\(\int \frac {b+a x^2}{(b-a x^2) \sqrt [4]{b x^3+a x^5}} \, dx\) [2807]

Optimal result

Integrand size = 33, antiderivative size = 275 \[ \int \frac {b+a x^2}{\left (b-a x^2\right ) \sqrt [4]{b x^3+a x^5}} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [8]{a} \sqrt [8]{b} x}{\sqrt [4]{b x^3+a x^5}}\right )}{\sqrt [4]{2} \sqrt [8]{a} \sqrt [8]{b}}-\frac {\arctan \left (\frac {2^{3/4} \sqrt [8]{a} \sqrt [8]{b} x \sqrt [4]{b x^3+a x^5}}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2-\sqrt {b x^3+a x^5}}\right )}{2^{3/4} \sqrt [8]{a} \sqrt [8]{b}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [8]{a} \sqrt [8]{b} x}{\sqrt [4]{b x^3+a x^5}}\right )}{\sqrt [4]{2} \sqrt [8]{a} \sqrt [8]{b}}+\frac {\text {arctanh}\left (\frac {\frac {\sqrt [8]{a} \sqrt [8]{b} x^2}{\sqrt [4]{2}}+\frac {\sqrt {b x^3+a x^5}}{2^{3/4} \sqrt [8]{a} \sqrt [8]{b}}}{x \sqrt [4]{b x^3+a x^5}}\right )}{2^{3/4} \sqrt [8]{a} \sqrt [8]{b}} \] Output:

1/2*arctan(2^(1/4)*a^(1/8)*b^(1/8)*x/(a*x^5+b*x^3)^(1/4))*2^(3/4)/a^(1/8)/ 
b^(1/8)-1/2*arctan(2^(3/4)*a^(1/8)*b^(1/8)*x*(a*x^5+b*x^3)^(1/4)/(2^(1/2)* 
a^(1/4)*b^(1/4)*x^2-(a*x^5+b*x^3)^(1/2)))*2^(1/4)/a^(1/8)/b^(1/8)+1/2*arct 
anh(2^(1/4)*a^(1/8)*b^(1/8)*x/(a*x^5+b*x^3)^(1/4))*2^(3/4)/a^(1/8)/b^(1/8) 
+1/2*arctanh((1/2*a^(1/8)*b^(1/8)*x^2*2^(3/4)+1/2*(a*x^5+b*x^3)^(1/2)*2^(1 
/4)/a^(1/8)/b^(1/8))/x/(a*x^5+b*x^3)^(1/4))*2^(1/4)/a^(1/8)/b^(1/8)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 2.12 (sec) , antiderivative size = 264, normalized size of antiderivative = 0.96 \[ \int \frac {b+a x^2}{\left (b-a x^2\right ) \sqrt [4]{b x^3+a x^5}} \, dx=\frac {x^{3/4} \sqrt [4]{b+a x^2} \left (\sqrt {2} \arctan \left (\frac {\sqrt [4]{2} \sqrt [8]{a} \sqrt [8]{b} \sqrt [4]{x}}{\sqrt [4]{b+a x^2}}\right )-\arctan \left (\frac {2^{3/4} \sqrt [8]{a} \sqrt [8]{b} \sqrt [4]{x} \sqrt [4]{b+a x^2}}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}-\sqrt {b+a x^2}}\right )+\sqrt {2} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [8]{a} \sqrt [8]{b} \sqrt [4]{x}}{\sqrt [4]{b+a x^2}}\right )+\text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b+a x^2}}{2^{3/4} \sqrt [8]{a} \sqrt [8]{b} \sqrt [4]{x} \sqrt [4]{b+a x^2}}\right )\right )}{2^{3/4} \sqrt [8]{a} \sqrt [8]{b} \sqrt [4]{x^3 \left (b+a x^2\right )}} \] Input:

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

Output:

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

Rubi [C] (warning: unable to verify)

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

Time = 0.38 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.25, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {2467, 368, 937, 936}

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.

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

\(\Big \downarrow \) 2467

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

\(\Big \downarrow \) 368

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

\(\Big \downarrow \) 937

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

\(\Big \downarrow \) 936

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 368

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) 
, x_Symbol] :> With[{k = Denominator[m]}, Simp[k/e   Subst[Int[x^(k*(m + 1) 
 - 1)*(a + b*(x^(k*2)/e^2))^p*(c + d*(x^(k*2)/e^2))^q, x], x, (e*x)^(1/k)], 
 x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && FractionQ[m 
] && IntegerQ[p]
 

rule 936

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[a^p*c^q*x*AppellF1[1/n, -p, -q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c) 
], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] 
 && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 937

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]) 
  Int[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, p, q 
}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])
 

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 [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.15

method	result	size
pseudoelliptic	\(-\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-4 a b \right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\left (x^{3} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}}\right )}{2}\)	\(42\)

Input:

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

Output:

-1/2*sum(ln((-_R*x+(x^3*(a*x^2+b))^(1/4))/x)/_R,_R=RootOf(_Z^8-4*a*b))
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate((a*x^2+b)/(-a*x^2+b)/(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)
 

Giac [F]

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

integrate((a*x^2+b)/(-a*x^2+b)/(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)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int((a*x^2+b)/(-a*x^2+b)/(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)*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)*b)