Integrand size = 19, antiderivative size = 91 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{10/3}} \, dx=\frac {(b c-a d) x}{7 a b \left (a+b x^3\right )^{7/3}}+\frac {(6 b c+a d) x}{28 a^2 b \left (a+b x^3\right )^{4/3}}+\frac {3 (6 b c+a d) x}{28 a^3 b \sqrt [3]{a+b x^3}} \]
1/7*(-a*d+b*c)*x/a/b/(b*x^3+a)^(7/3)+1/28*(a*d+6*b*c)*x/a^2/b/(b*x^3+a)^(4 /3)+3/28*(a*d+6*b*c)*x/a^3/b/(b*x^3+a)^(1/3)
Time = 0.52 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.66 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{10/3}} \, dx=\frac {28 a^2 c x+42 a b c x^4+7 a^2 d x^4+18 b^2 c x^7+3 a b d x^7}{28 a^3 \left (a+b x^3\right )^{7/3}} \]
(28*a^2*c*x + 42*a*b*c*x^4 + 7*a^2*d*x^4 + 18*b^2*c*x^7 + 3*a*b*d*x^7)/(28 *a^3*(a + b*x^3)^(7/3))
Time = 0.21 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {910, 749, 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.
\(\displaystyle \int \frac {c+d x^3}{\left (a+b x^3\right )^{10/3}} \, dx\) |
\(\Big \downarrow \) 910 |
\(\displaystyle \frac {(a d+6 b c) \int \frac {1}{\left (b x^3+a\right )^{7/3}}dx}{7 a b}+\frac {x (b c-a d)}{7 a b \left (a+b x^3\right )^{7/3}}\) |
\(\Big \downarrow \) 749 |
\(\displaystyle \frac {(a d+6 b c) \left (\frac {3 \int \frac {1}{\left (b x^3+a\right )^{4/3}}dx}{4 a}+\frac {x}{4 a \left (a+b x^3\right )^{4/3}}\right )}{7 a b}+\frac {x (b c-a d)}{7 a b \left (a+b x^3\right )^{7/3}}\) |
\(\Big \downarrow \) 746 |
\(\displaystyle \frac {\left (\frac {3 x}{4 a^2 \sqrt [3]{a+b x^3}}+\frac {x}{4 a \left (a+b x^3\right )^{4/3}}\right ) (a d+6 b c)}{7 a b}+\frac {x (b c-a d)}{7 a b \left (a+b x^3\right )^{7/3}}\) |
((b*c - a*d)*x)/(7*a*b*(a + b*x^3)^(7/3)) + ((6*b*c + a*d)*(x/(4*a*(a + b* x^3)^(4/3)) + (3*x)/(4*a^2*(a + b*x^3)^(1/3))))/(7*a*b)
3.1.61.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Simp[(n*(p + 1) + 1)/(a*n*(p + 1)) Int[(a + b*x^ n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (Inte gerQ[2*p] || Denominator[p + 1/n] < Denominator[p])
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Si mp[(-(b*c - a*d))*x*((a + b*x^n)^(p + 1)/(a*b*n*(p + 1))), x] - Simp[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)) Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/ n + p, 0])
Time = 3.93 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.57
method | result | size |
pseudoelliptic | \(\frac {x \left (\left (\frac {d \,x^{3}}{4}+c \right ) a^{2}+\frac {3 x^{3} \left (\frac {d \,x^{3}}{14}+c \right ) b a}{2}+\frac {9 b^{2} c \,x^{6}}{14}\right )}{\left (b \,x^{3}+a \right )^{\frac {7}{3}} a^{3}}\) | \(52\) |
gosper | \(\frac {x \left (3 a b d \,x^{6}+18 b^{2} c \,x^{6}+7 a^{2} d \,x^{3}+42 a b c \,x^{3}+28 a^{2} c \right )}{28 \left (b \,x^{3}+a \right )^{\frac {7}{3}} a^{3}}\) | \(57\) |
trager | \(\frac {x \left (3 a b d \,x^{6}+18 b^{2} c \,x^{6}+7 a^{2} d \,x^{3}+42 a b c \,x^{3}+28 a^{2} c \right )}{28 \left (b \,x^{3}+a \right )^{\frac {7}{3}} a^{3}}\) | \(57\) |
Time = 0.32 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.96 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{10/3}} \, dx=\frac {{\left (3 \, {\left (6 \, b^{2} c + a b d\right )} x^{7} + 7 \, {\left (6 \, a b c + a^{2} d\right )} x^{4} + 28 \, a^{2} c x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{28 \, {\left (a^{3} b^{3} x^{9} + 3 \, a^{4} b^{2} x^{6} + 3 \, a^{5} b x^{3} + a^{6}\right )}} \]
1/28*(3*(6*b^2*c + a*b*d)*x^7 + 7*(6*a*b*c + a^2*d)*x^4 + 28*a^2*c*x)*(b*x ^3 + a)^(2/3)/(a^3*b^3*x^9 + 3*a^4*b^2*x^6 + 3*a^5*b*x^3 + a^6)
Leaf count of result is larger than twice the leaf count of optimal. 709 vs. \(2 (83) = 166\).
Time = 100.01 (sec) , antiderivative size = 709, normalized size of antiderivative = 7.79 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{10/3}} \, dx=c \left (\frac {28 a^{5} x \Gamma \left (\frac {1}{3}\right )}{27 a^{\frac {25}{3}} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 81 a^{\frac {22}{3}} b x^{3} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 81 a^{\frac {19}{3}} b^{2} x^{6} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 27 a^{\frac {16}{3}} b^{3} x^{9} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right )} + \frac {70 a^{4} b x^{4} \Gamma \left (\frac {1}{3}\right )}{27 a^{\frac {25}{3}} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 81 a^{\frac {22}{3}} b x^{3} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 81 a^{\frac {19}{3}} b^{2} x^{6} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 27 a^{\frac {16}{3}} b^{3} x^{9} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right )} + \frac {60 a^{3} b^{2} x^{7} \Gamma \left (\frac {1}{3}\right )}{27 a^{\frac {25}{3}} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 81 a^{\frac {22}{3}} b x^{3} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 81 a^{\frac {19}{3}} b^{2} x^{6} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 27 a^{\frac {16}{3}} b^{3} x^{9} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right )} + \frac {18 a^{2} b^{3} x^{10} \Gamma \left (\frac {1}{3}\right )}{27 a^{\frac {25}{3}} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 81 a^{\frac {22}{3}} b x^{3} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 81 a^{\frac {19}{3}} b^{2} x^{6} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 27 a^{\frac {16}{3}} b^{3} x^{9} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right )}\right ) + d \left (\frac {7 a x^{4} \Gamma \left (\frac {4}{3}\right )}{9 a^{\frac {13}{3}} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 18 a^{\frac {10}{3}} b x^{3} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 9 a^{\frac {7}{3}} b^{2} x^{6} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right )} + \frac {3 b x^{7} \Gamma \left (\frac {4}{3}\right )}{9 a^{\frac {13}{3}} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 18 a^{\frac {10}{3}} b x^{3} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 9 a^{\frac {7}{3}} b^{2} x^{6} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right )}\right ) \]
c*(28*a**5*x*gamma(1/3)/(27*a**(25/3)*(1 + b*x**3/a)**(1/3)*gamma(10/3) + 81*a**(22/3)*b*x**3*(1 + b*x**3/a)**(1/3)*gamma(10/3) + 81*a**(19/3)*b**2* x**6*(1 + b*x**3/a)**(1/3)*gamma(10/3) + 27*a**(16/3)*b**3*x**9*(1 + b*x** 3/a)**(1/3)*gamma(10/3)) + 70*a**4*b*x**4*gamma(1/3)/(27*a**(25/3)*(1 + b* x**3/a)**(1/3)*gamma(10/3) + 81*a**(22/3)*b*x**3*(1 + b*x**3/a)**(1/3)*gam ma(10/3) + 81*a**(19/3)*b**2*x**6*(1 + b*x**3/a)**(1/3)*gamma(10/3) + 27*a **(16/3)*b**3*x**9*(1 + b*x**3/a)**(1/3)*gamma(10/3)) + 60*a**3*b**2*x**7* gamma(1/3)/(27*a**(25/3)*(1 + b*x**3/a)**(1/3)*gamma(10/3) + 81*a**(22/3)* b*x**3*(1 + b*x**3/a)**(1/3)*gamma(10/3) + 81*a**(19/3)*b**2*x**6*(1 + b*x **3/a)**(1/3)*gamma(10/3) + 27*a**(16/3)*b**3*x**9*(1 + b*x**3/a)**(1/3)*g amma(10/3)) + 18*a**2*b**3*x**10*gamma(1/3)/(27*a**(25/3)*(1 + b*x**3/a)** (1/3)*gamma(10/3) + 81*a**(22/3)*b*x**3*(1 + b*x**3/a)**(1/3)*gamma(10/3) + 81*a**(19/3)*b**2*x**6*(1 + b*x**3/a)**(1/3)*gamma(10/3) + 27*a**(16/3)* b**3*x**9*(1 + b*x**3/a)**(1/3)*gamma(10/3))) + d*(7*a*x**4*gamma(4/3)/(9* a**(13/3)*(1 + b*x**3/a)**(1/3)*gamma(10/3) + 18*a**(10/3)*b*x**3*(1 + b*x **3/a)**(1/3)*gamma(10/3) + 9*a**(7/3)*b**2*x**6*(1 + b*x**3/a)**(1/3)*gam ma(10/3)) + 3*b*x**7*gamma(4/3)/(9*a**(13/3)*(1 + b*x**3/a)**(1/3)*gamma(1 0/3) + 18*a**(10/3)*b*x**3*(1 + b*x**3/a)**(1/3)*gamma(10/3) + 9*a**(7/3)* b**2*x**6*(1 + b*x**3/a)**(1/3)*gamma(10/3)))
Time = 0.20 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.95 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{10/3}} \, dx=-\frac {{\left (4 \, b - \frac {7 \, {\left (b x^{3} + a\right )}}{x^{3}}\right )} d x^{7}}{28 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} a^{2}} + \frac {{\left (2 \, b^{2} - \frac {7 \, {\left (b x^{3} + a\right )} b}{x^{3}} + \frac {14 \, {\left (b x^{3} + a\right )}^{2}}{x^{6}}\right )} c x^{7}}{14 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} a^{3}} \]
-1/28*(4*b - 7*(b*x^3 + a)/x^3)*d*x^7/((b*x^3 + a)^(7/3)*a^2) + 1/14*(2*b^ 2 - 7*(b*x^3 + a)*b/x^3 + 14*(b*x^3 + a)^2/x^6)*c*x^7/((b*x^3 + a)^(7/3)*a ^3)
\[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{10/3}} \, dx=\int { \frac {d x^{3} + c}{{\left (b x^{3} + a\right )}^{\frac {10}{3}}} \,d x } \]
Time = 5.50 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.96 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{10/3}} \, dx=\frac {3\,a\,d\,x\,{\left (b\,x^3+a\right )}^2-4\,a^3\,d\,x+18\,b\,c\,x\,{\left (b\,x^3+a\right )}^2+a^2\,d\,x\,\left (b\,x^3+a\right )+4\,a^2\,b\,c\,x+6\,a\,b\,c\,x\,\left (b\,x^3+a\right )}{28\,a^3\,b\,{\left (b\,x^3+a\right )}^{7/3}} \]