Integrand size = 21, antiderivative size = 117 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{10/3}} \, dx=\frac {(b c-a d)^2 x}{7 a b^2 \left (a+b x^3\right )^{7/3}}+\frac {(b c-a d) (3 b c+4 a d) x}{14 a^2 b^2 \left (a+b x^3\right )^{4/3}}+\frac {\left (9 b^2 c^2+3 a b c d+2 a^2 d^2\right ) x}{14 a^3 b^2 \sqrt [3]{a+b x^3}} \] Output:
1/7*(-a*d+b*c)^2*x/a/b^2/(b*x^3+a)^(7/3)+1/14*(-a*d+b*c)*(4*a*d+3*b*c)*x/a ^2/b^2/(b*x^3+a)^(4/3)+1/14*(2*a^2*d^2+3*a*b*c*d+9*b^2*c^2)*x/a^3/b^2/(b*x ^3+a)^(1/3)
Time = 0.90 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.62 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{10/3}} \, dx=\frac {9 b^2 c^2 x^7+3 a b c x^4 \left (7 c+d x^3\right )+a^2 \left (14 c^2 x+7 c d x^4+2 d^2 x^7\right )}{14 a^3 \left (a+b x^3\right )^{7/3}} \] Input:
Integrate[(c + d*x^3)^2/(a + b*x^3)^(10/3),x]
Output:
(9*b^2*c^2*x^7 + 3*a*b*c*x^4*(7*c + d*x^3) + a^2*(14*c^2*x + 7*c*d*x^4 + 2 *d^2*x^7))/(14*a^3*(a + b*x^3)^(7/3))
Time = 0.33 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.72, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {903, 903, 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.
\(\displaystyle \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{10/3}} \, dx\) |
\(\Big \downarrow \) 903 |
\(\displaystyle \frac {6 c \int \frac {d x^3+c}{\left (b x^3+a\right )^{7/3}}dx}{7 a}+\frac {x \left (c+d x^3\right )^2}{7 a \left (a+b x^3\right )^{7/3}}\) |
\(\Big \downarrow \) 903 |
\(\displaystyle \frac {6 c \left (\frac {3 c \int \frac {1}{\left (b x^3+a\right )^{4/3}}dx}{4 a}+\frac {x \left (c+d x^3\right )}{4 a \left (a+b x^3\right )^{4/3}}\right )}{7 a}+\frac {x \left (c+d x^3\right )^2}{7 a \left (a+b x^3\right )^{7/3}}\) |
\(\Big \downarrow \) 746 |
\(\displaystyle \frac {6 c \left (\frac {3 c x}{4 a^2 \sqrt [3]{a+b x^3}}+\frac {x \left (c+d x^3\right )}{4 a \left (a+b x^3\right )^{4/3}}\right )}{7 a}+\frac {x \left (c+d x^3\right )^2}{7 a \left (a+b x^3\right )^{7/3}}\) |
Input:
Int[(c + d*x^3)^2/(a + b*x^3)^(10/3),x]
Output:
(x*(c + d*x^3)^2)/(7*a*(a + b*x^3)^(7/3)) + (6*c*((3*c*x)/(4*a^2*(a + b*x^ 3)^(1/3)) + (x*(c + d*x^3))/(4*a*(a + b*x^3)^(4/3))))/(7*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[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Simp[(-x)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(a*n*(p + 1))), x] - Simp[ c*(q/(a*(p + 1))) Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1), x], x] /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 0] && GtQ[q, 0] && NeQ[p, -1]
Time = 1.00 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.61
method | result | size |
pseudoelliptic | \(\frac {\left (2 a^{2} d^{2}+3 a b c d +9 b^{2} c^{2}\right ) x^{7}+7 \left (a^{2} c d +3 b \,c^{2} a \right ) x^{4}+14 a^{2} c^{2} x}{14 \left (b \,x^{3}+a \right )^{\frac {7}{3}} a^{3}}\) | \(71\) |
gosper | \(\frac {x \left (2 a^{2} d^{2} x^{6}+3 a b c d \,x^{6}+9 b^{2} c^{2} x^{6}+7 a^{2} c d \,x^{3}+21 a b \,c^{2} x^{3}+14 a^{2} c^{2}\right )}{14 \left (b \,x^{3}+a \right )^{\frac {7}{3}} a^{3}}\) | \(76\) |
trager | \(\frac {x \left (2 a^{2} d^{2} x^{6}+3 a b c d \,x^{6}+9 b^{2} c^{2} x^{6}+7 a^{2} c d \,x^{3}+21 a b \,c^{2} x^{3}+14 a^{2} c^{2}\right )}{14 \left (b \,x^{3}+a \right )^{\frac {7}{3}} a^{3}}\) | \(76\) |
orering | \(\frac {x \left (2 a^{2} d^{2} x^{6}+3 a b c d \,x^{6}+9 b^{2} c^{2} x^{6}+7 a^{2} c d \,x^{3}+21 a b \,c^{2} x^{3}+14 a^{2} c^{2}\right )}{14 \left (b \,x^{3}+a \right )^{\frac {7}{3}} a^{3}}\) | \(76\) |
Input:
int((d*x^3+c)^2/(b*x^3+a)^(10/3),x,method=_RETURNVERBOSE)
Output:
1/14*((2*a^2*d^2+3*a*b*c*d+9*b^2*c^2)*x^7+7*(a^2*c*d+3*a*b*c^2)*x^4+14*a^2 *c^2*x)/(b*x^3+a)^(7/3)/a^3
Time = 0.08 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.88 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{10/3}} \, dx=\frac {{\left ({\left (9 \, b^{2} c^{2} + 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{7} + 14 \, a^{2} c^{2} x + 7 \, {\left (3 \, a b c^{2} + a^{2} c d\right )} x^{4}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{14 \, {\left (a^{3} b^{3} x^{9} + 3 \, a^{4} b^{2} x^{6} + 3 \, a^{5} b x^{3} + a^{6}\right )}} \] Input:
integrate((d*x^3+c)^2/(b*x^3+a)^(10/3),x, algorithm="fricas")
Output:
1/14*((9*b^2*c^2 + 3*a*b*c*d + 2*a^2*d^2)*x^7 + 14*a^2*c^2*x + 7*(3*a*b*c^ 2 + a^2*c*d)*x^4)*(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)
Timed out. \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{10/3}} \, dx=\text {Timed out} \] Input:
integrate((d*x**3+c)**2/(b*x**3+a)**(10/3),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.93 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{10/3}} \, dx=-\frac {{\left (4 \, b - \frac {7 \, {\left (b x^{3} + a\right )}}{x^{3}}\right )} c d x^{7}}{14 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} a^{2}} + \frac {d^{2} x^{7}}{7 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} a} + \frac {{\left (2 \, b^{2} - \frac {7 \, {\left (b x^{3} + a\right )} b}{x^{3}} + \frac {14 \, {\left (b x^{3} + a\right )}^{2}}{x^{6}}\right )} c^{2} x^{7}}{14 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} a^{3}} \] Input:
integrate((d*x^3+c)^2/(b*x^3+a)^(10/3),x, algorithm="maxima")
Output:
-1/14*(4*b - 7*(b*x^3 + a)/x^3)*c*d*x^7/((b*x^3 + a)^(7/3)*a^2) + 1/7*d^2* x^7/((b*x^3 + a)^(7/3)*a) + 1/14*(2*b^2 - 7*(b*x^3 + a)*b/x^3 + 14*(b*x^3 + a)^2/x^6)*c^2*x^7/((b*x^3 + a)^(7/3)*a^3)
\[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{10/3}} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {10}{3}}} \,d x } \] Input:
integrate((d*x^3+c)^2/(b*x^3+a)^(10/3),x, algorithm="giac")
Output:
integrate((d*x^3 + c)^2/(b*x^3 + a)^(10/3), x)
Time = 0.74 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.26 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{10/3}} \, dx=\frac {2\,a^4\,d^2\,x+2\,a^2\,d^2\,x\,{\left (b\,x^3+a\right )}^2+9\,b^2\,c^2\,x\,{\left (b\,x^3+a\right )}^2+2\,a^2\,b^2\,c^2\,x-4\,a^3\,d^2\,x\,\left (b\,x^3+a\right )+3\,a\,b^2\,c^2\,x\,\left (b\,x^3+a\right )-4\,a^3\,b\,c\,d\,x+3\,a\,b\,c\,d\,x\,{\left (b\,x^3+a\right )}^2+a^2\,b\,c\,d\,x\,\left (b\,x^3+a\right )}{14\,a^3\,b^2\,{\left (b\,x^3+a\right )}^{7/3}} \] Input:
int((c + d*x^3)^2/(a + b*x^3)^(10/3),x)
Output:
(2*a^4*d^2*x + 2*a^2*d^2*x*(a + b*x^3)^2 + 9*b^2*c^2*x*(a + b*x^3)^2 + 2*a ^2*b^2*c^2*x - 4*a^3*d^2*x*(a + b*x^3) + 3*a*b^2*c^2*x*(a + b*x^3) - 4*a^3 *b*c*d*x + 3*a*b*c*d*x*(a + b*x^3)^2 + a^2*b*c*d*x*(a + b*x^3))/(14*a^3*b^ 2*(a + b*x^3)^(7/3))
\[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{10/3}} \, dx=\left (\int \frac {x^{6}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{3}+3 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} b \,x^{3}+3 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,b^{2} x^{6}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{3} x^{9}}d x \right ) d^{2}+2 \left (\int \frac {x^{3}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{3}+3 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} b \,x^{3}+3 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,b^{2} x^{6}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{3} x^{9}}d x \right ) c d +\left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{3}+3 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} b \,x^{3}+3 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,b^{2} x^{6}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{3} x^{9}}d x \right ) c^{2} \] Input:
int((d*x^3+c)^2/(b*x^3+a)^(10/3),x)
Output:
int(x**6/((a + b*x**3)**(1/3)*a**3 + 3*(a + b*x**3)**(1/3)*a**2*b*x**3 + 3 *(a + b*x**3)**(1/3)*a*b**2*x**6 + (a + b*x**3)**(1/3)*b**3*x**9),x)*d**2 + 2*int(x**3/((a + b*x**3)**(1/3)*a**3 + 3*(a + b*x**3)**(1/3)*a**2*b*x**3 + 3*(a + b*x**3)**(1/3)*a*b**2*x**6 + (a + b*x**3)**(1/3)*b**3*x**9),x)*c *d + int(1/((a + b*x**3)**(1/3)*a**3 + 3*(a + b*x**3)**(1/3)*a**2*b*x**3 + 3*(a + b*x**3)**(1/3)*a*b**2*x**6 + (a + b*x**3)**(1/3)*b**3*x**9),x)*c** 2