\(\int \frac {1}{(a+c x^4)^2} \, dx\) [147]

Optimal result

Integrand size = 9, antiderivative size = 202 \[ \int \frac {1}{\left (a+c x^4\right )^2} \, dx=\frac {x}{4 a \left (a+c x^4\right )}-\frac {3 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} \sqrt [4]{c}}+\frac {3 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} \sqrt [4]{c}}-\frac {3 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} \sqrt [4]{c}}+\frac {3 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} \sqrt [4]{c}} \]

[Out]

1/4*x/a/(c*x^4+a)+3/16*arctan(-1+c^(1/4)*x*2^(1/2)/a^(1/4))/a^(7/4)/c^(1/4)*2^(1/2)+3/16*arctan(1+c^(1/4)*x*2^
(1/2)/a^(1/4))/a^(7/4)/c^(1/4)*2^(1/2)-3/32*ln(-a^(1/4)*c^(1/4)*x*2^(1/2)+a^(1/2)+x^2*c^(1/2))/a^(7/4)/c^(1/4)
*2^(1/2)+3/32*ln(a^(1/4)*c^(1/4)*x*2^(1/2)+a^(1/2)+x^2*c^(1/2))/a^(7/4)/c^(1/4)*2^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.778, Rules used = {205, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {1}{\left (a+c x^4\right )^2} \, dx=-\frac {3 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} \sqrt [4]{c}}+\frac {3 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{8 \sqrt {2} a^{7/4} \sqrt [4]{c}}-\frac {3 \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} \sqrt [4]{c}}+\frac {3 \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} \sqrt [4]{c}}+\frac {x}{4 a \left (a+c x^4\right )} \]

[In]

Int[(a + c*x^4)^(-2),x]

[Out]

x/(4*a*(a + c*x^4)) - (3*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(8*Sqrt[2]*a^(7/4)*c^(1/4)) + (3*ArcTan[1 +
(Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(8*Sqrt[2]*a^(7/4)*c^(1/4)) - (3*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[
c]*x^2])/(16*Sqrt[2]*a^(7/4)*c^(1/4)) + (3*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(16*Sqrt[2]
*a^(7/4)*c^(1/4))

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
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 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 {x}{4 a \left (a+c x^4\right )}+\frac {3 \int \frac {1}{a+c x^4} \, dx}{4 a} \\ & = \frac {x}{4 a \left (a+c x^4\right )}+\frac {3 \int \frac {\sqrt {a}-\sqrt {c} x^2}{a+c x^4} \, dx}{8 a^{3/2}}+\frac {3 \int \frac {\sqrt {a}+\sqrt {c} x^2}{a+c x^4} \, dx}{8 a^{3/2}} \\ & = \frac {x}{4 a \left (a+c x^4\right )}+\frac {3 \int \frac {1}{\frac {\sqrt {a}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+x^2} \, dx}{16 a^{3/2} \sqrt {c}}+\frac {3 \int \frac {1}{\frac {\sqrt {a}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+x^2} \, dx}{16 a^{3/2} \sqrt {c}}-\frac {3 \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}{16 \sqrt {2} a^{7/4} \sqrt [4]{c}}-\frac {3 \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}{16 \sqrt {2} a^{7/4} \sqrt [4]{c}} \\ & = \frac {x}{4 a \left (a+c x^4\right )}-\frac {3 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} \sqrt [4]{c}}+\frac {3 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} \sqrt [4]{c}}+\frac {3 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} \sqrt [4]{c}}-\frac {3 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} \sqrt [4]{c}} \\ & = \frac {x}{4 a \left (a+c x^4\right )}-\frac {3 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} \sqrt [4]{c}}+\frac {3 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} \sqrt [4]{c}}-\frac {3 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} \sqrt [4]{c}}+\frac {3 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} \sqrt [4]{c}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (a+c x^4\right )^2} \, dx=\frac {\frac {8 a^{3/4} x}{a+c x^4}-\frac {6 \sqrt {2} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{\sqrt [4]{c}}+\frac {6 \sqrt {2} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{\sqrt [4]{c}}-\frac {3 \sqrt {2} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{\sqrt [4]{c}}+\frac {3 \sqrt {2} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{\sqrt [4]{c}}}{32 a^{7/4}} \]

[In]

Integrate[(a + c*x^4)^(-2),x]

[Out]

((8*a^(3/4)*x)/(a + c*x^4) - (6*Sqrt[2]*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/c^(1/4) + (6*Sqrt[2]*ArcTan[1
+ (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/c^(1/4) - (3*Sqrt[2]*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/
c^(1/4) + (3*Sqrt[2]*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/c^(1/4))/(32*a^(7/4))

Maple [C] (verified)

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

Time = 0.22 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.23

method	result	size
risch	\(\frac {x}{4 a \left (c \,x^{4}+a \right )}+\frac {3 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{4}+a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}\right )}{16 a c}\)	\(46\)
default	\(\frac {x}{4 a \left (c \,x^{4}+a \right )}+\frac {3 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 a^{2}}\)	\(118\)

[In]

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

[Out]

1/4*x/a/(c*x^4+a)+3/16/a/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 = 0.29 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (a+c x^4\right )^2} \, dx=\frac {3 \, {\left (a c x^{4} + a^{2}\right )} \left (-\frac {1}{a^{7} c}\right )^{\frac {1}{4}} \log \left (a^{2} \left (-\frac {1}{a^{7} c}\right )^{\frac {1}{4}} + x\right ) - 3 \, {\left (-i \, a c x^{4} - i \, a^{2}\right )} \left (-\frac {1}{a^{7} c}\right )^{\frac {1}{4}} \log \left (i \, a^{2} \left (-\frac {1}{a^{7} c}\right )^{\frac {1}{4}} + x\right ) - 3 \, {\left (i \, a c x^{4} + i \, a^{2}\right )} \left (-\frac {1}{a^{7} c}\right )^{\frac {1}{4}} \log \left (-i \, a^{2} \left (-\frac {1}{a^{7} c}\right )^{\frac {1}{4}} + x\right ) - 3 \, {\left (a c x^{4} + a^{2}\right )} \left (-\frac {1}{a^{7} c}\right )^{\frac {1}{4}} \log \left (-a^{2} \left (-\frac {1}{a^{7} c}\right )^{\frac {1}{4}} + x\right ) + 4 \, x}{16 \, {\left (a c x^{4} + a^{2}\right )}} \]

[In]

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

[Out]

1/16*(3*(a*c*x^4 + a^2)*(-1/(a^7*c))^(1/4)*log(a^2*(-1/(a^7*c))^(1/4) + x) - 3*(-I*a*c*x^4 - I*a^2)*(-1/(a^7*c
))^(1/4)*log(I*a^2*(-1/(a^7*c))^(1/4) + x) - 3*(I*a*c*x^4 + I*a^2)*(-1/(a^7*c))^(1/4)*log(-I*a^2*(-1/(a^7*c))^
(1/4) + x) - 3*(a*c*x^4 + a^2)*(-1/(a^7*c))^(1/4)*log(-a^2*(-1/(a^7*c))^(1/4) + x) + 4*x)/(a*c*x^4 + a^2)

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.19 \[ \int \frac {1}{\left (a+c x^4\right )^2} \, dx=\frac {x}{4 a^{2} + 4 a c x^{4}} + \operatorname {RootSum} {\left (65536 t^{4} a^{7} c + 81, \left ( t \mapsto t \log {\left (\frac {16 t a^{2}}{3} + x \right )} \right )\right )} \]

[In]

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

[Out]

x/(4*a**2 + 4*a*c*x**4) + RootSum(65536*_t**4*a**7*c + 81, Lambda(_t, _t*log(16*_t*a**2/3 + x)))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.94 \[ \int \frac {1}{\left (a+c x^4\right )^2} \, dx=\frac {x}{4 \, {\left (a c x^{4} + a^{2}\right )}} + \frac {3 \, {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {c}}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {c}}} + \frac {\sqrt {2} \log \left (\sqrt {c} x^{2} + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {1}{4}}} - \frac {\sqrt {2} \log \left (\sqrt {c} x^{2} - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {1}{4}}}\right )}}{32 \, a} \]

[In]

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

[Out]

1/4*x/(a*c*x^4 + a^2) + 3/32*(2*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))) + 2*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))) + sqrt(2)*log(sqrt(c)*x^2 + sqrt(2)*a^(1/4)*c^(1/4
)*x + sqrt(a))/(a^(3/4)*c^(1/4)) - sqrt(2)*log(sqrt(c)*x^2 - sqrt(2)*a^(1/4)*c^(1/4)*x + sqrt(a))/(a^(3/4)*c^(
1/4)))/a

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (a+c x^4\right )^2} \, dx=\frac {x}{4 \, {\left (c x^{4} + a\right )} a} + \frac {3 \, \sqrt {2} \left (a c^{3}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{16 \, a^{2} c} + \frac {3 \, \sqrt {2} \left (a c^{3}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{16 \, a^{2} c} + \frac {3 \, \sqrt {2} \left (a c^{3}\right )^{\frac {1}{4}} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{32 \, a^{2} c} - \frac {3 \, \sqrt {2} \left (a c^{3}\right )^{\frac {1}{4}} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{32 \, a^{2} c} \]

[In]

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

[Out]

1/4*x/((c*x^4 + a)*a) + 3/16*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^2*c) + 3/16*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^2*c) + 3/
32*sqrt(2)*(a*c^3)^(1/4)*log(x^2 + sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(a^2*c) - 3/32*sqrt(2)*(a*c^3)^(1/4)*log
(x^2 - sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(a^2*c)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.29 \[ \int \frac {1}{\left (a+c x^4\right )^2} \, dx=\frac {x}{4\,a\,\left (c\,x^4+a\right )}+\frac {3\,\mathrm {atan}\left (\frac {c^{1/4}\,x}{{\left (-a\right )}^{1/4}}\right )}{8\,{\left (-a\right )}^{7/4}\,c^{1/4}}+\frac {3\,\mathrm {atanh}\left (\frac {c^{1/4}\,x}{{\left (-a\right )}^{1/4}}\right )}{8\,{\left (-a\right )}^{7/4}\,c^{1/4}} \]

[In]

int(1/(a + c*x^4)^2,x)

[Out]

x/(4*a*(a + c*x^4)) + (3*atan((c^(1/4)*x)/(-a)^(1/4)))/(8*(-a)^(7/4)*c^(1/4)) + (3*atanh((c^(1/4)*x)/(-a)^(1/4
)))/(8*(-a)^(7/4)*c^(1/4))