\(\int \frac {1}{x^9 (a+b x^3)^{4/3} (c+d x^3)} \, dx\) [775]

Optimal result

Integrand size = 24, antiderivative size = 351 \[ \int \frac {1}{x^9 \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\frac {b}{a (b c-a d) x^8 \sqrt [3]{a+b x^3}}-\frac {(9 b c-a d) \left (a+b x^3\right )^{2/3}}{8 a^2 c (b c-a d) x^8}+\frac {(9 b c-4 a d) (3 b c+a d) \left (a+b x^3\right )^{2/3}}{20 a^3 c^2 (b c-a d) x^5}-\frac {\left (81 b^3 c^3-9 a b^2 c^2 d-12 a^2 b c d^2-20 a^3 d^3\right ) \left (a+b x^3\right )^{2/3}}{40 a^4 c^3 (b c-a d) x^2}+\frac {d^4 \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} c^{11/3} (b c-a d)^{4/3}}+\frac {d^4 \log \left (c+d x^3\right )}{6 c^{11/3} (b c-a d)^{4/3}}-\frac {d^4 \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{11/3} (b c-a d)^{4/3}} \] Output:

b/a/(-a*d+b*c)/x^8/(b*x^3+a)^(1/3)-1/8*(-a*d+9*b*c)*(b*x^3+a)^(2/3)/a^2/c/ 
(-a*d+b*c)/x^8+1/20*(-4*a*d+9*b*c)*(a*d+3*b*c)*(b*x^3+a)^(2/3)/a^3/c^2/(-a 
*d+b*c)/x^5-1/40*(-20*a^3*d^3-12*a^2*b*c*d^2-9*a*b^2*c^2*d+81*b^3*c^3)*(b* 
x^3+a)^(2/3)/a^4/c^3/(-a*d+b*c)/x^2+1/3*d^4*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)/(-a*d+b*c)^(4/3)+1 
/6*d^4*ln(d*x^3+c)/c^(11/3)/(-a*d+b*c)^(4/3)-1/2*d^4*ln((-a*d+b*c)^(1/3)*x 
/c^(1/3)-(b*x^3+a)^(1/3))/c^(11/3)/(-a*d+b*c)^(4/3)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 7.34 (sec) , antiderivative size = 459, normalized size of antiderivative = 1.31 \[ \int \frac {1}{x^9 \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\frac {-\frac {3 c^{2/3} \left (-81 b^4 c^3 x^9+9 a b^3 c^2 x^6 \left (-3 c+d x^3\right )+3 a^2 b^2 c x^3 \left (3 c^2+c d x^3+4 d^2 x^6\right )+a^4 d \left (5 c^2-8 c d x^3+20 d^2 x^6\right )+a^3 b \left (-5 c^3-c^2 d x^3+4 c d^2 x^6+20 d^3 x^9\right )\right )}{a^4 (-b c+a d) x^8 \sqrt [3]{a+b x^3}}-\frac {20 \sqrt {-6+6 i \sqrt {3}} d^4 \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{(b c-a d)^{4/3}}+\frac {20 \left (1+i \sqrt {3}\right ) d^4 \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{(b c-a d)^{4/3}}-\frac {10 i \left (-i+\sqrt {3}\right ) d^4 \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{(b c-a d)^{4/3}}}{120 c^{11/3}} \] Input:

Integrate[1/(x^9*(a + b*x^3)^(4/3)*(c + d*x^3)),x]
 

Output:

((-3*c^(2/3)*(-81*b^4*c^3*x^9 + 9*a*b^3*c^2*x^6*(-3*c + d*x^3) + 3*a^2*b^2 
*c*x^3*(3*c^2 + c*d*x^3 + 4*d^2*x^6) + a^4*d*(5*c^2 - 8*c*d*x^3 + 20*d^2*x 
^6) + a^3*b*(-5*c^3 - c^2*d*x^3 + 4*c*d^2*x^6 + 20*d^3*x^9)))/(a^4*(-(b*c) 
 + a*d)*x^8*(a + b*x^3)^(1/3)) - (20*Sqrt[-6 + (6*I)*Sqrt[3]]*d^4*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)^(4/3) + (20*(1 + I*Sqrt[3])*d^4*Log[2 
*(b*c - a*d)^(1/3)*x + (1 + I*Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3)])/(b*c - 
a*d)^(4/3) - ((10*I)*(-I + Sqrt[3])*d^4*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)^(4/3))/(120*c^(11/3))
 

Rubi [A] (verified)

Time = 1.04 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {972, 25, 1053, 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.

\(\displaystyle \int \frac {1}{x^9 \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx\)

\(\Big \downarrow \) 972

\(\displaystyle \frac {b}{a x^8 \sqrt [3]{a+b x^3} (b c-a d)}-\frac {\int -\frac {9 b d x^3+9 b c-a d}{x^9 \sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{a (b c-a d)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {9 b d x^3+9 b c-a d}{x^9 \sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{a (b c-a d)}+\frac {b}{a x^8 \sqrt [3]{a+b x^3} (b c-a d)}\)

\(\Big \downarrow \) 1053

\(\displaystyle \frac {-\frac {\int \frac {2 \left (3 b d (9 b c-a d) x^3+(9 b c-4 a d) (3 b c+a d)\right )}{x^6 \sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{8 a c}-\frac {\left (a+b x^3\right )^{2/3} (9 b c-a d)}{8 a c x^8}}{a (b c-a d)}+\frac {b}{a x^8 \sqrt [3]{a+b x^3} (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\int \frac {3 b d (9 b c-a d) x^3+(9 b c-4 a d) (3 b c+a d)}{x^6 \sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{4 a c}-\frac {\left (a+b x^3\right )^{2/3} (9 b c-a d)}{8 a c x^8}}{a (b c-a d)}+\frac {b}{a x^8 \sqrt [3]{a+b x^3} (b c-a d)}\)

\(\Big \downarrow \) 1053

\(\displaystyle \frac {-\frac {-\frac {\int \frac {81 b^3 c^3-9 a b^2 d c^2-12 a^2 b d^2 c-20 a^3 d^3+3 b d (9 b c-4 a d) (3 b c+a d) x^3}{x^3 \sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{5 a c}-\frac {\left (a+b x^3\right )^{2/3} (9 b c-4 a d) (a d+3 b c)}{5 a c x^5}}{4 a c}-\frac {\left (a+b x^3\right )^{2/3} (9 b c-a d)}{8 a c x^8}}{a (b c-a d)}+\frac {b}{a x^8 \sqrt [3]{a+b x^3} (b c-a d)}\)

\(\Big \downarrow \) 1053

\(\displaystyle \frac {-\frac {-\frac {-\frac {\int -\frac {40 a^4 d^4}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{2 a c}-\frac {\left (a+b x^3\right )^{2/3} \left (-20 a^3 d^3-12 a^2 b c d^2-9 a b^2 c^2 d+81 b^3 c^3\right )}{2 a c x^2}}{5 a c}-\frac {\left (a+b x^3\right )^{2/3} (9 b c-4 a d) (a d+3 b c)}{5 a c x^5}}{4 a c}-\frac {\left (a+b x^3\right )^{2/3} (9 b c-a d)}{8 a c x^8}}{a (b c-a d)}+\frac {b}{a x^8 \sqrt [3]{a+b x^3} (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {-\frac {\frac {20 a^3 d^4 \int \frac {1}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{c}-\frac {\left (a+b x^3\right )^{2/3} \left (-20 a^3 d^3-12 a^2 b c d^2-9 a b^2 c^2 d+81 b^3 c^3\right )}{2 a c x^2}}{5 a c}-\frac {\left (a+b x^3\right )^{2/3} (9 b c-4 a d) (a d+3 b c)}{5 a c x^5}}{4 a c}-\frac {\left (a+b x^3\right )^{2/3} (9 b c-a d)}{8 a c x^8}}{a (b c-a d)}+\frac {b}{a x^8 \sqrt [3]{a+b x^3} (b c-a d)}\)

\(\Big \downarrow \) 901

\(\displaystyle \frac {-\frac {-\frac {\frac {20 a^3 d^4 \left (\frac {\arctan \left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} c^{2/3} \sqrt [3]{b c-a d}}+\frac {\log \left (c+d x^3\right )}{6 c^{2/3} \sqrt [3]{b c-a d}}-\frac {\log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{2/3} \sqrt [3]{b c-a d}}\right )}{c}-\frac {\left (a+b x^3\right )^{2/3} \left (-20 a^3 d^3-12 a^2 b c d^2-9 a b^2 c^2 d+81 b^3 c^3\right )}{2 a c x^2}}{5 a c}-\frac {\left (a+b x^3\right )^{2/3} (9 b c-4 a d) (a d+3 b c)}{5 a c x^5}}{4 a c}-\frac {\left (a+b x^3\right )^{2/3} (9 b c-a d)}{8 a c x^8}}{a (b c-a d)}+\frac {b}{a x^8 \sqrt [3]{a+b x^3} (b c-a d)}\)

Input:

Int[1/(x^9*(a + b*x^3)^(4/3)*(c + d*x^3)),x]
 

Output:

b/(a*(b*c - a*d)*x^8*(a + b*x^3)^(1/3)) + (-1/8*((9*b*c - a*d)*(a + b*x^3) 
^(2/3))/(a*c*x^8) - (-1/5*((9*b*c - 4*a*d)*(3*b*c + a*d)*(a + b*x^3)^(2/3) 
)/(a*c*x^5) - (-1/2*((81*b^3*c^3 - 9*a*b^2*c^2*d - 12*a^2*b*c*d^2 - 20*a^3 
*d^3)*(a + b*x^3)^(2/3))/(a*c*x^2) + (20*a^3*d^4*(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*a*c))/(a*(b*c - a*d))
 

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 972

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[(-b)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x 
^n)^(q + 1)/(a*e*n*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*n*(b*c - a*d)*(p + 
 1))   Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*( 
b*c - a*d)*(p + 1) + d*b*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{ 
a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -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 = 1.69 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.01

method	result	size
pseudoelliptic	\(\frac {-\frac {3 c \left (d \left (4 d^{2} x^{6}-\frac {8}{5} c d \,x^{3}+c^{2}\right ) a^{4}-\left (-4 d^{3} x^{9}-\frac {4}{5} c \,d^{2} x^{6}+\frac {1}{5} c^{2} d \,x^{3}+c^{3}\right ) b \,a^{3}+\frac {9 c \left (\frac {4}{3} d^{2} x^{6}+\frac {1}{3} c d \,x^{3}+c^{2}\right ) b^{2} x^{3} a^{2}}{5}-\frac {27 c^{2} \left (-\frac {d \,x^{3}}{3}+c \right ) b^{3} x^{6} a}{5}-\frac {81 b^{4} c^{3} x^{9}}{5}\right ) \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}{4}+a^{4} d^{4} x^{8} \left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x -2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x}\right )+\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )-2 \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )\right )}{6 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{8} c^{4} \left (a d -b c \right ) a^{4}}\)	\(354\)

Input:

int(1/x^9/(b*x^3+a)^(4/3)/(d*x^3+c),x,method=_RETURNVERBOSE)
 

Output:

1/6/((a*d-b*c)/c)^(1/3)*(-3/4*c*(d*(4*d^2*x^6-8/5*c*d*x^3+c^2)*a^4-(-4*d^3 
*x^9-4/5*c*d^2*x^6+1/5*c^2*d*x^3+c^3)*b*a^3+9/5*c*(4/3*d^2*x^6+1/3*c*d*x^3 
+c^2)*b^2*x^3*a^2-27/5*c^2*(-1/3*d*x^3+c)*b^3*x^6*a-81/5*b^4*c^3*x^9)*((a* 
d-b*c)/c)^(1/3)+a^4*d^4*x^8*(b*x^3+a)^(1/3)*(-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)))/(b*x^3+a)^(1/3)/x^ 
8/c^4/(a*d-b*c)/a^4
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{x^9 \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\text {Timed out} \] Input:

integrate(1/x^9/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {1}{x^9 \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\int \frac {1}{x^{9} \left (a + b x^{3}\right )^{\frac {4}{3}} \left (c + d x^{3}\right )}\, dx \] Input:

integrate(1/x**9/(b*x**3+a)**(4/3)/(d*x**3+c),x)
 

Output:

Integral(1/(x**9*(a + b*x**3)**(4/3)*(c + d*x**3)), x)
 

Maxima [F]

\[ \int \frac {1}{x^9 \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {4}{3}} {\left (d x^{3} + c\right )} x^{9}} \,d x } \] Input:

integrate(1/x^9/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="maxima")
 

Output:

integrate(1/((b*x^3 + a)^(4/3)*(d*x^3 + c)*x^9), x)
 

Giac [F]

\[ \int \frac {1}{x^9 \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {4}{3}} {\left (d x^{3} + c\right )} x^{9}} \,d x } \] Input:

integrate(1/x^9/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="giac")
 

Output:

integrate(1/((b*x^3 + a)^(4/3)*(d*x^3 + c)*x^9), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^9 \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\int \frac {1}{x^9\,{\left (b\,x^3+a\right )}^{4/3}\,\left (d\,x^3+c\right )} \,d x \] Input:

int(1/(x^9*(a + b*x^3)^(4/3)*(c + d*x^3)),x)
 

Output:

int(1/(x^9*(a + b*x^3)^(4/3)*(c + d*x^3)), x)
 

Reduce [F]

\[ \int \frac {1}{x^9 \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a c \,x^{9}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} a d \,x^{12}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b c \,x^{12}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b d \,x^{15}}d x \] Input:

int(1/x^9/(b*x^3+a)^(4/3)/(d*x^3+c),x)
 

Output:

int(1/((a + b*x**3)**(1/3)*a*c*x**9 + (a + b*x**3)**(1/3)*a*d*x**12 + (a + 
 b*x**3)**(1/3)*b*c*x**12 + (a + b*x**3)**(1/3)*b*d*x**15),x)