3.402 \(\int \frac{(d+e x)^2}{(a+c x^4)^2} \, dx\)

Optimal. Leaf size=322 \[ -\frac{\left (3 \sqrt{c} d^2-\sqrt{a} e^2\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )}{16 \sqrt{2} a^{7/4} c^{3/4}}+\frac{\left (3 \sqrt{c} d^2-\sqrt{a} e^2\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )}{16 \sqrt{2} a^{7/4} c^{3/4}}-\frac{\left (\sqrt{a} e^2+3 \sqrt{c} d^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt{2} a^{7/4} c^{3/4}}+\frac{\left (\sqrt{a} e^2+3 \sqrt{c} d^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{8 \sqrt{2} a^{7/4} c^{3/4}}+\frac{d e \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{2 a^{3/2} \sqrt{c}}+\frac{x (d+e x)^2}{4 a \left (a+c x^4\right )} \]

[Out]

(x*(d + e*x)^2)/(4*a*(a + c*x^4)) + (d*e*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(2*a^(3/2)*Sqrt[c]) - ((3*Sqrt[c]*d^2
+ Sqrt[a]*e^2)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(8*Sqrt[2]*a^(7/4)*c^(3/4)) + ((3*Sqrt[c]*d^2 + Sqrt[a
]*e^2)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(8*Sqrt[2]*a^(7/4)*c^(3/4)) - ((3*Sqrt[c]*d^2 - Sqrt[a]*e^2)*L
og[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(16*Sqrt[2]*a^(7/4)*c^(3/4)) + ((3*Sqrt[c]*d^2 - Sqrt[a
]*e^2)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(16*Sqrt[2]*a^(7/4)*c^(3/4))

________________________________________________________________________________________

Rubi [A]  time = 0.269161, antiderivative size = 322, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.588, Rules used = {1855, 1876, 275, 205, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{\left (3 \sqrt{c} d^2-\sqrt{a} e^2\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )}{16 \sqrt{2} a^{7/4} c^{3/4}}+\frac{\left (3 \sqrt{c} d^2-\sqrt{a} e^2\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )}{16 \sqrt{2} a^{7/4} c^{3/4}}-\frac{\left (\sqrt{a} e^2+3 \sqrt{c} d^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt{2} a^{7/4} c^{3/4}}+\frac{\left (\sqrt{a} e^2+3 \sqrt{c} d^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{8 \sqrt{2} a^{7/4} c^{3/4}}+\frac{d e \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{2 a^{3/2} \sqrt{c}}+\frac{x (d+e x)^2}{4 a \left (a+c x^4\right )} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2/(a + c*x^4)^2,x]

[Out]

(x*(d + e*x)^2)/(4*a*(a + c*x^4)) + (d*e*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(2*a^(3/2)*Sqrt[c]) - ((3*Sqrt[c]*d^2
+ Sqrt[a]*e^2)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(8*Sqrt[2]*a^(7/4)*c^(3/4)) + ((3*Sqrt[c]*d^2 + Sqrt[a
]*e^2)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(8*Sqrt[2]*a^(7/4)*c^(3/4)) - ((3*Sqrt[c]*d^2 - Sqrt[a]*e^2)*L
og[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(16*Sqrt[2]*a^(7/4)*c^(3/4)) + ((3*Sqrt[c]*d^2 - Sqrt[a
]*e^2)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(16*Sqrt[2]*a^(7/4)*c^(3/4))

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

Mathematica [A]  time = 0.325009, size = 321, normalized size = 1. \[ \frac{\frac{\sqrt{2} \left (a^{3/4} e^2-3 \sqrt [4]{a} \sqrt{c} d^2\right ) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )}{c^{3/4}}+\frac{\sqrt{2} \left (3 \sqrt [4]{a} \sqrt{c} d^2-a^{3/4} e^2\right ) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )}{c^{3/4}}-\frac{2 \sqrt [4]{a} \left (8 \sqrt [4]{a} \sqrt [4]{c} d e+\sqrt{2} \sqrt{a} e^2+3 \sqrt{2} \sqrt{c} d^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{c^{3/4}}+\frac{2 \sqrt [4]{a} \left (-8 \sqrt [4]{a} \sqrt [4]{c} d e+\sqrt{2} \sqrt{a} e^2+3 \sqrt{2} \sqrt{c} d^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{c^{3/4}}+\frac{8 a x (d+e x)^2}{a+c x^4}}{32 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a*x*(d + e*x)^2)/(a + c*x^4) - (2*a^(1/4)*(3*Sqrt[2]*Sqrt[c]*d^2 + 8*a^(1/4)*c^(1/4)*d*e + Sqrt[2]*Sqrt[a]
*e^2)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/c^(3/4) + (2*a^(1/4)*(3*Sqrt[2]*Sqrt[c]*d^2 - 8*a^(1/4)*c^(1/4)
*d*e + Sqrt[2]*Sqrt[a]*e^2)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/c^(3/4) + (Sqrt[2]*(-3*a^(1/4)*Sqrt[c]*d^
2 + a^(3/4)*e^2)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/c^(3/4) + (Sqrt[2]*(3*a^(1/4)*Sqrt[c]
*d^2 - a^(3/4)*e^2)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/c^(3/4))/(32*a^2)

________________________________________________________________________________________

Maple [A]  time = 0.004, size = 362, normalized size = 1.1 \begin{align*}{\frac{{d}^{2}x}{4\,a \left ( c{x}^{4}+a \right ) }}+{\frac{3\,{d}^{2}\sqrt{2}}{32\,{a}^{2}}\sqrt [4]{{\frac{a}{c}}}\ln \left ({ \left ({x}^{2}+\sqrt [4]{{\frac{a}{c}}}x\sqrt{2}+\sqrt{{\frac{a}{c}}} \right ) \left ({x}^{2}-\sqrt [4]{{\frac{a}{c}}}x\sqrt{2}+\sqrt{{\frac{a}{c}}} \right ) ^{-1}} \right ) }+{\frac{3\,{d}^{2}\sqrt{2}}{16\,{a}^{2}}\sqrt [4]{{\frac{a}{c}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}+1 \right ) }+{\frac{3\,{d}^{2}\sqrt{2}}{16\,{a}^{2}}\sqrt [4]{{\frac{a}{c}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}-1 \right ) }+{\frac{de{x}^{2}}{2\,a \left ( c{x}^{4}+a \right ) }}+{\frac{de}{2\,a}\arctan \left ({x}^{2}\sqrt{{\frac{c}{a}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{{e}^{2}{x}^{3}}{4\,a \left ( c{x}^{4}+a \right ) }}+{\frac{{e}^{2}\sqrt{2}}{32\,ac}\ln \left ({ \left ({x}^{2}-\sqrt [4]{{\frac{a}{c}}}x\sqrt{2}+\sqrt{{\frac{a}{c}}} \right ) \left ({x}^{2}+\sqrt [4]{{\frac{a}{c}}}x\sqrt{2}+\sqrt{{\frac{a}{c}}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}+{\frac{{e}^{2}\sqrt{2}}{16\,ac}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}+{\frac{{e}^{2}\sqrt{2}}{16\,ac}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*d^2*x/a/(c*x^4+a)+3/32*d^2/a^2*(a/c)^(1/4)*2^(1/2)*ln((x^2+(a/c)^(1/4)*x*2^(1/2)+(a/c)^(1/2))/(x^2-(a/c)^(
1/4)*x*2^(1/2)+(a/c)^(1/2)))+3/16*d^2/a^2*(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x+1)+3/16*d^2/a^2*(a/
c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x-1)+1/2*d*e*x^2/a/(c*x^4+a)+1/2*d*e/a/(a*c)^(1/2)*arctan(x^2*(1/a
*c)^(1/2))+1/4*e^2*x^3/a/(c*x^4+a)+1/32*e^2/a/c/(a/c)^(1/4)*2^(1/2)*ln((x^2-(a/c)^(1/4)*x*2^(1/2)+(a/c)^(1/2))
/(x^2+(a/c)^(1/4)*x*2^(1/2)+(a/c)^(1/2)))+1/16*e^2/a/c/(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x+1)+1/1
6*e^2/a/c/(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x-1)

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

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((e*x+d)^2/(c*x^4+a)^2,x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [A]  time = 3.24013, size = 318, normalized size = 0.99 \begin{align*} \operatorname{RootSum}{\left (65536 t^{4} a^{7} c^{3} + 11264 t^{2} a^{4} c^{2} d^{2} e^{2} + t \left (256 a^{3} c d e^{5} - 2304 a^{2} c^{2} d^{5} e\right ) + a^{2} e^{8} + 82 a c d^{4} e^{4} + 81 c^{2} d^{8}, \left ( t \mapsto t \log{\left (x + \frac{4096 t^{3} a^{7} c^{2} e^{6} + 356352 t^{3} a^{6} c^{3} d^{4} e^{2} - 23552 t^{2} a^{5} c^{2} d^{3} e^{5} + 27648 t^{2} a^{4} c^{3} d^{7} e + 912 t a^{4} c d^{2} e^{8} + 43584 t a^{3} c^{2} d^{6} e^{4} + 3888 t a^{2} c^{3} d^{10} + 12 a^{3} d e^{11} - 1088 a^{2} c d^{5} e^{7} - 7020 a c^{2} d^{9} e^{3}}{a^{3} e^{12} - 649 a^{2} c d^{4} e^{8} - 5841 a c^{2} d^{8} e^{4} + 729 c^{3} d^{12}} \right )} \right )\right )} + \frac{d^{2} x + 2 d e x^{2} + e^{2} x^{3}}{4 a^{2} + 4 a c x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(65536*_t**4*a**7*c**3 + 11264*_t**2*a**4*c**2*d**2*e**2 + _t*(256*a**3*c*d*e**5 - 2304*a**2*c**2*d**5*
e) + a**2*e**8 + 82*a*c*d**4*e**4 + 81*c**2*d**8, Lambda(_t, _t*log(x + (4096*_t**3*a**7*c**2*e**6 + 356352*_t
**3*a**6*c**3*d**4*e**2 - 23552*_t**2*a**5*c**2*d**3*e**5 + 27648*_t**2*a**4*c**3*d**7*e + 912*_t*a**4*c*d**2*
e**8 + 43584*_t*a**3*c**2*d**6*e**4 + 3888*_t*a**2*c**3*d**10 + 12*a**3*d*e**11 - 1088*a**2*c*d**5*e**7 - 7020
*a*c**2*d**9*e**3)/(a**3*e**12 - 649*a**2*c*d**4*e**8 - 5841*a*c**2*d**8*e**4 + 729*c**3*d**12)))) + (d**2*x +
2*d*e*x**2 + e**2*x**3)/(4*a**2 + 4*a*c*x**4)

________________________________________________________________________________________

Giac [A]  time = 1.15916, size = 436, normalized size = 1.35 \begin{align*} \frac{x^{3} e^{2} + 2 \, d x^{2} e + d^{2} x}{4 \,{\left (c x^{4} + a\right )} a} + \frac{\sqrt{2}{\left (4 \, \sqrt{2} \sqrt{a c} c^{2} d e + 3 \, \left (a c^{3}\right )^{\frac{1}{4}} c^{2} d^{2} + \left (a c^{3}\right )^{\frac{3}{4}} e^{2}\right )} \arctan \left (\frac{\sqrt{2}{\left (2 \, x + \sqrt{2} \left (\frac{a}{c}\right )^{\frac{1}{4}}\right )}}{2 \, \left (\frac{a}{c}\right )^{\frac{1}{4}}}\right )}{16 \, a^{2} c^{3}} + \frac{\sqrt{2}{\left (4 \, \sqrt{2} \sqrt{a c} c^{2} d e + 3 \, \left (a c^{3}\right )^{\frac{1}{4}} c^{2} d^{2} + \left (a c^{3}\right )^{\frac{3}{4}} e^{2}\right )} \arctan \left (\frac{\sqrt{2}{\left (2 \, x - \sqrt{2} \left (\frac{a}{c}\right )^{\frac{1}{4}}\right )}}{2 \, \left (\frac{a}{c}\right )^{\frac{1}{4}}}\right )}{16 \, a^{2} c^{3}} + \frac{\sqrt{2}{\left (3 \, \left (a c^{3}\right )^{\frac{1}{4}} c^{2} d^{2} - \left (a c^{3}\right )^{\frac{3}{4}} e^{2}\right )} \log \left (x^{2} + \sqrt{2} x \left (\frac{a}{c}\right )^{\frac{1}{4}} + \sqrt{\frac{a}{c}}\right )}{32 \, a^{2} c^{3}} - \frac{\sqrt{2}{\left (3 \, \left (a c^{3}\right )^{\frac{1}{4}} c^{2} d^{2} - \left (a c^{3}\right )^{\frac{3}{4}} e^{2}\right )} \log \left (x^{2} - \sqrt{2} x \left (\frac{a}{c}\right )^{\frac{1}{4}} + \sqrt{\frac{a}{c}}\right )}{32 \, a^{2} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(x^3*e^2 + 2*d*x^2*e + d^2*x)/((c*x^4 + a)*a) + 1/16*sqrt(2)*(4*sqrt(2)*sqrt(a*c)*c^2*d*e + 3*(a*c^3)^(1/4
)*c^2*d^2 + (a*c^3)^(3/4)*e^2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(a/c)^(1/4))/(a/c)^(1/4))/(a^2*c^3) + 1/16*sq
rt(2)*(4*sqrt(2)*sqrt(a*c)*c^2*d*e + 3*(a*c^3)^(1/4)*c^2*d^2 + (a*c^3)^(3/4)*e^2)*arctan(1/2*sqrt(2)*(2*x - sq
rt(2)*(a/c)^(1/4))/(a/c)^(1/4))/(a^2*c^3) + 1/32*sqrt(2)*(3*(a*c^3)^(1/4)*c^2*d^2 - (a*c^3)^(3/4)*e^2)*log(x^2
+ sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(a^2*c^3) - 1/32*sqrt(2)*(3*(a*c^3)^(1/4)*c^2*d^2 - (a*c^3)^(3/4)*e^2)*l
og(x^2 - sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(a^2*c^3)