3.7.70 \(\int \frac {1}{x^2 (a+c x^4)^2} \, dx\) [670]

Optimal. Leaf size=214 \[ -\frac {5}{4 a^2 x}+\frac {1}{4 a x \left (a+c x^4\right )}+\frac {5 \sqrt [4]{c} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{9/4}}-\frac {5 \sqrt [4]{c} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{9/4}}-\frac {5 \sqrt [4]{c} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{9/4}}+\frac {5 \sqrt [4]{c} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{9/4}} \]

[Out]

-5/4/a^2/x+1/4/a/x/(c*x^4+a)-5/16*c^(1/4)*arctan(-1+c^(1/4)*x*2^(1/2)/a^(1/4))/a^(9/4)*2^(1/2)-5/16*c^(1/4)*ar
ctan(1+c^(1/4)*x*2^(1/2)/a^(1/4))/a^(9/4)*2^(1/2)-5/32*c^(1/4)*ln(-a^(1/4)*c^(1/4)*x*2^(1/2)+a^(1/2)+x^2*c^(1/
2))/a^(9/4)*2^(1/2)+5/32*c^(1/4)*ln(a^(1/4)*c^(1/4)*x*2^(1/2)+a^(1/2)+x^2*c^(1/2))/a^(9/4)*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, 303, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {5 \sqrt [4]{c} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{9/4}}-\frac {5 \sqrt [4]{c} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{8 \sqrt {2} a^{9/4}}-\frac {5 \sqrt [4]{c} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{9/4}}+\frac {5 \sqrt [4]{c} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{9/4}}-\frac {5}{4 a^2 x}+\frac {1}{4 a x \left (a+c x^4\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + c*x^4)^2),x]

[Out]

-5/(4*a^2*x) + 1/(4*a*x*(a + c*x^4)) + (5*c^(1/4)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(8*Sqrt[2]*a^(9/4))
- (5*c^(1/4)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(8*Sqrt[2]*a^(9/4)) - (5*c^(1/4)*Log[Sqrt[a] - Sqrt[2]*
a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(16*Sqrt[2]*a^(9/4)) + (5*c^(1/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x +
Sqrt[c]*x^2])/(16*Sqrt[2]*a^(9/4))

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 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 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*} \int \frac {1}{x^2 \left (a+c x^4\right )^2} \, dx &=\frac {1}{4 a x \left (a+c x^4\right )}+\frac {5 \int \frac {1}{x^2 \left (a+c x^4\right )} \, dx}{4 a}\\ &=-\frac {5}{4 a^2 x}+\frac {1}{4 a x \left (a+c x^4\right )}-\frac {(5 c) \int \frac {x^2}{a+c x^4} \, dx}{4 a^2}\\ &=-\frac {5}{4 a^2 x}+\frac {1}{4 a x \left (a+c x^4\right )}+\frac {\left (5 \sqrt {c}\right ) \int \frac {\sqrt {a}-\sqrt {c} x^2}{a+c x^4} \, dx}{8 a^2}-\frac {\left (5 \sqrt {c}\right ) \int \frac {\sqrt {a}+\sqrt {c} x^2}{a+c x^4} \, dx}{8 a^2}\\ &=-\frac {5}{4 a^2 x}+\frac {1}{4 a x \left (a+c x^4\right )}-\frac {5 \int \frac {1}{\frac {\sqrt {a}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+x^2} \, dx}{16 a^2}-\frac {5 \int \frac {1}{\frac {\sqrt {a}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+x^2} \, dx}{16 a^2}-\frac {\left (5 \sqrt [4]{c}\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^{9/4}}-\frac {\left (5 \sqrt [4]{c}\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^{9/4}}\\ &=-\frac {5}{4 a^2 x}+\frac {1}{4 a x \left (a+c x^4\right )}-\frac {5 \sqrt [4]{c} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{9/4}}+\frac {5 \sqrt [4]{c} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{9/4}}-\frac {\left (5 \sqrt [4]{c}\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^{9/4}}+\frac {\left (5 \sqrt [4]{c}\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^{9/4}}\\ &=-\frac {5}{4 a^2 x}+\frac {1}{4 a x \left (a+c x^4\right )}+\frac {5 \sqrt [4]{c} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{9/4}}-\frac {5 \sqrt [4]{c} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{9/4}}-\frac {5 \sqrt [4]{c} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{9/4}}+\frac {5 \sqrt [4]{c} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{9/4}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 196, normalized size = 0.92 \begin {gather*} \frac {-\frac {32 \sqrt [4]{a}}{x}-\frac {8 \sqrt [4]{a} c x^3}{a+c x^4}+10 \sqrt {2} \sqrt [4]{c} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )-10 \sqrt {2} \sqrt [4]{c} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )-5 \sqrt {2} \sqrt [4]{c} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )+5 \sqrt {2} \sqrt [4]{c} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{32 a^{9/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a + c*x^4)^2),x]

[Out]

((-32*a^(1/4))/x - (8*a^(1/4)*c*x^3)/(a + c*x^4) + 10*Sqrt[2]*c^(1/4)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)]
- 10*Sqrt[2]*c^(1/4)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)] - 5*Sqrt[2]*c^(1/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)
*c^(1/4)*x + Sqrt[c]*x^2] + 5*Sqrt[2]*c^(1/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(32*a^(9
/4))

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Maple [A]
time = 0.14, size = 132, normalized size = 0.62

method	result	size
risch	\(\frac {-\frac {5 c \,x^{4}}{4 a^{2}}-\frac {1}{a}}{x \left (x^{4} c +a \right )}+\frac {5 \left (\munderset {\textit {\_R} =\RootOf \left (a^{9} \textit {\_Z}^{4}+c \right )}{\sum }\textit {\_R} \ln \left (\left (5 \textit {\_R}^{4} a^{9}+4 c \right ) x +a^{7} \textit {\_R}^{3}\right )\right )}{16}\)	\(70\)
default	\(-\frac {c \left (\frac {x^{3}}{4 x^{4} c +4 a}+\frac {5 \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 c \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{a^{2}}-\frac {1}{a^{2} x}\)	\(132\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 203, normalized size = 0.95 \begin {gather*} -\frac {5 \, c x^{4} + 4 \, a}{4 \, {\left (a^{2} c x^{5} + a^{3} x\right )}} - \frac {5 \, 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 )}}{32 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(5*c*x^4 + 4*a)/(a^2*c*x^5 + a^3*x) - 5/32*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 + s
qrt(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^2

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Fricas [A]
time = 0.40, size = 198, normalized size = 0.93 \begin {gather*} -\frac {20 \, c x^{4} - 20 \, {\left (a^{2} c x^{5} + a^{3} x\right )} \left (-\frac {c}{a^{9}}\right )^{\frac {1}{4}} \arctan \left (-\frac {125 \, a^{2} c x \left (-\frac {c}{a^{9}}\right )^{\frac {1}{4}} - \sqrt {-15625 \, a^{5} c \sqrt {-\frac {c}{a^{9}}} + 15625 \, c^{2} x^{2}} a^{2} \left (-\frac {c}{a^{9}}\right )^{\frac {1}{4}}}{125 \, c}\right ) + 5 \, {\left (a^{2} c x^{5} + a^{3} x\right )} \left (-\frac {c}{a^{9}}\right )^{\frac {1}{4}} \log \left (125 \, a^{7} \left (-\frac {c}{a^{9}}\right )^{\frac {3}{4}} + 125 \, c x\right ) - 5 \, {\left (a^{2} c x^{5} + a^{3} x\right )} \left (-\frac {c}{a^{9}}\right )^{\frac {1}{4}} \log \left (-125 \, a^{7} \left (-\frac {c}{a^{9}}\right )^{\frac {3}{4}} + 125 \, c x\right ) + 16 \, a}{16 \, {\left (a^{2} c x^{5} + a^{3} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(20*c*x^4 - 20*(a^2*c*x^5 + a^3*x)*(-c/a^9)^(1/4)*arctan(-1/125*(125*a^2*c*x*(-c/a^9)^(1/4) - sqrt(-1562
5*a^5*c*sqrt(-c/a^9) + 15625*c^2*x^2)*a^2*(-c/a^9)^(1/4))/c) + 5*(a^2*c*x^5 + a^3*x)*(-c/a^9)^(1/4)*log(125*a^
7*(-c/a^9)^(3/4) + 125*c*x) - 5*(a^2*c*x^5 + a^3*x)*(-c/a^9)^(1/4)*log(-125*a^7*(-c/a^9)^(3/4) + 125*c*x) + 16
*a)/(a^2*c*x^5 + a^3*x)

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Sympy [A]
time = 0.17, size = 56, normalized size = 0.26 \begin {gather*} \frac {- 4 a - 5 c x^{4}}{4 a^{3} x + 4 a^{2} c x^{5}} + \operatorname {RootSum} {\left (65536 t^{4} a^{9} + 625 c, \left ( t \mapsto t \log {\left (- \frac {4096 t^{3} a^{7}}{125 c} + x \right )} \right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-4*a - 5*c*x**4)/(4*a**3*x + 4*a**2*c*x**5) + RootSum(65536*_t**4*a**9 + 625*c, Lambda(_t, _t*log(-4096*_t**3
*a**7/(125*c) + x)))

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Giac [A]
time = 1.34, size = 205, normalized size = 0.96 \begin {gather*} -\frac {5 \, c x^{4} + 4 \, a}{4 \, {\left (c x^{5} + a x\right )} a^{2}} - \frac {5 \, \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 )}{16 \, a^{3} c^{2}} - \frac {5 \, \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 )}{16 \, a^{3} c^{2}} + \frac {5 \, \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 )}{32 \, a^{3} c^{2}} - \frac {5 \, \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 )}{32 \, a^{3} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.03, size = 69, normalized size = 0.32 \begin {gather*} \frac {5\,{\left (-c\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (-c\right )}^{1/4}\,x}{a^{1/4}}\right )}{8\,a^{9/4}}-\frac {5\,{\left (-c\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{1/4}\,x}{a^{1/4}}\right )}{8\,a^{9/4}}-\frac {\frac {1}{a}+\frac {5\,c\,x^4}{4\,a^2}}{c\,x^5+a\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a + c*x^4)^2),x)

[Out]

(5*(-c)^(1/4)*atanh(((-c)^(1/4)*x)/a^(1/4)))/(8*a^(9/4)) - (5*(-c)^(1/4)*atan(((-c)^(1/4)*x)/a^(1/4)))/(8*a^(9
/4)) - (1/a + (5*c*x^4)/(4*a^2))/(a*x + c*x^5)

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