\(\int \frac {1}{x^2 (a+c x^4)} \, dx\) [656]

Optimal result

Integrand size = 13, antiderivative size = 193 \[ \int \frac {1}{x^2 \left (a+c x^4\right )} \, dx=-\frac {1}{a x}+\frac {\sqrt [4]{c} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{5/4}}-\frac {\sqrt [4]{c} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{5/4}}-\frac {\sqrt [4]{c} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{5/4}}+\frac {\sqrt [4]{c} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{5/4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {331, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {1}{x^2 \left (a+c x^4\right )} \, dx=\frac {\sqrt [4]{c} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{5/4}}-\frac {\sqrt [4]{c} \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} a^{5/4}}-\frac {\sqrt [4]{c} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{5/4}}+\frac {\sqrt [4]{c} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{5/4}}-\frac {1}{a x} \]

[In]

Int[1/(x^2*(a + c*x^4)),x]

[Out]

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

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 303

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

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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}& = -\frac {1}{a x}-\frac {c \int \frac {x^2}{a+c x^4} \, dx}{a} \\ & = -\frac {1}{a x}+\frac {\sqrt {c} \int \frac {\sqrt {a}-\sqrt {c} x^2}{a+c x^4} \, dx}{2 a}-\frac {\sqrt {c} \int \frac {\sqrt {a}+\sqrt {c} x^2}{a+c x^4} \, dx}{2 a} \\ & = -\frac {1}{a x}-\frac {\int \frac {1}{\frac {\sqrt {a}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+x^2} \, dx}{4 a}-\frac {\int \frac {1}{\frac {\sqrt {a}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+x^2} \, dx}{4 a}-\frac {\sqrt [4]{c} \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^{5/4}}-\frac {\sqrt [4]{c} \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^{5/4}} \\ & = -\frac {1}{a x}-\frac {\sqrt [4]{c} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{5/4}}+\frac {\sqrt [4]{c} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{5/4}}-\frac {\sqrt [4]{c} \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^{5/4}}+\frac {\sqrt [4]{c} \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^{5/4}} \\ & = -\frac {1}{a x}+\frac {\sqrt [4]{c} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{5/4}}-\frac {\sqrt [4]{c} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{5/4}}-\frac {\sqrt [4]{c} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{5/4}}+\frac {\sqrt [4]{c} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{5/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x^2 \left (a+c x^4\right )} \, dx=\frac {-8 \sqrt [4]{a}+2 \sqrt {2} \sqrt [4]{c} x \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )-2 \sqrt {2} \sqrt [4]{c} x \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )-\sqrt {2} \sqrt [4]{c} x \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )+\sqrt {2} \sqrt [4]{c} x \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{8 a^{5/4} x} \]

[In]

Integrate[1/(x^2*(a + c*x^4)),x]

[Out]

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

Maple [C] (verified)

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

Time = 3.90 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.26

method	result	size
risch	\(-\frac {1}{a x}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{5} \textit {\_Z}^{4}+c \right )}{\sum }\textit {\_R} \ln \left (\left (5 \textit {\_R}^{4} a^{5}+4 c \right ) x +a^{4} \textit {\_R}^{3}\right )\right )}{4}\)	\(50\)
default	\(-\frac {1}{a x}-\frac {\sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{8 a \left (\frac {a}{c}\right )^{\frac {1}{4}}}\)	\(111\)

[In]

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

[Out]

-1/a/x+1/4*sum(_R*ln((5*_R^4*a^5+4*c)*x+a^4*_R^3),_R=RootOf(_Z^4*a^5+c))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.66 \[ \int \frac {1}{x^2 \left (a+c x^4\right )} \, dx=-\frac {a x \left (-\frac {c}{a^{5}}\right )^{\frac {1}{4}} \log \left (a^{4} \left (-\frac {c}{a^{5}}\right )^{\frac {3}{4}} + c x\right ) - i \, a x \left (-\frac {c}{a^{5}}\right )^{\frac {1}{4}} \log \left (i \, a^{4} \left (-\frac {c}{a^{5}}\right )^{\frac {3}{4}} + c x\right ) + i \, a x \left (-\frac {c}{a^{5}}\right )^{\frac {1}{4}} \log \left (-i \, a^{4} \left (-\frac {c}{a^{5}}\right )^{\frac {3}{4}} + c x\right ) - a x \left (-\frac {c}{a^{5}}\right )^{\frac {1}{4}} \log \left (-a^{4} \left (-\frac {c}{a^{5}}\right )^{\frac {3}{4}} + c x\right ) + 4}{4 \, a x} \]

[In]

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

[Out]

-1/4*(a*x*(-c/a^5)^(1/4)*log(a^4*(-c/a^5)^(3/4) + c*x) - I*a*x*(-c/a^5)^(1/4)*log(I*a^4*(-c/a^5)^(3/4) + c*x)
+ I*a*x*(-c/a^5)^(1/4)*log(-I*a^4*(-c/a^5)^(3/4) + c*x) - a*x*(-c/a^5)^(1/4)*log(-a^4*(-c/a^5)^(3/4) + c*x) +
4)/(a*x)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.15 \[ \int \frac {1}{x^2 \left (a+c x^4\right )} \, dx=\operatorname {RootSum} {\left (256 t^{4} a^{5} + c, \left ( t \mapsto t \log {\left (- \frac {64 t^{3} a^{4}}{c} + x \right )} \right )\right )} - \frac {1}{a x} \]

[In]

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

[Out]

RootSum(256*_t**4*a**5 + c, Lambda(_t, _t*log(-64*_t**3*a**4/c + x))) - 1/(a*x)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x^2 \left (a+c x^4\right )} \, dx=-\frac {c {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {\sqrt {a} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {\sqrt {a} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {c} x^{2} + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {c} x^{2} - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{8 \, a} - \frac {1}{a x} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.97 \[ \int \frac {1}{x^2 \left (a+c x^4\right )} \, dx=-\frac {\sqrt {2} \left (a c^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{4 \, a^{2} c^{2}} - \frac {\sqrt {2} \left (a c^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{4 \, a^{2} c^{2}} + \frac {\sqrt {2} \left (a c^{3}\right )^{\frac {3}{4}} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{8 \, a^{2} c^{2}} - \frac {\sqrt {2} \left (a c^{3}\right )^{\frac {3}{4}} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{8 \, a^{2} c^{2}} - \frac {1}{a x} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.26 \[ \int \frac {1}{x^2 \left (a+c x^4\right )} \, dx=\frac {{\left (-c\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (-c\right )}^{1/4}\,x}{a^{1/4}}\right )}{2\,a^{5/4}}-\frac {{\left (-c\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{1/4}\,x}{a^{1/4}}\right )}{2\,a^{5/4}}-\frac {1}{a\,x} \]

[In]

int(1/(x^2*(a + c*x^4)),x)

[Out]

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