\(\int \frac {1}{(c+d x) \sqrt [3]{b c^3-b d^3 x^3}} \, dx\) [137]

Optimal result

Integrand size = 27, antiderivative size = 168 \[ \int \frac {1}{(c+d x) \sqrt [3]{b c^3-b d^3 x^3}} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {1+\frac {\sqrt [3]{2} \sqrt [3]{b} (c-d x)}{\sqrt [3]{b c^3-b d^3 x^3}}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt [3]{b} c d}-\frac {\log \left ((c-d x) (c+d x)^2\right )}{4 \sqrt [3]{2} \sqrt [3]{b} c d}+\frac {3 \log \left (-\sqrt [3]{b} d (c-d x)+2^{2/3} d \sqrt [3]{b c^3-b d^3 x^3}\right )}{4 \sqrt [3]{2} \sqrt [3]{b} c d} \] Output:

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.48 \[ \int \frac {1}{(c+d x) \sqrt [3]{b c^3-b d^3 x^3}} \, dx=\frac {\sqrt [3]{c^3-d^3 x^3} \left (2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{c^3-d^3 x^3}}{\sqrt [3]{2} c-\sqrt [3]{2} d x+\sqrt [3]{c^3-d^3 x^3}}\right )+2 \log \left (\sqrt {c} \sqrt {d} \left (\sqrt [3]{2} c-\sqrt [3]{2} d x-2 \sqrt [3]{c^3-d^3 x^3}\right )\right )-\log \left (c d \left (2^{2/3} c^2-2\ 2^{2/3} c d x+2^{2/3} d^2 x^2+2 \sqrt [3]{2} (c-d x) \sqrt [3]{c^3-d^3 x^3}+4 \left (c^3-d^3 x^3\right )^{2/3}\right )\right )\right )}{4 \sqrt [3]{2} c d \sqrt [3]{b \left (c^3-d^3 x^3\right )}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, 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.

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

\(\Big \downarrow \) 2574

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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