3.170 \(\int \frac {1+x+x^2+x^3}{a+b x^4} \, dx\)

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

[Out]

1/4*ln(b*x^4+a)/b+1/8*ln(-a^(1/4)*b^(1/4)*x*2^(1/2)+a^(1/2)+x^2*b^(1/2))*(a^(1/2)-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))*(a^(1/2)-b^(1/2))/a^(3/4)/b^(3/4)*2^(1/2)+1/2*arcta
n(x^2*b^(1/2)/a^(1/2))/a^(1/2)/b^(1/2)+1/4*arctan(-1+b^(1/4)*x*2^(1/2)/a^(1/4))*(a^(1/2)+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))*(a^(1/2)+b^(1/2))/a^(3/4)/b^(3/4)*2^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.20, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.579, Rules used = {1876, 1168, 1162, 617, 204, 1165, 628, 1248, 635, 205, 260} \[ \frac {\left (\sqrt {a}-\sqrt {b}\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{4 \sqrt {2} a^{3/4} b^{3/4}}-\frac {\left (\sqrt {a}-\sqrt {b}\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{4 \sqrt {2} a^{3/4} b^{3/4}}-\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} b^{3/4}}+\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} a^{3/4} b^{3/4}}+\frac {\log \left (a+b x^4\right )}{4 b}+\frac {\tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + x + x^2 + x^3)/(a + b*x^4),x]

[Out]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.24, size = 283, normalized size = 1.02 \[ \frac {\sqrt {2} \sqrt [4]{b} \left (a^{3/4}-\sqrt [4]{a} \sqrt {b}\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )+\sqrt {2} \sqrt [4]{b} \left (\sqrt [4]{a} \sqrt {b}-a^{3/4}\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )+2 a \log \left (a+b x^4\right )-2 \sqrt [4]{a} \sqrt [4]{b} \left (2 \sqrt [4]{a} \sqrt [4]{b}+\sqrt {2} \sqrt {a}+\sqrt {2} \sqrt {b}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )+2 \sqrt [4]{a} \sqrt [4]{b} \left (-2 \sqrt [4]{a} \sqrt [4]{b}+\sqrt {2} \sqrt {a}+\sqrt {2} \sqrt {b}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{8 a b} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x + x^2 + x^3)/(a + b*x^4),x]

[Out]

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

________________________________________________________________________________________

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((x^3+x^2+x+1)/(b*x^4+a),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

giac [A]  time = 0.21, size = 270, normalized size = 0.97 \[ \frac {\log \left ({\left | b x^{4} + a \right |}\right )}{4 \, b} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {1}{4}} b^{2} - \sqrt {2} \sqrt {a b^{3}} b + \left (a b^{3}\right )^{\frac {3}{4}}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{4 \, a b^{3}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {1}{4}} b^{2} + \sqrt {2} \sqrt {a b^{3}} b + \left (a b^{3}\right )^{\frac {3}{4}}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{4 \, a b^{3}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {1}{4}} b^{2} - \left (a b^{3}\right )^{\frac {3}{4}}\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{8 \, a b^{3}} - \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {1}{4}} b^{2} - \left (a b^{3}\right )^{\frac {3}{4}}\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{8 \, a b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+x^2+x+1)/(b*x^4+a),x, algorithm="giac")

[Out]

1/4*log(abs(b*x^4 + a))/b + 1/4*sqrt(2)*((a*b^3)^(1/4)*b^2 - sqrt(2)*sqrt(a*b^3)*b + (a*b^3)^(3/4))*arctan(1/2
*sqrt(2)*(2*x + sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/(a*b^3) + 1/4*sqrt(2)*((a*b^3)^(1/4)*b^2 + sqrt(2)*sqrt(a*b^
3)*b + (a*b^3)^(3/4))*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 - (a*b^3)^(3/4))*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 - (a*b^3)^(3/4))*log(x^2 - sqrt(2)*x*(a/b)^(1/4) + sqrt(a/b))/(a*b^3)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 286, normalized size = 1.03 \[ \frac {\arctan \left (\sqrt {\frac {b}{a}}\, x^{2}\right )}{2 \sqrt {a b}}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{4 a}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{4 a}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}{x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}\right )}{8 a}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{4 \left (\frac {a}{b}\right )^{\frac {1}{4}} b}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{4 \left (\frac {a}{b}\right )^{\frac {1}{4}} b}+\frac {\sqrt {2}\, \ln \left (\frac {x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}{x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}\right )}{8 \left (\frac {a}{b}\right )^{\frac {1}{4}} b}+\frac {\ln \left (b \,x^{4}+a \right )}{4 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3+x^2+x+1)/(b*x^4+a),x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 2.98, size = 296, normalized size = 1.07 \[ \frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {3}{4}} b^{\frac {1}{4}} - \sqrt {a} \sqrt {b} + b\right )} \log \left (\sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{8 \, a^{\frac {3}{4}} b^{\frac {5}{4}}} + \frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {3}{4}} b^{\frac {1}{4}} + \sqrt {a} \sqrt {b} - b\right )} \log \left (\sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{8 \, a^{\frac {3}{4}} b^{\frac {5}{4}}} + \frac {{\left ({\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {a}\right )} b + {\left (\sqrt {2} a^{\frac {3}{4}} b^{\frac {1}{4}} + 2 \, a\right )} \sqrt {b} - 2 \, a \sqrt {b}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{4 \, a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {b}} b^{\frac {5}{4}}} + \frac {{\left ({\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {a}\right )} b + {\left (\sqrt {2} a^{\frac {3}{4}} b^{\frac {1}{4}} - 2 \, a\right )} \sqrt {b} + 2 \, a \sqrt {b}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{4 \, a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {b}} b^{\frac {5}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+x^2+x+1)/(b*x^4+a),x, algorithm="maxima")

[Out]

1/8*sqrt(2)*(sqrt(2)*a^(3/4)*b^(1/4) - sqrt(a)*sqrt(b) + b)*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) + sqrt(a)*sqrt(b) - b)*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^(1/4) - 2*sqrt(a))*b + (sqrt(2)*a^(3
/4)*b^(1/4) + 2*a)*sqrt(b) - 2*a*sqrt(b))*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^(1/4) + 2*sqrt(a))*b + (sqrt(2
)*a^(3/4)*b^(1/4) - 2*a)*sqrt(b) + 2*a*sqrt(b))*arctan(1/2*sqrt(2)*(2*sqrt(b)*x - sqrt(2)*a^(1/4)*b^(1/4))/sqr
t(sqrt(a)*sqrt(b)))/(a^(3/4)*sqrt(sqrt(a)*sqrt(b))*b^(5/4))

________________________________________________________________________________________

mupad [B]  time = 5.04, size = 305, normalized size = 1.10 \[ \sum _{k=1}^4\ln \left (-\mathrm {root}\left (256\,a^3\,b^4\,z^4-256\,a^3\,b^3\,z^3+96\,a^3\,b^2\,z^2+96\,a^2\,b^3\,z^2-16\,a^3\,b\,z-16\,a\,b^3\,z-32\,a^2\,b^2\,z+3\,a^2\,b+3\,a\,b^2+b^3+a^3,z,k\right )\,\left (\mathrm {root}\left (256\,a^3\,b^4\,z^4-256\,a^3\,b^3\,z^3+96\,a^3\,b^2\,z^2+96\,a^2\,b^3\,z^2-16\,a^3\,b\,z-16\,a\,b^3\,z-32\,a^2\,b^2\,z+3\,a^2\,b+3\,a\,b^2+b^3+a^3,z,k\right )\,\left (16\,a\,b^3-16\,a\,b^3\,x\right )+x\,\left (4\,b^3+4\,a\,b^2\right )\right )\right )\,\mathrm {root}\left (256\,a^3\,b^4\,z^4-256\,a^3\,b^3\,z^3+96\,a^3\,b^2\,z^2+96\,a^2\,b^3\,z^2-16\,a^3\,b\,z-16\,a\,b^3\,z-32\,a^2\,b^2\,z+3\,a^2\,b+3\,a\,b^2+b^3+a^3,z,k\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + x^2 + x^3 + 1)/(a + b*x^4),x)

[Out]

symsum(log(-root(256*a^3*b^4*z^4 - 256*a^3*b^3*z^3 + 96*a^3*b^2*z^2 + 96*a^2*b^3*z^2 - 16*a^3*b*z - 16*a*b^3*z
- 32*a^2*b^2*z + 3*a^2*b + 3*a*b^2 + b^3 + a^3, z, k)*(root(256*a^3*b^4*z^4 - 256*a^3*b^3*z^3 + 96*a^3*b^2*z^
2 + 96*a^2*b^3*z^2 - 16*a^3*b*z - 16*a*b^3*z - 32*a^2*b^2*z + 3*a^2*b + 3*a*b^2 + b^3 + a^3, z, k)*(16*a*b^3 -
16*a*b^3*x) + x*(4*a*b^2 + 4*b^3)))*root(256*a^3*b^4*z^4 - 256*a^3*b^3*z^3 + 96*a^3*b^2*z^2 + 96*a^2*b^3*z^2
- 16*a^3*b*z - 16*a*b^3*z - 32*a^2*b^2*z + 3*a^2*b + 3*a*b^2 + b^3 + a^3, z, k), k, 1, 4)

________________________________________________________________________________________

sympy [A]  time = 2.27, size = 187, normalized size = 0.68 \[ \operatorname {RootSum} {\left (256 t^{4} a^{3} b^{4} - 256 t^{3} a^{3} b^{3} + t^{2} \left (96 a^{3} b^{2} + 96 a^{2} b^{3}\right ) + t \left (- 16 a^{3} b - 32 a^{2} b^{2} - 16 a b^{3}\right ) + a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}, \left (t \mapsto t \log {\left (x + \frac {64 t^{3} a^{3} b^{3} - 48 t^{2} a^{3} b^{2} + 16 t^{2} a^{2} b^{3} + 12 t a^{3} b + 16 t a^{2} b^{2} + 4 t a b^{3} - a^{3} - 2 a^{2} b - a b^{2}}{a^{2} b + 2 a b^{2} + b^{3}} \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3+x**2+x+1)/(b*x**4+a),x)

[Out]

RootSum(256*_t**4*a**3*b**4 - 256*_t**3*a**3*b**3 + _t**2*(96*a**3*b**2 + 96*a**2*b**3) + _t*(-16*a**3*b - 32*
a**2*b**2 - 16*a*b**3) + a**3 + 3*a**2*b + 3*a*b**2 + b**3, Lambda(_t, _t*log(x + (64*_t**3*a**3*b**3 - 48*_t*
*2*a**3*b**2 + 16*_t**2*a**2*b**3 + 12*_t*a**3*b + 16*_t*a**2*b**2 + 4*_t*a*b**3 - a**3 - 2*a**2*b - a*b**2)/(
a**2*b + 2*a*b**2 + b**3))))

________________________________________________________________________________________