\(\int \frac {(a+b x^3)^{2/3}}{x^6 (c+d x^3)} \, dx\) [703]

Optimal result

Integrand size = 24, antiderivative size = 206 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^6 \left (c+d x^3\right )} \, dx=-\frac {\left (a+b x^3\right )^{2/3}}{5 c x^5}-\frac {(2 b c-5 a d) \left (a+b x^3\right )^{2/3}}{10 a c^2 x^2}-\frac {d (b c-a d)^{2/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} c^{8/3}}-\frac {d (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 c^{8/3}}+\frac {d (b c-a d)^{2/3} \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{8/3}} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 2.02 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.62 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^6 \left (c+d x^3\right )} \, dx=\frac {\frac {6 c^{2/3} \left (a+b x^3\right )^{2/3} \left (-2 a c-2 b c x^3+5 a d x^3\right )}{a x^5}+10 \sqrt {-6+6 i \sqrt {3}} d (b c-a d)^{2/3} \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )-10 i \left (-i+\sqrt {3}\right ) d (b c-a d)^{2/3} \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )+5 \left (1+i \sqrt {3}\right ) d (b c-a d)^{2/3} \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{60 c^{8/3}} \] Input:

Integrate[(a + b*x^3)^(2/3)/(x^6*(c + d*x^3)),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.09, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {975, 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.

\(\displaystyle \int \frac {\left (a+b x^3\right )^{2/3}}{x^6 \left (c+d x^3\right )} \, dx\)

\(\Big \downarrow \) 975

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

\(\Big \downarrow \) 1053

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 901

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

Input:

Int[(a + b*x^3)^(2/3)/(x^6*(c + d*x^3)),x]
 

Output:

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

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 975

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/ 
(a*e*(m + 1))), x] - Simp[1/(a*e^n*(m + 1))   Int[(e*x)^(m + n)*(a + b*x^n) 
^p*(c + d*x^n)^(q - 1)*Simp[c*b*(m + 1) + n*(b*c*(p + 1) + a*d*q) + d*(b*(m 
 + 1) + b*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && 
 NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[0, q, 1] && LtQ[m, -1] && IntBinomi 
alQ[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 = 1.85 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.12

method	result	size
pseudoelliptic	\(\frac {\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) a \left (a d -b c \right ) d \,x^{5}-\frac {3 c \left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (\left (-\frac {5 a d}{2}+b c \right ) x^{3}+a c \right ) \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}{5}+a \left (a d -b c \right ) \left (\arctan \left (\frac {\sqrt {3}\, \left (-\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}+x \right )}{3 x}\right ) \sqrt {3}-\frac {\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{2}\right ) d \,x^{5}}{3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} c^{3} x^{5} a}\)	\(230\)

Input:

int((b*x^3+a)^(2/3)/x^6/(d*x^3+c),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^6 \left (c+d x^3\right )} \, dx=\text {Timed out} \] Input:

integrate((b*x^3+a)^(2/3)/x^6/(d*x^3+c),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

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

integrate((b*x**3+a)**(2/3)/x**6/(d*x**3+c),x)
 

Output:

Integral((a + b*x**3)**(2/3)/(x**6*(c + d*x**3)), x)
 

Maxima [F]

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

integrate((b*x^3+a)^(2/3)/x^6/(d*x^3+c),x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a)^(2/3)/((d*x^3 + c)*x^6), x)
 

Giac [F]

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

integrate((b*x^3+a)^(2/3)/x^6/(d*x^3+c),x, algorithm="giac")
 

Output:

integrate((b*x^3 + a)^(2/3)/((d*x^3 + c)*x^6), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^6 \left (c+d x^3\right )} \, dx=\int \frac {{\left (b\,x^3+a\right )}^{2/3}}{x^6\,\left (d\,x^3+c\right )} \,d x \] Input:

int((a + b*x^3)^(2/3)/(x^6*(c + d*x^3)),x)
 

Output:

int((a + b*x^3)^(2/3)/(x^6*(c + d*x^3)), x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^6 \left (c+d x^3\right )} \, dx=\frac {-2 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a +3 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b \,x^{3}-10 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{b d \,x^{9}+a d \,x^{6}+b c \,x^{6}+a c \,x^{3}}d x \right ) a^{2} d \,x^{5}+10 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{b d \,x^{9}+a d \,x^{6}+b c \,x^{6}+a c \,x^{3}}d x \right ) a b c \,x^{5}}{10 a c \,x^{5}} \] Input:

int((b*x^3+a)^(2/3)/x^6/(d*x^3+c),x)
 

Output:

( - 2*(a + b*x**3)**(2/3)*a + 3*(a + b*x**3)**(2/3)*b*x**3 - 10*int((a + b 
*x**3)**(2/3)/(a*c*x**3 + a*d*x**6 + b*c*x**6 + b*d*x**9),x)*a**2*d*x**5 + 
 10*int((a + b*x**3)**(2/3)/(a*c*x**3 + a*d*x**6 + b*c*x**6 + b*d*x**9),x) 
*a*b*c*x**5)/(10*a*c*x**5)