Integrand size = 26, antiderivative size = 101 \[ \int (c+d x)^{5/3} \sqrt [3]{c^2-d^2 x^2} \, dx=-\frac {27 c^2 \left (c^2-d^2 x^2\right )^{4/3}}{35 d (c+d x)^{4/3}}-\frac {18 c \left (c^2-d^2 x^2\right )^{4/3}}{35 d \sqrt [3]{c+d x}}-\frac {3 (c+d x)^{2/3} \left (c^2-d^2 x^2\right )^{4/3}}{10 d} \] Output:
-27/35*c^2*(-d^2*x^2+c^2)^(4/3)/d/(d*x+c)^(4/3)-18/35*c*(-d^2*x^2+c^2)^(4/ 3)/d/(d*x+c)^(1/3)-3/10*(d*x+c)^(2/3)*(-d^2*x^2+c^2)^(4/3)/d
Time = 0.12 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.57 \[ \int (c+d x)^{5/3} \sqrt [3]{c^2-d^2 x^2} \, dx=\frac {3 (-c+d x) \sqrt [3]{c^2-d^2 x^2} \left (37 c^2+26 c d x+7 d^2 x^2\right )}{70 d \sqrt [3]{c+d x}} \] Input:
Integrate[(c + d*x)^(5/3)*(c^2 - d^2*x^2)^(1/3),x]
Output:
(3*(-c + d*x)*(c^2 - d^2*x^2)^(1/3)*(37*c^2 + 26*c*d*x + 7*d^2*x^2))/(70*d *(c + d*x)^(1/3))
Time = 0.34 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {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.
\(\displaystyle \int (c+d x)^{5/3} \sqrt [3]{c^2-d^2 x^2} \, dx\) |
\(\Big \downarrow \) 459 |
\(\displaystyle \frac {6}{5} c \int (c+d x)^{2/3} \sqrt [3]{c^2-d^2 x^2}dx-\frac {3 (c+d x)^{2/3} \left (c^2-d^2 x^2\right )^{4/3}}{10 d}\) |
\(\Big \downarrow \) 459 |
\(\displaystyle \frac {6}{5} c \left (\frac {6}{7} c \int \frac {\sqrt [3]{c^2-d^2 x^2}}{\sqrt [3]{c+d x}}dx-\frac {3 \left (c^2-d^2 x^2\right )^{4/3}}{7 d \sqrt [3]{c+d x}}\right )-\frac {3 (c+d x)^{2/3} \left (c^2-d^2 x^2\right )^{4/3}}{10 d}\) |
\(\Big \downarrow \) 458 |
\(\displaystyle \frac {6}{5} c \left (-\frac {3 \left (c^2-d^2 x^2\right )^{4/3}}{7 d \sqrt [3]{c+d x}}-\frac {9 c \left (c^2-d^2 x^2\right )^{4/3}}{14 d (c+d x)^{4/3}}\right )-\frac {3 (c+d x)^{2/3} \left (c^2-d^2 x^2\right )^{4/3}}{10 d}\) |
Input:
Int[(c + d*x)^(5/3)*(c^2 - d^2*x^2)^(1/3),x]
Output:
(-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))/(14*d*(c + d*x)^(4/3)) - (3*(c^2 - d^2*x^2)^(4/3))/(7*d*(c + d*x)^(1/3))))/5
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 = 52, normalized size of antiderivative = 0.51
method | result | size |
gosper | \(-\frac {3 \left (-d x +c \right ) \left (7 d^{2} x^{2}+26 c d x +37 c^{2}\right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}}}{70 d \left (d x +c \right )^{\frac {1}{3}}}\) | \(52\) |
orering | \(-\frac {3 \left (-d x +c \right ) \left (7 d^{2} x^{2}+26 c d x +37 c^{2}\right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}}}{70 d \left (d x +c \right )^{\frac {1}{3}}}\) | \(52\) |
risch | \(-\frac {3 {\left (\left (-d^{2} x^{2}+c^{2}\right )^{2}\right )}^{\frac {1}{3}} {\left (\frac {\left (d^{2} x^{2}-c^{2}\right )^{2}}{\left (d x +c \right )^{2}}\right )}^{\frac {1}{3}} \left (d x +c \right )^{\frac {2}{3}} \left (-7 d^{3} x^{3}-19 c \,d^{2} x^{2}-11 c^{2} d x +37 c^{3}\right ) \left (-d x +c \right )}{70 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {2}{3}} {\left (\left (d^{2} x^{2}-c^{2}\right )^{2}\right )}^{\frac {1}{3}} d \left (\left (d x -c \right )^{2}\right )^{\frac {1}{3}}}\) | \(132\) |
Input:
int((d*x+c)^(5/3)*(-d^2*x^2+c^2)^(1/3),x,method=_RETURNVERBOSE)
Output:
-3/70*(-d*x+c)*(7*d^2*x^2+26*c*d*x+37*c^2)*(-d^2*x^2+c^2)^(1/3)/d/(d*x+c)^ (1/3)
Time = 0.07 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.63 \[ \int (c+d x)^{5/3} \sqrt [3]{c^2-d^2 x^2} \, dx=\frac {3 \, {\left (7 \, d^{3} x^{3} + 19 \, c d^{2} x^{2} + 11 \, c^{2} d x - 37 \, c^{3}\right )} {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{70 \, {\left (d^{2} x + c d\right )}} \] Input:
integrate((d*x+c)^(5/3)*(-d^2*x^2+c^2)^(1/3),x, algorithm="fricas")
Output:
3/70*(7*d^3*x^3 + 19*c*d^2*x^2 + 11*c^2*d*x - 37*c^3)*(-d^2*x^2 + c^2)^(1/ 3)*(d*x + c)^(2/3)/(d^2*x + c*d)
\[ \int (c+d x)^{5/3} \sqrt [3]{c^2-d^2 x^2} \, dx=\int \sqrt [3]{- \left (- c + d x\right ) \left (c + d x\right )} \left (c + d x\right )^{\frac {5}{3}}\, dx \] Input:
integrate((d*x+c)**(5/3)*(-d**2*x**2+c**2)**(1/3),x)
Output:
Integral((-(-c + d*x)*(c + d*x))**(1/3)*(c + d*x)**(5/3), x)
Time = 0.04 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.55 \[ \int (c+d x)^{5/3} \sqrt [3]{c^2-d^2 x^2} \, dx=\frac {3 \, {\left (7 \, d^{3} x^{3} + 19 \, c d^{2} x^{2} + 11 \, c^{2} d x - 37 \, c^{3}\right )} {\left (d x + c\right )} {\left (-d x + c\right )}^{\frac {1}{3}}}{70 \, {\left (d^{2} x + c d\right )}} \] Input:
integrate((d*x+c)^(5/3)*(-d^2*x^2+c^2)^(1/3),x, algorithm="maxima")
Output:
3/70*(7*d^3*x^3 + 19*c*d^2*x^2 + 11*c^2*d*x - 37*c^3)*(d*x + c)*(-d*x + c) ^(1/3)/(d^2*x + c*d)
\[ \int (c+d x)^{5/3} \sqrt [3]{c^2-d^2 x^2} \, dx=\int { {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {5}{3}} \,d x } \] Input:
integrate((d*x+c)^(5/3)*(-d^2*x^2+c^2)^(1/3),x, algorithm="giac")
Output:
integrate((-d^2*x^2 + c^2)^(1/3)*(d*x + c)^(5/3), x)
Time = 6.37 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.81 \[ \int (c+d x)^{5/3} \sqrt [3]{c^2-d^2 x^2} \, dx=\frac {{\left (c^2-d^2\,x^2\right )}^{1/3}\,\left (\frac {57\,c\,x^2\,{\left (c+d\,x\right )}^{2/3}}{70}-\frac {111\,c^3\,{\left (c+d\,x\right )}^{2/3}}{70\,d^2}+\frac {3\,d\,x^3\,{\left (c+d\,x\right )}^{2/3}}{10}+\frac {33\,c^2\,x\,{\left (c+d\,x\right )}^{2/3}}{70\,d}\right )}{x+\frac {c}{d}} \] Input:
int((c^2 - d^2*x^2)^(1/3)*(c + d*x)^(5/3),x)
Output:
((c^2 - d^2*x^2)^(1/3)*((57*c*x^2*(c + d*x)^(2/3))/70 - (111*c^3*(c + d*x) ^(2/3))/(70*d^2) + (3*d*x^3*(c + d*x)^(2/3))/10 + (33*c^2*x*(c + d*x)^(2/3 ))/(70*d)))/(x + c/d)
Time = 0.32 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.43 \[ \int (c+d x)^{5/3} \sqrt [3]{c^2-d^2 x^2} \, dx=\frac {3 \left (-d x +c \right )^{\frac {1}{3}} \left (7 d^{3} x^{3}+19 c \,d^{2} x^{2}+11 c^{2} d x -37 c^{3}\right )}{70 d} \] Input:
int((d*x+c)^(5/3)*(-d^2*x^2+c^2)^(1/3),x)
Output:
(3*(c - d*x)**(1/3)*( - 37*c**3 + 11*c**2*d*x + 19*c*d**2*x**2 + 7*d**3*x* *3))/(70*d)