3.394 \(\int \frac {(d+e x)^3}{a+c x^4} \, dx\)

Optimal. Leaf size=320 \[ -\frac {d \left (\sqrt {c} d^2-3 \sqrt {a} e^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} c^{3/4}}+\frac {d \left (\sqrt {c} d^2-3 \sqrt {a} e^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} c^{3/4}}-\frac {d \left (3 \sqrt {a} e^2+\sqrt {c} d^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} c^{3/4}}+\frac {d \left (3 \sqrt {a} e^2+\sqrt {c} d^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} a^{3/4} c^{3/4}}+\frac {3 d^2 e \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {c}}+\frac {e^3 \log \left (a+c x^4\right )}{4 c} \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 320, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.647, Rules used = {1876, 1168, 1162, 617, 204, 1165, 628, 1248, 635, 205, 260} \[ -\frac {d \left (\sqrt {c} d^2-3 \sqrt {a} e^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} c^{3/4}}+\frac {d \left (\sqrt {c} d^2-3 \sqrt {a} e^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} c^{3/4}}-\frac {d \left (3 \sqrt {a} e^2+\sqrt {c} d^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} c^{3/4}}+\frac {d \left (3 \sqrt {a} e^2+\sqrt {c} d^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} a^{3/4} c^{3/4}}+\frac {3 d^2 e \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {c}}+\frac {e^3 \log \left (a+c x^4\right )}{4 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3/(a + c*x^4),x]

[Out]

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

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

Rubi steps

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

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Mathematica [A]  time = 0.25, size = 322, normalized size = 1.01 \[ \frac {-\sqrt {2} \sqrt [4]{c} \left (\sqrt [4]{a} \sqrt {c} d^3-3 a^{3/4} d e^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )+\sqrt {2} \sqrt [4]{c} \left (\sqrt [4]{a} \sqrt {c} d^3-3 a^{3/4} d e^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )-2 \sqrt [4]{a} \sqrt [4]{c} d \left (6 \sqrt [4]{a} \sqrt [4]{c} d e+3 \sqrt {2} \sqrt {a} e^2+\sqrt {2} \sqrt {c} d^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+2 \sqrt [4]{a} \sqrt [4]{c} d \left (-6 \sqrt [4]{a} \sqrt [4]{c} d e+3 \sqrt {2} \sqrt {a} e^2+\sqrt {2} \sqrt {c} d^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )+2 a e^3 \log \left (a+c x^4\right )}{8 a c} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^3/(a + c*x^4),x]

[Out]

(-2*a^(1/4)*c^(1/4)*d*(Sqrt[2]*Sqrt[c]*d^2 + 6*a^(1/4)*c^(1/4)*d*e + 3*Sqrt[2]*Sqrt[a]*e^2)*ArcTan[1 - (Sqrt[2
]*c^(1/4)*x)/a^(1/4)] + 2*a^(1/4)*c^(1/4)*d*(Sqrt[2]*Sqrt[c]*d^2 - 6*a^(1/4)*c^(1/4)*d*e + 3*Sqrt[2]*Sqrt[a]*e
^2)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)] - Sqrt[2]*c^(1/4)*(a^(1/4)*Sqrt[c]*d^3 - 3*a^(3/4)*d*e^2)*Log[Sqrt
[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2] + Sqrt[2]*c^(1/4)*(a^(1/4)*Sqrt[c]*d^3 - 3*a^(3/4)*d*e^2)*Log[S
qrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2] + 2*a*e^3*Log[a + c*x^4])/(8*a*c)

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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)^3/(c*x^4+a),x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 0.36, size = 311, normalized size = 0.97 \[ \frac {e^{3} \log \left ({\left | c x^{4} + a \right |}\right )}{4 \, c} + \frac {\sqrt {2} {\left (3 \, \sqrt {2} \sqrt {a c} c^{2} d^{2} e + \left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{3} + 3 \, \left (a c^{3}\right )^{\frac {3}{4}} d e^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{4 \, a c^{3}} + \frac {\sqrt {2} {\left (3 \, \sqrt {2} \sqrt {a c} c^{2} d^{2} e + \left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{3} + 3 \, \left (a c^{3}\right )^{\frac {3}{4}} d e^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{4 \, a c^{3}} + \frac {\sqrt {2} {\left (\left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{3} - 3 \, \left (a c^{3}\right )^{\frac {3}{4}} d e^{2}\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{8 \, a c^{3}} - \frac {\sqrt {2} {\left (\left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{3} - 3 \, \left (a c^{3}\right )^{\frac {3}{4}} d e^{2}\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{8 \, a c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*e^3*log(abs(c*x^4 + a))/c + 1/4*sqrt(2)*(3*sqrt(2)*sqrt(a*c)*c^2*d^2*e + (a*c^3)^(1/4)*c^2*d^3 + 3*(a*c^3)
^(3/4)*d*e^2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(a/c)^(1/4))/(a/c)^(1/4))/(a*c^3) + 1/4*sqrt(2)*(3*sqrt(2)*sqr
t(a*c)*c^2*d^2*e + (a*c^3)^(1/4)*c^2*d^3 + 3*(a*c^3)^(3/4)*d*e^2)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(a/c)^(1/4
))/(a/c)^(1/4))/(a*c^3) + 1/8*sqrt(2)*((a*c^3)^(1/4)*c^2*d^3 - 3*(a*c^3)^(3/4)*d*e^2)*log(x^2 + sqrt(2)*x*(a/c
)^(1/4) + sqrt(a/c))/(a*c^3) - 1/8*sqrt(2)*((a*c^3)^(1/4)*c^2*d^3 - 3*(a*c^3)^(3/4)*d*e^2)*log(x^2 - sqrt(2)*x
*(a/c)^(1/4) + sqrt(a/c))/(a*c^3)

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maple [A]  time = 0.01, size = 314, normalized size = 0.98 \[ \frac {3 d^{2} e \arctan \left (\sqrt {\frac {c}{a}}\, x^{2}\right )}{2 \sqrt {a c}}+\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, d^{3} \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{4 a}+\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, d^{3} \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{4 a}+\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, d^{3} \ln \left (\frac {x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{c}}}{x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{c}}}\right )}{8 a}+\frac {3 \sqrt {2}\, d \,e^{2} \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{4 \left (\frac {a}{c}\right )^{\frac {1}{4}} c}+\frac {3 \sqrt {2}\, d \,e^{2} \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{4 \left (\frac {a}{c}\right )^{\frac {1}{4}} c}+\frac {3 \sqrt {2}\, d \,e^{2} \ln \left (\frac {x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{c}}}{x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{c}}}\right )}{8 \left (\frac {a}{c}\right )^{\frac {1}{4}} c}+\frac {e^{3} \ln \left (c \,x^{4}+a \right )}{4 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3/(c*x^4+a),x)

[Out]

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

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maxima [A]  time = 1.96, size = 310, normalized size = 0.97 \[ \frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {3}{4}} c^{\frac {1}{4}} e^{3} + c d^{3} - 3 \, \sqrt {a} \sqrt {c} d e^{2}\right )} \log \left (\sqrt {c} x^{2} + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{8 \, a^{\frac {3}{4}} c^{\frac {5}{4}}} + \frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {3}{4}} c^{\frac {1}{4}} e^{3} - c d^{3} + 3 \, \sqrt {a} \sqrt {c} d e^{2}\right )} \log \left (\sqrt {c} x^{2} - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{8 \, a^{\frac {3}{4}} c^{\frac {5}{4}}} + \frac {{\left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {5}{4}} d^{3} + 3 \, \sqrt {2} a^{\frac {3}{4}} c^{\frac {3}{4}} d e^{2} - 6 \, \sqrt {a} c d^{2} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{4 \, a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {c}} c^{\frac {5}{4}}} + \frac {{\left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {5}{4}} d^{3} + 3 \, \sqrt {2} a^{\frac {3}{4}} c^{\frac {3}{4}} d e^{2} + 6 \, \sqrt {a} c d^{2} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{4 \, a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {c}} c^{\frac {5}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*x^4+a),x, algorithm="maxima")

[Out]

1/8*sqrt(2)*(sqrt(2)*a^(3/4)*c^(1/4)*e^3 + c*d^3 - 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)) + 1/8*sqrt(2)*(sqrt(2)*a^(3/4)*c^(1/4)*e^3 - c*d^3 + 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)) + 1/4*(sqrt(2)*a^(1/4)*c^(5/4)
*d^3 + 3*sqrt(2)*a^(3/4)*c^(3/4)*d*e^2 - 6*sqrt(a)*c*d^2*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^(5/4)) + 1/4*(sqrt(2)*a^(1/4)*c^(5/4)*d^3 + 3
*sqrt(2)*a^(3/4)*c^(3/4)*d*e^2 + 6*sqrt(a)*c*d^2*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^(5/4))

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mupad [B]  time = 2.84, size = 894, normalized size = 2.79 \[ \sum _{k=1}^4\ln \left (-c\,d^2\,\left (-3\,c\,d^5\,e^2+5\,a\,d\,e^6+3\,a\,e^7\,x+{\mathrm {root}\left (256\,a^3\,c^4\,z^4-256\,a^3\,c^3\,e^3\,z^3+480\,a^2\,c^3\,d^4\,e^2\,z^2+96\,a^3\,c^2\,e^6\,z^2+192\,a^2\,c^2\,d^4\,e^5\,z-48\,a\,c^3\,d^8\,e\,z-16\,a^3\,c\,e^9\,z+3\,a^2\,c\,d^4\,e^8+3\,a\,c^2\,d^8\,e^4+c^3\,d^{12}+a^3\,e^{12},z,k\right )}^2\,a\,c^2\,d\,8+\mathrm {root}\left (256\,a^3\,c^4\,z^4-256\,a^3\,c^3\,e^3\,z^3+480\,a^2\,c^3\,d^4\,e^2\,z^2+96\,a^3\,c^2\,e^6\,z^2+192\,a^2\,c^2\,d^4\,e^5\,z-48\,a\,c^3\,d^8\,e\,z-16\,a^3\,c\,e^9\,z+3\,a^2\,c\,d^4\,e^8+3\,a\,c^2\,d^8\,e^4+c^3\,d^{12}+a^3\,e^{12},z,k\right )\,c^2\,d^4\,x\,2-5\,c\,d^4\,e^3\,x-{\mathrm {root}\left (256\,a^3\,c^4\,z^4-256\,a^3\,c^3\,e^3\,z^3+480\,a^2\,c^3\,d^4\,e^2\,z^2+96\,a^3\,c^2\,e^6\,z^2+192\,a^2\,c^2\,d^4\,e^5\,z-48\,a\,c^3\,d^8\,e\,z-16\,a^3\,c\,e^9\,z+3\,a^2\,c\,d^4\,e^8+3\,a\,c^2\,d^8\,e^4+c^3\,d^{12}+a^3\,e^{12},z,k\right )}^2\,a\,c^2\,e\,x\,24+\mathrm {root}\left (256\,a^3\,c^4\,z^4-256\,a^3\,c^3\,e^3\,z^3+480\,a^2\,c^3\,d^4\,e^2\,z^2+96\,a^3\,c^2\,e^6\,z^2+192\,a^2\,c^2\,d^4\,e^5\,z-48\,a\,c^3\,d^8\,e\,z-16\,a^3\,c\,e^9\,z+3\,a^2\,c\,d^4\,e^8+3\,a\,c^2\,d^8\,e^4+c^3\,d^{12}+a^3\,e^{12},z,k\right )\,a\,c\,d\,e^3\,32-\mathrm {root}\left (256\,a^3\,c^4\,z^4-256\,a^3\,c^3\,e^3\,z^3+480\,a^2\,c^3\,d^4\,e^2\,z^2+96\,a^3\,c^2\,e^6\,z^2+192\,a^2\,c^2\,d^4\,e^5\,z-48\,a\,c^3\,d^8\,e\,z-16\,a^3\,c\,e^9\,z+3\,a^2\,c\,d^4\,e^8+3\,a\,c^2\,d^8\,e^4+c^3\,d^{12}+a^3\,e^{12},z,k\right )\,a\,c\,e^4\,x\,6\right )\,2\right )\,\mathrm {root}\left (256\,a^3\,c^4\,z^4-256\,a^3\,c^3\,e^3\,z^3+480\,a^2\,c^3\,d^4\,e^2\,z^2+96\,a^3\,c^2\,e^6\,z^2+192\,a^2\,c^2\,d^4\,e^5\,z-48\,a\,c^3\,d^8\,e\,z-16\,a^3\,c\,e^9\,z+3\,a^2\,c\,d^4\,e^8+3\,a\,c^2\,d^8\,e^4+c^3\,d^{12}+a^3\,e^{12},z,k\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^3/(a + c*x^4),x)

[Out]

symsum(log(-2*c*d^2*(5*a*d*e^6 - 3*c*d^5*e^2 + 3*a*e^7*x + 8*root(256*a^3*c^4*z^4 - 256*a^3*c^3*e^3*z^3 + 480*
a^2*c^3*d^4*e^2*z^2 + 96*a^3*c^2*e^6*z^2 + 192*a^2*c^2*d^4*e^5*z - 48*a*c^3*d^8*e*z - 16*a^3*c*e^9*z + 3*a^2*c
*d^4*e^8 + 3*a*c^2*d^8*e^4 + c^3*d^12 + a^3*e^12, z, k)^2*a*c^2*d + 2*root(256*a^3*c^4*z^4 - 256*a^3*c^3*e^3*z
^3 + 480*a^2*c^3*d^4*e^2*z^2 + 96*a^3*c^2*e^6*z^2 + 192*a^2*c^2*d^4*e^5*z - 48*a*c^3*d^8*e*z - 16*a^3*c*e^9*z
+ 3*a^2*c*d^4*e^8 + 3*a*c^2*d^8*e^4 + c^3*d^12 + a^3*e^12, z, k)*c^2*d^4*x - 5*c*d^4*e^3*x - 24*root(256*a^3*c
^4*z^4 - 256*a^3*c^3*e^3*z^3 + 480*a^2*c^3*d^4*e^2*z^2 + 96*a^3*c^2*e^6*z^2 + 192*a^2*c^2*d^4*e^5*z - 48*a*c^3
*d^8*e*z - 16*a^3*c*e^9*z + 3*a^2*c*d^4*e^8 + 3*a*c^2*d^8*e^4 + c^3*d^12 + a^3*e^12, z, k)^2*a*c^2*e*x + 32*ro
ot(256*a^3*c^4*z^4 - 256*a^3*c^3*e^3*z^3 + 480*a^2*c^3*d^4*e^2*z^2 + 96*a^3*c^2*e^6*z^2 + 192*a^2*c^2*d^4*e^5*
z - 48*a*c^3*d^8*e*z - 16*a^3*c*e^9*z + 3*a^2*c*d^4*e^8 + 3*a*c^2*d^8*e^4 + c^3*d^12 + a^3*e^12, z, k)*a*c*d*e
^3 - 6*root(256*a^3*c^4*z^4 - 256*a^3*c^3*e^3*z^3 + 480*a^2*c^3*d^4*e^2*z^2 + 96*a^3*c^2*e^6*z^2 + 192*a^2*c^2
*d^4*e^5*z - 48*a*c^3*d^8*e*z - 16*a^3*c*e^9*z + 3*a^2*c*d^4*e^8 + 3*a*c^2*d^8*e^4 + c^3*d^12 + a^3*e^12, z, k
)*a*c*e^4*x))*root(256*a^3*c^4*z^4 - 256*a^3*c^3*e^3*z^3 + 480*a^2*c^3*d^4*e^2*z^2 + 96*a^3*c^2*e^6*z^2 + 192*
a^2*c^2*d^4*e^5*z - 48*a*c^3*d^8*e*z - 16*a^3*c*e^9*z + 3*a^2*c*d^4*e^8 + 3*a*c^2*d^8*e^4 + c^3*d^12 + a^3*e^1
2, z, k), k, 1, 4)

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sympy [A]  time = 5.29, size = 384, normalized size = 1.20 \[ \operatorname {RootSum} {\left (256 t^{4} a^{3} c^{4} - 256 t^{3} a^{3} c^{3} e^{3} + t^{2} \left (96 a^{3} c^{2} e^{6} + 480 a^{2} c^{3} d^{4} e^{2}\right ) + t \left (- 16 a^{3} c e^{9} + 192 a^{2} c^{2} d^{4} e^{5} - 48 a c^{3} d^{8} e\right ) + a^{3} e^{12} + 3 a^{2} c d^{4} e^{8} + 3 a c^{2} d^{8} e^{4} + c^{3} d^{12}, \left (t \mapsto t \log {\left (x + \frac {1728 t^{3} a^{4} c^{3} e^{6} + 960 t^{3} a^{3} c^{4} d^{4} e^{2} - 1296 t^{2} a^{4} c^{2} e^{9} - 2016 t^{2} a^{3} c^{3} d^{4} e^{5} + 48 t^{2} a^{2} c^{4} d^{8} e + 324 t a^{4} c e^{12} + 4716 t a^{3} c^{2} d^{4} e^{8} + 1452 t a^{2} c^{3} d^{8} e^{4} + 4 t a c^{4} d^{12} - 27 a^{4} e^{15} + 1119 a^{3} c d^{4} e^{11} - 609 a^{2} c^{2} d^{8} e^{7} - 91 a c^{3} d^{12} e^{3}}{729 a^{3} c d^{3} e^{12} - 1053 a^{2} c^{2} d^{7} e^{8} - 117 a c^{3} d^{11} e^{4} + c^{4} d^{15}} \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3/(c*x**4+a),x)

[Out]

RootSum(256*_t**4*a**3*c**4 - 256*_t**3*a**3*c**3*e**3 + _t**2*(96*a**3*c**2*e**6 + 480*a**2*c**3*d**4*e**2) +
_t*(-16*a**3*c*e**9 + 192*a**2*c**2*d**4*e**5 - 48*a*c**3*d**8*e) + a**3*e**12 + 3*a**2*c*d**4*e**8 + 3*a*c**
2*d**8*e**4 + c**3*d**12, Lambda(_t, _t*log(x + (1728*_t**3*a**4*c**3*e**6 + 960*_t**3*a**3*c**4*d**4*e**2 - 1
296*_t**2*a**4*c**2*e**9 - 2016*_t**2*a**3*c**3*d**4*e**5 + 48*_t**2*a**2*c**4*d**8*e + 324*_t*a**4*c*e**12 +
4716*_t*a**3*c**2*d**4*e**8 + 1452*_t*a**2*c**3*d**8*e**4 + 4*_t*a*c**4*d**12 - 27*a**4*e**15 + 1119*a**3*c*d*
*4*e**11 - 609*a**2*c**2*d**8*e**7 - 91*a*c**3*d**12*e**3)/(729*a**3*c*d**3*e**12 - 1053*a**2*c**2*d**7*e**8 -
117*a*c**3*d**11*e**4 + c**4*d**15))))

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