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

Optimal. Leaf size=339 \[ -\frac {2 c^2 \left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \tan ^{-1}\left (\frac {x \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{a \sqrt {b-\sqrt {b^2-4 a c}} \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )^{3/2}}-\frac {2 c^2 \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \tan ^{-1}\left (\frac {x \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}{\sqrt {\sqrt {b^2-4 a c}+b} \sqrt {d+e x^2}}\right )}{a \sqrt {\sqrt {b^2-4 a c}+b} \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )^{3/2}}+\frac {e x (c d-b e)}{a d \sqrt {d+e x^2} \left (e (a e-b d)+c d^2\right )}+\frac {-d-2 e x^2}{a d^2 x \sqrt {d+e x^2}} \]

[Out]

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

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Rubi [A]  time = 2.84, antiderivative size = 462, normalized size of antiderivative = 1.36, number of steps used = 12, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {1301, 271, 191, 6728, 264, 1692, 377, 205} \[ -\frac {c \left (\frac {2 a c e+b^2 (-e)+b c d}{\sqrt {b^2-4 a c}}-b e+c d\right ) \tan ^{-1}\left (\frac {x \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{a \sqrt {b-\sqrt {b^2-4 a c}} \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )} \left (a e^2-b d e+c d^2\right )}-\frac {c \left (-\frac {2 a c e+b^2 (-e)+b c d}{\sqrt {b^2-4 a c}}-b e+c d\right ) \tan ^{-1}\left (\frac {x \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}{\sqrt {\sqrt {b^2-4 a c}+b} \sqrt {d+e x^2}}\right )}{a \sqrt {\sqrt {b^2-4 a c}+b} \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )} \left (a e^2-b d e+c d^2\right )}-\frac {2 e^3 x}{d^2 \sqrt {d+e x^2} \left (a e^2-b d e+c d^2\right )}-\frac {e^2}{d x \sqrt {d+e x^2} \left (a e^2-b d e+c d^2\right )}-\frac {\sqrt {d+e x^2} (c d-b e)}{a d x \left (a e^2-b d e+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(e^2/(d*(c*d^2 - b*d*e + a*e^2)*x*Sqrt[d + e*x^2])) - (2*e^3*x)/(d^2*(c*d^2 - b*d*e + a*e^2)*Sqrt[d + e*x^2])
- ((c*d - b*e)*Sqrt[d + e*x^2])/(a*d*(c*d^2 - b*d*e + a*e^2)*x) - (c*(c*d - b*e + (b*c*d - b^2*e + 2*a*c*e)/S
qrt[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])])/(a*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*(c
*d - b*e - (b*c*d - b^2*e + 2*a*c*e)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*x)/(Sq
rt[b + Sqrt[b^2 - 4*a*c]]*Sqrt[d + e*x^2])])/(a*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 191

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 205

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

Rule 264

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

Rule 271

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

Rule 377

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 1301

Int[(((f_.)*(x_))^(m_.)*((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[(f*x)^m*(d + e*x^2)^q, x], x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[((f*x)^m*(d
+ e*x^2)^(q + 1)*Simp[c*d - b*e - c*e*x^2, x])/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x]
&& NeQ[b^2 - 4*a*c, 0] &&  !IntegerQ[q] && LtQ[q, -1]

Rule 1692

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]

Rule 6728

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*} \int \frac {1}{x^2 \left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )} \, dx &=\frac {\int \frac {c d-b e-c e x^2}{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}{x^2 \left (d+e x^2\right )^{3/2}} \, dx}{c d^2-b d e+a e^2}\\ &=-\frac {e^2}{d \left (c d^2-b d e+a e^2\right ) x \sqrt {d+e x^2}}+\frac {\int \left (\frac {c d-b e}{a x^2 \sqrt {d+e x^2}}+\frac {-b c d+b^2 e-a c e-c (c d-b e) x^2}{a \sqrt {d+e x^2} \left (a+b x^2+c x^4\right )}\right ) \, dx}{c d^2-b d e+a e^2}-\frac {\left (2 e^3\right ) \int \frac {1}{\left (d+e x^2\right )^{3/2}} \, dx}{d \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {e^2}{d \left (c d^2-b d e+a e^2\right ) x \sqrt {d+e x^2}}-\frac {2 e^3 x}{d^2 \left (c d^2-b d e+a e^2\right ) \sqrt {d+e x^2}}+\frac {\int \frac {-b c d+b^2 e-a c e-c (c d-b e) x^2}{\sqrt {d+e x^2} \left (a+b x^2+c x^4\right )} \, dx}{a \left (c d^2-b d e+a e^2\right )}+\frac {(c d-b e) \int \frac {1}{x^2 \sqrt {d+e x^2}} \, dx}{a \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {e^2}{d \left (c d^2-b d e+a e^2\right ) x \sqrt {d+e x^2}}-\frac {2 e^3 x}{d^2 \left (c d^2-b d e+a e^2\right ) \sqrt {d+e x^2}}-\frac {(c d-b e) \sqrt {d+e x^2}}{a d \left (c d^2-b d e+a e^2\right ) x}+\frac {\int \left (\frac {-c (c d-b e)+\frac {c \left (-b c d+b^2 e-2 a c e\right )}{\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 (c d-b e)-\frac {c \left (-b c d+b^2 e-2 a c e\right )}{\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}{a \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {e^2}{d \left (c d^2-b d e+a e^2\right ) x \sqrt {d+e x^2}}-\frac {2 e^3 x}{d^2 \left (c d^2-b d e+a e^2\right ) \sqrt {d+e x^2}}-\frac {(c d-b e) \sqrt {d+e x^2}}{a d \left (c d^2-b d e+a e^2\right ) x}-\frac {\left (c \left (c d-b e-\frac {b c d-b^2 e+2 a c 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}{a \left (c d^2-b d e+a e^2\right )}-\frac {\left (c \left (c d-b e+\frac {b c d-b^2 e+2 a c 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}{a \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {e^2}{d \left (c d^2-b d e+a e^2\right ) x \sqrt {d+e x^2}}-\frac {2 e^3 x}{d^2 \left (c d^2-b d e+a e^2\right ) \sqrt {d+e x^2}}-\frac {(c d-b e) \sqrt {d+e x^2}}{a d \left (c d^2-b d e+a e^2\right ) x}-\frac {\left (c \left (c d-b e-\frac {b c d-b^2 e+2 a c e}{\sqrt {b^2-4 a c}}\right )\right ) \operatorname {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 )}{a \left (c d^2-b d e+a e^2\right )}-\frac {\left (c \left (c d-b e+\frac {b c d-b^2 e+2 a c e}{\sqrt {b^2-4 a c}}\right )\right ) \operatorname {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 )}{a \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {e^2}{d \left (c d^2-b d e+a e^2\right ) x \sqrt {d+e x^2}}-\frac {2 e^3 x}{d^2 \left (c d^2-b d e+a e^2\right ) \sqrt {d+e x^2}}-\frac {(c d-b e) \sqrt {d+e x^2}}{a d \left (c d^2-b d e+a e^2\right ) x}-\frac {c \left (c d-b e+\frac {b c d-b^2 e+2 a c 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 )}{a \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 (c d-b e-\frac {b c d-b^2 e+2 a c 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 )}{a \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*}

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Mathematica [C]  time = 6.77, size = 2158, normalized size = 6.37 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((d + 2*e*x^2)/(a*d^2*x*Sqrt[d + e*x^2])) - ((c + (b*c)/Sqrt[b^2 - 4*a*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 - 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 + Sq
rt[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 + 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)))])/d))/(15*a*(b - 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)*(1 - (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c]))*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))]) + ((-c + (b*c)/Sqrt[b^2 - 4*a*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 + 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)) - (3
0*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 + Sqr
t[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 + 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))]*Hypergeometric2
F1[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*a*(b +
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)*
(1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]))*Sqrt[d + e*x^2]*Sqrt[((b + Sqrt[b^2 - 4*a*c])*(d + e*x^2))/(d*(b + Sqr
t[b^2 - 4*a*c] + 2*c*x^2))])

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [C]  time = 0.04, size = 387, normalized size = 1.14 \[ -\frac {8 b \,e^{\frac {3}{2}}}{\left (4 a \,e^{2}-4 d e b +4 c \,d^{2}\right ) \left (2 e \,x^{2}-2 \sqrt {e \,x^{2}+d}\, \sqrt {e}\, x +2 d \right ) a}+\frac {8 c d \sqrt {e}}{\left (4 a \,e^{2}-4 d e b +4 c \,d^{2}\right ) \left (2 e \,x^{2}-2 \sqrt {e \,x^{2}+d}\, \sqrt {e}\, x +2 d \right ) a}-\frac {2 \sqrt {e}\, \left (b c \,d^{2} e -c^{2} d^{3}+\left (b e -c d \right ) \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right )^{2} c +2 \left (-2 a c \,e^{2}+2 b^{2} e^{2}-3 b c d e +c^{2} d^{2}\right ) \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right )\right ) \ln \left (-\RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right )+\left (-\sqrt {e}\, x +\sqrt {e \,x^{2}+d}\right )^{2}\right )}{\left (4 a \,e^{2}-4 d e b +4 c \,d^{2}\right ) a \left (\RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right )^{3} c +3 \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right )^{2} b e -3 \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right )^{2} c d +8 \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right ) a \,e^{2}-4 \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right ) b d e +3 \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right ) c \,d^{2}+b \,d^{2} e -c \,d^{3}\right )}-\frac {2 e x}{\sqrt {e \,x^{2}+d}\, a \,d^{2}}-\frac {1}{\sqrt {e \,x^{2}+d}\, a d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a/d/x/(e*x^2+d)^(1/2)-2/a*e/d^2*x/(e*x^2+d)^(1/2)-2/a*e^(1/2)/(4*a*e^2-4*b*d*e+4*c*d^2)*sum((c*(b*e-c*d)*_R
^2+2*(-2*a*c*e^2+2*b^2*e^2-3*b*c*d*e+c^2*d^2)*_R+b*c*d^2*e-c^2*d^3)/(_R^3*c+3*_R^2*b*e-3*_R^2*c*d+8*_R*a*e^2-4
*_R*b*d*e+3*_R*c*d^2+b*d^2*e-c*d^3)*ln(-_R+(-e^(1/2)*x+(e*x^2+d)^(1/2))^2),_R=RootOf(_Z^4*c+c*d^4+(4*b*e-4*c*d
)*_Z^3+(16*a*e^2-8*b*d*e+6*c*d^2)*_Z^2+(4*b*d^2*e-4*c*d^3)*_Z))-8/a*e^(3/2)/(4*a*e^2-4*b*d*e+4*c*d^2)/(2*e*x^2
-2*(e*x^2+d)^(1/2)*e^(1/2)*x+2*d)*b+8/a*e^(1/2)/(4*a*e^2-4*b*d*e+4*c*d^2)/(2*e*x^2-2*(e*x^2+d)^(1/2)*e^(1/2)*x
+2*d)*c*d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{4} + b x^{2} + a\right )} {\left (e x^{2} + d\right )}^{\frac {3}{2}} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(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^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{x^2\,{\left (e\,x^2+d\right )}^{3/2}\,\left (c\,x^4+b\,x^2+a\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{2} \left (d + e x^{2}\right )^{\frac {3}{2}} \left (a + b x^{2} + c x^{4}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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