\(\int \frac {(-q+p x^4) \sqrt {q+p x^4}}{x^2 (a q+b x^2+a p x^4)} \, dx\) [3087]

Optimal result

Integrand size = 41, antiderivative size = 524 \[ \int \frac {\left (-q+p x^4\right ) \sqrt {q+p x^4}}{x^2 \left (a q+b x^2+a p x^4\right )} \, dx=\frac {\sqrt {q+p x^4}}{a x}+\frac {\left (\sqrt {2} b \sqrt {b+i \sqrt {b} \sqrt {-b+2 a \sqrt {p} \sqrt {q}}-a \sqrt {p} \sqrt {q}}-i \sqrt {2} \sqrt {b} \sqrt {b+i \sqrt {b} \sqrt {-b+2 a \sqrt {p} \sqrt {q}}-a \sqrt {p} \sqrt {q}} \sqrt {-b+2 a \sqrt {p} \sqrt {q}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {b+i \sqrt {b} \sqrt {-b+2 a \sqrt {p} \sqrt {q}}-a \sqrt {p} \sqrt {q}} x}{\sqrt {a} \left (\sqrt {q}+\sqrt {p} x^2+\sqrt {q+p x^4}\right )}\right )}{2 a^{5/2} \sqrt {p} \sqrt {q}}-\frac {i \left (-i \sqrt {2} b \sqrt {-b+i \sqrt {b} \sqrt {-b+2 a \sqrt {p} \sqrt {q}}+a \sqrt {p} \sqrt {q}}+\sqrt {2} \sqrt {b} \sqrt {-b+i \sqrt {b} \sqrt {-b+2 a \sqrt {p} \sqrt {q}}+a \sqrt {p} \sqrt {q}} \sqrt {-b+2 a \sqrt {p} \sqrt {q}}\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {-b+i \sqrt {b} \sqrt {-b+2 a \sqrt {p} \sqrt {q}}+a \sqrt {p} \sqrt {q}} x}{\sqrt {a} \left (\sqrt {q}+\sqrt {p} x^2+\sqrt {q+p x^4}\right )}\right )}{2 a^{5/2} \sqrt {p} \sqrt {q}} \]

[Out]

(p*x^4+q)^(1/2)/a/x+1/2*(2^(1/2)*b*(b+I*b^(1/2)*(-b+2*a*p^(1/2)*q^(1/2))^(1/2)-a*p^(1/2)*q^(1/2))^(1/2)-I*2^(1
/2)*b^(1/2)*(b+I*b^(1/2)*(-b+2*a*p^(1/2)*q^(1/2))^(1/2)-a*p^(1/2)*q^(1/2))^(1/2)*(-b+2*a*p^(1/2)*q^(1/2))^(1/2
))*arctan(2^(1/2)*(b+I*b^(1/2)*(-b+2*a*p^(1/2)*q^(1/2))^(1/2)-a*p^(1/2)*q^(1/2))^(1/2)*x/a^(1/2)/(q^(1/2)+p^(1
/2)*x^2+(p*x^4+q)^(1/2)))/a^(5/2)/p^(1/2)/q^(1/2)-1/2*I*(-I*2^(1/2)*b*(-b+I*b^(1/2)*(-b+2*a*p^(1/2)*q^(1/2))^(
1/2)+a*p^(1/2)*q^(1/2))^(1/2)+2^(1/2)*b^(1/2)*(-b+I*b^(1/2)*(-b+2*a*p^(1/2)*q^(1/2))^(1/2)+a*p^(1/2)*q^(1/2))^
(1/2)*(-b+2*a*p^(1/2)*q^(1/2))^(1/2))*arctanh(2^(1/2)*(-b+I*b^(1/2)*(-b+2*a*p^(1/2)*q^(1/2))^(1/2)+a*p^(1/2)*q
^(1/2))^(1/2)*x/a^(1/2)/(q^(1/2)+p^(1/2)*x^2+(p*x^4+q)^(1/2)))/a^(5/2)/p^(1/2)/q^(1/2)

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 1.98 (sec) , antiderivative size = 1029, normalized size of antiderivative = 1.96, number of steps used = 22, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.220, Rules used = {6860, 283, 311, 226, 1210, 1223, 1212, 1231, 1721} \[ \int \frac {\left (-q+p x^4\right ) \sqrt {q+p x^4}}{x^2 \left (a q+b x^2+a p x^4\right )} \, dx=\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a} \sqrt {p x^4+q}}\right )}{a^{3/2}}-\frac {\left (-2 \sqrt {p} \sqrt {q} a+b-\sqrt {b^2-4 a^2 p q}\right ) \left (\sqrt {p} x^2+\sqrt {q}\right ) \sqrt {\frac {p x^4+q}{\left (\sqrt {p} x^2+\sqrt {q}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{p} x}{\sqrt [4]{q}}\right ),\frac {1}{2}\right )}{4 a^2 \sqrt [4]{p} \sqrt [4]{q} \sqrt {p x^4+q}}-\frac {\left (-2 \sqrt {p} \sqrt {q} a+b+\sqrt {b^2-4 a^2 p q}\right ) \left (\sqrt {p} x^2+\sqrt {q}\right ) \sqrt {\frac {p x^4+q}{\left (\sqrt {p} x^2+\sqrt {q}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{p} x}{\sqrt [4]{q}}\right ),\frac {1}{2}\right )}{4 a^2 \sqrt [4]{p} \sqrt [4]{q} \sqrt {p x^4+q}}-\frac {\sqrt [4]{p} \sqrt [4]{q} \left (\sqrt {p} x^2+\sqrt {q}\right ) \sqrt {\frac {p x^4+q}{\left (\sqrt {p} x^2+\sqrt {q}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{p} x}{\sqrt [4]{q}}\right ),\frac {1}{2}\right )}{a \sqrt {p x^4+q}}+\frac {b \left (b-\sqrt {b^2-4 a^2 p q}\right ) \left (\sqrt {p} x^2+\sqrt {q}\right ) \sqrt {\frac {p x^4+q}{\left (\sqrt {p} x^2+\sqrt {q}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{p} x}{\sqrt [4]{q}}\right ),\frac {1}{2}\right )}{2 a^2 \sqrt [4]{p} \sqrt [4]{q} \left (-2 \sqrt {p} \sqrt {q} a+b-\sqrt {b^2-4 a^2 p q}\right ) \sqrt {p x^4+q}}+\frac {b \left (b+\sqrt {b^2-4 a^2 p q}\right ) \left (\sqrt {p} x^2+\sqrt {q}\right ) \sqrt {\frac {p x^4+q}{\left (\sqrt {p} x^2+\sqrt {q}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{p} x}{\sqrt [4]{q}}\right ),\frac {1}{2}\right )}{2 a^2 \sqrt [4]{p} \sqrt [4]{q} \left (-2 \sqrt {p} \sqrt {q} a+b+\sqrt {b^2-4 a^2 p q}\right ) \sqrt {p x^4+q}}-\frac {b \left (2 \sqrt {p} \sqrt {q} a+b-\sqrt {b^2-4 a^2 p q}\right ) \left (\sqrt {p} x^2+\sqrt {q}\right ) \sqrt {\frac {p x^4+q}{\left (\sqrt {p} x^2+\sqrt {q}\right )^2}} \operatorname {EllipticPi}\left (\frac {1}{4} \left (2-\frac {b}{a \sqrt {p} \sqrt {q}}\right ),2 \arctan \left (\frac {\sqrt [4]{p} x}{\sqrt [4]{q}}\right ),\frac {1}{2}\right )}{4 a^2 \sqrt [4]{p} \sqrt [4]{q} \left (-2 \sqrt {p} \sqrt {q} a+b-\sqrt {b^2-4 a^2 p q}\right ) \sqrt {p x^4+q}}-\frac {b \left (2 \sqrt {p} \sqrt {q} a+b+\sqrt {b^2-4 a^2 p q}\right ) \left (\sqrt {p} x^2+\sqrt {q}\right ) \sqrt {\frac {p x^4+q}{\left (\sqrt {p} x^2+\sqrt {q}\right )^2}} \operatorname {EllipticPi}\left (\frac {1}{4} \left (2-\frac {b}{a \sqrt {p} \sqrt {q}}\right ),2 \arctan \left (\frac {\sqrt [4]{p} x}{\sqrt [4]{q}}\right ),\frac {1}{2}\right )}{4 a^2 \sqrt [4]{p} \sqrt [4]{q} \left (-2 \sqrt {p} \sqrt {q} a+b+\sqrt {b^2-4 a^2 p q}\right ) \sqrt {p x^4+q}}+\frac {\sqrt {p x^4+q}}{a x} \]

[In]

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

[Out]

Sqrt[q + p*x^4]/(a*x) + (Sqrt[b]*ArcTan[(Sqrt[b]*x)/(Sqrt[a]*Sqrt[q + p*x^4])])/a^(3/2) - (p^(1/4)*q^(1/4)*(Sq
rt[q] + Sqrt[p]*x^2)*Sqrt[(q + p*x^4)/(Sqrt[q] + Sqrt[p]*x^2)^2]*EllipticF[2*ArcTan[(p^(1/4)*x)/q^(1/4)], 1/2]
)/(a*Sqrt[q + p*x^4]) + (b*(b - Sqrt[b^2 - 4*a^2*p*q])*(Sqrt[q] + Sqrt[p]*x^2)*Sqrt[(q + p*x^4)/(Sqrt[q] + Sqr
t[p]*x^2)^2]*EllipticF[2*ArcTan[(p^(1/4)*x)/q^(1/4)], 1/2])/(2*a^2*p^(1/4)*q^(1/4)*(b - 2*a*Sqrt[p]*Sqrt[q] -
Sqrt[b^2 - 4*a^2*p*q])*Sqrt[q + p*x^4]) - ((b - 2*a*Sqrt[p]*Sqrt[q] - Sqrt[b^2 - 4*a^2*p*q])*(Sqrt[q] + Sqrt[p
]*x^2)*Sqrt[(q + p*x^4)/(Sqrt[q] + Sqrt[p]*x^2)^2]*EllipticF[2*ArcTan[(p^(1/4)*x)/q^(1/4)], 1/2])/(4*a^2*p^(1/
4)*q^(1/4)*Sqrt[q + p*x^4]) + (b*(b + Sqrt[b^2 - 4*a^2*p*q])*(Sqrt[q] + Sqrt[p]*x^2)*Sqrt[(q + p*x^4)/(Sqrt[q]
+ Sqrt[p]*x^2)^2]*EllipticF[2*ArcTan[(p^(1/4)*x)/q^(1/4)], 1/2])/(2*a^2*p^(1/4)*q^(1/4)*(b - 2*a*Sqrt[p]*Sqrt
[q] + Sqrt[b^2 - 4*a^2*p*q])*Sqrt[q + p*x^4]) - ((b - 2*a*Sqrt[p]*Sqrt[q] + Sqrt[b^2 - 4*a^2*p*q])*(Sqrt[q] +
Sqrt[p]*x^2)*Sqrt[(q + p*x^4)/(Sqrt[q] + Sqrt[p]*x^2)^2]*EllipticF[2*ArcTan[(p^(1/4)*x)/q^(1/4)], 1/2])/(4*a^2
*p^(1/4)*q^(1/4)*Sqrt[q + p*x^4]) - (b*(b + 2*a*Sqrt[p]*Sqrt[q] - Sqrt[b^2 - 4*a^2*p*q])*(Sqrt[q] + Sqrt[p]*x^
2)*Sqrt[(q + p*x^4)/(Sqrt[q] + Sqrt[p]*x^2)^2]*EllipticPi[(2 - b/(a*Sqrt[p]*Sqrt[q]))/4, 2*ArcTan[(p^(1/4)*x)/
q^(1/4)], 1/2])/(4*a^2*p^(1/4)*q^(1/4)*(b - 2*a*Sqrt[p]*Sqrt[q] - Sqrt[b^2 - 4*a^2*p*q])*Sqrt[q + p*x^4]) - (b
*(b + 2*a*Sqrt[p]*Sqrt[q] + Sqrt[b^2 - 4*a^2*p*q])*(Sqrt[q] + Sqrt[p]*x^2)*Sqrt[(q + p*x^4)/(Sqrt[q] + Sqrt[p]
*x^2)^2]*EllipticPi[(2 - b/(a*Sqrt[p]*Sqrt[q]))/4, 2*ArcTan[(p^(1/4)*x)/q^(1/4)], 1/2])/(4*a^2*p^(1/4)*q^(1/4)
*(b - 2*a*Sqrt[p]*Sqrt[q] + Sqrt[b^2 - 4*a^2*p*q])*Sqrt[q + p*x^4])

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 1231

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

Rule 1721

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

Rule 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {\sqrt {q+p x^4}}{a x^2}+\frac {\left (b+2 a p x^2\right ) \sqrt {q+p x^4}}{a \left (a q+b x^2+a p x^4\right )}\right ) \, dx \\ & = -\frac {\int \frac {\sqrt {q+p x^4}}{x^2} \, dx}{a}+\frac {\int \frac {\left (b+2 a p x^2\right ) \sqrt {q+p x^4}}{a q+b x^2+a p x^4} \, dx}{a} \\ & = \frac {\sqrt {q+p x^4}}{a x}+\frac {\int \left (\frac {2 a p \sqrt {q+p x^4}}{b-\sqrt {b^2-4 a^2 p q}+2 a p x^2}+\frac {2 a p \sqrt {q+p x^4}}{b+\sqrt {b^2-4 a^2 p q}+2 a p x^2}\right ) \, dx}{a}-\frac {(2 p) \int \frac {x^2}{\sqrt {q+p x^4}} \, dx}{a} \\ & = \frac {\sqrt {q+p x^4}}{a x}+(2 p) \int \frac {\sqrt {q+p x^4}}{b-\sqrt {b^2-4 a^2 p q}+2 a p x^2} \, dx+(2 p) \int \frac {\sqrt {q+p x^4}}{b+\sqrt {b^2-4 a^2 p q}+2 a p x^2} \, dx-\frac {\left (2 \sqrt {p} \sqrt {q}\right ) \int \frac {1}{\sqrt {q+p x^4}} \, dx}{a}+\frac {\left (2 \sqrt {p} \sqrt {q}\right ) \int \frac {1-\frac {\sqrt {p} x^2}{\sqrt {q}}}{\sqrt {q+p x^4}} \, dx}{a} \\ & = \frac {\sqrt {q+p x^4}}{a x}-\frac {2 \sqrt {p} x \sqrt {q+p x^4}}{a \left (\sqrt {q}+\sqrt {p} x^2\right )}+\frac {2 \sqrt [4]{p} \sqrt [4]{q} \left (\sqrt {q}+\sqrt {p} x^2\right ) \sqrt {\frac {q+p x^4}{\left (\sqrt {q}+\sqrt {p} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{p} x}{\sqrt [4]{q}}\right )|\frac {1}{2}\right )}{a \sqrt {q+p x^4}}-\frac {\sqrt [4]{p} \sqrt [4]{q} \left (\sqrt {q}+\sqrt {p} x^2\right ) \sqrt {\frac {q+p x^4}{\left (\sqrt {q}+\sqrt {p} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{p} x}{\sqrt [4]{q}}\right ),\frac {1}{2}\right )}{a \sqrt {q+p x^4}}-\frac {\int \frac {p \left (b-\sqrt {b^2-4 a^2 p q}\right )-2 a p^2 x^2}{\sqrt {q+p x^4}} \, dx}{2 a^2 p}-\frac {\int \frac {p \left (b+\sqrt {b^2-4 a^2 p q}\right )-2 a p^2 x^2}{\sqrt {q+p x^4}} \, dx}{2 a^2 p}+\frac {\left (b \left (b-\sqrt {b^2-4 a^2 p q}\right )\right ) \int \frac {1}{\left (b-\sqrt {b^2-4 a^2 p q}+2 a p x^2\right ) \sqrt {q+p x^4}} \, dx}{a^2}+\frac {\left (b \left (b+\sqrt {b^2-4 a^2 p q}\right )\right ) \int \frac {1}{\left (b+\sqrt {b^2-4 a^2 p q}+2 a p x^2\right ) \sqrt {q+p x^4}} \, dx}{a^2} \\ & = \frac {\sqrt {q+p x^4}}{a x}-\frac {2 \sqrt {p} x \sqrt {q+p x^4}}{a \left (\sqrt {q}+\sqrt {p} x^2\right )}+\frac {2 \sqrt [4]{p} \sqrt [4]{q} \left (\sqrt {q}+\sqrt {p} x^2\right ) \sqrt {\frac {q+p x^4}{\left (\sqrt {q}+\sqrt {p} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{p} x}{\sqrt [4]{q}}\right )|\frac {1}{2}\right )}{a \sqrt {q+p x^4}}-\frac {\sqrt [4]{p} \sqrt [4]{q} \left (\sqrt {q}+\sqrt {p} x^2\right ) \sqrt {\frac {q+p x^4}{\left (\sqrt {q}+\sqrt {p} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{p} x}{\sqrt [4]{q}}\right ),\frac {1}{2}\right )}{a \sqrt {q+p x^4}}-2 \frac {\left (\sqrt {p} \sqrt {q}\right ) \int \frac {1-\frac {\sqrt {p} x^2}{\sqrt {q}}}{\sqrt {q+p x^4}} \, dx}{a}+\frac {\left (b \left (b-\sqrt {b^2-4 a^2 p q}\right )\right ) \int \frac {1}{\sqrt {q+p x^4}} \, dx}{a^2 \left (b-2 a \sqrt {p} \sqrt {q}-\sqrt {b^2-4 a^2 p q}\right )}-\frac {\left (2 b \sqrt {p} \sqrt {q} \left (b-\sqrt {b^2-4 a^2 p q}\right )\right ) \int \frac {1+\frac {\sqrt {p} x^2}{\sqrt {q}}}{\left (b-\sqrt {b^2-4 a^2 p q}+2 a p x^2\right ) \sqrt {q+p x^4}} \, dx}{a \left (b-2 a \sqrt {p} \sqrt {q}-\sqrt {b^2-4 a^2 p q}\right )}-\frac {\left (b-2 a \sqrt {p} \sqrt {q}-\sqrt {b^2-4 a^2 p q}\right ) \int \frac {1}{\sqrt {q+p x^4}} \, dx}{2 a^2}+\frac {\left (b \left (b+\sqrt {b^2-4 a^2 p q}\right )\right ) \int \frac {1}{\sqrt {q+p x^4}} \, dx}{a^2 \left (b-2 a \sqrt {p} \sqrt {q}+\sqrt {b^2-4 a^2 p q}\right )}-\frac {\left (2 b \sqrt {p} \sqrt {q} \left (b+\sqrt {b^2-4 a^2 p q}\right )\right ) \int \frac {1+\frac {\sqrt {p} x^2}{\sqrt {q}}}{\left (b+\sqrt {b^2-4 a^2 p q}+2 a p x^2\right ) \sqrt {q+p x^4}} \, dx}{a \left (b-2 a \sqrt {p} \sqrt {q}+\sqrt {b^2-4 a^2 p q}\right )}-\frac {\left (b-2 a \sqrt {p} \sqrt {q}+\sqrt {b^2-4 a^2 p q}\right ) \int \frac {1}{\sqrt {q+p x^4}} \, dx}{2 a^2} \\ & = \text {Too large to display} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.10 \[ \int \frac {\left (-q+p x^4\right ) \sqrt {q+p x^4}}{x^2 \left (a q+b x^2+a p x^4\right )} \, dx=\frac {\sqrt {q+p x^4}}{a x}+\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a} \sqrt {q+p x^4}}\right )}{a^{3/2}} \]

[In]

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

[Out]

Sqrt[q + p*x^4]/(a*x) + (Sqrt[b]*ArcTan[(Sqrt[b]*x)/(Sqrt[a]*Sqrt[q + p*x^4])])/a^(3/2)

Maple [A] (verified)

Time = 1.39 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.09

method	result	size
risch	\(\frac {\sqrt {p \,x^{4}+q}}{a x}-\frac {b \arctan \left (\frac {a \sqrt {p \,x^{4}+q}}{x \sqrt {a b}}\right )}{a \sqrt {a b}}\)	\(49\)
default	\(\frac {-b \arctan \left (\frac {a \sqrt {p \,x^{4}+q}}{x \sqrt {a b}}\right ) x +\sqrt {p \,x^{4}+q}\, \sqrt {a b}}{a x \sqrt {a b}}\)	\(53\)
pseudoelliptic	\(\frac {-b \arctan \left (\frac {a \sqrt {p \,x^{4}+q}}{x \sqrt {a b}}\right ) x +\sqrt {p \,x^{4}+q}\, \sqrt {a b}}{a x \sqrt {a b}}\)	\(53\)
elliptic	\(\frac {\left (\frac {\sqrt {p \,x^{4}+q}\, \sqrt {2}}{a x}-\frac {b \sqrt {2}\, \arctan \left (\frac {a \sqrt {p \,x^{4}+q}}{x \sqrt {a b}}\right )}{a \sqrt {a b}}\right ) \sqrt {2}}{2}\)	\(60\)

[In]

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

[Out]

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

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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