3.360 \(\int \frac{\sqrt{d+e x^2}}{x^5 (a+b x^2+c x^4)} \, dx\)

Optimal. Leaf size=552 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right ) \left (-a b e-a c d+b^2 d\right )}{a^3 \sqrt{d}}+\frac{\sqrt{c} \left (b^2 \left (d \sqrt{b^2-4 a c}-a e\right )-a b \left (e \sqrt{b^2-4 a c}+3 c d\right )-a c \left (d \sqrt{b^2-4 a c}-2 a e\right )+b^3 d\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2} a^3 \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}-\frac{\sqrt{c} \left (-b^2 \left (d \sqrt{b^2-4 a c}+a e\right )-a b \left (3 c d-e \sqrt{b^2-4 a c}\right )+a c \left (d \sqrt{b^2-4 a c}+2 a e\right )+b^3 d\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} a^3 \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}-\frac{e (b d-a e) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{2 a^2 d^{3/2}}+\frac{\sqrt{d+e x^2} (b d-a e)}{2 a^2 d x^2}-\frac{3 e^2 \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{8 a d^{3/2}}+\frac{3 e \sqrt{d+e x^2}}{8 a d x^2}-\frac{\sqrt{d+e x^2}}{4 a x^4} \]

[Out]

-Sqrt[d + e*x^2]/(4*a*x^4) + (3*e*Sqrt[d + e*x^2])/(8*a*d*x^2) + ((b*d - a*e)*Sqrt[d + e*x^2])/(2*a^2*d*x^2) -
(3*e^2*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(8*a*d^(3/2)) - (e*(b*d - a*e)*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(2*
a^2*d^(3/2)) - ((b^2*d - a*c*d - a*b*e)*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(a^3*Sqrt[d]) + (Sqrt[c]*(b^3*d - a*
c*(Sqrt[b^2 - 4*a*c]*d - 2*a*e) + b^2*(Sqrt[b^2 - 4*a*c]*d - a*e) - a*b*(3*c*d + Sqrt[b^2 - 4*a*c]*e))*ArcTanh
[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x^2])/Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]])/(Sqrt[2]*a^3*Sqrt[b^2 - 4*a*c]*Sq
rt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]) - (Sqrt[c]*(b^3*d - b^2*(Sqrt[b^2 - 4*a*c]*d + a*e) + a*c*(Sqrt[b^2 - 4
*a*c]*d + 2*a*e) - a*b*(3*c*d - Sqrt[b^2 - 4*a*c]*e))*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x^2])/Sqrt[2*c*d - (
b + Sqrt[b^2 - 4*a*c])*e]])/(Sqrt[2]*a^3*Sqrt[b^2 - 4*a*c]*Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e])

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Rubi [A]  time = 4.24389, antiderivative size = 552, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {1251, 897, 1287, 199, 206, 1166, 208} \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right ) \left (-a b e-a c d+b^2 d\right )}{a^3 \sqrt{d}}+\frac{\sqrt{c} \left (b^2 \left (d \sqrt{b^2-4 a c}-a e\right )-a b \left (e \sqrt{b^2-4 a c}+3 c d\right )-a c \left (d \sqrt{b^2-4 a c}-2 a e\right )+b^3 d\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2} a^3 \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}-\frac{\sqrt{c} \left (-b^2 \left (d \sqrt{b^2-4 a c}+a e\right )-a b \left (3 c d-e \sqrt{b^2-4 a c}\right )+a c \left (d \sqrt{b^2-4 a c}+2 a e\right )+b^3 d\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} a^3 \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}-\frac{e (b d-a e) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{2 a^2 d^{3/2}}+\frac{\sqrt{d+e x^2} (b d-a e)}{2 a^2 d x^2}-\frac{3 e^2 \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{8 a d^{3/2}}+\frac{3 e \sqrt{d+e x^2}}{8 a d x^2}-\frac{\sqrt{d+e x^2}}{4 a x^4} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[d + e*x^2]/(x^5*(a + b*x^2 + c*x^4)),x]

[Out]

-Sqrt[d + e*x^2]/(4*a*x^4) + (3*e*Sqrt[d + e*x^2])/(8*a*d*x^2) + ((b*d - a*e)*Sqrt[d + e*x^2])/(2*a^2*d*x^2) -
(3*e^2*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(8*a*d^(3/2)) - (e*(b*d - a*e)*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(2*
a^2*d^(3/2)) - ((b^2*d - a*c*d - a*b*e)*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(a^3*Sqrt[d]) + (Sqrt[c]*(b^3*d - a*
c*(Sqrt[b^2 - 4*a*c]*d - 2*a*e) + b^2*(Sqrt[b^2 - 4*a*c]*d - a*e) - a*b*(3*c*d + Sqrt[b^2 - 4*a*c]*e))*ArcTanh
[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x^2])/Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]])/(Sqrt[2]*a^3*Sqrt[b^2 - 4*a*c]*Sq
rt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]) - (Sqrt[c]*(b^3*d - b^2*(Sqrt[b^2 - 4*a*c]*d + a*e) + a*c*(Sqrt[b^2 - 4
*a*c]*d + 2*a*e) - a*b*(3*c*d - Sqrt[b^2 - 4*a*c]*e))*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x^2])/Sqrt[2*c*d - (
b + Sqrt[b^2 - 4*a*c])*e]])/(Sqrt[2]*a^3*Sqrt[b^2 - 4*a*c]*Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e])

Rule 1251

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

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1287

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.))/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[Ex
pandIntegrand[((f*x)^m*(d + e*x^2)^q)/(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] && IntegerQ[m]

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 1166

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

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\sqrt{d+e x^2}}{x^5 \left (a+b x^2+c x^4\right )} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{\sqrt{d+e x}}{x^3 \left (a+b x+c x^2\right )} \, dx,x,x^2\right )\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{\left (-\frac{d}{e}+\frac{x^2}{e}\right )^3 \left (\frac{c d^2-b d e+a e^2}{e^2}-\frac{(2 c d-b e) x^2}{e^2}+\frac{c x^4}{e^2}\right )} \, dx,x,\sqrt{d+e x^2}\right )}{e}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{d e^3}{a \left (d-x^2\right )^3}+\frac{e^2 (-b d+a e)}{a^2 \left (d-x^2\right )^2}+\frac{e \left (-b^2 d+a c d+a b e\right )}{a^3 \left (d-x^2\right )}+\frac{e \left (\left (b^2-a c\right ) \left (c d^2-b d e+a e^2\right )-c \left (b^2 d-a c d-a b e\right ) x^2\right )}{a^3 \left (c d^2-b d e+a e^2-(2 c d-b e) x^2+c x^4\right )}\right ) \, dx,x,\sqrt{d+e x^2}\right )}{e}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (b^2-a c\right ) \left (c d^2-b d e+a e^2\right )-c \left (b^2 d-a c d-a b e\right ) x^2}{c d^2-b d e+a e^2+(-2 c d+b e) x^2+c x^4} \, dx,x,\sqrt{d+e x^2}\right )}{a^3}-\frac{\left (d e^2\right ) \operatorname{Subst}\left (\int \frac{1}{\left (d-x^2\right )^3} \, dx,x,\sqrt{d+e x^2}\right )}{a}-\frac{(e (b d-a e)) \operatorname{Subst}\left (\int \frac{1}{\left (d-x^2\right )^2} \, dx,x,\sqrt{d+e x^2}\right )}{a^2}-\frac{\left (b^2 d-a c d-a b e\right ) \operatorname{Subst}\left (\int \frac{1}{d-x^2} \, dx,x,\sqrt{d+e x^2}\right )}{a^3}\\ &=-\frac{\sqrt{d+e x^2}}{4 a x^4}+\frac{(b d-a e) \sqrt{d+e x^2}}{2 a^2 d x^2}-\frac{\left (b^2 d-a c d-a b e\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{a^3 \sqrt{d}}-\frac{\left (3 e^2\right ) \operatorname{Subst}\left (\int \frac{1}{\left (d-x^2\right )^2} \, dx,x,\sqrt{d+e x^2}\right )}{4 a}-\frac{(e (b d-a e)) \operatorname{Subst}\left (\int \frac{1}{d-x^2} \, dx,x,\sqrt{d+e x^2}\right )}{2 a^2 d}+\frac{\left (c \left (b^3 d-b^2 \left (\sqrt{b^2-4 a c} d+a e\right )+a c \left (\sqrt{b^2-4 a c} d+2 a e\right )-a b \left (3 c d-\sqrt{b^2-4 a c} e\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{2} \sqrt{b^2-4 a c} e+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x^2}\right )}{2 a^3 \sqrt{b^2-4 a c}}-\frac{\left (c \left (b^3 d-a c \left (\sqrt{b^2-4 a c} d-2 a e\right )+b^2 \left (\sqrt{b^2-4 a c} d-a e\right )-a b \left (3 c d+\sqrt{b^2-4 a c} e\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{2} \sqrt{b^2-4 a c} e+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x^2}\right )}{2 a^3 \sqrt{b^2-4 a c}}\\ &=-\frac{\sqrt{d+e x^2}}{4 a x^4}+\frac{3 e \sqrt{d+e x^2}}{8 a d x^2}+\frac{(b d-a e) \sqrt{d+e x^2}}{2 a^2 d x^2}-\frac{e (b d-a e) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{2 a^2 d^{3/2}}-\frac{\left (b^2 d-a c d-a b e\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{a^3 \sqrt{d}}+\frac{\sqrt{c} \left (b^3 d-a c \left (\sqrt{b^2-4 a c} d-2 a e\right )+b^2 \left (\sqrt{b^2-4 a c} d-a e\right )-a b \left (3 c d+\sqrt{b^2-4 a c} e\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}}\right )}{\sqrt{2} a^3 \sqrt{b^2-4 a c} \sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}}-\frac{\sqrt{c} \left (b^3 d-b^2 \left (\sqrt{b^2-4 a c} d+a e\right )+a c \left (\sqrt{b^2-4 a c} d+2 a e\right )-a b \left (3 c d-\sqrt{b^2-4 a c} e\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}}\right )}{\sqrt{2} a^3 \sqrt{b^2-4 a c} \sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}}-\frac{\left (3 e^2\right ) \operatorname{Subst}\left (\int \frac{1}{d-x^2} \, dx,x,\sqrt{d+e x^2}\right )}{8 a d}\\ &=-\frac{\sqrt{d+e x^2}}{4 a x^4}+\frac{3 e \sqrt{d+e x^2}}{8 a d x^2}+\frac{(b d-a e) \sqrt{d+e x^2}}{2 a^2 d x^2}-\frac{3 e^2 \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{8 a d^{3/2}}-\frac{e (b d-a e) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{2 a^2 d^{3/2}}-\frac{\left (b^2 d-a c d-a b e\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{a^3 \sqrt{d}}+\frac{\sqrt{c} \left (b^3 d-a c \left (\sqrt{b^2-4 a c} d-2 a e\right )+b^2 \left (\sqrt{b^2-4 a c} d-a e\right )-a b \left (3 c d+\sqrt{b^2-4 a c} e\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}}\right )}{\sqrt{2} a^3 \sqrt{b^2-4 a c} \sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}}-\frac{\sqrt{c} \left (b^3 d-b^2 \left (\sqrt{b^2-4 a c} d+a e\right )+a c \left (\sqrt{b^2-4 a c} d+2 a e\right )-a b \left (3 c d-\sqrt{b^2-4 a c} e\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}}\right )}{\sqrt{2} a^3 \sqrt{b^2-4 a c} \sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}}\\ \end{align*}

Mathematica [A]  time = 2.04494, size = 466, normalized size = 0.84 \[ \frac{\frac{\log \left (\sqrt{d} \sqrt{d+e x^2}+d\right ) \left (4 a b d e+a \left (a e^2+8 c d^2\right )-8 b^2 d^2\right )}{d^{3/2}}-\frac{\log (x) \left (4 a b d e+a \left (a e^2+8 c d^2\right )-8 b^2 d^2\right )}{d^{3/2}}-\frac{4 \sqrt{2} \sqrt{c} \left (\frac{\left (b^2 \left (a e-d \sqrt{b^2-4 a c}\right )+a b \left (e \sqrt{b^2-4 a c}+3 c d\right )+a c \left (d \sqrt{b^2-4 a c}-2 a e\right )+b^3 (-d)\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{e \sqrt{b^2-4 a c}-b e+2 c d}}\right )}{\sqrt{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}}+\frac{\left (-b^2 \left (d \sqrt{b^2-4 a c}+a e\right )+a b \left (e \sqrt{b^2-4 a c}-3 c d\right )+a c \left (d \sqrt{b^2-4 a c}+2 a e\right )+b^3 d\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{b^2-4 a c}}+\frac{a \sqrt{d+e x^2} \left (4 b d x^2-a \left (2 d+e x^2\right )\right )}{d x^4}}{8 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d + e*x^2]/(x^5*(a + b*x^2 + c*x^4)),x]

[Out]

((a*Sqrt[d + e*x^2]*(4*b*d*x^2 - a*(2*d + e*x^2)))/(d*x^4) - (4*Sqrt[2]*Sqrt[c]*(((-(b^3*d) + a*c*(Sqrt[b^2 -
4*a*c]*d - 2*a*e) + b^2*(-(Sqrt[b^2 - 4*a*c]*d) + a*e) + a*b*(3*c*d + Sqrt[b^2 - 4*a*c]*e))*ArcTanh[(Sqrt[2]*S
qrt[c]*Sqrt[d + e*x^2])/Sqrt[2*c*d - b*e + Sqrt[b^2 - 4*a*c]*e]])/Sqrt[2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e] + (
(b^3*d - b^2*(Sqrt[b^2 - 4*a*c]*d + a*e) + a*c*(Sqrt[b^2 - 4*a*c]*d + 2*a*e) + a*b*(-3*c*d + Sqrt[b^2 - 4*a*c]
*e))*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x^2])/Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]])/Sqrt[2*c*d - (b + Sqr
t[b^2 - 4*a*c])*e]))/Sqrt[b^2 - 4*a*c] - ((-8*b^2*d^2 + 4*a*b*d*e + a*(8*c*d^2 + a*e^2))*Log[x])/d^(3/2) + ((-
8*b^2*d^2 + 4*a*b*d*e + a*(8*c*d^2 + a*e^2))*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])/d^(3/2))/(8*a^3)

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Maple [C]  time = 0.033, size = 655, normalized size = 1.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^(1/2)/x^5/(c*x^4+b*x^2+a),x)

[Out]

-1/4/a/d/x^4*(e*x^2+d)^(3/2)+1/8/a*e/d^2/x^2*(e*x^2+d)^(3/2)+1/8/a*e^2/d^(3/2)*ln((2*d+2*d^(1/2)*(e*x^2+d)^(1/
2))/x)-1/8/a*e^2/d^2*(e*x^2+d)^(1/2)-1/2/a^2*e^(1/2)*x*c-1/2/a^2*(e*x^2+d)^(1/2)*c+1/2/a^3*e^(1/2)*x*b^2+1/2/a
^3*(e*x^2+d)^(1/2)*b^2+1/2/a^2*d/((e*x^2+d)^(1/2)-e^(1/2)*x)*c-1/2/a^3*d/((e*x^2+d)^(1/2)-e^(1/2)*x)*b^2+1/4/a
^3*sum((c*(a*b*e+a*c*d-b^2*d)*_R^6+(-4*a^2*c*e^2+4*a*b^2*e^2+5*a*b*c*d*e-3*a*c^2*d^2-4*b^3*d*e+3*b^2*c*d^2)*_R
^4+d*(4*a^2*c*e^2-4*a*b^2*e^2-5*a*b*c*d*e+3*a*c^2*d^2+4*b^3*d*e-3*b^2*c*d^2)*_R^2-a*b*c*d^3*e-a*c^2*d^4+b^2*c*
d^4)/(_R^7*c+3*_R^5*b*e-3*_R^5*c*d+8*_R^3*a*e^2-4*_R^3*b*d*e+3*_R^3*c*d^2+_R*b*d^2*e-_R*c*d^3)*ln((e*x^2+d)^(1
/2)-e^(1/2)*x-_R),_R=RootOf(c*_Z^8+(4*b*e-4*c*d)*_Z^6+(16*a*e^2-8*b*d*e+6*c*d^2)*_Z^4+(4*b*d^2*e-4*c*d^3)*_Z^2
+c*d^4))+1/a^2*d^(1/2)*ln((2*d+2*d^(1/2)*(e*x^2+d)^(1/2))/x)*c-1/a^3*d^(1/2)*ln((2*d+2*d^(1/2)*(e*x^2+d)^(1/2)
)/x)*b^2+1/2/a^2*b/d/x^2*(e*x^2+d)^(3/2)+1/2/a^2*b*e/d^(1/2)*ln((2*d+2*d^(1/2)*(e*x^2+d)^(1/2))/x)-1/2/a^2*b*e
/d*(e*x^2+d)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{e x^{2} + d}}{{\left (c x^{4} + b x^{2} + a\right )} x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^(1/2)/x^5/(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

integrate(sqrt(e*x^2 + d)/((c*x^4 + b*x^2 + a)*x^5), x)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^(1/2)/x^5/(c*x^4+b*x^2+a),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**(1/2)/x**5/(c*x**4+b*x**2+a),x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^(1/2)/x^5/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

Timed out