Integrand size = 22, antiderivative size = 67 \[ \int (c+d x) \sqrt [8]{c^2-d^2 x^2} \, dx=\frac {8 \sqrt [8]{2} \left (c^2-d^2 x^2\right )^{17/8} \operatorname {Hypergeometric2F1}\left (-\frac {1}{8},\frac {17}{8},\frac {25}{8},\frac {c+d x}{2 c}\right )}{17 c^2 d \left (\frac {c-d x}{c}\right )^{17/8}} \] Output:
8/17*2^(1/8)*(-d^2*x^2+c^2)^(17/8)*hypergeom([-1/8, 17/8],[25/8],1/2*(d*x+ c)/c)/c^2/d/((-d*x+c)/c)^(17/8)
Time = 8.87 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.19 \[ \int (c+d x) \sqrt [8]{c^2-d^2 x^2} \, dx=-\frac {4 \left (c^2-d^2 x^2\right )^{9/8}}{9 d}+\frac {c x \sqrt [8]{c^2-d^2 x^2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{8},\frac {1}{2},\frac {3}{2},\frac {d^2 x^2}{c^2}\right )}{\sqrt [8]{1-\frac {d^2 x^2}{c^2}}} \] Input:
Integrate[(c + d*x)*(c^2 - d^2*x^2)^(1/8),x]
Output:
(-4*(c^2 - d^2*x^2)^(9/8))/(9*d) + (c*x*(c^2 - d^2*x^2)^(1/8)*Hypergeometr ic2F1[-1/8, 1/2, 3/2, (d^2*x^2)/c^2])/(1 - (d^2*x^2)/c^2)^(1/8)
Time = 0.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.19, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {455, 238, 237}
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.
\(\displaystyle \int (c+d x) \sqrt [8]{c^2-d^2 x^2} \, dx\) |
\(\Big \downarrow \) 455 |
\(\displaystyle c \int \sqrt [8]{c^2-d^2 x^2}dx-\frac {4 \left (c^2-d^2 x^2\right )^{9/8}}{9 d}\) |
\(\Big \downarrow \) 238 |
\(\displaystyle \frac {c \sqrt [8]{c^2-d^2 x^2} \int \sqrt [8]{1-\frac {d^2 x^2}{c^2}}dx}{\sqrt [8]{1-\frac {d^2 x^2}{c^2}}}-\frac {4 \left (c^2-d^2 x^2\right )^{9/8}}{9 d}\) |
\(\Big \downarrow \) 237 |
\(\displaystyle \frac {c x \sqrt [8]{c^2-d^2 x^2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{8},\frac {1}{2},\frac {3}{2},\frac {d^2 x^2}{c^2}\right )}{\sqrt [8]{1-\frac {d^2 x^2}{c^2}}}-\frac {4 \left (c^2-d^2 x^2\right )^{9/8}}{9 d}\) |
Input:
Int[(c + d*x)*(c^2 - d^2*x^2)^(1/8),x]
Output:
(-4*(c^2 - d^2*x^2)^(9/8))/(9*d) + (c*x*(c^2 - d^2*x^2)^(1/8)*Hypergeometr ic2F1[-1/8, 1/2, 3/2, (d^2*x^2)/c^2])/(1 - (d^2*x^2)/c^2)^(1/8)
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] && !IntegerQ[2*p ] && GtQ[a, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2) ^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[(1 + b*(x^2/a))^p, x], x] / ; FreeQ[{a, b, p}, x] && !IntegerQ[2*p] && !GtQ[a, 0]
Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
\[\int \left (d x +c \right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{8}}d x\]
Input:
int((d*x+c)*(-d^2*x^2+c^2)^(1/8),x)
Output:
int((d*x+c)*(-d^2*x^2+c^2)^(1/8),x)
\[ \int (c+d x) \sqrt [8]{c^2-d^2 x^2} \, dx=\int { {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{8}} {\left (d x + c\right )} \,d x } \] Input:
integrate((d*x+c)*(-d^2*x^2+c^2)^(1/8),x, algorithm="fricas")
Output:
integral((-d^2*x^2 + c^2)^(1/8)*(d*x + c), x)
Time = 1.15 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.01 \[ \int (c+d x) \sqrt [8]{c^2-d^2 x^2} \, dx=c^{\frac {5}{4}} x {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{8}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {d^{2} x^{2} e^{2 i \pi }}{c^{2}}} \right )} + d \left (\begin {cases} \frac {x^{2} \sqrt [8]{c^{2}}}{2} & \text {for}\: d^{2} = 0 \\- \frac {4 \left (c^{2} - d^{2} x^{2}\right )^{\frac {9}{8}}}{9 d^{2}} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((d*x+c)*(-d**2*x**2+c**2)**(1/8),x)
Output:
c**(5/4)*x*hyper((-1/8, 1/2), (3/2,), d**2*x**2*exp_polar(2*I*pi)/c**2) + d*Piecewise((x**2*(c**2)**(1/8)/2, Eq(d**2, 0)), (-4*(c**2 - d**2*x**2)**( 9/8)/(9*d**2), True))
\[ \int (c+d x) \sqrt [8]{c^2-d^2 x^2} \, dx=\int { {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{8}} {\left (d x + c\right )} \,d x } \] Input:
integrate((d*x+c)*(-d^2*x^2+c^2)^(1/8),x, algorithm="maxima")
Output:
integrate((-d^2*x^2 + c^2)^(1/8)*(d*x + c), x)
\[ \int (c+d x) \sqrt [8]{c^2-d^2 x^2} \, dx=\int { {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{8}} {\left (d x + c\right )} \,d x } \] Input:
integrate((d*x+c)*(-d^2*x^2+c^2)^(1/8),x, algorithm="giac")
Output:
integrate((-d^2*x^2 + c^2)^(1/8)*(d*x + c), x)
Time = 6.52 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int (c+d x) \sqrt [8]{c^2-d^2 x^2} \, dx=\frac {c\,x\,{\left (c^2-d^2\,x^2\right )}^{1/8}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{8},\frac {1}{2};\ \frac {3}{2};\ \frac {d^2\,x^2}{c^2}\right )}{{\left (1-\frac {d^2\,x^2}{c^2}\right )}^{1/8}}-\frac {4\,{\left (c^2-d^2\,x^2\right )}^{9/8}}{9\,d} \] Input:
int((c^2 - d^2*x^2)^(1/8)*(c + d*x),x)
Output:
(c*x*(c^2 - d^2*x^2)^(1/8)*hypergeom([-1/8, 1/2], 3/2, (d^2*x^2)/c^2))/(1 - (d^2*x^2)/c^2)^(1/8) - (4*(c^2 - d^2*x^2)^(9/8))/(9*d)
\[ \int (c+d x) \sqrt [8]{c^2-d^2 x^2} \, dx=\frac {-4 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{8}} c^{2}+4 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{8}} d^{2} x^{2}+9 \left (\int \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{8}}d x \right ) c d}{9 d} \] Input:
int((d*x+c)*(-d^2*x^2+c^2)^(1/8),x)
Output:
( - 4*(c**2 - d**2*x**2)**(1/8)*c**2 + 4*(c**2 - d**2*x**2)**(1/8)*d**2*x* *2 + 9*int((c**2 - d**2*x**2)**(1/8),x)*c*d)/(9*d)