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

Optimal. Leaf size=202 \[ -\frac{3 \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )}{16 \sqrt{2} a^{7/4} \sqrt [4]{c}}+\frac{3 \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )}{16 \sqrt{2} a^{7/4} \sqrt [4]{c}}-\frac{3 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt{2} a^{7/4} \sqrt [4]{c}}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{8 \sqrt{2} a^{7/4} \sqrt [4]{c}}+\frac{x}{4 a \left (a+c x^4\right )} \]

[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))

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Rubi [A]  time = 0.115676, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.778, Rules used = {199, 211, 1165, 628, 1162, 617, 204} \[ -\frac{3 \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )}{16 \sqrt{2} a^{7/4} \sqrt [4]{c}}+\frac{3 \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )}{16 \sqrt{2} a^{7/4} \sqrt [4]{c}}-\frac{3 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt{2} a^{7/4} \sqrt [4]{c}}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{8 \sqrt{2} a^{7/4} \sqrt [4]{c}}+\frac{x}{4 a \left (a+c x^4\right )} \]

Antiderivative was successfully verified.

[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 199

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] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 211

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 1165

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]

Rule 628

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 1162

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 617

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 204

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

Rubi steps

\begin{align*} \int \frac{1}{\left (a+c x^4\right )^2} \, dx &=\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 \operatorname{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 \operatorname{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]  time = 0.10742, size = 183, normalized size = 0.91 \[ \frac{\frac{8 a^{3/4} x}{a+c x^4}-\frac{3 \sqrt{2} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )}{\sqrt [4]{c}}+\frac{3 \sqrt{2} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )}{\sqrt [4]{c}}-\frac{6 \sqrt{2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{\sqrt [4]{c}}+\frac{6 \sqrt{2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{\sqrt [4]{c}}}{32 a^{7/4}} \]

Antiderivative was successfully verified.

[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))

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Maple [A]  time = 0.004, size = 143, normalized size = 0.7 \begin{align*}{\frac{x}{4\,a \left ( c{x}^{4}+a \right ) }}+{\frac{3\,\sqrt{2}}{32\,{a}^{2}}\sqrt [4]{{\frac{a}{c}}}\ln \left ({ \left ({x}^{2}+\sqrt [4]{{\frac{a}{c}}}x\sqrt{2}+\sqrt{{\frac{a}{c}}} \right ) \left ({x}^{2}-\sqrt [4]{{\frac{a}{c}}}x\sqrt{2}+\sqrt{{\frac{a}{c}}} \right ) ^{-1}} \right ) }+{\frac{3\,\sqrt{2}}{16\,{a}^{2}}\sqrt [4]{{\frac{a}{c}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}+1 \right ) }+{\frac{3\,\sqrt{2}}{16\,{a}^{2}}\sqrt [4]{{\frac{a}{c}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}-1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(1/(c*x^4+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.996213, size = 414, normalized size = 2.05 \begin{align*} \frac{12 \,{\left (a c x^{4} + a^{2}\right )} \left (-\frac{1}{a^{7} c}\right )^{\frac{1}{4}} \arctan \left (-a^{5} c x \left (-\frac{1}{a^{7} c}\right )^{\frac{3}{4}} + \sqrt{a^{4} \sqrt{-\frac{1}{a^{7} c}} + x^{2}} a^{5} c \left (-\frac{1}{a^{7} c}\right )^{\frac{3}{4}}\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 ) - 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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(12*(a*c*x^4 + a^2)*(-1/(a^7*c))^(1/4)*arctan(-a^5*c*x*(-1/(a^7*c))^(3/4) + sqrt(a^4*sqrt(-1/(a^7*c)) + x
^2)*a^5*c*(-1/(a^7*c))^(3/4)) + 3*(a*c*x^4 + a^2)*(-1/(a^7*c))^(1/4)*log(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)

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Sympy [A]  time = 0.469475, size = 39, normalized size = 0.19 \begin{align*} \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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Giac [A]  time = 1.11891, size = 262, normalized size = 1.3 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)