Integrand size = 19, antiderivative size = 101 \[ \int \frac {1}{(a+b x)^{10/3} (c+d x)^{2/3}} \, dx=-\frac {3 \sqrt [3]{c+d x}}{7 (b c-a d) (a+b x)^{7/3}}+\frac {9 d \sqrt [3]{c+d x}}{14 (b c-a d)^2 (a+b x)^{4/3}}-\frac {27 d^2 \sqrt [3]{c+d x}}{14 (b c-a d)^3 \sqrt [3]{a+b x}} \] Output:
-3/7*(d*x+c)^(1/3)/(-a*d+b*c)/(b*x+a)^(7/3)+9/14*d*(d*x+c)^(1/3)/(-a*d+b*c )^2/(b*x+a)^(4/3)-27/14*d^2*(d*x+c)^(1/3)/(-a*d+b*c)^3/(b*x+a)^(1/3)
Time = 0.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.74 \[ \int \frac {1}{(a+b x)^{10/3} (c+d x)^{2/3}} \, dx=-\frac {3 \sqrt [3]{c+d x} \left (14 a^2 d^2-7 a b d (c-3 d x)+b^2 \left (2 c^2-3 c d x+9 d^2 x^2\right )\right )}{14 (b c-a d)^3 (a+b x)^{7/3}} \] Input:
Integrate[1/((a + b*x)^(10/3)*(c + d*x)^(2/3)),x]
Output:
(-3*(c + d*x)^(1/3)*(14*a^2*d^2 - 7*a*b*d*(c - 3*d*x) + b^2*(2*c^2 - 3*c*d *x + 9*d^2*x^2)))/(14*(b*c - a*d)^3*(a + b*x)^(7/3))
Time = 0.18 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.13, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {55, 55, 48}
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.
\(\displaystyle \int \frac {1}{(a+b x)^{10/3} (c+d x)^{2/3}} \, dx\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {6 d \int \frac {1}{(a+b x)^{7/3} (c+d x)^{2/3}}dx}{7 (b c-a d)}-\frac {3 \sqrt [3]{c+d x}}{7 (a+b x)^{7/3} (b c-a d)}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {6 d \left (-\frac {3 d \int \frac {1}{(a+b x)^{4/3} (c+d x)^{2/3}}dx}{4 (b c-a d)}-\frac {3 \sqrt [3]{c+d x}}{4 (a+b x)^{4/3} (b c-a d)}\right )}{7 (b c-a d)}-\frac {3 \sqrt [3]{c+d x}}{7 (a+b x)^{7/3} (b c-a d)}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle -\frac {6 d \left (\frac {9 d \sqrt [3]{c+d x}}{4 \sqrt [3]{a+b x} (b c-a d)^2}-\frac {3 \sqrt [3]{c+d x}}{4 (a+b x)^{4/3} (b c-a d)}\right )}{7 (b c-a d)}-\frac {3 \sqrt [3]{c+d x}}{7 (a+b x)^{7/3} (b c-a d)}\) |
Input:
Int[1/((a + b*x)^(10/3)*(c + d*x)^(2/3)),x]
Output:
(-3*(c + d*x)^(1/3))/(7*(b*c - a*d)*(a + b*x)^(7/3)) - (6*d*((-3*(c + d*x) ^(1/3))/(4*(b*c - a*d)*(a + b*x)^(4/3)) + (9*d*(c + d*x)^(1/3))/(4*(b*c - a*d)^2*(a + b*x)^(1/3))))/(7*(b*c - a*d))
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
Time = 0.16 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.04
method | result | size |
gosper | \(\frac {3 \left (x d +c \right )^{\frac {1}{3}} \left (9 d^{2} x^{2} b^{2}+21 x a b \,d^{2}-3 x \,b^{2} c d +14 a^{2} d^{2}-7 a b c d +2 b^{2} c^{2}\right )}{14 \left (b x +a \right )^{\frac {7}{3}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(105\) |
orering | \(\frac {3 \left (x d +c \right )^{\frac {1}{3}} \left (9 d^{2} x^{2} b^{2}+21 x a b \,d^{2}-3 x \,b^{2} c d +14 a^{2} d^{2}-7 a b c d +2 b^{2} c^{2}\right )}{14 \left (b x +a \right )^{\frac {7}{3}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(105\) |
Input:
int(1/(b*x+a)^(10/3)/(d*x+c)^(2/3),x,method=_RETURNVERBOSE)
Output:
3/14*(d*x+c)^(1/3)*(9*b^2*d^2*x^2+21*a*b*d^2*x-3*b^2*c*d*x+14*a^2*d^2-7*a* b*c*d+2*b^2*c^2)/(b*x+a)^(7/3)/(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^ 3)
Leaf count of result is larger than twice the leaf count of optimal. 251 vs. \(2 (83) = 166\).
Time = 0.08 (sec) , antiderivative size = 251, normalized size of antiderivative = 2.49 \[ \int \frac {1}{(a+b x)^{10/3} (c+d x)^{2/3}} \, dx=-\frac {3 \, {\left (9 \, b^{2} d^{2} x^{2} + 2 \, b^{2} c^{2} - 7 \, a b c d + 14 \, a^{2} d^{2} - 3 \, {\left (b^{2} c d - 7 \, a b d^{2}\right )} x\right )} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}}}{14 \, {\left (a^{3} b^{3} c^{3} - 3 \, a^{4} b^{2} c^{2} d + 3 \, a^{5} b c d^{2} - a^{6} d^{3} + {\left (b^{6} c^{3} - 3 \, a b^{5} c^{2} d + 3 \, a^{2} b^{4} c d^{2} - a^{3} b^{3} d^{3}\right )} x^{3} + 3 \, {\left (a b^{5} c^{3} - 3 \, a^{2} b^{4} c^{2} d + 3 \, a^{3} b^{3} c d^{2} - a^{4} b^{2} d^{3}\right )} x^{2} + 3 \, {\left (a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d + 3 \, a^{4} b^{2} c d^{2} - a^{5} b d^{3}\right )} x\right )}} \] Input:
integrate(1/(b*x+a)^(10/3)/(d*x+c)^(2/3),x, algorithm="fricas")
Output:
-3/14*(9*b^2*d^2*x^2 + 2*b^2*c^2 - 7*a*b*c*d + 14*a^2*d^2 - 3*(b^2*c*d - 7 *a*b*d^2)*x)*(b*x + a)^(2/3)*(d*x + c)^(1/3)/(a^3*b^3*c^3 - 3*a^4*b^2*c^2* d + 3*a^5*b*c*d^2 - a^6*d^3 + (b^6*c^3 - 3*a*b^5*c^2*d + 3*a^2*b^4*c*d^2 - a^3*b^3*d^3)*x^3 + 3*(a*b^5*c^3 - 3*a^2*b^4*c^2*d + 3*a^3*b^3*c*d^2 - a^4 *b^2*d^3)*x^2 + 3*(a^2*b^4*c^3 - 3*a^3*b^3*c^2*d + 3*a^4*b^2*c*d^2 - a^5*b *d^3)*x)
\[ \int \frac {1}{(a+b x)^{10/3} (c+d x)^{2/3}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {10}{3}} \left (c + d x\right )^{\frac {2}{3}}}\, dx \] Input:
integrate(1/(b*x+a)**(10/3)/(d*x+c)**(2/3),x)
Output:
Integral(1/((a + b*x)**(10/3)*(c + d*x)**(2/3)), x)
\[ \int \frac {1}{(a+b x)^{10/3} (c+d x)^{2/3}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {10}{3}} {\left (d x + c\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate(1/(b*x+a)^(10/3)/(d*x+c)^(2/3),x, algorithm="maxima")
Output:
integrate(1/((b*x + a)^(10/3)*(d*x + c)^(2/3)), x)
\[ \int \frac {1}{(a+b x)^{10/3} (c+d x)^{2/3}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {10}{3}} {\left (d x + c\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate(1/(b*x+a)^(10/3)/(d*x+c)^(2/3),x, algorithm="giac")
Output:
integrate(1/((b*x + a)^(10/3)*(d*x + c)^(2/3)), x)
Time = 0.57 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.32 \[ \int \frac {1}{(a+b x)^{10/3} (c+d x)^{2/3}} \, dx=\frac {{\left (c+d\,x\right )}^{1/3}\,\left (\frac {27\,d^2\,x^2}{14\,{\left (a\,d-b\,c\right )}^3}+\frac {42\,a^2\,d^2-21\,a\,b\,c\,d+6\,b^2\,c^2}{14\,b^2\,{\left (a\,d-b\,c\right )}^3}+\frac {9\,d\,x\,\left (7\,a\,d-b\,c\right )}{14\,b\,{\left (a\,d-b\,c\right )}^3}\right )}{x^2\,{\left (a+b\,x\right )}^{1/3}+\frac {a^2\,{\left (a+b\,x\right )}^{1/3}}{b^2}+\frac {2\,a\,x\,{\left (a+b\,x\right )}^{1/3}}{b}} \] Input:
int(1/((a + b*x)^(10/3)*(c + d*x)^(2/3)),x)
Output:
((c + d*x)^(1/3)*((27*d^2*x^2)/(14*(a*d - b*c)^3) + (42*a^2*d^2 + 6*b^2*c^ 2 - 21*a*b*c*d)/(14*b^2*(a*d - b*c)^3) + (9*d*x*(7*a*d - b*c))/(14*b*(a*d - b*c)^3)))/(x^2*(a + b*x)^(1/3) + (a^2*(a + b*x)^(1/3))/b^2 + (2*a*x*(a + b*x)^(1/3))/b)
\[ \int \frac {1}{(a+b x)^{10/3} (c+d x)^{2/3}} \, dx=\int \frac {1}{\left (d x +c \right )^{\frac {2}{3}} \left (b x +a \right )^{\frac {1}{3}} a^{3}+3 \left (d x +c \right )^{\frac {2}{3}} \left (b x +a \right )^{\frac {1}{3}} a^{2} b x +3 \left (d x +c \right )^{\frac {2}{3}} \left (b x +a \right )^{\frac {1}{3}} a \,b^{2} x^{2}+\left (d x +c \right )^{\frac {2}{3}} \left (b x +a \right )^{\frac {1}{3}} b^{3} x^{3}}d x \] Input:
int(1/(b*x+a)^(10/3)/(d*x+c)^(2/3),x)
Output:
int(1/((c + d*x)**(2/3)*(a + b*x)**(1/3)*a**3 + 3*(c + d*x)**(2/3)*(a + b* x)**(1/3)*a**2*b*x + 3*(c + d*x)**(2/3)*(a + b*x)**(1/3)*a*b**2*x**2 + (c + d*x)**(2/3)*(a + b*x)**(1/3)*b**3*x**3),x)