\(\int \frac {x^4 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx\) [681]

Optimal result

Integrand size = 24, antiderivative size = 276 \[ \int \frac {x^4 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\frac {x^2 \sqrt [3]{a+b x^3}}{3 d}+\frac {(3 b c-a d) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{2/3} d^2}-\frac {c^{2/3} \sqrt [3]{b c-a d} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} d^2}+\frac {c^{2/3} \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^2}+\frac {(3 b c-a d) \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{6 b^{2/3} d^2}-\frac {c^{2/3} \sqrt [3]{b c-a d} \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 d^2} \] Output:

1/3*x^2*(b*x^3+a)^(1/3)/d+1/9*(-a*d+3*b*c)*arctan(1/3*(1+2*b^(1/3)*x/(b*x^ 
3+a)^(1/3))*3^(1/2))*3^(1/2)/b^(2/3)/d^2-1/3*c^(2/3)*(-a*d+b*c)^(1/3)*arct 
an(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)/d 
^2+1/6*c^(2/3)*(-a*d+b*c)^(1/3)*ln(d*x^3+c)/d^2+1/6*(-a*d+3*b*c)*ln(b^(1/3 
)*x-(b*x^3+a)^(1/3))/b^(2/3)/d^2-1/2*c^(2/3)*(-a*d+b*c)^(1/3)*ln((-a*d+b*c 
)^(1/3)*x/c^(1/3)-(b*x^3+a)^(1/3))/d^2
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.98 (sec) , antiderivative size = 467, normalized size of antiderivative = 1.69 \[ \int \frac {x^4 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\frac {12 d x^2 \sqrt [3]{a+b x^3}+\frac {4 \sqrt {3} (3 b c-a d) \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )}{b^{2/3}}+6 \sqrt {-6-6 i \sqrt {3}} c^{2/3} \sqrt [3]{b c-a d} \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )+\frac {4 (3 b c-a d) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{b^{2/3}}+6 \left (1-i \sqrt {3}\right ) c^{2/3} \sqrt [3]{b c-a d} \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )+\frac {2 (-3 b c+a d) \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{b^{2/3}}+3 i \left (i+\sqrt {3}\right ) c^{2/3} \sqrt [3]{b c-a d} \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{36 d^2} \] Input:

Integrate[(x^4*(a + b*x^3)^(1/3))/(c + d*x^3),x]
 

Output:

(12*d*x^2*(a + b*x^3)^(1/3) + (4*Sqrt[3]*(3*b*c - a*d)*ArcTan[(Sqrt[3]*b^( 
1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3))])/b^(2/3) + 6*Sqrt[-6 - (6*I)*Sq 
rt[3]]*c^(2/3)*(b*c - a*d)^(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))] + (4*(3*b 
*c - a*d)*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/b^(2/3) + 6*(1 - I*Sqrt[3 
])*c^(2/3)*(b*c - a*d)^(1/3)*Log[2*(b*c - a*d)^(1/3)*x + (1 + I*Sqrt[3])*c 
^(1/3)*(a + b*x^3)^(1/3)] + (2*(-3*b*c + a*d)*Log[b^(2/3)*x^2 + b^(1/3)*x* 
(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)])/b^(2/3) + (3*I)*(I + Sqrt[3])*c^(2 
/3)*(b*c - a*d)^(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)])/(36*d^2)
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.02, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {978, 1054, 2009}

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.

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

\(\Big \downarrow \) 978

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

\(\Big \downarrow \) 1054

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

\(\Big \downarrow \) 2009

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

Input:

Int[(x^4*(a + b*x^3)^(1/3))/(c + d*x^3),x]
 

Output:

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

Defintions of rubi rules used

rule 978

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/(b*(m + n*(p + q) + 1))), x] - Simp[e^n/(b*(m + n*(p + q) + 
1))   Int[(e*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[a*c*(m - n + 
 1) + (a*d*(m - n + 1) - n*q*(b*c - a*d))*x^n, x], x], x] /; FreeQ[{a, b, c 
, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && GtQ[q, 0] && GtQ[m - n 
 + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
 

rule 1054

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n 
_)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a 
+ b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, 
 m, p}, x] && IGtQ[n, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 2.02 (sec) , antiderivative size = 379, normalized size of antiderivative = 1.37

method	result	size
pseudoelliptic	\(\frac {\frac {\left (-a d \,b^{\frac {2}{3}}+b^{\frac {5}{3}} c \right ) \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{2}-\sqrt {3}\, \left (a d \,b^{\frac {2}{3}}-b^{\frac {5}{3}} c \right ) \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x -2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x}\right )+\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} \left (a d -3 b c \right ) \ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{6}+\left (a d \,b^{\frac {2}{3}}-b^{\frac {5}{3}} c \right ) \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )-\frac {\left (-\sqrt {3}\, \left (a d -3 b c \right ) \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right )+\left (a d -3 b c \right ) \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )-3 b^{\frac {2}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} d \,x^{2}\right ) \left (\frac {a d -b c}{c}\right )^{\frac {2}{3}}}{3}}{3 \left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} b^{\frac {2}{3}} d^{2}}\)	\(379\)

Input:

int(x^4*(b*x^3+a)^(1/3)/(d*x^3+c),x,method=_RETURNVERBOSE)
 

Output:

1/3/((a*d-b*c)/c)^(2/3)*(1/2*(-a*d*b^(2/3)+b^(5/3)*c)*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)-3^(1/2 
)*(a*d*b^(2/3)-b^(5/3)*c)*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)+1/6*((a*d-b*c)/c)^(2/3)*(a*d-3*b*c)*ln 
((b^(2/3)*x^2+b^(1/3)*(b*x^3+a)^(1/3)*x+(b*x^3+a)^(2/3))/x^2)+(a*d*b^(2/3) 
-b^(5/3)*c)*ln((((a*d-b*c)/c)^(1/3)*x+(b*x^3+a)^(1/3))/x)-1/3*(-3^(1/2)*(a 
*d-3*b*c)*arctan(1/3*3^(1/2)*(b^(1/3)*x+2*(b*x^3+a)^(1/3))/b^(1/3)/x)+(a*d 
-3*b*c)*ln((-b^(1/3)*x+(b*x^3+a)^(1/3))/x)-3*b^(2/3)*(b*x^3+a)^(1/3)*d*x^2 
)*((a*d-b*c)/c)^(2/3))/b^(2/3)/d^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 449 vs. \(2 (222) = 444\).

Time = 0.25 (sec) , antiderivative size = 449, normalized size of antiderivative = 1.63 \[ \int \frac {x^4 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\frac {6 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{2} d x^{2} + 6 \, \sqrt {3} {\left (-b c^{3} + a c^{2} d\right )}^{\frac {1}{3}} b^{2} \arctan \left (-\frac {\sqrt {3} {\left (b c^{2} - a c d\right )} x + 2 \, \sqrt {3} {\left (-b c^{3} + a c^{2} d\right )}^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{3 \, {\left (b c^{2} - a c d\right )} x}\right ) + 6 \, {\left (-b c^{3} + a c^{2} d\right )}^{\frac {1}{3}} b^{2} \log \left (\frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} c + {\left (-b c^{3} + a c^{2} d\right )}^{\frac {1}{3}} x}{x}\right ) - 3 \, {\left (-b c^{3} + a c^{2} d\right )}^{\frac {1}{3}} b^{2} \log \left (\frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}} c^{2} - {\left (-b c^{3} + a c^{2} d\right )}^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} c x + {\left (-b c^{3} + a c^{2} d\right )}^{\frac {2}{3}} x^{2}}{x^{2}}\right ) - 6 \, \sqrt {\frac {1}{3}} {\left (3 \, b^{2} c - a b d\right )} \sqrt {-\left (-b^{2}\right )^{\frac {1}{3}}} \arctan \left (-\frac {\sqrt {\frac {1}{3}} {\left (\left (-b^{2}\right )^{\frac {1}{3}} b x - 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {2}{3}}\right )} \sqrt {-\left (-b^{2}\right )^{\frac {1}{3}}}}{b^{2} x}\right ) + 2 \, \left (-b^{2}\right )^{\frac {2}{3}} {\left (3 \, b c - a d\right )} \log \left (-\frac {\left (-b^{2}\right )^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x}\right ) - \left (-b^{2}\right )^{\frac {2}{3}} {\left (3 \, b c - a d\right )} \log \left (-\frac {\left (-b^{2}\right )^{\frac {1}{3}} b x^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {2}{3}} b}{x^{2}}\right )}{18 \, b^{2} d^{2}} \] Input:

integrate(x^4*(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="fricas")
 

Output:

1/18*(6*(b*x^3 + a)^(1/3)*b^2*d*x^2 + 6*sqrt(3)*(-b*c^3 + a*c^2*d)^(1/3)*b 
^2*arctan(-1/3*(sqrt(3)*(b*c^2 - a*c*d)*x + 2*sqrt(3)*(-b*c^3 + a*c^2*d)^( 
2/3)*(b*x^3 + a)^(1/3))/((b*c^2 - a*c*d)*x)) + 6*(-b*c^3 + a*c^2*d)^(1/3)* 
b^2*log(((b*x^3 + a)^(1/3)*c + (-b*c^3 + a*c^2*d)^(1/3)*x)/x) - 3*(-b*c^3 
+ a*c^2*d)^(1/3)*b^2*log(((b*x^3 + a)^(2/3)*c^2 - (-b*c^3 + a*c^2*d)^(1/3) 
*(b*x^3 + a)^(1/3)*c*x + (-b*c^3 + a*c^2*d)^(2/3)*x^2)/x^2) - 6*sqrt(1/3)* 
(3*b^2*c - a*b*d)*sqrt(-(-b^2)^(1/3))*arctan(-sqrt(1/3)*((-b^2)^(1/3)*b*x 
- 2*(b*x^3 + a)^(1/3)*(-b^2)^(2/3))*sqrt(-(-b^2)^(1/3))/(b^2*x)) + 2*(-b^2 
)^(2/3)*(3*b*c - a*d)*log(-((-b^2)^(2/3)*x - (b*x^3 + a)^(1/3)*b)/x) - (-b 
^2)^(2/3)*(3*b*c - a*d)*log(-((-b^2)^(1/3)*b*x^2 - (b*x^3 + a)^(1/3)*(-b^2 
)^(2/3)*x - (b*x^3 + a)^(2/3)*b)/x^2))/(b^2*d^2)
 

Sympy [F]

\[ \int \frac {x^4 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\int \frac {x^{4} \sqrt [3]{a + b x^{3}}}{c + d x^{3}}\, dx \] Input:

integrate(x**4*(b*x**3+a)**(1/3)/(d*x**3+c),x)
 

Output:

Integral(x**4*(a + b*x**3)**(1/3)/(c + d*x**3), x)
 

Maxima [F]

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

integrate(x^4*(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a)^(1/3)*x^4/(d*x^3 + c), x)
 

Giac [F]

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

integrate(x^4*(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="giac")
 

Output:

integrate((b*x^3 + a)^(1/3)*x^4/(d*x^3 + c), x)
 

Mupad [F(-1)]

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

int((x^4*(a + b*x^3)^(1/3))/(c + d*x^3),x)
 

Output:

int((x^4*(a + b*x^3)^(1/3))/(c + d*x^3), x)
 

Reduce [F]

\[ \int \frac {x^4 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{2}+\left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{4}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a d -3 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{4}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) b c -2 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a c}{3 d} \] Input:

int(x^4*(b*x^3+a)^(1/3)/(d*x^3+c),x)
                                                                                    
                                                                                    
 

Output:

((a + b*x**3)**(1/3)*x**2 + int(((a + b*x**3)**(1/3)*x**4)/(a*c + a*d*x**3 
 + b*c*x**3 + b*d*x**6),x)*a*d - 3*int(((a + b*x**3)**(1/3)*x**4)/(a*c + a 
*d*x**3 + b*c*x**3 + b*d*x**6),x)*b*c - 2*int(((a + b*x**3)**(1/3)*x)/(a*c 
 + a*d*x**3 + b*c*x**3 + b*d*x**6),x)*a*c)/(3*d)