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

Optimal. Leaf size=382 \[ -\frac{\left (77 \sqrt{b} c-15 \sqrt{a} e\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right )}{512 \sqrt{2} a^{15/4} b^{3/4}}+\frac{\left (77 \sqrt{b} c-15 \sqrt{a} e\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right )}{512 \sqrt{2} a^{15/4} b^{3/4}}-\frac{\left (15 \sqrt{a} e+77 \sqrt{b} c\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{256 \sqrt{2} a^{15/4} b^{3/4}}+\frac{\left (15 \sqrt{a} e+77 \sqrt{b} c\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{256 \sqrt{2} a^{15/4} b^{3/4}}+\frac{x \left (11 c+10 d x+9 e x^2\right )}{96 a^2 \left (a+b x^4\right )^2}+\frac{x \left (77 c+60 d x+45 e x^2\right )}{384 a^3 \left (a+b x^4\right )}+\frac{5 d \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{32 a^{7/2} \sqrt{b}}-\frac{a f-b x \left (c+d x+e x^2\right )}{12 a b \left (a+b x^4\right )^3} \]

[Out]

(x*(11*c + 10*d*x + 9*e*x^2))/(96*a^2*(a + b*x^4)^2) + (x*(77*c + 60*d*x + 45*e*x^2))/(384*a^3*(a + b*x^4)) -
(a*f - b*x*(c + d*x + e*x^2))/(12*a*b*(a + b*x^4)^3) + (5*d*ArcTan[(Sqrt[b]*x^2)/Sqrt[a]])/(32*a^(7/2)*Sqrt[b]
) - ((77*Sqrt[b]*c + 15*Sqrt[a]*e)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(256*Sqrt[2]*a^(15/4)*b^(3/4)) + (
(77*Sqrt[b]*c + 15*Sqrt[a]*e)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(256*Sqrt[2]*a^(15/4)*b^(3/4)) - ((77*S
qrt[b]*c - 15*Sqrt[a]*e)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(512*Sqrt[2]*a^(15/4)*b^(3/4)
) + ((77*Sqrt[b]*c - 15*Sqrt[a]*e)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(512*Sqrt[2]*a^(15/
4)*b^(3/4))

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Rubi [A]  time = 0.405867, antiderivative size = 382, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 11, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.44, Rules used = {1854, 1855, 1876, 275, 205, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{\left (77 \sqrt{b} c-15 \sqrt{a} e\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right )}{512 \sqrt{2} a^{15/4} b^{3/4}}+\frac{\left (77 \sqrt{b} c-15 \sqrt{a} e\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right )}{512 \sqrt{2} a^{15/4} b^{3/4}}-\frac{\left (15 \sqrt{a} e+77 \sqrt{b} c\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{256 \sqrt{2} a^{15/4} b^{3/4}}+\frac{\left (15 \sqrt{a} e+77 \sqrt{b} c\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{256 \sqrt{2} a^{15/4} b^{3/4}}+\frac{x \left (11 c+10 d x+9 e x^2\right )}{96 a^2 \left (a+b x^4\right )^2}+\frac{x \left (77 c+60 d x+45 e x^2\right )}{384 a^3 \left (a+b x^4\right )}+\frac{5 d \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{32 a^{7/2} \sqrt{b}}-\frac{a f-b x \left (c+d x+e x^2\right )}{12 a b \left (a+b x^4\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(11*c + 10*d*x + 9*e*x^2))/(96*a^2*(a + b*x^4)^2) + (x*(77*c + 60*d*x + 45*e*x^2))/(384*a^3*(a + b*x^4)) -
(a*f - b*x*(c + d*x + e*x^2))/(12*a*b*(a + b*x^4)^3) + (5*d*ArcTan[(Sqrt[b]*x^2)/Sqrt[a]])/(32*a^(7/2)*Sqrt[b]
) - ((77*Sqrt[b]*c + 15*Sqrt[a]*e)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(256*Sqrt[2]*a^(15/4)*b^(3/4)) + (
(77*Sqrt[b]*c + 15*Sqrt[a]*e)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(256*Sqrt[2]*a^(15/4)*b^(3/4)) - ((77*S
qrt[b]*c - 15*Sqrt[a]*e)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(512*Sqrt[2]*a^(15/4)*b^(3/4)
) + ((77*Sqrt[b]*c - 15*Sqrt[a]*e)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(512*Sqrt[2]*a^(15/
4)*b^(3/4))

Rule 1854

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

Rule 1855

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(x*Pq*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Di
st[1/(a*n*(p + 1)), Int[ExpandToSum[n*(p + 1)*Pq + D[x*Pq, x], x]*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b},
x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1] && LtQ[Expon[Pq, x], n - 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}{\left (a+b x^4\right )^4} \, dx &=-\frac{a f-b x \left (c+d x+e x^2\right )}{12 a b \left (a+b x^4\right )^3}-\frac{\int \frac{-11 c-10 d x-9 e x^2}{\left (a+b x^4\right )^3} \, dx}{12 a}\\ &=\frac{x \left (11 c+10 d x+9 e x^2\right )}{96 a^2 \left (a+b x^4\right )^2}-\frac{a f-b x \left (c+d x+e x^2\right )}{12 a b \left (a+b x^4\right )^3}+\frac{\int \frac{77 c+60 d x+45 e x^2}{\left (a+b x^4\right )^2} \, dx}{96 a^2}\\ &=\frac{x \left (11 c+10 d x+9 e x^2\right )}{96 a^2 \left (a+b x^4\right )^2}+\frac{x \left (77 c+60 d x+45 e x^2\right )}{384 a^3 \left (a+b x^4\right )}-\frac{a f-b x \left (c+d x+e x^2\right )}{12 a b \left (a+b x^4\right )^3}-\frac{\int \frac{-231 c-120 d x-45 e x^2}{a+b x^4} \, dx}{384 a^3}\\ &=\frac{x \left (11 c+10 d x+9 e x^2\right )}{96 a^2 \left (a+b x^4\right )^2}+\frac{x \left (77 c+60 d x+45 e x^2\right )}{384 a^3 \left (a+b x^4\right )}-\frac{a f-b x \left (c+d x+e x^2\right )}{12 a b \left (a+b x^4\right )^3}-\frac{\int \left (-\frac{120 d x}{a+b x^4}+\frac{-231 c-45 e x^2}{a+b x^4}\right ) \, dx}{384 a^3}\\ &=\frac{x \left (11 c+10 d x+9 e x^2\right )}{96 a^2 \left (a+b x^4\right )^2}+\frac{x \left (77 c+60 d x+45 e x^2\right )}{384 a^3 \left (a+b x^4\right )}-\frac{a f-b x \left (c+d x+e x^2\right )}{12 a b \left (a+b x^4\right )^3}-\frac{\int \frac{-231 c-45 e x^2}{a+b x^4} \, dx}{384 a^3}+\frac{(5 d) \int \frac{x}{a+b x^4} \, dx}{16 a^3}\\ &=\frac{x \left (11 c+10 d x+9 e x^2\right )}{96 a^2 \left (a+b x^4\right )^2}+\frac{x \left (77 c+60 d x+45 e x^2\right )}{384 a^3 \left (a+b x^4\right )}-\frac{a f-b x \left (c+d x+e x^2\right )}{12 a b \left (a+b x^4\right )^3}+\frac{(5 d) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,x^2\right )}{32 a^3}+\frac{\left (\frac{77 \sqrt{b} c}{\sqrt{a}}-15 e\right ) \int \frac{\sqrt{a} \sqrt{b}-b x^2}{a+b x^4} \, dx}{256 a^3 b}+\frac{\left (\frac{77 \sqrt{b} c}{\sqrt{a}}+15 e\right ) \int \frac{\sqrt{a} \sqrt{b}+b x^2}{a+b x^4} \, dx}{256 a^3 b}\\ &=\frac{x \left (11 c+10 d x+9 e x^2\right )}{96 a^2 \left (a+b x^4\right )^2}+\frac{x \left (77 c+60 d x+45 e x^2\right )}{384 a^3 \left (a+b x^4\right )}-\frac{a f-b x \left (c+d x+e x^2\right )}{12 a b \left (a+b x^4\right )^3}+\frac{5 d \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{32 a^{7/2} \sqrt{b}}+\frac{\left (\frac{77 \sqrt{b} c}{\sqrt{a}}+15 e\right ) \int \frac{1}{\frac{\sqrt{a}}{\sqrt{b}}-\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx}{512 a^3 b}+\frac{\left (\frac{77 \sqrt{b} c}{\sqrt{a}}+15 e\right ) \int \frac{1}{\frac{\sqrt{a}}{\sqrt{b}}+\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx}{512 a^3 b}-\frac{\left (77 \sqrt{b} c-15 \sqrt{a} e\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}{512 \sqrt{2} a^{15/4} b^{3/4}}-\frac{\left (77 \sqrt{b} c-15 \sqrt{a} e\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}{512 \sqrt{2} a^{15/4} b^{3/4}}\\ &=\frac{x \left (11 c+10 d x+9 e x^2\right )}{96 a^2 \left (a+b x^4\right )^2}+\frac{x \left (77 c+60 d x+45 e x^2\right )}{384 a^3 \left (a+b x^4\right )}-\frac{a f-b x \left (c+d x+e x^2\right )}{12 a b \left (a+b x^4\right )^3}+\frac{5 d \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{32 a^{7/2} \sqrt{b}}-\frac{\left (77 \sqrt{b} c-15 \sqrt{a} e\right ) \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{b} x^2\right )}{512 \sqrt{2} a^{15/4} b^{3/4}}+\frac{\left (77 \sqrt{b} c-15 \sqrt{a} e\right ) \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{b} x^2\right )}{512 \sqrt{2} a^{15/4} b^{3/4}}+\frac{\left (77 \sqrt{b} c+15 \sqrt{a} e\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{256 \sqrt{2} a^{15/4} b^{3/4}}-\frac{\left (77 \sqrt{b} c+15 \sqrt{a} e\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{256 \sqrt{2} a^{15/4} b^{3/4}}\\ &=\frac{x \left (11 c+10 d x+9 e x^2\right )}{96 a^2 \left (a+b x^4\right )^2}+\frac{x \left (77 c+60 d x+45 e x^2\right )}{384 a^3 \left (a+b x^4\right )}-\frac{a f-b x \left (c+d x+e x^2\right )}{12 a b \left (a+b x^4\right )^3}+\frac{5 d \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{32 a^{7/2} \sqrt{b}}-\frac{\left (77 \sqrt{b} c+15 \sqrt{a} e\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{256 \sqrt{2} a^{15/4} b^{3/4}}+\frac{\left (77 \sqrt{b} c+15 \sqrt{a} e\right ) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{256 \sqrt{2} a^{15/4} b^{3/4}}-\frac{\left (77 \sqrt{b} c-15 \sqrt{a} e\right ) \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{b} x^2\right )}{512 \sqrt{2} a^{15/4} b^{3/4}}+\frac{\left (77 \sqrt{b} c-15 \sqrt{a} e\right ) \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{b} x^2\right )}{512 \sqrt{2} a^{15/4} b^{3/4}}\\ \end{align*}

Mathematica [A]  time = 0.408928, size = 379, normalized size = 0.99 \[ \frac{\frac{3 \sqrt{2} \left (15 a^{3/4} e-77 \sqrt [4]{a} \sqrt{b} c\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right )}{b^{3/4}}+\frac{3 \sqrt{2} \left (77 \sqrt [4]{a} \sqrt{b} c-15 a^{3/4} e\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right )}{b^{3/4}}-\frac{256 a^3 (a f-b x (c+x (d+e x)))}{b \left (a+b x^4\right )^3}+\frac{32 a^2 x (11 c+x (10 d+9 e x))}{\left (a+b x^4\right )^2}-\frac{6 \sqrt [4]{a} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (80 \sqrt [4]{a} \sqrt [4]{b} d+15 \sqrt{2} \sqrt{a} e+77 \sqrt{2} \sqrt{b} c\right )}{b^{3/4}}+\frac{6 \sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right ) \left (-80 \sqrt [4]{a} \sqrt [4]{b} d+15 \sqrt{2} \sqrt{a} e+77 \sqrt{2} \sqrt{b} c\right )}{b^{3/4}}+\frac{8 a x (77 c+15 x (4 d+3 e x))}{a+b x^4}}{3072 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a*x*(77*c + 15*x*(4*d + 3*e*x)))/(a + b*x^4) + (32*a^2*x*(11*c + x*(10*d + 9*e*x)))/(a + b*x^4)^2 - (256*a
^3*(a*f - b*x*(c + x*(d + e*x))))/(b*(a + b*x^4)^3) - (6*a^(1/4)*(77*Sqrt[2]*Sqrt[b]*c + 80*a^(1/4)*b^(1/4)*d
+ 15*Sqrt[2]*Sqrt[a]*e)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/b^(3/4) + (6*a^(1/4)*(77*Sqrt[2]*Sqrt[b]*c -
80*a^(1/4)*b^(1/4)*d + 15*Sqrt[2]*Sqrt[a]*e)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/b^(3/4) + (3*Sqrt[2]*(-7
7*a^(1/4)*Sqrt[b]*c + 15*a^(3/4)*e)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/b^(3/4) + (3*Sqrt[
2]*(77*a^(1/4)*Sqrt[b]*c - 15*a^(3/4)*e)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/b^(3/4))/(307
2*a^4)

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Maple [A]  time = 0.014, size = 400, normalized size = 1.1 \begin{align*}{\frac{1}{ \left ( b{x}^{4}+a \right ) ^{3}} \left ({\frac{15\,{b}^{2}e{x}^{11}}{128\,{a}^{3}}}+{\frac{5\,{b}^{2}d{x}^{10}}{32\,{a}^{3}}}+{\frac{77\,{b}^{2}c{x}^{9}}{384\,{a}^{3}}}+{\frac{21\,be{x}^{7}}{64\,{a}^{2}}}+{\frac{5\,bd{x}^{6}}{12\,{a}^{2}}}+{\frac{33\,bc{x}^{5}}{64\,{a}^{2}}}+{\frac{113\,e{x}^{3}}{384\,a}}+{\frac{11\,d{x}^{2}}{32\,a}}+{\frac{51\,cx}{128\,a}}-{\frac{f}{12\,b}} \right ) }+{\frac{77\,c\sqrt{2}}{1024\,{a}^{4}}\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{77\,c\sqrt{2}}{512\,{a}^{4}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ) }+{\frac{77\,c\sqrt{2}}{512\,{a}^{4}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ) }+{\frac{5\,d}{32\,{a}^{3}}\arctan \left ({x}^{2}\sqrt{{\frac{b}{a}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{15\,e\sqrt{2}}{1024\,{a}^{3}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{15\,e\sqrt{2}}{512\,{a}^{3}b}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+{\frac{15\,e\sqrt{2}}{512\,{a}^{3}b}\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((f*x^3+e*x^2+d*x+c)/(b*x^4+a)^4,x)

[Out]

(15/128*e/a^3*b^2*x^11+5/32*d/a^3*b^2*x^10+77/384*c/a^3*b^2*x^9+21/64/a^2*b*e*x^7+5/12/a^2*d*b*x^6+33/64/a^2*c
*b*x^5+113/384/a*e*x^3+11/32*d/a*x^2+51/128*c/a*x-1/12*f/b)/(b*x^4+a)^3+77/1024/a^4*c*(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)))+77/512/a^4*c*(1/b*a)
^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x+1)+77/512/a^4*c*(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/
4)*x-1)+5/32/a^3*d/(a*b)^(1/2)*arctan(x^2*(b/a)^(1/2))+15/1024/a^3*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)))+15/512/a^3*e/b/(1/b*a)^(1/4)*2^(1/2
)*arctan(2^(1/2)/(1/b*a)^(1/4)*x+1)+15/512/a^3*e/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((f*x^3+e*x^2+d*x+c)/(b*x^4+a)^4,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((f*x^3+e*x^2+d*x+c)/(b*x^4+a)^4,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((f*x**3+e*x**2+d*x+c)/(b*x**4+a)**4,x)

[Out]

Timed out

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Giac [A]  time = 1.08056, size = 528, normalized size = 1.38 \begin{align*} \frac{\sqrt{2}{\left (40 \, \sqrt{2} \sqrt{a b} b^{2} d + 77 \, \left (a b^{3}\right )^{\frac{1}{4}} b^{2} c + 15 \, \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 )}{512 \, a^{4} b^{3}} + \frac{\sqrt{2}{\left (40 \, \sqrt{2} \sqrt{a b} b^{2} d + 77 \, \left (a b^{3}\right )^{\frac{1}{4}} b^{2} c + 15 \, \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 )}{512 \, a^{4} b^{3}} + \frac{\sqrt{2}{\left (77 \, \left (a b^{3}\right )^{\frac{1}{4}} b^{2} c - 15 \, \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 )}{1024 \, a^{4} b^{3}} - \frac{\sqrt{2}{\left (77 \, \left (a b^{3}\right )^{\frac{1}{4}} b^{2} c - 15 \, \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 )}{1024 \, a^{4} b^{3}} + \frac{45 \, b^{3} x^{11} e + 60 \, b^{3} d x^{10} + 77 \, b^{3} c x^{9} + 126 \, a b^{2} x^{7} e + 160 \, a b^{2} d x^{6} + 198 \, a b^{2} c x^{5} + 113 \, a^{2} b x^{3} e + 132 \, a^{2} b d x^{2} + 153 \, a^{2} b c x - 32 \, a^{3} f}{384 \,{\left (b x^{4} + a\right )}^{3} a^{3} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/512*sqrt(2)*(40*sqrt(2)*sqrt(a*b)*b^2*d + 77*(a*b^3)^(1/4)*b^2*c + 15*(a*b^3)^(3/4)*e)*arctan(1/2*sqrt(2)*(2
*x + sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/(a^4*b^3) + 1/512*sqrt(2)*(40*sqrt(2)*sqrt(a*b)*b^2*d + 77*(a*b^3)^(1/4
)*b^2*c + 15*(a*b^3)^(3/4)*e)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/(a^4*b^3) + 1/1024*s
qrt(2)*(77*(a*b^3)^(1/4)*b^2*c - 15*(a*b^3)^(3/4)*e)*log(x^2 + sqrt(2)*x*(a/b)^(1/4) + sqrt(a/b))/(a^4*b^3) -
1/1024*sqrt(2)*(77*(a*b^3)^(1/4)*b^2*c - 15*(a*b^3)^(3/4)*e)*log(x^2 - sqrt(2)*x*(a/b)^(1/4) + sqrt(a/b))/(a^4
*b^3) + 1/384*(45*b^3*x^11*e + 60*b^3*d*x^10 + 77*b^3*c*x^9 + 126*a*b^2*x^7*e + 160*a*b^2*d*x^6 + 198*a*b^2*c*
x^5 + 113*a^2*b*x^3*e + 132*a^2*b*d*x^2 + 153*a^2*b*c*x - 32*a^3*f)/((b*x^4 + a)^3*a^3*b)