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

Optimal. Leaf size=241 \[ -\frac {3 d \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} \sqrt [4]{c}}+\frac {3 d \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} \sqrt [4]{c}}-\frac {3 d \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} \sqrt [4]{c}}+\frac {3 d \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{8 \sqrt {2} a^{7/4} \sqrt [4]{c}}+\frac {e \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{4 a^{3/2} \sqrt {c}}+\frac {x (d+e x)}{4 a \left (a+c x^4\right )} \]

[Out]

1/4*x*(e*x+d)/a/(c*x^4+a)+3/16*d*arctan(-1+c^(1/4)*x*2^(1/2)/a^(1/4))/a^(7/4)/c^(1/4)*2^(1/2)+3/16*d*arctan(1+
c^(1/4)*x*2^(1/2)/a^(1/4))/a^(7/4)/c^(1/4)*2^(1/2)-3/32*d*ln(-a^(1/4)*c^(1/4)*x*2^(1/2)+a^(1/2)+x^2*c^(1/2))/a
^(7/4)/c^(1/4)*2^(1/2)+3/32*d*ln(a^(1/4)*c^(1/4)*x*2^(1/2)+a^(1/2)+x^2*c^(1/2))/a^(7/4)/c^(1/4)*2^(1/2)+1/4*e*
arctan(x^2*c^(1/2)/a^(1/2))/a^(3/2)/c^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 241, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {1855, 1876, 211, 1165, 628, 1162, 617, 204, 275, 205} \[ -\frac {3 d \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} \sqrt [4]{c}}+\frac {3 d \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} \sqrt [4]{c}}-\frac {3 d \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} \sqrt [4]{c}}+\frac {3 d \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{8 \sqrt {2} a^{7/4} \sqrt [4]{c}}+\frac {e \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{4 a^{3/2} \sqrt {c}}+\frac {x (d+e x)}{4 a \left (a+c x^4\right )} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)/(a + c*x^4)^2,x]

[Out]

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

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 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 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

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 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 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 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 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 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

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 224, normalized size = 0.93 \[ \frac {\frac {8 a^{3/4} x (d+e x)}{a+c x^4}-\frac {2 \left (4 \sqrt [4]{a} e+3 \sqrt {2} \sqrt [4]{c} d\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{\sqrt {c}}+\frac {2 \left (3 \sqrt {2} \sqrt [4]{c} d-4 \sqrt [4]{a} e\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{\sqrt {c}}-\frac {3 \sqrt {2} d \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{\sqrt [4]{c}}+\frac {3 \sqrt {2} d \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{\sqrt [4]{c}}}{32 a^{7/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)/(a + c*x^4)^2,x]

[Out]

((8*a^(3/4)*x*(d + e*x))/(a + c*x^4) - (2*(3*Sqrt[2]*c^(1/4)*d + 4*a^(1/4)*e)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a
^(1/4)])/Sqrt[c] + (2*(3*Sqrt[2]*c^(1/4)*d - 4*a^(1/4)*e)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/Sqrt[c] - (
3*Sqrt[2]*d*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/c^(1/4) + (3*Sqrt[2]*d*Log[Sqrt[a] + Sqrt[
2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/c^(1/4))/(32*a^(7/4))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [A]  time = 0.34, size = 241, normalized size = 1.00 \[ \frac {3 \, \sqrt {2} \left (a c^{3}\right )^{\frac {1}{4}} d \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{32 \, a^{2} c} - \frac {3 \, \sqrt {2} \left (a c^{3}\right )^{\frac {1}{4}} d \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{32 \, a^{2} c} + \frac {x^{2} e + d x}{4 \, {\left (c x^{4} + a\right )} a} + \frac {\sqrt {2} {\left (2 \, \sqrt {2} \sqrt {a c} c e + 3 \, \left (a c^{3}\right )^{\frac {1}{4}} c d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{16 \, a^{2} c^{2}} + \frac {\sqrt {2} {\left (2 \, \sqrt {2} \sqrt {a c} c e + 3 \, \left (a c^{3}\right )^{\frac {1}{4}} c d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{16 \, a^{2} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 188, normalized size = 0.78 \[ \frac {e \,x^{2}}{4 \left (c \,x^{4}+a \right ) a}+\frac {d x}{4 \left (c \,x^{4}+a \right ) a}+\frac {e \arctan \left (\sqrt {\frac {c}{a}}\, x^{2}\right )}{4 \sqrt {a c}\, a}+\frac {3 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, d \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{16 a^{2}}+\frac {3 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, d \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{16 a^{2}}+\frac {3 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, d \ln \left (\frac {x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{c}}}{x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{c}}}\right )}{32 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.43, size = 238, normalized size = 0.99 \[ \frac {e x^{2} + d x}{4 \, {\left (a c x^{4} + a^{2}\right )}} + \frac {\frac {3 \, \sqrt {2} d \log \left (\sqrt {c} x^{2} + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {1}{4}}} - \frac {3 \, \sqrt {2} d \log \left (\sqrt {c} x^{2} - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {1}{4}}} + \frac {2 \, {\left (3 \, \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} d - 4 \, \sqrt {a} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {c}} c^{\frac {1}{4}}} + \frac {2 \, {\left (3 \, \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} d + 4 \, \sqrt {a} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {c}} c^{\frac {1}{4}}}}{32 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.27, size = 282, normalized size = 1.17 \[ \left (\sum _{k=1}^4\ln \left (\frac {c^2\,\left (3\,d\,e^2+2\,e^3\,x-{\mathrm {root}\left (65536\,a^7\,c^2\,z^4+2048\,a^4\,c\,e^2\,z^2-1152\,a^2\,c\,d^2\,e\,z+81\,c\,d^4+16\,a\,e^4,z,k\right )}^2\,a^3\,c\,d\,192+{\mathrm {root}\left (65536\,a^7\,c^2\,z^4+2048\,a^4\,c\,e^2\,z^2-1152\,a^2\,c\,d^2\,e\,z+81\,c\,d^4+16\,a\,e^4,z,k\right )}^2\,a^3\,c\,e\,x\,128-\mathrm {root}\left (65536\,a^7\,c^2\,z^4+2048\,a^4\,c\,e^2\,z^2-1152\,a^2\,c\,d^2\,e\,z+81\,c\,d^4+16\,a\,e^4,z,k\right )\,a\,c\,d^2\,x\,36\right )}{a^3\,16}\right )\,\mathrm {root}\left (65536\,a^7\,c^2\,z^4+2048\,a^4\,c\,e^2\,z^2-1152\,a^2\,c\,d^2\,e\,z+81\,c\,d^4+16\,a\,e^4,z,k\right )\right )+\frac {\frac {e\,x^2}{4\,a}+\frac {d\,x}{4\,a}}{c\,x^4+a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)/(a + c*x^4)^2,x)

[Out]

symsum(log((c^2*(3*d*e^2 + 2*e^3*x - 192*root(65536*a^7*c^2*z^4 + 2048*a^4*c*e^2*z^2 - 1152*a^2*c*d^2*e*z + 81
*c*d^4 + 16*a*e^4, z, k)^2*a^3*c*d + 128*root(65536*a^7*c^2*z^4 + 2048*a^4*c*e^2*z^2 - 1152*a^2*c*d^2*e*z + 81
*c*d^4 + 16*a*e^4, z, k)^2*a^3*c*e*x - 36*root(65536*a^7*c^2*z^4 + 2048*a^4*c*e^2*z^2 - 1152*a^2*c*d^2*e*z + 8
1*c*d^4 + 16*a*e^4, z, k)*a*c*d^2*x))/(16*a^3))*root(65536*a^7*c^2*z^4 + 2048*a^4*c*e^2*z^2 - 1152*a^2*c*d^2*e
*z + 81*c*d^4 + 16*a*e^4, z, k), k, 1, 4) + ((e*x^2)/(4*a) + (d*x)/(4*a))/(a + c*x^4)

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sympy [A]  time = 1.09, size = 155, normalized size = 0.64 \[ \operatorname {RootSum} {\left (65536 t^{4} a^{7} c^{2} + 2048 t^{2} a^{4} c e^{2} - 1152 t a^{2} c d^{2} e + 16 a e^{4} + 81 c d^{4}, \left (t \mapsto t \log {\left (x + \frac {- 32768 t^{3} a^{6} c e^{2} - 4608 t^{2} a^{4} c d^{2} e - 512 t a^{3} e^{4} - 1296 t a^{2} c d^{4} + 360 a d^{2} e^{3}}{192 a d e^{4} - 243 c d^{5}} \right )} \right )\right )} + \frac {d x + e x^{2}}{4 a^{2} + 4 a c x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(65536*_t**4*a**7*c**2 + 2048*_t**2*a**4*c*e**2 - 1152*_t*a**2*c*d**2*e + 16*a*e**4 + 81*c*d**4, Lambda
(_t, _t*log(x + (-32768*_t**3*a**6*c*e**2 - 4608*_t**2*a**4*c*d**2*e - 512*_t*a**3*e**4 - 1296*_t*a**2*c*d**4
+ 360*a*d**2*e**3)/(192*a*d*e**4 - 243*c*d**5)))) + (d*x + e*x**2)/(4*a**2 + 4*a*c*x**4)

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