\(\int \frac {1}{a+\frac {b}{x^4}} \, dx\) [2057]

Optimal result

Integrand size = 9, antiderivative size = 190 \[ \int \frac {1}{a+\frac {b}{x^4}} \, dx=\frac {x}{a}+\frac {\sqrt [4]{b} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}\right )}{2 \sqrt {2} a^{5/4}}-\frac {\sqrt [4]{b} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}\right )}{2 \sqrt {2} a^{5/4}}+\frac {\sqrt [4]{b} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a} x^2\right )}{4 \sqrt {2} a^{5/4}}-\frac {\sqrt [4]{b} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a} x^2\right )}{4 \sqrt {2} a^{5/4}} \]

[Out]

x/a-1/4*b^(1/4)*arctan(-1+a^(1/4)*x*2^(1/2)/b^(1/4))/a^(5/4)*2^(1/2)-1/4*b^(1/4)*arctan(1+a^(1/4)*x*2^(1/2)/b^
(1/4))/a^(5/4)*2^(1/2)+1/8*b^(1/4)*ln(-a^(1/4)*b^(1/4)*x*2^(1/2)+x^2*a^(1/2)+b^(1/2))/a^(5/4)*2^(1/2)-1/8*b^(1
/4)*ln(a^(1/4)*b^(1/4)*x*2^(1/2)+x^2*a^(1/2)+b^(1/2))/a^(5/4)*2^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.889, Rules used = {199, 327, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {1}{a+\frac {b}{x^4}} \, dx=\frac {\sqrt [4]{b} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}\right )}{2 \sqrt {2} a^{5/4}}-\frac {\sqrt [4]{b} \arctan \left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+1\right )}{2 \sqrt {2} a^{5/4}}+\frac {\sqrt [4]{b} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a} x^2+\sqrt {b}\right )}{4 \sqrt {2} a^{5/4}}-\frac {\sqrt [4]{b} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a} x^2+\sqrt {b}\right )}{4 \sqrt {2} a^{5/4}}+\frac {x}{a} \]

[In]

Int[(a + b/x^4)^(-1),x]

[Out]

x/a + (b^(1/4)*ArcTan[1 - (Sqrt[2]*a^(1/4)*x)/b^(1/4)])/(2*Sqrt[2]*a^(5/4)) - (b^(1/4)*ArcTan[1 + (Sqrt[2]*a^(
1/4)*x)/b^(1/4)])/(2*Sqrt[2]*a^(5/4)) + (b^(1/4)*Log[Sqrt[b] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[a]*x^2])/(4*Sq
rt[2]*a^(5/4)) - (b^(1/4)*Log[Sqrt[b] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[a]*x^2])/(4*Sqrt[2]*a^(5/4))

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b}, x] && LtQ[n, 0]
&& IntegerQ[p]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 631

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 642

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 1176

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 1179

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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^4}{b+a x^4} \, dx \\ & = \frac {x}{a}-\frac {b \int \frac {1}{b+a x^4} \, dx}{a} \\ & = \frac {x}{a}-\frac {\sqrt {b} \int \frac {\sqrt {b}-\sqrt {a} x^2}{b+a x^4} \, dx}{2 a}-\frac {\sqrt {b} \int \frac {\sqrt {b}+\sqrt {a} x^2}{b+a x^4} \, dx}{2 a} \\ & = \frac {x}{a}+\frac {\sqrt [4]{b} \int \frac {\frac {\sqrt {2} \sqrt [4]{b}}{\sqrt [4]{a}}+2 x}{-\frac {\sqrt {b}}{\sqrt {a}}-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}-x^2} \, dx}{4 \sqrt {2} a^{5/4}}+\frac {\sqrt [4]{b} \int \frac {\frac {\sqrt {2} \sqrt [4]{b}}{\sqrt [4]{a}}-2 x}{-\frac {\sqrt {b}}{\sqrt {a}}+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}-x^2} \, dx}{4 \sqrt {2} a^{5/4}}-\frac {\sqrt {b} \int \frac {1}{\frac {\sqrt {b}}{\sqrt {a}}-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+x^2} \, dx}{4 a^{3/2}}-\frac {\sqrt {b} \int \frac {1}{\frac {\sqrt {b}}{\sqrt {a}}+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+x^2} \, dx}{4 a^{3/2}} \\ & = \frac {x}{a}+\frac {\sqrt [4]{b} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a} x^2\right )}{4 \sqrt {2} a^{5/4}}-\frac {\sqrt [4]{b} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a} x^2\right )}{4 \sqrt {2} a^{5/4}}-\frac {\sqrt [4]{b} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}\right )}{2 \sqrt {2} a^{5/4}}+\frac {\sqrt [4]{b} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}\right )}{2 \sqrt {2} a^{5/4}} \\ & = \frac {x}{a}+\frac {\sqrt [4]{b} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}\right )}{2 \sqrt {2} a^{5/4}}-\frac {\sqrt [4]{b} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}\right )}{2 \sqrt {2} a^{5/4}}+\frac {\sqrt [4]{b} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a} x^2\right )}{4 \sqrt {2} a^{5/4}}-\frac {\sqrt [4]{b} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a} x^2\right )}{4 \sqrt {2} a^{5/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.91 \[ \int \frac {1}{a+\frac {b}{x^4}} \, dx=\frac {8 \sqrt [4]{a} x+2 \sqrt {2} \sqrt [4]{b} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}\right )-2 \sqrt {2} \sqrt [4]{b} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}\right )+\sqrt {2} \sqrt [4]{b} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a} x^2\right )-\sqrt {2} \sqrt [4]{b} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a} x^2\right )}{8 a^{5/4}} \]

[In]

Integrate[(a + b/x^4)^(-1),x]

[Out]

(8*a^(1/4)*x + 2*Sqrt[2]*b^(1/4)*ArcTan[1 - (Sqrt[2]*a^(1/4)*x)/b^(1/4)] - 2*Sqrt[2]*b^(1/4)*ArcTan[1 + (Sqrt[
2]*a^(1/4)*x)/b^(1/4)] + Sqrt[2]*b^(1/4)*Log[Sqrt[b] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[a]*x^2] - Sqrt[2]*b^(1
/4)*Log[Sqrt[b] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[a]*x^2])/(8*a^(5/4))

Maple [C] (verified)

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

Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.18

method	result	size
risch	\(\frac {x}{a}-\frac {b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{4}+b \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}\right )}{4 a^{2}}\)	\(34\)
default	\(\frac {x}{a}-\frac {\left (\frac {b}{a}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {b}{a}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {b}{a}}}{x^{2}-\left (\frac {b}{a}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {b}{a}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {b}{a}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {b}{a}\right )^{\frac {1}{4}}}-1\right )\right )}{8 a}\)	\(108\)

[In]

int(1/(a+b/x^4),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.56 \[ \int \frac {1}{a+\frac {b}{x^4}} \, dx=-\frac {a \left (-\frac {b}{a^{5}}\right )^{\frac {1}{4}} \log \left (a \left (-\frac {b}{a^{5}}\right )^{\frac {1}{4}} + x\right ) + i \, a \left (-\frac {b}{a^{5}}\right )^{\frac {1}{4}} \log \left (i \, a \left (-\frac {b}{a^{5}}\right )^{\frac {1}{4}} + x\right ) - i \, a \left (-\frac {b}{a^{5}}\right )^{\frac {1}{4}} \log \left (-i \, a \left (-\frac {b}{a^{5}}\right )^{\frac {1}{4}} + x\right ) - a \left (-\frac {b}{a^{5}}\right )^{\frac {1}{4}} \log \left (-a \left (-\frac {b}{a^{5}}\right )^{\frac {1}{4}} + x\right ) - 4 \, x}{4 \, a} \]

[In]

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

[Out]

-1/4*(a*(-b/a^5)^(1/4)*log(a*(-b/a^5)^(1/4) + x) + I*a*(-b/a^5)^(1/4)*log(I*a*(-b/a^5)^(1/4) + x) - I*a*(-b/a^
5)^(1/4)*log(-I*a*(-b/a^5)^(1/4) + x) - a*(-b/a^5)^(1/4)*log(-a*(-b/a^5)^(1/4) + x) - 4*x)/a

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.12 \[ \int \frac {1}{a+\frac {b}{x^4}} \, dx=\operatorname {RootSum} {\left (256 t^{4} a^{5} + b, \left ( t \mapsto t \log {\left (- 4 t a + x \right )} \right )\right )} + \frac {x}{a} \]

[In]

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

[Out]

RootSum(256*_t**4*a**5 + b, Lambda(_t, _t*log(-4*_t*a + x))) + x/a

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.94 \[ \int \frac {1}{a+\frac {b}{x^4}} \, dx=-\frac {\frac {2 \, \sqrt {2} \sqrt {b} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {a} x + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}}} + \frac {2 \, \sqrt {2} \sqrt {b} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {a} x - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}}} + \frac {\sqrt {2} b^{\frac {1}{4}} \log \left (\sqrt {a} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {b}\right )}{a^{\frac {1}{4}}} - \frac {\sqrt {2} b^{\frac {1}{4}} \log \left (\sqrt {a} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {b}\right )}{a^{\frac {1}{4}}}}{8 \, a} + \frac {x}{a} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.91 \[ \int \frac {1}{a+\frac {b}{x^4}} \, dx=\frac {x}{a} - \frac {\sqrt {2} \left (a^{3} b\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {b}{a}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {b}{a}\right )^{\frac {1}{4}}}\right )}{4 \, a^{2}} - \frac {\sqrt {2} \left (a^{3} b\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {b}{a}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {b}{a}\right )^{\frac {1}{4}}}\right )}{4 \, a^{2}} - \frac {\sqrt {2} \left (a^{3} b\right )^{\frac {1}{4}} \log \left (x^{2} + \sqrt {2} x \left (\frac {b}{a}\right )^{\frac {1}{4}} + \sqrt {\frac {b}{a}}\right )}{8 \, a^{2}} + \frac {\sqrt {2} \left (a^{3} b\right )^{\frac {1}{4}} \log \left (x^{2} - \sqrt {2} x \left (\frac {b}{a}\right )^{\frac {1}{4}} + \sqrt {\frac {b}{a}}\right )}{8 \, a^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.87 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.25 \[ \int \frac {1}{a+\frac {b}{x^4}} \, dx=\frac {x}{a}-\frac {{\left (-b\right )}^{1/4}\,\mathrm {atan}\left (\frac {a^{1/4}\,x}{{\left (-b\right )}^{1/4}}\right )}{2\,a^{5/4}}-\frac {{\left (-b\right )}^{1/4}\,\mathrm {atanh}\left (\frac {a^{1/4}\,x}{{\left (-b\right )}^{1/4}}\right )}{2\,a^{5/4}} \]

[In]

int(1/(a + b/x^4),x)

[Out]

x/a - ((-b)^(1/4)*atan((a^(1/4)*x)/(-b)^(1/4)))/(2*a^(5/4)) - ((-b)^(1/4)*atanh((a^(1/4)*x)/(-b)^(1/4)))/(2*a^
(5/4))