\(\int \frac {1}{(c+d x) \sqrt [3]{-c^3+d^3 x^3}} \, dx\) [136]

Optimal result

Integrand size = 25, antiderivative size = 139 \[ \int \frac {1}{(c+d x) \sqrt [3]{-c^3+d^3 x^3}} \, dx=\frac {\sqrt {3} \arctan \left (\frac {1-\frac {\sqrt [3]{2} (c-d x)}{\sqrt [3]{-c^3+d^3 x^3}}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} c d}+\frac {\log \left ((c-d x) (c+d x)^2\right )}{4 \sqrt [3]{2} c d}-\frac {3 \log \left (d (c-d x)+2^{2/3} d \sqrt [3]{-c^3+d^3 x^3}\right )}{4 \sqrt [3]{2} c d} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.91 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.24 \[ \int \frac {1}{(c+d x) \sqrt [3]{-c^3+d^3 x^3}} \, dx=\frac {\sqrt [3]{-\frac {1}{2}} \left (2 i \sqrt {3} \text {arctanh}\left (\frac {\sqrt [3]{2} \left (3+i \sqrt {3}\right ) c+\sqrt [3]{2} \left (-3-i \sqrt {3}\right ) d x+2 i \sqrt {3} \sqrt [3]{-c^3+d^3 x^3}}{6 \sqrt [3]{-c^3+d^3 x^3}}\right )+2 \log \left (\sqrt {c} \sqrt {d} \left (-c+i \sqrt {3} c+d x-i \sqrt {3} d x+2\ 2^{2/3} \sqrt [3]{-c^3+d^3 x^3}\right )\right )-\log \left (-c d \left (\left (1+i \sqrt {3}\right ) c^2+\left (1+i \sqrt {3}\right ) d^2 x^2-2 (-2)^{2/3} d x \sqrt [3]{-c^3+d^3 x^3}-4 \sqrt [3]{2} \left (-c^3+d^3 x^3\right )^{2/3}+2 c \left (\left (-1-i \sqrt {3}\right ) d x+(-2)^{2/3} \sqrt [3]{-c^3+d^3 x^3}\right )\right )\right )\right )}{4 c d} \] Input:

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

Output:

((-1/2)^(1/3)*((2*I)*Sqrt[3]*ArcTanh[(2^(1/3)*(3 + I*Sqrt[3])*c + 2^(1/3)* 
(-3 - I*Sqrt[3])*d*x + (2*I)*Sqrt[3]*(-c^3 + d^3*x^3)^(1/3))/(6*(-c^3 + d^ 
3*x^3)^(1/3))] + 2*Log[Sqrt[c]*Sqrt[d]*(-c + I*Sqrt[3]*c + d*x - I*Sqrt[3] 
*d*x + 2*2^(2/3)*(-c^3 + d^3*x^3)^(1/3))] - Log[-(c*d*((1 + I*Sqrt[3])*c^2 
 + (1 + I*Sqrt[3])*d^2*x^2 - 2*(-2)^(2/3)*d*x*(-c^3 + d^3*x^3)^(1/3) - 4*2 
^(1/3)*(-c^3 + d^3*x^3)^(2/3) + 2*c*((-1 - I*Sqrt[3])*d*x + (-2)^(2/3)*(-c 
^3 + d^3*x^3)^(1/3))))]))/(4*c*d)
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2574}

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.

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

\(\Big \downarrow \) 2574

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2574

Int[1/(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^3)^(1/3)), x_Symbol] :> Simp[ 
Sqrt[3]*(ArcTan[(1 - 2^(1/3)*Rt[b, 3]*((c - d*x)/(d*(a + b*x^3)^(1/3))))/Sq 
rt[3]]/(2^(4/3)*Rt[b, 3]*c)), x] + (Simp[Log[(c + d*x)^2*(c - d*x)]/(2^(7/3 
)*Rt[b, 3]*c), x] - Simp[(3*Log[Rt[b, 3]*(c - d*x) + 2^(2/3)*d*(a + b*x^3)^ 
(1/3)])/(2^(7/3)*Rt[b, 3]*c), x]) /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^3 + 
a*d^3, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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