3.487 \(\int \frac {c+d x+e x^2+f x^3}{a+b x^4} \, dx\)

Optimal. Leaf size=293 \[ -\frac {\left (\sqrt {b} c-\sqrt {a} e\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{4 \sqrt {2} a^{3/4} b^{3/4}}+\frac {\left (\sqrt {b} c-\sqrt {a} e\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{4 \sqrt {2} a^{3/4} b^{3/4}}-\frac {\left (\sqrt {a} e+\sqrt {b} c\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} b^{3/4}}+\frac {\left (\sqrt {a} e+\sqrt {b} c\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} a^{3/4} b^{3/4}}+\frac {d \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b}}+\frac {f \log \left (a+b x^4\right )}{4 b} \]

[Out]

1/4*f*ln(b*x^4+a)/b+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 = 0.22, antiderivative size = 293, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {1876, 1168, 1162, 617, 204, 1165, 628, 1248, 635, 205, 260} \[ -\frac {\left (\sqrt {b} c-\sqrt {a} e\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{4 \sqrt {2} a^{3/4} b^{3/4}}+\frac {\left (\sqrt {b} c-\sqrt {a} e\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{4 \sqrt {2} a^{3/4} b^{3/4}}-\frac {\left (\sqrt {a} e+\sqrt {b} c\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} b^{3/4}}+\frac {\left (\sqrt {a} e+\sqrt {b} c\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} a^{3/4} b^{3/4}}+\frac {d \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b}}+\frac {f \log \left (a+b x^4\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x + e*x^2 + f*x^3)/(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)) + (f*Log[a + b*x^4])/(4*b)

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

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Mathematica [A]  time = 0.23, size = 296, normalized size = 1.01 \[ \frac {-\sqrt {2} \sqrt [4]{b} \left (\sqrt [4]{a} \sqrt {b} c-a^{3/4} e\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )+\sqrt {2} \sqrt [4]{b} \left (\sqrt [4]{a} \sqrt {b} c-a^{3/4} e\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )-2 \sqrt [4]{a} \sqrt [4]{b} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (2 \sqrt [4]{a} \sqrt [4]{b} d+\sqrt {2} \sqrt {a} e+\sqrt {2} \sqrt {b} c\right )+2 \sqrt [4]{a} \sqrt [4]{b} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right ) \left (-2 \sqrt [4]{a} \sqrt [4]{b} d+\sqrt {2} \sqrt {a} e+\sqrt {2} \sqrt {b} c\right )+2 a f \log \left (a+b x^4\right )}{8 a b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x + e*x^2 + f*x^3)/(a + b*x^4),x]

[Out]

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

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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((f*x^3+e*x^2+d*x+c)/(b*x^4+a),x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 0.18, size = 290, normalized size = 0.99 \[ \frac {f \log \left ({\left | b x^{4} + a \right |}\right )}{4 \, b} - \frac {\sqrt {2} {\left (\sqrt {2} \sqrt {a b} b^{2} d - \left (a b^{3}\right )^{\frac {1}{4}} b^{2} c - \left (a b^{3}\right )^{\frac {3}{4}} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{4 \, a b^{3}} - \frac {\sqrt {2} {\left (\sqrt {2} \sqrt {a b} b^{2} d - \left (a b^{3}\right )^{\frac {1}{4}} b^{2} c - \left (a b^{3}\right )^{\frac {3}{4}} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{4 \, a b^{3}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {1}{4}} b^{2} c - \left (a b^{3}\right )^{\frac {3}{4}} e\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{8 \, a b^{3}} - \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {1}{4}} b^{2} c - \left (a b^{3}\right )^{\frac {3}{4}} e\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{8 \, a b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^3+e*x^2+d*x+c)/(b*x^4+a),x, algorithm="giac")

[Out]

1/4*f*log(abs(b*x^4 + a))/b - 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)*ar
ctan(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*sqr
t(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 = 294, normalized size = 1.00 \[ \frac {d \arctan \left (\sqrt {\frac {b}{a}}\, x^{2}\right )}{2 \sqrt {a b}}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, c \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{4 a}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, c \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{4 a}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, c \ln \left (\frac {x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}{x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}\right )}{8 a}+\frac {\sqrt {2}\, e \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{4 \left (\frac {a}{b}\right )^{\frac {1}{4}} b}+\frac {\sqrt {2}\, e \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{4 \left (\frac {a}{b}\right )^{\frac {1}{4}} b}+\frac {\sqrt {2}\, e \ln \left (\frac {x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}{x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}\right )}{8 \left (\frac {a}{b}\right )^{\frac {1}{4}} b}+\frac {f \ln \left (b \,x^{4}+a \right )}{4 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x^3+e*x^2+d*x+c)/(b*x^4+a),x)

[Out]

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

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maxima [A]  time = 3.03, size = 277, normalized size = 0.95 \[ \frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {3}{4}} b^{\frac {1}{4}} f + b c - \sqrt {a} \sqrt {b} e\right )} \log \left (\sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{8 \, a^{\frac {3}{4}} b^{\frac {5}{4}}} + \frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {3}{4}} b^{\frac {1}{4}} f - b c + \sqrt {a} \sqrt {b} e\right )} \log \left (\sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{8 \, a^{\frac {3}{4}} b^{\frac {5}{4}}} + \frac {{\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {5}{4}} c + \sqrt {2} a^{\frac {3}{4}} b^{\frac {3}{4}} e - 2 \, \sqrt {a} b d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{4 \, a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {b}} b^{\frac {5}{4}}} + \frac {{\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {5}{4}} c + \sqrt {2} a^{\frac {3}{4}} b^{\frac {3}{4}} e + 2 \, \sqrt {a} b d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{4 \, a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {b}} b^{\frac {5}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^3+e*x^2+d*x+c)/(b*x^4+a),x, algorithm="maxima")

[Out]

1/8*sqrt(2)*(sqrt(2)*a^(3/4)*b^(1/4)*f + b*c - sqrt(a)*sqrt(b)*e)*log(sqrt(b)*x^2 + sqrt(2)*a^(1/4)*b^(1/4)*x
+ sqrt(a))/(a^(3/4)*b^(5/4)) + 1/8*sqrt(2)*(sqrt(2)*a^(3/4)*b^(1/4)*f - b*c + sqrt(a)*sqrt(b)*e)*log(sqrt(b)*x
^2 - sqrt(2)*a^(1/4)*b^(1/4)*x + sqrt(a))/(a^(3/4)*b^(5/4)) + 1/4*(sqrt(2)*a^(1/4)*b^(5/4)*c + sqrt(2)*a^(3/4)
*b^(3/4)*e - 2*sqrt(a)*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^(5/4)) + 1/4*(sqrt(2)*a^(1/4)*b^(5/4)*c + sqrt(2)*a^(3/4)*b^(3/4)*e + 2*sqrt(
a)*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^(5/4))

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mupad [B]  time = 0.93, size = 1952, normalized size = 6.66 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x + e*x^2 + f*x^3)/(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 - a*b*c*f^2 - 16*root(256*a^3*b^4*z^4 - 256*a^3*b^3*f*z
^3 + 64*a^2*b^3*c*e*z^2 + 96*a^3*b^2*f^2*z^2 + 32*a^2*b^3*d^2*z^2 - 32*a^2*b^2*c*e*f*z - 16*a^2*b^2*d^2*f*z +
16*a^2*b^2*d*e^2*z - 16*a*b^3*c^2*d*z - 16*a^3*b*f^3*z - 4*a^2*b*d*e^2*f + 4*a^2*b*c*e*f^2 + 4*a*b^2*c^2*d*f -
4*a*b^2*c*d^2*e + 2*a^2*b*d^2*f^2 + 2*a*b^2*c^2*e^2 + a^2*b*e^4 + a*b^2*d^4 + a^3*f^4 + b^3*c^4, z, k)^2*a*b^
3*c - 4*root(256*a^3*b^4*z^4 - 256*a^3*b^3*f*z^3 + 64*a^2*b^3*c*e*z^2 + 96*a^3*b^2*f^2*z^2 + 32*a^2*b^3*d^2*z^
2 - 32*a^2*b^2*c*e*f*z - 16*a^2*b^2*d^2*f*z + 16*a^2*b^2*d*e^2*z - 16*a*b^3*c^2*d*z - 16*a^3*b*f^3*z - 4*a^2*b
*d*e^2*f + 4*a^2*b*c*e*f^2 + 4*a*b^2*c^2*d*f - 4*a*b^2*c*d^2*e + 2*a^2*b*d^2*f^2 + 2*a*b^2*c^2*e^2 + a^2*b*e^4
+ a*b^2*d^4 + a^3*f^4 + b^3*c^4, z, k)*b^3*c^2*x + b^2*c^2*f*x + 16*root(256*a^3*b^4*z^4 - 256*a^3*b^3*f*z^3
+ 64*a^2*b^3*c*e*z^2 + 96*a^3*b^2*f^2*z^2 + 32*a^2*b^3*d^2*z^2 - 32*a^2*b^2*c*e*f*z - 16*a^2*b^2*d^2*f*z + 16*
a^2*b^2*d*e^2*z - 16*a*b^3*c^2*d*z - 16*a^3*b*f^3*z - 4*a^2*b*d*e^2*f + 4*a^2*b*c*e*f^2 + 4*a*b^2*c^2*d*f - 4*
a*b^2*c*d^2*e + 2*a^2*b*d^2*f^2 + 2*a*b^2*c^2*e^2 + a^2*b*e^4 + a*b^2*d^4 + a^3*f^4 + b^3*c^4, z, k)^2*a*b^3*d
*x + 4*root(256*a^3*b^4*z^4 - 256*a^3*b^3*f*z^3 + 64*a^2*b^3*c*e*z^2 + 96*a^3*b^2*f^2*z^2 + 32*a^2*b^3*d^2*z^2
- 32*a^2*b^2*c*e*f*z - 16*a^2*b^2*d^2*f*z + 16*a^2*b^2*d*e^2*z - 16*a*b^3*c^2*d*z - 16*a^3*b*f^3*z - 4*a^2*b*
d*e^2*f + 4*a^2*b*c*e*f^2 + 4*a*b^2*c^2*d*f - 4*a*b^2*c*d^2*e + 2*a^2*b*d^2*f^2 + 2*a*b^2*c^2*e^2 + a^2*b*e^4
+ a*b^2*d^4 + a^3*f^4 + b^3*c^4, z, k)*a*b^2*e^2*x + 2*a*b*d*e*f + 8*root(256*a^3*b^4*z^4 - 256*a^3*b^3*f*z^3
+ 64*a^2*b^3*c*e*z^2 + 96*a^3*b^2*f^2*z^2 + 32*a^2*b^3*d^2*z^2 - 32*a^2*b^2*c*e*f*z - 16*a^2*b^2*d^2*f*z + 16*
a^2*b^2*d*e^2*z - 16*a*b^3*c^2*d*z - 16*a^3*b*f^3*z - 4*a^2*b*d*e^2*f + 4*a^2*b*c*e*f^2 + 4*a*b^2*c^2*d*f - 4*
a*b^2*c*d^2*e + 2*a^2*b*d^2*f^2 + 2*a*b^2*c^2*e^2 + a^2*b*e^4 + a*b^2*d^4 + a^3*f^4 + b^3*c^4, z, k)*a*b^2*c*f
- 8*root(256*a^3*b^4*z^4 - 256*a^3*b^3*f*z^3 + 64*a^2*b^3*c*e*z^2 + 96*a^3*b^2*f^2*z^2 + 32*a^2*b^3*d^2*z^2 -
32*a^2*b^2*c*e*f*z - 16*a^2*b^2*d^2*f*z + 16*a^2*b^2*d*e^2*z - 16*a*b^3*c^2*d*z - 16*a^3*b*f^3*z - 4*a^2*b*d*
e^2*f + 4*a^2*b*c*e*f^2 + 4*a*b^2*c^2*d*f - 4*a*b^2*c*d^2*e + 2*a^2*b*d^2*f^2 + 2*a*b^2*c^2*e^2 + a^2*b*e^4 +
a*b^2*d^4 + a^3*f^4 + b^3*c^4, z, k)*a*b^2*d*e + a*b*d*f^2*x - a*b*e^2*f*x - 2*b^2*c*d*e*x - 8*root(256*a^3*b^
4*z^4 - 256*a^3*b^3*f*z^3 + 64*a^2*b^3*c*e*z^2 + 96*a^3*b^2*f^2*z^2 + 32*a^2*b^3*d^2*z^2 - 32*a^2*b^2*c*e*f*z
- 16*a^2*b^2*d^2*f*z + 16*a^2*b^2*d*e^2*z - 16*a*b^3*c^2*d*z - 16*a^3*b*f^3*z - 4*a^2*b*d*e^2*f + 4*a^2*b*c*e*
f^2 + 4*a*b^2*c^2*d*f - 4*a*b^2*c*d^2*e + 2*a^2*b*d^2*f^2 + 2*a*b^2*c^2*e^2 + a^2*b*e^4 + a*b^2*d^4 + a^3*f^4
+ b^3*c^4, z, k)*a*b^2*d*f*x)*root(256*a^3*b^4*z^4 - 256*a^3*b^3*f*z^3 + 64*a^2*b^3*c*e*z^2 + 96*a^3*b^2*f^2*z
^2 + 32*a^2*b^3*d^2*z^2 - 32*a^2*b^2*c*e*f*z - 16*a^2*b^2*d^2*f*z + 16*a^2*b^2*d*e^2*z - 16*a*b^3*c^2*d*z - 16
*a^3*b*f^3*z - 4*a^2*b*d*e^2*f + 4*a^2*b*c*e*f^2 + 4*a*b^2*c^2*d*f - 4*a*b^2*c*d^2*e + 2*a^2*b*d^2*f^2 + 2*a*b
^2*c^2*e^2 + a^2*b*e^4 + a*b^2*d^4 + a^3*f^4 + b^3*c^4, z, k), k, 1, 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((f*x**3+e*x**2+d*x+c)/(b*x**4+a),x)

[Out]

Timed out

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