Integrand size = 40, antiderivative size = 238 \[ \int \frac {\sqrt [4]{b x^2+a x^4} \left (-b-a x^4+x^8\right )}{-b+a x^4} \, dx=\frac {\sqrt [4]{b x^2+a x^4} \left (-96 a^3 x+96 a b x-7 b^2 x+4 a b x^3+32 a^2 x^5\right )}{192 a^3}+\frac {\left (32 a^3 b-32 a b^2-7 b^3\right ) \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{128 a^{15/4}}+\frac {\left (-32 a^3 b+32 a b^2+7 b^3\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{128 a^{15/4}}+\frac {\left (-2 a^2 b+b^2\right ) \text {RootSum}\left [a^2-a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x) \text {$\#$1}+\log \left (\sqrt [4]{b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}}{-a+\text {$\#$1}^4}\&\right ]}{4 a^2} \] Output:
Unintegrable
Time = 0.00 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.09 \[ \int \frac {\sqrt [4]{b x^2+a x^4} \left (-b-a x^4+x^8\right )}{-b+a x^4} \, dx=\frac {x^{3/2} \left (b+a x^2\right )^{3/4} \left (2 a^{3/4} x^{3/2} \sqrt [4]{b+a x^2} \left (-96 a^3-7 b^2+32 a^2 x^4+4 a b \left (24+x^2\right )\right )-3 b \left (-32 a^3+32 a b+7 b^2\right ) \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )+3 b \left (-32 a^3+32 a b+7 b^2\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )-96 a^{7/4} \left (2 a^2-b\right ) b \text {RootSum}\left [a^2-a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log \left (\sqrt {x}\right ) \text {$\#$1}+\log \left (\sqrt [4]{b+a x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}}{-a+\text {$\#$1}^4}\&\right ]\right )}{384 a^{15/4} \left (x^2 \left (b+a x^2\right )\right )^{3/4}} \] Input:
Integrate[((b*x^2 + a*x^4)^(1/4)*(-b - a*x^4 + x^8))/(-b + a*x^4),x]
Output:
(x^(3/2)*(b + a*x^2)^(3/4)*(2*a^(3/4)*x^(3/2)*(b + a*x^2)^(1/4)*(-96*a^3 - 7*b^2 + 32*a^2*x^4 + 4*a*b*(24 + x^2)) - 3*b*(-32*a^3 + 32*a*b + 7*b^2)*A rcTan[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)] + 3*b*(-32*a^3 + 32*a*b + 7*b^2 )*ArcTanh[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)] - 96*a^(7/4)*(2*a^2 - b)*b* RootSum[a^2 - a*b - 2*a*#1^4 + #1^8 & , (-(Log[Sqrt[x]]*#1) + Log[(b + a*x ^2)^(1/4) - Sqrt[x]*#1]*#1)/(-a + #1^4) & ]))/(384*a^(15/4)*(x^2*(b + a*x^ 2))^(3/4))
Time = 1.39 (sec) , antiderivative size = 448, normalized size of antiderivative = 1.88, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2467, 2035, 7276, 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.
\(\displaystyle \int \frac {\sqrt [4]{a x^4+b x^2} \left (-a x^4-b+x^8\right )}{a x^4-b} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt [4]{a x^4+b x^2} \int \frac {\sqrt {x} \sqrt [4]{a x^2+b} \left (-x^8+a x^4+b\right )}{b-a x^4}dx}{\sqrt {x} \sqrt [4]{a x^2+b}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle \frac {2 \sqrt [4]{a x^4+b x^2} \int \frac {x \sqrt [4]{a x^2+b} \left (-x^8+a x^4+b\right )}{b-a x^4}d\sqrt {x}}{\sqrt {x} \sqrt [4]{a x^2+b}}\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \frac {2 \sqrt [4]{a x^4+b x^2} \int \left (\frac {\sqrt [4]{a x^2+b} x^5}{a}-\left (1-\frac {b}{a^2}\right ) \sqrt [4]{a x^2+b} x-\frac {\left (b^2-2 a^2 b\right ) \sqrt [4]{a x^2+b} x}{a^2 \left (b-a x^4\right )}\right )d\sqrt {x}}{\sqrt {x} \sqrt [4]{a x^2+b}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \sqrt [4]{a x^4+b x^2} \left (-\frac {7 b^3 \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{256 a^{15/4}}+\frac {7 b^3 \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{256 a^{15/4}}-\frac {7 b^2 x^{3/2} \sqrt [4]{a x^2+b}}{384 a^3}+\frac {x^{3/2} \left (2 a^2-b\right ) \sqrt [4]{a x^2+b} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},-\frac {\sqrt {a} x^2}{\sqrt {b}},-\frac {a x^2}{b}\right )}{6 a^2 \sqrt [4]{\frac {a x^2}{b}+1}}+\frac {x^{3/2} \left (2 a^2-b\right ) \sqrt [4]{a x^2+b} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},\frac {\sqrt {a} x^2}{\sqrt {b}},-\frac {a x^2}{b}\right )}{6 a^2 \sqrt [4]{\frac {a x^2}{b}+1}}+\frac {b x^{7/2} \sqrt [4]{a x^2+b}}{96 a^2}-\frac {1}{4} x^{3/2} \left (1-\frac {b}{a^2}\right ) \sqrt [4]{a x^2+b}+\frac {b \left (1-\frac {b}{a^2}\right ) \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{8 a^{3/4}}-\frac {b \left (1-\frac {b}{a^2}\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{8 a^{3/4}}+\frac {x^{11/2} \sqrt [4]{a x^2+b}}{12 a}\right )}{\sqrt {x} \sqrt [4]{a x^2+b}}\) |
Input:
Int[((b*x^2 + a*x^4)^(1/4)*(-b - a*x^4 + x^8))/(-b + a*x^4),x]
Output:
(2*(b*x^2 + a*x^4)^(1/4)*((-7*b^2*x^(3/2)*(b + a*x^2)^(1/4))/(384*a^3) - ( (1 - b/a^2)*x^(3/2)*(b + a*x^2)^(1/4))/4 + (b*x^(7/2)*(b + a*x^2)^(1/4))/( 96*a^2) + (x^(11/2)*(b + a*x^2)^(1/4))/(12*a) + ((2*a^2 - b)*x^(3/2)*(b + a*x^2)^(1/4)*AppellF1[3/4, 1, -1/4, 7/4, -((Sqrt[a]*x^2)/Sqrt[b]), -((a*x^ 2)/b)])/(6*a^2*(1 + (a*x^2)/b)^(1/4)) + ((2*a^2 - b)*x^(3/2)*(b + a*x^2)^( 1/4)*AppellF1[3/4, 1, -1/4, 7/4, (Sqrt[a]*x^2)/Sqrt[b], -((a*x^2)/b)])/(6* a^2*(1 + (a*x^2)/b)^(1/4)) - (7*b^3*ArcTan[(a^(1/4)*Sqrt[x])/(b + a*x^2)^( 1/4)])/(256*a^(15/4)) + (b*(1 - b/a^2)*ArcTan[(a^(1/4)*Sqrt[x])/(b + a*x^2 )^(1/4)])/(8*a^(3/4)) + (7*b^3*ArcTanh[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4) ])/(256*a^(15/4)) - (b*(1 - b/a^2)*ArcTanh[(a^(1/4)*Sqrt[x])/(b + a*x^2)^( 1/4)])/(8*a^(3/4))))/(Sqrt[x]*(b + a*x^2)^(1/4))
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]
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]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 0.00 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.93
method | result | size |
pseudoelliptic | \(-\frac {\left (-a^{\frac {7}{4}} b^{2}+2 a^{\frac {15}{4}} b \right ) \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-2 a \,\textit {\_Z}^{4}+a^{2}-a b \right )}{\sum }\frac {\textit {\_R} \ln \left (\frac {-\textit {\_R} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}^{4}-a}\right )+\frac {\left (a^{3}-a b -\frac {7}{32} b^{2}\right ) b \ln \left (\frac {a^{\frac {1}{4}} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{-a^{\frac {1}{4}} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}\right )}{2}+\left (a^{3}-a b -\frac {7}{32} b^{2}\right ) b \arctan \left (\frac {\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )-\frac {2 \left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}} x \left (b \left (\frac {x^{2}}{8}+3\right ) a^{\frac {7}{4}}+a^{\frac {11}{4}} x^{4}-\frac {7 b^{2} a^{\frac {3}{4}}}{32}-3 a^{\frac {15}{4}}\right )}{3}}{4 a^{\frac {15}{4}}}\) | \(222\) |
Input:
int((a*x^4+b*x^2)^(1/4)*(x^8-a*x^4-b)/(a*x^4-b),x,method=_RETURNVERBOSE)
Output:
-1/4*((-a^(7/4)*b^2+2*a^(15/4)*b)*sum(_R*ln((-_R*x+(x^2*(a*x^2+b))^(1/4))/ x)/(_R^4-a),_R=RootOf(_Z^8-2*_Z^4*a+a^2-a*b))+1/2*(a^3-a*b-7/32*b^2)*b*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)))+(a^3 -a*b-7/32*b^2)*b*arctan(1/a^(1/4)/x*(x^2*(a*x^2+b))^(1/4))-2/3*(x^2*(a*x^2 +b))^(1/4)*x*(b*(1/8*x^2+3)*a^(7/4)+a^(11/4)*x^4-7/32*b^2*a^(3/4)-3*a^(15/ 4)))/a^(15/4)
Timed out. \[ \int \frac {\sqrt [4]{b x^2+a x^4} \left (-b-a x^4+x^8\right )}{-b+a x^4} \, dx=\text {Timed out} \] Input:
integrate((a*x^4+b*x^2)^(1/4)*(x^8-a*x^4-b)/(a*x^4-b),x, algorithm="fricas ")
Output:
Timed out
Not integrable
Time = 52.51 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.13 \[ \int \frac {\sqrt [4]{b x^2+a x^4} \left (-b-a x^4+x^8\right )}{-b+a x^4} \, dx=\int \frac {\sqrt [4]{x^{2} \left (a x^{2} + b\right )} \left (- a x^{4} - b + x^{8}\right )}{a x^{4} - b}\, dx \] Input:
integrate((a*x**4+b*x**2)**(1/4)*(x**8-a*x**4-b)/(a*x**4-b),x)
Output:
Integral((x**2*(a*x**2 + b))**(1/4)*(-a*x**4 - b + x**8)/(a*x**4 - b), x)
Not integrable
Time = 0.09 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.17 \[ \int \frac {\sqrt [4]{b x^2+a x^4} \left (-b-a x^4+x^8\right )}{-b+a x^4} \, dx=\int { \frac {{\left (x^{8} - a x^{4} - b\right )} {\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}}}{a x^{4} - b} \,d x } \] Input:
integrate((a*x^4+b*x^2)^(1/4)*(x^8-a*x^4-b)/(a*x^4-b),x, algorithm="maxima ")
Output:
integrate((x^8 - a*x^4 - b)*(a*x^4 + b*x^2)^(1/4)/(a*x^4 - b), x)
Timed out. \[ \int \frac {\sqrt [4]{b x^2+a x^4} \left (-b-a x^4+x^8\right )}{-b+a x^4} \, dx=\text {Timed out} \] Input:
integrate((a*x^4+b*x^2)^(1/4)*(x^8-a*x^4-b)/(a*x^4-b),x, algorithm="giac")
Output:
Timed out
Not integrable
Time = 0.00 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.16 \[ \int \frac {\sqrt [4]{b x^2+a x^4} \left (-b-a x^4+x^8\right )}{-b+a x^4} \, dx=\int \frac {{\left (a\,x^4+b\,x^2\right )}^{1/4}\,\left (-x^8+a\,x^4+b\right )}{b-a\,x^4} \,d x \] Input:
int(((a*x^4 + b*x^2)^(1/4)*(b + a*x^4 - x^8))/(b - a*x^4),x)
Output:
int(((a*x^4 + b*x^2)^(1/4)*(b + a*x^4 - x^8))/(b - a*x^4), x)
Not integrable
Time = 22.52 (sec) , antiderivative size = 497, normalized size of antiderivative = 2.09 \[ \int \frac {\sqrt [4]{b x^2+a x^4} \left (-b-a x^4+x^8\right )}{-b+a x^4} \, dx=\frac {-192 \sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}} a^{3} x +64 \sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}} a^{2} x^{5}+8 \sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}} a b \,x^{3}+192 \sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}} a b x -14 \sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}} b^{2} x -96 \left (\int \frac {\sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}} x^{4}}{a^{2} x^{6}+a b \,x^{4}-a b \,x^{2}-b^{2}}d x \right ) a^{4} b +96 \left (\int \frac {\sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}} x^{4}}{a^{2} x^{6}+a b \,x^{4}-a b \,x^{2}-b^{2}}d x \right ) a^{2} b^{2}+21 \left (\int \frac {\sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}} x^{4}}{a^{2} x^{6}+a b \,x^{4}-a b \,x^{2}-b^{2}}d x \right ) a \,b^{3}-768 \left (\int \frac {\sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}} x^{2}}{a^{2} x^{6}+a b \,x^{4}-a b \,x^{2}-b^{2}}d x \right ) a^{4} b +384 \left (\int \frac {\sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}} x^{2}}{a^{2} x^{6}+a b \,x^{4}-a b \,x^{2}-b^{2}}d x \right ) a^{2} b^{2}-672 \left (\int \frac {\sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}}}{a^{2} x^{6}+a b \,x^{4}-a b \,x^{2}-b^{2}}d x \right ) a^{3} b^{2}+288 \left (\int \frac {\sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}}}{a^{2} x^{6}+a b \,x^{4}-a b \,x^{2}-b^{2}}d x \right ) a \,b^{3}-21 \left (\int \frac {\sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}}}{a^{2} x^{6}+a b \,x^{4}-a b \,x^{2}-b^{2}}d x \right ) b^{4}}{384 a^{3}} \] Input:
int((a*x^4+b*x^2)^(1/4)*(x^8-a*x^4-b)/(a*x^4-b),x)
Output:
( - 192*sqrt(x)*(a*x**2 + b)**(1/4)*a**3*x + 64*sqrt(x)*(a*x**2 + b)**(1/4 )*a**2*x**5 + 8*sqrt(x)*(a*x**2 + b)**(1/4)*a*b*x**3 + 192*sqrt(x)*(a*x**2 + b)**(1/4)*a*b*x - 14*sqrt(x)*(a*x**2 + b)**(1/4)*b**2*x - 96*int((sqrt( x)*(a*x**2 + b)**(1/4)*x**4)/(a**2*x**6 + a*b*x**4 - a*b*x**2 - b**2),x)*a **4*b + 96*int((sqrt(x)*(a*x**2 + b)**(1/4)*x**4)/(a**2*x**6 + a*b*x**4 - a*b*x**2 - b**2),x)*a**2*b**2 + 21*int((sqrt(x)*(a*x**2 + b)**(1/4)*x**4)/ (a**2*x**6 + a*b*x**4 - a*b*x**2 - b**2),x)*a*b**3 - 768*int((sqrt(x)*(a*x **2 + b)**(1/4)*x**2)/(a**2*x**6 + a*b*x**4 - a*b*x**2 - b**2),x)*a**4*b + 384*int((sqrt(x)*(a*x**2 + b)**(1/4)*x**2)/(a**2*x**6 + a*b*x**4 - a*b*x* *2 - b**2),x)*a**2*b**2 - 672*int((sqrt(x)*(a*x**2 + b)**(1/4))/(a**2*x**6 + a*b*x**4 - a*b*x**2 - b**2),x)*a**3*b**2 + 288*int((sqrt(x)*(a*x**2 + b )**(1/4))/(a**2*x**6 + a*b*x**4 - a*b*x**2 - b**2),x)*a*b**3 - 21*int((sqr t(x)*(a*x**2 + b)**(1/4))/(a**2*x**6 + a*b*x**4 - a*b*x**2 - b**2),x)*b**4 )/(384*a**3)