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

Optimal result

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

[Out]

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)

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.78, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2081, 470, 335, 338, 304, 209, 212} \[ \int \frac {\left (b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{-b+a x^2} \, dx=-\frac {7 b \sqrt [4]{a x^4-b x^2} \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2-b}}\right )}{4 a^{3/4} \sqrt {x} \sqrt [4]{a x^2-b}}+\frac {7 b \sqrt [4]{a x^4-b x^2} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2-b}}\right )}{4 a^{3/4} \sqrt {x} \sqrt [4]{a x^2-b}}+\frac {1}{2} x \sqrt [4]{a x^4-b x^2} \]

[In]

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

[Out]

(x*(-(b*x^2) + a*x^4)^(1/4))/2 - (7*b*(-(b*x^2) + a*x^4)^(1/4)*ArcTan[(a^(1/4)*Sqrt[x])/(-b + a*x^2)^(1/4)])/(
4*a^(3/4)*Sqrt[x]*(-b + a*x^2)^(1/4)) + (7*b*(-(b*x^2) + a*x^4)^(1/4)*ArcTanh[(a^(1/4)*Sqrt[x])/(-b + a*x^2)^(
1/4)])/(4*a^(3/4)*Sqrt[x]*(-b + a*x^2)^(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 304

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

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 338

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[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 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 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 )}{\left (-b+a x^2\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{-b+a x^2}} \\ & = \frac {1}{2} x \sqrt [4]{-b x^2+a x^4}+\frac {\left (7 b \sqrt [4]{-b x^2+a x^4}\right ) \int \frac {\sqrt {x}}{\left (-b+a x^2\right )^{3/4}} \, dx}{4 \sqrt {x} \sqrt [4]{-b+a x^2}} \\ & = \frac {1}{2} x \sqrt [4]{-b x^2+a x^4}+\frac {\left (7 b \sqrt [4]{-b x^2+a x^4}\right ) \text {Subst}\left (\int \frac {x^2}{\left (-b+a x^4\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{2 \sqrt {x} \sqrt [4]{-b+a x^2}} \\ & = \frac {1}{2} x \sqrt [4]{-b x^2+a x^4}+\frac {\left (7 b \sqrt [4]{-b x^2+a x^4}\right ) \text {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{2 \sqrt {x} \sqrt [4]{-b+a x^2}} \\ & = \frac {1}{2} x \sqrt [4]{-b x^2+a x^4}+\frac {\left (7 b \sqrt [4]{-b x^2+a x^4}\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{4 \sqrt {a} \sqrt {x} \sqrt [4]{-b+a x^2}}-\frac {\left (7 b \sqrt [4]{-b x^2+a x^4}\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{4 \sqrt {a} \sqrt {x} \sqrt [4]{-b+a x^2}} \\ & = \frac {1}{2} x \sqrt [4]{-b x^2+a x^4}-\frac {7 b \sqrt [4]{-b x^2+a x^4} \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{4 a^{3/4} \sqrt {x} \sqrt [4]{-b+a x^2}}+\frac {7 b \sqrt [4]{-b x^2+a x^4} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{4 a^{3/4} \sqrt {x} \sqrt [4]{-b+a x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.39 \[ \int \frac {\left (b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{-b+a x^2} \, dx=\frac {\sqrt [4]{-b x^2+a x^4} \left (2 a^{3/4} x^{3/2} \sqrt [4]{-b+a x^2}-7 b \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )+7 b \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )\right )}{4 a^{3/4} \sqrt {x} \sqrt [4]{-b+a x^2}} \]

[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) - 7*b*ArcTan[(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))

Maple [A] (verified)

Time = 2.49 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.16

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

[In]

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

[Out]

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

[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. 222 vs. \(2 (70) = 140\).

Time = 0.30 (sec) , antiderivative size = 222, normalized size of antiderivative = 2.47 \[ \int \frac {\left (b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{-b+a x^2} \, dx=\frac {8 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} b x^{2} + \frac {14 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a} + \frac {14 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a} + \frac {7 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{2}}}\right )}{a} + \frac {7 \, \sqrt {2} b^{2} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{2}}}\right )}{\left (-a\right )^{\frac {3}{4}}}}{16 \, b} \]

[In]

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

[Out]

1/16*(8*(a - b/x^2)^(1/4)*b*x^2 + 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 + 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. \[ \int \frac {\left (b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{-b+a x^2} \, dx=\int -\frac {\left (a\,x^2+b\right )\,{\left (a\,x^4-b\,x^2\right )}^{1/4}}{b-a\,x^2} \,d x \]

[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)