\(\int \frac {1}{(d+e x^2)^{3/2} (a+b x^2+c x^4)} \, dx\) [397]

Optimal result

Integrand size = 26, antiderivative size = 341 \[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )} \, dx=\frac {e^2 x}{d \left (c d^2-b d e+a e^2\right ) \sqrt {d+e x^2}}-\frac {c \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e} x}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e} \left (c d^2-b d e+a e^2\right )}-\frac {c \left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e} x}{\sqrt {b+\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{\sqrt {b+\sqrt {b^2-4 a c}} \sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e} \left (c d^2-b d e+a e^2\right )} \]

[Out]

e^2*x/d/(a*e^2-b*d*e+c*d^2)/(e*x^2+d)^(1/2)-c*arctan(x*(2*c*d-e*(b-(-4*a*c+b^2)^(1/2)))^(1/2)/(e*x^2+d)^(1/2)/
(b-(-4*a*c+b^2)^(1/2))^(1/2))*(e+(b*e-2*c*d)/(-4*a*c+b^2)^(1/2))/(a*e^2-b*d*e+c*d^2)/(2*c*d-e*(b-(-4*a*c+b^2)^
(1/2)))^(1/2)/(b-(-4*a*c+b^2)^(1/2))^(1/2)-c*arctan(x*(2*c*d-e*(b+(-4*a*c+b^2)^(1/2)))^(1/2)/(e*x^2+d)^(1/2)/(
b+(-4*a*c+b^2)^(1/2))^(1/2))*(e+(-b*e+2*c*d)/(-4*a*c+b^2)^(1/2))/(a*e^2-b*d*e+c*d^2)/(b+(-4*a*c+b^2)^(1/2))^(1
/2)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2)))^(1/2)

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1186, 197, 1706, 385, 211} \[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )} \, dx=-\frac {c \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {x \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )} \left (a e^2-b d e+c d^2\right )}-\frac {c \left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right ) \arctan \left (\frac {x \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}{\sqrt {\sqrt {b^2-4 a c}+b} \sqrt {d+e x^2}}\right )}{\sqrt {\sqrt {b^2-4 a c}+b} \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )} \left (a e^2-b d e+c d^2\right )}+\frac {e^2 x}{d \sqrt {d+e x^2} \left (a e^2-b d e+c d^2\right )} \]

[In]

Int[1/((d + e*x^2)^(3/2)*(a + b*x^2 + c*x^4)),x]

[Out]

(e^2*x)/(d*(c*d^2 - b*d*e + a*e^2)*Sqrt[d + e*x^2]) - (c*(e - (2*c*d - b*e)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2*
c*d - (b - Sqrt[b^2 - 4*a*c])*e]*x)/(Sqrt[b - Sqrt[b^2 - 4*a*c]]*Sqrt[d + e*x^2])])/(Sqrt[b - Sqrt[b^2 - 4*a*c
]]*Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]*(c*d^2 - b*d*e + a*e^2)) - (c*(e + (2*c*d - b*e)/Sqrt[b^2 - 4*a*c])
*ArcTan[(Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*x)/(Sqrt[b + Sqrt[b^2 - 4*a*c]]*Sqrt[d + e*x^2])])/(Sqrt[b +
Sqrt[b^2 - 4*a*c]]*Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*(c*d^2 - b*d*e + a*e^2))

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 211

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

Rule 385

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

Rule 1186

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

Rule 1706

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandInteg
rand[Px*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, q}, x] && PolyQ[Px, x^2] && NeQ[b
^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p]

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 18.25 (sec) , antiderivative size = 2061, normalized size of antiderivative = 6.04 \[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )} \, dx=\text {Result too large to show} \]

[In]

Integrate[1/((d + e*x^2)^(3/2)*(a + b*x^2 + c*x^4)),x]

[Out]

(2*c*x*(45*Sqrt[-(((-b + Sqrt[b^2 - 4*a*c])*(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)*x^2*(d + e*x^2))/(d^2*(b - Sq
rt[b^2 - 4*a*c] + 2*c*x^2)^2))] + (30*e*x^2*Sqrt[-(((-b + Sqrt[b^2 - 4*a*c])*(2*c*d + (-b + Sqrt[b^2 - 4*a*c])
*e)*x^2*(d + e*x^2))/(d^2*(b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)^2))])/d - 45*ArcSin[Sqrt[-(((2*c*d + (-b + Sqrt[b^
2 - 4*a*c])*e)*x^2)/(d*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2)))]] - (30*e*x^2*ArcSin[Sqrt[-(((2*c*d + (-b + Sqrt[b
^2 - 4*a*c])*e)*x^2)/(d*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2)))]])/d - (45*(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)*x
^2*ArcSin[Sqrt[-(((2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)*x^2)/(d*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2)))]])/(d*(-b
+ Sqrt[b^2 - 4*a*c] - 2*c*x^2)) - (30*e*(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)*x^4*ArcSin[Sqrt[-(((2*c*d + (-b +
Sqrt[b^2 - 4*a*c])*e)*x^2)/(d*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2)))]])/(d^2*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2)
) + 4*(-(((2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)*x^2)/(d*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2))))^(5/2)*Sqrt[((-b +
Sqrt[b^2 - 4*a*c])*(d + e*x^2))/(d*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2))]*Hypergeometric2F1[2, 2, 7/2, -(((2*c*
d + (-b + Sqrt[b^2 - 4*a*c])*e)*x^2)/(d*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2)))] + (4*e*x^2*(-(((2*c*d + (-b + Sq
rt[b^2 - 4*a*c])*e)*x^2)/(d*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2))))^(5/2)*Sqrt[((-b + Sqrt[b^2 - 4*a*c])*(d + e*
x^2))/(d*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2))]*Hypergeometric2F1[2, 2, 7/2, -(((2*c*d + (-b + Sqrt[b^2 - 4*a*c]
)*e)*x^2)/(d*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2)))])/d))/(15*Sqrt[b^2 - 4*a*c]*d*(-(((2*c*d + (-b + Sqrt[b^2 -
4*a*c])*e)*x^2)/(d*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2))))^(3/2)*(b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)*Sqrt[d + e*x^
2]*Sqrt[((-b + Sqrt[b^2 - 4*a*c])*(d + e*x^2))/(d*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2))]) - (2*c*x*Sqrt[((b + Sq
rt[b^2 - 4*a*c])*(d + e*x^2))/(d*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2))]*(45*Sqrt[-(((b + Sqrt[b^2 - 4*a*c])*(-2*c
*d + (b + Sqrt[b^2 - 4*a*c])*e)*x^2*(d + e*x^2))/(d^2*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)^2))] + (30*e*x^2*Sqrt[
-(((b + Sqrt[b^2 - 4*a*c])*(-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)*x^2*(d + e*x^2))/(d^2*(b + Sqrt[b^2 - 4*a*c] +
2*c*x^2)^2))])/d - 45*ArcSin[Sqrt[((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*x^2)/(d*(b + Sqrt[b^2 - 4*a*c] + 2*c*x
^2))]] - (30*e*x^2*ArcSin[Sqrt[((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*x^2)/(d*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2))
]])/d + (45*(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*x^2*ArcSin[Sqrt[((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*x^2)/(d*(
b + Sqrt[b^2 - 4*a*c] + 2*c*x^2))]])/(d*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)) - (30*e*(-2*c*d + (b + Sqrt[b^2 - 4
*a*c])*e)*x^4*ArcSin[Sqrt[((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*x^2)/(d*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2))]])/(
d^2*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)) + 4*(((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*x^2)/(d*(b + Sqrt[b^2 - 4*a*c
] + 2*c*x^2)))^(5/2)*Sqrt[((b + Sqrt[b^2 - 4*a*c])*(d + e*x^2))/(d*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2))]*Hyperge
ometric2F1[2, 2, 7/2, ((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*x^2)/(d*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2))] + (4*e*
x^2*(((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*x^2)/(d*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)))^(5/2)*Sqrt[((b + Sqrt[b^
2 - 4*a*c])*(d + e*x^2))/(d*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2))]*Hypergeometric2F1[2, 2, 7/2, ((2*c*d - (b + Sq
rt[b^2 - 4*a*c])*e)*x^2)/(d*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2))])/d))/(15*Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a
*c])*(((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*x^2)/(d*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)))^(3/2)*(d + e*x^2)^(3/2)
)

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 429, normalized size of antiderivative = 1.26

method	result	size
default	\(-\frac {\left (\frac {\left (b e -c d \right ) \sqrt {-4 d^{2} \left (a c -\frac {b^{2}}{4}\right )}}{2}+\left (\frac {b c d}{2}+e \left (a c -\frac {b^{2}}{2}\right )\right ) d \right ) \sqrt {2}\, d \sqrt {\left (-2 a e +b d +\sqrt {-4 d^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) a}\, \sqrt {e \,x^{2}+d}\, \operatorname {arctanh}\left (\frac {a \sqrt {e \,x^{2}+d}\, \sqrt {2}}{x \sqrt {\left (2 a e -b d +\sqrt {-4 d^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) a}}\right )+\left (\sqrt {2}\, \left (\frac {\left (-b e +c d \right ) \sqrt {-4 d^{2} \left (a c -\frac {b^{2}}{4}\right )}}{2}+\left (\frac {b c d}{2}+e \left (a c -\frac {b^{2}}{2}\right )\right ) d \right ) d \sqrt {e \,x^{2}+d}\, \arctan \left (\frac {a \sqrt {e \,x^{2}+d}\, \sqrt {2}}{x \sqrt {\left (-2 a e +b d +\sqrt {-4 d^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) a}}\right )-e^{2} x \sqrt {-4 d^{2} \left (a c -\frac {b^{2}}{4}\right )}\, \sqrt {\left (-2 a e +b d +\sqrt {-4 d^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) a}\right ) \sqrt {\left (2 a e -b d +\sqrt {-4 d^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) a}}{\sqrt {\left (-2 a e +b d +\sqrt {-4 d^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) a}\, \sqrt {e \,x^{2}+d}\, \sqrt {\left (2 a e -b d +\sqrt {-4 d^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) a}\, \sqrt {-4 d^{2} \left (a c -\frac {b^{2}}{4}\right )}\, \left (a \,e^{2}-b d e +c \,d^{2}\right ) d}\)	\(429\)
pseudoelliptic	\(-\frac {\left (\frac {\left (b e -c d \right ) \sqrt {-4 d^{2} \left (a c -\frac {b^{2}}{4}\right )}}{2}+\left (\frac {b c d}{2}+e \left (a c -\frac {b^{2}}{2}\right )\right ) d \right ) \sqrt {2}\, d \sqrt {\left (-2 a e +b d +\sqrt {-4 d^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) a}\, \sqrt {e \,x^{2}+d}\, \operatorname {arctanh}\left (\frac {a \sqrt {e \,x^{2}+d}\, \sqrt {2}}{x \sqrt {\left (2 a e -b d +\sqrt {-4 d^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) a}}\right )+\left (\sqrt {2}\, \left (\frac {\left (-b e +c d \right ) \sqrt {-4 d^{2} \left (a c -\frac {b^{2}}{4}\right )}}{2}+\left (\frac {b c d}{2}+e \left (a c -\frac {b^{2}}{2}\right )\right ) d \right ) d \sqrt {e \,x^{2}+d}\, \arctan \left (\frac {a \sqrt {e \,x^{2}+d}\, \sqrt {2}}{x \sqrt {\left (-2 a e +b d +\sqrt {-4 d^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) a}}\right )-e^{2} x \sqrt {-4 d^{2} \left (a c -\frac {b^{2}}{4}\right )}\, \sqrt {\left (-2 a e +b d +\sqrt {-4 d^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) a}\right ) \sqrt {\left (2 a e -b d +\sqrt {-4 d^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) a}}{\sqrt {\left (-2 a e +b d +\sqrt {-4 d^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) a}\, \sqrt {e \,x^{2}+d}\, \sqrt {\left (2 a e -b d +\sqrt {-4 d^{2} \left (a c -\frac {b^{2}}{4}\right )}\right ) a}\, \sqrt {-4 d^{2} \left (a c -\frac {b^{2}}{4}\right )}\, \left (a \,e^{2}-b d e +c \,d^{2}\right ) d}\)	\(429\)

[In]

int(1/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x,method=_RETURNVERBOSE)

[Out]

-1/((-2*a*e+b*d+(-4*d^2*(a*c-1/4*b^2))^(1/2))*a)^(1/2)/(e*x^2+d)^(1/2)/((2*a*e-b*d+(-4*d^2*(a*c-1/4*b^2))^(1/2
))*a)^(1/2)/(-4*d^2*(a*c-1/4*b^2))^(1/2)*((1/2*(b*e-c*d)*(-4*d^2*(a*c-1/4*b^2))^(1/2)+(1/2*b*c*d+e*(a*c-1/2*b^
2))*d)*2^(1/2)*d*((-2*a*e+b*d+(-4*d^2*(a*c-1/4*b^2))^(1/2))*a)^(1/2)*(e*x^2+d)^(1/2)*arctanh(a/x*(e*x^2+d)^(1/
2)*2^(1/2)/((2*a*e-b*d+(-4*d^2*(a*c-1/4*b^2))^(1/2))*a)^(1/2))+(2^(1/2)*(1/2*(-b*e+c*d)*(-4*d^2*(a*c-1/4*b^2))
^(1/2)+(1/2*b*c*d+e*(a*c-1/2*b^2))*d)*d*(e*x^2+d)^(1/2)*arctan(a/x*(e*x^2+d)^(1/2)*2^(1/2)/((-2*a*e+b*d+(-4*d^
2*(a*c-1/4*b^2))^(1/2))*a)^(1/2))-e^2*x*(-4*d^2*(a*c-1/4*b^2))^(1/2)*((-2*a*e+b*d+(-4*d^2*(a*c-1/4*b^2))^(1/2)
)*a)^(1/2))*((2*a*e-b*d+(-4*d^2*(a*c-1/4*b^2))^(1/2))*a)^(1/2))/(a*e^2-b*d*e+c*d^2)/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 17249 vs. \(2 (299) = 598\).

Time = 217.47 (sec) , antiderivative size = 17249, normalized size of antiderivative = 50.58 \[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )} \, dx=\text {Too large to display} \]

[In]

integrate(1/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x, algorithm="fricas")

[Out]

Too large to include

Sympy [F]

\[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )} \, dx=\int \frac {1}{\left (d + e x^{2}\right )^{\frac {3}{2}} \left (a + b x^{2} + c x^{4}\right )}\, dx \]

[In]

integrate(1/(e*x**2+d)**(3/2)/(c*x**4+b*x**2+a),x)

[Out]

Integral(1/((d + e*x**2)**(3/2)*(a + b*x**2 + c*x**4)), x)

Maxima [F]

\[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )} \, dx=\int { \frac {1}{{\left (c x^{4} + b x^{2} + a\right )} {\left (e x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

integrate(1/((c*x^4 + b*x^2 + a)*(e*x^2 + d)^(3/2)), x)

Giac [F(-1)]

Timed out. \[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )} \, dx=\text {Timed out} \]

[In]

integrate(1/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )} \, dx=\int \frac {1}{{\left (e\,x^2+d\right )}^{3/2}\,\left (c\,x^4+b\,x^2+a\right )} \,d x \]

[In]

int(1/((d + e*x^2)^(3/2)*(a + b*x^2 + c*x^4)),x)

[Out]

int(1/((d + e*x^2)^(3/2)*(a + b*x^2 + c*x^4)), x)