\(\int \frac {x}{\sqrt [3]{c+d x^3} (a c+2 a d x^3)} \, dx\) [213]

Optimal result

Integrand size = 26, antiderivative size = 214 \[ \int \frac {x}{\sqrt [3]{c+d x^3} \left (a c+2 a d x^3\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [3]{d}-\frac {\sqrt [3]{2} c^{2/3}}{x \sqrt [3]{c+d x^3}}}{\sqrt {3} \sqrt [3]{d}}\right )}{2 \sqrt [3]{2} \sqrt {3} a c^{2/3} d^{2/3}}-\frac {\log \left (2^{2/3} \sqrt [3]{d}+\frac {c^{2/3}}{x \sqrt [3]{c+d x^3}}\right )}{6 \sqrt [3]{2} a c^{2/3} d^{2/3}}+\frac {\log \left (2 \sqrt [3]{2} d^{2/3}+\frac {c^{4/3}}{x^2 \left (c+d x^3\right )^{2/3}}-\frac {2^{2/3} c^{2/3} \sqrt [3]{d}}{x \sqrt [3]{c+d x^3}}\right )}{12 \sqrt [3]{2} a c^{2/3} d^{2/3}} \] Output:

1/12*arctan(1/3*(d^(1/3)-2^(1/3)*c^(2/3)/x/(d*x^3+c)^(1/3))*3^(1/2)/d^(1/3 
))*2^(2/3)*3^(1/2)/a/c^(2/3)/d^(2/3)-1/12*ln(2^(2/3)*d^(1/3)+c^(2/3)/x/(d* 
x^3+c)^(1/3))*2^(2/3)/a/c^(2/3)/d^(2/3)+1/24*ln(2*2^(1/3)*d^(2/3)+c^(4/3)/ 
x^2/(d*x^3+c)^(2/3)-2^(2/3)*c^(2/3)*d^(1/3)/x/(d*x^3+c)^(1/3))*2^(2/3)/a/c 
^(2/3)/d^(2/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 5.77 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.02 \[ \int \frac {x}{\sqrt [3]{c+d x^3} \left (a c+2 a d x^3\right )} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{d} x \sqrt [3]{c+d x^3}}{\sqrt [3]{2} c^{2/3}-\sqrt [3]{d} x \sqrt [3]{c+d x^3}}\right )+2 \log \left (x \sqrt [3]{c+d x^3}\right )-\log \left (x^2 \left (c+d x^3\right )^{2/3}\right )-2 \log \left (\sqrt [3]{2} c^{2/3}+2 \sqrt [3]{d} x \sqrt [3]{c+d x^3}\right )+\log \left (2^{2/3} c^{4/3}-2 \sqrt [3]{2} c^{2/3} \sqrt [3]{d} x \sqrt [3]{c+d x^3}+4 d^{2/3} x^2 \left (c+d x^3\right )^{2/3}\right )}{12 \sqrt [3]{2} a c^{2/3} d^{2/3}} \] Input:

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

Output:

(2*Sqrt[3]*ArcTan[(Sqrt[3]*d^(1/3)*x*(c + d*x^3)^(1/3))/(2^(1/3)*c^(2/3) - 
 d^(1/3)*x*(c + d*x^3)^(1/3))] + 2*Log[x*(c + d*x^3)^(1/3)] - Log[x^2*(c + 
 d*x^3)^(2/3)] - 2*Log[2^(1/3)*c^(2/3) + 2*d^(1/3)*x*(c + d*x^3)^(1/3)] + 
Log[2^(2/3)*c^(4/3) - 2*2^(1/3)*c^(2/3)*d^(1/3)*x*(c + d*x^3)^(1/3) + 4*d^ 
(2/3)*x^2*(c + d*x^3)^(2/3)])/(12*2^(1/3)*a*c^(2/3)*d^(2/3))
 

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 0.34 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.31, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1013, 27, 1012}

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.

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

\(\Big \downarrow \) 1013

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1012

\(\displaystyle \frac {x^2 \sqrt [3]{\frac {d x^3}{c}+1} \operatorname {AppellF1}\left (\frac {2}{3},1,\frac {1}{3},\frac {5}{3},-\frac {2 d x^3}{c},-\frac {d x^3}{c}\right )}{2 a c \sqrt [3]{c+d x^3}}\)

Input:

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

Output:

(x^2*(1 + (d*x^3)/c)^(1/3)*AppellF1[2/3, 1, 1/3, 5/3, (-2*d*x^3)/c, -((d*x 
^3)/c)])/(2*a*c*(c + d*x^3)^(1/3))
 

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 1012

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m 
+ 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, 
 b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n 
 - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 1013

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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