\(\int \frac {1}{(d+e x) (a+c x^4)} \, dx\) [398]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [C] (verified)
   Fricas [C] (verification not implemented)
   Sympy [F(-1)]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 17, antiderivative size = 416 \[ \int \frac {1}{(d+e x) \left (a+c x^4\right )} \, dx=-\frac {\sqrt {c} d^2 e \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \left (c d^4+a e^4\right )}-\frac {\sqrt [4]{c} d \left (\sqrt {c} d^2+\sqrt {a} e^2\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} \left (c d^4+a e^4\right )}+\frac {\sqrt [4]{c} d \left (\sqrt {c} d^2+\sqrt {a} e^2\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} \left (c d^4+a e^4\right )}+\frac {e^3 \log (d+e x)}{c d^4+a e^4}-\frac {\sqrt [4]{c} d \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} \left (c d^4+a e^4\right )}+\frac {\sqrt [4]{c} d \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} \left (c d^4+a e^4\right )}-\frac {e^3 \log \left (a+c x^4\right )}{4 \left (c d^4+a e^4\right )} \]

[Out]

e^3*ln(e*x+d)/(a*e^4+c*d^4)-1/4*e^3*ln(c*x^4+a)/(a*e^4+c*d^4)-1/2*d^2*e*arctan(x^2*c^(1/2)/a^(1/2))*c^(1/2)/(a
*e^4+c*d^4)/a^(1/2)-1/8*c^(1/4)*d*ln(-a^(1/4)*c^(1/4)*x*2^(1/2)+a^(1/2)+x^2*c^(1/2))*(-e^2*a^(1/2)+d^2*c^(1/2)
)/a^(3/4)/(a*e^4+c*d^4)*2^(1/2)+1/8*c^(1/4)*d*ln(a^(1/4)*c^(1/4)*x*2^(1/2)+a^(1/2)+x^2*c^(1/2))*(-e^2*a^(1/2)+
d^2*c^(1/2))/a^(3/4)/(a*e^4+c*d^4)*2^(1/2)+1/4*c^(1/4)*d*arctan(-1+c^(1/4)*x*2^(1/2)/a^(1/4))*(e^2*a^(1/2)+d^2
*c^(1/2))/a^(3/4)/(a*e^4+c*d^4)*2^(1/2)+1/4*c^(1/4)*d*arctan(1+c^(1/4)*x*2^(1/2)/a^(1/4))*(e^2*a^(1/2)+d^2*c^(
1/2))/a^(3/4)/(a*e^4+c*d^4)*2^(1/2)

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.706, Rules used = {6857, 1890, 1182, 1176, 631, 210, 1179, 642, 1262, 649, 211, 266} \[ \int \frac {1}{(d+e x) \left (a+c x^4\right )} \, dx=-\frac {\sqrt [4]{c} d \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right ) \left (\sqrt {a} e^2+\sqrt {c} d^2\right )}{2 \sqrt {2} a^{3/4} \left (a e^4+c d^4\right )}+\frac {\sqrt [4]{c} d \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right ) \left (\sqrt {a} e^2+\sqrt {c} d^2\right )}{2 \sqrt {2} a^{3/4} \left (a e^4+c d^4\right )}-\frac {\sqrt [4]{c} d \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} \left (a e^4+c d^4\right )}+\frac {\sqrt [4]{c} d \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} \left (a e^4+c d^4\right )}-\frac {\sqrt {c} d^2 e \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \left (a e^4+c d^4\right )}-\frac {e^3 \log \left (a+c x^4\right )}{4 \left (a e^4+c d^4\right )}+\frac {e^3 \log (d+e x)}{a e^4+c d^4} \]

[In]

Int[1/((d + e*x)*(a + c*x^4)),x]

[Out]

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

Rule 210

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

Rule 211

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

Rule 266

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]

Rule 631

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 642

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 649

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 1176

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 1179

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 1182

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 1262

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

Rule 1890

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 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
 /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 404, normalized size of antiderivative = 0.97 \[ \int \frac {1}{(d+e x) \left (a+c x^4\right )} \, dx=\frac {-2 \sqrt [4]{c} d \left (\sqrt {2} \sqrt {c} d^2-2 \sqrt [4]{a} \sqrt [4]{c} d e+\sqrt {2} \sqrt {a} e^2\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+2 \sqrt [4]{c} d \left (\sqrt {2} \sqrt {c} d^2+2 \sqrt [4]{a} \sqrt [4]{c} d e+\sqrt {2} \sqrt {a} e^2\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+8 a^{3/4} e^3 \log (d+e x)-\sqrt {2} c^{3/4} d^3 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )+\sqrt {2} \sqrt {a} \sqrt [4]{c} d e^2 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )+\sqrt {2} c^{3/4} d^3 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )-\sqrt {2} \sqrt {a} \sqrt [4]{c} d e^2 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )-2 a^{3/4} e^3 \log \left (a+c x^4\right )}{8 a^{3/4} \left (c d^4+a e^4\right )} \]

[In]

Integrate[1/((d + e*x)*(a + c*x^4)),x]

[Out]

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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.89 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.51

method result size
risch \(\frac {e^{3} \ln \left (e x +d \right )}{e^{4} a +d^{4} c}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (1+\left (a^{4} e^{4}+a^{3} c \,d^{4}\right ) \textit {\_Z}^{4}+4 a^{3} e^{3} \textit {\_Z}^{3}+6 \textit {\_Z}^{2} a^{2} e^{2}+4 \textit {\_Z} a e \right )}{\sum }\textit {\_R} \ln \left (\left (\left (5 e^{6} a^{3}-3 a^{2} d^{4} e^{2} c \right ) \textit {\_R}^{3}+\left (15 a^{2} e^{5}-3 a c \,d^{4} e \right ) \textit {\_R}^{2}+\left (15 e^{4} a -d^{4} c \right ) \textit {\_R} +5 e^{3}\right ) x +\left (6 a^{3} d \,e^{5}-2 a^{2} c \,d^{5} e \right ) \textit {\_R}^{3}+\left (13 a^{2} e^{4} d -a c \,d^{5}\right ) \textit {\_R}^{2}+8 a d \,e^{3} \textit {\_R} +d \,e^{2}\right )\right )}{4}\) \(211\)
default \(\frac {c \left (\frac {d^{3} \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{8 a}-\frac {d^{2} e \arctan \left (x^{2} \sqrt {\frac {c}{a}}\right )}{2 \sqrt {a c}}+\frac {d \,e^{2} \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{8 c \left (\frac {a}{c}\right )^{\frac {1}{4}}}-\frac {e^{3} \ln \left (c \,x^{4}+a \right )}{4 c}\right )}{e^{4} a +d^{4} c}+\frac {e^{3} \ln \left (e x +d \right )}{e^{4} a +d^{4} c}\) \(289\)

[In]

int(1/(e*x+d)/(c*x^4+a),x,method=_RETURNVERBOSE)

[Out]

e^3*ln(e*x+d)/(a*e^4+c*d^4)+1/4*sum(_R*ln(((5*a^3*e^6-3*a^2*c*d^4*e^2)*_R^3+(15*a^2*e^5-3*a*c*d^4*e)*_R^2+(15*
a*e^4-c*d^4)*_R+5*e^3)*x+(6*a^3*d*e^5-2*a^2*c*d^5*e)*_R^3+(13*a^2*d*e^4-a*c*d^5)*_R^2+8*a*d*e^3*_R+d*e^2),_R=R
ootOf(1+(a^4*e^4+a^3*c*d^4)*_Z^4+4*a^3*e^3*_Z^3+6*_Z^2*a^2*e^2+4*_Z*a*e))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 97.11 (sec) , antiderivative size = 352864, normalized size of antiderivative = 848.23 \[ \int \frac {1}{(d+e x) \left (a+c x^4\right )} \, dx=\text {Too large to display} \]

[In]

integrate(1/(e*x+d)/(c*x^4+a),x, algorithm="fricas")

[Out]

Too large to include

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(d+e x) \left (a+c x^4\right )} \, dx=\text {Timed out} \]

[In]

integrate(1/(e*x+d)/(c*x**4+a),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 345, normalized size of antiderivative = 0.83 \[ \int \frac {1}{(d+e x) \left (a+c x^4\right )} \, dx=\frac {e^{3} \log \left (e x + d\right )}{c d^{4} + a e^{4}} - \frac {c {\left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {3}{4}} c^{\frac {1}{4}} e^{3} - c d^{3} + \sqrt {a} \sqrt {c} d e^{2}\right )} \log \left (\sqrt {c} x^{2} + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {5}{4}}} + \frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {3}{4}} c^{\frac {1}{4}} e^{3} + c d^{3} - \sqrt {a} \sqrt {c} d e^{2}\right )} \log \left (\sqrt {c} x^{2} - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {5}{4}}} - \frac {2 \, {\left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {5}{4}} d^{3} + \sqrt {2} a^{\frac {3}{4}} c^{\frac {3}{4}} d e^{2} + 2 \, \sqrt {a} c d^{2} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {c}} c^{\frac {5}{4}}} - \frac {2 \, {\left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {5}{4}} d^{3} + \sqrt {2} a^{\frac {3}{4}} c^{\frac {3}{4}} d e^{2} - 2 \, \sqrt {a} c d^{2} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {c}} c^{\frac {5}{4}}}\right )}}{8 \, {\left (c d^{4} + a e^{4}\right )}} \]

[In]

integrate(1/(e*x+d)/(c*x^4+a),x, algorithm="maxima")

[Out]

e^3*log(e*x + d)/(c*d^4 + a*e^4) - 1/8*c*(sqrt(2)*(sqrt(2)*a^(3/4)*c^(1/4)*e^3 - c*d^3 + sqrt(a)*sqrt(c)*d*e^2
)*log(sqrt(c)*x^2 + sqrt(2)*a^(1/4)*c^(1/4)*x + sqrt(a))/(a^(3/4)*c^(5/4)) + sqrt(2)*(sqrt(2)*a^(3/4)*c^(1/4)*
e^3 + c*d^3 - sqrt(a)*sqrt(c)*d*e^2)*log(sqrt(c)*x^2 - sqrt(2)*a^(1/4)*c^(1/4)*x + sqrt(a))/(a^(3/4)*c^(5/4))
- 2*(sqrt(2)*a^(1/4)*c^(5/4)*d^3 + sqrt(2)*a^(3/4)*c^(3/4)*d*e^2 + 2*sqrt(a)*c*d^2*e)*arctan(1/2*sqrt(2)*(2*sq
rt(c)*x + sqrt(2)*a^(1/4)*c^(1/4))/sqrt(sqrt(a)*sqrt(c)))/(a^(3/4)*sqrt(sqrt(a)*sqrt(c))*c^(5/4)) - 2*(sqrt(2)
*a^(1/4)*c^(5/4)*d^3 + sqrt(2)*a^(3/4)*c^(3/4)*d*e^2 - 2*sqrt(a)*c*d^2*e)*arctan(1/2*sqrt(2)*(2*sqrt(c)*x - sq
rt(2)*a^(1/4)*c^(1/4))/sqrt(sqrt(a)*sqrt(c)))/(a^(3/4)*sqrt(sqrt(a)*sqrt(c))*c^(5/4)))/(c*d^4 + a*e^4)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 377, normalized size of antiderivative = 0.91 \[ \int \frac {1}{(d+e x) \left (a+c x^4\right )} \, dx=\frac {e^{4} \log \left ({\left | e x + d \right |}\right )}{c d^{4} e + a e^{5}} - \frac {e^{3} \log \left ({\left | c x^{4} + a \right |}\right )}{4 \, {\left (c d^{4} + a e^{4}\right )}} + \frac {\left (a c^{3}\right )^{\frac {1}{4}} c d \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{2 \, {\left (\sqrt {2} a c^{2} d^{2} + \sqrt {2} \sqrt {a c} a c e^{2} - 2 \, \left (a c^{3}\right )^{\frac {1}{4}} a c d e\right )}} + \frac {\left (a c^{3}\right )^{\frac {1}{4}} c d \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{2 \, {\left (\sqrt {2} a c^{2} d^{2} + \sqrt {2} \sqrt {a c} a c e^{2} + 2 \, \left (a c^{3}\right )^{\frac {1}{4}} a c d e\right )}} + \frac {{\left (\left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{3} - \left (a c^{3}\right )^{\frac {3}{4}} d e^{2}\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{4 \, {\left (\sqrt {2} a c^{3} d^{4} + \sqrt {2} a^{2} c^{2} e^{4}\right )}} - \frac {{\left (\left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{3} - \left (a c^{3}\right )^{\frac {3}{4}} d e^{2}\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{4 \, {\left (\sqrt {2} a c^{3} d^{4} + \sqrt {2} a^{2} c^{2} e^{4}\right )}} \]

[In]

integrate(1/(e*x+d)/(c*x^4+a),x, algorithm="giac")

[Out]

e^4*log(abs(e*x + d))/(c*d^4*e + a*e^5) - 1/4*e^3*log(abs(c*x^4 + a))/(c*d^4 + a*e^4) + 1/2*(a*c^3)^(1/4)*c*d*
arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(a/c)^(1/4))/(a/c)^(1/4))/(sqrt(2)*a*c^2*d^2 + sqrt(2)*sqrt(a*c)*a*c*e^2 - 2
*(a*c^3)^(1/4)*a*c*d*e) + 1/2*(a*c^3)^(1/4)*c*d*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(a/c)^(1/4))/(a/c)^(1/4))/(s
qrt(2)*a*c^2*d^2 + sqrt(2)*sqrt(a*c)*a*c*e^2 + 2*(a*c^3)^(1/4)*a*c*d*e) + 1/4*((a*c^3)^(1/4)*c^2*d^3 - (a*c^3)
^(3/4)*d*e^2)*log(x^2 + sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(sqrt(2)*a*c^3*d^4 + sqrt(2)*a^2*c^2*e^4) - 1/4*((a
*c^3)^(1/4)*c^2*d^3 - (a*c^3)^(3/4)*d*e^2)*log(x^2 - sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(sqrt(2)*a*c^3*d^4 + s
qrt(2)*a^2*c^2*e^4)

Mupad [B] (verification not implemented)

Time = 9.35 (sec) , antiderivative size = 874, normalized size of antiderivative = 2.10 \[ \int \frac {1}{(d+e x) \left (a+c x^4\right )} \, dx=\left (\sum _{k=1}^4\ln \left (\mathrm {root}\left (256\,a^3\,c\,d^4\,z^4+256\,a^4\,e^4\,z^4+256\,a^3\,e^3\,z^3+96\,a^2\,e^2\,z^2+16\,a\,e\,z+1,z,k\right )\,c^4\,e\,\left (d\,e^2+5\,e^3\,x+{\mathrm {root}\left (256\,a^3\,c\,d^4\,z^4+256\,a^4\,e^4\,z^4+256\,a^3\,e^3\,z^3+96\,a^2\,e^2\,z^2+16\,a\,e\,z+1,z,k\right )}^2\,a^2\,e^5\,x\,240+{\mathrm {root}\left (256\,a^3\,c\,d^4\,z^4+256\,a^4\,e^4\,z^4+256\,a^3\,e^3\,z^3+96\,a^2\,e^2\,z^2+16\,a\,e\,z+1,z,k\right )}^3\,a^3\,e^6\,x\,320+\mathrm {root}\left (256\,a^3\,c\,d^4\,z^4+256\,a^4\,e^4\,z^4+256\,a^3\,e^3\,z^3+96\,a^2\,e^2\,z^2+16\,a\,e\,z+1,z,k\right )\,a\,d\,e^3\,32+\mathrm {root}\left (256\,a^3\,c\,d^4\,z^4+256\,a^4\,e^4\,z^4+256\,a^3\,e^3\,z^3+96\,a^2\,e^2\,z^2+16\,a\,e\,z+1,z,k\right )\,a\,e^4\,x\,60-\mathrm {root}\left (256\,a^3\,c\,d^4\,z^4+256\,a^4\,e^4\,z^4+256\,a^3\,e^3\,z^3+96\,a^2\,e^2\,z^2+16\,a\,e\,z+1,z,k\right )\,c\,d^4\,x\,4-{\mathrm {root}\left (256\,a^3\,c\,d^4\,z^4+256\,a^4\,e^4\,z^4+256\,a^3\,e^3\,z^3+96\,a^2\,e^2\,z^2+16\,a\,e\,z+1,z,k\right )}^2\,a\,c\,d^5\,16+{\mathrm {root}\left (256\,a^3\,c\,d^4\,z^4+256\,a^4\,e^4\,z^4+256\,a^3\,e^3\,z^3+96\,a^2\,e^2\,z^2+16\,a\,e\,z+1,z,k\right )}^2\,a^2\,d\,e^4\,208+{\mathrm {root}\left (256\,a^3\,c\,d^4\,z^4+256\,a^4\,e^4\,z^4+256\,a^3\,e^3\,z^3+96\,a^2\,e^2\,z^2+16\,a\,e\,z+1,z,k\right )}^3\,a^3\,d\,e^5\,384-{\mathrm {root}\left (256\,a^3\,c\,d^4\,z^4+256\,a^4\,e^4\,z^4+256\,a^3\,e^3\,z^3+96\,a^2\,e^2\,z^2+16\,a\,e\,z+1,z,k\right )}^3\,a^2\,c\,d^5\,e\,128-{\mathrm {root}\left (256\,a^3\,c\,d^4\,z^4+256\,a^4\,e^4\,z^4+256\,a^3\,e^3\,z^3+96\,a^2\,e^2\,z^2+16\,a\,e\,z+1,z,k\right )}^3\,a^2\,c\,d^4\,e^2\,x\,192-{\mathrm {root}\left (256\,a^3\,c\,d^4\,z^4+256\,a^4\,e^4\,z^4+256\,a^3\,e^3\,z^3+96\,a^2\,e^2\,z^2+16\,a\,e\,z+1,z,k\right )}^2\,a\,c\,d^4\,e\,x\,48\right )\right )\,\mathrm {root}\left (256\,a^3\,c\,d^4\,z^4+256\,a^4\,e^4\,z^4+256\,a^3\,e^3\,z^3+96\,a^2\,e^2\,z^2+16\,a\,e\,z+1,z,k\right )\right )+\frac {e^3\,\ln \left (d+e\,x\right )}{c\,d^4+a\,e^4} \]

[In]

int(1/((a + c*x^4)*(d + e*x)),x)

[Out]

symsum(log(root(256*a^3*c*d^4*z^4 + 256*a^4*e^4*z^4 + 256*a^3*e^3*z^3 + 96*a^2*e^2*z^2 + 16*a*e*z + 1, z, k)*c
^4*e*(d*e^2 + 5*e^3*x + 240*root(256*a^3*c*d^4*z^4 + 256*a^4*e^4*z^4 + 256*a^3*e^3*z^3 + 96*a^2*e^2*z^2 + 16*a
*e*z + 1, z, k)^2*a^2*e^5*x + 320*root(256*a^3*c*d^4*z^4 + 256*a^4*e^4*z^4 + 256*a^3*e^3*z^3 + 96*a^2*e^2*z^2
+ 16*a*e*z + 1, z, k)^3*a^3*e^6*x + 32*root(256*a^3*c*d^4*z^4 + 256*a^4*e^4*z^4 + 256*a^3*e^3*z^3 + 96*a^2*e^2
*z^2 + 16*a*e*z + 1, z, k)*a*d*e^3 + 60*root(256*a^3*c*d^4*z^4 + 256*a^4*e^4*z^4 + 256*a^3*e^3*z^3 + 96*a^2*e^
2*z^2 + 16*a*e*z + 1, z, k)*a*e^4*x - 4*root(256*a^3*c*d^4*z^4 + 256*a^4*e^4*z^4 + 256*a^3*e^3*z^3 + 96*a^2*e^
2*z^2 + 16*a*e*z + 1, z, k)*c*d^4*x - 16*root(256*a^3*c*d^4*z^4 + 256*a^4*e^4*z^4 + 256*a^3*e^3*z^3 + 96*a^2*e
^2*z^2 + 16*a*e*z + 1, z, k)^2*a*c*d^5 + 208*root(256*a^3*c*d^4*z^4 + 256*a^4*e^4*z^4 + 256*a^3*e^3*z^3 + 96*a
^2*e^2*z^2 + 16*a*e*z + 1, z, k)^2*a^2*d*e^4 + 384*root(256*a^3*c*d^4*z^4 + 256*a^4*e^4*z^4 + 256*a^3*e^3*z^3
+ 96*a^2*e^2*z^2 + 16*a*e*z + 1, z, k)^3*a^3*d*e^5 - 128*root(256*a^3*c*d^4*z^4 + 256*a^4*e^4*z^4 + 256*a^3*e^
3*z^3 + 96*a^2*e^2*z^2 + 16*a*e*z + 1, z, k)^3*a^2*c*d^5*e - 192*root(256*a^3*c*d^4*z^4 + 256*a^4*e^4*z^4 + 25
6*a^3*e^3*z^3 + 96*a^2*e^2*z^2 + 16*a*e*z + 1, z, k)^3*a^2*c*d^4*e^2*x - 48*root(256*a^3*c*d^4*z^4 + 256*a^4*e
^4*z^4 + 256*a^3*e^3*z^3 + 96*a^2*e^2*z^2 + 16*a*e*z + 1, z, k)^2*a*c*d^4*e*x))*root(256*a^3*c*d^4*z^4 + 256*a
^4*e^4*z^4 + 256*a^3*e^3*z^3 + 96*a^2*e^2*z^2 + 16*a*e*z + 1, z, k), k, 1, 4) + (e^3*log(d + e*x))/(a*e^4 + c*
d^4)