\(\int \frac {(a+b x^3)^{2/3}}{x^6 (a d-b d x^3)} \, dx\) [811]

Optimal result

Integrand size = 28, antiderivative size = 182 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^6 \left (a d-b d x^3\right )} \, dx=-\frac {\left (a+b x^3\right )^{2/3}}{5 a d x^5}-\frac {7 b \left (a+b x^3\right )^{2/3}}{10 a^2 d x^2}+\frac {2^{2/3} b^{5/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{2} \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} a^2 d}+\frac {b^{5/3} \log \left (a d-b d x^3\right )}{3 \sqrt [3]{2} a^2 d}-\frac {b^{5/3} \log \left (\sqrt [3]{2} \sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{\sqrt [3]{2} a^2 d} \] Output:

-1/5*(b*x^3+a)^(2/3)/a/d/x^5-7/10*b*(b*x^3+a)^(2/3)/a^2/d/x^2+1/3*2^(2/3)* 
b^(5/3)*arctan(1/3*(1+2*2^(1/3)*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))*3^(1/2 
)/a^2/d+1/6*b^(5/3)*ln(-b*d*x^3+a*d)*2^(2/3)/a^2/d-1/2*b^(5/3)*ln(2^(1/3)* 
b^(1/3)*x-(b*x^3+a)^(1/3))*2^(2/3)/a^2/d
 

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.19 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^6 \left (a d-b d x^3\right )} \, dx=-\frac {\left (a+b x^3\right )^{2/3} \left (2 a+7 b x^3\right )}{10 a^2 d x^5}+\frac {2^{2/3} b^{5/3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2^{2/3} \sqrt [3]{a+b x^3}}\right )}{\sqrt {3} a^2 d}-\frac {2^{2/3} b^{5/3} \log \left (-2 \sqrt [3]{b} x+2^{2/3} \sqrt [3]{a+b x^3}\right )}{3 a^2 d}+\frac {b^{5/3} \log \left (2 b^{2/3} x^2+2^{2/3} \sqrt [3]{b} x \sqrt [3]{a+b x^3}+\sqrt [3]{2} \left (a+b x^3\right )^{2/3}\right )}{3 \sqrt [3]{2} a^2 d} \] Input:

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

Output:

-1/10*((a + b*x^3)^(2/3)*(2*a + 7*b*x^3))/(a^2*d*x^5) + (2^(2/3)*b^(5/3)*A 
rcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2^(2/3)*(a + b*x^3)^(1/3))])/(Sqrt[ 
3]*a^2*d) - (2^(2/3)*b^(5/3)*Log[-2*b^(1/3)*x + 2^(2/3)*(a + b*x^3)^(1/3)] 
)/(3*a^2*d) + (b^(5/3)*Log[2*b^(2/3)*x^2 + 2^(2/3)*b^(1/3)*x*(a + b*x^3)^( 
1/3) + 2^(1/3)*(a + b*x^3)^(2/3)])/(3*2^(1/3)*a^2*d)
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {975, 27, 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.

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

\(\Big \downarrow \) 975

\(\displaystyle \frac {\int \frac {b \left (3 b x^3+7 a\right )}{x^3 \left (a-b x^3\right ) \sqrt [3]{b x^3+a}}dx}{5 a d}-\frac {\left (a+b x^3\right )^{2/3}}{5 a d x^5}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b \int \frac {3 b x^3+7 a}{x^3 \left (a-b x^3\right ) \sqrt [3]{b x^3+a}}dx}{5 a d}-\frac {\left (a+b x^3\right )^{2/3}}{5 a d x^5}\)

\(\Big \downarrow \) 1053

\(\displaystyle \frac {b \left (-\frac {\int -\frac {20 a^2 b}{\left (a-b x^3\right ) \sqrt [3]{b x^3+a}}dx}{2 a^2}-\frac {7 \left (a+b x^3\right )^{2/3}}{2 a x^2}\right )}{5 a d}-\frac {\left (a+b x^3\right )^{2/3}}{5 a d x^5}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b \left (10 b \int \frac {1}{\left (a-b x^3\right ) \sqrt [3]{b x^3+a}}dx-\frac {7 \left (a+b x^3\right )^{2/3}}{2 a x^2}\right )}{5 a d}-\frac {\left (a+b x^3\right )^{2/3}}{5 a d x^5}\)

\(\Big \downarrow \) 901

\(\displaystyle \frac {b \left (10 b \left (\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{2} \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3} a \sqrt [3]{b}}+\frac {\log \left (a-b x^3\right )}{6 \sqrt [3]{2} a \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{2} \sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{2} a \sqrt [3]{b}}\right )-\frac {7 \left (a+b x^3\right )^{2/3}}{2 a x^2}\right )}{5 a d}-\frac {\left (a+b x^3\right )^{2/3}}{5 a d x^5}\)

Input:

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

Output:

-1/5*(a + b*x^3)^(2/3)/(a*d*x^5) + (b*((-7*(a + b*x^3)^(2/3))/(2*a*x^2) + 
10*b*(ArcTan[(1 + (2*2^(1/3)*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(2^(1/ 
3)*Sqrt[3]*a*b^(1/3)) + Log[a - b*x^3]/(6*2^(1/3)*a*b^(1/3)) - Log[2^(1/3) 
*b^(1/3)*x - (a + b*x^3)^(1/3)]/(2*2^(1/3)*a*b^(1/3)))))/(5*a*d)
 

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 = 1.74 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.82

method	result	size
pseudoelliptic	\(\frac {5 x^{5} 2^{\frac {2}{3}} \left (-2 \arctan \left (\frac {\sqrt {3}\, \left (2^{\frac {2}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+b^{\frac {1}{3}} x \right )}{3 b^{\frac {1}{3}} x}\right ) \sqrt {3}+\ln \left (\frac {b^{\frac {2}{3}} 2^{\frac {2}{3}} x^{2}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{\frac {1}{3}} 2^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )-2 \ln \left (\frac {-2^{\frac {1}{3}} b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )\right ) b^{\frac {5}{3}}-3 \left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (7 b \,x^{3}+2 a \right )}{30 x^{5} d \,a^{2}}\)	\(150\)

Input:

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

Output:

1/30*(5*x^5*2^(2/3)*(-2*arctan(1/3*3^(1/2)*(2^(2/3)*(b*x^3+a)^(1/3)+b^(1/3 
)*x)/b^(1/3)/x)*3^(1/2)+ln((b^(2/3)*2^(2/3)*x^2+(b*x^3+a)^(1/3)*b^(1/3)*2^ 
(1/3)*x+(b*x^3+a)^(2/3))/x^2)-2*ln((-2^(1/3)*b^(1/3)*x+(b*x^3+a)^(1/3))/x) 
)*b^(5/3)-3*(b*x^3+a)^(2/3)*(7*b*x^3+2*a))/x^5/d/a^2
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

-Integral((a + b*x**3)**(2/3)/(-a*x**6 + b*x**9), x)/d
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^6 \left (a d-b d x^3\right )} \, dx=\frac {-2 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a +3 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b \,x^{3}+20 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{-b^{2} x^{9}+a^{2} x^{3}}d x \right ) a^{2} b \,x^{5}}{10 a^{2} d \,x^{5}} \] Input:

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

Output:

( - 2*(a + b*x**3)**(2/3)*a + 3*(a + b*x**3)**(2/3)*b*x**3 + 20*int((a + b 
*x**3)**(2/3)/(a**2*x**3 - b**2*x**9),x)*a**2*b*x**5)/(10*a**2*d*x**5)