3.24 \(\int \frac{d+e x+f x^2+g x^3+h x^4+i x^5}{a+b x^2+c x^4} \, dx\)

Optimal. Leaf size=321 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right ) \left (\frac{-c (2 a h+b f)+b^2 h+2 c^2 d}{\sqrt{b^2-4 a c}}-b h+c f\right )}{\sqrt{2} c^{3/2} \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right ) \left (-\frac{-2 a c h+b^2 h-b c f+2 c^2 d}{\sqrt{b^2-4 a c}}-b h+c f\right )}{\sqrt{2} c^{3/2} \sqrt{\sqrt{b^2-4 a c}+b}}-\frac{\tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right ) \left (-2 a c i+b^2 i-b c g+2 c^2 e\right )}{2 c^2 \sqrt{b^2-4 a c}}+\frac{(c g-b i) \log \left (a+b x^2+c x^4\right )}{4 c^2}+\frac{h x}{c}+\frac{i x^2}{2 c} \]

[Out]

(h*x)/c + (i*x^2)/(2*c) + ((c*f - b*h + (2*c^2*d + b^2*h - c*(b*f + 2*a*h))/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]
*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*c^(3/2)*Sqrt[b - Sqrt[b^2 - 4*a*c]]) + ((c*f - b*h - (2*c^2
*d - b*c*f + b^2*h - 2*a*c*h)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqr
t[2]*c^(3/2)*Sqrt[b + Sqrt[b^2 - 4*a*c]]) - ((2*c^2*e - b*c*g + b^2*i - 2*a*c*i)*ArcTanh[(b + 2*c*x^2)/Sqrt[b^
2 - 4*a*c]])/(2*c^2*Sqrt[b^2 - 4*a*c]) + ((c*g - b*i)*Log[a + b*x^2 + c*x^4])/(4*c^2)

________________________________________________________________________________________

Rubi [A]  time = 0.533943, antiderivative size = 321, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1673, 1676, 1166, 205, 1663, 1657, 634, 618, 206, 628} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right ) \left (\frac{-c (2 a h+b f)+b^2 h+2 c^2 d}{\sqrt{b^2-4 a c}}-b h+c f\right )}{\sqrt{2} c^{3/2} \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right ) \left (-\frac{-2 a c h+b^2 h-b c f+2 c^2 d}{\sqrt{b^2-4 a c}}-b h+c f\right )}{\sqrt{2} c^{3/2} \sqrt{\sqrt{b^2-4 a c}+b}}-\frac{\tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right ) \left (-2 a c i+b^2 i-b c g+2 c^2 e\right )}{2 c^2 \sqrt{b^2-4 a c}}+\frac{(c g-b i) \log \left (a+b x^2+c x^4\right )}{4 c^2}+\frac{h x}{c}+\frac{i x^2}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x + f*x^2 + g*x^3 + h*x^4 + i*x^5)/(a + b*x^2 + c*x^4),x]

[Out]

(h*x)/c + (i*x^2)/(2*c) + ((c*f - b*h + (2*c^2*d + b^2*h - c*(b*f + 2*a*h))/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]
*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*c^(3/2)*Sqrt[b - Sqrt[b^2 - 4*a*c]]) + ((c*f - b*h - (2*c^2
*d - b*c*f + b^2*h - 2*a*c*h)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqr
t[2]*c^(3/2)*Sqrt[b + Sqrt[b^2 - 4*a*c]]) - ((2*c^2*e - b*c*g + b^2*i - 2*a*c*i)*ArcTanh[(b + 2*c*x^2)/Sqrt[b^
2 - 4*a*c]])/(2*c^2*Sqrt[b^2 - 4*a*c]) + ((c*g - b*i)*Log[a + b*x^2 + c*x^4])/(4*c^2)

Rule 1673

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], k}, Int[Sum[Coeff[
Pq, x, 2*k]*x^(2*k), {k, 0, q/2}]*(a + b*x^2 + c*x^4)^p, x] + Int[x*Sum[Coeff[Pq, x, 2*k + 1]*x^(2*k), {k, 0,
(q - 1)/2}]*(a + b*x^2 + c*x^4)^p, x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] &&  !PolyQ[Pq, x^2]

Rule 1676

Int[(Pq_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^2 + c*x^4), x], x
] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x^2] && Expon[Pq, x^2] > 1

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 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 1663

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)
*SubstFor[x^2, Pq, x]*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x^2] && Inte
gerQ[(m - 1)/2]

Rule 1657

Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 634

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

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{d+e x+f x^2+g x^3+h x^4+24 x^5}{a+b x^2+c x^4} \, dx &=\int \frac{x \left (e+g x^2+24 x^4\right )}{a+b x^2+c x^4} \, dx+\int \frac{d+f x^2+h x^4}{a+b x^2+c x^4} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{e+g x+24 x^2}{a+b x+c x^2} \, dx,x,x^2\right )+\int \left (\frac{h}{c}+\frac{c d-a h+(c f-b h) x^2}{c \left (a+b x^2+c x^4\right )}\right ) \, dx\\ &=\frac{h x}{c}+\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{24}{c}-\frac{24 a-c e+(24 b-c g) x}{c \left (a+b x+c x^2\right )}\right ) \, dx,x,x^2\right )+\frac{\int \frac{c d-a h+(c f-b h) x^2}{a+b x^2+c x^4} \, dx}{c}\\ &=\frac{h x}{c}+\frac{12 x^2}{c}-\frac{\operatorname{Subst}\left (\int \frac{24 a-c e+(24 b-c g) x}{a+b x+c x^2} \, dx,x,x^2\right )}{2 c}+\frac{\left (c f-b h-\frac{2 c^2 d-b c f+b^2 h-2 a c h}{\sqrt{b^2-4 a c}}\right ) \int \frac{1}{\frac{b}{2}+\frac{1}{2} \sqrt{b^2-4 a c}+c x^2} \, dx}{2 c}+\frac{\left (c f-b h+\frac{2 c^2 d+b^2 h-c (b f+2 a h)}{\sqrt{b^2-4 a c}}\right ) \int \frac{1}{\frac{b}{2}-\frac{1}{2} \sqrt{b^2-4 a c}+c x^2} \, dx}{2 c}\\ &=\frac{h x}{c}+\frac{12 x^2}{c}+\frac{\left (c f-b h+\frac{2 c^2 d+b^2 h-c (b f+2 a h)}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} c^{3/2} \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{\left (c f-b h-\frac{2 c^2 d-b c f+b^2 h-2 a c h}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b+\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} c^{3/2} \sqrt{b+\sqrt{b^2-4 a c}}}-\frac{(24 b-c g) \operatorname{Subst}\left (\int \frac{b+2 c x}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^2}-\frac{(2 c (24 a-c e)-b (24 b-c g)) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^2}\\ &=\frac{h x}{c}+\frac{12 x^2}{c}+\frac{\left (c f-b h+\frac{2 c^2 d+b^2 h-c (b f+2 a h)}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} c^{3/2} \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{\left (c f-b h-\frac{2 c^2 d-b c f+b^2 h-2 a c h}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b+\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} c^{3/2} \sqrt{b+\sqrt{b^2-4 a c}}}-\frac{(24 b-c g) \log \left (a+b x^2+c x^4\right )}{4 c^2}+\frac{(2 c (24 a-c e)-b (24 b-c g)) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^2\right )}{2 c^2}\\ &=\frac{h x}{c}+\frac{12 x^2}{c}+\frac{\left (c f-b h+\frac{2 c^2 d+b^2 h-c (b f+2 a h)}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} c^{3/2} \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{\left (c f-b h-\frac{2 c^2 d-b c f+b^2 h-2 a c h}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b+\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} c^{3/2} \sqrt{b+\sqrt{b^2-4 a c}}}-\frac{\left (24 b^2-2 c (24 a-c e)-b c g\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 c^2 \sqrt{b^2-4 a c}}-\frac{(24 b-c g) \log \left (a+b x^2+c x^4\right )}{4 c^2}\\ \end{align*}

Mathematica [A]  time = 0.797854, size = 441, normalized size = 1.37 \[ \frac{\frac{2 \sqrt{2} \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right ) \left (c \left (f \sqrt{b^2-4 a c}-2 a h-b f\right )+b h \left (b-\sqrt{b^2-4 a c}\right )+2 c^2 d\right )}{\sqrt{b^2-4 a c} \sqrt{b-\sqrt{b^2-4 a c}}}-\frac{2 \sqrt{2} \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right ) \left (-c \left (f \sqrt{b^2-4 a c}+2 a h+b f\right )+b h \left (\sqrt{b^2-4 a c}+b\right )+2 c^2 d\right )}{\sqrt{b^2-4 a c} \sqrt{\sqrt{b^2-4 a c}+b}}+\frac{\log \left (\sqrt{b^2-4 a c}-b-2 c x^2\right ) \left (c \left (g \sqrt{b^2-4 a c}-2 a i-b g\right )+b i \left (b-\sqrt{b^2-4 a c}\right )+2 c^2 e\right )}{\sqrt{b^2-4 a c}}-\frac{\log \left (\sqrt{b^2-4 a c}+b+2 c x^2\right ) \left (-c \left (g \sqrt{b^2-4 a c}+2 a i+b g\right )+b i \left (\sqrt{b^2-4 a c}+b\right )+2 c^2 e\right )}{\sqrt{b^2-4 a c}}+4 c h x+2 c i x^2}{4 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x + f*x^2 + g*x^3 + h*x^4 + i*x^5)/(a + b*x^2 + c*x^4),x]

[Out]

(4*c*h*x + 2*c*i*x^2 + (2*Sqrt[2]*Sqrt[c]*(2*c^2*d + b*(b - Sqrt[b^2 - 4*a*c])*h + c*(-(b*f) + Sqrt[b^2 - 4*a*
c]*f - 2*a*h))*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[b^2 - 4*a*c]*Sqrt[b - Sqrt[b^2 -
4*a*c]]) - (2*Sqrt[2]*Sqrt[c]*(2*c^2*d + b*(b + Sqrt[b^2 - 4*a*c])*h - c*(b*f + Sqrt[b^2 - 4*a*c]*f + 2*a*h))
*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[b^2 - 4*a*c]*Sqrt[b + Sqrt[b^2 - 4*a*c]]) + ((
2*c^2*e + b*(b - Sqrt[b^2 - 4*a*c])*i + c*(-(b*g) + Sqrt[b^2 - 4*a*c]*g - 2*a*i))*Log[-b + Sqrt[b^2 - 4*a*c] -
2*c*x^2])/Sqrt[b^2 - 4*a*c] - ((2*c^2*e + b*(b + Sqrt[b^2 - 4*a*c])*i - c*(b*g + Sqrt[b^2 - 4*a*c]*g + 2*a*i)
)*Log[b + Sqrt[b^2 - 4*a*c] + 2*c*x^2])/Sqrt[b^2 - 4*a*c])/(4*c^2)

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Maple [B]  time = 0.029, size = 1435, normalized size = 4.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((i*x^5+h*x^4+g*x^3+f*x^2+e*x+d)/(c*x^4+b*x^2+a),x)

[Out]

1/2*i*x^2/c+h*x/c+1/2*(-4*a*c+b^2)/(4*a*c-b^2)/c*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctan(c*x*2^(1/2)/(
(b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*b*h+1/2*(-4*a*c+b^2)^(1/2)/(4*a*c-b^2)/c*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(
1/2)*arctan(c*x*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*b^2*h-1/2*(-4*a*c+b^2)/(4*a*c-b^2)/c*2^(1/2)/(((-4*a
*c+b^2)^(1/2)-b)*c)^(1/2)*arctanh(c*x*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2))*b*h+1/2*(-4*a*c+b^2)^(1/2)/(4*
a*c-b^2)/c*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2)*arctanh(c*x*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2))*b^2*
h-1/2*(-4*a*c+b^2)^(1/2)/(4*a*c-b^2)*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2)*arctanh(c*x*2^(1/2)/(((-4*a*c+b^
2)^(1/2)-b)*c)^(1/2))*b*f+c*(-4*a*c+b^2)^(1/2)/(4*a*c-b^2)*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2)*arctanh(c*
x*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2))*d-1/2*(-4*a*c+b^2)^(1/2)/(4*a*c-b^2)*2^(1/2)/((b+(-4*a*c+b^2)^(1/2
))*c)^(1/2)*arctan(c*x*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*b*f+c*(-4*a*c+b^2)^(1/2)/(4*a*c-b^2)*2^(1/2)/
((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctan(c*x*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*d-1/4*(-4*a*c+b^2)/(4*a*
c-b^2)/c*ln(-2*c*x^2+(-4*a*c+b^2)^(1/2)-b)*g-1/4*(-4*a*c+b^2)/(4*a*c-b^2)/c*ln(2*c*x^2+(-4*a*c+b^2)^(1/2)+b)*g
+1/4*(-4*a*c+b^2)/(4*a*c-b^2)/c^2*ln(2*c*x^2+(-4*a*c+b^2)^(1/2)+b)*b*i-1/2*(-4*a*c+b^2)^(1/2)/(4*a*c-b^2)/c*ln
(2*c*x^2+(-4*a*c+b^2)^(1/2)+b)*a*i+1/4*(-4*a*c+b^2)^(1/2)/(4*a*c-b^2)/c^2*ln(2*c*x^2+(-4*a*c+b^2)^(1/2)+b)*b^2
*i+1/4*(-4*a*c+b^2)/(4*a*c-b^2)/c^2*ln(-2*c*x^2+(-4*a*c+b^2)^(1/2)-b)*b*i+1/2*(-4*a*c+b^2)^(1/2)/(4*a*c-b^2)/c
*ln(-2*c*x^2+(-4*a*c+b^2)^(1/2)-b)*a*i-1/4*(-4*a*c+b^2)^(1/2)/(4*a*c-b^2)/c^2*ln(-2*c*x^2+(-4*a*c+b^2)^(1/2)-b
)*b^2*i+1/2*(-4*a*c+b^2)/(4*a*c-b^2)*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2)*arctanh(c*x*2^(1/2)/(((-4*a*c+b^
2)^(1/2)-b)*c)^(1/2))*f-1/2*(-4*a*c+b^2)/(4*a*c-b^2)*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctan(c*x*2^(1/
2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*f+1/4*(-4*a*c+b^2)^(1/2)/(4*a*c-b^2)/c*ln(-2*c*x^2+(-4*a*c+b^2)^(1/2)-b)*
b*g-1/4*(-4*a*c+b^2)^(1/2)/(4*a*c-b^2)/c*ln(2*c*x^2+(-4*a*c+b^2)^(1/2)+b)*b*g-1/2*(-4*a*c+b^2)^(1/2)/(4*a*c-b^
2)*e*ln(-2*c*x^2+(-4*a*c+b^2)^(1/2)-b)+1/2*(-4*a*c+b^2)^(1/2)/(4*a*c-b^2)*e*ln(2*c*x^2+(-4*a*c+b^2)^(1/2)+b)-(
-4*a*c+b^2)^(1/2)/(4*a*c-b^2)*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2)*arctanh(c*x*2^(1/2)/(((-4*a*c+b^2)^(1/2
)-b)*c)^(1/2))*a*h-(-4*a*c+b^2)^(1/2)/(4*a*c-b^2)*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctan(c*x*2^(1/2)/
((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*a*h

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{i x^{2} + 2 \, h x}{2 \, c} - \frac{-\int \frac{{\left (c g - b i\right )} x^{3} +{\left (c f - b h\right )} x^{2} + c d - a h +{\left (c e - a i\right )} x}{c x^{4} + b x^{2} + a}\,{d x}}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x^5+h*x^4+g*x^3+f*x^2+e*x+d)/(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

1/2*(i*x^2 + 2*h*x)/c - integrate(-((c*g - b*i)*x^3 + (c*f - b*h)*x^2 + c*d - a*h + (c*e - a*i)*x)/(c*x^4 + b*
x^2 + a), x)/c

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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((i*x^5+h*x^4+g*x^3+f*x^2+e*x+d)/(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((i*x**5+h*x**4+g*x**3+f*x**2+e*x+d)/(c*x**4+b*x**2+a),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x^5+h*x^4+g*x^3+f*x^2+e*x+d)/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError