\(\int \frac {1+x+x^2+x^3}{a-b x^4} \, dx\) [169]

Optimal result

Integrand size = 20, antiderivative size = 124 \[ \int \frac {1+x+x^2+x^3}{a-b x^4} \, dx=-\frac {\left (\sqrt {a}-\sqrt {b}\right ) \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{3/4}}+\frac {\left (\sqrt {a}+\sqrt {b}\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b}}-\frac {\log \left (a-b x^4\right )}{4 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1890, 1181, 211, 214, 1262, 649, 266} \[ \int \frac {1+x+x^2+x^3}{a-b x^4} \, dx=-\frac {\left (\sqrt {a}-\sqrt {b}\right ) \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{3/4}}+\frac {\left (\sqrt {a}+\sqrt {b}\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b}}-\frac {\log \left (a-b x^4\right )}{4 b} \]

[In]

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

[Out]

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 266

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 649

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 1181

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[e/2 + c*(d/(2*q))
, Int[1/(-q + c*x^2), x], x] + Dist[e/2 - c*(d/(2*q)), Int[1/(q + c*x^2), x], x]] /; FreeQ[{a, c, d, e}, x] &&
NeQ[c*d^2 - a*e^2, 0] && PosQ[(-a)*c]

Rule 1262

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 1890

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.64 \[ \int \frac {1+x+x^2+x^3}{a-b x^4} \, dx=\frac {\left (-a^{3/4}+\sqrt [4]{a} \sqrt {b}\right ) \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 a b^{3/4}}-\frac {\left (a^{3/4}+\sqrt {a} \sqrt [4]{b}+\sqrt [4]{a} \sqrt {b}\right ) \log \left (\sqrt [4]{a}-\sqrt [4]{b} x\right )}{4 a b^{3/4}}-\frac {\left (-a^{3/4}+\sqrt {a} \sqrt [4]{b}-\sqrt [4]{a} \sqrt {b}\right ) \log \left (\sqrt [4]{a}+\sqrt [4]{b} x\right )}{4 a b^{3/4}}+\frac {\log \left (\sqrt {a}+\sqrt {b} x^2\right )}{4 \sqrt {a} \sqrt {b}}-\frac {\log \left (a-b x^4\right )}{4 b} \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.48 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.31

method	result	size
risch	\(-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4} b -a \right )}{\sum }\frac {\left (\textit {\_R}^{3}+\textit {\_R}^{2}+\textit {\_R} +1\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}}{4 b}\)	\(38\)
default	\(\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \left (\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )\right )}{4 a}+\frac {\ln \left (\frac {a +x^{2} \sqrt {a b}}{a -x^{2} \sqrt {a b}}\right )}{4 \sqrt {a b}}-\frac {2 \arctan \left (\frac {x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )-\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{4 b \left (\frac {a}{b}\right )^{\frac {1}{4}}}-\frac {\ln \left (-b \,x^{4}+a \right )}{4 b}\)	\(150\)

[In]

int((x^3+x^2+x+1)/(-b*x^4+a),x,method=_RETURNVERBOSE)

[Out]

-1/4/b*sum((_R^3+_R^2+_R+1)/_R^3*ln(x-_R),_R=RootOf(_Z^4*b-a))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.28 (sec) , antiderivative size = 91748, normalized size of antiderivative = 739.90 \[ \int \frac {1+x+x^2+x^3}{a-b x^4} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [A] (verification not implemented)

Time = 0.97 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.51 \[ \int \frac {1+x+x^2+x^3}{a-b x^4} \, dx=- \operatorname {RootSum} {\left (256 t^{4} a^{3} b^{4} - 256 t^{3} a^{3} b^{3} + t^{2} \cdot \left (96 a^{3} b^{2} - 96 a^{2} b^{3}\right ) + t \left (- 16 a^{3} b + 32 a^{2} b^{2} - 16 a b^{3}\right ) + a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}, \left ( t \mapsto t \log {\left (x + \frac {- 64 t^{3} a^{3} b^{3} + 48 t^{2} a^{3} b^{2} + 16 t^{2} a^{2} b^{3} - 12 t a^{3} b + 16 t a^{2} b^{2} - 4 t a b^{3} + a^{3} - 2 a^{2} b + a b^{2}}{a^{2} b - 2 a b^{2} + b^{3}} \right )} \right )\right )} \]

[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))))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.29 \[ \int \frac {1+x+x^2+x^3}{a-b x^4} \, dx=-\frac {{\left (\sqrt {a} - \sqrt {b}\right )} \arctan \left (\frac {\sqrt {b} x}{\sqrt {\sqrt {a} \sqrt {b}}}\right )}{2 \, \sqrt {a} \sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {{\left (\sqrt {a} - \sqrt {b}\right )} \log \left (\sqrt {b} x^{2} + \sqrt {a}\right )}{4 \, \sqrt {a} b} - \frac {{\left (\sqrt {a} + \sqrt {b}\right )} \log \left (\sqrt {b} x^{2} - \sqrt {a}\right )}{4 \, \sqrt {a} b} - \frac {{\left (\sqrt {a} + \sqrt {b}\right )} \log \left (\frac {\sqrt {b} x - \sqrt {\sqrt {a} \sqrt {b}}}{\sqrt {b} x + \sqrt {\sqrt {a} \sqrt {b}}}\right )}{4 \, \sqrt {a} \sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 290 vs. \(2 (84) = 168\).

Time = 0.27 (sec) , antiderivative size = 290, normalized size of antiderivative = 2.34 \[ \int \frac {1+x+x^2+x^3}{a-b x^4} \, dx=-\frac {\log \left ({\left | b x^{4} - a \right |}\right )}{4 \, b} + \frac {\sqrt {2} {\left (\left (-a b^{3}\right )^{\frac {1}{4}} b^{2} - \sqrt {2} \sqrt {-a b^{3}} b + \left (-a b^{3}\right )^{\frac {3}{4}}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{4 \, a b^{3}} + \frac {\sqrt {2} {\left (\left (-a b^{3}\right )^{\frac {1}{4}} b^{2} + \sqrt {2} \sqrt {-a b^{3}} b + \left (-a b^{3}\right )^{\frac {3}{4}}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{4 \, a b^{3}} + \frac {\sqrt {2} {\left (\left (-a b^{3}\right )^{\frac {1}{4}} b^{2} - \left (-a b^{3}\right )^{\frac {3}{4}}\right )} \log \left (x^{2} + \sqrt {2} x \left (-\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {-\frac {a}{b}}\right )}{8 \, a b^{3}} - \frac {\sqrt {2} {\left (\left (-a b^{3}\right )^{\frac {1}{4}} b^{2} - \left (-a b^{3}\right )^{\frac {3}{4}}\right )} \log \left (x^{2} - \sqrt {2} x \left (-\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {-\frac {a}{b}}\right )}{8 \, a b^{3}} \]

[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)*sq
rt(-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*sqr
t(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)

Mupad [B] (verification not implemented)

Time = 9.46 (sec) , antiderivative size = 312, normalized size of antiderivative = 2.52 \[ \int \frac {1+x+x^2+x^3}{a-b x^4} \, dx=\sum _{k=1}^4\ln \left (-\mathrm {root}\left (256\,a^3\,b^4\,z^4+256\,a^3\,b^3\,z^3+96\,a^3\,b^2\,z^2-96\,a^2\,b^3\,z^2+16\,a^3\,b\,z+16\,a\,b^3\,z-32\,a^2\,b^2\,z-3\,a^2\,b+3\,a\,b^2-b^3+a^3,z,k\right )\,\left (\mathrm {root}\left (256\,a^3\,b^4\,z^4+256\,a^3\,b^3\,z^3+96\,a^3\,b^2\,z^2-96\,a^2\,b^3\,z^2+16\,a^3\,b\,z+16\,a\,b^3\,z-32\,a^2\,b^2\,z-3\,a^2\,b+3\,a\,b^2-b^3+a^3,z,k\right )\,\left (16\,a\,b^3-16\,a\,b^3\,x\right )-x\,\left (4\,a\,b^2-4\,b^3\right )\right )\right )\,\mathrm {root}\left (256\,a^3\,b^4\,z^4+256\,a^3\,b^3\,z^3+96\,a^3\,b^2\,z^2-96\,a^2\,b^3\,z^2+16\,a^3\,b\,z+16\,a\,b^3\,z-32\,a^2\,b^2\,z-3\,a^2\,b+3\,a\,b^2-b^3+a^3,z,k\right ) \]

[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)