\(\int \frac {(b+a x^2) \sqrt [4]{-b x^2+a x^4}}{-b+a x^2} \, dx\) [1240]

Optimal result

Integrand size = 35, antiderivative size = 90 \[ \int \frac {\left (b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{-b+a x^2} \, dx=\frac {1}{2} x \sqrt [4]{-b x^2+a x^4}-\frac {7 b \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^2+a x^4}}\right )}{4 a^{3/4}}+\frac {7 b \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^2+a x^4}}\right )}{4 a^{3/4}} \] Output:

1/2*x*(a*x^4-b*x^2)^(1/4)-7/4*b*arctan(a^(1/4)*x/(a*x^4-b*x^2)^(1/4))/a^(3 
/4)+7/4*b*arctanh(a^(1/4)*x/(a*x^4-b*x^2)^(1/4))/a^(3/4)
 

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.39 \[ \int \frac {\left (b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{-b+a x^2} \, dx=\frac {\sqrt [4]{-b x^2+a x^4} \left (2 a^{3/4} x^{3/2} \sqrt [4]{-b+a x^2}-7 b \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )+7 b \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )\right )}{4 a^{3/4} \sqrt {x} \sqrt [4]{-b+a x^2}} \] Input:

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

Output:

((-(b*x^2) + a*x^4)^(1/4)*(2*a^(3/4)*x^(3/2)*(-b + a*x^2)^(1/4) - 7*b*ArcT 
an[(a^(1/4)*Sqrt[x])/(-b + a*x^2)^(1/4)] + 7*b*ArcTanh[(a^(1/4)*Sqrt[x])/( 
-b + a*x^2)^(1/4)]))/(4*a^(3/4)*Sqrt[x]*(-b + a*x^2)^(1/4))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.47, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2467, 363, 266, 854, 827, 216, 219}

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.

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

\(\Big \downarrow \) 2467

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

\(\Big \downarrow \) 363

\(\displaystyle \frac {\sqrt [4]{a x^4-b x^2} \left (\frac {7}{4} b \int \frac {\sqrt {x}}{\left (a x^2-b\right )^{3/4}}dx+\frac {1}{2} x^{3/2} \sqrt [4]{a x^2-b}\right )}{\sqrt {x} \sqrt [4]{a x^2-b}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {\sqrt [4]{a x^4-b x^2} \left (\frac {7}{2} b \int \frac {x}{\left (a x^2-b\right )^{3/4}}d\sqrt {x}+\frac {1}{2} x^{3/2} \sqrt [4]{a x^2-b}\right )}{\sqrt {x} \sqrt [4]{a x^2-b}}\)

\(\Big \downarrow \) 854

\(\displaystyle \frac {\sqrt [4]{a x^4-b x^2} \left (\frac {7}{2} b \int \frac {x}{1-a x^2}d\frac {\sqrt {x}}{\sqrt [4]{a x^2-b}}+\frac {1}{2} x^{3/2} \sqrt [4]{a x^2-b}\right )}{\sqrt {x} \sqrt [4]{a x^2-b}}\)

\(\Big \downarrow \) 827

\(\displaystyle \frac {\sqrt [4]{a x^4-b x^2} \left (\frac {7}{2} b \left (\frac {\int \frac {1}{1-\sqrt {a} x}d\frac {\sqrt {x}}{\sqrt [4]{a x^2-b}}}{2 \sqrt {a}}-\frac {\int \frac {1}{\sqrt {a} x+1}d\frac {\sqrt {x}}{\sqrt [4]{a x^2-b}}}{2 \sqrt {a}}\right )+\frac {1}{2} x^{3/2} \sqrt [4]{a x^2-b}\right )}{\sqrt {x} \sqrt [4]{a x^2-b}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\sqrt [4]{a x^4-b x^2} \left (\frac {7}{2} b \left (\frac {\int \frac {1}{1-\sqrt {a} x}d\frac {\sqrt {x}}{\sqrt [4]{a x^2-b}}}{2 \sqrt {a}}-\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2-b}}\right )}{2 a^{3/4}}\right )+\frac {1}{2} x^{3/2} \sqrt [4]{a x^2-b}\right )}{\sqrt {x} \sqrt [4]{a x^2-b}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sqrt [4]{a x^4-b x^2} \left (\frac {7}{2} b \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2-b}}\right )}{2 a^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2-b}}\right )}{2 a^{3/4}}\right )+\frac {1}{2} x^{3/2} \sqrt [4]{a x^2-b}\right )}{\sqrt {x} \sqrt [4]{a x^2-b}}\)

Input:

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

Output:

((-(b*x^2) + a*x^4)^(1/4)*((x^(3/2)*(-b + a*x^2)^(1/4))/2 + (7*b*(-1/2*Arc 
Tan[(a^(1/4)*Sqrt[x])/(-b + a*x^2)^(1/4)]/a^(3/4) + ArcTanh[(a^(1/4)*Sqrt[ 
x])/(-b + a*x^2)^(1/4)]/(2*a^(3/4))))/2))/(Sqrt[x]*(-b + a*x^2)^(1/4))
 

Defintions of rubi rules used

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 363

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), 
 x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3))   Int[(e*x)^ 
m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d 
, 0] && NeQ[m + 2*p + 3, 0]
 

rule 827

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ 
[a/b, 0]
 

rule 854

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + (m + 
 1)/n)   Subst[Int[x^m/(1 - b*x^n)^(p + (m + 1)/n + 1), x], x, x/(a + b*x^n 
)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, - 
2^(-1)] && IntegersQ[m, p + (m + 1)/n]
 

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 = 0.44 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.16

method	result	size
pseudoelliptic	\(\frac {4 \left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}} x \,a^{\frac {3}{4}}+7 \ln \left (\frac {a^{\frac {1}{4}} x +\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}{-a^{\frac {1}{4}} x +\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}\right ) b +14 \arctan \left (\frac {\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) b}{8 a^{\frac {3}{4}}}\)	\(104\)

Input:

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

Output:

1/8*(4*(x^2*(a*x^2-b))^(1/4)*x*a^(3/4)+7*ln((a^(1/4)*x+(x^2*(a*x^2-b))^(1/ 
4))/(-a^(1/4)*x+(x^2*(a*x^2-b))^(1/4)))*b+14*arctan(1/a^(1/4)/x*(x^2*(a*x^ 
2-b))^(1/4))*b)/a^(3/4)
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 222 vs. \(2 (70) = 140\).

Time = 0.35 (sec) , antiderivative size = 222, normalized size of antiderivative = 2.47 \[ \int \frac {\left (b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{-b+a x^2} \, dx=\frac {8 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} b x^{2} + \frac {14 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a} + \frac {14 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a} + \frac {7 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{2}}}\right )}{a} + \frac {7 \, \sqrt {2} b^{2} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{2}}}\right )}{\left (-a\right )^{\frac {3}{4}}}}{16 \, b} \] Input:

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

Output:

1/16*(8*(a - b/x^2)^(1/4)*b*x^2 + 14*sqrt(2)*(-a)^(1/4)*b^2*arctan(1/2*sqr 
t(2)*(sqrt(2)*(-a)^(1/4) + 2*(a - b/x^2)^(1/4))/(-a)^(1/4))/a + 14*sqrt(2) 
*(-a)^(1/4)*b^2*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a - b/x^2)^(1 
/4))/(-a)^(1/4))/a + 7*sqrt(2)*(-a)^(1/4)*b^2*log(sqrt(2)*(-a)^(1/4)*(a - 
b/x^2)^(1/4) + sqrt(-a) + sqrt(a - b/x^2))/a + 7*sqrt(2)*b^2*log(-sqrt(2)* 
(-a)^(1/4)*(a - b/x^2)^(1/4) + sqrt(-a) + sqrt(a - b/x^2))/(-a)^(3/4))/b
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*sqrt(x)*(a*x**2 - b)**(1/4)*x + 7*int((sqrt(x)*(a*x**2 - b)**(1/4))/(a* 
x**2 - b),x)*b)/4