\(\int \frac {1}{x^5 (a+b x^3)^{2/3} (c+d x^3)} \, dx\) [757]

Optimal result

Integrand size = 24, antiderivative size = 215 \[ \int \frac {1}{x^5 \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=-\frac {\sqrt [3]{a+b x^3}}{4 a c x^4}+\frac {(3 b c+4 a d) \sqrt [3]{a+b x^3}}{4 a^2 c^2 x}-\frac {d^2 \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} c^{7/3} (b c-a d)^{2/3}}+\frac {d^2 \log \left (c+d x^3\right )}{6 c^{7/3} (b c-a d)^{2/3}}-\frac {d^2 \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{7/3} (b c-a d)^{2/3}} \] Output:

-1/4*(b*x^3+a)^(1/3)/a/c/x^4+1/4*(4*a*d+3*b*c)*(b*x^3+a)^(1/3)/a^2/c^2/x-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^(7/3)/(-a*d+b*c)^(2/3)+1/6*d^2*ln(d*x^3+c)/c^(7/3)/(-a*d+b*c)^ 
(2/3)-1/2*d^2*ln((-a*d+b*c)^(1/3)*x/c^(1/3)-(b*x^3+a)^(1/3))/c^(7/3)/(-a*d 
+b*c)^(2/3)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.31 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.58 \[ \int \frac {1}{x^5 \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\frac {\frac {3 \sqrt [3]{c} \sqrt [3]{a+b x^3} \left (-a c+3 b c x^3+4 a d x^3\right )}{a^2 x^4}+\frac {2 \sqrt {-6-6 i \sqrt {3}} d^2 \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{(b c-a d)^{2/3}}+\frac {2 \left (1-i \sqrt {3}\right ) d^2 \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{(b c-a d)^{2/3}}+\frac {i \left (i+\sqrt {3}\right ) d^2 \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{(b c-a d)^{2/3}}}{12 c^{7/3}} \] Input:

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

Output:

((3*c^(1/3)*(a + b*x^3)^(1/3)*(-(a*c) + 3*b*c*x^3 + 4*a*d*x^3))/(a^2*x^4) 
+ (2*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)^(2/3) + (2*(1 - I*Sqrt[3])*d^2*Log[2*(b*c - a*d)^(1/3)*x + (1 + I*Sqr 
t[3])*c^(1/3)*(a + b*x^3)^(1/3)])/(b*c - a*d)^(2/3) + (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)^(2/3))/(12*c^(7/3))
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.05, 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, 992}

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.

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

\(\Big \downarrow \) 980

\(\displaystyle \frac {\int -\frac {3 b d x^3+3 b c+4 a d}{x^2 \left (b x^3+a\right )^{2/3} \left (d x^3+c\right )}dx}{4 a c}-\frac {\sqrt [3]{a+b x^3}}{4 a c x^4}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int \frac {3 b d x^3+3 b c+4 a d}{x^2 \left (b x^3+a\right )^{2/3} \left (d x^3+c\right )}dx}{4 a c}-\frac {\sqrt [3]{a+b x^3}}{4 a c x^4}\)

\(\Big \downarrow \) 1053

\(\displaystyle -\frac {-\frac {\int \frac {4 a^2 d^2 x}{\left (b x^3+a\right )^{2/3} \left (d x^3+c\right )}dx}{a c}-\frac {\sqrt [3]{a+b x^3} (4 a d+3 b c)}{a c x}}{4 a c}-\frac {\sqrt [3]{a+b x^3}}{4 a c x^4}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {4 a d^2 \int \frac {x}{\left (b x^3+a\right )^{2/3} \left (d x^3+c\right )}dx}{c}-\frac {\sqrt [3]{a+b x^3} (4 a d+3 b c)}{a c x}}{4 a c}-\frac {\sqrt [3]{a+b x^3}}{4 a c x^4}\)

\(\Big \downarrow \) 992

\(\displaystyle -\frac {-\frac {4 a d^2 \left (-\frac {\arctan \left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{c} (b c-a d)^{2/3}}+\frac {\log \left (c+d x^3\right )}{6 \sqrt [3]{c} (b c-a d)^{2/3}}-\frac {\log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{c} (b c-a d)^{2/3}}\right )}{c}-\frac {\sqrt [3]{a+b x^3} (4 a d+3 b c)}{a c x}}{4 a c}-\frac {\sqrt [3]{a+b x^3}}{4 a c x^4}\)

Input:

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

Output:

-1/4*(a + b*x^3)^(1/3)/(a*c*x^4) - (-(((3*b*c + 4*a*d)*(a + b*x^3)^(1/3))/ 
(a*c*x)) - (4*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^(1/3)*(b*c - a*d)^(2/3))) + Log[c + d*x^ 
3]/(6*c^(1/3)*(b*c - a*d)^(2/3)) - Log[((b*c - a*d)^(1/3)*x)/c^(1/3) - (a 
+ b*x^3)^(1/3)]/(2*c^(1/3)*(b*c - a*d)^(2/3))))/c)/(4*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 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 992

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

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 = 1.80 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.08

method	result	size
pseudoelliptic	\(-\frac {-\frac {\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) a^{2} d^{2} x^{4}}{2}-d^{2} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (-\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}+x \right )}{3 x}\right ) a^{2} x^{4}+\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) a^{2} d^{2} x^{4}+\frac {3 \left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} c \left (\left (-4 d \,x^{3}+c \right ) a -3 x^{3} b c \right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{4}}{3 \left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} a^{2} c^{3} x^{4}}\)	\(232\)

Input:

int(1/x^5/(b*x^3+a)^(2/3)/(d*x^3+c),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{x^5 \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\text {Timed out} \] Input:

integrate(1/x^5/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {1}{x^5 \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\int \frac {1}{x^{5} \left (a + b x^{3}\right )^{\frac {2}{3}} \left (c + d x^{3}\right )}\, dx \] Input:

integrate(1/x**5/(b*x**3+a)**(2/3)/(d*x**3+c),x)
 

Output:

Integral(1/(x**5*(a + b*x**3)**(2/3)*(c + d*x**3)), x)
 

Maxima [F]

\[ \int \frac {1}{x^5 \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (d x^{3} + c\right )} x^{5}} \,d x } \] Input:

integrate(1/x^5/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {1}{x^5 \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (d x^{3} + c\right )} x^{5}} \,d x } \] Input:

integrate(1/x^5/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^5 \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\int \frac {1}{x^5\,{\left (b\,x^3+a\right )}^{2/3}\,\left (d\,x^3+c\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{x^5 \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} c \,x^{5}+\left (b \,x^{3}+a \right )^{\frac {2}{3}} d \,x^{8}}d x \] Input:

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

Output:

int(1/((a + b*x**3)**(2/3)*c*x**5 + (a + b*x**3)**(2/3)*d*x**8),x)