Integrand size = 23, antiderivative size = 67 \[ \int \frac {c-d x}{\left (c^2-d^2 x^2\right )^{7/8}} \, dx=-\frac {4 \sqrt [8]{2} \left (c^2-d^2 x^2\right )^{9/8} \operatorname {Hypergeometric2F1}\left (\frac {7}{8},\frac {9}{8},\frac {17}{8},\frac {c-d x}{2 c}\right )}{9 c^2 d \left (\frac {c+d x}{c}\right )^{9/8}} \] Output:
-4/9*2^(1/8)*(-d^2*x^2+c^2)^(9/8)*hypergeom([7/8, 9/8],[17/8],1/2*(-d*x+c) /c)/c^2/d/((d*x+c)/c)^(9/8)
Time = 10.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.12 \[ \int \frac {c-d x}{\left (c^2-d^2 x^2\right )^{7/8}} \, dx=\frac {4 c^2-4 d^2 x^2+c d x \left (1-\frac {d^2 x^2}{c^2}\right )^{7/8} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{8},\frac {3}{2},\frac {d^2 x^2}{c^2}\right )}{d \left (c^2-d^2 x^2\right )^{7/8}} \] Input:
Integrate[(c - d*x)/(c^2 - d^2*x^2)^(7/8),x]
Output:
(4*c^2 - 4*d^2*x^2 + c*d*x*(1 - (d^2*x^2)/c^2)^(7/8)*Hypergeometric2F1[1/2 , 7/8, 3/2, (d^2*x^2)/c^2])/(d*(c^2 - d^2*x^2)^(7/8))
Time = 0.29 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.16, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, 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.
\(\displaystyle \int \frac {c-d x}{\left (c^2-d^2 x^2\right )^{7/8}} \, dx\) |
\(\Big \downarrow \) 455 |
\(\displaystyle c \int \frac {1}{\left (c^2-d^2 x^2\right )^{7/8}}dx+\frac {4 \sqrt [8]{c^2-d^2 x^2}}{d}\) |
\(\Big \downarrow \) 238 |
\(\displaystyle \frac {c \left (1-\frac {d^2 x^2}{c^2}\right )^{7/8} \int \frac {1}{\left (1-\frac {d^2 x^2}{c^2}\right )^{7/8}}dx}{\left (c^2-d^2 x^2\right )^{7/8}}+\frac {4 \sqrt [8]{c^2-d^2 x^2}}{d}\) |
\(\Big \downarrow \) 237 |
\(\displaystyle \frac {c x \left (1-\frac {d^2 x^2}{c^2}\right )^{7/8} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{8},\frac {3}{2},\frac {d^2 x^2}{c^2}\right )}{\left (c^2-d^2 x^2\right )^{7/8}}+\frac {4 \sqrt [8]{c^2-d^2 x^2}}{d}\) |
Input:
Int[(c - d*x)/(c^2 - d^2*x^2)^(7/8),x]
Output:
(4*(c^2 - d^2*x^2)^(1/8))/d + (c*x*(1 - (d^2*x^2)/c^2)^(7/8)*Hypergeometri c2F1[1/2, 7/8, 3/2, (d^2*x^2)/c^2])/(c^2 - d^2*x^2)^(7/8)
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]
\[\int \frac {-d x +c}{\left (-d^{2} x^{2}+c^{2}\right )^{\frac {7}{8}}}d x\]
Input:
int((-d*x+c)/(-d^2*x^2+c^2)^(7/8),x)
Output:
int((-d*x+c)/(-d^2*x^2+c^2)^(7/8),x)
\[ \int \frac {c-d x}{\left (c^2-d^2 x^2\right )^{7/8}} \, dx=\int { -\frac {d x - c}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {7}{8}}} \,d x } \] Input:
integrate((-d*x+c)/(-d^2*x^2+c^2)^(7/8),x, algorithm="fricas")
Output:
integral((-d^2*x^2 + c^2)^(1/8)/(d*x + c), x)
Time = 1.30 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.97 \[ \int \frac {c-d x}{\left (c^2-d^2 x^2\right )^{7/8}} \, dx=- d \left (\begin {cases} \frac {x^{2}}{2 \left (c^{2}\right )^{\frac {7}{8}}} & \text {for}\: d^{2} = 0 \\- \frac {4 \sqrt [8]{c^{2} - d^{2} x^{2}}}{d^{2}} & \text {otherwise} \end {cases}\right ) + \frac {x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{8} \\ \frac {3}{2} \end {matrix}\middle | {\frac {d^{2} x^{2} e^{2 i \pi }}{c^{2}}} \right )}}{c^{\frac {3}{4}}} \] Input:
integrate((-d*x+c)/(-d**2*x**2+c**2)**(7/8),x)
Output:
-d*Piecewise((x**2/(2*(c**2)**(7/8)), Eq(d**2, 0)), (-4*(c**2 - d**2*x**2) **(1/8)/d**2, True)) + x*hyper((1/2, 7/8), (3/2,), d**2*x**2*exp_polar(2*I *pi)/c**2)/c**(3/4)
\[ \int \frac {c-d x}{\left (c^2-d^2 x^2\right )^{7/8}} \, dx=\int { -\frac {d x - c}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {7}{8}}} \,d x } \] Input:
integrate((-d*x+c)/(-d^2*x^2+c^2)^(7/8),x, algorithm="maxima")
Output:
-integrate((d*x - c)/(-d^2*x^2 + c^2)^(7/8), x)
\[ \int \frac {c-d x}{\left (c^2-d^2 x^2\right )^{7/8}} \, dx=\int { -\frac {d x - c}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {7}{8}}} \,d x } \] Input:
integrate((-d*x+c)/(-d^2*x^2+c^2)^(7/8),x, algorithm="giac")
Output:
integrate(-(d*x - c)/(-d^2*x^2 + c^2)^(7/8), x)
Time = 6.57 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int \frac {c-d x}{\left (c^2-d^2 x^2\right )^{7/8}} \, dx=\frac {4\,{\left (c^2-d^2\,x^2\right )}^{1/8}}{d}+\frac {c\,x\,{\left (1-\frac {d^2\,x^2}{c^2}\right )}^{7/8}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{8};\ \frac {3}{2};\ \frac {d^2\,x^2}{c^2}\right )}{{\left (c^2-d^2\,x^2\right )}^{7/8}} \] Input:
int((c - d*x)/(c^2 - d^2*x^2)^(7/8),x)
Output:
(4*(c^2 - d^2*x^2)^(1/8))/d + (c*x*(1 - (d^2*x^2)/c^2)^(7/8)*hypergeom([1/ 2, 7/8], 3/2, (d^2*x^2)/c^2))/(c^2 - d^2*x^2)^(7/8)
\[ \int \frac {c-d x}{\left (c^2-d^2 x^2\right )^{7/8}} \, dx=\frac {-4 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {7}{8}}+5 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{4}} \left (\int \frac {\left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{4}}}{\left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{8}} c^{2}-\left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{8}} d^{2} x^{2}}d x \right ) c d}{5 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{4}} d} \] Input:
int((-d*x+c)/(-d^2*x^2+c^2)^(7/8),x)
Output:
( - 4*(c**2 - d**2*x**2)**(7/8) + 5*(c**2 - d**2*x**2)**(3/4)*int((c**2 - d**2*x**2)**(3/4)/((c**2 - d**2*x**2)**(5/8)*c**2 - (c**2 - d**2*x**2)**(5 /8)*d**2*x**2),x)*c*d)/(5*(c**2 - d**2*x**2)**(3/4)*d)