3.7.84 \(\int \frac {x^2}{(a+c x^4)^3} \, dx\) [684]

Optimal. Leaf size=223 \[ \frac {x^3}{8 a \left (a+c x^4\right )^2}+\frac {5 x^3}{32 a^2 \left (a+c x^4\right )}-\frac {5 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{64 \sqrt {2} a^{9/4} c^{3/4}}+\frac {5 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{64 \sqrt {2} a^{9/4} c^{3/4}}+\frac {5 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{128 \sqrt {2} a^{9/4} c^{3/4}}-\frac {5 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{128 \sqrt {2} a^{9/4} c^{3/4}} \]

[Out]

1/8*x^3/a/(c*x^4+a)^2+5/32*x^3/a^2/(c*x^4+a)+5/128*arctan(-1+c^(1/4)*x*2^(1/2)/a^(1/4))/a^(9/4)/c^(3/4)*2^(1/2
)+5/128*arctan(1+c^(1/4)*x*2^(1/2)/a^(1/4))/a^(9/4)/c^(3/4)*2^(1/2)+5/256*ln(-a^(1/4)*c^(1/4)*x*2^(1/2)+a^(1/2
)+x^2*c^(1/2))/a^(9/4)/c^(3/4)*2^(1/2)-5/256*ln(a^(1/4)*c^(1/4)*x*2^(1/2)+a^(1/2)+x^2*c^(1/2))/a^(9/4)/c^(3/4)
*2^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {296, 303, 1176, 631, 210, 1179, 642} \begin {gather*} -\frac {5 \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{64 \sqrt {2} a^{9/4} c^{3/4}}+\frac {5 \text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{64 \sqrt {2} a^{9/4} c^{3/4}}+\frac {5 \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{128 \sqrt {2} a^{9/4} c^{3/4}}-\frac {5 \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{128 \sqrt {2} a^{9/4} c^{3/4}}+\frac {5 x^3}{32 a^2 \left (a+c x^4\right )}+\frac {x^3}{8 a \left (a+c x^4\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(a + c*x^4)^3,x]

[Out]

x^3/(8*a*(a + c*x^4)^2) + (5*x^3)/(32*a^2*(a + c*x^4)) - (5*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(64*Sqrt[
2]*a^(9/4)*c^(3/4)) + (5*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(64*Sqrt[2]*a^(9/4)*c^(3/4)) + (5*Log[Sqrt[a
] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(128*Sqrt[2]*a^(9/4)*c^(3/4)) - (5*Log[Sqrt[a] + Sqrt[2]*a^(1/4)
*c^(1/4)*x + Sqrt[c]*x^2])/(128*Sqrt[2]*a^(9/4)*c^(3/4))

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 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 {x^2}{\left (a+c x^4\right )^3} \, dx &=\frac {x^3}{8 a \left (a+c x^4\right )^2}+\frac {5 \int \frac {x^2}{\left (a+c x^4\right )^2} \, dx}{8 a}\\ &=\frac {x^3}{8 a \left (a+c x^4\right )^2}+\frac {5 x^3}{32 a^2 \left (a+c x^4\right )}+\frac {5 \int \frac {x^2}{a+c x^4} \, dx}{32 a^2}\\ &=\frac {x^3}{8 a \left (a+c x^4\right )^2}+\frac {5 x^3}{32 a^2 \left (a+c x^4\right )}-\frac {5 \int \frac {\sqrt {a}-\sqrt {c} x^2}{a+c x^4} \, dx}{64 a^2 \sqrt {c}}+\frac {5 \int \frac {\sqrt {a}+\sqrt {c} x^2}{a+c x^4} \, dx}{64 a^2 \sqrt {c}}\\ &=\frac {x^3}{8 a \left (a+c x^4\right )^2}+\frac {5 x^3}{32 a^2 \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}{128 a^2 c}+\frac {5 \int \frac {1}{\frac {\sqrt {a}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+x^2} \, dx}{128 a^2 c}+\frac {5 \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}{128 \sqrt {2} a^{9/4} c^{3/4}}+\frac {5 \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}{128 \sqrt {2} a^{9/4} c^{3/4}}\\ &=\frac {x^3}{8 a \left (a+c x^4\right )^2}+\frac {5 x^3}{32 a^2 \left (a+c x^4\right )}+\frac {5 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{128 \sqrt {2} a^{9/4} c^{3/4}}-\frac {5 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{128 \sqrt {2} a^{9/4} c^{3/4}}+\frac {5 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{64 \sqrt {2} a^{9/4} c^{3/4}}-\frac {5 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{64 \sqrt {2} a^{9/4} c^{3/4}}\\ &=\frac {x^3}{8 a \left (a+c x^4\right )^2}+\frac {5 x^3}{32 a^2 \left (a+c x^4\right )}-\frac {5 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{64 \sqrt {2} a^{9/4} c^{3/4}}+\frac {5 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{64 \sqrt {2} a^{9/4} c^{3/4}}+\frac {5 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{128 \sqrt {2} a^{9/4} c^{3/4}}-\frac {5 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{128 \sqrt {2} a^{9/4} c^{3/4}}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 204, normalized size = 0.91 \begin {gather*} \frac {\frac {32 a^{5/4} x^3}{\left (a+c x^4\right )^2}+\frac {40 \sqrt [4]{a} x^3}{a+c x^4}-\frac {10 \sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{c^{3/4}}+\frac {10 \sqrt {2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{c^{3/4}}+\frac {5 \sqrt {2} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{c^{3/4}}-\frac {5 \sqrt {2} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{c^{3/4}}}{256 a^{9/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + c*x^4)^3,x]

[Out]

((32*a^(5/4)*x^3)/(a + c*x^4)^2 + (40*a^(1/4)*x^3)/(a + c*x^4) - (10*Sqrt[2]*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^
(1/4)])/c^(3/4) + (10*Sqrt[2]*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/c^(3/4) + (5*Sqrt[2]*Log[Sqrt[a] - Sqrt
[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/c^(3/4) - (5*Sqrt[2]*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x
^2])/c^(3/4))/(256*a^(9/4))

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Maple [A]
time = 0.14, size = 146, normalized size = 0.65

method result size
risch \(\frac {\frac {5 c \,x^{7}}{32 a^{2}}+\frac {9 x^{3}}{32 a}}{\left (x^{4} c +a \right )^{2}}+\frac {5 \left (\munderset {\textit {\_R} =\RootOf \left (c \,\textit {\_Z}^{4}+a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}}\right )}{128 a^{2} c}\) \(59\)
default \(\frac {x^{3}}{8 a \left (x^{4} c +a \right )^{2}}+\frac {\frac {5 x^{3}}{32 a \left (x^{4} c +a \right )}+\frac {5 \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{256 a c \left (\frac {a}{c}\right )^{\frac {1}{4}}}}{a}\) \(146\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(c*x^4+a)^3,x,method=_RETURNVERBOSE)

[Out]

1/8*x^3/a/(c*x^4+a)^2+5/8/a*(1/4*x^3/a/(c*x^4+a)+1/32/a/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)))

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Maxima [A]
time = 0.51, size = 214, normalized size = 0.96 \begin {gather*} \frac {5 \, c x^{7} + 9 \, a x^{3}}{32 \, {\left (a^{2} c^{2} x^{8} + 2 \, a^{3} c x^{4} + a^{4}\right )}} + \frac {5 \, {\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 )}}{256 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(c*x^4+a)^3,x, algorithm="maxima")

[Out]

1/32*(5*c*x^7 + 9*a*x^3)/(a^2*c^2*x^8 + 2*a^3*c*x^4 + a^4) + 5/256*(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)*lo
g(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^2

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Fricas [A]
time = 0.36, size = 242, normalized size = 1.09 \begin {gather*} \frac {20 \, c x^{7} + 36 \, a x^{3} - 20 \, {\left (a^{2} c^{2} x^{8} + 2 \, a^{3} c x^{4} + a^{4}\right )} \left (-\frac {1}{a^{9} c^{3}}\right )^{\frac {1}{4}} \arctan \left (-a^{2} c x \left (-\frac {1}{a^{9} c^{3}}\right )^{\frac {1}{4}} + \sqrt {-a^{5} c \sqrt {-\frac {1}{a^{9} c^{3}}} + x^{2}} a^{2} c \left (-\frac {1}{a^{9} c^{3}}\right )^{\frac {1}{4}}\right ) + 5 \, {\left (a^{2} c^{2} x^{8} + 2 \, a^{3} c x^{4} + a^{4}\right )} \left (-\frac {1}{a^{9} c^{3}}\right )^{\frac {1}{4}} \log \left (a^{7} c^{2} \left (-\frac {1}{a^{9} c^{3}}\right )^{\frac {3}{4}} + x\right ) - 5 \, {\left (a^{2} c^{2} x^{8} + 2 \, a^{3} c x^{4} + a^{4}\right )} \left (-\frac {1}{a^{9} c^{3}}\right )^{\frac {1}{4}} \log \left (-a^{7} c^{2} \left (-\frac {1}{a^{9} c^{3}}\right )^{\frac {3}{4}} + x\right )}{128 \, {\left (a^{2} c^{2} x^{8} + 2 \, a^{3} c x^{4} + a^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(c*x^4+a)^3,x, algorithm="fricas")

[Out]

1/128*(20*c*x^7 + 36*a*x^3 - 20*(a^2*c^2*x^8 + 2*a^3*c*x^4 + a^4)*(-1/(a^9*c^3))^(1/4)*arctan(-a^2*c*x*(-1/(a^
9*c^3))^(1/4) + sqrt(-a^5*c*sqrt(-1/(a^9*c^3)) + x^2)*a^2*c*(-1/(a^9*c^3))^(1/4)) + 5*(a^2*c^2*x^8 + 2*a^3*c*x
^4 + a^4)*(-1/(a^9*c^3))^(1/4)*log(a^7*c^2*(-1/(a^9*c^3))^(3/4) + x) - 5*(a^2*c^2*x^8 + 2*a^3*c*x^4 + a^4)*(-1
/(a^9*c^3))^(1/4)*log(-a^7*c^2*(-1/(a^9*c^3))^(3/4) + x))/(a^2*c^2*x^8 + 2*a^3*c*x^4 + a^4)

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Sympy [A]
time = 0.21, size = 71, normalized size = 0.32 \begin {gather*} \frac {9 a x^{3} + 5 c x^{7}}{32 a^{4} + 64 a^{3} c x^{4} + 32 a^{2} c^{2} x^{8}} + \operatorname {RootSum} {\left (268435456 t^{4} a^{9} c^{3} + 625, \left ( t \mapsto t \log {\left (\frac {2097152 t^{3} a^{7} c^{2}}{125} + x \right )} \right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(c*x**4+a)**3,x)

[Out]

(9*a*x**3 + 5*c*x**7)/(32*a**4 + 64*a**3*c*x**4 + 32*a**2*c**2*x**8) + RootSum(268435456*_t**4*a**9*c**3 + 625
, Lambda(_t, _t*log(2097152*_t**3*a**7*c**2/125 + x)))

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Giac [A]
time = 0.76, size = 206, normalized size = 0.92 \begin {gather*} \frac {5 \, c x^{7} + 9 \, a x^{3}}{32 \, {\left (c x^{4} + a\right )}^{2} 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 )}{128 \, a^{3} c^{3}} + \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 )}{128 \, a^{3} c^{3}} - \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 )}{256 \, a^{3} c^{3}} + \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 )}{256 \, a^{3} c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(c*x^4+a)^3,x, algorithm="giac")

[Out]

1/32*(5*c*x^7 + 9*a*x^3)/((c*x^4 + a)^2*a^2) + 5/128*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^3) + 5/128*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^3) - 5/256*sqrt(2)*(a*c^3)^(3/4)*log(x^2 + sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(a^3*c^3)
+ 5/256*sqrt(2)*(a*c^3)^(3/4)*log(x^2 - sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(a^3*c^3)

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Mupad [B]
time = 1.07, size = 82, normalized size = 0.37 \begin {gather*} \frac {\frac {9\,x^3}{32\,a}+\frac {5\,c\,x^7}{32\,a^2}}{a^2+2\,a\,c\,x^4+c^2\,x^8}+\frac {5\,\mathrm {atan}\left (\frac {c^{1/4}\,x}{{\left (-a\right )}^{1/4}}\right )}{64\,{\left (-a\right )}^{9/4}\,c^{3/4}}-\frac {5\,\mathrm {atanh}\left (\frac {c^{1/4}\,x}{{\left (-a\right )}^{1/4}}\right )}{64\,{\left (-a\right )}^{9/4}\,c^{3/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + c*x^4)^3,x)

[Out]

((9*x^3)/(32*a) + (5*c*x^7)/(32*a^2))/(a^2 + c^2*x^8 + 2*a*c*x^4) + (5*atan((c^(1/4)*x)/(-a)^(1/4)))/(64*(-a)^
(9/4)*c^(3/4)) - (5*atanh((c^(1/4)*x)/(-a)^(1/4)))/(64*(-a)^(9/4)*c^(3/4))

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