\(\int \frac {x^3}{(a+b x^3)^{4/3} (c+d x^3)} \, dx\) [771]

Optimal result

Integrand size = 24, antiderivative size = 172 \[ \int \frac {x^3}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=-\frac {x}{(b c-a d) \sqrt [3]{a+b x^3}}+\frac {\sqrt [3]{c} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} (b c-a d)^{4/3}}+\frac {\sqrt [3]{c} \log \left (c+d x^3\right )}{6 (b c-a d)^{4/3}}-\frac {\sqrt [3]{c} \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 (b c-a d)^{4/3}} \] Output:

-x/(-a*d+b*c)/(b*x^3+a)^(1/3)+1/3*c^(1/3)*arctan(1/3*(1+2*(-a*d+b*c)^(1/3) 
*x/c^(1/3)/(b*x^3+a)^(1/3))*3^(1/2))*3^(1/2)/(-a*d+b*c)^(4/3)+1/6*c^(1/3)* 
ln(d*x^3+c)/(-a*d+b*c)^(4/3)-1/2*c^(1/3)*ln((-a*d+b*c)^(1/3)*x/c^(1/3)-(b* 
x^3+a)^(1/3))/(-a*d+b*c)^(4/3)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.26 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.87 \[ \int \frac {x^3}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\frac {1}{12} \left (-\frac {12 x}{(b c-a d) \sqrt [3]{a+b x^3}}-\frac {2 \sqrt {-6+6 i \sqrt {3}} \sqrt [3]{c} \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{(b c-a d)^{4/3}}+\frac {2 \left (1+i \sqrt {3}\right ) \sqrt [3]{c} \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{(b c-a d)^{4/3}}-\frac {i \left (-i+\sqrt {3}\right ) \sqrt [3]{c} \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{(b c-a d)^{4/3}}\right ) \] Input:

Integrate[x^3/((a + b*x^3)^(4/3)*(c + d*x^3)),x]
 

Output:

((-12*x)/((b*c - a*d)*(a + b*x^3)^(1/3)) - (2*Sqrt[-6 + (6*I)*Sqrt[3]]*c^( 
1/3)*ArcTan[(3*(b*c - a*d)^(1/3)*x)/(Sqrt[3]*(b*c - a*d)^(1/3)*x - (3*I + 
Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3))])/(b*c - a*d)^(4/3) + (2*(1 + I*Sqrt[3 
])*c^(1/3)*Log[2*(b*c - a*d)^(1/3)*x + (1 + I*Sqrt[3])*c^(1/3)*(a + b*x^3) 
^(1/3)])/(b*c - a*d)^(4/3) - (I*(-I + Sqrt[3])*c^(1/3)*Log[2*(b*c - a*d)^( 
2/3)*x^2 + (-1 - I*Sqrt[3])*c^(1/3)*(b*c - a*d)^(1/3)*x*(a + b*x^3)^(1/3) 
+ I*(I + Sqrt[3])*c^(2/3)*(a + b*x^3)^(2/3)])/(b*c - a*d)^(4/3))/12
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.08, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {971, 27, 901}

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.

\(\displaystyle \int \frac {x^3}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx\)

\(\Big \downarrow \) 971

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 901

\(\displaystyle \frac {c \left (\frac {\arctan \left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} c^{2/3} \sqrt [3]{b c-a d}}+\frac {\log \left (c+d x^3\right )}{6 c^{2/3} \sqrt [3]{b c-a d}}-\frac {\log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{2/3} \sqrt [3]{b c-a d}}\right )}{b c-a d}-\frac {x}{\sqrt [3]{a+b x^3} (b c-a d)}\)

Input:

Int[x^3/((a + b*x^3)^(4/3)*(c + d*x^3)),x]
 

Output:

-(x/((b*c - a*d)*(a + b*x^3)^(1/3))) + (c*(ArcTan[(1 + (2*(b*c - a*d)^(1/3 
)*x)/(c^(1/3)*(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*c^(2/3)*(b*c - a*d)^(1 
/3)) + Log[c + d*x^3]/(6*c^(2/3)*(b*c - a*d)^(1/3)) - Log[((b*c - a*d)^(1/ 
3)*x)/c^(1/3) - (a + b*x^3)^(1/3)]/(2*c^(2/3)*(b*c - a*d)^(1/3))))/(b*c - 
a*d)
 

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 901

Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> Wit 
h[{q = Rt[(b*c - a*d)/c, 3]}, Simp[ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/S 
qrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q), x] 
 + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0]
 

rule 971

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

Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.26

method	result	size
pseudoelliptic	\(-\frac {2 \left (-3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\frac {\left (2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x -2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x}\right )+2 \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )-\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )\right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{2}\right )}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (6 a d -6 b c \right )}\)	\(217\)

Input:

int(x^3/(b*x^3+a)^(4/3)/(d*x^3+c),x,method=_RETURNVERBOSE)
 

Output:

-2/((a*d-b*c)/c)^(1/3)*(-3*((a*d-b*c)/c)^(1/3)*x+1/2*(2*3^(1/2)*arctan(1/3 
*3^(1/2)*(((a*d-b*c)/c)^(1/3)*x-2*(b*x^3+a)^(1/3))/((a*d-b*c)/c)^(1/3)/x)+ 
2*ln((((a*d-b*c)/c)^(1/3)*x+(b*x^3+a)^(1/3))/x)-ln((((a*d-b*c)/c)^(2/3)*x^ 
2-((a*d-b*c)/c)^(1/3)*(b*x^3+a)^(1/3)*x+(b*x^3+a)^(2/3))/x^2))*(b*x^3+a)^( 
1/3))/(b*x^3+a)^(1/3)/(6*a*d-6*b*c)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {x^3}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\text {Timed out} \] Input:

integrate(x^3/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {x^3}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\int \frac {x^{3}}{\left (a + b x^{3}\right )^{\frac {4}{3}} \left (c + d x^{3}\right )}\, dx \] Input:

integrate(x**3/(b*x**3+a)**(4/3)/(d*x**3+c),x)
 

Output:

Integral(x**3/((a + b*x**3)**(4/3)*(c + d*x**3)), x)
 

Maxima [F]

\[ \int \frac {x^3}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\int { \frac {x^{3}}{{\left (b x^{3} + a\right )}^{\frac {4}{3}} {\left (d x^{3} + c\right )}} \,d x } \] Input:

integrate(x^3/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="maxima")
 

Output:

integrate(x^3/((b*x^3 + a)^(4/3)*(d*x^3 + c)), x)
 

Giac [F]

\[ \int \frac {x^3}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\int { \frac {x^{3}}{{\left (b x^{3} + a\right )}^{\frac {4}{3}} {\left (d x^{3} + c\right )}} \,d x } \] Input:

integrate(x^3/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="giac")
 

Output:

integrate(x^3/((b*x^3 + a)^(4/3)*(d*x^3 + c)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\int \frac {x^3}{{\left (b\,x^3+a\right )}^{4/3}\,\left (d\,x^3+c\right )} \,d x \] Input:

int(x^3/((a + b*x^3)^(4/3)*(c + d*x^3)),x)
 

Output:

int(x^3/((a + b*x^3)^(4/3)*(c + d*x^3)), x)
 

Reduce [F]

\[ \int \frac {x^3}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\int \frac {x^{3}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a c +\left (b \,x^{3}+a \right )^{\frac {1}{3}} a d \,x^{3}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b c \,x^{3}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b d \,x^{6}}d x \] Input:

int(x^3/(b*x^3+a)^(4/3)/(d*x^3+c),x)
 

Output:

int(x**3/((a + b*x**3)**(1/3)*a*c + (a + b*x**3)**(1/3)*a*d*x**3 + (a + b* 
x**3)**(1/3)*b*c*x**3 + (a + b*x**3)**(1/3)*b*d*x**6),x)