Integrand size = 26, antiderivative size = 136 \[ \int (c+d x)^{8/3} \sqrt [3]{c^2-d^2 x^2} \, dx=-\frac {486 c^3 \left (c^2-d^2 x^2\right )^{4/3}}{455 d (c+d x)^{4/3}}-\frac {324 c^2 \left (c^2-d^2 x^2\right )^{4/3}}{455 d \sqrt [3]{c+d x}}-\frac {27 c (c+d x)^{2/3} \left (c^2-d^2 x^2\right )^{4/3}}{65 d}-\frac {3 (c+d x)^{5/3} \left (c^2-d^2 x^2\right )^{4/3}}{13 d} \] Output:
-486/455*c^3*(-d^2*x^2+c^2)^(4/3)/d/(d*x+c)^(4/3)-324/455*c^2*(-d^2*x^2+c^ 2)^(4/3)/d/(d*x+c)^(1/3)-27/65*c*(d*x+c)^(2/3)*(-d^2*x^2+c^2)^(4/3)/d-3/13 *(d*x+c)^(5/3)*(-d^2*x^2+c^2)^(4/3)/d
Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.51 \[ \int (c+d x)^{8/3} \sqrt [3]{c^2-d^2 x^2} \, dx=\frac {3 (-c+d x) \sqrt [3]{c^2-d^2 x^2} \left (368 c^3+339 c^2 d x+168 c d^2 x^2+35 d^3 x^3\right )}{455 d \sqrt [3]{c+d x}} \] Input:
Integrate[(c + d*x)^(8/3)*(c^2 - d^2*x^2)^(1/3),x]
Output:
(3*(-c + d*x)*(c^2 - d^2*x^2)^(1/3)*(368*c^3 + 339*c^2*d*x + 168*c*d^2*x^2 + 35*d^3*x^3))/(455*d*(c + d*x)^(1/3))
Time = 0.41 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {459, 459, 459, 458}
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.
\(\displaystyle \int (c+d x)^{8/3} \sqrt [3]{c^2-d^2 x^2} \, dx\) |
\(\Big \downarrow \) 459 |
\(\displaystyle \frac {18}{13} c \int (c+d x)^{5/3} \sqrt [3]{c^2-d^2 x^2}dx-\frac {3 (c+d x)^{5/3} \left (c^2-d^2 x^2\right )^{4/3}}{13 d}\) |
\(\Big \downarrow \) 459 |
\(\displaystyle \frac {18}{13} c \left (\frac {6}{5} c \int (c+d x)^{2/3} \sqrt [3]{c^2-d^2 x^2}dx-\frac {3 (c+d x)^{2/3} \left (c^2-d^2 x^2\right )^{4/3}}{10 d}\right )-\frac {3 (c+d x)^{5/3} \left (c^2-d^2 x^2\right )^{4/3}}{13 d}\) |
\(\Big \downarrow \) 459 |
\(\displaystyle \frac {18}{13} c \left (\frac {6}{5} c \left (\frac {6}{7} c \int \frac {\sqrt [3]{c^2-d^2 x^2}}{\sqrt [3]{c+d x}}dx-\frac {3 \left (c^2-d^2 x^2\right )^{4/3}}{7 d \sqrt [3]{c+d x}}\right )-\frac {3 (c+d x)^{2/3} \left (c^2-d^2 x^2\right )^{4/3}}{10 d}\right )-\frac {3 (c+d x)^{5/3} \left (c^2-d^2 x^2\right )^{4/3}}{13 d}\) |
\(\Big \downarrow \) 458 |
\(\displaystyle \frac {18}{13} c \left (\frac {6}{5} c \left (-\frac {3 \left (c^2-d^2 x^2\right )^{4/3}}{7 d \sqrt [3]{c+d x}}-\frac {9 c \left (c^2-d^2 x^2\right )^{4/3}}{14 d (c+d x)^{4/3}}\right )-\frac {3 (c+d x)^{2/3} \left (c^2-d^2 x^2\right )^{4/3}}{10 d}\right )-\frac {3 (c+d x)^{5/3} \left (c^2-d^2 x^2\right )^{4/3}}{13 d}\) |
Input:
Int[(c + d*x)^(8/3)*(c^2 - d^2*x^2)^(1/3),x]
Output:
(-3*(c + d*x)^(5/3)*(c^2 - d^2*x^2)^(4/3))/(13*d) + (18*c*((-3*(c + d*x)^( 2/3)*(c^2 - d^2*x^2)^(4/3))/(10*d) + (6*c*((-9*c*(c^2 - d^2*x^2)^(4/3))/(1 4*d*(c + d*x)^(4/3)) - (3*(c^2 - d^2*x^2)^(4/3))/(7*d*(c + d*x)^(1/3))))/5 ))/13
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(p + 1))), x] /; FreeQ[{a, b, c , d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && EqQ[n + p, 0]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[2*c* (Simplify[n + p]/(n + 2*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] && IGtQ[Simplif y[n + p], 0]
Time = 0.21 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.46
method | result | size |
gosper | \(-\frac {3 \left (-d x +c \right ) \left (35 d^{3} x^{3}+168 c \,d^{2} x^{2}+339 c^{2} d x +368 c^{3}\right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}}}{455 d \left (d x +c \right )^{\frac {1}{3}}}\) | \(63\) |
orering | \(-\frac {3 \left (-d x +c \right ) \left (35 d^{3} x^{3}+168 c \,d^{2} x^{2}+339 c^{2} d x +368 c^{3}\right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}}}{455 d \left (d x +c \right )^{\frac {1}{3}}}\) | \(63\) |
risch | \(-\frac {3 {\left (\left (-d^{2} x^{2}+c^{2}\right )^{2}\right )}^{\frac {1}{3}} {\left (\frac {\left (d^{2} x^{2}-c^{2}\right )^{2}}{\left (d x +c \right )^{2}}\right )}^{\frac {1}{3}} \left (d x +c \right )^{\frac {2}{3}} \left (-35 d^{4} x^{4}-133 d^{3} c \,x^{3}-171 c^{2} x^{2} d^{2}-29 c^{3} d x +368 c^{4}\right ) \left (-d x +c \right )}{455 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {2}{3}} {\left (\left (d^{2} x^{2}-c^{2}\right )^{2}\right )}^{\frac {1}{3}} d \left (\left (d x -c \right )^{2}\right )^{\frac {1}{3}}}\) | \(143\) |
Input:
int((d*x+c)^(8/3)*(-d^2*x^2+c^2)^(1/3),x,method=_RETURNVERBOSE)
Output:
-3/455*(-d*x+c)*(35*d^3*x^3+168*c*d^2*x^2+339*c^2*d*x+368*c^3)*(-d^2*x^2+c ^2)^(1/3)/d/(d*x+c)^(1/3)
Time = 0.09 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.55 \[ \int (c+d x)^{8/3} \sqrt [3]{c^2-d^2 x^2} \, dx=\frac {3 \, {\left (35 \, d^{4} x^{4} + 133 \, c d^{3} x^{3} + 171 \, c^{2} d^{2} x^{2} + 29 \, c^{3} d x - 368 \, c^{4}\right )} {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{455 \, {\left (d^{2} x + c d\right )}} \] Input:
integrate((d*x+c)^(8/3)*(-d^2*x^2+c^2)^(1/3),x, algorithm="fricas")
Output:
3/455*(35*d^4*x^4 + 133*c*d^3*x^3 + 171*c^2*d^2*x^2 + 29*c^3*d*x - 368*c^4 )*(-d^2*x^2 + c^2)^(1/3)*(d*x + c)^(2/3)/(d^2*x + c*d)
\[ \int (c+d x)^{8/3} \sqrt [3]{c^2-d^2 x^2} \, dx=\int \sqrt [3]{- \left (- c + d x\right ) \left (c + d x\right )} \left (c + d x\right )^{\frac {8}{3}}\, dx \] Input:
integrate((d*x+c)**(8/3)*(-d**2*x**2+c**2)**(1/3),x)
Output:
Integral((-(-c + d*x)*(c + d*x))**(1/3)*(c + d*x)**(8/3), x)
Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.49 \[ \int (c+d x)^{8/3} \sqrt [3]{c^2-d^2 x^2} \, dx=\frac {3 \, {\left (35 \, d^{4} x^{4} + 133 \, c d^{3} x^{3} + 171 \, c^{2} d^{2} x^{2} + 29 \, c^{3} d x - 368 \, c^{4}\right )} {\left (d x + c\right )} {\left (-d x + c\right )}^{\frac {1}{3}}}{455 \, {\left (d^{2} x + c d\right )}} \] Input:
integrate((d*x+c)^(8/3)*(-d^2*x^2+c^2)^(1/3),x, algorithm="maxima")
Output:
3/455*(35*d^4*x^4 + 133*c*d^3*x^3 + 171*c^2*d^2*x^2 + 29*c^3*d*x - 368*c^4 )*(d*x + c)*(-d*x + c)^(1/3)/(d^2*x + c*d)
\[ \int (c+d x)^{8/3} \sqrt [3]{c^2-d^2 x^2} \, dx=\int { {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {8}{3}} \,d x } \] Input:
integrate((d*x+c)^(8/3)*(-d^2*x^2+c^2)^(1/3),x, algorithm="giac")
Output:
integrate((-d^2*x^2 + c^2)^(1/3)*(d*x + c)^(8/3), x)
Time = 6.85 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.74 \[ \int (c+d x)^{8/3} \sqrt [3]{c^2-d^2 x^2} \, dx=\frac {{\left (c^2-d^2\,x^2\right )}^{1/3}\,\left (\frac {513\,c^2\,x^2\,{\left (c+d\,x\right )}^{2/3}}{455}-\frac {1104\,c^4\,{\left (c+d\,x\right )}^{2/3}}{455\,d^2}+\frac {3\,d^2\,x^4\,{\left (c+d\,x\right )}^{2/3}}{13}+\frac {57\,c\,d\,x^3\,{\left (c+d\,x\right )}^{2/3}}{65}+\frac {87\,c^3\,x\,{\left (c+d\,x\right )}^{2/3}}{455\,d}\right )}{x+\frac {c}{d}} \] Input:
int((c^2 - d^2*x^2)^(1/3)*(c + d*x)^(8/3),x)
Output:
((c^2 - d^2*x^2)^(1/3)*((513*c^2*x^2*(c + d*x)^(2/3))/455 - (1104*c^4*(c + d*x)^(2/3))/(455*d^2) + (3*d^2*x^4*(c + d*x)^(2/3))/13 + (57*c*d*x^3*(c + d*x)^(2/3))/65 + (87*c^3*x*(c + d*x)^(2/3))/(455*d)))/(x + c/d)
Time = 0.35 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.40 \[ \int (c+d x)^{8/3} \sqrt [3]{c^2-d^2 x^2} \, dx=\frac {3 \left (-d x +c \right )^{\frac {1}{3}} \left (35 d^{4} x^{4}+133 c \,d^{3} x^{3}+171 c^{2} d^{2} x^{2}+29 c^{3} d x -368 c^{4}\right )}{455 d} \] Input:
int((d*x+c)^(8/3)*(-d^2*x^2+c^2)^(1/3),x)
Output:
(3*(c - d*x)**(1/3)*( - 368*c**4 + 29*c**3*d*x + 171*c**2*d**2*x**2 + 133* c*d**3*x**3 + 35*d**4*x**4))/(455*d)