\(\int \frac {x^6}{\sqrt [3]{a+b x^3} (c+d x^3)} \, dx\) [736]

Optimal result

Integrand size = 24, antiderivative size = 273 \[ \int \frac {x^6}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\frac {x \left (a+b x^3\right )^{2/3}}{3 b d}-\frac {(3 b c+a d) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{4/3} d^2}+\frac {c^{4/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} d^2 \sqrt [3]{b c-a d}}+\frac {c^{4/3} \log \left (c+d x^3\right )}{6 d^2 \sqrt [3]{b c-a d}}-\frac {c^{4/3} \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 d^2 \sqrt [3]{b c-a d}}+\frac {(3 b c+a d) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{6 b^{4/3} d^2} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 3.87 (sec) , antiderivative size = 466, normalized size of antiderivative = 1.71 \[ \int \frac {x^6}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\frac {\frac {12 d x \left (a+b x^3\right )^{2/3}}{b}-\frac {4 \sqrt {3} (3 b c+a d) \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )}{b^{4/3}}-\frac {6 \sqrt {-6+6 i \sqrt {3}} c^{4/3} \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{\sqrt [3]{b c-a d}}+\frac {4 (3 b c+a d) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{b^{4/3}}+\frac {6 \left (1+i \sqrt {3}\right ) c^{4/3} \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{\sqrt [3]{b c-a d}}-\frac {2 (3 b c+a d) \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{b^{4/3}}-\frac {3 i \left (-i+\sqrt {3}\right ) c^{4/3} \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{\sqrt [3]{b c-a d}}}{36 d^2} \] Input:

Integrate[x^6/((a + b*x^3)^(1/3)*(c + d*x^3)),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {979, 1026, 769, 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.

\(\displaystyle \int \frac {x^6}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx\)

\(\Big \downarrow \) 979

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

\(\Big \downarrow \) 1026

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

\(\Big \downarrow \) 769

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

\(\Big \downarrow \) 901

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

Input:

Int[x^6/((a + b*x^3)^(1/3)*(c + d*x^3)),x]
 

Output:

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

Defintions of rubi rules used

rule 769

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]* 
(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^ 
3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, 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 979

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[e^(2*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 
 1)*((c + d*x^n)^(q + 1)/(b*d*(m + n*(p + q) + 1))), x] - Simp[e^(2*n)/(b*d 
*(m + n*(p + q) + 1))   Int[(e*x)^(m - 2*n)*(a + b*x^n)^p*(c + d*x^n)^q*Sim 
p[a*c*(m - 2*n + 1) + (a*d*(m + n*(q - 1) + 1) + b*c*(m + n*(p - 1) + 1))*x 
^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && I 
GtQ[n, 0] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x 
]
 

rule 1026

Int[(((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)* 
(x_)^(n_)), x_Symbol] :> Simp[f/d   Int[(a + b*x^n)^p, x], x] + Simp[(d*e - 
 c*f)/d   Int[(a + b*x^n)^p/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, f, 
 p, n}, x]
 

Maple [A] (verified)

Time = 2.11 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.15

method	result	size
pseudoelliptic	\(-\frac {\left (-6 d x \left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{\frac {1}{3}}+\left (a d +3 b c \right ) \left (-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right )+\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )-2 \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )\right )\right ) \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}-6 c \,b^{\frac {4}{3}} \left (\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x -2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x}\right )+\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )-\frac {\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{2}\right )}{18 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} b^{\frac {4}{3}} d^{2}}\)	\(313\)

Input:

int(x^6/(b*x^3+a)^(1/3)/(d*x^3+c),x,method=_RETURNVERBOSE)
 

Output:

-1/18/((a*d-b*c)/c)^(1/3)*((-6*d*x*(b*x^3+a)^(2/3)*b^(1/3)+(a*d+3*b*c)*(-2 
*3^(1/2)*arctan(1/3*3^(1/2)*(b^(1/3)*x+2*(b*x^3+a)^(1/3))/b^(1/3)/x)+ln((b 
^(2/3)*x^2+b^(1/3)*(b*x^3+a)^(1/3)*x+(b*x^3+a)^(2/3))/x^2)-2*ln((-b^(1/3)* 
x+(b*x^3+a)^(1/3))/x)))*((a*d-b*c)/c)^(1/3)-6*c*b^(4/3)*(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) 
+ln((((a*d-b*c)/c)^(1/3)*x+(b*x^3+a)^(1/3))/x)-1/2*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^(4/3) 
/d^2
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 826, normalized size of antiderivative = 3.03 \[ \int \frac {x^6}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\text {Too large to display} \] Input:

integrate(x^6/(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="fricas")
 

Output:

[-1/18*(6*sqrt(3)*b^2*c*(-c/(b*c - a*d))^(1/3)*arctan(-1/3*(sqrt(3)*x - 2* 
sqrt(3)*(b*x^3 + a)^(1/3)*(-c/(b*c - a*d))^(1/3))/x) - 6*b^2*c*(-c/(b*c - 
a*d))^(1/3)*log(-((b*c - a*d)*x*(-c/(b*c - a*d))^(2/3) - (b*x^3 + a)^(1/3) 
*c)/x) + 3*b^2*c*(-c/(b*c - a*d))^(1/3)*log(-((b*c - a*d)*x^2*(-c/(b*c - a 
*d))^(1/3) - (b*x^3 + a)^(1/3)*(b*c - a*d)*x*(-c/(b*c - a*d))^(2/3) - (b*x 
^3 + a)^(2/3)*c)/x^2) - 6*(b*x^3 + a)^(2/3)*b*d*x - 3*sqrt(1/3)*(3*b^2*c + 
 a*b*d)*sqrt(-1/b^(2/3))*log(3*b*x^3 - 3*(b*x^3 + a)^(1/3)*b^(2/3)*x^2 - 3 
*sqrt(1/3)*(b^(4/3)*x^3 + (b*x^3 + a)^(1/3)*b*x^2 - 2*(b*x^3 + a)^(2/3)*b^ 
(2/3)*x)*sqrt(-1/b^(2/3)) + 2*a) - 2*(3*b*c + a*d)*b^(2/3)*log(-(b^(1/3)*x 
 - (b*x^3 + a)^(1/3))/x) + (3*b*c + a*d)*b^(2/3)*log((b^(2/3)*x^2 + (b*x^3 
 + a)^(1/3)*b^(1/3)*x + (b*x^3 + a)^(2/3))/x^2))/(b^2*d^2), -1/18*(6*sqrt( 
3)*b^2*c*(-c/(b*c - a*d))^(1/3)*arctan(-1/3*(sqrt(3)*x - 2*sqrt(3)*(b*x^3 
+ a)^(1/3)*(-c/(b*c - a*d))^(1/3))/x) - 6*b^2*c*(-c/(b*c - a*d))^(1/3)*log 
(-((b*c - a*d)*x*(-c/(b*c - a*d))^(2/3) - (b*x^3 + a)^(1/3)*c)/x) + 3*b^2* 
c*(-c/(b*c - a*d))^(1/3)*log(-((b*c - a*d)*x^2*(-c/(b*c - a*d))^(1/3) - (b 
*x^3 + a)^(1/3)*(b*c - a*d)*x*(-c/(b*c - a*d))^(2/3) - (b*x^3 + a)^(2/3)*c 
)/x^2) - 6*(b*x^3 + a)^(2/3)*b*d*x - 2*(3*b*c + a*d)*b^(2/3)*log(-(b^(1/3) 
*x - (b*x^3 + a)^(1/3))/x) + (3*b*c + a*d)*b^(2/3)*log((b^(2/3)*x^2 + (b*x 
^3 + a)^(1/3)*b^(1/3)*x + (b*x^3 + a)^(2/3))/x^2) - 6*sqrt(1/3)*(3*b^2*c + 
 a*b*d)*arctan(sqrt(1/3)*(b^(1/3)*x + 2*(b*x^3 + a)^(1/3))/(b^(1/3)*x))...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(x^6/(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(x^6/(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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