Integrand size = 20, antiderivative size = 117 \[ \int \frac {(a-b x)^{2/3}}{(a+b x)^{17/3}} \, dx=-\frac {3 (a-b x)^{5/3}}{28 a b (a+b x)^{14/3}}-\frac {27 (a-b x)^{5/3}}{616 a^2 b (a+b x)^{11/3}}-\frac {81 (a-b x)^{5/3}}{4928 a^3 b (a+b x)^{8/3}}-\frac {243 (a-b x)^{5/3}}{49280 a^4 b (a+b x)^{5/3}} \] Output:
-3/28*(-b*x+a)^(5/3)/a/b/(b*x+a)^(14/3)-27/616*(-b*x+a)^(5/3)/a^2/b/(b*x+a )^(11/3)-81/4928*(-b*x+a)^(5/3)/a^3/b/(b*x+a)^(8/3)-243/49280*(-b*x+a)^(5/ 3)/a^4/b/(b*x+a)^(5/3)
Time = 0.19 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.50 \[ \int \frac {(a-b x)^{2/3}}{(a+b x)^{17/3}} \, dx=-\frac {3 (a-b x)^{5/3} \left (2831 a^3+1503 a^2 b x+513 a b^2 x^2+81 b^3 x^3\right )}{49280 a^4 b (a+b x)^{14/3}} \] Input:
Integrate[(a - b*x)^(2/3)/(a + b*x)^(17/3),x]
Output:
(-3*(a - b*x)^(5/3)*(2831*a^3 + 1503*a^2*b*x + 513*a*b^2*x^2 + 81*b^3*x^3) )/(49280*a^4*b*(a + b*x)^(14/3))
Time = 0.19 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.14, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {55, 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.
\(\displaystyle \int \frac {(a-b x)^{2/3}}{(a+b x)^{17/3}} \, dx\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {9 \int \frac {(a-b x)^{2/3}}{(a+b x)^{14/3}}dx}{28 a}-\frac {3 (a-b x)^{5/3}}{28 a b (a+b x)^{14/3}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {9 \left (\frac {3 \int \frac {(a-b x)^{2/3}}{(a+b x)^{11/3}}dx}{11 a}-\frac {3 (a-b x)^{5/3}}{22 a b (a+b x)^{11/3}}\right )}{28 a}-\frac {3 (a-b x)^{5/3}}{28 a b (a+b x)^{14/3}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {9 \left (\frac {3 \left (\frac {3 \int \frac {(a-b x)^{2/3}}{(a+b x)^{8/3}}dx}{16 a}-\frac {3 (a-b x)^{5/3}}{16 a b (a+b x)^{8/3}}\right )}{11 a}-\frac {3 (a-b x)^{5/3}}{22 a b (a+b x)^{11/3}}\right )}{28 a}-\frac {3 (a-b x)^{5/3}}{28 a b (a+b x)^{14/3}}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {9 \left (\frac {3 \left (-\frac {9 (a-b x)^{5/3}}{160 a^2 b (a+b x)^{5/3}}-\frac {3 (a-b x)^{5/3}}{16 a b (a+b x)^{8/3}}\right )}{11 a}-\frac {3 (a-b x)^{5/3}}{22 a b (a+b x)^{11/3}}\right )}{28 a}-\frac {3 (a-b x)^{5/3}}{28 a b (a+b x)^{14/3}}\) |
Input:
Int[(a - b*x)^(2/3)/(a + b*x)^(17/3),x]
Output:
(-3*(a - b*x)^(5/3))/(28*a*b*(a + b*x)^(14/3)) + (9*((-3*(a - b*x)^(5/3))/ (22*a*b*(a + b*x)^(11/3)) + (3*((-3*(a - b*x)^(5/3))/(16*a*b*(a + b*x)^(8/ 3)) - (9*(a - b*x)^(5/3))/(160*a^2*b*(a + b*x)^(5/3))))/(11*a)))/(28*a)
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 = 0.13 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.46
method | result | size |
gosper | \(-\frac {3 \left (-b x +a \right )^{\frac {5}{3}} \left (81 b^{3} x^{3}+513 a \,b^{2} x^{2}+1503 a^{2} b x +2831 a^{3}\right )}{49280 \left (b x +a \right )^{\frac {14}{3}} a^{4} b}\) | \(54\) |
orering | \(-\frac {3 \left (-b x +a \right )^{\frac {5}{3}} \left (81 b^{3} x^{3}+513 a \,b^{2} x^{2}+1503 a^{2} b x +2831 a^{3}\right )}{49280 \left (b x +a \right )^{\frac {14}{3}} a^{4} b}\) | \(54\) |
Input:
int((-b*x+a)^(2/3)/(b*x+a)^(17/3),x,method=_RETURNVERBOSE)
Output:
-3/49280*(-b*x+a)^(5/3)*(81*b^3*x^3+513*a*b^2*x^2+1503*a^2*b*x+2831*a^3)/( b*x+a)^(14/3)/a^4/b
Time = 0.11 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.01 \[ \int \frac {(a-b x)^{2/3}}{(a+b x)^{17/3}} \, dx=\frac {3 \, {\left (81 \, b^{4} x^{4} + 432 \, a b^{3} x^{3} + 990 \, a^{2} b^{2} x^{2} + 1328 \, a^{3} b x - 2831 \, a^{4}\right )} {\left (b x + a\right )}^{\frac {1}{3}} {\left (-b x + a\right )}^{\frac {2}{3}}}{49280 \, {\left (a^{4} b^{6} x^{5} + 5 \, a^{5} b^{5} x^{4} + 10 \, a^{6} b^{4} x^{3} + 10 \, a^{7} b^{3} x^{2} + 5 \, a^{8} b^{2} x + a^{9} b\right )}} \] Input:
integrate((-b*x+a)^(2/3)/(b*x+a)^(17/3),x, algorithm="fricas")
Output:
3/49280*(81*b^4*x^4 + 432*a*b^3*x^3 + 990*a^2*b^2*x^2 + 1328*a^3*b*x - 283 1*a^4)*(b*x + a)^(1/3)*(-b*x + a)^(2/3)/(a^4*b^6*x^5 + 5*a^5*b^5*x^4 + 10* a^6*b^4*x^3 + 10*a^7*b^3*x^2 + 5*a^8*b^2*x + a^9*b)
Timed out. \[ \int \frac {(a-b x)^{2/3}}{(a+b x)^{17/3}} \, dx=\text {Timed out} \] Input:
integrate((-b*x+a)**(2/3)/(b*x+a)**(17/3),x)
Output:
Timed out
\[ \int \frac {(a-b x)^{2/3}}{(a+b x)^{17/3}} \, dx=\int { \frac {{\left (-b x + a\right )}^{\frac {2}{3}}}{{\left (b x + a\right )}^{\frac {17}{3}}} \,d x } \] Input:
integrate((-b*x+a)^(2/3)/(b*x+a)^(17/3),x, algorithm="maxima")
Output:
integrate((-b*x + a)^(2/3)/(b*x + a)^(17/3), x)
\[ \int \frac {(a-b x)^{2/3}}{(a+b x)^{17/3}} \, dx=\int { \frac {{\left (-b x + a\right )}^{\frac {2}{3}}}{{\left (b x + a\right )}^{\frac {17}{3}}} \,d x } \] Input:
integrate((-b*x+a)^(2/3)/(b*x+a)^(17/3),x, algorithm="giac")
Output:
integrate((-b*x + a)^(2/3)/(b*x + a)^(17/3), x)
Timed out. \[ \int \frac {(a-b x)^{2/3}}{(a+b x)^{17/3}} \, dx=\int \frac {{\left (a-b\,x\right )}^{2/3}}{{\left (a+b\,x\right )}^{17/3}} \,d x \] Input:
int((a - b*x)^(2/3)/(a + b*x)^(17/3),x)
Output:
int((a - b*x)^(2/3)/(a + b*x)^(17/3), x)
Time = 0.24 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.89 \[ \int \frac {(a-b x)^{2/3}}{(a+b x)^{17/3}} \, dx=\frac {3 \left (-b x +a \right )^{\frac {2}{3}} \left (81 b^{4} x^{4}+432 a \,b^{3} x^{3}+990 a^{2} b^{2} x^{2}+1328 a^{3} b x -2831 a^{4}\right )}{49280 \left (b x +a \right )^{\frac {2}{3}} a^{4} b \left (b^{4} x^{4}+4 a \,b^{3} x^{3}+6 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4}\right )} \] Input:
int((-b*x+a)^(2/3)/(b*x+a)^(17/3),x)
Output:
(3*(a - b*x)**(2/3)*( - 2831*a**4 + 1328*a**3*b*x + 990*a**2*b**2*x**2 + 4 32*a*b**3*x**3 + 81*b**4*x**4))/(49280*(a + b*x)**(2/3)*a**4*b*(a**4 + 4*a **3*b*x + 6*a**2*b**2*x**2 + 4*a*b**3*x**3 + b**4*x**4))