Integrand size = 20, antiderivative size = 117 \[ \int \frac {(a-b x)^{4/3}}{(a+b x)^{19/3}} \, dx=-\frac {3 (a-b x)^{7/3}}{32 a b (a+b x)^{16/3}}-\frac {27 (a-b x)^{7/3}}{832 a^2 b (a+b x)^{13/3}}-\frac {81 (a-b x)^{7/3}}{8320 a^3 b (a+b x)^{10/3}}-\frac {243 (a-b x)^{7/3}}{116480 a^4 b (a+b x)^{7/3}} \] Output:
-3/32*(-b*x+a)^(7/3)/a/b/(b*x+a)^(16/3)-27/832*(-b*x+a)^(7/3)/a^2/b/(b*x+a )^(13/3)-81/8320*(-b*x+a)^(7/3)/a^3/b/(b*x+a)^(10/3)-243/116480*(-b*x+a)^( 7/3)/a^4/b/(b*x+a)^(7/3)
Time = 0.24 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.50 \[ \int \frac {(a-b x)^{4/3}}{(a+b x)^{19/3}} \, dx=-\frac {3 (a-b x)^{7/3} \left (5359 a^3+2259 a^2 b x+621 a b^2 x^2+81 b^3 x^3\right )}{116480 a^4 b (a+b x)^{16/3}} \] Input:
Integrate[(a - b*x)^(4/3)/(a + b*x)^(19/3),x]
Output:
(-3*(a - b*x)^(7/3)*(5359*a^3 + 2259*a^2*b*x + 621*a*b^2*x^2 + 81*b^3*x^3) )/(116480*a^4*b*(a + b*x)^(16/3))
Time = 0.18 (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.
\(\displaystyle \int \frac {(a-b x)^{4/3}}{(a+b x)^{19/3}} \, dx\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {9 \int \frac {(a-b x)^{4/3}}{(a+b x)^{16/3}}dx}{32 a}-\frac {3 (a-b x)^{7/3}}{32 a b (a+b x)^{16/3}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {9 \left (\frac {3 \int \frac {(a-b x)^{4/3}}{(a+b x)^{13/3}}dx}{13 a}-\frac {3 (a-b x)^{7/3}}{26 a b (a+b x)^{13/3}}\right )}{32 a}-\frac {3 (a-b x)^{7/3}}{32 a b (a+b x)^{16/3}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {9 \left (\frac {3 \left (\frac {3 \int \frac {(a-b x)^{4/3}}{(a+b x)^{10/3}}dx}{20 a}-\frac {3 (a-b x)^{7/3}}{20 a b (a+b x)^{10/3}}\right )}{13 a}-\frac {3 (a-b x)^{7/3}}{26 a b (a+b x)^{13/3}}\right )}{32 a}-\frac {3 (a-b x)^{7/3}}{32 a b (a+b x)^{16/3}}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {9 \left (\frac {3 \left (-\frac {9 (a-b x)^{7/3}}{280 a^2 b (a+b x)^{7/3}}-\frac {3 (a-b x)^{7/3}}{20 a b (a+b x)^{10/3}}\right )}{13 a}-\frac {3 (a-b x)^{7/3}}{26 a b (a+b x)^{13/3}}\right )}{32 a}-\frac {3 (a-b x)^{7/3}}{32 a b (a+b x)^{16/3}}\) |
Input:
Int[(a - b*x)^(4/3)/(a + b*x)^(19/3),x]
Output:
(-3*(a - b*x)^(7/3))/(32*a*b*(a + b*x)^(16/3)) + (9*((-3*(a - b*x)^(7/3))/ (26*a*b*(a + b*x)^(13/3)) + (3*((-3*(a - b*x)^(7/3))/(20*a*b*(a + b*x)^(10 /3)) - (9*(a - b*x)^(7/3))/(280*a^2*b*(a + b*x)^(7/3))))/(13*a)))/(32*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 = 0.14 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.46
method | result | size |
gosper | \(-\frac {3 \left (-b x +a \right )^{\frac {7}{3}} \left (81 b^{3} x^{3}+621 a \,b^{2} x^{2}+2259 a^{2} b x +5359 a^{3}\right )}{116480 \left (b x +a \right )^{\frac {16}{3}} a^{4} b}\) | \(54\) |
orering | \(-\frac {3 \left (-b x +a \right )^{\frac {7}{3}} \left (81 b^{3} x^{3}+621 a \,b^{2} x^{2}+2259 a^{2} b x +5359 a^{3}\right )}{116480 \left (b x +a \right )^{\frac {16}{3}} a^{4} b}\) | \(54\) |
Input:
int((-b*x+a)^(4/3)/(b*x+a)^(19/3),x,method=_RETURNVERBOSE)
Output:
-3/116480*(-b*x+a)^(7/3)*(81*b^3*x^3+621*a*b^2*x^2+2259*a^2*b*x+5359*a^3)/ (b*x+a)^(16/3)/a^4/b
Time = 0.13 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.20 \[ \int \frac {(a-b x)^{4/3}}{(a+b x)^{19/3}} \, dx=-\frac {3 \, {\left (81 \, b^{5} x^{5} + 459 \, a b^{4} x^{4} + 1098 \, a^{2} b^{3} x^{3} + 1462 \, a^{3} b^{2} x^{2} - 8459 \, a^{4} b x + 5359 \, a^{5}\right )} {\left (b x + a\right )}^{\frac {2}{3}} {\left (-b x + a\right )}^{\frac {1}{3}}}{116480 \, {\left (a^{4} b^{7} x^{6} + 6 \, a^{5} b^{6} x^{5} + 15 \, a^{6} b^{5} x^{4} + 20 \, a^{7} b^{4} x^{3} + 15 \, a^{8} b^{3} x^{2} + 6 \, a^{9} b^{2} x + a^{10} b\right )}} \] Input:
integrate((-b*x+a)^(4/3)/(b*x+a)^(19/3),x, algorithm="fricas")
Output:
-3/116480*(81*b^5*x^5 + 459*a*b^4*x^4 + 1098*a^2*b^3*x^3 + 1462*a^3*b^2*x^ 2 - 8459*a^4*b*x + 5359*a^5)*(b*x + a)^(2/3)*(-b*x + a)^(1/3)/(a^4*b^7*x^6 + 6*a^5*b^6*x^5 + 15*a^6*b^5*x^4 + 20*a^7*b^4*x^3 + 15*a^8*b^3*x^2 + 6*a^ 9*b^2*x + a^10*b)
Timed out. \[ \int \frac {(a-b x)^{4/3}}{(a+b x)^{19/3}} \, dx=\text {Timed out} \] Input:
integrate((-b*x+a)**(4/3)/(b*x+a)**(19/3),x)
Output:
Timed out
\[ \int \frac {(a-b x)^{4/3}}{(a+b x)^{19/3}} \, dx=\int { \frac {{\left (-b x + a\right )}^{\frac {4}{3}}}{{\left (b x + a\right )}^{\frac {19}{3}}} \,d x } \] Input:
integrate((-b*x+a)^(4/3)/(b*x+a)^(19/3),x, algorithm="maxima")
Output:
integrate((-b*x + a)^(4/3)/(b*x + a)^(19/3), x)
\[ \int \frac {(a-b x)^{4/3}}{(a+b x)^{19/3}} \, dx=\int { \frac {{\left (-b x + a\right )}^{\frac {4}{3}}}{{\left (b x + a\right )}^{\frac {19}{3}}} \,d x } \] Input:
integrate((-b*x+a)^(4/3)/(b*x+a)^(19/3),x, algorithm="giac")
Output:
integrate((-b*x + a)^(4/3)/(b*x + a)^(19/3), x)
Time = 0.34 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.39 \[ \int \frac {(a-b x)^{4/3}}{(a+b x)^{19/3}} \, dx=-\frac {{\left (a-b\,x\right )}^{1/3}\,\left (\frac {16077\,a}{116480\,b^6}-\frac {25377\,x}{116480\,b^5}+\frac {2193\,x^2}{58240\,a\,b^4}+\frac {1647\,x^3}{58240\,a^2\,b^3}+\frac {1377\,x^4}{116480\,a^3\,b^2}+\frac {243\,x^5}{116480\,a^4\,b}\right )}{x^5\,{\left (a+b\,x\right )}^{1/3}+\frac {a^5\,{\left (a+b\,x\right )}^{1/3}}{b^5}+\frac {10\,a^2\,x^3\,{\left (a+b\,x\right )}^{1/3}}{b^2}+\frac {10\,a^3\,x^2\,{\left (a+b\,x\right )}^{1/3}}{b^3}+\frac {5\,a\,x^4\,{\left (a+b\,x\right )}^{1/3}}{b}+\frac {5\,a^4\,x\,{\left (a+b\,x\right )}^{1/3}}{b^4}} \] Input:
int((a - b*x)^(4/3)/(a + b*x)^(19/3),x)
Output:
-((a - b*x)^(1/3)*((16077*a)/(116480*b^6) - (25377*x)/(116480*b^5) + (2193 *x^2)/(58240*a*b^4) + (1647*x^3)/(58240*a^2*b^3) + (1377*x^4)/(116480*a^3* b^2) + (243*x^5)/(116480*a^4*b)))/(x^5*(a + b*x)^(1/3) + (a^5*(a + b*x)^(1 /3))/b^5 + (10*a^2*x^3*(a + b*x)^(1/3))/b^2 + (10*a^3*x^2*(a + b*x)^(1/3)) /b^3 + (5*a*x^4*(a + b*x)^(1/3))/b + (5*a^4*x*(a + b*x)^(1/3))/b^4)
Time = 0.52 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.08 \[ \int \frac {(a-b x)^{4/3}}{(a+b x)^{19/3}} \, dx=\frac {3 \left (-b x +a \right )^{\frac {1}{3}} \left (-81 b^{5} x^{5}-459 a \,b^{4} x^{4}-1098 a^{2} b^{3} x^{3}-1462 a^{3} b^{2} x^{2}+8459 a^{4} b x -5359 a^{5}\right )}{116480 \left (b x +a \right )^{\frac {1}{3}} a^{4} b \left (b^{5} x^{5}+5 a \,b^{4} x^{4}+10 a^{2} b^{3} x^{3}+10 a^{3} b^{2} x^{2}+5 a^{4} b x +a^{5}\right )} \] Input:
int((-b*x+a)^(4/3)/(b*x+a)^(19/3),x)
Output:
(3*(a - b*x)**(1/3)*( - 5359*a**5 + 8459*a**4*b*x - 1462*a**3*b**2*x**2 - 1098*a**2*b**3*x**3 - 459*a*b**4*x**4 - 81*b**5*x**5))/(116480*(a + b*x)** (1/3)*a**4*b*(a**5 + 5*a**4*b*x + 10*a**3*b**2*x**2 + 10*a**2*b**3*x**3 + 5*a*b**4*x**4 + b**5*x**5))