\(\int \frac {\sqrt [3]{b x+a x^3} (b+a x^4)}{x^4} \, dx\) [2308]

Optimal result

Integrand size = 24, antiderivative size = 177 \[ \int \frac {\sqrt [3]{b x+a x^3} \left (b+a x^4\right )}{x^4} \, dx=\frac {\sqrt [3]{b x+a x^3} \left (-3 b-3 a x^2+4 a x^4\right )}{8 x^3}-\frac {\sqrt [3]{a} b \arctan \left (\frac {\sqrt {3} \sqrt [3]{a} x}{\sqrt [3]{a} x+2 \sqrt [3]{b x+a x^3}}\right )}{2 \sqrt {3}}-\frac {1}{6} \sqrt [3]{a} b \log \left (-\sqrt [3]{a} x+\sqrt [3]{b x+a x^3}\right )+\frac {1}{12} \sqrt [3]{a} b \log \left (a^{2/3} x^2+\sqrt [3]{a} x \sqrt [3]{b x+a x^3}+\left (b x+a x^3\right )^{2/3}\right ) \] Output:

1/8*(a*x^3+b*x)^(1/3)*(4*a*x^4-3*a*x^2-3*b)/x^3-1/6*a^(1/3)*b*arctan(3^(1/ 
2)*a^(1/3)*x/(a^(1/3)*x+2*(a*x^3+b*x)^(1/3)))*3^(1/2)-1/6*a^(1/3)*b*ln(-a^ 
(1/3)*x+(a*x^3+b*x)^(1/3))+1/12*a^(1/3)*b*ln(a^(2/3)*x^2+a^(1/3)*x*(a*x^3+ 
b*x)^(1/3)+(a*x^3+b*x)^(2/3))
 

Mathematica [A] (verified)

Time = 2.72 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.35 \[ \int \frac {\sqrt [3]{b x+a x^3} \left (b+a x^4\right )}{x^4} \, dx=-\frac {\sqrt [3]{x \left (b+a x^2\right )} \left (9 b \sqrt [3]{b+a x^2}+9 a x^2 \sqrt [3]{b+a x^2}-12 a x^4 \sqrt [3]{b+a x^2}+4 \sqrt {3} \sqrt [3]{a} b x^{8/3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{a} x^{2/3}}{\sqrt [3]{a} x^{2/3}+2 \sqrt [3]{b+a x^2}}\right )+4 \sqrt [3]{a} b x^{8/3} \log \left (-\sqrt [3]{a} x^{2/3}+\sqrt [3]{b+a x^2}\right )-2 \sqrt [3]{a} b x^{8/3} \log \left (a^{2/3} x^{4/3}+\sqrt [3]{a} x^{2/3} \sqrt [3]{b+a x^2}+\left (b+a x^2\right )^{2/3}\right )\right )}{24 x^3 \sqrt [3]{b+a x^2}} \] Input:

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

Output:

-1/24*((x*(b + a*x^2))^(1/3)*(9*b*(b + a*x^2)^(1/3) + 9*a*x^2*(b + a*x^2)^ 
(1/3) - 12*a*x^4*(b + a*x^2)^(1/3) + 4*Sqrt[3]*a^(1/3)*b*x^(8/3)*ArcTan[(S 
qrt[3]*a^(1/3)*x^(2/3))/(a^(1/3)*x^(2/3) + 2*(b + a*x^2)^(1/3))] + 4*a^(1/ 
3)*b*x^(8/3)*Log[-(a^(1/3)*x^(2/3)) + (b + a*x^2)^(1/3)] - 2*a^(1/3)*b*x^( 
8/3)*Log[a^(2/3)*x^(4/3) + a^(1/3)*x^(2/3)*(b + a*x^2)^(1/3) + (b + a*x^2) 
^(2/3)]))/(x^3*(b + a*x^2)^(1/3))
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.02, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2449, 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.

\(\displaystyle \int \frac {\sqrt [3]{a x^3+b x} \left (a x^4+b\right )}{x^4} \, dx\)

\(\Big \downarrow \) 2449

\(\displaystyle \int \left (a \sqrt [3]{a x^3+b x}+\frac {b \sqrt [3]{a x^3+b x}}{x^4}\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(a*x*(b*x + a*x^3)^(1/3))/2 - (3*(b*x + a*x^3)^(4/3))/(8*x^4) - (a^(1/3)*b 
*x^(2/3)*(b + a*x^2)^(2/3)*ArcTan[(1 + (2*a^(1/3)*x^(2/3))/(b + a*x^2)^(1/ 
3))/Sqrt[3]])/(2*Sqrt[3]*(b*x + a*x^3)^(2/3)) - (a^(1/3)*b*x^(2/3)*(b + a* 
x^2)^(2/3)*Log[a^(1/3)*x^(2/3) - (b + a*x^2)^(1/3)])/(4*(b*x + a*x^3)^(2/3 
))
 

Defintions of rubi rules used

rule 2009

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

rule 2449

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_S 
ymbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a*x^j + b*x^n)^p, x], x] /; FreeQ 
[{a, b, c, j, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !Integer 
Q[p] && NeQ[n, j]
 

Maple [A] (verified)

Time = 0.75 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.84

method	result	size
pseudoelliptic	\(\frac {\left (12 a \,x^{4}-9 a \,x^{2}-9 b \right ) {\left (x \left (a \,x^{2}+b \right )\right )}^{\frac {1}{3}}-2 b \,x^{3} a^{\frac {1}{3}} \left (-2 \arctan \left (\frac {\sqrt {3}\, \left (a^{\frac {1}{3}} x +2 {\left (x \left (a \,x^{2}+b \right )\right )}^{\frac {1}{3}}\right )}{3 a^{\frac {1}{3}} x}\right ) \sqrt {3}+2 \ln \left (\frac {-a^{\frac {1}{3}} x +{\left (x \left (a \,x^{2}+b \right )\right )}^{\frac {1}{3}}}{x}\right )-\ln \left (\frac {a^{\frac {2}{3}} x^{2}+a^{\frac {1}{3}} {\left (x \left (a \,x^{2}+b \right )\right )}^{\frac {1}{3}} x +{\left (x \left (a \,x^{2}+b \right )\right )}^{\frac {2}{3}}}{x^{2}}\right )\right )}{24 x^{3}}\)	\(148\)

Input:

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

Output:

1/24*((12*a*x^4-9*a*x^2-9*b)*(x*(a*x^2+b))^(1/3)-2*b*x^3*a^(1/3)*(-2*arcta 
n(1/3*3^(1/2)*(a^(1/3)*x+2*(x*(a*x^2+b))^(1/3))/a^(1/3)/x)*3^(1/2)+2*ln((- 
a^(1/3)*x+(x*(a*x^2+b))^(1/3))/x)-ln((a^(2/3)*x^2+a^(1/3)*(x*(a*x^2+b))^(1 
/3)*x+(x*(a*x^2+b))^(2/3))/x^2)))/x^3
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

Integral((x*(a*x**2 + b))**(1/3)*(a*x**4 + b)/x**4, x)
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 16.28 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.69 \[ \int \frac {\sqrt [3]{b x+a x^3} \left (b+a x^4\right )}{x^4} \, dx=\frac {1}{24} \, {\left (\frac {12 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{3}} a x^{2}}{b} + 4 \, \sqrt {3} a^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right ) + 2 \, a^{\frac {1}{3}} \log \left ({\left (a + \frac {b}{x^{2}}\right )}^{\frac {2}{3}} + {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) - 4 \, a^{\frac {1}{3}} \log \left ({\left | {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right ) - \frac {9 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {4}{3}}}{b}\right )} b \] Input:

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

Output:

1/24*(12*(a + b/x^2)^(1/3)*a*x^2/b + 4*sqrt(3)*a^(1/3)*arctan(1/3*sqrt(3)* 
(2*(a + b/x^2)^(1/3) + a^(1/3))/a^(1/3)) + 2*a^(1/3)*log((a + b/x^2)^(2/3) 
 + (a + b/x^2)^(1/3)*a^(1/3) + a^(2/3)) - 4*a^(1/3)*log(abs((a + b/x^2)^(1 
/3) - a^(1/3))) - 9*(a + b/x^2)^(4/3)/b)*b
 

Mupad [B] (verification not implemented)

Time = 9.39 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.37 \[ \int \frac {\sqrt [3]{b x+a x^3} \left (b+a x^4\right )}{x^4} \, dx=\frac {3\,a\,x\,{\left (a\,x^3+b\,x\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{3},\frac {2}{3};\ \frac {5}{3};\ -\frac {a\,x^2}{b}\right )}{4\,{\left (\frac {a\,x^2}{b}+1\right )}^{1/3}}-\frac {3\,{\left (a\,x^3+b\,x\right )}^{1/3}\,\left (a\,x^2+b\right )}{8\,x^3} \] Input:

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

Output:

(3*a*x*(b*x + a*x^3)^(1/3)*hypergeom([-1/3, 2/3], 5/3, -(a*x^2)/b))/(4*((a 
*x^2)/b + 1)^(1/3)) - (3*(b*x + a*x^3)^(1/3)*(b + a*x^2))/(8*x^3)
 

Reduce [F]

\[ \int \frac {\sqrt [3]{b x+a x^3} \left (b+a x^4\right )}{x^4} \, dx=\frac {12 \left (a \,x^{2}+b \right )^{\frac {1}{3}} a \,x^{4}-9 \left (a \,x^{2}+b \right )^{\frac {1}{3}} a \,x^{2}-9 \left (a \,x^{2}+b \right )^{\frac {1}{3}} b +8 x^{\frac {8}{3}} \left (\int \frac {x^{\frac {1}{3}}}{\left (a \,x^{2}+b \right )^{\frac {2}{3}}}d x \right ) a b}{24 x^{\frac {8}{3}}} \] Input:

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

Output:

(12*(a*x**2 + b)**(1/3)*a*x**4 - 9*(a*x**2 + b)**(1/3)*a*x**2 - 9*(a*x**2 
+ b)**(1/3)*b + 8*x**(2/3)*int((x**(1/3)*(a*x**2 + b)**(1/3))/(a*x**2 + b) 
,x)*a*b*x**2)/(24*x**(2/3)*x**2)