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

Optimal. Leaf size=337 \[ -\frac{\log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right ) \left (-\sqrt{a} \sqrt{b} e-a g+b c\right )}{4 \sqrt{2} a^{3/4} b^{5/4}}+\frac{\log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right ) \left (-\sqrt{a} \sqrt{b} e-a g+b c\right )}{4 \sqrt{2} a^{3/4} b^{5/4}}-\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (\sqrt{a} \sqrt{b} e-a g+b c\right )}{2 \sqrt{2} a^{3/4} b^{5/4}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right ) \left (\sqrt{a} \sqrt{b} e-a g+b c\right )}{2 \sqrt{2} a^{3/4} b^{5/4}}+\frac{(b d-a h) \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{2 \sqrt{a} b^{3/2}}+\frac{f \log \left (a+b x^4\right )}{4 b}+\frac{g x}{b}+\frac{h x^2}{2 b} \]

[Out]

(g*x)/b + (h*x^2)/(2*b) + ((b*d - a*h)*ArcTan[(Sqrt[b]*x^2)/Sqrt[a]])/(2*Sqrt[a]*b^(3/2)) - ((b*c + Sqrt[a]*Sq
rt[b]*e - a*g)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^(3/4)*b^(5/4)) + ((b*c + Sqrt[a]*Sqrt[b]*
e - a*g)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^(3/4)*b^(5/4)) - ((b*c - Sqrt[a]*Sqrt[b]*e - a*
g)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(4*Sqrt[2]*a^(3/4)*b^(5/4)) + ((b*c - Sqrt[a]*Sqrt[
b]*e - a*g)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(4*Sqrt[2]*a^(3/4)*b^(5/4)) + (f*Log[a + b
*x^4])/(4*b)

________________________________________________________________________________________

Rubi [A]  time = 0.399356, antiderivative size = 337, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 13, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.371, Rules used = {1885, 1887, 1168, 1162, 617, 204, 1165, 628, 1819, 1810, 635, 205, 260} \[ -\frac{\log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right ) \left (-\sqrt{a} \sqrt{b} e-a g+b c\right )}{4 \sqrt{2} a^{3/4} b^{5/4}}+\frac{\log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right ) \left (-\sqrt{a} \sqrt{b} e-a g+b c\right )}{4 \sqrt{2} a^{3/4} b^{5/4}}-\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (\sqrt{a} \sqrt{b} e-a g+b c\right )}{2 \sqrt{2} a^{3/4} b^{5/4}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right ) \left (\sqrt{a} \sqrt{b} e-a g+b c\right )}{2 \sqrt{2} a^{3/4} b^{5/4}}+\frac{(b d-a h) \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{2 \sqrt{a} b^{3/2}}+\frac{f \log \left (a+b x^4\right )}{4 b}+\frac{g x}{b}+\frac{h x^2}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(g*x)/b + (h*x^2)/(2*b) + ((b*d - a*h)*ArcTan[(Sqrt[b]*x^2)/Sqrt[a]])/(2*Sqrt[a]*b^(3/2)) - ((b*c + Sqrt[a]*Sq
rt[b]*e - a*g)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^(3/4)*b^(5/4)) + ((b*c + Sqrt[a]*Sqrt[b]*
e - a*g)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^(3/4)*b^(5/4)) - ((b*c - Sqrt[a]*Sqrt[b]*e - a*
g)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(4*Sqrt[2]*a^(3/4)*b^(5/4)) + ((b*c - Sqrt[a]*Sqrt[
b]*e - a*g)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(4*Sqrt[2]*a^(3/4)*b^(5/4)) + (f*Log[a + b
*x^4])/(4*b)

Rule 1885

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

Rule 1887

Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^n), x], x] /; FreeQ[{a, b}, x
] && PolyQ[Pq, x] && IntegerQ[n]

Rule 1168

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

Rule 1162

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

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

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]

Rule 1819

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

Rule 1810

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

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(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 260

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

Rubi steps

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

Mathematica [A]  time = 0.324556, size = 342, normalized size = 1.01 \[ \frac{-2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (-2 a^{5/4} h+\sqrt{2} \sqrt{a} b^{3/4} e+2 \sqrt [4]{a} b d-\sqrt{2} a \sqrt [4]{b} g+\sqrt{2} b^{5/4} c\right )+2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right ) \left (2 a^{5/4} h+\sqrt{2} \sqrt{a} b^{3/4} e-2 \sqrt [4]{a} b d-\sqrt{2} a \sqrt [4]{b} g+\sqrt{2} b^{5/4} c\right )+\sqrt [4]{b} \left (2 a^{3/4} \sqrt [4]{b} \left (f \log \left (a+b x^4\right )+2 x (2 g+h x)\right )+\sqrt{2} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right ) \left (\sqrt{a} \sqrt{b} e+a g-b c\right )+\sqrt{2} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right ) \left (-\sqrt{a} \sqrt{b} e-a g+b c\right )\right )}{8 a^{3/4} b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(Sqrt[2]*b^(5/4)*c + 2*a^(1/4)*b*d + Sqrt[2]*Sqrt[a]*b^(3/4)*e - Sqrt[2]*a*b^(1/4)*g - 2*a^(5/4)*h)*ArcTan
[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)] + 2*(Sqrt[2]*b^(5/4)*c - 2*a^(1/4)*b*d + Sqrt[2]*Sqrt[a]*b^(3/4)*e - Sqrt[2]
*a*b^(1/4)*g + 2*a^(5/4)*h)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)] + b^(1/4)*(Sqrt[2]*(-(b*c) + Sqrt[a]*Sqrt[
b]*e + a*g)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2] + Sqrt[2]*(b*c - Sqrt[a]*Sqrt[b]*e - a*g)*L
og[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2] + 2*a^(3/4)*b^(1/4)*(2*x*(2*g + h*x) + f*Log[a + b*x^4])
))/(8*a^(3/4)*b^(3/2))

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Maple [A]  time = 0.003, size = 462, normalized size = 1.4 \begin{align*}{\frac{h{x}^{2}}{2\,b}}+{\frac{gx}{b}}-{\frac{\sqrt{2}g}{4\,b}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ) }+{\frac{c\sqrt{2}}{4\,a}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ) }-{\frac{\sqrt{2}g}{4\,b}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ) }+{\frac{c\sqrt{2}}{4\,a}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ) }-{\frac{\sqrt{2}g}{8\,b}\sqrt [4]{{\frac{a}{b}}}\ln \left ({ \left ({x}^{2}+\sqrt [4]{{\frac{a}{b}}}x\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) \left ({x}^{2}-\sqrt [4]{{\frac{a}{b}}}x\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) ^{-1}} \right ) }+{\frac{c\sqrt{2}}{8\,a}\sqrt [4]{{\frac{a}{b}}}\ln \left ({ \left ({x}^{2}+\sqrt [4]{{\frac{a}{b}}}x\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) \left ({x}^{2}-\sqrt [4]{{\frac{a}{b}}}x\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) ^{-1}} \right ) }-{\frac{ah}{2\,b}\arctan \left ({x}^{2}\sqrt{{\frac{b}{a}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{d}{2}\arctan \left ({x}^{2}\sqrt{{\frac{b}{a}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{e\sqrt{2}}{8\,b}\ln \left ({ \left ({x}^{2}-\sqrt [4]{{\frac{a}{b}}}x\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) \left ({x}^{2}+\sqrt [4]{{\frac{a}{b}}}x\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+{\frac{e\sqrt{2}}{4\,b}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+{\frac{e\sqrt{2}}{4\,b}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+{\frac{f\ln \left ( b{x}^{4}+a \right ) }{4\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*h*x^2/b+g*x/b-1/4/b*(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x+1)*g+1/4*c*(1/b*a)^(1/4)/a*2^(1/2
)*arctan(2^(1/2)/(1/b*a)^(1/4)*x+1)-1/4/b*(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x-1)*g+1/4*c*(1/b
*a)^(1/4)/a*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x-1)-1/8/b*(1/b*a)^(1/4)*2^(1/2)*ln((x^2+(1/b*a)^(1/4)*x*2^(1
/2)+(1/b*a)^(1/2))/(x^2-(1/b*a)^(1/4)*x*2^(1/2)+(1/b*a)^(1/2)))*g+1/8*c*(1/b*a)^(1/4)/a*2^(1/2)*ln((x^2+(1/b*a
)^(1/4)*x*2^(1/2)+(1/b*a)^(1/2))/(x^2-(1/b*a)^(1/4)*x*2^(1/2)+(1/b*a)^(1/2)))-1/2/b/(a*b)^(1/2)*arctan(x^2*(b/
a)^(1/2))*a*h+1/2*d/(a*b)^(1/2)*arctan(x^2*(b/a)^(1/2))+1/8*e/b/(1/b*a)^(1/4)*2^(1/2)*ln((x^2-(1/b*a)^(1/4)*x*
2^(1/2)+(1/b*a)^(1/2))/(x^2+(1/b*a)^(1/4)*x*2^(1/2)+(1/b*a)^(1/2)))+1/4*e/b/(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/
2)/(1/b*a)^(1/4)*x+1)+1/4*e/b/(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x-1)+1/4*f*ln(b*x^4+a)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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

[Out]

Timed out

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Giac [A]  time = 1.08285, size = 506, normalized size = 1.5 \begin{align*} \frac{f \log \left ({\left | b x^{4} + a \right |}\right )}{4 \, b} + \frac{b h x^{2} + 2 \, b g x}{2 \, b^{2}} + \frac{\sqrt{2}{\left (\sqrt{2} \sqrt{a b} b^{2} d + \sqrt{2} \sqrt{a b} a b h + \left (a b^{3}\right )^{\frac{1}{4}} b^{2} c - \left (a b^{3}\right )^{\frac{1}{4}} a b g + \left (a b^{3}\right )^{\frac{3}{4}} e\right )} \arctan \left (\frac{\sqrt{2}{\left (2 \, x + \sqrt{2} \left (\frac{a}{b}\right )^{\frac{1}{4}}\right )}}{2 \, \left (\frac{a}{b}\right )^{\frac{1}{4}}}\right )}{4 \, a b^{3}} + \frac{\sqrt{2}{\left (\sqrt{2} \sqrt{a b} b^{2} d + \sqrt{2} \sqrt{a b} a b h + \left (a b^{3}\right )^{\frac{1}{4}} b^{2} c - \left (a b^{3}\right )^{\frac{1}{4}} a b g + \left (a b^{3}\right )^{\frac{3}{4}} e\right )} \arctan \left (\frac{\sqrt{2}{\left (2 \, x - \sqrt{2} \left (\frac{a}{b}\right )^{\frac{1}{4}}\right )}}{2 \, \left (\frac{a}{b}\right )^{\frac{1}{4}}}\right )}{4 \, a b^{3}} + \frac{\sqrt{2}{\left (\left (a b^{3}\right )^{\frac{1}{4}} b^{2} c - \left (a b^{3}\right )^{\frac{1}{4}} a b g - \left (a b^{3}\right )^{\frac{3}{4}} e\right )} \log \left (x^{2} + \sqrt{2} x \left (\frac{a}{b}\right )^{\frac{1}{4}} + \sqrt{\frac{a}{b}}\right )}{8 \, a b^{3}} - \frac{\sqrt{2}{\left (\left (a b^{3}\right )^{\frac{1}{4}} b^{2} c - \left (a b^{3}\right )^{\frac{1}{4}} a b g - \left (a b^{3}\right )^{\frac{3}{4}} e\right )} \log \left (x^{2} - \sqrt{2} x \left (\frac{a}{b}\right )^{\frac{1}{4}} + \sqrt{\frac{a}{b}}\right )}{8 \, a b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*f*log(abs(b*x^4 + a))/b + 1/2*(b*h*x^2 + 2*b*g*x)/b^2 + 1/4*sqrt(2)*(sqrt(2)*sqrt(a*b)*b^2*d + sqrt(2)*sqr
t(a*b)*a*b*h + (a*b^3)^(1/4)*b^2*c - (a*b^3)^(1/4)*a*b*g + (a*b^3)^(3/4)*e)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*
(a/b)^(1/4))/(a/b)^(1/4))/(a*b^3) + 1/4*sqrt(2)*(sqrt(2)*sqrt(a*b)*b^2*d + sqrt(2)*sqrt(a*b)*a*b*h + (a*b^3)^(
1/4)*b^2*c - (a*b^3)^(1/4)*a*b*g + (a*b^3)^(3/4)*e)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4)
)/(a*b^3) + 1/8*sqrt(2)*((a*b^3)^(1/4)*b^2*c - (a*b^3)^(1/4)*a*b*g - (a*b^3)^(3/4)*e)*log(x^2 + sqrt(2)*x*(a/b
)^(1/4) + sqrt(a/b))/(a*b^3) - 1/8*sqrt(2)*((a*b^3)^(1/4)*b^2*c - (a*b^3)^(1/4)*a*b*g - (a*b^3)^(3/4)*e)*log(x
^2 - sqrt(2)*x*(a/b)^(1/4) + sqrt(a/b))/(a*b^3)