\(\int \frac {1}{a+c x^4} \, dx\) [141]

Optimal result

Integrand size = 9, antiderivative size = 185 \[ \int \frac {1}{a+c x^4} \, dx=-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{c}}+\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{c}}-\frac {\log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} \sqrt [4]{c}}+\frac {\log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} \sqrt [4]{c}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {217, 1179, 642, 1176, 631, 210} \[ \int \frac {1}{a+c x^4} \, dx=-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{c}}+\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{c}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} \sqrt [4]{c}}+\frac {\log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} \sqrt [4]{c}} \]

[In]

Int[(a + c*x^4)^(-1),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.72 \[ \int \frac {1}{a+c x^4} \, dx=\frac {-2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+2 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )-\log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )+\log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} \sqrt [4]{c}} \]

[In]

Integrate[(a + c*x^4)^(-1),x]

[Out]

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

Maple [C] (verified)

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

Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.15

method	result	size
risch	\(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{4}+a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}}{4 c}\)	\(27\)
default	\(\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{8 a}\)	\(102\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.61 \[ \int \frac {1}{a+c x^4} \, dx=\frac {1}{4} \, \left (-\frac {1}{a^{3} c}\right )^{\frac {1}{4}} \log \left (a \left (-\frac {1}{a^{3} c}\right )^{\frac {1}{4}} + x\right ) + \frac {1}{4} i \, \left (-\frac {1}{a^{3} c}\right )^{\frac {1}{4}} \log \left (i \, a \left (-\frac {1}{a^{3} c}\right )^{\frac {1}{4}} + x\right ) - \frac {1}{4} i \, \left (-\frac {1}{a^{3} c}\right )^{\frac {1}{4}} \log \left (-i \, a \left (-\frac {1}{a^{3} c}\right )^{\frac {1}{4}} + x\right ) - \frac {1}{4} \, \left (-\frac {1}{a^{3} c}\right )^{\frac {1}{4}} \log \left (-a \left (-\frac {1}{a^{3} c}\right )^{\frac {1}{4}} + x\right ) \]

[In]

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

[Out]

1/4*(-1/(a^3*c))^(1/4)*log(a*(-1/(a^3*c))^(1/4) + x) + 1/4*I*(-1/(a^3*c))^(1/4)*log(I*a*(-1/(a^3*c))^(1/4) + x
) - 1/4*I*(-1/(a^3*c))^(1/4)*log(-I*a*(-1/(a^3*c))^(1/4) + x) - 1/4*(-1/(a^3*c))^(1/4)*log(-a*(-1/(a^3*c))^(1/
4) + x)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.11 \[ \int \frac {1}{a+c x^4} \, dx=\operatorname {RootSum} {\left (256 t^{4} a^{3} c + 1, \left ( t \mapsto t \log {\left (4 t a + x \right )} \right )\right )} \]

[In]

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

[Out]

RootSum(256*_t**4*a**3*c + 1, Lambda(_t, _t*log(4*_t*a + x)))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.91 \[ \int \frac {1}{a+c x^4} \, dx=\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{4 \, \sqrt {a} \sqrt {\sqrt {a} \sqrt {c}}} + \frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{4 \, \sqrt {a} \sqrt {\sqrt {a} \sqrt {c}}} + \frac {\sqrt {2} \log \left (\sqrt {c} x^{2} + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{8 \, a^{\frac {3}{4}} c^{\frac {1}{4}}} - \frac {\sqrt {2} \log \left (\sqrt {c} x^{2} - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{8 \, a^{\frac {3}{4}} c^{\frac {1}{4}}} \]

[In]

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

[Out]

1/4*sqrt(2)*arctan(1/2*sqrt(2)*(2*sqrt(c)*x + sqrt(2)*a^(1/4)*c^(1/4))/sqrt(sqrt(a)*sqrt(c)))/(sqrt(a)*sqrt(sq
rt(a)*sqrt(c))) + 1/4*sqrt(2)*arctan(1/2*sqrt(2)*(2*sqrt(c)*x - sqrt(2)*a^(1/4)*c^(1/4))/sqrt(sqrt(a)*sqrt(c))
)/(sqrt(a)*sqrt(sqrt(a)*sqrt(c))) + 1/8*sqrt(2)*log(sqrt(c)*x^2 + sqrt(2)*a^(1/4)*c^(1/4)*x + sqrt(a))/(a^(3/4
)*c^(1/4)) - 1/8*sqrt(2)*log(sqrt(c)*x^2 - sqrt(2)*a^(1/4)*c^(1/4)*x + sqrt(a))/(a^(3/4)*c^(1/4))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.97 \[ \int \frac {1}{a+c x^4} \, dx=\frac {\sqrt {2} \left (a c^{3}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{4 \, a c} + \frac {\sqrt {2} \left (a c^{3}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{4 \, a c} + \frac {\sqrt {2} \left (a c^{3}\right )^{\frac {1}{4}} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{8 \, a c} - \frac {\sqrt {2} \left (a c^{3}\right )^{\frac {1}{4}} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{8 \, a c} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.58 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.18 \[ \int \frac {1}{a+c x^4} \, dx=-\frac {\mathrm {atan}\left (\frac {c^{1/4}\,x}{{\left (-a\right )}^{1/4}}\right )+\mathrm {atanh}\left (\frac {c^{1/4}\,x}{{\left (-a\right )}^{1/4}}\right )}{2\,{\left (-a\right )}^{3/4}\,c^{1/4}} \]

[In]

int(1/(a + c*x^4),x)

[Out]

-(atan((c^(1/4)*x)/(-a)^(1/4)) + atanh((c^(1/4)*x)/(-a)^(1/4)))/(2*(-a)^(3/4)*c^(1/4))