\(\int \frac {1}{(c+d x) (c^2-d^2 x^2)^{2/3}} \, dx\) [267]

Optimal result

Integrand size = 24, antiderivative size = 325 \[ \int \frac {1}{(c+d x) \left (c^2-d^2 x^2\right )^{2/3}} \, dx=-\frac {3 \sqrt [3]{c^2-d^2 x^2}}{4 c d (c+d x)}+\frac {3^{3/4} \sqrt {2-\sqrt {3}} \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right ) \sqrt {\frac {c^{4/3}+c^{2/3} \sqrt [3]{c^2-d^2 x^2}+\left (c^2-d^2 x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}{\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}\right ),-7+4 \sqrt {3}\right )}{4 c d^2 x \sqrt {-\frac {c^{2/3} \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}}} \] Output:

-3/4*(-d^2*x^2+c^2)^(1/3)/c/d/(d*x+c)+1/4*3^(3/4)*(1/2*6^(1/2)-1/2*2^(1/2) 
)*(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)*El 
lipticF(((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))/c/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 = 5.74 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.22 \[ \int \frac {1}{(c+d x) \left (c^2-d^2 x^2\right )^{2/3}} \, dx=-\frac {3 (c-d x) \left (1+\frac {d x}{c}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {5}{3},\frac {4}{3},\frac {c-d x}{2 c}\right )}{2\ 2^{2/3} c d \left (c^2-d^2 x^2\right )^{2/3}} \] Input:

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

Output:

(-3*(c - d*x)*(1 + (d*x)/c)^(2/3)*Hypergeometric2F1[1/3, 5/3, 4/3, (c - d* 
x)/(2*c)])/(2*2^(2/3)*c*d*(c^2 - d^2*x^2)^(2/3))
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {504, 215, 234, 241, 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.

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

\(\Big \downarrow \) 504

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

\(\Big \downarrow \) 215

\(\displaystyle c \left (\frac {\int \frac {1}{\left (c^2-d^2 x^2\right )^{2/3}}dx}{4 c^2}+\frac {3 x}{4 c^2 \left (c^2-d^2 x^2\right )^{2/3}}\right )-d \int \frac {x}{\left (c^2-d^2 x^2\right )^{5/3}}dx\)

\(\Big \downarrow \) 234

\(\displaystyle c \left (\frac {3 x}{4 c^2 \left (c^2-d^2 x^2\right )^{2/3}}-\frac {3 \sqrt {-d^2 x^2} \int \frac {1}{\sqrt {-d^2 x^2}}d\sqrt [3]{c^2-d^2 x^2}}{8 c^2 d^2 x}\right )-d \int \frac {x}{\left (c^2-d^2 x^2\right )^{5/3}}dx\)

\(\Big \downarrow \) 241

\(\displaystyle c \left (\frac {3 x}{4 c^2 \left (c^2-d^2 x^2\right )^{2/3}}-\frac {3 \sqrt {-d^2 x^2} \int \frac {1}{\sqrt {-d^2 x^2}}d\sqrt [3]{c^2-d^2 x^2}}{8 c^2 d^2 x}\right )-\frac {3}{4 d \left (c^2-d^2 x^2\right )^{2/3}}\)

\(\Big \downarrow \) 760

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

Input:

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

Output:

-3/(4*d*(c^2 - d^2*x^2)^(2/3)) + c*((3*x)/(4*c^2*(c^2 - d^2*x^2)^(2/3)) + 
(3^(3/4)*Sqrt[2 - Sqrt[3]]*(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 + 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]])/(4*c^2*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 215

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 
*p])
 

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 241

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

rule 504

Int[((a_) + (b_.)*(x_)^2)^(p_)/((c_) + (d_.)*(x_)), x_Symbol] :> Simp[c   I 
nt[(a + b*x^2)^p/(c^2 - d^2*x^2), x], x] - Simp[d   Int[x*((a + b*x^2)^p/(c 
^2 - d^2*x^2)), x], x] /; FreeQ[{a, b, c, d, 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(1/(d*x+c)/(-d^2*x^2+c^2)^(2/3),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(1/(d*x+c)/(-d^2*x^2+c^2)^(2/3),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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