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

Optimal. Leaf size=124 \[ -\frac{\left (\sqrt{a}-\sqrt{b}\right ) \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{3/4}}+\frac{\left (\sqrt{a}+\sqrt{b}\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{3/4}}-\frac{\log \left (a-b x^4\right )}{4 b}+\frac{\tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{2 \sqrt{a} \sqrt{b}} \]

[Out]

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

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Rubi [A]  time = 0.0870212, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {1876, 1167, 205, 208, 1248, 635, 260} \[ -\frac{\left (\sqrt{a}-\sqrt{b}\right ) \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{3/4}}+\frac{\left (\sqrt{a}+\sqrt{b}\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{3/4}}-\frac{\log \left (a-b x^4\right )}{4 b}+\frac{\tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{2 \sqrt{a} \sqrt{b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1167

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 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 208

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

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 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 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]

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{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} \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 (\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]  time = 0.0499505, size = 203, normalized size = 1.64 \[ -\frac{\left (a^{3/4}+\sqrt{a} \sqrt [4]{b}+\sqrt [4]{a} \sqrt{b}\right ) \log \left (\sqrt [4]{a}-\sqrt [4]{b} x\right )}{4 a b^{3/4}}-\frac{\left (-a^{3/4}+\sqrt{a} \sqrt [4]{b}-\sqrt [4]{a} \sqrt{b}\right ) \log \left (\sqrt [4]{a}+\sqrt [4]{b} x\right )}{4 a b^{3/4}}+\frac{\left (\sqrt [4]{a} \sqrt{b}-a^{3/4}\right ) \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 a b^{3/4}}+\frac{\log \left (\sqrt{a}+\sqrt{b} x^2\right )}{4 \sqrt{a} \sqrt{b}}-\frac{\log \left (a-b x^4\right )}{4 b} \]

Antiderivative was successfully verified.

[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)

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Maple [B]  time = 0.004, size = 171, normalized size = 1.4 \begin{align*}{\frac{1}{4\,a}\sqrt [4]{{\frac{a}{b}}}\ln \left ({ \left ( x+\sqrt [4]{{\frac{a}{b}}} \right ) \left ( x-\sqrt [4]{{\frac{a}{b}}} \right ) ^{-1}} \right ) }+{\frac{1}{2\,a}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({x{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}} \right ) }-{\frac{1}{4}\ln \left ({ \left ( -a+{x}^{2}\sqrt{ab} \right ) \left ( -a-{x}^{2}\sqrt{ab} \right ) ^{-1}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{1}{2\,b}\arctan \left ({x{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}} \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+{\frac{1}{4\,b}\ln \left ({ \left ( x+\sqrt [4]{{\frac{a}{b}}} \right ) \left ( x-\sqrt [4]{{\frac{a}{b}}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-{\frac{\ln \left ( b{x}^{4}-a \right ) }{4\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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

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Sympy [A]  time = 1.02863, size = 187, normalized size = 1.51 \begin{align*} - \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 )} \end{align*}

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))))

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Giac [B]  time = 1.09042, size = 392, normalized size = 3.16 \begin{align*} -\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}} \end{align*}

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)*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)