Integrand size = 22, antiderivative size = 83 \[ \int \frac {A+B x^3}{x^6 \left (a+b x^3\right )^{4/3}} \, dx=-\frac {A}{5 a x^5 \sqrt [3]{a+b x^3}}-\frac {6 A b-5 a B}{5 a^2 x^2 \sqrt [3]{a+b x^3}}+\frac {3 (6 A b-5 a B) \left (a+b x^3\right )^{2/3}}{10 a^3 x^2} \] Output:
-1/5*A/a/x^5/(b*x^3+a)^(1/3)-1/5*(6*A*b-5*B*a)/a^2/x^2/(b*x^3+a)^(1/3)+3/1 0*(6*A*b-5*B*a)*(b*x^3+a)^(2/3)/a^3/x^2
Time = 0.44 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.75 \[ \int \frac {A+B x^3}{x^6 \left (a+b x^3\right )^{4/3}} \, dx=\frac {-2 a^2 A+6 a A b x^3-5 a^2 B x^3+18 A b^2 x^6-15 a b B x^6}{10 a^3 x^5 \sqrt [3]{a+b x^3}} \] Input:
Integrate[(A + B*x^3)/(x^6*(a + b*x^3)^(4/3)),x]
Output:
(-2*a^2*A + 6*a*A*b*x^3 - 5*a^2*B*x^3 + 18*A*b^2*x^6 - 15*a*b*B*x^6)/(10*a ^3*x^5*(a + b*x^3)^(1/3))
Time = 0.34 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {955, 803, 746}
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.
\(\displaystyle \int \frac {A+B x^3}{x^6 \left (a+b x^3\right )^{4/3}} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle -\frac {(6 A b-5 a B) \int \frac {1}{x^3 \left (b x^3+a\right )^{4/3}}dx}{5 a}-\frac {A}{5 a x^5 \sqrt [3]{a+b x^3}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {(6 A b-5 a B) \left (-\frac {3 b \int \frac {1}{\left (b x^3+a\right )^{4/3}}dx}{2 a}-\frac {1}{2 a x^2 \sqrt [3]{a+b x^3}}\right )}{5 a}-\frac {A}{5 a x^5 \sqrt [3]{a+b x^3}}\) |
\(\Big \downarrow \) 746 |
\(\displaystyle -\frac {\left (-\frac {3 b x}{2 a^2 \sqrt [3]{a+b x^3}}-\frac {1}{2 a x^2 \sqrt [3]{a+b x^3}}\right ) (6 A b-5 a B)}{5 a}-\frac {A}{5 a x^5 \sqrt [3]{a+b x^3}}\) |
Input:
Int[(A + B*x^3)/(x^6*(a + b*x^3)^(4/3)),x]
Output:
-1/5*A/(a*x^5*(a + b*x^3)^(1/3)) - ((6*A*b - 5*a*B)*(-1/2*1/(a*x^2*(a + b* x^3)^(1/3)) - (3*b*x)/(2*a^2*(a + b*x^3)^(1/3))))/(5*a)
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1) /a), x] /; FreeQ[{a, b, n, p}, x] && EqQ[1/n + p + 1, 0]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*(( a + b*x^n)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + n*(p + 1) + 1)/(a*(m + 1 ))) Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x] && I LtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)) Int[(e *x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b* c - a*d, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) && !ILtQ[p, -1]
Time = 0.86 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.69
method | result | size |
pseudoelliptic | \(\frac {\left (-5 B \,x^{3}-2 A \right ) a^{2}+6 b \left (-\frac {5 B \,x^{3}}{2}+A \right ) x^{3} a +18 A \,b^{2} x^{6}}{10 \left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{5} a^{3}}\) | \(57\) |
gosper | \(-\frac {-18 A \,b^{2} x^{6}+15 B a b \,x^{6}-6 a A b \,x^{3}+5 B \,a^{2} x^{3}+2 a^{2} A}{10 x^{5} \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{3}}\) | \(59\) |
trager | \(-\frac {-18 A \,b^{2} x^{6}+15 B a b \,x^{6}-6 a A b \,x^{3}+5 B \,a^{2} x^{3}+2 a^{2} A}{10 x^{5} \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{3}}\) | \(59\) |
orering | \(-\frac {-18 A \,b^{2} x^{6}+15 B a b \,x^{6}-6 a A b \,x^{3}+5 B \,a^{2} x^{3}+2 a^{2} A}{10 x^{5} \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{3}}\) | \(59\) |
risch | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (-8 A b \,x^{3}+5 B a \,x^{3}+2 A a \right )}{10 a^{3} x^{5}}+\frac {x \left (A b -B a \right ) b}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{3}}\) | \(61\) |
Input:
int((B*x^3+A)/x^6/(b*x^3+a)^(4/3),x,method=_RETURNVERBOSE)
Output:
1/10*((-5*B*x^3-2*A)*a^2+6*b*(-5/2*B*x^3+A)*x^3*a+18*A*b^2*x^6)/(b*x^3+a)^ (1/3)/x^5/a^3
Time = 0.09 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.83 \[ \int \frac {A+B x^3}{x^6 \left (a+b x^3\right )^{4/3}} \, dx=-\frac {{\left (3 \, {\left (5 \, B a b - 6 \, A b^{2}\right )} x^{6} + {\left (5 \, B a^{2} - 6 \, A a b\right )} x^{3} + 2 \, A a^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{10 \, {\left (a^{3} b x^{8} + a^{4} x^{5}\right )}} \] Input:
integrate((B*x^3+A)/x^6/(b*x^3+a)^(4/3),x, algorithm="fricas")
Output:
-1/10*(3*(5*B*a*b - 6*A*b^2)*x^6 + (5*B*a^2 - 6*A*a*b)*x^3 + 2*A*a^2)*(b*x ^3 + a)^(2/3)/(a^3*b*x^8 + a^4*x^5)
Leaf count of result is larger than twice the leaf count of optimal. 396 vs. \(2 (75) = 150\).
Time = 21.63 (sec) , antiderivative size = 396, normalized size of antiderivative = 4.77 \[ \int \frac {A+B x^3}{x^6 \left (a+b x^3\right )^{4/3}} \, dx=A \left (- \frac {2 a^{3} b^{\frac {14}{3}} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {5}{3}\right )}{27 a^{5} b^{4} x^{3} \Gamma \left (\frac {4}{3}\right ) + 54 a^{4} b^{5} x^{6} \Gamma \left (\frac {4}{3}\right ) + 27 a^{3} b^{6} x^{9} \Gamma \left (\frac {4}{3}\right )} + \frac {4 a^{2} b^{\frac {17}{3}} x^{3} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {5}{3}\right )}{27 a^{5} b^{4} x^{3} \Gamma \left (\frac {4}{3}\right ) + 54 a^{4} b^{5} x^{6} \Gamma \left (\frac {4}{3}\right ) + 27 a^{3} b^{6} x^{9} \Gamma \left (\frac {4}{3}\right )} + \frac {24 a b^{\frac {20}{3}} x^{6} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {5}{3}\right )}{27 a^{5} b^{4} x^{3} \Gamma \left (\frac {4}{3}\right ) + 54 a^{4} b^{5} x^{6} \Gamma \left (\frac {4}{3}\right ) + 27 a^{3} b^{6} x^{9} \Gamma \left (\frac {4}{3}\right )} + \frac {18 b^{\frac {23}{3}} x^{9} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {5}{3}\right )}{27 a^{5} b^{4} x^{3} \Gamma \left (\frac {4}{3}\right ) + 54 a^{4} b^{5} x^{6} \Gamma \left (\frac {4}{3}\right ) + 27 a^{3} b^{6} x^{9} \Gamma \left (\frac {4}{3}\right )}\right ) + B \left (\frac {\Gamma \left (- \frac {2}{3}\right )}{9 a \sqrt [3]{b} x^{3} \sqrt [3]{\frac {a}{b x^{3}} + 1} \Gamma \left (\frac {4}{3}\right )} + \frac {b^{\frac {2}{3}} \Gamma \left (- \frac {2}{3}\right )}{3 a^{2} \sqrt [3]{\frac {a}{b x^{3}} + 1} \Gamma \left (\frac {4}{3}\right )}\right ) \] Input:
integrate((B*x**3+A)/x**6/(b*x**3+a)**(4/3),x)
Output:
A*(-2*a**3*b**(14/3)*(a/(b*x**3) + 1)**(2/3)*gamma(-5/3)/(27*a**5*b**4*x** 3*gamma(4/3) + 54*a**4*b**5*x**6*gamma(4/3) + 27*a**3*b**6*x**9*gamma(4/3) ) + 4*a**2*b**(17/3)*x**3*(a/(b*x**3) + 1)**(2/3)*gamma(-5/3)/(27*a**5*b** 4*x**3*gamma(4/3) + 54*a**4*b**5*x**6*gamma(4/3) + 27*a**3*b**6*x**9*gamma (4/3)) + 24*a*b**(20/3)*x**6*(a/(b*x**3) + 1)**(2/3)*gamma(-5/3)/(27*a**5* b**4*x**3*gamma(4/3) + 54*a**4*b**5*x**6*gamma(4/3) + 27*a**3*b**6*x**9*ga mma(4/3)) + 18*b**(23/3)*x**9*(a/(b*x**3) + 1)**(2/3)*gamma(-5/3)/(27*a**5 *b**4*x**3*gamma(4/3) + 54*a**4*b**5*x**6*gamma(4/3) + 27*a**3*b**6*x**9*g amma(4/3))) + B*(gamma(-2/3)/(9*a*b**(1/3)*x**3*(a/(b*x**3) + 1)**(1/3)*ga mma(4/3)) + b**(2/3)*gamma(-2/3)/(3*a**2*(a/(b*x**3) + 1)**(1/3)*gamma(4/3 )))
Time = 0.04 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.12 \[ \int \frac {A+B x^3}{x^6 \left (a+b x^3\right )^{4/3}} \, dx=-\frac {1}{2} \, B {\left (\frac {2 \, b x}{{\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{2}} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{a^{2} x^{2}}\right )} + \frac {1}{5} \, A {\left (\frac {5 \, b^{2} x}{{\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{3}} + \frac {\frac {5 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b}{x^{2}} - \frac {{\left (b x^{3} + a\right )}^{\frac {5}{3}}}{x^{5}}}{a^{3}}\right )} \] Input:
integrate((B*x^3+A)/x^6/(b*x^3+a)^(4/3),x, algorithm="maxima")
Output:
-1/2*B*(2*b*x/((b*x^3 + a)^(1/3)*a^2) + (b*x^3 + a)^(2/3)/(a^2*x^2)) + 1/5 *A*(5*b^2*x/((b*x^3 + a)^(1/3)*a^3) + (5*(b*x^3 + a)^(2/3)*b/x^2 - (b*x^3 + a)^(5/3)/x^5)/a^3)
\[ \int \frac {A+B x^3}{x^6 \left (a+b x^3\right )^{4/3}} \, dx=\int { \frac {B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac {4}{3}} x^{6}} \,d x } \] Input:
integrate((B*x^3+A)/x^6/(b*x^3+a)^(4/3),x, algorithm="giac")
Output:
integrate((B*x^3 + A)/((b*x^3 + a)^(4/3)*x^6), x)
Time = 0.95 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.84 \[ \int \frac {A+B x^3}{x^6 \left (a+b x^3\right )^{4/3}} \, dx=\frac {18\,A\,{\left (b\,x^3+a\right )}^2+10\,A\,a^2+10\,B\,a^2\,x^3-30\,A\,a\,\left (b\,x^3+a\right )-15\,B\,a\,x^3\,\left (b\,x^3+a\right )}{10\,a^3\,x^5\,{\left (b\,x^3+a\right )}^{1/3}} \] Input:
int((A + B*x^3)/(x^6*(a + b*x^3)^(4/3)),x)
Output:
(18*A*(a + b*x^3)^2 + 10*A*a^2 + 10*B*a^2*x^3 - 30*A*a*(a + b*x^3) - 15*B* a*x^3*(a + b*x^3))/(10*a^3*x^5*(a + b*x^3)^(1/3))
\[ \int \frac {A+B x^3}{x^6 \left (a+b x^3\right )^{4/3}} \, dx=\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{6}}d x \] Input:
int((B*x^3+A)/x^6/(b*x^3+a)^(4/3),x)
Output:
int(1/((a + b*x**3)**(1/3)*x**6),x)