Integrand size = 26, antiderivative size = 141 \[ \int \frac {\left (c^2-d^2 x^2\right )^{2/3}}{(c+d x)^{19/3}} \, dx=-\frac {3 \left (c^2-d^2 x^2\right )^{5/3}}{28 c d (c+d x)^{19/3}}-\frac {27 \left (c^2-d^2 x^2\right )^{5/3}}{616 c^2 d (c+d x)^{16/3}}-\frac {81 \left (c^2-d^2 x^2\right )^{5/3}}{4928 c^3 d (c+d x)^{13/3}}-\frac {243 \left (c^2-d^2 x^2\right )^{5/3}}{49280 c^4 d (c+d x)^{10/3}} \] Output:
-3/28*(-d^2*x^2+c^2)^(5/3)/c/d/(d*x+c)^(19/3)-27/616*(-d^2*x^2+c^2)^(5/3)/ c^2/d/(d*x+c)^(16/3)-81/4928*(-d^2*x^2+c^2)^(5/3)/c^3/d/(d*x+c)^(13/3)-243 /49280*(-d^2*x^2+c^2)^(5/3)/c^4/d/(d*x+c)^(10/3)
Time = 3.28 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.50 \[ \int \frac {\left (c^2-d^2 x^2\right )^{2/3}}{(c+d x)^{19/3}} \, dx=-\frac {3 (c-d x) \left (c^2-d^2 x^2\right )^{2/3} \left (2831 c^3+1503 c^2 d x+513 c d^2 x^2+81 d^3 x^3\right )}{49280 c^4 d (c+d x)^{16/3}} \] Input:
Integrate[(c^2 - d^2*x^2)^(2/3)/(c + d*x)^(19/3),x]
Output:
(-3*(c - d*x)*(c^2 - d^2*x^2)^(2/3)*(2831*c^3 + 1503*c^2*d*x + 513*c*d^2*x ^2 + 81*d^3*x^3))/(49280*c^4*d*(c + d*x)^(16/3))
Time = 0.42 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {461, 461, 461, 460}
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.
\(\displaystyle \int \frac {\left (c^2-d^2 x^2\right )^{2/3}}{(c+d x)^{19/3}} \, dx\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {9 \int \frac {\left (c^2-d^2 x^2\right )^{2/3}}{(c+d x)^{16/3}}dx}{28 c}-\frac {3 \left (c^2-d^2 x^2\right )^{5/3}}{28 c d (c+d x)^{19/3}}\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {9 \left (\frac {3 \int \frac {\left (c^2-d^2 x^2\right )^{2/3}}{(c+d x)^{13/3}}dx}{11 c}-\frac {3 \left (c^2-d^2 x^2\right )^{5/3}}{22 c d (c+d x)^{16/3}}\right )}{28 c}-\frac {3 \left (c^2-d^2 x^2\right )^{5/3}}{28 c d (c+d x)^{19/3}}\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {9 \left (\frac {3 \left (\frac {3 \int \frac {\left (c^2-d^2 x^2\right )^{2/3}}{(c+d x)^{10/3}}dx}{16 c}-\frac {3 \left (c^2-d^2 x^2\right )^{5/3}}{16 c d (c+d x)^{13/3}}\right )}{11 c}-\frac {3 \left (c^2-d^2 x^2\right )^{5/3}}{22 c d (c+d x)^{16/3}}\right )}{28 c}-\frac {3 \left (c^2-d^2 x^2\right )^{5/3}}{28 c d (c+d x)^{19/3}}\) |
\(\Big \downarrow \) 460 |
\(\displaystyle \frac {9 \left (\frac {3 \left (-\frac {9 \left (c^2-d^2 x^2\right )^{5/3}}{160 c^2 d (c+d x)^{10/3}}-\frac {3 \left (c^2-d^2 x^2\right )^{5/3}}{16 c d (c+d x)^{13/3}}\right )}{11 c}-\frac {3 \left (c^2-d^2 x^2\right )^{5/3}}{22 c d (c+d x)^{16/3}}\right )}{28 c}-\frac {3 \left (c^2-d^2 x^2\right )^{5/3}}{28 c d (c+d x)^{19/3}}\) |
Input:
Int[(c^2 - d^2*x^2)^(2/3)/(c + d*x)^(19/3),x]
Output:
(-3*(c^2 - d^2*x^2)^(5/3))/(28*c*d*(c + d*x)^(19/3)) + (9*((-3*(c^2 - d^2* x^2)^(5/3))/(22*c*d*(c + d*x)^(16/3)) + (3*((-3*(c^2 - d^2*x^2)^(5/3))/(16 *c*d*(c + d*x)^(13/3)) - (9*(c^2 - d^2*x^2)^(5/3))/(160*c^2*d*(c + d*x)^(1 0/3))))/(11*c)))/(28*c)
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(b*c*n)), x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && EqQ[n + 2*p + 2, 0]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[Simpl ify[n + 2*p + 2]/(2*c*(n + p + 1)) Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x ], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && ILtQ[Simp lify[n + 2*p + 2], 0] && (LtQ[n, -1] || GtQ[n + p, 0])
Time = 0.21 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.47
method | result | size |
gosper | \(-\frac {3 \left (-d x +c \right ) \left (81 d^{3} x^{3}+513 c \,d^{2} x^{2}+1503 c^{2} d x +2831 c^{3}\right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {2}{3}}}{49280 \left (d x +c \right )^{\frac {16}{3}} c^{4} d}\) | \(66\) |
orering | \(-\frac {3 \left (-d x +c \right ) \left (81 d^{3} x^{3}+513 c \,d^{2} x^{2}+1503 c^{2} d x +2831 c^{3}\right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {2}{3}}}{49280 \left (d x +c \right )^{\frac {16}{3}} c^{4} d}\) | \(66\) |
Input:
int((-d^2*x^2+c^2)^(2/3)/(d*x+c)^(19/3),x,method=_RETURNVERBOSE)
Output:
-3/49280*(-d*x+c)*(81*d^3*x^3+513*c*d^2*x^2+1503*c^2*d*x+2831*c^3)*(-d^2*x ^2+c^2)^(2/3)/(d*x+c)^(16/3)/c^4/d
Time = 0.10 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.96 \[ \int \frac {\left (c^2-d^2 x^2\right )^{2/3}}{(c+d x)^{19/3}} \, dx=\frac {3 \, {\left (81 \, d^{4} x^{4} + 432 \, c d^{3} x^{3} + 990 \, c^{2} d^{2} x^{2} + 1328 \, c^{3} d x - 2831 \, c^{4}\right )} {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{49280 \, {\left (c^{4} d^{7} x^{6} + 6 \, c^{5} d^{6} x^{5} + 15 \, c^{6} d^{5} x^{4} + 20 \, c^{7} d^{4} x^{3} + 15 \, c^{8} d^{3} x^{2} + 6 \, c^{9} d^{2} x + c^{10} d\right )}} \] Input:
integrate((-d^2*x^2+c^2)^(2/3)/(d*x+c)^(19/3),x, algorithm="fricas")
Output:
3/49280*(81*d^4*x^4 + 432*c*d^3*x^3 + 990*c^2*d^2*x^2 + 1328*c^3*d*x - 283 1*c^4)*(-d^2*x^2 + c^2)^(2/3)*(d*x + c)^(2/3)/(c^4*d^7*x^6 + 6*c^5*d^6*x^5 + 15*c^6*d^5*x^4 + 20*c^7*d^4*x^3 + 15*c^8*d^3*x^2 + 6*c^9*d^2*x + c^10*d )
Timed out. \[ \int \frac {\left (c^2-d^2 x^2\right )^{2/3}}{(c+d x)^{19/3}} \, dx=\text {Timed out} \] Input:
integrate((-d**2*x**2+c**2)**(2/3)/(d*x+c)**(19/3),x)
Output:
Timed out
\[ \int \frac {\left (c^2-d^2 x^2\right )^{2/3}}{(c+d x)^{19/3}} \, dx=\int { \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {2}{3}}}{{\left (d x + c\right )}^{\frac {19}{3}}} \,d x } \] Input:
integrate((-d^2*x^2+c^2)^(2/3)/(d*x+c)^(19/3),x, algorithm="maxima")
Output:
integrate((-d^2*x^2 + c^2)^(2/3)/(d*x + c)^(19/3), x)
Time = 0.15 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.52 \[ \int \frac {\left (c^2-d^2 x^2\right )^{2/3}}{(c+d x)^{19/3}} \, dx=-\frac {3 \, {\left (220 \, {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {14}{3}} + 840 \, {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {11}{3}} + 1155 \, {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {8}{3}} + 616 \, {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {5}{3}}\right )}}{49280 \, c^{4} d} \] Input:
integrate((-d^2*x^2+c^2)^(2/3)/(d*x+c)^(19/3),x, algorithm="giac")
Output:
-3/49280*(220*(2*c/(d*x + c) - 1)^(14/3) + 840*(2*c/(d*x + c) - 1)^(11/3) + 1155*(2*c/(d*x + c) - 1)^(8/3) + 616*(2*c/(d*x + c) - 1)^(5/3))/(c^4*d)
Time = 7.39 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.13 \[ \int \frac {\left (c^2-d^2 x^2\right )^{2/3}}{(c+d x)^{19/3}} \, dx=\frac {{\left (c^2-d^2\,x^2\right )}^{2/3}\,\left (\frac {249\,x}{3080\,c\,d^5}-\frac {8493}{49280\,d^6}+\frac {27\,x^2}{448\,c^2\,d^4}+\frac {81\,x^3}{3080\,c^3\,d^3}+\frac {243\,x^4}{49280\,c^4\,d^2}\right )}{x^5\,{\left (c+d\,x\right )}^{1/3}+\frac {c^5\,{\left (c+d\,x\right )}^{1/3}}{d^5}+\frac {10\,c^2\,x^3\,{\left (c+d\,x\right )}^{1/3}}{d^2}+\frac {10\,c^3\,x^2\,{\left (c+d\,x\right )}^{1/3}}{d^3}+\frac {5\,c\,x^4\,{\left (c+d\,x\right )}^{1/3}}{d}+\frac {5\,c^4\,x\,{\left (c+d\,x\right )}^{1/3}}{d^4}} \] Input:
int((c^2 - d^2*x^2)^(2/3)/(c + d*x)^(19/3),x)
Output:
((c^2 - d^2*x^2)^(2/3)*((249*x)/(3080*c*d^5) - 8493/(49280*d^6) + (27*x^2) /(448*c^2*d^4) + (81*x^3)/(3080*c^3*d^3) + (243*x^4)/(49280*c^4*d^2)))/(x^ 5*(c + d*x)^(1/3) + (c^5*(c + d*x)^(1/3))/d^5 + (10*c^2*x^3*(c + d*x)^(1/3 ))/d^2 + (10*c^3*x^2*(c + d*x)^(1/3))/d^3 + (5*c*x^4*(c + d*x)^(1/3))/d + (5*c^4*x*(c + d*x)^(1/3))/d^4)
Time = 0.24 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.74 \[ \int \frac {\left (c^2-d^2 x^2\right )^{2/3}}{(c+d x)^{19/3}} \, dx=\frac {3 \left (-d x +c \right )^{\frac {2}{3}} \left (81 d^{4} x^{4}+432 c \,d^{3} x^{3}+990 c^{2} d^{2} x^{2}+1328 c^{3} d x -2831 c^{4}\right )}{49280 \left (d x +c \right )^{\frac {2}{3}} c^{4} d \left (d^{4} x^{4}+4 c \,d^{3} x^{3}+6 c^{2} d^{2} x^{2}+4 c^{3} d x +c^{4}\right )} \] Input:
int((-d^2*x^2+c^2)^(2/3)/(d*x+c)^(19/3),x)
Output:
(3*(c - d*x)**(2/3)*( - 2831*c**4 + 1328*c**3*d*x + 990*c**2*d**2*x**2 + 4 32*c*d**3*x**3 + 81*d**4*x**4))/(49280*(c + d*x)**(2/3)*c**4*d*(c**4 + 4*c **3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4))