\(\int \frac {-b+2 a x^2}{(b+a x^2) \sqrt [4]{b x^2+a x^4}} \, dx\) [1241]

Optimal result

Integrand size = 35, antiderivative size = 90 \[ \int \frac {-b+2 a x^2}{\left (b+a x^2\right ) \sqrt [4]{b x^2+a x^4}} \, dx=-\frac {6 \left (b x^2+a x^4\right )^{3/4}}{x \left (b+a x^2\right )}+\frac {2 \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{\sqrt [4]{a}}+\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{\sqrt [4]{a}} \]

[Out]

-6*(a*x^4+b*x^2)^(3/4)/x/(a*x^2+b)+2*arctan(a^(1/4)*x/(a*x^4+b*x^2)^(1/4))/a^(1/4)+2*arctanh(a^(1/4)*x/(a*x^4+
b*x^2)^(1/4))/a^(1/4)

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.57, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2081, 463, 335, 246, 218, 212, 209} \[ \int \frac {-b+2 a x^2}{\left (b+a x^2\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\frac {2 \sqrt {x} \sqrt [4]{a x^2+b} \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{\sqrt [4]{a} \sqrt [4]{a x^4+b x^2}}+\frac {2 \sqrt {x} \sqrt [4]{a x^2+b} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{\sqrt [4]{a} \sqrt [4]{a x^4+b x^2}}-\frac {6 x}{\sqrt [4]{a x^4+b x^2}} \]

[In]

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

[Out]

(-6*x)/(b*x^2 + a*x^4)^(1/4) + (2*Sqrt[x]*(b + a*x^2)^(1/4)*ArcTan[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(a^(1
/4)*(b*x^2 + a*x^4)^(1/4)) + (2*Sqrt[x]*(b + a*x^2)^(1/4)*ArcTanh[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(a^(1/
4)*(b*x^2 + a*x^4)^(1/4))

Rule 209

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

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 218

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

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 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)] && IntegerQ[p
+ 1/n]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 463

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

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 {\left (\sqrt {x} \sqrt [4]{b+a x^2}\right ) \int \frac {-b+2 a x^2}{\sqrt {x} \left (b+a x^2\right )^{5/4}} \, dx}{\sqrt [4]{b x^2+a x^4}} \\ & = -\frac {6 x}{\sqrt [4]{b x^2+a x^4}}+\frac {\left (2 \sqrt {x} \sqrt [4]{b+a x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt [4]{b+a x^2}} \, dx}{\sqrt [4]{b x^2+a x^4}} \\ & = -\frac {6 x}{\sqrt [4]{b x^2+a x^4}}+\frac {\left (4 \sqrt {x} \sqrt [4]{b+a x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{b+a x^4}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^4}} \\ & = -\frac {6 x}{\sqrt [4]{b x^2+a x^4}}+\frac {\left (4 \sqrt {x} \sqrt [4]{b+a x^2}\right ) \text {Subst}\left (\int \frac {1}{1-a x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\sqrt [4]{b x^2+a x^4}} \\ & = -\frac {6 x}{\sqrt [4]{b x^2+a x^4}}+\frac {\left (2 \sqrt {x} \sqrt [4]{b+a x^2}\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\sqrt [4]{b x^2+a x^4}}+\frac {\left (2 \sqrt {x} \sqrt [4]{b+a x^2}\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\sqrt [4]{b x^2+a x^4}} \\ & = -\frac {6 x}{\sqrt [4]{b x^2+a x^4}}+\frac {2 \sqrt {x} \sqrt [4]{b+a x^2} \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\sqrt [4]{a} \sqrt [4]{b x^2+a x^4}}+\frac {2 \sqrt {x} \sqrt [4]{b+a x^2} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\sqrt [4]{a} \sqrt [4]{b x^2+a x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.22 \[ \int \frac {-b+2 a x^2}{\left (b+a x^2\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\frac {2 \sqrt {x} \left (-3 \sqrt [4]{a} \sqrt {x}+\sqrt [4]{b+a x^2} \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )+\sqrt [4]{b+a x^2} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )\right )}{\sqrt [4]{a} \sqrt [4]{x^2 \left (b+a x^2\right )}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.41 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.22

method	result	size
pseudoelliptic	\(-\frac {2 \left (3 a^{\frac {1}{4}} x +\frac {\left (2 \arctan \left (\frac {\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )-\ln \left (\frac {a^{\frac {1}{4}} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{-a^{\frac {1}{4}} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}\right )\right ) \left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{2}\right )}{a^{\frac {1}{4}} \left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}\)	\(110\)

[In]

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

[Out]

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

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 195 vs. \(2 (76) = 152\).

Time = 0.30 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.17 \[ \int \frac {-b+2 a x^2}{\left (b+a x^2\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a} + \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a} - \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{2}}}\right )}{2 \, a} + \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{2}}}\right )}{2 \, a} - \frac {6}{{\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}} \]

[In]

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

[Out]

sqrt(2)*(-a)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b/x^2)^(1/4))/(-a)^(1/4))/a + sqrt(2)*(-a)^
(3/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a + b/x^2)^(1/4))/(-a)^(1/4))/a - 1/2*sqrt(2)*(-a)^(3/4)*lo
g(sqrt(2)*(-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt(-a) + sqrt(a + b/x^2))/a + 1/2*sqrt(2)*(-a)^(3/4)*log(-sqrt(2)*(
-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt(-a) + sqrt(a + b/x^2))/a - 6/(a + b/x^2)^(1/4)

Mupad [F(-1)]

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

[In]

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

[Out]

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