3.666 \(\int \frac {x^6}{(a+c x^4)^2} \, dx\)

Optimal. Leaf size=204 \[ \frac {3 \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} \sqrt [4]{a} c^{7/4}}-\frac {3 \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} \sqrt [4]{a} c^{7/4}}-\frac {3 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} \sqrt [4]{a} c^{7/4}}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{8 \sqrt {2} \sqrt [4]{a} c^{7/4}}-\frac {x^3}{4 c \left (a+c x^4\right )} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 204, 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, 297, 1162, 617, 204, 1165, 628} \[ \frac {3 \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} \sqrt [4]{a} c^{7/4}}-\frac {3 \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} \sqrt [4]{a} c^{7/4}}-\frac {3 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} \sqrt [4]{a} c^{7/4}}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{8 \sqrt {2} \sqrt [4]{a} c^{7/4}}-\frac {x^3}{4 c \left (a+c x^4\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^6/(a + c*x^4)^2,x]

[Out]

-x^3/(4*c*(a + c*x^4)) - (3*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(8*Sqrt[2]*a^(1/4)*c^(7/4)) + (3*ArcTan[1
+ (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(8*Sqrt[2]*a^(1/4)*c^(7/4)) + (3*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sq
rt[c]*x^2])/(16*Sqrt[2]*a^(1/4)*c^(7/4)) - (3*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(16*Sqrt
[2]*a^(1/4)*c^(7/4))

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 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 297

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 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^6}{\left (a+c x^4\right )^2} \, dx &=-\frac {x^3}{4 c \left (a+c x^4\right )}+\frac {3 \int \frac {x^2}{a+c x^4} \, dx}{4 c}\\ &=-\frac {x^3}{4 c \left (a+c x^4\right )}-\frac {3 \int \frac {\sqrt {a}-\sqrt {c} x^2}{a+c x^4} \, dx}{8 c^{3/2}}+\frac {3 \int \frac {\sqrt {a}+\sqrt {c} x^2}{a+c x^4} \, dx}{8 c^{3/2}}\\ &=-\frac {x^3}{4 c \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 c^2}+\frac {3 \int \frac {1}{\frac {\sqrt {a}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+x^2} \, dx}{16 c^2}+\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} \sqrt [4]{a} c^{7/4}}+\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} \sqrt [4]{a} c^{7/4}}\\ &=-\frac {x^3}{4 c \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} \sqrt [4]{a} c^{7/4}}-\frac {3 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} \sqrt [4]{a} c^{7/4}}+\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} \sqrt [4]{a} c^{7/4}}-\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} \sqrt [4]{a} c^{7/4}}\\ &=-\frac {x^3}{4 c \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} \sqrt [4]{a} c^{7/4}}+\frac {3 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} \sqrt [4]{a} c^{7/4}}+\frac {3 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} \sqrt [4]{a} c^{7/4}}-\frac {3 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} \sqrt [4]{a} c^{7/4}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 185, normalized size = 0.91 \[ \frac {-\frac {8 c^{3/4} x^3}{a+c x^4}+\frac {3 \sqrt {2} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{\sqrt [4]{a}}-\frac {3 \sqrt {2} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{\sqrt [4]{a}}-\frac {6 \sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{\sqrt [4]{a}}+\frac {6 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{\sqrt [4]{a}}}{32 c^{7/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(a + c*x^4)^2,x]

[Out]

((-8*c^(3/4)*x^3)/(a + c*x^4) - (6*Sqrt[2]*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/a^(1/4) + (6*Sqrt[2]*ArcTa
n[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/a^(1/4) + (3*Sqrt[2]*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2
])/a^(1/4) - (3*Sqrt[2]*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/a^(1/4))/(32*c^(7/4))

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fricas [A]  time = 0.73, size = 181, normalized size = 0.89 \[ -\frac {4 \, x^{3} + 12 \, {\left (c^{2} x^{4} + a c\right )} \left (-\frac {1}{a c^{7}}\right )^{\frac {1}{4}} \arctan \left (-c^{2} x \left (-\frac {1}{a c^{7}}\right )^{\frac {1}{4}} + \sqrt {-a c^{3} \sqrt {-\frac {1}{a c^{7}}} + x^{2}} c^{2} \left (-\frac {1}{a c^{7}}\right )^{\frac {1}{4}}\right ) - 3 \, {\left (c^{2} x^{4} + a c\right )} \left (-\frac {1}{a c^{7}}\right )^{\frac {1}{4}} \log \left (a c^{5} \left (-\frac {1}{a c^{7}}\right )^{\frac {3}{4}} + x\right ) + 3 \, {\left (c^{2} x^{4} + a c\right )} \left (-\frac {1}{a c^{7}}\right )^{\frac {1}{4}} \log \left (-a c^{5} \left (-\frac {1}{a c^{7}}\right )^{\frac {3}{4}} + x\right )}{16 \, {\left (c^{2} x^{4} + a c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(c*x^4+a)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 196, normalized size = 0.96 \[ -\frac {x^{3}}{4 \, {\left (c x^{4} + a\right )} c} + \frac {3 \, \sqrt {2} \left (a c^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{16 \, a c^{4}} + \frac {3 \, \sqrt {2} \left (a c^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{16 \, a c^{4}} - \frac {3 \, \sqrt {2} \left (a c^{3}\right )^{\frac {3}{4}} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{32 \, a c^{4}} + \frac {3 \, \sqrt {2} \left (a c^{3}\right )^{\frac {3}{4}} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{32 \, a c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(c*x^4+a)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 145, normalized size = 0.71 \[ -\frac {x^{3}}{4 \left (c \,x^{4}+a \right ) c}+\frac {3 \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{16 \left (\frac {a}{c}\right )^{\frac {1}{4}} c^{2}}+\frac {3 \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{16 \left (\frac {a}{c}\right )^{\frac {1}{4}} c^{2}}+\frac {3 \sqrt {2}\, \ln \left (\frac {x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{c}}}{x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{c}}}\right )}{32 \left (\frac {a}{c}\right )^{\frac {1}{4}} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(c*x^4+a)^2,x)

[Out]

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

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maxima [A]  time = 3.01, size = 192, normalized size = 0.94 \[ -\frac {x^{3}}{4 \, {\left (c^{2} x^{4} + a c\right )}} + \frac {3 \, {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {\sqrt {a} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {\sqrt {a} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {c} x^{2} + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {c} x^{2} - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{32 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(c*x^4+a)^2,x, algorithm="maxima")

[Out]

-1/4*x^3/(c^2*x^4 + a*c) + 3/32*(2*sqrt(2)*arctan(1/2*sqrt(2)*(2*sqrt(c)*x + sqrt(2)*a^(1/4)*c^(1/4))/sqrt(sqr
t(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 + 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)))/c

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mupad [B]  time = 1.08, size = 60, normalized size = 0.29 \[ \frac {3\,\mathrm {atan}\left (\frac {c^{1/4}\,x}{{\left (-a\right )}^{1/4}}\right )}{8\,{\left (-a\right )}^{1/4}\,c^{7/4}}-\frac {3\,\mathrm {atanh}\left (\frac {c^{1/4}\,x}{{\left (-a\right )}^{1/4}}\right )}{8\,{\left (-a\right )}^{1/4}\,c^{7/4}}-\frac {x^3}{4\,c\,\left (c\,x^4+a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(a + c*x^4)^2,x)

[Out]

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

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sympy [A]  time = 0.37, size = 44, normalized size = 0.22 \[ - \frac {x^{3}}{4 a c + 4 c^{2} x^{4}} + \operatorname {RootSum} {\left (65536 t^{4} a c^{7} + 81, \left (t \mapsto t \log {\left (\frac {4096 t^{3} a c^{5}}{27} + x \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(c*x**4+a)**2,x)

[Out]

-x**3/(4*a*c + 4*c**2*x**4) + RootSum(65536*_t**4*a*c**7 + 81, Lambda(_t, _t*log(4096*_t**3*a*c**5/27 + x)))

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