} - 6 \, B a b^{2} x^{3} + 2 \, A b^{3} x^{3}}{6 \, b^{6}} - \frac {18 \, B a^{2} b^{2} x^{6} - 9 \, A a b^{3} x^{6} + 28 \, B a^{3} b x^{3} - 12 \, A a^{2} b^{2} x^{3} + 11 \, B a^{4} - 4 \, A a^{3} b}{6 \, {\left (b x^{3} + a\right )}^{2} b^{5}} \] Input:
integrate(x^11*(B*x^3+A)/(b*x^3+a)^3,x, algorithm="giac")
Output:
(2*B*a^2 - A*a*b)*log(abs(b*x^3 + a))/b^5 + 1/6*(B*b^3*x^6 - 6*B*a*b^2*x^3 + 2*A*b^3*x^3)/b^6 - 1/6*(18*B*a^2*b^2*x^6 - 9*A*a*b^3*x^6 + 28*B*a^3*b*x ^3 - 12*A*a^2*b^2*x^3 + 11*B*a^4 - 4*A*a^3*b)/((b*x^3 + a)^2*b^5)
Time = 0.13 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.09 \[ \int \frac {x^{11} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {\frac {7\,B\,a^4-5\,A\,a^3\,b}{6\,b}+x^3\,\left (\frac {4\,B\,a^3}{3}-A\,a^2\,b\right )}{a^2\,b^4+2\,a\,b^5\,x^3+b^6\,x^6}+x^3\,\left (\frac {A}{3\,b^3}-\frac {B\,a}{b^4}\right )+\frac {\ln \left (b\,x^3+a\right )\,\left (2\,B\,a^2-A\,a\,b\right )}{b^5}+\frac {B\,x^6}{6\,b^3} \] Input:
int((x^11*(A + B*x^3))/(a + b*x^3)^3,x)
Output:
((7*B*a^4 - 5*A*a^3*b)/(6*b) + x^3*((4*B*a^3)/3 - A*a^2*b))/(a^2*b^4 + b^6 *x^6 + 2*a*b^5*x^3) + x^3*(A/(3*b^3) - (B*a)/b^4) + (log(a + b*x^3)*(2*B*a ^2 - A*a*b))/b^5 + (B*x^6)/(6*b^3)
Time = 0.22 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.21 \[ \int \frac {x^{11} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {6 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a^{3}+6 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a^{2} b \,x^{3}+6 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a^{3}+6 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a^{2} b \,x^{3}-6 a^{2} b \,x^{3}-3 a \,b^{2} x^{6}+b^{3} x^{9}}{6 b^{4} \left (b \,x^{3}+a \right )} \] Input:
int(x^11*(B*x^3+A)/(b*x^3+a)^3,x)
Output:
(6*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**3 + 6*log(a**(2/ 3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**2*b*x**3 + 6*log(a**(1/3) + b **(1/3)*x)*a**3 + 6*log(a**(1/3) + b**(1/3)*x)*a**2*b*x**3 - 6*a**2*b*x**3 - 3*a*b**2*x**6 + b**3*x**9)/(6*b**4*(a + b*x**3))