\(\int \frac {(a+b x^3)^{2/3}}{x^9 (c+d x^3)} \, dx\) [704]

Optimal result

Integrand size = 24, antiderivative size = 257 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^9 \left (c+d x^3\right )} \, dx=-\frac {\left (a+b x^3\right )^{2/3}}{8 c x^8}-\frac {(b c-4 a d) \left (a+b x^3\right )^{2/3}}{20 a c^2 x^5}+\frac {\left (3 b^2 c^2+8 a b c d-20 a^2 d^2\right ) \left (a+b x^3\right )^{2/3}}{40 a^2 c^3 x^2}+\frac {d^2 (b c-a d)^{2/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} c^{11/3}}+\frac {d^2 (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 c^{11/3}}-\frac {d^2 (b c-a d)^{2/3} \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{11/3}} \] Output:

-1/8*(b*x^3+a)^(2/3)/c/x^8-1/20*(-4*a*d+b*c)*(b*x^3+a)^(2/3)/a/c^2/x^5+1/4 
0*(-20*a^2*d^2+8*a*b*c*d+3*b^2*c^2)*(b*x^3+a)^(2/3)/a^2/c^3/x^2+1/3*d^2*(- 
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^(11/3)+1/6*d^2*(-a*d+b*c)^(2/3)*ln(d*x^3+c)/c^(11/3)-1 
/2*d^2*(-a*d+b*c)^(2/3)*ln((-a*d+b*c)^(1/3)*x/c^(1/3)-(b*x^3+a)^(1/3))/c^( 
11/3)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 2.28 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.46 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^9 \left (c+d x^3\right )} \, dx=\frac {-\frac {3 c^{2/3} \left (a+b x^3\right )^{2/3} \left (-3 b^2 c^2 x^6+2 a b c x^3 \left (c-4 d x^3\right )+a^2 \left (5 c^2-8 c d x^3+20 d^2 x^6\right )\right )}{a^2 x^8}-20 \sqrt {-6+6 i \sqrt {3}} d^2 (b c-a d)^{2/3} \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )+20 \left (1+i \sqrt {3}\right ) d^2 (b c-a d)^{2/3} \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )-10 i \left (-i+\sqrt {3}\right ) d^2 (b c-a d)^{2/3} \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{120 c^{11/3}} \] Input:

Integrate[(a + b*x^3)^(2/3)/(x^9*(c + d*x^3)),x]
 

Output:

((-3*c^(2/3)*(a + b*x^3)^(2/3)*(-3*b^2*c^2*x^6 + 2*a*b*c*x^3*(c - 4*d*x^3) 
 + a^2*(5*c^2 - 8*c*d*x^3 + 20*d^2*x^6)))/(a^2*x^8) - 20*Sqrt[-6 + (6*I)*S 
qrt[3]]*d^2*(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))] + 20*(1 + I* 
Sqrt[3])*d^2*(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)] - (10*I)*(-I + Sqrt[3])*d^2*(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)])/(120*c^(1 
1/3))
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {975, 27, 1053, 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.

\(\displaystyle \int \frac {\left (a+b x^3\right )^{2/3}}{x^9 \left (c+d x^3\right )} \, dx\)

\(\Big \downarrow \) 975

\(\displaystyle \frac {\int \frac {2 \left (-3 b d x^3+b c-4 a d\right )}{x^6 \sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{8 c}-\frac {\left (a+b x^3\right )^{2/3}}{8 c x^8}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {-3 b d x^3+b c-4 a d}{x^6 \sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{4 c}-\frac {\left (a+b x^3\right )^{2/3}}{8 c x^8}\)

\(\Big \downarrow \) 1053

\(\displaystyle \frac {-\frac {\int \frac {3 b d (b c-4 a d) x^3+3 b^2 c^2-20 a^2 d^2+8 a b c d}{x^3 \sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{5 a c}-\frac {\left (a+b x^3\right )^{2/3} (b c-4 a d)}{5 a c x^5}}{4 c}-\frac {\left (a+b x^3\right )^{2/3}}{8 c x^8}\)

\(\Big \downarrow \) 1053

\(\displaystyle \frac {-\frac {-\frac {\int \frac {40 a^2 d^2 (b c-a d)}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{2 a c}-\frac {\left (a+b x^3\right )^{2/3} \left (\frac {3 b^2 c}{a}-\frac {20 a d^2}{c}+8 b d\right )}{2 x^2}}{5 a c}-\frac {\left (a+b x^3\right )^{2/3} (b c-4 a d)}{5 a c x^5}}{4 c}-\frac {\left (a+b x^3\right )^{2/3}}{8 c x^8}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {-\frac {20 a d^2 (b c-a d) \int \frac {1}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{c}-\frac {\left (a+b x^3\right )^{2/3} \left (\frac {3 b^2 c}{a}-\frac {20 a d^2}{c}+8 b d\right )}{2 x^2}}{5 a c}-\frac {\left (a+b x^3\right )^{2/3} (b c-4 a d)}{5 a c x^5}}{4 c}-\frac {\left (a+b x^3\right )^{2/3}}{8 c x^8}\)

\(\Big \downarrow \) 901

\(\displaystyle \frac {-\frac {-\frac {20 a d^2 (b c-a d) \left (\frac {\arctan \left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} c^{2/3} \sqrt [3]{b c-a d}}+\frac {\log \left (c+d x^3\right )}{6 c^{2/3} \sqrt [3]{b c-a d}}-\frac {\log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{2/3} \sqrt [3]{b c-a d}}\right )}{c}-\frac {\left (a+b x^3\right )^{2/3} \left (\frac {3 b^2 c}{a}-\frac {20 a d^2}{c}+8 b d\right )}{2 x^2}}{5 a c}-\frac {\left (a+b x^3\right )^{2/3} (b c-4 a d)}{5 a c x^5}}{4 c}-\frac {\left (a+b x^3\right )^{2/3}}{8 c x^8}\)

Input:

Int[(a + b*x^3)^(2/3)/(x^9*(c + d*x^3)),x]
 

Output:

-1/8*(a + b*x^3)^(2/3)/(c*x^8) + (-1/5*((b*c - 4*a*d)*(a + b*x^3)^(2/3))/( 
a*c*x^5) - (-1/2*(((3*b^2*c)/a + 8*b*d - (20*a*d^2)/c)*(a + b*x^3)^(2/3))/ 
x^2 - (20*a*d^2*(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*a*c))/(4*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 = 1.86 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.07

method	result	size
pseudoelliptic	\(\frac {-3 c \left (\left (4 a^{2} d^{2}-\frac {8}{5} a b c d -\frac {3}{5} b^{2} c^{2}\right ) x^{6}+\frac {2 \left (-4 a^{2} c d +b \,c^{2} a \right ) x^{3}}{5}+a^{2} c^{2}\right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}+4 a^{2} d^{2} x^{8} \left (a d -b c \right ) \left (-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x -2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x}\right )+\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )-2 \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )\right )}{24 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x^{8} c^{4} a^{2}}\)	\(275\)

Input:

int((b*x^3+a)^(2/3)/x^9/(d*x^3+c),x,method=_RETURNVERBOSE)
 

Output:

1/24/((a*d-b*c)/c)^(1/3)*(-3*c*((4*a^2*d^2-8/5*a*b*c*d-3/5*b^2*c^2)*x^6+2/ 
5*(-4*a^2*c*d+a*b*c^2)*x^3+a^2*c^2)*(b*x^3+a)^(2/3)*((a*d-b*c)/c)^(1/3)+4* 
a^2*d^2*x^8*(a*d-b*c)*(-2*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)^(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)-2*ln((((a*d-b*c) 
/c)^(1/3)*x+(b*x^3+a)^(1/3))/x)))/x^8/c^4/a^2
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^9 \left (c+d x^3\right )} \, dx=\text {Timed out} \] Input:

integrate((b*x^3+a)^(2/3)/x^9/(d*x^3+c),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

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

integrate((b*x**3+a)**(2/3)/x**9/(d*x**3+c),x)
 

Output:

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

Maxima [F]

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

integrate((b*x^3+a)^(2/3)/x^9/(d*x^3+c),x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a)^(2/3)/((d*x^3 + c)*x^9), x)
 

Giac [F]

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

integrate((b*x^3+a)^(2/3)/x^9/(d*x^3+c),x, algorithm="giac")
 

Output:

integrate((b*x^3 + a)^(2/3)/((d*x^3 + c)*x^9), x)
 

Mupad [F(-1)]

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

int((a + b*x^3)^(2/3)/(x^9*(c + d*x^3)),x)
 

Output:

int((a + b*x^3)^(2/3)/(x^9*(c + d*x^3)), x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^9 \left (c+d x^3\right )} \, dx=\frac {-5 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a^{2} c +8 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a^{2} d \,x^{3}-2 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a b c \,x^{3}-12 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a b d \,x^{6}+3 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{2} c \,x^{6}+40 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{b d \,x^{9}+a d \,x^{6}+b c \,x^{6}+a c \,x^{3}}d x \right ) a^{3} d^{2} x^{8}-40 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{b d \,x^{9}+a d \,x^{6}+b c \,x^{6}+a c \,x^{3}}d x \right ) a^{2} b c d \,x^{8}}{40 a^{2} c^{2} x^{8}} \] Input:

int((b*x^3+a)^(2/3)/x^9/(d*x^3+c),x)
 

Output:

( - 5*(a + b*x**3)**(2/3)*a**2*c + 8*(a + b*x**3)**(2/3)*a**2*d*x**3 - 2*( 
a + b*x**3)**(2/3)*a*b*c*x**3 - 12*(a + b*x**3)**(2/3)*a*b*d*x**6 + 3*(a + 
 b*x**3)**(2/3)*b**2*c*x**6 + 40*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**3*d**2*x**8 - 40*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*b*c*d*x**8)/(40*a**2*c 
**2*x**8)