\(\int \frac {c+d x}{(c^2-d^2 x^2)^{2/3}} \, dx\) [265]

Optimal result

Integrand size = 22, antiderivative size = 310 \[ \int \frac {c+d x}{\left (c^2-d^2 x^2\right )^{2/3}} \, dx=-\frac {3 \sqrt [3]{c^2-d^2 x^2}}{2 d}+\frac {3^{3/4} \sqrt {2-\sqrt {3}} c \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right ) \sqrt {\frac {c^{4/3}+c^{2/3} \sqrt [3]{c^2-d^2 x^2}+\left (c^2-d^2 x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}{\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}\right ),-7+4 \sqrt {3}\right )}{d^2 x \sqrt {-\frac {c^{2/3} \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}}} \] Output:

-3/2*(-d^2*x^2+c^2)^(1/3)/d+3^(3/4)*(1/2*6^(1/2)-1/2*2^(1/2))*c*(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))/((1-3^(1/2))*c^(2/3)-(-d^2*x^2+c^2)^(1/3))^2)^(1/2)*EllipticF(((1+ 
3^(1/2))*c^(2/3)-(-d^2*x^2+c^2)^(1/3))/((1-3^(1/2))*c^(2/3)-(-d^2*x^2+c^2) 
^(1/3)),2*I-I*3^(1/2))/d^2/x/(-c^(2/3)*(c^(2/3)-(-d^2*x^2+c^2)^(1/3))/((1- 
3^(1/2))*c^(2/3)-(-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 = 9.80 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.26 \[ \int \frac {c+d x}{\left (c^2-d^2 x^2\right )^{2/3}} \, dx=-\frac {3 \sqrt [3]{c^2-d^2 x^2}}{2 d}+\frac {c x \left (1-\frac {d^2 x^2}{c^2}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {d^2 x^2}{c^2}\right )}{\left (c^2-d^2 x^2\right )^{2/3}} \] Input:

Integrate[(c + d*x)/(c^2 - d^2*x^2)^(2/3),x]
 

Output:

(-3*(c^2 - d^2*x^2)^(1/3))/(2*d) + (c*x*(1 - (d^2*x^2)/c^2)^(2/3)*Hypergeo 
metric2F1[1/2, 2/3, 3/2, (d^2*x^2)/c^2])/(c^2 - d^2*x^2)^(2/3)
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {455, 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.

\(\displaystyle \int \frac {c+d x}{\left (c^2-d^2 x^2\right )^{2/3}} \, dx\)

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 234

\(\displaystyle -\frac {3 c \sqrt {-d^2 x^2} \int \frac {1}{\sqrt {-d^2 x^2}}d\sqrt [3]{c^2-d^2 x^2}}{2 d^2 x}-\frac {3 \sqrt [3]{c^2-d^2 x^2}}{2 d}\)

\(\Big \downarrow \) 760

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

Input:

Int[(c + d*x)/(c^2 - d^2*x^2)^(2/3),x]
 

Output:

(-3*(c^2 - d^2*x^2)^(1/3))/(2*d) + (3^(3/4)*Sqrt[2 - Sqrt[3]]*c*(c^(2/3) - 
 (c^2 - d^2*x^2)^(1/3))*Sqrt[(c^(4/3) + c^(2/3)*(c^2 - d^2*x^2)^(1/3) + (c 
^2 - d^2*x^2)^(2/3))/((1 - Sqrt[3])*c^(2/3) - (c^2 - d^2*x^2)^(1/3))^2]*El 
lipticF[ArcSin[((1 + Sqrt[3])*c^(2/3) - (c^2 - d^2*x^2)^(1/3))/((1 - Sqrt[ 
3])*c^(2/3) - (c^2 - d^2*x^2)^(1/3))], -7 + 4*Sqrt[3]])/(d^2*x*Sqrt[-((c^( 
2/3)*(c^(2/3) - (c^2 - d^2*x^2)^(1/3)))/((1 - Sqrt[3])*c^(2/3) - (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 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]

\[\int \frac {d x +c}{\left (-d^{2} x^{2}+c^{2}\right )^{\frac {2}{3}}}d x\]

Input:

int((d*x+c)/(-d^2*x^2+c^2)^(2/3),x)
 

Output:

int((d*x+c)/(-d^2*x^2+c^2)^(2/3),x)
 

Fricas [F]

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

integrate((d*x+c)/(-d^2*x^2+c^2)^(2/3),x, algorithm="fricas")
 

Output:

integral(-(-d^2*x^2 + c^2)^(1/3)/(d*x - c), x)
 

Sympy [A] (verification not implemented)

Time = 0.98 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.21 \[ \int \frac {c+d x}{\left (c^2-d^2 x^2\right )^{2/3}} \, dx=d \left (\begin {cases} \frac {x^{2}}{2 \left (c^{2}\right )^{\frac {2}{3}}} & \text {for}\: d^{2} = 0 \\- \frac {3 \sqrt [3]{c^{2} - d^{2} x^{2}}}{2 d^{2}} & \text {otherwise} \end {cases}\right ) + \frac {x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {2}{3} \\ \frac {3}{2} \end {matrix}\middle | {\frac {d^{2} x^{2} e^{2 i \pi }}{c^{2}}} \right )}}{\sqrt [3]{c}} \] Input:

integrate((d*x+c)/(-d**2*x**2+c**2)**(2/3),x)
 

Output:

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

Maxima [F]

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

integrate((d*x+c)/(-d^2*x^2+c^2)^(2/3),x, algorithm="maxima")
 

Output:

integrate((d*x + c)/(-d^2*x^2 + c^2)^(2/3), x)
 

Giac [F]

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

integrate((d*x+c)/(-d^2*x^2+c^2)^(2/3),x, algorithm="giac")
 

Output:

integrate((d*x + c)/(-d^2*x^2 + c^2)^(2/3), x)
 

Mupad [B] (verification not implemented)

Time = 7.35 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.22 \[ \int \frac {c+d x}{\left (c^2-d^2 x^2\right )^{2/3}} \, dx=\frac {c\,x\,{\left (1-\frac {d^2\,x^2}{c^2}\right )}^{2/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {2}{3};\ \frac {3}{2};\ \frac {d^2\,x^2}{c^2}\right )}{{\left (c^2-d^2\,x^2\right )}^{2/3}}-\frac {3\,{\left (c^2-d^2\,x^2\right )}^{1/3}}{2\,d} \] Input:

int((c + d*x)/(c^2 - d^2*x^2)^(2/3),x)
 

Output:

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

Reduce [F]

\[ \int \frac {c+d x}{\left (c^2-d^2 x^2\right )^{2/3}} \, dx=\frac {-3 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}}+2 \left (\int \frac {1}{\left (-d^{2} x^{2}+c^{2}\right )^{\frac {2}{3}}}d x \right ) c d}{2 d} \] Input:

int((d*x+c)/(-d^2*x^2+c^2)^(2/3),x)
 

Output:

( - 3*(c**2 - d**2*x**2)**(1/3) + 2*int(1/(c**2 - d**2*x**2)**(2/3),x)*c*d 
)/(2*d)