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

Optimal. Leaf size=353 \[ -\frac{\log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right ) \left (-\sqrt{a} \sqrt{b} e+a g+3 b c\right )}{16 \sqrt{2} a^{7/4} b^{5/4}}+\frac{\log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right ) \left (-\sqrt{a} \sqrt{b} e+a g+3 b c\right )}{16 \sqrt{2} a^{7/4} b^{5/4}}-\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (\sqrt{a} \sqrt{b} e+a g+3 b c\right )}{8 \sqrt{2} a^{7/4} b^{5/4}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right ) \left (\sqrt{a} \sqrt{b} e+a g+3 b c\right )}{8 \sqrt{2} a^{7/4} b^{5/4}}+\frac{(a h+b d) \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{4 a^{3/2} b^{3/2}}+\frac{x \left (x (b d-a h)-a g+b c+b e x^2+b f x^3\right )}{4 a b \left (a+b x^4\right )} \]

[Out]

(x*(b*c - a*g + (b*d - a*h)*x + b*e*x^2 + b*f*x^3))/(4*a*b*(a + b*x^4)) + ((b*d + a*h)*ArcTan[(Sqrt[b]*x^2)/Sq
rt[a]])/(4*a^(3/2)*b^(3/2)) - ((3*b*c + Sqrt[a]*Sqrt[b]*e + a*g)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(8*S
qrt[2]*a^(7/4)*b^(5/4)) + ((3*b*c + Sqrt[a]*Sqrt[b]*e + a*g)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(8*Sqrt[
2]*a^(7/4)*b^(5/4)) - ((3*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
])/(16*Sqrt[2]*a^(7/4)*b^(5/4)) + ((3*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])/(16*Sqrt[2]*a^(7/4)*b^(5/4))

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Rubi [A]  time = 0.340409, antiderivative size = 353, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {1858, 1876, 275, 205, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{\log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right ) \left (-\sqrt{a} \sqrt{b} e+a g+3 b c\right )}{16 \sqrt{2} a^{7/4} b^{5/4}}+\frac{\log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right ) \left (-\sqrt{a} \sqrt{b} e+a g+3 b c\right )}{16 \sqrt{2} a^{7/4} b^{5/4}}-\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (\sqrt{a} \sqrt{b} e+a g+3 b c\right )}{8 \sqrt{2} a^{7/4} b^{5/4}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right ) \left (\sqrt{a} \sqrt{b} e+a g+3 b c\right )}{8 \sqrt{2} a^{7/4} b^{5/4}}+\frac{(a h+b d) \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{4 a^{3/2} b^{3/2}}+\frac{x \left (x (b d-a h)-a g+b c+b e x^2+b f x^3\right )}{4 a b \left (a+b x^4\right )} \]

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)^2,x]

[Out]

(x*(b*c - a*g + (b*d - a*h)*x + b*e*x^2 + b*f*x^3))/(4*a*b*(a + b*x^4)) + ((b*d + a*h)*ArcTan[(Sqrt[b]*x^2)/Sq
rt[a]])/(4*a^(3/2)*b^(3/2)) - ((3*b*c + Sqrt[a]*Sqrt[b]*e + a*g)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(8*S
qrt[2]*a^(7/4)*b^(5/4)) + ((3*b*c + Sqrt[a]*Sqrt[b]*e + a*g)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(8*Sqrt[
2]*a^(7/4)*b^(5/4)) - ((3*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
])/(16*Sqrt[2]*a^(7/4)*b^(5/4)) + ((3*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])/(16*Sqrt[2]*a^(7/4)*b^(5/4))

Rule 1858

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, Module[{Q = PolynomialQuotient
[b^(Floor[(q - 1)/n] + 1)*Pq, a + b*x^n, x], R = PolynomialRemainder[b^(Floor[(q - 1)/n] + 1)*Pq, a + b*x^n, x
]}, Dist[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), Int[(a + b*x^n)^(p + 1)*ExpandToSum[a*n*(p + 1)*Q + n*(p +
1)*R + D[x*R, x], x], x], x] - Simp[(x*R*(a + b*x^n)^(p + 1))/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), x]] /; G
eQ[q, n]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1]

Rule 1876

Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = Sum[(x^ii*(Coeff[Pq, x, ii] + Coeff[Pq, x, n/2 + ii
]*x^(n/2)))/(a + b*x^n), {ii, 0, n/2 - 1}]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ
[n/2, 0] && Expon[Pq, x] < n

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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 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]

Rubi steps

\begin{align*} \int \frac{c+d x+e x^2+f x^3+g x^4+h x^5}{\left (a+b x^4\right )^2} \, dx &=\frac{x \left (b c-a g+(b d-a h) x+b e x^2+b f x^3\right )}{4 a b \left (a+b x^4\right )}-\frac{\int \frac{-b (3 b c+a g)-2 b (b d+a h) x-b^2 e x^2}{a+b x^4} \, dx}{4 a b^2}\\ &=\frac{x \left (b c-a g+(b d-a h) x+b e x^2+b f x^3\right )}{4 a b \left (a+b x^4\right )}-\frac{\int \left (-\frac{2 b (b d+a h) x}{a+b x^4}+\frac{-b (3 b c+a g)-b^2 e x^2}{a+b x^4}\right ) \, dx}{4 a b^2}\\ &=\frac{x \left (b c-a g+(b d-a h) x+b e x^2+b f x^3\right )}{4 a b \left (a+b x^4\right )}-\frac{\int \frac{-b (3 b c+a g)-b^2 e x^2}{a+b x^4} \, dx}{4 a b^2}+\frac{(b d+a h) \int \frac{x}{a+b x^4} \, dx}{2 a b}\\ &=\frac{x \left (b c-a g+(b d-a h) x+b e x^2+b f x^3\right )}{4 a b \left (a+b x^4\right )}+\frac{\left (3 b c-\sqrt{a} \sqrt{b} e+a g\right ) \int \frac{\sqrt{a} \sqrt{b}-b x^2}{a+b x^4} \, dx}{8 a^{3/2} b^{3/2}}+\frac{\left (3 b c+\sqrt{a} \sqrt{b} e+a g\right ) \int \frac{\sqrt{a} \sqrt{b}+b x^2}{a+b x^4} \, dx}{8 a^{3/2} b^{3/2}}+\frac{(b d+a h) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,x^2\right )}{4 a b}\\ &=\frac{x \left (b c-a g+(b d-a h) x+b e x^2+b f x^3\right )}{4 a b \left (a+b x^4\right )}+\frac{(b d+a h) \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{4 a^{3/2} b^{3/2}}-\frac{\left (3 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}{16 \sqrt{2} a^{7/4} b^{5/4}}-\frac{\left (3 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}{16 \sqrt{2} a^{7/4} b^{5/4}}+\frac{\left (3 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}{16 a^{3/2} b^{3/2}}+\frac{\left (3 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}{16 a^{3/2} b^{3/2}}\\ &=\frac{x \left (b c-a g+(b d-a h) x+b e x^2+b f x^3\right )}{4 a b \left (a+b x^4\right )}+\frac{(b d+a h) \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{4 a^{3/2} b^{3/2}}-\frac{\left (3 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 )}{16 \sqrt{2} a^{7/4} b^{5/4}}+\frac{\left (3 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 )}{16 \sqrt{2} a^{7/4} b^{5/4}}+\frac{\left (3 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 )}{8 \sqrt{2} a^{7/4} b^{5/4}}-\frac{\left (3 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 )}{8 \sqrt{2} a^{7/4} b^{5/4}}\\ &=\frac{x \left (b c-a g+(b d-a h) x+b e x^2+b f x^3\right )}{4 a b \left (a+b x^4\right )}+\frac{(b d+a h) \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{4 a^{3/2} b^{3/2}}-\frac{\left (3 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 )}{8 \sqrt{2} a^{7/4} b^{5/4}}+\frac{\left (3 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 )}{8 \sqrt{2} a^{7/4} b^{5/4}}-\frac{\left (3 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 )}{16 \sqrt{2} a^{7/4} b^{5/4}}+\frac{\left (3 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 )}{16 \sqrt{2} a^{7/4} b^{5/4}}\\ \end{align*}

Mathematica [A]  time = 0.241713, size = 359, normalized size = 1.02 \[ \frac{-2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (4 a^{5/4} h+\sqrt{2} \sqrt{a} b^{3/4} e+4 \sqrt [4]{a} b d+\sqrt{2} a \sqrt [4]{b} g+3 \sqrt{2} b^{5/4} c\right )+2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right ) \left (-4 a^{5/4} h+\sqrt{2} \sqrt{a} b^{3/4} e-4 \sqrt [4]{a} b d+\sqrt{2} a \sqrt [4]{b} g+3 \sqrt{2} b^{5/4} c\right )-\frac{8 a^{3/4} \sqrt{b} (a (f+x (g+h x))-b x (c+x (d+e x)))}{a+b x^4}+\sqrt{2} \sqrt [4]{b} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right ) \left (\sqrt{a} \sqrt{b} e-a g-3 b c\right )+\sqrt{2} \sqrt [4]{b} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right ) \left (-\sqrt{a} \sqrt{b} e+a g+3 b c\right )}{32 a^{7/4} b^{3/2}} \]

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)^2,x]

[Out]

((-8*a^(3/4)*Sqrt[b]*(-(b*x*(c + x*(d + e*x))) + a*(f + x*(g + h*x))))/(a + b*x^4) - 2*(3*Sqrt[2]*b^(5/4)*c +
4*a^(1/4)*b*d + Sqrt[2]*Sqrt[a]*b^(3/4)*e + Sqrt[2]*a*b^(1/4)*g + 4*a^(5/4)*h)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/
a^(1/4)] + 2*(3*Sqrt[2]*b^(5/4)*c - 4*a^(1/4)*b*d + Sqrt[2]*Sqrt[a]*b^(3/4)*e + Sqrt[2]*a*b^(1/4)*g - 4*a^(5/4
)*h)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)] + Sqrt[2]*b^(1/4)*(-3*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^(1/4)*(3*b*c - Sqrt[a]*Sqrt[b]*e + a*g)*Log[Sqrt[a] + S
qrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(32*a^(7/4)*b^(3/2))

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Maple [A]  time = 0.009, size = 515, normalized size = 1.5 \begin{align*}{\frac{1}{b{x}^{4}+a} \left ({\frac{e{x}^{3}}{4\,a}}-{\frac{ \left ( ah-bd \right ){x}^{2}}{4\,ab}}-{\frac{ \left ( ag-bc \right ) x}{4\,ab}}-{\frac{f}{4\,b}} \right ) }+{\frac{\sqrt{2}g}{16\,ab}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ) }+{\frac{3\,c\sqrt{2}}{16\,{a}^{2}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ) }+{\frac{\sqrt{2}g}{16\,ab}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ) }+{\frac{3\,c\sqrt{2}}{16\,{a}^{2}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ) }+{\frac{\sqrt{2}g}{32\,ab}\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{3\,c\sqrt{2}}{32\,{a}^{2}}\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{h}{4\,b}\arctan \left ({x}^{2}\sqrt{{\frac{b}{a}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{d}{4\,a}\arctan \left ({x}^{2}\sqrt{{\frac{b}{a}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{e\sqrt{2}}{32\,ab}\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}}{16\,ab}\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}}{16\,ab}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{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)^2,x)

[Out]

(1/4/a*e*x^3-1/4*(a*h-b*d)/a/b*x^2-1/4*(a*g-b*c)/a/b*x-1/4*f/b)/(b*x^4+a)+1/16/b/a*(1/b*a)^(1/4)*2^(1/2)*arcta
n(2^(1/2)/(1/b*a)^(1/4)*x+1)*g+3/16*c/a^2*(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x+1)+1/16/b/a*(1/
b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x-1)*g+3/16*c/a^2*(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)
^(1/4)*x-1)+1/32/b/a*(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+3/32*c/a^2*(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/b/(a*b)^(1/2)*arctan(x^2*(b/a)^(1/2))*h+1/4*d/a/(a*b)^(1/2)*arct
an(x^2*(b/a)^(1/2))+1/32*e/a/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)))+1/16*e/a/b/(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x+1)+1/16*e/a
/b/(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x-1)

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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)^2,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)^2,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)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.10694, size = 537, normalized size = 1.52 \begin{align*} \frac{b x^{3} e + b d x^{2} - a h x^{2} + b c x - a g x - a f}{4 \,{\left (b x^{4} + a\right )} a b} + \frac{\sqrt{2}{\left (2 \, \sqrt{2} \sqrt{a b} b^{2} d + 2 \, \sqrt{2} \sqrt{a b} a b h + 3 \, \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 )}{16 \, a^{2} b^{3}} + \frac{\sqrt{2}{\left (2 \, \sqrt{2} \sqrt{a b} b^{2} d + 2 \, \sqrt{2} \sqrt{a b} a b h + 3 \, \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 )}{16 \, a^{2} b^{3}} + \frac{\sqrt{2}{\left (3 \, \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 )}{32 \, a^{2} b^{3}} - \frac{\sqrt{2}{\left (3 \, \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 )}{32 \, a^{2} 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)^2,x, algorithm="giac")

[Out]

1/4*(b*x^3*e + b*d*x^2 - a*h*x^2 + b*c*x - a*g*x - a*f)/((b*x^4 + a)*a*b) + 1/16*sqrt(2)*(2*sqrt(2)*sqrt(a*b)*
b^2*d + 2*sqrt(2)*sqrt(a*b)*a*b*h + 3*(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^2*b^3) + 1/16*sqrt(2)*(2*sqrt(2)*sqrt(a*b)*b^2*d + 2*sqrt(
2)*sqrt(a*b)*a*b*h + 3*(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^2*b^3) + 1/32*sqrt(2)*(3*(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^2*b^3) - 1/32*sqrt(2)*(3*(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^2*b^3)