\(\int \frac {b+a x^6}{x^3 (b+a x^3) \sqrt [4]{b x+a x^4}} \, dx\) [1818]

Optimal result

Integrand size = 33, antiderivative size = 123 \[ \int \frac {b+a x^6}{x^3 \left (b+a x^3\right ) \sqrt [4]{b x+a x^4}} \, dx=-\frac {4 \left (b+4 a x^3+3 b x^3\right ) \left (b x+a x^4\right )^{3/4}}{9 b x^3 \left (b+a x^3\right )}+\frac {2 \arctan \left (\frac {\sqrt [4]{a} \left (b x+a x^4\right )^{3/4}}{b+a x^3}\right )}{3 \sqrt [4]{a}}+\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{a} \left (b x+a x^4\right )^{3/4}}{b+a x^3}\right )}{3 \sqrt [4]{a}} \] Output:

-4/9*(4*a*x^3+3*b*x^3+b)*(a*x^4+b*x)^(3/4)/b/x^3/(a*x^3+b)+2/3*arctan(a^(1 
/4)*(a*x^4+b*x)^(3/4)/(a*x^3+b))/a^(1/4)+2/3*arctanh(a^(1/4)*(a*x^4+b*x)^( 
3/4)/(a*x^3+b))/a^(1/4)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 10.10 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.59 \[ \int \frac {b+a x^6}{x^3 \left (b+a x^3\right ) \sqrt [4]{b x+a x^4}} \, dx=\frac {4 \left (-5 \left (b+4 a x^3\right )+3 a x^6 \sqrt [4]{1+\frac {a x^3}{b}} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {5}{4},\frac {9}{4},-\frac {a x^3}{b}\right )\right )}{45 b x^2 \sqrt [4]{x \left (b+a x^3\right )}} \] Input:

Integrate[(b + a*x^6)/(x^3*(b + a*x^3)*(b*x + a*x^4)^(1/4)),x]
 

Output:

(4*(-5*(b + 4*a*x^3) + 3*a*x^6*(1 + (a*x^3)/b)^(1/4)*Hypergeometric2F1[5/4 
, 5/4, 9/4, -((a*x^3)/b)]))/(45*b*x^2*(x*(b + a*x^3))^(1/4))
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.29, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2467, 1813, 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.

\(\displaystyle \int \frac {a x^6+b}{x^3 \left (a x^3+b\right ) \sqrt [4]{a x^4+b x}} \, dx\)

\(\Big \downarrow \) 2467

\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{a x^3+b} \int \frac {a x^6+b}{x^{13/4} \left (a x^3+b\right )^{5/4}}dx}{\sqrt [4]{a x^4+b x}}\)

\(\Big \downarrow \) 1813

\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{a x^3+b} \int \left (\frac {a x^{11/4}}{\left (a x^3+b\right )^{5/4}}+\frac {b}{\left (a x^3+b\right )^{5/4} x^{13/4}}\right )dx}{\sqrt [4]{a x^4+b x}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{a x^3+b} \left (\frac {2 \arctan \left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3+b}}\right )}{3 \sqrt [4]{a}}+\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3+b}}\right )}{3 \sqrt [4]{a}}-\frac {16 a x^{3/4}}{9 b \sqrt [4]{a x^3+b}}-\frac {4 x^{3/4}}{3 \sqrt [4]{a x^3+b}}-\frac {4}{9 x^{9/4} \sqrt [4]{a x^3+b}}\right )}{\sqrt [4]{a x^4+b x}}\)

Input:

Int[(b + a*x^6)/(x^3*(b + a*x^3)*(b*x + a*x^4)^(1/4)),x]
 

Output:

(x^(1/4)*(b + a*x^3)^(1/4)*(-4/(9*x^(9/4)*(b + a*x^3)^(1/4)) - (4*x^(3/4)) 
/(3*(b + a*x^3)^(1/4)) - (16*a*x^(3/4))/(9*b*(b + a*x^3)^(1/4)) + (2*ArcTa 
n[(a^(1/4)*x^(3/4))/(b + a*x^3)^(1/4)])/(3*a^(1/4)) + (2*ArcTanh[(a^(1/4)* 
x^(3/4))/(b + a*x^3)^(1/4)])/(3*a^(1/4))))/(b*x + a*x^4)^(1/4)
 

Defintions of rubi rules used

rule 1813

Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^ 
(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^n)^q*(a + c* 
x^(2*n))^p, x], x] /; FreeQ[{a, c, d, e, f, m, q}, x] && EqQ[n2, 2*n] && IG 
tQ[n, 0] && IGtQ[p, 0]
 

rule 2009

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

rule 2467

Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F 
racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p])   Int[x^( 
p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P 
olyQ[Px, x] &&  !IntegerQ[p] &&  !MonomialQ[Px, x] &&  !PolyQ[Fx, x]
 

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.06

method	result	size
pseudoelliptic	\(-\frac {2 \left (\frac {8 x^{3} a^{\frac {5}{4}}}{3}+\frac {b \left (\left (12 x^{3}+4\right ) a^{\frac {1}{4}}+3 {\left (\left (a \,x^{3}+b \right ) x \right )}^{\frac {1}{4}} x^{2} \left (2 \arctan \left (\frac {{\left (\left (a \,x^{3}+b \right ) x \right )}^{\frac {1}{4}}}{x \,a^{\frac {1}{4}}}\right )-\ln \left (\frac {-a^{\frac {1}{4}} x -{\left (\left (a \,x^{3}+b \right ) x \right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x -{\left (\left (a \,x^{3}+b \right ) x \right )}^{\frac {1}{4}}}\right )\right )\right )}{6}\right )}{3 a^{\frac {1}{4}} {\left (\left (a \,x^{3}+b \right ) x \right )}^{\frac {1}{4}} x^{2} b}\)	\(130\)

Input:

int((a*x^6+b)/x^3/(a*x^3+b)/(a*x^4+b*x)^(1/4),x,method=_RETURNVERBOSE)
 

Output:

-2/3/a^(1/4)/((a*x^3+b)*x)^(1/4)*(8/3*x^3*a^(5/4)+1/6*b*((12*x^3+4)*a^(1/4 
)+3*((a*x^3+b)*x)^(1/4)*x^2*(2*arctan(((a*x^3+b)*x)^(1/4)/x/a^(1/4))-ln((- 
a^(1/4)*x-((a*x^3+b)*x)^(1/4))/(a^(1/4)*x-((a*x^3+b)*x)^(1/4))))))/x^2/b
 

Fricas [F(-1)]

Timed out. \[ \int \frac {b+a x^6}{x^3 \left (b+a x^3\right ) \sqrt [4]{b x+a x^4}} \, dx=\text {Timed out} \] Input:

integrate((a*x^6+b)/x^3/(a*x^3+b)/(a*x^4+b*x)^(1/4),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {b+a x^6}{x^3 \left (b+a x^3\right ) \sqrt [4]{b x+a x^4}} \, dx=\int \frac {a x^{6} + b}{x^{3} \sqrt [4]{x \left (a x^{3} + b\right )} \left (a x^{3} + b\right )}\, dx \] Input:

integrate((a*x**6+b)/x**3/(a*x**3+b)/(a*x**4+b*x)**(1/4),x)
 

Output:

Integral((a*x**6 + b)/(x**3*(x*(a*x**3 + b))**(1/4)*(a*x**3 + b)), x)
 

Maxima [F]

\[ \int \frac {b+a x^6}{x^3 \left (b+a x^3\right ) \sqrt [4]{b x+a x^4}} \, dx=\int { \frac {a x^{6} + b}{{\left (a x^{4} + b x\right )}^{\frac {1}{4}} {\left (a x^{3} + b\right )} x^{3}} \,d x } \] Input:

integrate((a*x^6+b)/x^3/(a*x^3+b)/(a*x^4+b*x)^(1/4),x, algorithm="maxima")
 

Output:

integrate((a*x^6 + b)/((a*x^4 + b*x)^(1/4)*(a*x^3 + b)*x^3), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (103) = 206\).

Time = 0.19 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.76 \[ \int \frac {b+a x^6}{x^3 \left (b+a x^3\right ) \sqrt [4]{b x+a x^4}} \, dx=\frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{3 \, a} + \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{3 \, a} - \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{3}}}\right )}{6 \, a} + \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{3}}}\right )}{6 \, a} - \frac {4 \, {\left (a + b\right )}}{3 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}} b} - \frac {4 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {3}{4}}}{9 \, b} \] Input:

integrate((a*x^6+b)/x^3/(a*x^3+b)/(a*x^4+b*x)^(1/4),x, algorithm="giac")
 

Output:

1/3*sqrt(2)*(-a)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b/x 
^3)^(1/4))/(-a)^(1/4))/a + 1/3*sqrt(2)*(-a)^(3/4)*arctan(-1/2*sqrt(2)*(sqr 
t(2)*(-a)^(1/4) - 2*(a + b/x^3)^(1/4))/(-a)^(1/4))/a - 1/6*sqrt(2)*(-a)^(3 
/4)*log(sqrt(2)*(-a)^(1/4)*(a + b/x^3)^(1/4) + sqrt(-a) + sqrt(a + b/x^3)) 
/a + 1/6*sqrt(2)*(-a)^(3/4)*log(-sqrt(2)*(-a)^(1/4)*(a + b/x^3)^(1/4) + sq 
rt(-a) + sqrt(a + b/x^3))/a - 4/3*(a + b)/((a + b/x^3)^(1/4)*b) - 4/9*(a + 
 b/x^3)^(3/4)/b
 

Mupad [F(-1)]

Timed out. \[ \int \frac {b+a x^6}{x^3 \left (b+a x^3\right ) \sqrt [4]{b x+a x^4}} \, dx=\int \frac {a\,x^6+b}{x^3\,{\left (a\,x^4+b\,x\right )}^{1/4}\,\left (a\,x^3+b\right )} \,d x \] Input:

int((b + a*x^6)/(x^3*(b*x + a*x^4)^(1/4)*(b + a*x^3)),x)
 

Output:

int((b + a*x^6)/(x^3*(b*x + a*x^4)^(1/4)*(b + a*x^3)), x)
 

Reduce [F]

\[ \int \frac {b+a x^6}{x^3 \left (b+a x^3\right ) \sqrt [4]{b x+a x^4}} \, dx=\left (\int \frac {x^{3}}{x^{\frac {13}{4}} \left (a \,x^{3}+b \right )^{\frac {1}{4}} a +x^{\frac {1}{4}} \left (a \,x^{3}+b \right )^{\frac {1}{4}} b}d x \right ) a +\left (\int \frac {1}{x^{\frac {25}{4}} \left (a \,x^{3}+b \right )^{\frac {1}{4}} a +x^{\frac {13}{4}} \left (a \,x^{3}+b \right )^{\frac {1}{4}} b}d x \right ) b \] Input:

int((a*x^6+b)/x^3/(a*x^3+b)/(a*x^4+b*x)^(1/4),x)
 

Output:

int(x**3/(x**(1/4)*(a*x**3 + b)**(1/4)*a*x**3 + x**(1/4)*(a*x**3 + b)**(1/ 
4)*b),x)*a + int(1/(x**(1/4)*(a*x**3 + b)**(1/4)*a*x**6 + x**(1/4)*(a*x**3 
 + b)**(1/4)*b*x**3),x)*b