\(\int \frac {(b+a x^2) \sqrt {b^2+a^2 x^4}}{x^2 (-b+a x^2)} \, dx\) [921]

Optimal result

Integrand size = 37, antiderivative size = 70 \[ \int \frac {\left (b+a x^2\right ) \sqrt {b^2+a^2 x^4}}{x^2 \left (-b+a x^2\right )} \, dx=\frac {\sqrt {b^2+a^2 x^4}}{x}-\sqrt {2} \sqrt {a} \sqrt {b} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {b} x}{\sqrt {b^2+a^2 x^4}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.270, Rules used = {6857, 283, 311, 226, 1210, 1223, 1212, 1225, 1713, 214} \[ \int \frac {\left (b+a x^2\right ) \sqrt {b^2+a^2 x^4}}{x^2 \left (-b+a x^2\right )} \, dx=\frac {\sqrt {a^2 x^4+b^2}}{x}-\sqrt {2} \sqrt {a} \sqrt {b} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {b} x}{\sqrt {a^2 x^4+b^2}}\right ) \]

[In]

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

[Out]

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

Rule 214

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

Rule 226

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 283

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 311

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1210

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a +
c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 1212

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(e + d*q)/q, Int
[1/Sqrt[a + c*x^4], x], x] - Dist[e/q, Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a
, c, d, e}, x] && PosQ[c/a]

Rule 1223

Int[((a_) + (c_.)*(x_)^4)^(p_)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[-(e^2)^(-1), Int[(c*d - c*e*x^2)*(a +
c*x^4)^(p - 1), x], x] + Dist[(c*d^2 + a*e^2)/e^2, Int[(a + c*x^4)^(p - 1)/(d + e*x^2), x], x] /; FreeQ[{a, c,
d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p + 1/2, 0]

Rule 1225

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[1/(2*d), Int[1/Sqrt[a + c*x^4], x],
x] + Dist[1/(2*d), Int[(d - e*x^2)/((d + e*x^2)*Sqrt[a + c*x^4]), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d
^2 + a*e^2, 0] && EqQ[c*d^2 - a*e^2, 0]

Rule 1713

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[A, Subst[Int[1/
(d + 2*a*e*x^2), x], x, x/Sqrt[a + c*x^4]], x] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ
[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 0]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

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

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00 \[ \int \frac {\left (b+a x^2\right ) \sqrt {b^2+a^2 x^4}}{x^2 \left (-b+a x^2\right )} \, dx=\frac {\sqrt {b^2+a^2 x^4}}{x}-\sqrt {2} \sqrt {a} \sqrt {b} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {b} x}{\sqrt {b^2+a^2 x^4}}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 10.79 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.90

method	result	size
elliptic	\(\frac {\left (\frac {\sqrt {x^{4} a^{2}+b^{2}}\, \sqrt {2}}{x}-\frac {2 a b \,\operatorname {arctanh}\left (\frac {\sqrt {x^{4} a^{2}+b^{2}}\, \sqrt {2}}{2 x \sqrt {a b}}\right )}{\sqrt {a b}}\right ) \sqrt {2}}{2}\)	\(63\)
risch	\(\frac {\sqrt {x^{4} a^{2}+b^{2}}}{x}-\frac {b a \sqrt {2}\, \left (2 \ln \left (2\right )+\ln \left (-\frac {2 \left (-\frac {\sqrt {2}\, \sqrt {a b}\, \sqrt {x^{4} a^{2}+b^{2}}}{2}+\left (a \,x^{2}+b \right ) \sqrt {a b}+a b x \right ) a}{a \,x^{2}+2 x \sqrt {a b}+b}\right )+\ln \left (-\frac {2 a \left (-\frac {\sqrt {2}\, \sqrt {a b}\, \sqrt {x^{4} a^{2}+b^{2}}}{2}+\left (-a \,x^{2}-b \right ) \sqrt {a b}+a b x \right )}{a \,x^{2}-2 x \sqrt {a b}+b}\right )\right )}{2 \sqrt {a b}}\)	\(163\)
default	\(-\frac {2 x b a \sqrt {2}\, \ln \left (2\right )+x b a \sqrt {2}\, \ln \left (-\frac {2 \left (-\frac {\sqrt {2}\, \sqrt {a b}\, \sqrt {x^{4} a^{2}+b^{2}}}{2}+\left (a \,x^{2}+b \right ) \sqrt {a b}+a b x \right ) a}{a \,x^{2}+2 x \sqrt {a b}+b}\right )+x b a \sqrt {2}\, \ln \left (-\frac {2 a \left (-\frac {\sqrt {2}\, \sqrt {a b}\, \sqrt {x^{4} a^{2}+b^{2}}}{2}+\left (-a \,x^{2}-b \right ) \sqrt {a b}+a b x \right )}{a \,x^{2}-2 x \sqrt {a b}+b}\right )-2 \sqrt {x^{4} a^{2}+b^{2}}\, \sqrt {a b}}{2 \sqrt {a b}\, x}\)	\(183\)
pseudoelliptic	\(-\frac {2 x b a \sqrt {2}\, \ln \left (2\right )+x b a \sqrt {2}\, \ln \left (-\frac {2 \left (-\frac {\sqrt {2}\, \sqrt {a b}\, \sqrt {x^{4} a^{2}+b^{2}}}{2}+\left (a \,x^{2}+b \right ) \sqrt {a b}+a b x \right ) a}{a \,x^{2}+2 x \sqrt {a b}+b}\right )+x b a \sqrt {2}\, \ln \left (-\frac {2 a \left (-\frac {\sqrt {2}\, \sqrt {a b}\, \sqrt {x^{4} a^{2}+b^{2}}}{2}+\left (-a \,x^{2}-b \right ) \sqrt {a b}+a b x \right )}{a \,x^{2}-2 x \sqrt {a b}+b}\right )-2 \sqrt {x^{4} a^{2}+b^{2}}\, \sqrt {a b}}{2 \sqrt {a b}\, x}\)	\(183\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.70 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.27 \[ \int \frac {\left (b+a x^2\right ) \sqrt {b^2+a^2 x^4}}{x^2 \left (-b+a x^2\right )} \, dx=\left [\frac {\sqrt {2} \sqrt {a b} x \log \left (\frac {a^{2} x^{4} + 2 \, a b x^{2} - 2 \, \sqrt {2} \sqrt {a^{2} x^{4} + b^{2}} \sqrt {a b} x + b^{2}}{a^{2} x^{4} - 2 \, a b x^{2} + b^{2}}\right ) + 2 \, \sqrt {a^{2} x^{4} + b^{2}}}{2 \, x}, \frac {\sqrt {2} \sqrt {-a b} x \arctan \left (\frac {\sqrt {2} \sqrt {a^{2} x^{4} + b^{2}} \sqrt {-a b}}{2 \, a b x}\right ) + \sqrt {a^{2} x^{4} + b^{2}}}{x}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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