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

Optimal. Leaf size=291 \[ -\frac {\left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} c^{3/4}}+\frac {\left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} c^{3/4}}-\frac {\left (\sqrt {a} e^2+\sqrt {c} d^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} c^{3/4}}+\frac {\left (\sqrt {a} e^2+\sqrt {c} d^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} a^{3/4} c^{3/4}}+\frac {d e \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {c}} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 291, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {1876, 275, 205, 1168, 1162, 617, 204, 1165, 628} \[ -\frac {\left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} c^{3/4}}+\frac {\left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} c^{3/4}}-\frac {\left (\sqrt {a} e^2+\sqrt {c} d^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} c^{3/4}}+\frac {\left (\sqrt {a} e^2+\sqrt {c} d^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} a^{3/4} c^{3/4}}+\frac {d e \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2/(a + c*x^4),x]

[Out]

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

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

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Mathematica [A]  time = 0.10, size = 243, normalized size = 0.84 \[ \frac {-\sqrt {2} \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \left (\log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )-\log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )\right )-2 \left (4 \sqrt [4]{a} \sqrt [4]{c} d e+\sqrt {2} \sqrt {a} e^2+\sqrt {2} \sqrt {c} d^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+2 \left (-4 \sqrt [4]{a} \sqrt [4]{c} d e+\sqrt {2} \sqrt {a} e^2+\sqrt {2} \sqrt {c} d^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{8 a^{3/4} c^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^2/(a + c*x^4),x]

[Out]

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

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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)^2/(c*x^4+a),x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 0.34, size = 285, normalized size = 0.98 \[ \frac {\sqrt {2} {\left (2 \, \sqrt {2} \sqrt {a c} c^{2} d e + \left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{2} + \left (a c^{3}\right )^{\frac {3}{4}} e^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{4 \, a c^{3}} + \frac {\sqrt {2} {\left (2 \, \sqrt {2} \sqrt {a c} c^{2} d e + \left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{2} + \left (a c^{3}\right )^{\frac {3}{4}} e^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{4 \, a c^{3}} + \frac {\sqrt {2} {\left (\left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{2} - \left (a c^{3}\right )^{\frac {3}{4}} e^{2}\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{8 \, a c^{3}} - \frac {\sqrt {2} {\left (\left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{2} - \left (a c^{3}\right )^{\frac {3}{4}} e^{2}\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{8 \, a c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 292, normalized size = 1.00 \[ \frac {d e \arctan \left (\sqrt {\frac {c}{a}}\, x^{2}\right )}{\sqrt {a c}}+\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, d^{2} \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{4 a}+\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, d^{2} \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{4 a}+\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, d^{2} \ln \left (\frac {x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{c}}}{x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{c}}}\right )}{8 a}+\frac {\sqrt {2}\, e^{2} \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{4 \left (\frac {a}{c}\right )^{\frac {1}{4}} c}+\frac {\sqrt {2}\, e^{2} \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{4 \left (\frac {a}{c}\right )^{\frac {1}{4}} c}+\frac {\sqrt {2}\, e^{2} \ln \left (\frac {x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{c}}}{x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{c}}}\right )}{8 \left (\frac {a}{c}\right )^{\frac {1}{4}} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.05, size = 275, normalized size = 0.95 \[ \frac {\sqrt {2} {\left (\sqrt {c} d^{2} - \sqrt {a} e^{2}\right )} \log \left (\sqrt {c} x^{2} + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{8 \, a^{\frac {3}{4}} c^{\frac {3}{4}}} - \frac {\sqrt {2} {\left (\sqrt {c} d^{2} - \sqrt {a} e^{2}\right )} \log \left (\sqrt {c} x^{2} - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{8 \, a^{\frac {3}{4}} c^{\frac {3}{4}}} + \frac {{\left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {3}{4}} d^{2} + \sqrt {2} a^{\frac {3}{4}} c^{\frac {1}{4}} e^{2} - 4 \, \sqrt {a} \sqrt {c} d e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{4 \, a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {c}} c^{\frac {3}{4}}} + \frac {{\left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {3}{4}} d^{2} + \sqrt {2} a^{\frac {3}{4}} c^{\frac {1}{4}} e^{2} + 4 \, \sqrt {a} \sqrt {c} d e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{4 \, a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {c}} c^{\frac {3}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.66, size = 556, normalized size = 1.91 \[ \sum _{k=1}^4\ln \left (3\,c^2\,d^4\,e^2-a\,c\,e^6+4\,c^2\,d^3\,e^3\,x-\mathrm {root}\left (256\,a^3\,c^3\,z^4+192\,a^2\,c^2\,d^2\,e^2\,z^2+32\,a^2\,c\,d\,e^5\,z-32\,a\,c^2\,d^5\,e\,z+2\,a\,c\,d^4\,e^4+c^2\,d^8+a^2\,e^8,z,k\right )\,c^3\,d^4\,x\,4-{\mathrm {root}\left (256\,a^3\,c^3\,z^4+192\,a^2\,c^2\,d^2\,e^2\,z^2+32\,a^2\,c\,d\,e^5\,z-32\,a\,c^2\,d^5\,e\,z+2\,a\,c\,d^4\,e^4+c^2\,d^8+a^2\,e^8,z,k\right )}^2\,a\,c^3\,d^2\,16+\mathrm {root}\left (256\,a^3\,c^3\,z^4+192\,a^2\,c^2\,d^2\,e^2\,z^2+32\,a^2\,c\,d\,e^5\,z-32\,a\,c^2\,d^5\,e\,z+2\,a\,c\,d^4\,e^4+c^2\,d^8+a^2\,e^8,z,k\right )\,a\,c^2\,e^4\,x\,4-\mathrm {root}\left (256\,a^3\,c^3\,z^4+192\,a^2\,c^2\,d^2\,e^2\,z^2+32\,a^2\,c\,d\,e^5\,z-32\,a\,c^2\,d^5\,e\,z+2\,a\,c\,d^4\,e^4+c^2\,d^8+a^2\,e^8,z,k\right )\,a\,c^2\,d\,e^3\,16+{\mathrm {root}\left (256\,a^3\,c^3\,z^4+192\,a^2\,c^2\,d^2\,e^2\,z^2+32\,a^2\,c\,d\,e^5\,z-32\,a\,c^2\,d^5\,e\,z+2\,a\,c\,d^4\,e^4+c^2\,d^8+a^2\,e^8,z,k\right )}^2\,a\,c^3\,d\,e\,x\,32\right )\,\mathrm {root}\left (256\,a^3\,c^3\,z^4+192\,a^2\,c^2\,d^2\,e^2\,z^2+32\,a^2\,c\,d\,e^5\,z-32\,a\,c^2\,d^5\,e\,z+2\,a\,c\,d^4\,e^4+c^2\,d^8+a^2\,e^8,z,k\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^2/(a + c*x^4),x)

[Out]

symsum(log(3*c^2*d^4*e^2 - a*c*e^6 + 4*c^2*d^3*e^3*x - 4*root(256*a^3*c^3*z^4 + 192*a^2*c^2*d^2*e^2*z^2 + 32*a
^2*c*d*e^5*z - 32*a*c^2*d^5*e*z + 2*a*c*d^4*e^4 + c^2*d^8 + a^2*e^8, z, k)*c^3*d^4*x - 16*root(256*a^3*c^3*z^4
+ 192*a^2*c^2*d^2*e^2*z^2 + 32*a^2*c*d*e^5*z - 32*a*c^2*d^5*e*z + 2*a*c*d^4*e^4 + c^2*d^8 + a^2*e^8, z, k)^2*
a*c^3*d^2 + 4*root(256*a^3*c^3*z^4 + 192*a^2*c^2*d^2*e^2*z^2 + 32*a^2*c*d*e^5*z - 32*a*c^2*d^5*e*z + 2*a*c*d^4
*e^4 + c^2*d^8 + a^2*e^8, z, k)*a*c^2*e^4*x - 16*root(256*a^3*c^3*z^4 + 192*a^2*c^2*d^2*e^2*z^2 + 32*a^2*c*d*e
^5*z - 32*a*c^2*d^5*e*z + 2*a*c*d^4*e^4 + c^2*d^8 + a^2*e^8, z, k)*a*c^2*d*e^3 + 32*root(256*a^3*c^3*z^4 + 192
*a^2*c^2*d^2*e^2*z^2 + 32*a^2*c*d*e^5*z - 32*a*c^2*d^5*e*z + 2*a*c*d^4*e^4 + c^2*d^8 + a^2*e^8, z, k)^2*a*c^3*
d*e*x)*root(256*a^3*c^3*z^4 + 192*a^2*c^2*d^2*e^2*z^2 + 32*a^2*c*d*e^5*z - 32*a*c^2*d^5*e*z + 2*a*c*d^4*e^4 +
c^2*d^8 + a^2*e^8, z, k), k, 1, 4)

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sympy [A]  time = 2.68, size = 277, normalized size = 0.95 \[ \operatorname {RootSum} {\left (256 t^{4} a^{3} c^{3} + 192 t^{2} a^{2} c^{2} d^{2} e^{2} + t \left (32 a^{2} c d e^{5} - 32 a c^{2} d^{5} e\right ) + a^{2} e^{8} + 2 a c d^{4} e^{4} + c^{2} d^{8}, \left (t \mapsto t \log {\left (x + \frac {64 t^{3} a^{4} c^{2} e^{6} + 448 t^{3} a^{3} c^{3} d^{4} e^{2} - 160 t^{2} a^{3} c^{2} d^{3} e^{5} + 32 t^{2} a^{2} c^{3} d^{7} e + 60 t a^{3} c d^{2} e^{8} + 256 t a^{2} c^{2} d^{6} e^{4} + 4 t a c^{3} d^{10} + 6 a^{3} d e^{11} - 24 a^{2} c d^{5} e^{7} - 30 a c^{2} d^{9} e^{3}}{a^{3} e^{12} - 33 a^{2} c d^{4} e^{8} - 33 a c^{2} d^{8} e^{4} + c^{3} d^{12}} \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(256*_t**4*a**3*c**3 + 192*_t**2*a**2*c**2*d**2*e**2 + _t*(32*a**2*c*d*e**5 - 32*a*c**2*d**5*e) + a**2*
e**8 + 2*a*c*d**4*e**4 + c**2*d**8, Lambda(_t, _t*log(x + (64*_t**3*a**4*c**2*e**6 + 448*_t**3*a**3*c**3*d**4*
e**2 - 160*_t**2*a**3*c**2*d**3*e**5 + 32*_t**2*a**2*c**3*d**7*e + 60*_t*a**3*c*d**2*e**8 + 256*_t*a**2*c**2*d
**6*e**4 + 4*_t*a*c**3*d**10 + 6*a**3*d*e**11 - 24*a**2*c*d**5*e**7 - 30*a*c**2*d**9*e**3)/(a**3*e**12 - 33*a*
*2*c*d**4*e**8 - 33*a*c**2*d**8*e**4 + c**3*d**12))))

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