\(\int \frac {1}{\sqrt [3]{c+d x} (c^2-d^2 x^2)^{2/3}} \, dx\) [299]

Optimal result

Integrand size = 26, antiderivative size = 225 \[ \int \frac {1}{\sqrt [3]{c+d x} \left (c^2-d^2 x^2\right )^{2/3}} \, dx=-\frac {\sqrt {3} (c-d x)^{2/3} (c+d x)^{2/3} \arctan \left (\frac {\sqrt [3]{c}+2^{2/3} \sqrt [3]{c-d x}}{\sqrt {3} \sqrt [3]{c}}\right )}{2^{2/3} c^{2/3} d \left (c^2-d^2 x^2\right )^{2/3}}-\frac {(c-d x)^{2/3} (c+d x)^{2/3} \log (c+d x)}{2\ 2^{2/3} c^{2/3} d \left (c^2-d^2 x^2\right )^{2/3}}+\frac {3 (c-d x)^{2/3} (c+d x)^{2/3} \log \left (\sqrt [3]{2} \sqrt [3]{c}-\sqrt [3]{c-d x}\right )}{2\ 2^{2/3} c^{2/3} d \left (c^2-d^2 x^2\right )^{2/3}} \] Output:

-1/2*3^(1/2)*(-d*x+c)^(2/3)*(d*x+c)^(2/3)*arctan(1/3*(c^(1/3)+2^(2/3)*(-d* 
x+c)^(1/3))*3^(1/2)/c^(1/3))*2^(1/3)/c^(2/3)/d/(-d^2*x^2+c^2)^(2/3)-1/4*(- 
d*x+c)^(2/3)*(d*x+c)^(2/3)*ln(d*x+c)*2^(1/3)/c^(2/3)/d/(-d^2*x^2+c^2)^(2/3 
)+3/4*(-d*x+c)^(2/3)*(d*x+c)^(2/3)*ln(2^(1/3)*c^(1/3)-(-d*x+c)^(1/3))*2^(1 
/3)/c^(2/3)/d/(-d^2*x^2+c^2)^(2/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.90 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\sqrt [3]{c+d x} \left (c^2-d^2 x^2\right )^{2/3}} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{c} \sqrt [3]{c+d x}}{\sqrt [3]{c} \sqrt [3]{c+d x}+2^{2/3} \sqrt [3]{c^2-d^2 x^2}}\right )+2 \log \left (-2 \sqrt [3]{c} \sqrt [3]{c+d x}+2^{2/3} \sqrt [3]{c^2-d^2 x^2}\right )-\log \left (2 c^{2/3} (c+d x)^{2/3}+2^{2/3} \sqrt [3]{c} \sqrt [3]{c+d x} \sqrt [3]{c^2-d^2 x^2}+\sqrt [3]{2} \left (c^2-d^2 x^2\right )^{2/3}\right )}{2\ 2^{2/3} c^{2/3} d} \] Input:

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

Output:

(2*Sqrt[3]*ArcTan[(Sqrt[3]*c^(1/3)*(c + d*x)^(1/3))/(c^(1/3)*(c + d*x)^(1/ 
3) + 2^(2/3)*(c^2 - d^2*x^2)^(1/3))] + 2*Log[-2*c^(1/3)*(c + d*x)^(1/3) + 
2^(2/3)*(c^2 - d^2*x^2)^(1/3)] - Log[2*c^(2/3)*(c + d*x)^(2/3) + 2^(2/3)*c 
^(1/3)*(c + d*x)^(1/3)*(c^2 - d^2*x^2)^(1/3) + 2^(1/3)*(c^2 - d^2*x^2)^(2/ 
3)])/(2*2^(2/3)*c^(2/3)*d)
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.72, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {474, 473, 27, 69, 16, 1082, 217}

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.

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

\(\Big \downarrow \) 474

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

\(\Big \downarrow \) 473

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 69

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

(c*(c^2 - d^2*x^2)^(1/3)*(-((Sqrt[3]*ArcTan[(1 + (2^(2/3)*(c^2 - c*d*x)^(1 
/3))/c^(2/3))/Sqrt[3]])/(2^(2/3)*c^(4/3)*d)) - Log[c + d*x]/(2*2^(2/3)*c^( 
4/3)*d) + (3*Log[2^(1/3)*c^(2/3) - (c^2 - c*d*x)^(1/3)])/(2*2^(2/3)*c^(4/3 
)*d)))/((c + d*x)^(1/3)*(c^2 - c*d*x)^(1/3))
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

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 69

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ 
{q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2), 
 x] + (-Simp[3/(2*b*q)   Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1 
/3)], x] - Simp[3/(2*b*q^2)   Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], 
 x])] /; FreeQ[{a, b, c, d}, x] && PosQ[(b*c - a*d)/b]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 473

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
c^(n - 1)*((a + b*x^2)^(p + 1)/((1 + d*(x/c))^(p + 1)*(a/c + (b*x)/d)^(p + 
1)))   Int[(1 + d*(x/c))^(n + p)*(a/c + (b/d)*x)^p, x], x] /; FreeQ[{a, b, 
c, d, n}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[n] || GtQ[c, 0]) &&  !Gt 
Q[a, 0] &&  !(IntegerQ[n] && (IntegerQ[3*p] || IntegerQ[4*p]))
 

rule 474

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n])   Int[(1 + d 
*(x/c))^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c^2 + 
 a*d^2, 0] &&  !(IntegerQ[n] || GtQ[c, 0])
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.12 \[ \int \frac {1}{\sqrt [3]{c+d x} \left (c^2-d^2 x^2\right )^{2/3}} \, dx=-\frac {4 \, \sqrt {3} c \sqrt {4^{\frac {1}{3}} {\left (c^{2}\right )}^{\frac {1}{3}}} \arctan \left (\frac {\sqrt {3} {\left (4^{\frac {2}{3}} {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} {\left (c^{2}\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (c d x + c^{2}\right )} {\left (c^{2}\right )}^{\frac {1}{3}}\right )} \sqrt {4^{\frac {1}{3}} {\left (c^{2}\right )}^{\frac {1}{3}}}}{6 \, {\left (c^{2} d x + c^{3}\right )}}\right ) + 4^{\frac {2}{3}} {\left (c^{2}\right )}^{\frac {2}{3}} \log \left (\frac {4^{\frac {2}{3}} {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} {\left (c^{2}\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {2}{3}} + 2 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} c + 2 \cdot 4^{\frac {1}{3}} {\left (c d x + c^{2}\right )} {\left (c^{2}\right )}^{\frac {1}{3}}}{d x + c}\right ) - 2 \cdot 4^{\frac {2}{3}} {\left (c^{2}\right )}^{\frac {2}{3}} \log \left (-\frac {4^{\frac {2}{3}} {\left (c^{2}\right )}^{\frac {2}{3}} {\left (d x + c\right )} - 2 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} c}{d x + c}\right )}{8 \, c^{2} d} \] Input:

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

Output:

-1/8*(4*sqrt(3)*c*sqrt(4^(1/3)*(c^2)^(1/3))*arctan(1/6*sqrt(3)*(4^(2/3)*(- 
d^2*x^2 + c^2)^(1/3)*(c^2)^(2/3)*(d*x + c)^(2/3) + 4^(1/3)*(c*d*x + c^2)*( 
c^2)^(1/3))*sqrt(4^(1/3)*(c^2)^(1/3))/(c^2*d*x + c^3)) + 4^(2/3)*(c^2)^(2/ 
3)*log((4^(2/3)*(-d^2*x^2 + c^2)^(1/3)*(c^2)^(2/3)*(d*x + c)^(2/3) + 2*(-d 
^2*x^2 + c^2)^(2/3)*(d*x + c)^(1/3)*c + 2*4^(1/3)*(c*d*x + c^2)*(c^2)^(1/3 
))/(d*x + c)) - 2*4^(2/3)*(c^2)^(2/3)*log(-(4^(2/3)*(c^2)^(2/3)*(d*x + c) 
- 2*(-d^2*x^2 + c^2)^(1/3)*(d*x + c)^(2/3)*c)/(d*x + c)))/(c^2*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Giac [F]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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