\(\int \frac {(c+d x)^2}{(c^2-d^2 x^2)^{2/3}} \, dx\) [264]

Optimal result

Integrand size = 24, antiderivative size = 322 \[ \int \frac {(c+d x)^2}{\left (c^2-d^2 x^2\right )^{2/3}} \, dx=-\frac {3 (5 c+d x) \sqrt [3]{c^2-d^2 x^2}}{5 d}+\frac {8\ 3^{3/4} \sqrt {2-\sqrt {3}} c^2 \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right ) \sqrt {\frac {c^{4/3}+c^{2/3} \sqrt [3]{c^2-d^2 x^2}+\left (c^2-d^2 x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}{\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}\right ),-7+4 \sqrt {3}\right )}{5 d^2 x \sqrt {-\frac {c^{2/3} \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}}} \] Output:

-3/5*(d*x+5*c)*(-d^2*x^2+c^2)^(1/3)/d+8/5*3^(3/4)*(1/2*6^(1/2)-1/2*2^(1/2) 
)*c^2*(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 = 10.05 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.31 \[ \int \frac {(c+d x)^2}{\left (c^2-d^2 x^2\right )^{2/3}} \, dx=\frac {3 \left (-5 c^3-c^2 d x+5 c d^2 x^2+d^3 x^3\right )+8 c^2 d x \left (1-\frac {d^2 x^2}{c^2}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {d^2 x^2}{c^2}\right )}{5 d \left (c^2-d^2 x^2\right )^{2/3}} \] Input:

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

Output:

(3*(-5*c^3 - c^2*d*x + 5*c*d^2*x^2 + d^3*x^3) + 8*c^2*d*x*(1 - (d^2*x^2)/c 
^2)^(2/3)*Hypergeometric2F1[1/2, 2/3, 3/2, (d^2*x^2)/c^2])/(5*d*(c^2 - d^2 
*x^2)^(2/3))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {497, 27, 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.

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

\(\Big \downarrow \) 497

\(\displaystyle -\frac {3 \int -\frac {8 c d^2 (c+d x)}{3 \left (c^2-d^2 x^2\right )^{2/3}}dx}{5 d^2}-\frac {3 \sqrt [3]{c^2-d^2 x^2} (c+d x)}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {8}{5} c \int \frac {c+d x}{\left (c^2-d^2 x^2\right )^{2/3}}dx-\frac {3 (c+d x) \sqrt [3]{c^2-d^2 x^2}}{5 d}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {8}{5} c \left (c \int \frac {1}{\left (c^2-d^2 x^2\right )^{2/3}}dx-\frac {3 \sqrt [3]{c^2-d^2 x^2}}{2 d}\right )-\frac {3 (c+d x) \sqrt [3]{c^2-d^2 x^2}}{5 d}\)

\(\Big \downarrow \) 234

\(\displaystyle \frac {8}{5} c \left (-\frac {3 c \sqrt {-d^2 x^2} \int \frac {1}{\sqrt {-d^2 x^2}}d\sqrt [3]{c^2-d^2 x^2}}{2 d^2 x}-\frac {3 \sqrt [3]{c^2-d^2 x^2}}{2 d}\right )-\frac {3 (c+d x) \sqrt [3]{c^2-d^2 x^2}}{5 d}\)

\(\Big \downarrow \) 760

\(\displaystyle \frac {8}{5} c \left (\frac {3^{3/4} \sqrt {2-\sqrt {3}} c \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right ) \sqrt {\frac {c^{4/3}+\left (c^2-d^2 x^2\right )^{2/3}+c^{2/3} \sqrt [3]{c^2-d^2 x^2}}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}{\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}\right ),-7+4 \sqrt {3}\right )}{d^2 x \sqrt {-\frac {c^{2/3} \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}}}-\frac {3 \sqrt [3]{c^2-d^2 x^2}}{2 d}\right )-\frac {3 (c+d x) \sqrt [3]{c^2-d^2 x^2}}{5 d}\)

Input:

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

Output:

(-3*(c + d*x)*(c^2 - d^2*x^2)^(1/3))/(5*d) + (8*c*((-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]*EllipticF[ArcSin[((1 + Sq 
rt[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)])))/5
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 497

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[1/(b 
*(n + 2*p + 1))   Int[(c + d*x)^(n - 2)*(a + b*x^2)^p*Simp[b*c^2*(n + 2*p + 
 1) - a*d^2*(n - 1) + 2*b*c*d*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, n 
, p}, x] && If[RationalQ[n], GtQ[n, 1], SumSimplerQ[n, -2]] && NeQ[n + 2*p 
+ 1, 0] && IntQuadraticQ[a, 0, b, c, d, n, p, x]
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [A] (verification not implemented)

Time = 1.26 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.33 \[ \int \frac {(c+d x)^2}{\left (c^2-d^2 x^2\right )^{2/3}} \, dx=c^{\frac {2}{3}} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {2}{3} \\ \frac {3}{2} \end {matrix}\middle | {\frac {d^{2} x^{2} e^{2 i \pi }}{c^{2}}} \right )} + 2 c d \left (\begin {cases} \frac {x^{2}}{2 \left (c^{2}\right )^{\frac {2}{3}}} & \text {for}\: d^{2} = 0 \\- \frac {3 \sqrt [3]{c^{2} - d^{2} x^{2}}}{2 d^{2}} & \text {otherwise} \end {cases}\right ) + \frac {d^{2} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {d^{2} x^{2} e^{2 i \pi }}{c^{2}}} \right )}}{3 c^{\frac {4}{3}}} \] Input:

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

Output:

c**(2/3)*x*hyper((1/2, 2/3), (3/2,), d**2*x**2*exp_polar(2*I*pi)/c**2) + 2 
*c*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)) + d**2*x**3*hyper((2/3, 3/2), (5/2,), d**2*x**2 
*exp_polar(2*I*pi)/c**2)/(3*c**(4/3))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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