\(\int \frac {(-b+a x^2) \sqrt [4]{-b x^2+a x^4}}{b+a x^2} \, dx\) [2286]

Optimal result

Integrand size = 35, antiderivative size = 174 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b+a x^2} \, dx=\frac {1}{2} x \sqrt [4]{-b x^2+a x^4}+\frac {9 b \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^2+a x^4}}\right )}{4 a^{3/4}}-\frac {2 \sqrt [4]{2} b \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b x^2+a x^4}}\right )}{a^{3/4}}-\frac {9 b \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^2+a x^4}}\right )}{4 a^{3/4}}+\frac {2 \sqrt [4]{2} b \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b x^2+a x^4}}\right )}{a^{3/4}} \]

[Out]

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

Rubi [C] (verified)

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

Time = 0.15 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.37, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {2081, 477, 525, 524} \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b+a x^2} \, dx=-\frac {2 x \sqrt [4]{a x^4-b x^2} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {5}{4},\frac {7}{4},-\frac {a x^2}{b},\frac {a x^2}{b}\right )}{3 \sqrt [4]{1-\frac {a x^2}{b}}} \]

[In]

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

[Out]

(-2*x*(-(b*x^2) + a*x^4)^(1/4)*AppellF1[3/4, 1, -5/4, 7/4, -((a*x^2)/b), (a*x^2)/b])/(3*(1 - (a*x^2)/b)^(1/4))

Rule 477

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/e^n))^p*(c + d*(x^(k*n)/e^n))^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 524

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

Rule 525

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

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [4]{-b x^2+a x^4} \int \frac {\sqrt {x} \left (-b+a x^2\right )^{5/4}}{b+a x^2} \, dx}{\sqrt {x} \sqrt [4]{-b+a x^2}} \\ & = \frac {\left (2 \sqrt [4]{-b x^2+a x^4}\right ) \text {Subst}\left (\int \frac {x^2 \left (-b+a x^4\right )^{5/4}}{b+a x^4} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{-b+a x^2}} \\ & = -\frac {\left (2 b \sqrt [4]{-b x^2+a x^4}\right ) \text {Subst}\left (\int \frac {x^2 \left (1-\frac {a x^4}{b}\right )^{5/4}}{b+a x^4} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{1-\frac {a x^2}{b}}} \\ & = -\frac {2 x \sqrt [4]{-b x^2+a x^4} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {5}{4},\frac {7}{4},-\frac {a x^2}{b},\frac {a x^2}{b}\right )}{3 \sqrt [4]{1-\frac {a x^2}{b}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.16 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b+a x^2} \, dx=\frac {\sqrt [4]{-b x^2+a x^4} \left (2 a^{3/4} x^{3/2} \sqrt [4]{-b+a x^2}+9 b \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )-8 \sqrt [4]{2} b \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )-9 b \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )+8 \sqrt [4]{2} b \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )\right )}{4 a^{3/4} \sqrt {x} \sqrt [4]{-b+a x^2}} \]

[In]

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

[Out]

((-(b*x^2) + a*x^4)^(1/4)*(2*a^(3/4)*x^(3/2)*(-b + a*x^2)^(1/4) + 9*b*ArcTan[(a^(1/4)*Sqrt[x])/(-b + a*x^2)^(1
/4)] - 8*2^(1/4)*b*ArcTan[(2^(1/4)*a^(1/4)*Sqrt[x])/(-b + a*x^2)^(1/4)] - 9*b*ArcTanh[(a^(1/4)*Sqrt[x])/(-b +
a*x^2)^(1/4)] + 8*2^(1/4)*b*ArcTanh[(2^(1/4)*a^(1/4)*Sqrt[x])/(-b + a*x^2)^(1/4)]))/(4*a^(3/4)*Sqrt[x]*(-b + a
*x^2)^(1/4))

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.14

method	result	size
pseudoelliptic	\(\frac {\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}} x \,a^{\frac {3}{4}}+4 \,2^{\frac {1}{4}} \arctan \left (\frac {\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}} 2^{\frac {3}{4}}}{2 x \,a^{\frac {1}{4}}}\right ) b +2 \,2^{\frac {1}{4}} \ln \left (\frac {x 2^{\frac {1}{4}} a^{\frac {1}{4}}+\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}{-x 2^{\frac {1}{4}} a^{\frac {1}{4}}+\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}\right ) b -\frac {9 b \arctan \left (\frac {\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{2}-\frac {9 b \ln \left (\frac {-a^{\frac {1}{4}} x -\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}\right )}{4}}{2 a^{\frac {3}{4}}}\)	\(199\)

[In]

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

[Out]

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

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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. 403 vs. \(2 (134) = 268\).

Time = 0.31 (sec) , antiderivative size = 403, normalized size of antiderivative = 2.32 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b+a x^2} \, dx=\frac {1}{2} \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} x^{2} + \frac {2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} b \arctan \left (\frac {2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} \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 {2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} b \arctan \left (-\frac {2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} \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 {2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} b \log \left (2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {2} \sqrt {-a} + \sqrt {a - \frac {b}{x^{2}}}\right )}{2 \, a} - \frac {2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} b \log \left (-2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {2} \sqrt {-a} + \sqrt {a - \frac {b}{x^{2}}}\right )}{2 \, a} + \frac {9 \, \sqrt {2} b \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 )}{8 \, \left (-a\right )^{\frac {3}{4}}} + \frac {9 \, \sqrt {2} b \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 )}{8 \, \left (-a\right )^{\frac {3}{4}}} + \frac {9 \, \sqrt {2} b \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 )}{16 \, \left (-a\right )^{\frac {3}{4}}} + \frac {9 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b \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 )}{16 \, a} \]

[In]

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

[Out]

1/2*(a - b/x^2)^(1/4)*x^2 + 2^(3/4)*(-a)^(1/4)*b*arctan(1/2*2^(1/4)*(2^(3/4)*(-a)^(1/4) + 2*(a - b/x^2)^(1/4))
/(-a)^(1/4))/a + 2^(3/4)*(-a)^(1/4)*b*arctan(-1/2*2^(1/4)*(2^(3/4)*(-a)^(1/4) - 2*(a - b/x^2)^(1/4))/(-a)^(1/4
))/a + 1/2*2^(3/4)*(-a)^(1/4)*b*log(2^(3/4)*(-a)^(1/4)*(a - b/x^2)^(1/4) + sqrt(2)*sqrt(-a) + sqrt(a - b/x^2))
/a - 1/2*2^(3/4)*(-a)^(1/4)*b*log(-2^(3/4)*(-a)^(1/4)*(a - b/x^2)^(1/4) + sqrt(2)*sqrt(-a) + sqrt(a - b/x^2))/
a + 9/8*sqrt(2)*b*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a - b/x^2)^(1/4))/(-a)^(1/4))/(-a)^(3/4) + 9/8*s
qrt(2)*b*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a - b/x^2)^(1/4))/(-a)^(1/4))/(-a)^(3/4) + 9/16*sqrt(2)*
b*log(sqrt(2)*(-a)^(1/4)*(a - b/x^2)^(1/4) + sqrt(-a) + sqrt(a - b/x^2))/(-a)^(3/4) + 9/16*sqrt(2)*(-a)^(1/4)*
b*log(-sqrt(2)*(-a)^(1/4)*(a - b/x^2)^(1/4) + sqrt(-a) + sqrt(a - b/x^2))/a

Mupad [F(-1)]

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

[In]

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

[Out]

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