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

Optimal. Leaf size=277 \[ -\frac {\left (\sqrt {b} c-\sqrt {a} e\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{4 \sqrt {2} a^{3/4} b^{3/4}}+\frac {\left (\sqrt {b} c-\sqrt {a} e\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{4 \sqrt {2} a^{3/4} b^{3/4}}-\frac {\left (\sqrt {a} e+\sqrt {b} c\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} b^{3/4}}+\frac {\left (\sqrt {a} e+\sqrt {b} c\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} a^{3/4} b^{3/4}}+\frac {d \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b}} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {1876, 275, 205, 1168, 1162, 617, 204, 1165, 628} \[ -\frac {\left (\sqrt {b} c-\sqrt {a} e\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{4 \sqrt {2} a^{3/4} b^{3/4}}+\frac {\left (\sqrt {b} c-\sqrt {a} e\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{4 \sqrt {2} a^{3/4} b^{3/4}}-\frac {\left (\sqrt {a} e+\sqrt {b} c\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} b^{3/4}}+\frac {\left (\sqrt {a} e+\sqrt {b} c\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} a^{3/4} b^{3/4}}+\frac {d \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x + e*x^2)/(a + b*x^4),x]

[Out]

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

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Mathematica [A]  time = 0.12, size = 229, normalized size = 0.83 \[ \frac {-2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (2 \sqrt [4]{a} \sqrt [4]{b} d+\sqrt {2} \sqrt {a} e+\sqrt {2} \sqrt {b} c\right )+2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right ) \left (-2 \sqrt [4]{a} \sqrt [4]{b} d+\sqrt {2} \sqrt {a} e+\sqrt {2} \sqrt {b} c\right )-\sqrt {2} \left (\sqrt {b} c-\sqrt {a} e\right ) \left (\log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )-\log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )\right )}{8 a^{3/4} b^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x + e*x^2)/(a + b*x^4),x]

[Out]

(-2*(Sqrt[2]*Sqrt[b]*c + 2*a^(1/4)*b^(1/4)*d + Sqrt[2]*Sqrt[a]*e)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)] + 2*
(Sqrt[2]*Sqrt[b]*c - 2*a^(1/4)*b^(1/4)*d + Sqrt[2]*Sqrt[a]*e)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)] - Sqrt[2
]*(Sqrt[b]*c - Sqrt[a]*e)*(Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2] - Log[Sqrt[a] + Sqrt[2]*a^(1
/4)*b^(1/4)*x + Sqrt[b]*x^2]))/(8*a^(3/4)*b^(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^2+d*x+c)/(b*x^4+a),x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 0.17, size = 275, normalized size = 0.99 \[ -\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {a b} b^{2} d - \left (a b^{3}\right )^{\frac {1}{4}} b^{2} c - \left (a b^{3}\right )^{\frac {3}{4}} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{4 \, a b^{3}} - \frac {\sqrt {2} {\left (\sqrt {2} \sqrt {a b} b^{2} d - \left (a b^{3}\right )^{\frac {1}{4}} b^{2} c - \left (a b^{3}\right )^{\frac {3}{4}} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{4 \, a b^{3}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {1}{4}} b^{2} c - \left (a b^{3}\right )^{\frac {3}{4}} e\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{8 \, a b^{3}} - \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {1}{4}} b^{2} c - \left (a b^{3}\right )^{\frac {3}{4}} e\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{8 \, a b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 280, normalized size = 1.01 \[ \frac {d \arctan \left (\sqrt {\frac {b}{a}}\, x^{2}\right )}{2 \sqrt {a b}}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, c \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{4 a}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, c \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{4 a}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, c \ln \left (\frac {x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}{x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}\right )}{8 a}+\frac {\sqrt {2}\, e \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{4 \left (\frac {a}{b}\right )^{\frac {1}{4}} b}+\frac {\sqrt {2}\, e \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{4 \left (\frac {a}{b}\right )^{\frac {1}{4}} b}+\frac {\sqrt {2}\, e \ln \left (\frac {x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}{x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}\right )}{8 \left (\frac {a}{b}\right )^{\frac {1}{4}} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.04, size = 257, normalized size = 0.93 \[ \frac {\sqrt {2} {\left (\sqrt {b} c - \sqrt {a} e\right )} \log \left (\sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{8 \, a^{\frac {3}{4}} b^{\frac {3}{4}}} - \frac {\sqrt {2} {\left (\sqrt {b} c - \sqrt {a} e\right )} \log \left (\sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{8 \, a^{\frac {3}{4}} b^{\frac {3}{4}}} + \frac {{\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {3}{4}} c + \sqrt {2} a^{\frac {3}{4}} b^{\frac {1}{4}} e - 2 \, \sqrt {a} \sqrt {b} d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{4 \, a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {b}} b^{\frac {3}{4}}} + \frac {{\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {3}{4}} c + \sqrt {2} a^{\frac {3}{4}} b^{\frac {1}{4}} e + 2 \, \sqrt {a} \sqrt {b} d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{4 \, a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {b}} b^{\frac {3}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.09, size = 712, normalized size = 2.57 \[ \sum _{k=1}^4\ln \left (b^2\,c\,d^2-b^2\,c^2\,e+b^2\,d^3\,x-a\,b\,e^3-{\mathrm {root}\left (256\,a^3\,b^3\,z^4+64\,a^2\,b^2\,c\,e\,z^2+32\,a^2\,b^2\,d^2\,z^2+16\,a^2\,b\,d\,e^2\,z-16\,a\,b^2\,c^2\,d\,z-4\,a\,b\,c\,d^2\,e+2\,a\,b\,c^2\,e^2+a\,b\,d^4+a^2\,e^4+b^2\,c^4,z,k\right )}^2\,a\,b^3\,c\,16-\mathrm {root}\left (256\,a^3\,b^3\,z^4+64\,a^2\,b^2\,c\,e\,z^2+32\,a^2\,b^2\,d^2\,z^2+16\,a^2\,b\,d\,e^2\,z-16\,a\,b^2\,c^2\,d\,z-4\,a\,b\,c\,d^2\,e+2\,a\,b\,c^2\,e^2+a\,b\,d^4+a^2\,e^4+b^2\,c^4,z,k\right )\,b^3\,c^2\,x\,4+{\mathrm {root}\left (256\,a^3\,b^3\,z^4+64\,a^2\,b^2\,c\,e\,z^2+32\,a^2\,b^2\,d^2\,z^2+16\,a^2\,b\,d\,e^2\,z-16\,a\,b^2\,c^2\,d\,z-4\,a\,b\,c\,d^2\,e+2\,a\,b\,c^2\,e^2+a\,b\,d^4+a^2\,e^4+b^2\,c^4,z,k\right )}^2\,a\,b^3\,d\,x\,16+\mathrm {root}\left (256\,a^3\,b^3\,z^4+64\,a^2\,b^2\,c\,e\,z^2+32\,a^2\,b^2\,d^2\,z^2+16\,a^2\,b\,d\,e^2\,z-16\,a\,b^2\,c^2\,d\,z-4\,a\,b\,c\,d^2\,e+2\,a\,b\,c^2\,e^2+a\,b\,d^4+a^2\,e^4+b^2\,c^4,z,k\right )\,a\,b^2\,e^2\,x\,4-\mathrm {root}\left (256\,a^3\,b^3\,z^4+64\,a^2\,b^2\,c\,e\,z^2+32\,a^2\,b^2\,d^2\,z^2+16\,a^2\,b\,d\,e^2\,z-16\,a\,b^2\,c^2\,d\,z-4\,a\,b\,c\,d^2\,e+2\,a\,b\,c^2\,e^2+a\,b\,d^4+a^2\,e^4+b^2\,c^4,z,k\right )\,a\,b^2\,d\,e\,8-2\,b^2\,c\,d\,e\,x\right )\,\mathrm {root}\left (256\,a^3\,b^3\,z^4+64\,a^2\,b^2\,c\,e\,z^2+32\,a^2\,b^2\,d^2\,z^2+16\,a^2\,b\,d\,e^2\,z-16\,a\,b^2\,c^2\,d\,z-4\,a\,b\,c\,d^2\,e+2\,a\,b\,c^2\,e^2+a\,b\,d^4+a^2\,e^4+b^2\,c^4,z,k\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x + e*x^2)/(a + b*x^4),x)

[Out]

symsum(log(b^2*c*d^2 - b^2*c^2*e + b^2*d^3*x - a*b*e^3 - 16*root(256*a^3*b^3*z^4 + 64*a^2*b^2*c*e*z^2 + 32*a^2
*b^2*d^2*z^2 + 16*a^2*b*d*e^2*z - 16*a*b^2*c^2*d*z - 4*a*b*c*d^2*e + 2*a*b*c^2*e^2 + a*b*d^4 + a^2*e^4 + b^2*c
^4, z, k)^2*a*b^3*c - 4*root(256*a^3*b^3*z^4 + 64*a^2*b^2*c*e*z^2 + 32*a^2*b^2*d^2*z^2 + 16*a^2*b*d*e^2*z - 16
*a*b^2*c^2*d*z - 4*a*b*c*d^2*e + 2*a*b*c^2*e^2 + a*b*d^4 + a^2*e^4 + b^2*c^4, z, k)*b^3*c^2*x + 16*root(256*a^
3*b^3*z^4 + 64*a^2*b^2*c*e*z^2 + 32*a^2*b^2*d^2*z^2 + 16*a^2*b*d*e^2*z - 16*a*b^2*c^2*d*z - 4*a*b*c*d^2*e + 2*
a*b*c^2*e^2 + a*b*d^4 + a^2*e^4 + b^2*c^4, z, k)^2*a*b^3*d*x + 4*root(256*a^3*b^3*z^4 + 64*a^2*b^2*c*e*z^2 + 3
2*a^2*b^2*d^2*z^2 + 16*a^2*b*d*e^2*z - 16*a*b^2*c^2*d*z - 4*a*b*c*d^2*e + 2*a*b*c^2*e^2 + a*b*d^4 + a^2*e^4 +
b^2*c^4, z, k)*a*b^2*e^2*x - 8*root(256*a^3*b^3*z^4 + 64*a^2*b^2*c*e*z^2 + 32*a^2*b^2*d^2*z^2 + 16*a^2*b*d*e^2
*z - 16*a*b^2*c^2*d*z - 4*a*b*c*d^2*e + 2*a*b*c^2*e^2 + a*b*d^4 + a^2*e^4 + b^2*c^4, z, k)*a*b^2*d*e - 2*b^2*c
*d*e*x)*root(256*a^3*b^3*z^4 + 64*a^2*b^2*c*e*z^2 + 32*a^2*b^2*d^2*z^2 + 16*a^2*b*d*e^2*z - 16*a*b^2*c^2*d*z -
4*a*b*c*d^2*e + 2*a*b*c^2*e^2 + a*b*d^4 + a^2*e^4 + b^2*c^4, z, k), k, 1, 4)

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sympy [A]  time = 10.54, size = 466, normalized size = 1.68 \[ \operatorname {RootSum} {\left (256 t^{4} a^{3} b^{3} + t^{2} \left (64 a^{2} b^{2} c e + 32 a^{2} b^{2} d^{2}\right ) + t \left (16 a^{2} b d e^{2} - 16 a b^{2} c^{2} d\right ) + a^{2} e^{4} + 2 a b c^{2} e^{2} - 4 a b c d^{2} e + a b d^{4} + b^{2} c^{4}, \left (t \mapsto t \log {\left (x + \frac {64 t^{3} a^{4} b^{2} e^{3} - 64 t^{3} a^{3} b^{3} c^{2} e + 128 t^{3} a^{3} b^{3} c d^{2} + 48 t^{2} a^{3} b^{2} c d e^{2} - 32 t^{2} a^{3} b^{2} d^{3} e + 16 t^{2} a^{2} b^{3} c^{3} d + 12 t a^{3} b c e^{4} + 12 t a^{3} b d^{2} e^{3} - 16 t a^{2} b^{2} c^{3} e^{2} + 36 t a^{2} b^{2} c^{2} d^{2} e + 8 t a^{2} b^{2} c d^{4} + 4 t a b^{3} c^{5} + 3 a^{3} d e^{5} + 5 a^{2} b c d^{3} e^{2} - 2 a^{2} b d^{5} e + 5 a b^{2} c^{4} d e - 5 a b^{2} c^{3} d^{3}}{a^{3} e^{6} - a^{2} b c^{2} e^{4} + 8 a^{2} b c d^{2} e^{3} - 4 a^{2} b d^{4} e^{2} - a b^{2} c^{4} e^{2} + 8 a b^{2} c^{3} d^{2} e - 4 a b^{2} c^{2} d^{4} + b^{3} c^{6}} \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(256*_t**4*a**3*b**3 + _t**2*(64*a**2*b**2*c*e + 32*a**2*b**2*d**2) + _t*(16*a**2*b*d*e**2 - 16*a*b**2*
c**2*d) + a**2*e**4 + 2*a*b*c**2*e**2 - 4*a*b*c*d**2*e + a*b*d**4 + b**2*c**4, Lambda(_t, _t*log(x + (64*_t**3
*a**4*b**2*e**3 - 64*_t**3*a**3*b**3*c**2*e + 128*_t**3*a**3*b**3*c*d**2 + 48*_t**2*a**3*b**2*c*d*e**2 - 32*_t
**2*a**3*b**2*d**3*e + 16*_t**2*a**2*b**3*c**3*d + 12*_t*a**3*b*c*e**4 + 12*_t*a**3*b*d**2*e**3 - 16*_t*a**2*b
**2*c**3*e**2 + 36*_t*a**2*b**2*c**2*d**2*e + 8*_t*a**2*b**2*c*d**4 + 4*_t*a*b**3*c**5 + 3*a**3*d*e**5 + 5*a**
2*b*c*d**3*e**2 - 2*a**2*b*d**5*e + 5*a*b**2*c**4*d*e - 5*a*b**2*c**3*d**3)/(a**3*e**6 - a**2*b*c**2*e**4 + 8*
a**2*b*c*d**2*e**3 - 4*a**2*b*d**4*e**2 - a*b**2*c**4*e**2 + 8*a*b**2*c**3*d**2*e - 4*a*b**2*c**2*d**4 + b**3*
c**6))))

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