\(\int \frac {(a+b x^3)^{4/3}}{x^2 (c+d x^3)} \, dx\) [719]

Optimal result

Integrand size = 24, antiderivative size = 254 \[ \int \frac {\left (a+b x^3\right )^{4/3}}{x^2 \left (c+d x^3\right )} \, dx=-\frac {a \sqrt [3]{a+b x^3}}{c x}-\frac {b^{4/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} d}+\frac {(b c-a d)^{4/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} c^{4/3} d}-\frac {(b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 c^{4/3} d}-\frac {b^{4/3} \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 d}+\frac {(b c-a d)^{4/3} \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{4/3} d} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 3.67 (sec) , antiderivative size = 457, normalized size of antiderivative = 1.80 \[ \int \frac {\left (a+b x^3\right )^{4/3}}{x^2 \left (c+d x^3\right )} \, dx=\frac {-12 a \sqrt [3]{c} d \sqrt [3]{a+b x^3}-4 \sqrt {3} b^{4/3} c^{4/3} x \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )-2 \sqrt {-6-6 i \sqrt {3}} (b c-a d)^{4/3} x \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )-4 b^{4/3} c^{4/3} x \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )+2 i \left (i+\sqrt {3}\right ) (b c-a d)^{4/3} x \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )+2 b^{4/3} c^{4/3} x \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )+\left (1-i \sqrt {3}\right ) (b c-a d)^{4/3} x \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{12 c^{4/3} d x} \] Input:

Integrate[(a + b*x^3)^(4/3)/(x^2*(c + d*x^3)),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {974, 1054, 2009}

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.

\(\displaystyle \int \frac {\left (a+b x^3\right )^{4/3}}{x^2 \left (c+d x^3\right )} \, dx\)

\(\Big \downarrow \) 974

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

\(\Big \downarrow \) 1054

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

\(\Big \downarrow \) 2009

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

Input:

Int[(a + b*x^3)^(4/3)/(x^2*(c + d*x^3)),x]
 

Output:

-((a*(a + b*x^3)^(1/3))/(c*x)) + (-((b^(4/3)*c*ArcTan[(1 + (2*b^(1/3)*x)/( 
a + b*x^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*d)) + ((b*c - a*d)^(4/3)*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)*d) - ((b*c - a*d)^(4/3)*Log[c + d*x^3])/(6*c^(1/3)*d) - (b^(4/3)*c*L 
og[b^(1/3)*x - (a + b*x^3)^(1/3)])/(2*d) + ((b*c - a*d)^(4/3)*Log[((b*c - 
a*d)^(1/3)*x)/c^(1/3) - (a + b*x^3)^(1/3)])/(2*c^(1/3)*d))/c
 

Defintions of rubi rules used

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 1054

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n 
_)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a 
+ b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, 
 m, p}, x] && IGtQ[n, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 1.92 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.32

method	result	size
pseudoelliptic	\(-\frac {3 \left (b \,x^{3}+a \right )^{\frac {1}{3}} c a d \left (\frac {a d -b c}{c}\right )^{\frac {2}{3}}+\frac {\left (c^{2} \left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} \left (-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right )+2 \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )-\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )\right ) b^{\frac {4}{3}}+\left (a d -b c \right )^{2} \left (2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x -2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x}\right )+\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )-2 \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )\right )\right ) x}{2}}{3 \left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} c^{2} x d}\)	\(336\)

Input:

int((b*x^3+a)^(4/3)/x^2/(d*x^3+c),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3\right )^{4/3}}{x^2 \left (c+d x^3\right )} \, dx=\text {Timed out} \] Input:

integrate((b*x^3+a)^(4/3)/x^2/(d*x^3+c),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

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

integrate((b*x**3+a)**(4/3)/x**2/(d*x**3+c),x)
 

Output:

Integral((a + b*x**3)**(4/3)/(x**2*(c + d*x**3)), x)
 

Maxima [F]

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

integrate((b*x^3+a)^(4/3)/x^2/(d*x^3+c),x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a)^(4/3)/((d*x^3 + c)*x^2), x)
 

Giac [F]

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

integrate((b*x^3+a)^(4/3)/x^2/(d*x^3+c),x, algorithm="giac")
 

Output:

integrate((b*x^3 + a)^(4/3)/((d*x^3 + c)*x^2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3\right )^{4/3}}{x^2 \left (c+d x^3\right )} \, dx=\int \frac {{\left (b\,x^3+a\right )}^{4/3}}{x^2\,\left (d\,x^3+c\right )} \,d x \] Input:

int((a + b*x^3)^(4/3)/(x^2*(c + d*x^3)),x)
 

Output:

int((a + b*x^3)^(4/3)/(x^2*(c + d*x^3)), x)
 

Reduce [F]

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

int((b*x^3+a)^(4/3)/x^2/(d*x^3+c),x)
 

Output:

( - (a + b*x**3)**(1/3)*b + 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*d*x - 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*c*x + int(((a + b*x**3)**(1/3)*x)/(c 
+ d*x**3),x)*b*d*x)/(d*x)