\(\int \frac {1}{x^6 \sqrt [3]{a+b x^3} (c+d x^3)} \, dx\) [740]

Optimal result

Integrand size = 24, antiderivative size = 214 \[ \int \frac {1}{x^6 \sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=-\frac {\left (a+b x^3\right )^{2/3}}{5 a c x^5}+\frac {(3 b c+5 a d) \left (a+b x^3\right )^{2/3}}{10 a^2 c^2 x^2}+\frac {d^2 \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} c^{8/3} \sqrt [3]{b c-a d}}+\frac {d^2 \log \left (c+d x^3\right )}{6 c^{8/3} \sqrt [3]{b c-a d}}-\frac {d^2 \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{8/3} \sqrt [3]{b c-a d}} \] Output:

-1/5*(b*x^3+a)^(2/3)/a/c/x^5+1/10*(5*a*d+3*b*c)*(b*x^3+a)^(2/3)/a^2/c^2/x^ 
2+1/3*d^2*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)/c^(8/3)/(-a*d+b*c)^(1/3)+1/6*d^2*ln(d*x^3+c)/c^(8/3)/(-a*d+b* 
c)^(1/3)-1/2*d^2*ln((-a*d+b*c)^(1/3)*x/c^(1/3)-(b*x^3+a)^(1/3))/c^(8/3)/(- 
a*d+b*c)^(1/3)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.94 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.59 \[ \int \frac {1}{x^6 \sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\frac {\frac {6 c^{2/3} \left (a+b x^3\right )^{2/3} \left (-2 a c+3 b c x^3+5 a d x^3\right )}{a^2 x^5}-\frac {10 \sqrt {-6+6 i \sqrt {3}} d^2 \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{\sqrt [3]{b c-a d}}+\frac {10 \left (1+i \sqrt {3}\right ) d^2 \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{\sqrt [3]{b c-a d}}-\frac {5 i \left (-i+\sqrt {3}\right ) d^2 \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{\sqrt [3]{b c-a d}}}{60 c^{8/3}} \] Input:

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

Output:

((6*c^(2/3)*(a + b*x^3)^(2/3)*(-2*a*c + 3*b*c*x^3 + 5*a*d*x^3))/(a^2*x^5) 
- (10*Sqrt[-6 + (6*I)*Sqrt[3]]*d^2*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) + (10*(1 + I*Sqrt[3])*d^2*Log[2*(b*c - a*d)^(1/3)*x + (1 + I*S 
qrt[3])*c^(1/3)*(a + b*x^3)^(1/3)])/(b*c - a*d)^(1/3) - ((5*I)*(-I + Sqrt[ 
3])*d^2*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))/(60*c^(8/3))
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {980, 25, 1053, 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.

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

\(\Big \downarrow \) 980

\(\displaystyle \frac {\int -\frac {3 b d x^3+3 b c+5 a d}{x^3 \sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{5 a c}-\frac {\left (a+b x^3\right )^{2/3}}{5 a c x^5}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int \frac {3 b d x^3+3 b c+5 a d}{x^3 \sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{5 a c}-\frac {\left (a+b x^3\right )^{2/3}}{5 a c x^5}\)

\(\Big \downarrow \) 1053

\(\displaystyle -\frac {-\frac {\int \frac {10 a^2 d^2}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{2 a c}-\frac {\left (a+b x^3\right )^{2/3} (5 a d+3 b c)}{2 a c x^2}}{5 a c}-\frac {\left (a+b x^3\right )^{2/3}}{5 a c x^5}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {5 a d^2 \int \frac {1}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{c}-\frac {\left (a+b x^3\right )^{2/3} (5 a d+3 b c)}{2 a c x^2}}{5 a c}-\frac {\left (a+b x^3\right )^{2/3}}{5 a c x^5}\)

\(\Big \downarrow \) 901

\(\displaystyle -\frac {-\frac {5 a d^2 \left (\frac {\arctan \left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} c^{2/3} \sqrt [3]{b c-a d}}+\frac {\log \left (c+d x^3\right )}{6 c^{2/3} \sqrt [3]{b c-a d}}-\frac {\log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{2/3} \sqrt [3]{b c-a d}}\right )}{c}-\frac {\left (a+b x^3\right )^{2/3} (5 a d+3 b c)}{2 a c x^2}}{5 a c}-\frac {\left (a+b x^3\right )^{2/3}}{5 a c x^5}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 980

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

rule 1053

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

Maple [A] (verified)

Time = 2.00 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.08

method	result	size
pseudoelliptic	\(\frac {-\frac {d^{2} \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) a^{2} x^{5}}{2}+d^{2} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (-\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}+x \right )}{3 x}\right ) a^{2} x^{5}+d^{2} \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) a^{2} x^{5}-\frac {3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (\left (-\frac {5 d \,x^{3}}{2}+c \right ) a -\frac {3 x^{3} b c}{2}\right ) c \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{5}}{3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} a^{2} c^{3} x^{5}}\)	\(231\)

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{x^6 \sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} c \,x^{6}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} d \,x^{9}}d x \] Input:

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

Output:

int(1/((a + b*x**3)**(1/3)*c*x**6 + (a + b*x**3)**(1/3)*d*x**9),x)