\(\int \frac {(a+b x^3)^{4/3}}{x^8 (c+d x^3)} \, dx\) [721]

Optimal result

Integrand size = 24, antiderivative size = 250 \[ \int \frac {\left (a+b x^3\right )^{4/3}}{x^8 \left (c+d x^3\right )} \, dx=-\frac {a \sqrt [3]{a+b x^3}}{7 c x^7}-\frac {(8 b c-7 a d) \sqrt [3]{a+b x^3}}{28 c^2 x^4}-\frac {\left (4 b^2 c^2-35 a b c d+28 a^2 d^2\right ) \sqrt [3]{a+b x^3}}{28 a c^3 x}+\frac {d (b c-a d)^{4/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} c^{10/3}}-\frac {d (b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 c^{10/3}}+\frac {d (b c-a d)^{4/3} \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{10/3}} \] Output:

-1/7*a*(b*x^3+a)^(1/3)/c/x^7-1/28*(-7*a*d+8*b*c)*(b*x^3+a)^(1/3)/c^2/x^4-1 
/28*(28*a^2*d^2-35*a*b*c*d+4*b^2*c^2)*(b*x^3+a)^(1/3)/a/c^3/x+1/3*d*(-a*d+ 
b*c)^(4/3)*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^(10/3)-1/6*d*(-a*d+b*c)^(4/3)*ln(d*x^3+c)/c^(10/3)+1/2*d*( 
-a*d+b*c)^(4/3)*ln((-a*d+b*c)^(1/3)*x/c^(1/3)-(b*x^3+a)^(1/3))/c^(10/3)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 2.73 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.48 \[ \int \frac {\left (a+b x^3\right )^{4/3}}{x^8 \left (c+d x^3\right )} \, dx=\frac {-\frac {3 \sqrt [3]{c} \sqrt [3]{a+b x^3} \left (4 b^2 c^2 x^6+a b c x^3 \left (8 c-35 d x^3\right )+a^2 \left (4 c^2-7 c d x^3+28 d^2 x^6\right )\right )}{a x^7}-14 \sqrt {-6-6 i \sqrt {3}} d (b c-a d)^{4/3} \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )+14 i \left (i+\sqrt {3}\right ) d (b c-a d)^{4/3} \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )+7 \left (1-i \sqrt {3}\right ) d (b c-a d)^{4/3} \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{84 c^{10/3}} \] Input:

Integrate[(a + b*x^3)^(4/3)/(x^8*(c + d*x^3)),x]
 

Output:

((-3*c^(1/3)*(a + b*x^3)^(1/3)*(4*b^2*c^2*x^6 + a*b*c*x^3*(8*c - 35*d*x^3) 
 + a^2*(4*c^2 - 7*c*d*x^3 + 28*d^2*x^6)))/(a*x^7) - 14*Sqrt[-6 - (6*I)*Sqr 
t[3]]*d*(b*c - a*d)^(4/3)*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))] + (14*I)*(I + Sq 
rt[3])*d*(b*c - a*d)^(4/3)*Log[2*(b*c - a*d)^(1/3)*x + (1 + I*Sqrt[3])*c^( 
1/3)*(a + b*x^3)^(1/3)] + 7*(1 - I*Sqrt[3])*d*(b*c - a*d)^(4/3)*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)])/(84*c^(10/3))
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {974, 1053, 25, 27, 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.

\(\displaystyle \int \frac {\left (a+b x^3\right )^{4/3}}{x^8 \left (c+d x^3\right )} \, dx\)

\(\Big \downarrow \) 974

\(\displaystyle \frac {\int \frac {b (7 b c-6 a d) x^3+a (8 b c-7 a d)}{x^5 \left (b x^3+a\right )^{2/3} \left (d x^3+c\right )}dx}{7 c}-\frac {a \sqrt [3]{a+b x^3}}{7 c x^7}\)

\(\Big \downarrow \) 1053

\(\displaystyle \frac {-\frac {\int -\frac {a \left (-3 b d (8 b c-7 a d) x^3+4 b^2 c^2+28 a^2 d^2-35 a b c d\right )}{x^2 \left (b x^3+a\right )^{2/3} \left (d x^3+c\right )}dx}{4 a c}-\frac {\sqrt [3]{a+b x^3} (8 b c-7 a d)}{4 c x^4}}{7 c}-\frac {a \sqrt [3]{a+b x^3}}{7 c x^7}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {a \left (-3 b d (8 b c-7 a d) x^3+4 b^2 c^2+28 a^2 d^2-35 a b c d\right )}{x^2 \left (b x^3+a\right )^{2/3} \left (d x^3+c\right )}dx}{4 a c}-\frac {\sqrt [3]{a+b x^3} (8 b c-7 a d)}{4 c x^4}}{7 c}-\frac {a \sqrt [3]{a+b x^3}}{7 c x^7}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {-3 b d (8 b c-7 a d) x^3+4 b^2 c^2+28 a^2 d^2-35 a b c d}{x^2 \left (b x^3+a\right )^{2/3} \left (d x^3+c\right )}dx}{4 c}-\frac {\sqrt [3]{a+b x^3} (8 b c-7 a d)}{4 c x^4}}{7 c}-\frac {a \sqrt [3]{a+b x^3}}{7 c x^7}\)

\(\Big \downarrow \) 1053

\(\displaystyle \frac {\frac {-\frac {\int \frac {28 a d (b c-a d)^2 x}{\left (b x^3+a\right )^{2/3} \left (d x^3+c\right )}dx}{a c}-\frac {\sqrt [3]{a+b x^3} \left (\frac {4 b^2 c}{a}+\frac {28 a d^2}{c}-35 b d\right )}{x}}{4 c}-\frac {\sqrt [3]{a+b x^3} (8 b c-7 a d)}{4 c x^4}}{7 c}-\frac {a \sqrt [3]{a+b x^3}}{7 c x^7}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {28 d (b c-a d)^2 \int \frac {x}{\left (b x^3+a\right )^{2/3} \left (d x^3+c\right )}dx}{c}-\frac {\sqrt [3]{a+b x^3} \left (\frac {4 b^2 c}{a}+\frac {28 a d^2}{c}-35 b d\right )}{x}}{4 c}-\frac {\sqrt [3]{a+b x^3} (8 b c-7 a d)}{4 c x^4}}{7 c}-\frac {a \sqrt [3]{a+b x^3}}{7 c x^7}\)

\(\Big \downarrow \) 992

\(\displaystyle \frac {\frac {-\frac {28 d (b c-a d)^2 \left (-\frac {\arctan \left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{c} (b c-a d)^{2/3}}+\frac {\log \left (c+d x^3\right )}{6 \sqrt [3]{c} (b c-a d)^{2/3}}-\frac {\log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{c} (b c-a d)^{2/3}}\right )}{c}-\frac {\sqrt [3]{a+b x^3} \left (\frac {4 b^2 c}{a}+\frac {28 a d^2}{c}-35 b d\right )}{x}}{4 c}-\frac {\sqrt [3]{a+b x^3} (8 b c-7 a d)}{4 c x^4}}{7 c}-\frac {a \sqrt [3]{a+b x^3}}{7 c x^7}\)

Input:

Int[(a + b*x^3)^(4/3)/(x^8*(c + d*x^3)),x]
 

Output:

-1/7*(a*(a + b*x^3)^(1/3))/(c*x^7) + (-1/4*((8*b*c - 7*a*d)*(a + b*x^3)^(1 
/3))/(c*x^4) + (-((((4*b^2*c)/a - 35*b*d + (28*a*d^2)/c)*(a + b*x^3)^(1/3) 
)/x) - (28*d*(b*c - 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*c))/(7*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 974

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_) 
)^(q_), x_Symbol] :> Simp[c*(e*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^ 
(q - 1)/(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 - 2)*Simp[c*(c*b - a*d)*(m + 1) + c*n*(b*c*(p + 1 
) + a*d*(q - 1)) + d*((c*b - a*d)*(m + 1) + c*b*n*(p + q))*x^n, x], x], x] 
/; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && GtQ[q 
, 1] && 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 = 1.88 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.04

method	result	size
pseudoelliptic	\(-\frac {c \left (\left (b \,x^{3}+a \right )^{2} c^{2}-\frac {7 a d \,x^{3} \left (5 b \,x^{3}+a \right ) c}{4}+7 a^{2} d^{2} x^{6}\right ) \left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\frac {7 a d \,x^{7} \left (a d -b c \right )^{2} \left (2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x -2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x}\right )+\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )-2 \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )\right )}{6}}{7 \left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{7} c^{4} a}\)	\(259\)

Input:

int((b*x^3+a)^(4/3)/x^8/(d*x^3+c),x,method=_RETURNVERBOSE)
 

Output:

-1/7/((a*d-b*c)/c)^(2/3)*(c*((b*x^3+a)^2*c^2-7/4*a*d*x^3*(5*b*x^3+a)*c+7*a 
^2*d^2*x^6)*((a*d-b*c)/c)^(2/3)*(b*x^3+a)^(1/3)+7/6*a*d*x^7*(a*d-b*c)^2*(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^7/c^4/a
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3\right )^{4/3}}{x^8 \left (c+d x^3\right )} \, dx=\text {Timed out} \] Input:

integrate((b*x^3+a)^(4/3)/x^8/(d*x^3+c),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {\left (a+b x^3\right )^{4/3}}{x^8 \left (c+d x^3\right )} \, dx=\int \frac {\left (a + b x^{3}\right )^{\frac {4}{3}}}{x^{8} \left (c + d x^{3}\right )}\, dx \] Input:

integrate((b*x**3+a)**(4/3)/x**8/(d*x**3+c),x)
 

Output:

Integral((a + b*x**3)**(4/3)/(x**8*(c + d*x**3)), x)
 

Maxima [F]

\[ \int \frac {\left (a+b x^3\right )^{4/3}}{x^8 \left (c+d x^3\right )} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {4}{3}}}{{\left (d x^{3} + c\right )} x^{8}} \,d x } \] Input:

integrate((b*x^3+a)^(4/3)/x^8/(d*x^3+c),x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a)^(4/3)/((d*x^3 + c)*x^8), x)
 

Giac [F]

\[ \int \frac {\left (a+b x^3\right )^{4/3}}{x^8 \left (c+d x^3\right )} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {4}{3}}}{{\left (d x^{3} + c\right )} x^{8}} \,d x } \] Input:

integrate((b*x^3+a)^(4/3)/x^8/(d*x^3+c),x, algorithm="giac")
 

Output:

integrate((b*x^3 + a)^(4/3)/((d*x^3 + c)*x^8), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3\right )^{4/3}}{x^8 \left (c+d x^3\right )} \, dx=\int \frac {{\left (b\,x^3+a\right )}^{4/3}}{x^8\,\left (d\,x^3+c\right )} \,d x \] Input:

int((a + b*x^3)^(4/3)/(x^8*(c + d*x^3)),x)
 

Output:

int((a + b*x^3)^(4/3)/(x^8*(c + d*x^3)), x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^3\right )^{4/3}}{x^8 \left (c+d x^3\right )} \, dx=\frac {-4 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} c +7 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} d \,x^{3}-8 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a b c \,x^{3}-21 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a b d \,x^{6}+24 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} c \,x^{6}+28 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b d \,x^{8}+a d \,x^{5}+b c \,x^{5}+a c \,x^{2}}d x \right ) a^{3} d^{2} x^{7}-56 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b d \,x^{8}+a d \,x^{5}+b c \,x^{5}+a c \,x^{2}}d x \right ) a^{2} b c d \,x^{7}+28 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b d \,x^{8}+a d \,x^{5}+b c \,x^{5}+a c \,x^{2}}d x \right ) a \,b^{2} c^{2} x^{7}}{28 a \,c^{2} x^{7}} \] Input:

int((b*x^3+a)^(4/3)/x^8/(d*x^3+c),x)
 

Output:

( - 4*(a + b*x**3)**(1/3)*a**2*c + 7*(a + b*x**3)**(1/3)*a**2*d*x**3 - 8*( 
a + b*x**3)**(1/3)*a*b*c*x**3 - 21*(a + b*x**3)**(1/3)*a*b*d*x**6 + 24*(a 
+ b*x**3)**(1/3)*b**2*c*x**6 + 28*int((a + b*x**3)**(1/3)/(a*c*x**2 + a*d* 
x**5 + b*c*x**5 + b*d*x**8),x)*a**3*d**2*x**7 - 56*int((a + b*x**3)**(1/3) 
/(a*c*x**2 + a*d*x**5 + b*c*x**5 + b*d*x**8),x)*a**2*b*c*d*x**7 + 28*int(( 
a + b*x**3)**(1/3)/(a*c*x**2 + a*d*x**5 + b*c*x**5 + b*d*x**8),x)*a*b**2*c 
**2*x**7)/(28*a*c**2*x**7)