3.9.87 \(\int \frac {x^4}{\sqrt [4]{-1+x^4} (-1+x^8)} \, dx\)

Optimal. Leaf size=67 \[ -\frac {x}{2 \sqrt [4]{x^4-1}}+\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-1}}\right )}{4 \sqrt [4]{2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-1}}\right )}{4 \sqrt [4]{2}} \]

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Rubi [C]  time = 0.08, antiderivative size = 58, normalized size of antiderivative = 0.87, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {1479, 511, 510} \begin {gather*} -\frac {x^5 \sqrt [4]{1-x^4} \, _2F_1\left (\frac {5}{4},\frac {5}{4};\frac {9}{4};\frac {2 x^4}{x^4+1}\right )}{5 \sqrt [4]{x^4-1} \left (x^4+1\right )^{5/4}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Int[x^4/((-1 + x^4)^(1/4)*(-1 + x^8)),x]

[Out]

-1/5*(x^5*(1 - x^4)^(1/4)*Hypergeometric2F1[5/4, 5/4, 9/4, (2*x^4)/(1 + x^4)])/((-1 + x^4)^(1/4)*(1 + x^4)^(5/
4))

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 1479

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))^(q_.)*((a_) + (c_.)*(x_)^(n2_))^(p_.), x_Symbol] :> Int[(f*x)^
m*(d + e*x^n)^(q + p)*(a/d + (c*x^n)/e)^p, x] /; FreeQ[{a, c, d, e, f, q, m, n, q}, x] && EqQ[n2, 2*n] && EqQ[
c*d^2 + a*e^2, 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {x^4}{\sqrt [4]{-1+x^4} \left (-1+x^8\right )} \, dx &=\int \frac {x^4}{\left (-1+x^4\right )^{5/4} \left (1+x^4\right )} \, dx\\ &=-\frac {\sqrt [4]{1-x^4} \int \frac {x^4}{\left (1-x^4\right )^{5/4} \left (1+x^4\right )} \, dx}{\sqrt [4]{-1+x^4}}\\ &=-\frac {x^5 \sqrt [4]{1-x^4} \, _2F_1\left (\frac {5}{4},\frac {5}{4};\frac {9}{4};\frac {2 x^4}{1+x^4}\right )}{5 \sqrt [4]{-1+x^4} \left (1+x^4\right )^{5/4}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 91, normalized size = 1.36 \begin {gather*} \frac {-\log \left (1-\frac {\sqrt [4]{2} x}{\sqrt [4]{1-x^4}}\right )+\log \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1-x^4}}+1\right )+2 \tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1-x^4}}\right )}{8 \sqrt [4]{2}}-\frac {x}{2 \sqrt [4]{x^4-1}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[x^4/((-1 + x^4)^(1/4)*(-1 + x^8)),x]

[Out]

-1/2*x/(-1 + x^4)^(1/4) + (2*ArcTan[(2^(1/4)*x)/(1 - x^4)^(1/4)] - Log[1 - (2^(1/4)*x)/(1 - x^4)^(1/4)] + Log[
1 + (2^(1/4)*x)/(1 - x^4)^(1/4)])/(8*2^(1/4))

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IntegrateAlgebraic [A]  time = 0.32, size = 67, normalized size = 1.00 \begin {gather*} -\frac {x}{2 \sqrt [4]{-1+x^4}}+\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-1+x^4}}\right )}{4 \sqrt [4]{2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-1+x^4}}\right )}{4 \sqrt [4]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^4/((-1 + x^4)^(1/4)*(-1 + x^8)),x]

[Out]

-1/2*x/(-1 + x^4)^(1/4) + ArcTan[(2^(1/4)*x)/(-1 + x^4)^(1/4)]/(4*2^(1/4)) + ArcTanh[(2^(1/4)*x)/(-1 + x^4)^(1
/4)]/(4*2^(1/4))

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fricas [B]  time = 5.25, size = 242, normalized size = 3.61 \begin {gather*} -\frac {4 \cdot 2^{\frac {3}{4}} {\left (x^{4} - 1\right )} \arctan \left (\frac {4 \cdot 2^{\frac {3}{4}} {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 4 \cdot 2^{\frac {1}{4}} {\left (x^{4} - 1\right )}^{\frac {3}{4}} x + 2^{\frac {3}{4}} {\left (2 \cdot 2^{\frac {3}{4}} \sqrt {x^{4} - 1} x^{2} + 2^{\frac {1}{4}} {\left (3 \, x^{4} - 1\right )}\right )}}{2 \, {\left (x^{4} + 1\right )}}\right ) - 2^{\frac {3}{4}} {\left (x^{4} - 1\right )} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} - 1} x^{2} + 2^{\frac {3}{4}} {\left (3 \, x^{4} - 1\right )} + 4 \, {\left (x^{4} - 1\right )}^{\frac {3}{4}} x}{x^{4} + 1}\right ) + 2^{\frac {3}{4}} {\left (x^{4} - 1\right )} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} - 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} - 1} x^{2} - 2^{\frac {3}{4}} {\left (3 \, x^{4} - 1\right )} + 4 \, {\left (x^{4} - 1\right )}^{\frac {3}{4}} x}{x^{4} + 1}\right ) + 16 \, {\left (x^{4} - 1\right )}^{\frac {3}{4}} x}{32 \, {\left (x^{4} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(x^4-1)^(1/4)/(x^8-1),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4}}{{\left (x^{8} - 1\right )} {\left (x^{4} - 1\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(x^4-1)^(1/4)/(x^8-1),x, algorithm="giac")

[Out]

integrate(x^4/((x^8 - 1)*(x^4 - 1)^(1/4)), x)

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maple [C]  time = 2.08, size = 217, normalized size = 3.24

method	result	size
trager	\(-\frac {x}{2 \left (x^{4}-1\right )^{\frac {1}{4}}}+\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (-\frac {\sqrt {x^{4}-1}\, \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{2}+2 \left (x^{4}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{3}-3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{4}-4 \left (x^{4}-1\right )^{\frac {3}{4}} x +\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right )}{x^{4}+1}\right )}{16}+\frac {\RootOf \left (\textit {\_Z}^{4}-8\right ) \ln \left (\frac {\sqrt {x^{4}-1}\, \RootOf \left (\textit {\_Z}^{4}-8\right )^{3} x^{2}+2 \left (x^{4}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{3}+3 x^{4} \RootOf \left (\textit {\_Z}^{4}-8\right )+4 \left (x^{4}-1\right )^{\frac {3}{4}} x -\RootOf \left (\textit {\_Z}^{4}-8\right )}{x^{4}+1}\right )}{16}\)	\(217\)
risch	\(-\frac {x}{2 \left (x^{4}-1\right )^{\frac {1}{4}}}+\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (-\frac {\sqrt {x^{4}-1}\, \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{2}+2 \left (x^{4}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{3}-3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{4}-4 \left (x^{4}-1\right )^{\frac {3}{4}} x +\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right )}{x^{4}+1}\right )}{16}+\frac {\RootOf \left (\textit {\_Z}^{4}-8\right ) \ln \left (\frac {\sqrt {x^{4}-1}\, \RootOf \left (\textit {\_Z}^{4}-8\right )^{3} x^{2}+2 \left (x^{4}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{3}+3 x^{4} \RootOf \left (\textit {\_Z}^{4}-8\right )+4 \left (x^{4}-1\right )^{\frac {3}{4}} x -\RootOf \left (\textit {\_Z}^{4}-8\right )}{x^{4}+1}\right )}{16}\)	\(217\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(x^4-1)^(1/4)/(x^8-1),x,method=_RETURNVERBOSE)

[Out]

-1/2*x/(x^4-1)^(1/4)+1/16*RootOf(_Z^2+RootOf(_Z^4-8)^2)*ln(-((x^4-1)^(1/2)*RootOf(_Z^2+RootOf(_Z^4-8)^2)*RootO
f(_Z^4-8)^2*x^2+2*(x^4-1)^(1/4)*RootOf(_Z^4-8)^2*x^3-3*RootOf(_Z^2+RootOf(_Z^4-8)^2)*x^4-4*(x^4-1)^(3/4)*x+Roo
tOf(_Z^2+RootOf(_Z^4-8)^2))/(x^4+1))+1/16*RootOf(_Z^4-8)*ln(((x^4-1)^(1/2)*RootOf(_Z^4-8)^3*x^2+2*(x^4-1)^(1/4
)*RootOf(_Z^4-8)^2*x^3+3*x^4*RootOf(_Z^4-8)+4*(x^4-1)^(3/4)*x-RootOf(_Z^4-8))/(x^4+1))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4}}{{\left (x^{8} - 1\right )} {\left (x^{4} - 1\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(x^4-1)^(1/4)/(x^8-1),x, algorithm="maxima")

[Out]

integrate(x^4/((x^8 - 1)*(x^4 - 1)^(1/4)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^4}{{\left (x^4-1\right )}^{1/4}\,\left (x^8-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/((x^4 - 1)^(1/4)*(x^8 - 1)),x)

[Out]

int(x^4/((x^4 - 1)^(1/4)*(x^8 - 1)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4}}{\sqrt [4]{\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )} \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \left (x^{4} + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(x**4-1)**(1/4)/(x**8-1),x)

[Out]

Integral(x**4/(((x - 1)*(x + 1)*(x**2 + 1))**(1/4)*(x - 1)*(x + 1)*(x**2 + 1)*(x**4 + 1)), x)

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