3.7.91 \(\int \frac {(-1+x^4) \sqrt [4]{1+x^4}}{x^2} \, dx\)

Optimal. Leaf size=54 \[ \frac {\sqrt [4]{x^4+1} \left (x^4+4\right )}{4 x}+\frac {3}{8} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {3}{8} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 62, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {453, 279, 331, 298, 203, 206} \begin {gather*} \frac {\left (x^4+1\right )^{5/4}}{x}+\frac {3}{8} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {3}{8} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {3}{4} \sqrt [4]{x^4+1} x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-1 + x^4)*(1 + x^4)^(1/4))/x^2,x]

[Out]

(-3*x^3*(1 + x^4)^(1/4))/4 + (1 + x^4)^(5/4)/x + (3*ArcTan[x/(1 + x^4)^(1/4)])/8 - (3*ArcTanh[x/(1 + x^4)^(1/4
)])/8

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 279

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

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 453

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

Rubi steps

\begin {align*} \int \frac {\left (-1+x^4\right ) \sqrt [4]{1+x^4}}{x^2} \, dx &=\frac {\left (1+x^4\right )^{5/4}}{x}-3 \int x^2 \sqrt [4]{1+x^4} \, dx\\ &=-\frac {3}{4} x^3 \sqrt [4]{1+x^4}+\frac {\left (1+x^4\right )^{5/4}}{x}-\frac {3}{4} \int \frac {x^2}{\left (1+x^4\right )^{3/4}} \, dx\\ &=-\frac {3}{4} x^3 \sqrt [4]{1+x^4}+\frac {\left (1+x^4\right )^{5/4}}{x}-\frac {3}{4} \operatorname {Subst}\left (\int \frac {x^2}{1-x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )\\ &=-\frac {3}{4} x^3 \sqrt [4]{1+x^4}+\frac {\left (1+x^4\right )^{5/4}}{x}-\frac {3}{8} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {3}{8} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )\\ &=-\frac {3}{4} x^3 \sqrt [4]{1+x^4}+\frac {\left (1+x^4\right )^{5/4}}{x}+\frac {3}{8} \tan ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {3}{8} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 34, normalized size = 0.63 \begin {gather*} \frac {\left (x^4+1\right )^{5/4}}{x}-x^3 \, _2F_1\left (-\frac {1}{4},\frac {3}{4};\frac {7}{4};-x^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-1 + x^4)*(1 + x^4)^(1/4))/x^2,x]

[Out]

(1 + x^4)^(5/4)/x - x^3*Hypergeometric2F1[-1/4, 3/4, 7/4, -x^4]

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IntegrateAlgebraic [A]  time = 0.17, size = 54, normalized size = 1.00 \begin {gather*} \frac {\sqrt [4]{1+x^4} \left (4+x^4\right )}{4 x}+\frac {3}{8} \tan ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {3}{8} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-1 + x^4)*(1 + x^4)^(1/4))/x^2,x]

[Out]

((1 + x^4)^(1/4)*(4 + x^4))/(4*x) + (3*ArcTan[x/(1 + x^4)^(1/4)])/8 - (3*ArcTanh[x/(1 + x^4)^(1/4)])/8

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fricas [B]  time = 2.58, size = 92, normalized size = 1.70 \begin {gather*} \frac {3 \, x \arctan \left (2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 2 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x\right ) + 3 \, x \log \left (-2 \, x^{4} + 2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} - 2 \, \sqrt {x^{4} + 1} x^{2} + 2 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x - 1\right ) + 4 \, {\left (x^{4} + 4\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{16 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(3*x*arctan(2*(x^4 + 1)^(1/4)*x^3 + 2*(x^4 + 1)^(3/4)*x) + 3*x*log(-2*x^4 + 2*(x^4 + 1)^(1/4)*x^3 - 2*sqr
t(x^4 + 1)*x^2 + 2*(x^4 + 1)^(3/4)*x - 1) + 4*(x^4 + 4)*(x^4 + 1)^(1/4))/x

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giac [A]  time = 0.18, size = 70, normalized size = 1.30 \begin {gather*} \frac {1}{4} \, {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + \frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x} - \frac {3}{8} \, \arctan \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {3}{16} \, \log \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x} + 1\right ) + \frac {3}{16} \, \log \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(x^4 + 1)^(1/4)*x^3 + (x^4 + 1)^(1/4)/x - 3/8*arctan((x^4 + 1)^(1/4)/x) - 3/16*log((x^4 + 1)^(1/4)/x + 1)
+ 3/16*log((x^4 + 1)^(1/4)/x - 1)

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maple [C]  time = 4.21, size = 33, normalized size = 0.61

method	result	size
meijerg	\(\frac {\hypergeom \left (\left [-\frac {1}{4}, -\frac {1}{4}\right ], \left [\frac {3}{4}\right ], -x^{4}\right )}{x}+\frac {x^{3} \hypergeom \left (\left [-\frac {1}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], -x^{4}\right )}{3}\)	\(33\)
risch	\(\frac {x^{8}+5 x^{4}+4}{4 x \left (x^{4}+1\right )^{\frac {3}{4}}}-\frac {x^{3} \hypergeom \left (\left [\frac {3}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], -x^{4}\right )}{4}\)	\(40\)
trager	\(\frac {\left (x^{4}+1\right )^{\frac {1}{4}} \left (x^{4}+4\right )}{4 x}-\frac {3 \ln \left (-2 \left (x^{4}+1\right )^{\frac {3}{4}} x -2 x^{2} \sqrt {x^{4}+1}-2 x^{3} \left (x^{4}+1\right )^{\frac {1}{4}}-2 x^{4}-1\right )}{16}+\frac {3 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{4}+1}\, x^{2}-2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{4}-2 \left (x^{4}+1\right )^{\frac {3}{4}} x +2 x^{3} \left (x^{4}+1\right )^{\frac {1}{4}}-\RootOf \left (\textit {\_Z}^{2}+1\right )\right )}{16}\)	\(132\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/x*hypergeom([-1/4,-1/4],[3/4],-x^4)+1/3*x^3*hypergeom([-1/4,3/4],[7/4],-x^4)

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maxima [A]  time = 0.42, size = 83, normalized size = 1.54 \begin {gather*} \frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x} + \frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{4 \, x {\left (\frac {x^{4} + 1}{x^{4}} - 1\right )}} - \frac {3}{8} \, \arctan \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {3}{16} \, \log \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x} + 1\right ) + \frac {3}{16} \, \log \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^4 + 1)^(1/4)/x + 1/4*(x^4 + 1)^(1/4)/(x*((x^4 + 1)/x^4 - 1)) - 3/8*arctan((x^4 + 1)^(1/4)/x) - 3/16*log((x^
4 + 1)^(1/4)/x + 1) + 3/16*log((x^4 + 1)^(1/4)/x - 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\left (x^4-1\right )\,{\left (x^4+1\right )}^{1/4}}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^4 - 1)*(x^4 + 1)^(1/4))/x^2,x)

[Out]

int(((x^4 - 1)*(x^4 + 1)^(1/4))/x^2, x)

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sympy [C]  time = 2.46, size = 65, normalized size = 1.20 \begin {gather*} \frac {x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} - \frac {\Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**3*gamma(3/4)*hyper((-1/4, 3/4), (7/4,), x**4*exp_polar(I*pi))/(4*gamma(7/4)) - gamma(-1/4)*hyper((-1/4, -1/
4), (3/4,), x**4*exp_polar(I*pi))/(4*x*gamma(3/4))

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