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

Optimal. Leaf size=202 \[ -\frac {\log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{3/4} c^{5/4}}+\frac {\log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{3/4} c^{5/4}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{3/4} c^{5/4}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{8 \sqrt {2} a^{3/4} c^{5/4}}-\frac {x}{4 c \left (a+c x^4\right )} \]

[Out]

-1/4*x/c/(c*x^4+a)+1/16*arctan(-1+c^(1/4)*x*2^(1/2)/a^(1/4))/a^(3/4)/c^(5/4)*2^(1/2)+1/16*arctan(1+c^(1/4)*x*2
^(1/2)/a^(1/4))/a^(3/4)/c^(5/4)*2^(1/2)-1/32*ln(-a^(1/4)*c^(1/4)*x*2^(1/2)+a^(1/2)+x^2*c^(1/2))/a^(3/4)/c^(5/4
)*2^(1/2)+1/32*ln(a^(1/4)*c^(1/4)*x*2^(1/2)+a^(1/2)+x^2*c^(1/2))/a^(3/4)/c^(5/4)*2^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {288, 211, 1165, 628, 1162, 617, 204} \[ -\frac {\log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{3/4} c^{5/4}}+\frac {\log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{3/4} c^{5/4}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{3/4} c^{5/4}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{8 \sqrt {2} a^{3/4} c^{5/4}}-\frac {x}{4 c \left (a+c x^4\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(a + c*x^4)^2,x]

[Out]

-x/(4*c*(a + c*x^4)) - ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)]/(8*Sqrt[2]*a^(3/4)*c^(5/4)) + ArcTan[1 + (Sqrt[
2]*c^(1/4)*x)/a^(1/4)]/(8*Sqrt[2]*a^(3/4)*c^(5/4)) - Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2]/(1
6*Sqrt[2]*a^(3/4)*c^(5/4)) + Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2]/(16*Sqrt[2]*a^(3/4)*c^(5/4
))

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])

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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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 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 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]

Rubi steps

\begin {align*} \int \frac {x^4}{\left (a+c x^4\right )^2} \, dx &=-\frac {x}{4 c \left (a+c x^4\right )}+\frac {\int \frac {1}{a+c x^4} \, dx}{4 c}\\ &=-\frac {x}{4 c \left (a+c x^4\right )}+\frac {\int \frac {\sqrt {a}-\sqrt {c} x^2}{a+c x^4} \, dx}{8 \sqrt {a} c}+\frac {\int \frac {\sqrt {a}+\sqrt {c} x^2}{a+c x^4} \, dx}{8 \sqrt {a} c}\\ &=-\frac {x}{4 c \left (a+c x^4\right )}+\frac {\int \frac {1}{\frac {\sqrt {a}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+x^2} \, dx}{16 \sqrt {a} c^{3/2}}+\frac {\int \frac {1}{\frac {\sqrt {a}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+x^2} \, dx}{16 \sqrt {a} c^{3/2}}-\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}{16 \sqrt {2} a^{3/4} c^{5/4}}-\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}{16 \sqrt {2} a^{3/4} c^{5/4}}\\ &=-\frac {x}{4 c \left (a+c x^4\right )}-\frac {\log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{3/4} c^{5/4}}+\frac {\log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{3/4} c^{5/4}}+\frac {\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^{3/4} c^{5/4}}-\frac {\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^{3/4} c^{5/4}}\\ &=-\frac {x}{4 c \left (a+c x^4\right )}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{3/4} c^{5/4}}+\frac {\tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{3/4} c^{5/4}}-\frac {\log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{3/4} c^{5/4}}+\frac {\log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{3/4} c^{5/4}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 182, normalized size = 0.90 \[ \frac {-\frac {\sqrt {2} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{a^{3/4}}+\frac {\sqrt {2} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{a^{3/4}}-\frac {2 \sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{a^{3/4}}+\frac {2 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{a^{3/4}}-\frac {8 \sqrt [4]{c} x}{a+c x^4}}{32 c^{5/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(a + c*x^4)^2,x]

[Out]

((-8*c^(1/4)*x)/(a + c*x^4) - (2*Sqrt[2]*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/a^(3/4) + (2*Sqrt[2]*ArcTan[
1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/a^(3/4) - (Sqrt[2]*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/a
^(3/4) + (Sqrt[2]*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/a^(3/4))/(32*c^(5/4))

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fricas [A]  time = 0.80, size = 181, normalized size = 0.90 \[ \frac {4 \, {\left (c^{2} x^{4} + a c\right )} \left (-\frac {1}{a^{3} c^{5}}\right )^{\frac {1}{4}} \arctan \left (-a^{2} c^{4} x \left (-\frac {1}{a^{3} c^{5}}\right )^{\frac {3}{4}} + \sqrt {a^{2} c^{2} \sqrt {-\frac {1}{a^{3} c^{5}}} + x^{2}} a^{2} c^{4} \left (-\frac {1}{a^{3} c^{5}}\right )^{\frac {3}{4}}\right ) + {\left (c^{2} x^{4} + a c\right )} \left (-\frac {1}{a^{3} c^{5}}\right )^{\frac {1}{4}} \log \left (a c \left (-\frac {1}{a^{3} c^{5}}\right )^{\frac {1}{4}} + x\right ) - {\left (c^{2} x^{4} + a c\right )} \left (-\frac {1}{a^{3} c^{5}}\right )^{\frac {1}{4}} \log \left (-a c \left (-\frac {1}{a^{3} c^{5}}\right )^{\frac {1}{4}} + x\right ) - 4 \, x}{16 \, {\left (c^{2} x^{4} + a c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 194, normalized size = 0.96 \[ -\frac {x}{4 \, {\left (c x^{4} + a\right )} c} + \frac {\sqrt {2} \left (a c^{3}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{16 \, a c^{2}} + \frac {\sqrt {2} \left (a c^{3}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{16 \, a c^{2}} + \frac {\sqrt {2} \left (a c^{3}\right )^{\frac {1}{4}} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{32 \, a c^{2}} - \frac {\sqrt {2} \left (a c^{3}\right )^{\frac {1}{4}} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{32 \, a c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 152, normalized size = 0.75 \[ -\frac {x}{4 \left (c \,x^{4}+a \right ) c}+\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{16 a c}+\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{16 a c}+\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{c}}}{x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{c}}}\right )}{32 a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.96, size = 190, normalized size = 0.94 \[ -\frac {x}{4 \, {\left (c^{2} x^{4} + a c\right )}} + \frac {\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {c}}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {c}}} + \frac {\sqrt {2} \log \left (\sqrt {c} x^{2} + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {1}{4}}} - \frac {\sqrt {2} \log \left (\sqrt {c} x^{2} - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {1}{4}}}}{32 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x/(c^2*x^4 + a*c) + 1/32*(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)))/c

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mupad [B]  time = 1.06, size = 58, normalized size = 0.29 \[ -\frac {x}{4\,c\,\left (c\,x^4+a\right )}-\frac {\mathrm {atan}\left (\frac {c^{1/4}\,x}{{\left (-a\right )}^{1/4}}\right )}{8\,{\left (-a\right )}^{3/4}\,c^{5/4}}-\frac {\mathrm {atanh}\left (\frac {c^{1/4}\,x}{{\left (-a\right )}^{1/4}}\right )}{8\,{\left (-a\right )}^{3/4}\,c^{5/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a + c*x^4)^2,x)

[Out]

- x/(4*c*(a + c*x^4)) - atan((c^(1/4)*x)/(-a)^(1/4))/(8*(-a)^(3/4)*c^(5/4)) - atanh((c^(1/4)*x)/(-a)^(1/4))/(8
*(-a)^(3/4)*c^(5/4))

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sympy [A]  time = 0.54, size = 39, normalized size = 0.19 \[ - \frac {x}{4 a c + 4 c^{2} x^{4}} + \operatorname {RootSum} {\left (65536 t^{4} a^{3} c^{5} + 1, \left (t \mapsto t \log {\left (16 t a c + x \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x/(4*a*c + 4*c**2*x**4) + RootSum(65536*_t**4*a**3*c**5 + 1, Lambda(_t, _t*log(16*_t*a*c + x)))

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