3.386 \(\int \frac{x^8}{\sqrt{d+e x^2} (a+b x^2+c x^4)} \, dx\)

Optimal. Leaf size=479 \[ -\frac{\left (-\frac{2 a^2 c^2-4 a b^2 c+b^4}{\sqrt{b^2-4 a c}}-2 a b c+b^3\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{c^3 \sqrt{b-\sqrt{b^2-4 a c}} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}-\frac{\left (\frac{2 a^2 c^2-4 a b^2 c+b^4}{\sqrt{b^2-4 a c}}-2 a b c+b^3\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{d+e x^2}}\right )}{c^3 \sqrt{\sqrt{b^2-4 a c}+b} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}+\frac{\left (b^2-a c\right ) \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{c^3 \sqrt{e}}+\frac{b d \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{2 c^2 e^{3/2}}-\frac{b x \sqrt{d+e x^2}}{2 c^2 e}+\frac{3 d^2 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{8 c e^{5/2}}-\frac{3 d x \sqrt{d+e x^2}}{8 c e^2}+\frac{x^3 \sqrt{d+e x^2}}{4 c e} \]

[Out]

(-3*d*x*Sqrt[d + e*x^2])/(8*c*e^2) - (b*x*Sqrt[d + e*x^2])/(2*c^2*e) + (x^3*Sqrt[d + e*x^2])/(4*c*e) - ((b^3 -
2*a*b*c - (b^4 - 4*a*b^2*c + 2*a^2*c^2)/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])])/(c^3*Sqrt[b - Sqrt[b^2 - 4*a*c]]*Sqrt[2*c*d - (b - Sqrt[b^2 -
4*a*c])*e]) - ((b^3 - 2*a*b*c + (b^4 - 4*a*b^2*c + 2*a^2*c^2)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2*c*d - (b + Sq
rt[b^2 - 4*a*c])*e]*x)/(Sqrt[b + Sqrt[b^2 - 4*a*c]]*Sqrt[d + e*x^2])])/(c^3*Sqrt[b + Sqrt[b^2 - 4*a*c]]*Sqrt[2
*c*d - (b + Sqrt[b^2 - 4*a*c])*e]) + (3*d^2*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(8*c*e^(5/2)) + (b*d*ArcTanh
[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(2*c^2*e^(3/2)) + ((b^2 - a*c)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(c^3*Sqrt[
e])

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Rubi [A]  time = 1.85821, antiderivative size = 479, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {1303, 217, 206, 321, 1692, 377, 205} \[ -\frac{\left (-\frac{2 a^2 c^2-4 a b^2 c+b^4}{\sqrt{b^2-4 a c}}-2 a b c+b^3\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{c^3 \sqrt{b-\sqrt{b^2-4 a c}} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}-\frac{\left (\frac{2 a^2 c^2-4 a b^2 c+b^4}{\sqrt{b^2-4 a c}}-2 a b c+b^3\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{d+e x^2}}\right )}{c^3 \sqrt{\sqrt{b^2-4 a c}+b} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}+\frac{\left (b^2-a c\right ) \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{c^3 \sqrt{e}}+\frac{b d \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{2 c^2 e^{3/2}}-\frac{b x \sqrt{d+e x^2}}{2 c^2 e}+\frac{3 d^2 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{8 c e^{5/2}}-\frac{3 d x \sqrt{d+e x^2}}{8 c e^2}+\frac{x^3 \sqrt{d+e x^2}}{4 c e} \]

Antiderivative was successfully verified.

[In]

Int[x^8/(Sqrt[d + e*x^2]*(a + b*x^2 + c*x^4)),x]

[Out]

(-3*d*x*Sqrt[d + e*x^2])/(8*c*e^2) - (b*x*Sqrt[d + e*x^2])/(2*c^2*e) + (x^3*Sqrt[d + e*x^2])/(4*c*e) - ((b^3 -
2*a*b*c - (b^4 - 4*a*b^2*c + 2*a^2*c^2)/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])])/(c^3*Sqrt[b - Sqrt[b^2 - 4*a*c]]*Sqrt[2*c*d - (b - Sqrt[b^2 -
4*a*c])*e]) - ((b^3 - 2*a*b*c + (b^4 - 4*a*b^2*c + 2*a^2*c^2)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2*c*d - (b + Sq
rt[b^2 - 4*a*c])*e]*x)/(Sqrt[b + Sqrt[b^2 - 4*a*c]]*Sqrt[d + e*x^2])])/(c^3*Sqrt[b + Sqrt[b^2 - 4*a*c]]*Sqrt[2
*c*d - (b + Sqrt[b^2 - 4*a*c])*e]) + (3*d^2*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(8*c*e^(5/2)) + (b*d*ArcTanh
[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(2*c^2*e^(3/2)) + ((b^2 - a*c)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(c^3*Sqrt[
e])

Rule 1303

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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 321

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

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 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 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]

Rubi steps

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

Mathematica [A]  time = 1.93862, size = 461, normalized size = 0.96 \[ \frac{-\frac{8 \left (-\frac{2 a^2 c^2-4 a b^2 c+b^4}{\sqrt{b^2-4 a c}}-2 a b c+b^3\right ) \tan ^{-1}\left (\frac{x \sqrt{e \sqrt{b^2-4 a c}-b e+2 c d}}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{d+e x^2}}\right )}{\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}}-\frac{8 \left (\frac{2 a^2 c^2-4 a b^2 c+b^4}{\sqrt{b^2-4 a c}}-2 a b c+b^3\right ) \tan ^{-1}\left (\frac{x \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{d+e x^2}}\right )}{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}+\frac{8 \left (b^2-a c\right ) \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{\sqrt{e}}+\frac{4 b c d \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{e^{3/2}}-\frac{4 b c x \sqrt{d+e x^2}}{e}+\frac{3 c^2 d \left (d \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )-\sqrt{e} x \sqrt{d+e x^2}\right )}{e^{5/2}}+\frac{2 c^2 x^3 \sqrt{d+e x^2}}{e}}{8 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8/(Sqrt[d + e*x^2]*(a + b*x^2 + c*x^4)),x]

[Out]

((-4*b*c*x*Sqrt[d + e*x^2])/e + (2*c^2*x^3*Sqrt[d + e*x^2])/e - (8*(b^3 - 2*a*b*c - (b^4 - 4*a*b^2*c + 2*a^2*c
^2)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2*c*d - b*e + 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]) - (8*(b^3 - 2*a*b*c + (b^4
- 4*a*b^2*c + 2*a^2*c^2)/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]) + (4*b*
c*d*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/e^(3/2) + (8*(b^2 - a*c)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/Sqrt[
e] + (3*c^2*d*(-(Sqrt[e]*x*Sqrt[d + e*x^2]) + d*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]))/e^(5/2))/(8*c^3)

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Maple [C]  time = 0.029, size = 377, normalized size = 0.8 \begin{align*}{\frac{{x}^{3}}{4\,ce}\sqrt{e{x}^{2}+d}}-{\frac{3\,dx}{8\,c{e}^{2}}\sqrt{e{x}^{2}+d}}+{\frac{3\,{d}^{2}}{8\,c}\ln \left ( \sqrt{e}x+\sqrt{e{x}^{2}+d} \right ){e}^{-{\frac{5}{2}}}}-{\frac{bx}{2\,{c}^{2}e}\sqrt{e{x}^{2}+d}}+{\frac{bd}{2\,{c}^{2}}\ln \left ( \sqrt{e}x+\sqrt{e{x}^{2}+d} \right ){e}^{-{\frac{3}{2}}}}-{\frac{a}{{c}^{2}}\ln \left ( \sqrt{e}x+\sqrt{e{x}^{2}+d} \right ){\frac{1}{\sqrt{e}}}}+{\frac{{b}^{2}}{{c}^{3}}\ln \left ( \sqrt{e}x+\sqrt{e{x}^{2}+d} \right ){\frac{1}{\sqrt{e}}}}-{\frac{1}{2\,{c}^{3}}\sqrt{e}\sum _{{\it \_R}={\it RootOf} \left ( c{{\it \_Z}}^{4}+ \left ( 4\,be-4\,cd \right ){{\it \_Z}}^{3}+ \left ( 16\,a{e}^{2}-8\,deb+6\,c{d}^{2} \right ){{\it \_Z}}^{2}+ \left ( 4\,b{d}^{2}e-4\,c{d}^{3} \right ){\it \_Z}+c{d}^{4} \right ) }{\frac{b \left ( 2\,ac-{b}^{2} \right ){{\it \_R}}^{2}+2\, \left ( 2\,{a}^{2}ce-2\,a{b}^{2}e-2\,abcd+{b}^{3}d \right ){\it \_R}+2\,abc{d}^{2}-{b}^{3}{d}^{2}}{{{\it \_R}}^{3}c+3\,{{\it \_R}}^{2}be-3\,{{\it \_R}}^{2}cd+8\,{\it \_R}\,a{e}^{2}-4\,{\it \_R}\,bde+3\,{\it \_R}\,c{d}^{2}+b{d}^{2}e-c{d}^{3}}\ln \left ( \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x \right ) ^{2}-{\it \_R} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(c*x^4+b*x^2+a)/(e*x^2+d)^(1/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{8}}{{\left (c x^{4} + b x^{2} + a\right )} \sqrt{e x^{2} + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(c*x^4+b*x^2+a)/(e*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^8/((c*x^4 + b*x^2 + a)*sqrt(e*x^2 + d)), 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(x^8/(c*x^4+b*x^2+a)/(e*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{8}}{\sqrt{d + e x^{2}} \left (a + b x^{2} + c x^{4}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8/(c*x**4+b*x**2+a)/(e*x**2+d)**(1/2),x)

[Out]

Integral(x**8/(sqrt(d + e*x**2)*(a + b*x**2 + c*x**4)), x)

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Giac [A]  time = 1.23256, size = 142, normalized size = 0.3 \begin{align*} \frac{1}{8} \, \sqrt{x^{2} e + d}{\left (\frac{2 \, x^{2} e^{\left (-1\right )}}{c} - \frac{{\left (3 \, c^{5} d e + 4 \, b c^{4} e^{2}\right )} e^{\left (-3\right )}}{c^{6}}\right )} x - \frac{{\left (3 \, c^{2} d^{2} + 4 \, b c d e + 8 \, b^{2} e^{2} - 8 \, a c e^{2}\right )} e^{\left (-\frac{5}{2}\right )} \log \left ({\left (x e^{\frac{1}{2}} - \sqrt{x^{2} e + d}\right )}^{2}\right )}{16 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(c*x^4+b*x^2+a)/(e*x^2+d)^(1/2),x, algorithm="giac")

[Out]

1/8*sqrt(x^2*e + d)*(2*x^2*e^(-1)/c - (3*c^5*d*e + 4*b*c^4*e^2)*e^(-3)/c^6)*x - 1/16*(3*c^2*d^2 + 4*b*c*d*e +
8*b^2*e^2 - 8*a*c*e^2)*e^(-5/2)*log((x*e^(1/2) - sqrt(x^2*e + d))^2)/c^3