\(\int \frac {c-d x}{(c^2-d^2 x^2)^{5/6}} \, dx\) [327]

Optimal result

Integrand size = 23, antiderivative size = 323 \[ \int \frac {c-d x}{\left (c^2-d^2 x^2\right )^{5/6}} \, dx=\frac {3 \sqrt [6]{c^2-d^2 x^2}}{d}-\frac {3^{3/4} \sqrt [3]{c} \sqrt [6]{c^2-d^2 x^2} \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right ) \sqrt {\frac {c^{4/3}+c^{2/3} \sqrt [3]{c^2-d^2 x^2}+\left (c^2-d^2 x^2\right )^{2/3}}{\left (c^{2/3}-\left (1+\sqrt {3}\right ) \sqrt [3]{c^2-d^2 x^2}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {c^{2/3}-\left (1-\sqrt {3}\right ) \sqrt [3]{c^2-d^2 x^2}}{c^{2/3}-\left (1+\sqrt {3}\right ) \sqrt [3]{c^2-d^2 x^2}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2 d^2 x \sqrt {-\frac {\sqrt [3]{c^2-d^2 x^2} \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )}{\left (c^{2/3}-\left (1+\sqrt {3}\right ) \sqrt [3]{c^2-d^2 x^2}\right )^2}}} \] Output:

3*(-d^2*x^2+c^2)^(1/6)/d-1/2*3^(3/4)*c^(1/3)*(-d^2*x^2+c^2)^(1/6)*(c^(2/3) 
-(-d^2*x^2+c^2)^(1/3))*((c^(4/3)+c^(2/3)*(-d^2*x^2+c^2)^(1/3)+(-d^2*x^2+c^ 
2)^(2/3))/(c^(2/3)-(1+3^(1/2))*(-d^2*x^2+c^2)^(1/3))^2)^(1/2)*InverseJacob 
iAM(arccos((c^(2/3)-(1-3^(1/2))*(-d^2*x^2+c^2)^(1/3))/(c^(2/3)-(1+3^(1/2)) 
*(-d^2*x^2+c^2)^(1/3))),1/4*6^(1/2)+1/4*2^(1/2))/d^2/x/(-(-d^2*x^2+c^2)^(1 
/3)*(c^(2/3)-(-d^2*x^2+c^2)^(1/3))/(c^(2/3)-(1+3^(1/2))*(-d^2*x^2+c^2)^(1/ 
3))^2)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.23 \[ \int \frac {c-d x}{\left (c^2-d^2 x^2\right )^{5/6}} \, dx=\frac {3 c^2-3 d^2 x^2+c d x \left (1-\frac {d^2 x^2}{c^2}\right )^{5/6} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {d^2 x^2}{c^2}\right )}{d \left (c^2-d^2 x^2\right )^{5/6}} \] Input:

Integrate[(c - d*x)/(c^2 - d^2*x^2)^(5/6),x]
 

Output:

(3*c^2 - 3*d^2*x^2 + c*d*x*(1 - (d^2*x^2)/c^2)^(5/6)*Hypergeometric2F1[1/2 
, 5/6, 3/2, (d^2*x^2)/c^2])/(d*(c^2 - d^2*x^2)^(5/6))
 

Rubi [A] (warning: unable to verify)

Time = 0.52 (sec) , antiderivative size = 426, normalized size of antiderivative = 1.32, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {455, 236, 234, 760}

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.

\(\displaystyle \int \frac {c-d x}{\left (c^2-d^2 x^2\right )^{5/6}} \, dx\)

\(\Big \downarrow \) 455

\(\displaystyle c \int \frac {1}{\left (c^2-d^2 x^2\right )^{5/6}}dx+\frac {3 \sqrt [6]{c^2-d^2 x^2}}{d}\)

\(\Big \downarrow \) 236

\(\displaystyle \frac {c \int \frac {1}{\left (\frac {d^2 x^2}{c^2-d^2 x^2}+1\right )^{2/3}}d\frac {x}{\sqrt {c^2-d^2 x^2}}}{\sqrt [3]{\frac {c^2}{c^2-d^2 x^2}} \sqrt [3]{c^2-d^2 x^2}}+\frac {3 \sqrt [6]{c^2-d^2 x^2}}{d}\)

\(\Big \downarrow \) 234

\(\displaystyle \frac {3 c \sqrt {\frac {d^2 x^2}{c^2-d^2 x^2}} \sqrt [6]{c^2-d^2 x^2} \int \frac {1}{\sqrt {\frac {x^3}{\left (c^2-d^2 x^2\right )^{3/2}}-1}}d\sqrt [3]{\frac {d^2 x^2}{c^2-d^2 x^2}+1}}{2 d^2 x \sqrt [3]{\frac {c^2}{c^2-d^2 x^2}}}+\frac {3 \sqrt [6]{c^2-d^2 x^2}}{d}\)

\(\Big \downarrow \) 760

\(\displaystyle \frac {3 \sqrt [6]{c^2-d^2 x^2}}{d}-\frac {3^{3/4} \sqrt {2-\sqrt {3}} c \sqrt {\frac {d^2 x^2}{c^2-d^2 x^2}} \sqrt [6]{c^2-d^2 x^2} \left (1-\sqrt [3]{\frac {d^2 x^2}{c^2-d^2 x^2}+1}\right ) \sqrt {\frac {\frac {x^2}{c^2-d^2 x^2}+\sqrt [3]{\frac {d^2 x^2}{c^2-d^2 x^2}+1}+1}{\left (-\sqrt [3]{\frac {d^2 x^2}{c^2-d^2 x^2}+1}-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-\sqrt [3]{\frac {d^2 x^2}{c^2-d^2 x^2}+1}+\sqrt {3}+1}{-\sqrt [3]{\frac {d^2 x^2}{c^2-d^2 x^2}+1}-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{d^2 x \sqrt [3]{\frac {c^2}{c^2-d^2 x^2}} \sqrt {\frac {x^3}{\left (c^2-d^2 x^2\right )^{3/2}}-1} \sqrt {-\frac {1-\sqrt [3]{\frac {d^2 x^2}{c^2-d^2 x^2}+1}}{\left (-\sqrt [3]{\frac {d^2 x^2}{c^2-d^2 x^2}+1}-\sqrt {3}+1\right )^2}}}\)

Input:

Int[(c - d*x)/(c^2 - d^2*x^2)^(5/6),x]
 

Output:

(3*(c^2 - d^2*x^2)^(1/6))/d - (3^(3/4)*Sqrt[2 - Sqrt[3]]*c*Sqrt[(d^2*x^2)/ 
(c^2 - d^2*x^2)]*(c^2 - d^2*x^2)^(1/6)*(1 - (1 + (d^2*x^2)/(c^2 - d^2*x^2) 
)^(1/3))*Sqrt[(1 + x^2/(c^2 - d^2*x^2) + (1 + (d^2*x^2)/(c^2 - d^2*x^2))^( 
1/3))/(1 - Sqrt[3] - (1 + (d^2*x^2)/(c^2 - d^2*x^2))^(1/3))^2]*EllipticF[A 
rcSin[(1 + Sqrt[3] - (1 + (d^2*x^2)/(c^2 - d^2*x^2))^(1/3))/(1 - Sqrt[3] - 
 (1 + (d^2*x^2)/(c^2 - d^2*x^2))^(1/3))], -7 + 4*Sqrt[3]])/(d^2*x*(c^2/(c^ 
2 - d^2*x^2))^(1/3)*Sqrt[-1 + x^3/(c^2 - d^2*x^2)^(3/2)]*Sqrt[-((1 - (1 + 
(d^2*x^2)/(c^2 - d^2*x^2))^(1/3))/(1 - Sqrt[3] - (1 + (d^2*x^2)/(c^2 - d^2 
*x^2))^(1/3))^2)])
 

Defintions of rubi rules used

rule 234

Int[((a_) + (b_.)*(x_)^2)^(-2/3), x_Symbol] :> Simp[3*(Sqrt[b*x^2]/(2*b*x)) 
   Subst[Int[1/Sqrt[-a + x^3], x], x, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b 
}, x]
 

rule 236

Int[((a_) + (b_.)*(x_)^2)^(-5/6), x_Symbol] :> Simp[1/((a/(a + b*x^2))^(1/3 
)*(a + b*x^2)^(1/3))   Subst[Int[1/(1 - b*x^2)^(2/3), x], x, x/Sqrt[a + b*x 
^2]], x] /; FreeQ[{a, b}, x]
 

rule 455

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]
 

rule 760

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s 
*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(- 
s)*((s + r*x)/((1 - Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 + Sqrt[3]) 
*s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x 
] && NegQ[a]
 

Maple [F]

\[\int \frac {-d x +c}{\left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{6}}}d x\]

Input:

int((-d*x+c)/(-d^2*x^2+c^2)^(5/6),x)
 

Output:

int((-d*x+c)/(-d^2*x^2+c^2)^(5/6),x)
 

Fricas [F]

\[ \int \frac {c-d x}{\left (c^2-d^2 x^2\right )^{5/6}} \, dx=\int { -\frac {d x - c}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{6}}} \,d x } \] Input:

integrate((-d*x+c)/(-d^2*x^2+c^2)^(5/6),x, algorithm="fricas")
 

Output:

integral((-d^2*x^2 + c^2)^(1/6)/(d*x + c), x)
 

Sympy [A] (verification not implemented)

Time = 1.26 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.20 \[ \int \frac {c-d x}{\left (c^2-d^2 x^2\right )^{5/6}} \, dx=- d \left (\begin {cases} \frac {x^{2}}{2 \left (c^{2}\right )^{\frac {5}{6}}} & \text {for}\: d^{2} = 0 \\- \frac {3 \sqrt [6]{c^{2} - d^{2} x^{2}}}{d^{2}} & \text {otherwise} \end {cases}\right ) + \frac {x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{6} \\ \frac {3}{2} \end {matrix}\middle | {\frac {d^{2} x^{2} e^{2 i \pi }}{c^{2}}} \right )}}{c^{\frac {2}{3}}} \] Input:

integrate((-d*x+c)/(-d**2*x**2+c**2)**(5/6),x)
 

Output:

-d*Piecewise((x**2/(2*(c**2)**(5/6)), Eq(d**2, 0)), (-3*(c**2 - d**2*x**2) 
**(1/6)/d**2, True)) + x*hyper((1/2, 5/6), (3/2,), d**2*x**2*exp_polar(2*I 
*pi)/c**2)/c**(2/3)
 

Maxima [F]

\[ \int \frac {c-d x}{\left (c^2-d^2 x^2\right )^{5/6}} \, dx=\int { -\frac {d x - c}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{6}}} \,d x } \] Input:

integrate((-d*x+c)/(-d^2*x^2+c^2)^(5/6),x, algorithm="maxima")
 

Output:

-integrate((d*x - c)/(-d^2*x^2 + c^2)^(5/6), x)
 

Giac [F]

\[ \int \frac {c-d x}{\left (c^2-d^2 x^2\right )^{5/6}} \, dx=\int { -\frac {d x - c}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{6}}} \,d x } \] Input:

integrate((-d*x+c)/(-d^2*x^2+c^2)^(5/6),x, algorithm="giac")
 

Output:

integrate(-(d*x - c)/(-d^2*x^2 + c^2)^(5/6), x)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 6.48 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.21 \[ \int \frac {c-d x}{\left (c^2-d^2 x^2\right )^{5/6}} \, dx=\frac {3\,{\left (c^2-d^2\,x^2\right )}^{1/6}}{d}+\frac {c\,x\,{\left (1-\frac {d^2\,x^2}{c^2}\right )}^{5/6}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {5}{6};\ \frac {3}{2};\ \frac {d^2\,x^2}{c^2}\right )}{{\left (c^2-d^2\,x^2\right )}^{5/6}} \] Input:

int((c - d*x)/(c^2 - d^2*x^2)^(5/6),x)
 

Output:

(3*(c^2 - d^2*x^2)^(1/6))/d + (c*x*(1 - (d^2*x^2)/c^2)^(5/6)*hypergeom([1/ 
2, 5/6], 3/2, (d^2*x^2)/c^2))/(c^2 - d^2*x^2)^(5/6)
 

Reduce [F]

\[ \int \frac {c-d x}{\left (c^2-d^2 x^2\right )^{5/6}} \, dx=\frac {-\left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{6}}+\left (-d^{2} x^{2}+c^{2}\right )^{\frac {2}{3}} \left (\int \frac {\left (-d^{2} x^{2}+c^{2}\right )^{\frac {7}{6}}}{d^{4} x^{4}-2 c^{2} d^{2} x^{2}+c^{4}}d x \right ) c d}{\left (-d^{2} x^{2}+c^{2}\right )^{\frac {2}{3}} d} \] Input:

int((-d*x+c)/(-d^2*x^2+c^2)^(5/6),x)
 

Output:

( - (c**2 - d**2*x**2)**(5/6) + (c**2 - d**2*x**2)**(2/3)*int((c**2 - d**2 
*x**2)**(7/6)/(c**4 - 2*c**2*d**2*x**2 + d**4*x**4),x)*c*d)/((c**2 - d**2* 
x**2)**(2/3)*d)