3.7.71 \(\int \frac {1}{x^4 (a+c x^4)^2} \, dx\) [671]

Optimal. Leaf size=214 \[ -\frac {7}{12 a^2 x^3}+\frac {1}{4 a x^3 \left (a+c x^4\right )}+\frac {7 c^{3/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{11/4}}-\frac {7 c^{3/4} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{11/4}}+\frac {7 c^{3/4} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{11/4}}-\frac {7 c^{3/4} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{11/4}} \]

[Out]

-7/12/a^2/x^3+1/4/a/x^3/(c*x^4+a)-7/16*c^(3/4)*arctan(-1+c^(1/4)*x*2^(1/2)/a^(1/4))/a^(11/4)*2^(1/2)-7/16*c^(3
/4)*arctan(1+c^(1/4)*x*2^(1/2)/a^(1/4))/a^(11/4)*2^(1/2)+7/32*c^(3/4)*ln(-a^(1/4)*c^(1/4)*x*2^(1/2)+a^(1/2)+x^
2*c^(1/2))/a^(11/4)*2^(1/2)-7/32*c^(3/4)*ln(a^(1/4)*c^(1/4)*x*2^(1/2)+a^(1/2)+x^2*c^(1/2))/a^(11/4)*2^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {296, 331, 217, 1179, 642, 1176, 631, 210} \begin {gather*} \frac {7 c^{3/4} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{11/4}}-\frac {7 c^{3/4} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{8 \sqrt {2} a^{11/4}}+\frac {7 c^{3/4} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{11/4}}-\frac {7 c^{3/4} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{11/4}}-\frac {7}{12 a^2 x^3}+\frac {1}{4 a x^3 \left (a+c x^4\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + c*x^4)^2),x]

[Out]

-7/(12*a^2*x^3) + 1/(4*a*x^3*(a + c*x^4)) + (7*c^(3/4)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(8*Sqrt[2]*a^(
11/4)) - (7*c^(3/4)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(8*Sqrt[2]*a^(11/4)) + (7*c^(3/4)*Log[Sqrt[a] - S
qrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(16*Sqrt[2]*a^(11/4)) - (7*c^(3/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1
/4)*x + Sqrt[c]*x^2])/(16*Sqrt[2]*a^(11/4))

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 296

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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*} \int \frac {1}{x^4 \left (a+c x^4\right )^2} \, dx &=\frac {1}{4 a x^3 \left (a+c x^4\right )}+\frac {7 \int \frac {1}{x^4 \left (a+c x^4\right )} \, dx}{4 a}\\ &=-\frac {7}{12 a^2 x^3}+\frac {1}{4 a x^3 \left (a+c x^4\right )}-\frac {(7 c) \int \frac {1}{a+c x^4} \, dx}{4 a^2}\\ &=-\frac {7}{12 a^2 x^3}+\frac {1}{4 a x^3 \left (a+c x^4\right )}-\frac {(7 c) \int \frac {\sqrt {a}-\sqrt {c} x^2}{a+c x^4} \, dx}{8 a^{5/2}}-\frac {(7 c) \int \frac {\sqrt {a}+\sqrt {c} x^2}{a+c x^4} \, dx}{8 a^{5/2}}\\ &=-\frac {7}{12 a^2 x^3}+\frac {1}{4 a x^3 \left (a+c x^4\right )}-\frac {\left (7 \sqrt {c}\right ) \int \frac {1}{\frac {\sqrt {a}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+x^2} \, dx}{16 a^{5/2}}-\frac {\left (7 \sqrt {c}\right ) \int \frac {1}{\frac {\sqrt {a}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+x^2} \, dx}{16 a^{5/2}}+\frac {\left (7 c^{3/4}\right ) \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^{11/4}}+\frac {\left (7 c^{3/4}\right ) \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^{11/4}}\\ &=-\frac {7}{12 a^2 x^3}+\frac {1}{4 a x^3 \left (a+c x^4\right )}+\frac {7 c^{3/4} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{11/4}}-\frac {7 c^{3/4} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{11/4}}-\frac {\left (7 c^{3/4}\right ) \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^{11/4}}+\frac {\left (7 c^{3/4}\right ) \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^{11/4}}\\ &=-\frac {7}{12 a^2 x^3}+\frac {1}{4 a x^3 \left (a+c x^4\right )}+\frac {7 c^{3/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{11/4}}-\frac {7 c^{3/4} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{11/4}}+\frac {7 c^{3/4} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{11/4}}-\frac {7 c^{3/4} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{11/4}}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 194, normalized size = 0.91 \begin {gather*} \frac {-\frac {32 a^{3/4}}{x^3}-\frac {24 a^{3/4} c x}{a+c x^4}+42 \sqrt {2} c^{3/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )-42 \sqrt {2} c^{3/4} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+21 \sqrt {2} c^{3/4} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )-21 \sqrt {2} c^{3/4} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{96 a^{11/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a + c*x^4)^2),x]

[Out]

((-32*a^(3/4))/x^3 - (24*a^(3/4)*c*x)/(a + c*x^4) + 42*Sqrt[2]*c^(3/4)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)]
- 42*Sqrt[2]*c^(3/4)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)] + 21*Sqrt[2]*c^(3/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/
4)*c^(1/4)*x + Sqrt[c]*x^2] - 21*Sqrt[2]*c^(3/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(96*a
^(11/4))

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Maple [A]
time = 0.15, size = 130, normalized size = 0.61

method	result	size
risch	\(\frac {-\frac {7 c \,x^{4}}{12 a^{2}}-\frac {1}{3 a}}{x^{3} \left (x^{4} c +a \right )}+\frac {7 \left (\munderset {\textit {\_R} =\RootOf \left (a^{11} \textit {\_Z}^{4}+c^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (-5 \textit {\_R}^{4} a^{11}-4 c^{3}\right ) x -a^{3} c^{2} \textit {\_R} \right )\right )}{16}\)	\(76\)
default	\(-\frac {c \left (\frac {x}{4 x^{4} c +4 a}+\frac {7 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 a}\right )}{a^{2}}-\frac {1}{3 a^{2} x^{3}}\)	\(130\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.51, size = 206, normalized size = 0.96 \begin {gather*} -\frac {7 \, c x^{4} + 4 \, a}{12 \, {\left (a^{2} c x^{7} + a^{3} x^{3}\right )}} - \frac {7 \, {\left (\frac {2 \, \sqrt {2} c \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} c \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} c^{\frac {3}{4}} \log \left (\sqrt {c} x^{2} + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}}} - \frac {\sqrt {2} c^{\frac {3}{4}} \log \left (\sqrt {c} x^{2} - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}}}\right )}}{32 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(7*c*x^4 + 4*a)/(a^2*c*x^7 + a^3*x^3) - 7/32*(2*sqrt(2)*c*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)*c*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)*c^(3/4)*log(sq
rt(c)*x^2 + sqrt(2)*a^(1/4)*c^(1/4)*x + sqrt(a))/a^(3/4) - sqrt(2)*c^(3/4)*log(sqrt(c)*x^2 - sqrt(2)*a^(1/4)*c
^(1/4)*x + sqrt(a))/a^(3/4))/a^2

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Fricas [A]
time = 0.39, size = 218, normalized size = 1.02 \begin {gather*} -\frac {28 \, c x^{4} + 84 \, {\left (a^{2} c x^{7} + a^{3} x^{3}\right )} \left (-\frac {c^{3}}{a^{11}}\right )^{\frac {1}{4}} \arctan \left (-\frac {a^{8} c x \left (-\frac {c^{3}}{a^{11}}\right )^{\frac {3}{4}} - \sqrt {a^{6} \sqrt {-\frac {c^{3}}{a^{11}}} + c^{2} x^{2}} a^{8} \left (-\frac {c^{3}}{a^{11}}\right )^{\frac {3}{4}}}{c^{3}}\right ) + 21 \, {\left (a^{2} c x^{7} + a^{3} x^{3}\right )} \left (-\frac {c^{3}}{a^{11}}\right )^{\frac {1}{4}} \log \left (7 \, a^{3} \left (-\frac {c^{3}}{a^{11}}\right )^{\frac {1}{4}} + 7 \, c x\right ) - 21 \, {\left (a^{2} c x^{7} + a^{3} x^{3}\right )} \left (-\frac {c^{3}}{a^{11}}\right )^{\frac {1}{4}} \log \left (-7 \, a^{3} \left (-\frac {c^{3}}{a^{11}}\right )^{\frac {1}{4}} + 7 \, c x\right ) + 16 \, a}{48 \, {\left (a^{2} c x^{7} + a^{3} x^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(28*c*x^4 + 84*(a^2*c*x^7 + a^3*x^3)*(-c^3/a^11)^(1/4)*arctan(-(a^8*c*x*(-c^3/a^11)^(3/4) - sqrt(a^6*sqr
t(-c^3/a^11) + c^2*x^2)*a^8*(-c^3/a^11)^(3/4))/c^3) + 21*(a^2*c*x^7 + a^3*x^3)*(-c^3/a^11)^(1/4)*log(7*a^3*(-c
^3/a^11)^(1/4) + 7*c*x) - 21*(a^2*c*x^7 + a^3*x^3)*(-c^3/a^11)^(1/4)*log(-7*a^3*(-c^3/a^11)^(1/4) + 7*c*x) + 1
6*a)/(a^2*c*x^7 + a^3*x^3)

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Sympy [A]
time = 0.20, size = 58, normalized size = 0.27 \begin {gather*} \frac {- 4 a - 7 c x^{4}}{12 a^{3} x^{3} + 12 a^{2} c x^{7}} + \operatorname {RootSum} {\left (65536 t^{4} a^{11} + 2401 c^{3}, \left ( t \mapsto t \log {\left (- \frac {16 t a^{3}}{7 c} + x \right )} \right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-4*a - 7*c*x**4)/(12*a**3*x**3 + 12*a**2*c*x**7) + RootSum(65536*_t**4*a**11 + 2401*c**3, Lambda(_t, _t*log(-
16*_t*a**3/(7*c) + x)))

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Giac [A]
time = 0.84, size = 191, normalized size = 0.89 \begin {gather*} -\frac {c x}{4 \, {\left (c x^{4} + a\right )} a^{2}} - \frac {7 \, \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^{3}} - \frac {7 \, \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^{3}} - \frac {7 \, \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^{3}} + \frac {7 \, \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^{3}} - \frac {1}{3 \, a^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.11, size = 73, normalized size = 0.34 \begin {gather*} \frac {7\,{\left (-c\right )}^{3/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{1/4}\,x}{a^{1/4}}\right )}{8\,a^{11/4}}-\frac {\frac {1}{3\,a}+\frac {7\,c\,x^4}{12\,a^2}}{c\,x^7+a\,x^3}+\frac {7\,{\left (-c\right )}^{3/4}\,\mathrm {atanh}\left (\frac {{\left (-c\right )}^{1/4}\,x}{a^{1/4}}\right )}{8\,a^{11/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a + c*x^4)^2),x)

[Out]

(7*(-c)^(3/4)*atan(((-c)^(1/4)*x)/a^(1/4)))/(8*a^(11/4)) - (1/(3*a) + (7*c*x^4)/(12*a^2))/(a*x^3 + c*x^7) + (7
*(-c)^(3/4)*atanh(((-c)^(1/4)*x)/a^(1/4)))/(8*a^(11/4))

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