\(\int \frac {-b+a x^2}{(b+x+a x^2) \sqrt [4]{b x^3+a x^5}} \, dx\) [1660]

Optimal result

Integrand size = 35, antiderivative size = 112 \[ \int \frac {-b+a x^2}{\left (b+x+a x^2\right ) \sqrt [4]{b x^3+a x^5}} \, dx=-\sqrt {2} \arctan \left (\frac {\sqrt {2} x \sqrt [4]{b x^3+a x^5}}{-x^2+\sqrt {b x^3+a x^5}}\right )-\sqrt {2} \text {arctanh}\left (\frac {\frac {x^2}{\sqrt {2}}+\frac {\sqrt {b x^3+a x^5}}{\sqrt {2}}}{x \sqrt [4]{b x^3+a x^5}}\right ) \] Output:

-2^(1/2)*arctan(2^(1/2)*x*(a*x^5+b*x^3)^(1/4)/(-x^2+(a*x^5+b*x^3)^(1/2)))- 
2^(1/2)*arctanh((1/2*2^(1/2)*x^2+1/2*(a*x^5+b*x^3)^(1/2)*2^(1/2))/x/(a*x^5 
+b*x^3)^(1/4))
                                                                                    
                                                                                    
 

Mathematica [F]

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

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

Output:

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

Rubi [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 1.42 (sec) , antiderivative size = 469, normalized size of antiderivative = 4.19, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2467, 25, 2035, 7279, 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.

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

\(\Big \downarrow \) 2467

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2035

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

\(\Big \downarrow \) 7279

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {4 x^{3/4} \sqrt [4]{a x^2+b} \left (\frac {\sqrt [4]{x} \sqrt [4]{\frac {a x^2}{b}+1} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},\frac {2 a^2 x^2}{-2 a b-\sqrt {1-4 a b}+1},-\frac {a x^2}{b}\right )}{\sqrt [4]{a x^2+b}}+\frac {\sqrt [4]{x} \sqrt [4]{\frac {a x^2}{b}+1} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},\frac {2 a^2 x^2}{-2 a b+\sqrt {1-4 a b}+1},-\frac {a x^2}{b}\right )}{\sqrt [4]{a x^2+b}}-\frac {a x^{5/4} \left (1-\sqrt {1-4 a b}\right ) \sqrt [4]{\frac {a x^2}{b}+1} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},\frac {2 a^2 x^2}{-2 a b-\sqrt {1-4 a b}+1},-\frac {a x^2}{b}\right )}{5 \left (-2 a b-\sqrt {1-4 a b}+1\right ) \sqrt [4]{a x^2+b}}-\frac {a x^{5/4} \left (\sqrt {1-4 a b}+1\right ) \sqrt [4]{\frac {a x^2}{b}+1} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},\frac {2 a^2 x^2}{-2 a b+\sqrt {1-4 a b}+1},-\frac {a x^2}{b}\right )}{5 \left (-2 a b+\sqrt {1-4 a b}+1\right ) \sqrt [4]{a x^2+b}}-\frac {\sqrt [4]{x} \sqrt [4]{\frac {a x^2}{b}+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-\frac {a x^2}{b}\right )}{\sqrt [4]{a x^2+b}}\right )}{\sqrt [4]{a x^5+b x^3}}\)

Input:

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

Output:

(-4*x^(3/4)*(b + a*x^2)^(1/4)*((x^(1/4)*(1 + (a*x^2)/b)^(1/4)*AppellF1[1/8 
, 1, 1/4, 9/8, (2*a^2*x^2)/(1 - 2*a*b - Sqrt[1 - 4*a*b]), -((a*x^2)/b)])/( 
b + a*x^2)^(1/4) + (x^(1/4)*(1 + (a*x^2)/b)^(1/4)*AppellF1[1/8, 1, 1/4, 9/ 
8, (2*a^2*x^2)/(1 - 2*a*b + Sqrt[1 - 4*a*b]), -((a*x^2)/b)])/(b + a*x^2)^( 
1/4) - (a*(1 - Sqrt[1 - 4*a*b])*x^(5/4)*(1 + (a*x^2)/b)^(1/4)*AppellF1[5/8 
, 1, 1/4, 13/8, (2*a^2*x^2)/(1 - 2*a*b - Sqrt[1 - 4*a*b]), -((a*x^2)/b)])/ 
(5*(1 - 2*a*b - Sqrt[1 - 4*a*b])*(b + a*x^2)^(1/4)) - (a*(1 + Sqrt[1 - 4*a 
*b])*x^(5/4)*(1 + (a*x^2)/b)^(1/4)*AppellF1[5/8, 1, 1/4, 13/8, (2*a^2*x^2) 
/(1 - 2*a*b + Sqrt[1 - 4*a*b]), -((a*x^2)/b)])/(5*(1 - 2*a*b + Sqrt[1 - 4* 
a*b])*(b + a*x^2)^(1/4)) - (x^(1/4)*(1 + (a*x^2)/b)^(1/4)*Hypergeometric2F 
1[1/8, 1/4, 9/8, -((a*x^2)/b)])/(b + a*x^2)^(1/4)))/(b*x^3 + a*x^5)^(1/4)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009

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

rule 2035

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

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]
 

rule 7279

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ 
{v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su 
mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
 

Maple [A] (verified)

Time = 1.22 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.21

method	result	size
pseudoelliptic	\(\frac {\sqrt {2}\, \left (\ln \left (\frac {-\left (x^{3} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}} \sqrt {2}\, x +x^{2}+\sqrt {x^{3} \left (a \,x^{2}+b \right )}}{\left (x^{3} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}} \sqrt {2}\, x +x^{2}+\sqrt {x^{3} \left (a \,x^{2}+b \right )}}\right )+2 \arctan \left (\frac {\left (x^{3} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}} \sqrt {2}+x}{x}\right )+2 \arctan \left (\frac {\left (x^{3} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}} \sqrt {2}-x}{x}\right )\right )}{2}\)	\(136\)

Input:

int((a*x^2-b)/(a*x^2+b+x)/(a*x^5+b*x^3)^(1/4),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [F(-1)]

Timed out. \[ \int \frac {-b+a x^2}{\left (b+x+a x^2\right ) \sqrt [4]{b x^3+a x^5}} \, dx=\text {Timed out} \] Input:

integrate((a*x^2-b)/(a*x^2+b+x)/(a*x^5+b*x^3)^(1/4),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

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

integrate((a*x**2-b)/(a*x**2+b+x)/(a*x**5+b*x**3)**(1/4),x)
 

Output:

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

Maxima [F]

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

integrate((a*x^2-b)/(a*x^2+b+x)/(a*x^5+b*x^3)^(1/4),x, algorithm="maxima")
 

Output:

integrate((a*x^2 - b)/((a*x^5 + b*x^3)^(1/4)*(a*x^2 + b + x)), x)
 

Giac [F]

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

integrate((a*x^2-b)/(a*x^2+b+x)/(a*x^5+b*x^3)^(1/4),x, algorithm="giac")
 

Output:

integrate((a*x^2 - b)/((a*x^5 + b*x^3)^(1/4)*(a*x^2 + b + x)), x)
 

Mupad [F(-1)]

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

int(-(b - a*x^2)/((a*x^5 + b*x^3)^(1/4)*(b + x + a*x^2)),x)
 

Output:

int(-(b - a*x^2)/((a*x^5 + b*x^3)^(1/4)*(b + x + a*x^2)), x)
 

Reduce [F]

\[ \int \frac {-b+a x^2}{\left (b+x+a x^2\right ) \sqrt [4]{b x^3+a x^5}} \, dx=\left (\int \frac {x^{2}}{x^{\frac {11}{4}} \left (a \,x^{2}+b \right )^{\frac {1}{4}} a +x^{\frac {3}{4}} \left (a \,x^{2}+b \right )^{\frac {1}{4}} b +x^{\frac {7}{4}} \left (a \,x^{2}+b \right )^{\frac {1}{4}}}d x \right ) a -\left (\int \frac {1}{x^{\frac {11}{4}} \left (a \,x^{2}+b \right )^{\frac {1}{4}} a +x^{\frac {3}{4}} \left (a \,x^{2}+b \right )^{\frac {1}{4}} b +x^{\frac {7}{4}} \left (a \,x^{2}+b \right )^{\frac {1}{4}}}d x \right ) b \] Input:

int((a*x^2-b)/(a*x^2+b+x)/(a*x^5+b*x^3)^(1/4),x)
 

Output:

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