\(\int \frac {(a+b x^3)^{4/3}}{x^5 (c+d x^3)} \, dx\) [720]

Optimal result

Integrand size = 24, antiderivative size = 201 \[ \int \frac {\left (a+b x^3\right )^{4/3}}{x^5 \left (c+d x^3\right )} \, dx=-\frac {a \sqrt [3]{a+b x^3}}{4 c x^4}-\frac {(5 b c-4 a d) \sqrt [3]{a+b x^3}}{4 c^2 x}-\frac {(b c-a d)^{4/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} c^{7/3}}+\frac {(b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 c^{7/3}}-\frac {(b c-a d)^{4/3} \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{7/3}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.04 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.63 \[ \int \frac {\left (a+b x^3\right )^{4/3}}{x^5 \left (c+d x^3\right )} \, dx=\frac {\frac {3 \sqrt [3]{c} \sqrt [3]{a+b x^3} \left (-a c-5 b c x^3+4 a d x^3\right )}{x^4}+2 \sqrt {-6-6 i \sqrt {3}} (b c-a d)^{4/3} \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )+2 \left (1-i \sqrt {3}\right ) (b c-a d)^{4/3} \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )+i \left (i+\sqrt {3}\right ) (b c-a d)^{4/3} \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{12 c^{7/3}} \] Input:

Integrate[(a + b*x^3)^(4/3)/(x^5*(c + d*x^3)),x]
 

Output:

((3*c^(1/3)*(a + b*x^3)^(1/3)*(-(a*c) - 5*b*c*x^3 + 4*a*d*x^3))/x^4 + 2*Sq 
rt[-6 - (6*I)*Sqrt[3]]*(b*c - a*d)^(4/3)*ArcTan[(3*(b*c - a*d)^(1/3)*x)/(S 
qrt[3]*(b*c - a*d)^(1/3)*x - (3*I + Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3))] + 
 2*(1 - I*Sqrt[3])*(b*c - a*d)^(4/3)*Log[2*(b*c - a*d)^(1/3)*x + (1 + I*Sq 
rt[3])*c^(1/3)*(a + b*x^3)^(1/3)] + I*(I + Sqrt[3])*(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)])/(12*c^(7/3))
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {974, 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.

\(\displaystyle \int \frac {\left (a+b x^3\right )^{4/3}}{x^5 \left (c+d x^3\right )} \, dx\)

\(\Big \downarrow \) 974

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

\(\Big \downarrow \) 1053

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 992

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

Input:

Int[(a + b*x^3)^(4/3)/(x^5*(c + d*x^3)),x]
 

Output:

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

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 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 = 1.82 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.14

method	result	size
pseudoelliptic	\(\frac {-x^{4} \left (a d -b c \right )^{2} \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )-\frac {3 c \left (\left (-4 a d +5 b c \right ) x^{3}+a c \right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (\frac {a d -b c}{c}\right )^{\frac {2}{3}}}{4}+\left (a d -b c \right )^{2} \left (\arctan \left (\frac {\sqrt {3}\, \left (-\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}+x \right )}{3 x}\right ) \sqrt {3}+\frac {\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{2}\right ) x^{4}}{3 \left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} c^{3} x^{4}}\)	\(229\)

Input:

int((b*x^3+a)^(4/3)/x^5/(d*x^3+c),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3\right )^{4/3}}{x^5 \left (c+d x^3\right )} \, dx=\text {Timed out} \] Input:

integrate((b*x^3+a)^(4/3)/x^5/(d*x^3+c),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {\left (a+b x^3\right )^{4/3}}{x^5 \left (c+d x^3\right )} \, dx=\int \frac {\left (a + b x^{3}\right )^{\frac {4}{3}}}{x^{5} \left (c + d x^{3}\right )}\, dx \] Input:

integrate((b*x**3+a)**(4/3)/x**5/(d*x**3+c),x)
 

Output:

Integral((a + b*x**3)**(4/3)/(x**5*(c + d*x**3)), x)
 

Maxima [F]

\[ \int \frac {\left (a+b x^3\right )^{4/3}}{x^5 \left (c+d x^3\right )} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {4}{3}}}{{\left (d x^{3} + c\right )} x^{5}} \,d x } \] Input:

integrate((b*x^3+a)^(4/3)/x^5/(d*x^3+c),x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a)^(4/3)/((d*x^3 + c)*x^5), x)
 

Giac [F]

\[ \int \frac {\left (a+b x^3\right )^{4/3}}{x^5 \left (c+d x^3\right )} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {4}{3}}}{{\left (d x^{3} + c\right )} x^{5}} \,d x } \] Input:

integrate((b*x^3+a)^(4/3)/x^5/(d*x^3+c),x, algorithm="giac")
 

Output:

integrate((b*x^3 + a)^(4/3)/((d*x^3 + c)*x^5), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3\right )^{4/3}}{x^5 \left (c+d x^3\right )} \, dx=\int \frac {{\left (b\,x^3+a\right )}^{4/3}}{x^5\,\left (d\,x^3+c\right )} \,d x \] Input:

int((a + b*x^3)^(4/3)/(x^5*(c + d*x^3)),x)
 

Output:

int((a + b*x^3)^(4/3)/(x^5*(c + d*x^3)), x)
 

Reduce [F]

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

int((b*x^3+a)^(4/3)/x^5/(d*x^3+c),x)
 

Output:

( - (a + b*x**3)**(1/3)*a**2*d + 3*(a + b*x**3)**(1/3)*a*b*d*x**3 - 4*(a + 
 b*x**3)**(1/3)*b**2*c*x**3 - 4*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**4 + 8*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**4 - 4*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**4)/(4*a*c*d*x**4)