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

Optimal. Leaf size=416 \[ -\frac {\sqrt [4]{c} d \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} \left (a e^4+c d^4\right )}+\frac {\sqrt [4]{c} d \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} \left (a e^4+c d^4\right )}-\frac {\sqrt [4]{c} d \left (\sqrt {a} e^2+\sqrt {c} d^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} \left (a e^4+c d^4\right )}+\frac {\sqrt [4]{c} d \left (\sqrt {a} e^2+\sqrt {c} d^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} a^{3/4} \left (a e^4+c d^4\right )}-\frac {e^3 \log \left (a+c x^4\right )}{4 \left (a e^4+c d^4\right )}+\frac {e^3 \log (d+e x)}{a e^4+c d^4}-\frac {\sqrt {c} d^2 e \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \left (a e^4+c d^4\right )} \]

[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)

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Rubi [A]  time = 0.43, antiderivative size = 416, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 12, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.706, Rules used = {6725, 1876, 1168, 1162, 617, 204, 1165, 628, 1248, 635, 205, 260} \[ -\frac {\sqrt [4]{c} d \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} \left (a e^4+c d^4\right )}+\frac {\sqrt [4]{c} d \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} \left (a e^4+c d^4\right )}-\frac {\sqrt [4]{c} d \left (\sqrt {a} e^2+\sqrt {c} d^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} \left (a e^4+c d^4\right )}+\frac {\sqrt [4]{c} d \left (\sqrt {a} e^2+\sqrt {c} d^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} a^{3/4} \left (a e^4+c d^4\right )}-\frac {e^3 \log \left (a+c x^4\right )}{4 \left (a e^4+c d^4\right )}-\frac {\sqrt {c} d^2 e \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \left (a e^4+c d^4\right )}+\frac {e^3 \log (d+e x)}{a e^4+c d^4} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)*(a + c*x^4)),x]

[Out]

-(Sqrt[c]*d^2*e*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(2*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 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 260

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 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 635

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 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 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 1248

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 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 6725

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*} \int \frac {1}{(d+e x) \left (a+c x^4\right )} \, dx &=\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 \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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*}

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Mathematica [A]  time = 0.15, size = 404, normalized size = 0.97 \[ \frac {-2 a^{3/4} e^3 \log \left (a+c x^4\right )+8 a^{3/4} e^3 \log (d+e x)-\sqrt {2} c^{3/4} d^3 \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )+\sqrt {2} c^{3/4} d^3 \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )-2 \sqrt [4]{c} d \left (-2 \sqrt [4]{a} \sqrt [4]{c} d e+\sqrt {2} \sqrt {a} e^2+\sqrt {2} \sqrt {c} d^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+2 \sqrt [4]{c} d \left (2 \sqrt [4]{a} \sqrt [4]{c} d e+\sqrt {2} \sqrt {a} e^2+\sqrt {2} \sqrt {c} d^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )+\sqrt {2} \sqrt {a} \sqrt [4]{c} d e^2 \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )-\sqrt {2} \sqrt {a} \sqrt [4]{c} d e^2 \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{8 a^{3/4} \left (a e^4+c d^4\right )} \]

Antiderivative was successfully verified.

[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))

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

[Out]

Timed out

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giac [A]  time = 0.42, size = 371, normalized size = 0.89 \[ \frac {\left (a c^{3}\right )^{\frac {1}{4}} c d \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{2 \, {\left (\sqrt {2} a c^{2} d^{2} - 2 \, \left (a c^{3}\right )^{\frac {1}{4}} a c d e + \sqrt {2} \sqrt {a c} a c e^{2}\right )}} + \frac {\left (a c^{3}\right )^{\frac {1}{4}} c d \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{2 \, {\left (\sqrt {2} a c^{2} d^{2} + 2 \, \left (a c^{3}\right )^{\frac {1}{4}} a c d e + \sqrt {2} \sqrt {a c} a c e^{2}\right )}} + \frac {{\left (\left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{3} - \left (a c^{3}\right )^{\frac {3}{4}} d e^{2}\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{4 \, {\left (\sqrt {2} a c^{3} d^{4} + \sqrt {2} a^{2} c^{2} e^{4}\right )}} - \frac {{\left (\left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{3} - \left (a c^{3}\right )^{\frac {3}{4}} d e^{2}\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{4 \, {\left (\sqrt {2} a c^{3} d^{4} + \sqrt {2} a^{2} c^{2} e^{4}\right )}} - \frac {e^{3} \log \left ({\left | c x^{4} + a \right |}\right )}{4 \, {\left (c d^{4} + a e^{4}\right )}} + \frac {e^{4} \log \left ({\left | x e + d \right |}\right )}{c d^{4} e + a e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 433, normalized size = 1.04 \[ -\frac {c \,d^{2} e \arctan \left (\sqrt {\frac {c}{a}}\, x^{2}\right )}{2 \left (a \,e^{4}+c \,d^{4}\right ) \sqrt {a c}}+\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, c \,d^{3} \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{4 \left (a \,e^{4}+c \,d^{4}\right ) a}+\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, c \,d^{3} \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{4 \left (a \,e^{4}+c \,d^{4}\right ) a}+\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, c \,d^{3} \ln \left (\frac {x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{c}}}{x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{c}}}\right )}{8 \left (a \,e^{4}+c \,d^{4}\right ) a}+\frac {\sqrt {2}\, d \,e^{2} \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{4 \left (a \,e^{4}+c \,d^{4}\right ) \left (\frac {a}{c}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, d \,e^{2} \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{4 \left (a \,e^{4}+c \,d^{4}\right ) \left (\frac {a}{c}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, d \,e^{2} \ln \left (\frac {x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{c}}}{x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{c}}}\right )}{8 \left (a \,e^{4}+c \,d^{4}\right ) \left (\frac {a}{c}\right )^{\frac {1}{4}}}-\frac {e^{3} \ln \left (c \,x^{4}+a \right )}{4 \left (a \,e^{4}+c \,d^{4}\right )}+\frac {e^{3} \ln \left (e x +d \right )}{a \,e^{4}+c \,d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.73, size = 345, normalized size = 0.83 \[ \frac {e^{3} \log \left (e x + d\right )}{c d^{4} + a e^{4}} - \frac {c {\left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {3}{4}} c^{\frac {1}{4}} e^{3} - c d^{3} + \sqrt {a} \sqrt {c} d e^{2}\right )} \log \left (\sqrt {c} x^{2} + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {5}{4}}} + \frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {3}{4}} c^{\frac {1}{4}} e^{3} + c d^{3} - \sqrt {a} \sqrt {c} d e^{2}\right )} \log \left (\sqrt {c} x^{2} - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {5}{4}}} - \frac {2 \, {\left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {5}{4}} d^{3} + \sqrt {2} a^{\frac {3}{4}} c^{\frac {3}{4}} d e^{2} + 2 \, \sqrt {a} c d^{2} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {c}} c^{\frac {5}{4}}} - \frac {2 \, {\left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {5}{4}} d^{3} + \sqrt {2} a^{\frac {3}{4}} c^{\frac {3}{4}} d e^{2} - 2 \, \sqrt {a} c d^{2} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {c}} c^{\frac {5}{4}}}\right )}}{8 \, {\left (c d^{4} + a e^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.42, size = 874, normalized size = 2.10 \[ \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} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

[Out]

Timed out

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