3.11.91 \(\int \frac {1+2 x^4}{\sqrt [4]{1+x^4} (-2-x^4+x^8)} \, dx\)

Optimal. Leaf size=81 \[ \frac {x}{3 \sqrt [4]{x^4+1}}-\frac {5 \tan ^{-1}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{x^4+1}}\right )}{6\ 2^{3/4} \sqrt [4]{3}}-\frac {5 \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{x^4+1}}\right )}{6\ 2^{3/4} \sqrt [4]{3}} \]

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Rubi [A]  time = 0.08, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {1454, 527, 12, 377, 212, 206, 203} \begin {gather*} \frac {x}{3 \sqrt [4]{x^4+1}}-\frac {5 \tan ^{-1}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{x^4+1}}\right )}{6\ 2^{3/4} \sqrt [4]{3}}-\frac {5 \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{x^4+1}}\right )}{6\ 2^{3/4} \sqrt [4]{3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x^4)/((1 + x^4)^(1/4)*(-2 - x^4 + x^8)),x]

[Out]

x/(3*(1 + x^4)^(1/4)) - (5*ArcTan[((3/2)^(1/4)*x)/(1 + x^4)^(1/4)])/(6*2^(3/4)*3^(1/4)) - (5*ArcTanh[((3/2)^(1
/4)*x)/(1 + x^4)^(1/4)])/(6*2^(3/4)*3^(1/4))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 212

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

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 1454

Int[((d_) + (e_.)*(x_)^(n_))^(q_.)*((f_) + (g_.)*(x_)^(n_))^(r_.)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^
(p_.), x_Symbol] :> Int[(d + e*x^n)^(p + q)*(f + g*x^n)^r*(a/d + (c*x^n)/e)^p, x] /; FreeQ[{a, b, c, d, e, f,
g, n, q, r}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {1+2 x^4}{\sqrt [4]{1+x^4} \left (-2-x^4+x^8\right )} \, dx &=\int \frac {1+2 x^4}{\left (-2+x^4\right ) \left (1+x^4\right )^{5/4}} \, dx\\ &=\frac {x}{3 \sqrt [4]{1+x^4}}+\frac {1}{3} \int \frac {5}{\left (-2+x^4\right ) \sqrt [4]{1+x^4}} \, dx\\ &=\frac {x}{3 \sqrt [4]{1+x^4}}+\frac {5}{3} \int \frac {1}{\left (-2+x^4\right ) \sqrt [4]{1+x^4}} \, dx\\ &=\frac {x}{3 \sqrt [4]{1+x^4}}+\frac {5}{3} \operatorname {Subst}\left (\int \frac {1}{-2+3 x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )\\ &=\frac {x}{3 \sqrt [4]{1+x^4}}-\frac {5 \operatorname {Subst}\left (\int \frac {1}{\sqrt {2}-\sqrt {3} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt {2}}-\frac {5 \operatorname {Subst}\left (\int \frac {1}{\sqrt {2}+\sqrt {3} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt {2}}\\ &=\frac {x}{3 \sqrt [4]{1+x^4}}-\frac {5 \tan ^{-1}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{1+x^4}}\right )}{6\ 2^{3/4} \sqrt [4]{3}}-\frac {5 \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{1+x^4}}\right )}{6\ 2^{3/4} \sqrt [4]{3}}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 102, normalized size = 1.26 \begin {gather*} \frac {x}{3 \sqrt [4]{x^4+1}}-\frac {5 \left (-\log \left (2-\frac {2^{3/4} \sqrt [4]{3} x}{\sqrt [4]{x^4+1}}\right )+\log \left (\frac {2^{3/4} \sqrt [4]{3} x}{\sqrt [4]{x^4+1}}+2\right )+2 \tan ^{-1}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{x^4+1}}\right )\right )}{12\ 2^{3/4} \sqrt [4]{3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x^4)/((1 + x^4)^(1/4)*(-2 - x^4 + x^8)),x]

[Out]

x/(3*(1 + x^4)^(1/4)) - (5*(2*ArcTan[((3/2)^(1/4)*x)/(1 + x^4)^(1/4)] - Log[2 - (2^(3/4)*3^(1/4)*x)/(1 + x^4)^
(1/4)] + Log[2 + (2^(3/4)*3^(1/4)*x)/(1 + x^4)^(1/4)]))/(12*2^(3/4)*3^(1/4))

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IntegrateAlgebraic [A]  time = 0.39, size = 81, normalized size = 1.00 \begin {gather*} \frac {x}{3 \sqrt [4]{1+x^4}}-\frac {5 \tan ^{-1}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{1+x^4}}\right )}{6\ 2^{3/4} \sqrt [4]{3}}-\frac {5 \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{1+x^4}}\right )}{6\ 2^{3/4} \sqrt [4]{3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 + 2*x^4)/((1 + x^4)^(1/4)*(-2 - x^4 + x^8)),x]

[Out]

x/(3*(1 + x^4)^(1/4)) - (5*ArcTan[((3/2)^(1/4)*x)/(1 + x^4)^(1/4)])/(6*2^(3/4)*3^(1/4)) - (5*ArcTanh[((3/2)^(1
/4)*x)/(1 + x^4)^(1/4)])/(6*2^(3/4)*3^(1/4))

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fricas [B]  time = 6.36, size = 245, normalized size = 3.02 \begin {gather*} \frac {20 \cdot 24^{\frac {3}{4}} {\left (x^{4} + 1\right )} \arctan \left (\frac {3 \cdot 24^{\frac {3}{4}} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 12 \cdot 24^{\frac {1}{4}} {\left (x^{4} + 1\right )}^{\frac {3}{4}} x + 6^{\frac {1}{4}} \sqrt {3} {\left (24^{\frac {3}{4}} \sqrt {x^{4} + 1} x^{2} + 24^{\frac {1}{4}} {\left (5 \, x^{4} + 2\right )}\right )}}{6 \, {\left (x^{4} - 2\right )}}\right ) - 5 \cdot 24^{\frac {3}{4}} {\left (x^{4} + 1\right )} \log \left (\frac {24 \, \sqrt {6} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 24 \cdot 24^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} + 24^{\frac {3}{4}} {\left (5 \, x^{4} + 2\right )} + 48 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} - 2}\right ) + 5 \cdot 24^{\frac {3}{4}} {\left (x^{4} + 1\right )} \log \left (\frac {24 \, \sqrt {6} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} - 24 \cdot 24^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} - 24^{\frac {3}{4}} {\left (5 \, x^{4} + 2\right )} + 48 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} - 2}\right ) + 192 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{576 \, {\left (x^{4} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^4+1)/(x^4+1)^(1/4)/(x^8-x^4-2),x, algorithm="fricas")

[Out]

1/576*(20*24^(3/4)*(x^4 + 1)*arctan(1/6*(3*24^(3/4)*(x^4 + 1)^(1/4)*x^3 + 12*24^(1/4)*(x^4 + 1)^(3/4)*x + 6^(1
/4)*sqrt(3)*(24^(3/4)*sqrt(x^4 + 1)*x^2 + 24^(1/4)*(5*x^4 + 2)))/(x^4 - 2)) - 5*24^(3/4)*(x^4 + 1)*log((24*sqr
t(6)*(x^4 + 1)^(1/4)*x^3 + 24*24^(1/4)*sqrt(x^4 + 1)*x^2 + 24^(3/4)*(5*x^4 + 2) + 48*(x^4 + 1)^(3/4)*x)/(x^4 -
2)) + 5*24^(3/4)*(x^4 + 1)*log((24*sqrt(6)*(x^4 + 1)^(1/4)*x^3 - 24*24^(1/4)*sqrt(x^4 + 1)*x^2 - 24^(3/4)*(5*
x^4 + 2) + 48*(x^4 + 1)^(3/4)*x)/(x^4 - 2)) + 192*(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 {2 \, x^{4} + 1}{{\left (x^{8} - x^{4} - 2\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((2*x^4+1)/(x^4+1)^(1/4)/(x^8-x^4-2),x, algorithm="giac")

[Out]

integrate((2*x^4 + 1)/((x^8 - x^4 - 2)*(x^4 + 1)^(1/4)), x)

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maple [C]  time = 2.92, size = 222, normalized size = 2.74

method	result	size
trager	\(\frac {x}{3 \left (x^{4}+1\right )^{\frac {1}{4}}}-\frac {5 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-54\right )^{2}\right ) \ln \left (-\frac {-2 \sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{4}-54\right )^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-54\right )^{2}\right ) x^{2}-6 \left (x^{4}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-54\right )^{2} x^{3}+15 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-54\right )^{2}\right ) x^{4}+36 \left (x^{4}+1\right )^{\frac {3}{4}} x +6 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-54\right )^{2}\right )}{x^{4}-2}\right )}{72}-\frac {5 \RootOf \left (\textit {\_Z}^{4}-54\right ) \ln \left (-\frac {2 \sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{4}-54\right )^{3} x^{2}+6 \left (x^{4}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-54\right )^{2} x^{3}+15 \RootOf \left (\textit {\_Z}^{4}-54\right ) x^{4}+36 \left (x^{4}+1\right )^{\frac {3}{4}} x +6 \RootOf \left (\textit {\_Z}^{4}-54\right )}{x^{4}-2}\right )}{72}\)	\(222\)
risch	\(\frac {x}{3 \left (x^{4}+1\right )^{\frac {1}{4}}}+\frac {5 \RootOf \left (\textit {\_Z}^{4}-54\right ) \ln \left (-\frac {-2 \sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{4}-54\right )^{3} x^{2}+6 \left (x^{4}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-54\right )^{2} x^{3}-15 \RootOf \left (\textit {\_Z}^{4}-54\right ) x^{4}+36 \left (x^{4}+1\right )^{\frac {3}{4}} x -6 \RootOf \left (\textit {\_Z}^{4}-54\right )}{x^{4}-2}\right )}{72}-\frac {5 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-54\right )^{2}\right ) \ln \left (-\frac {-2 \sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{4}-54\right )^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-54\right )^{2}\right ) x^{2}-6 \left (x^{4}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-54\right )^{2} x^{3}+15 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-54\right )^{2}\right ) x^{4}+36 \left (x^{4}+1\right )^{\frac {3}{4}} x +6 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-54\right )^{2}\right )}{x^{4}-2}\right )}{72}\)	\(222\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^4+1)/(x^4+1)^(1/4)/(x^8-x^4-2),x,method=_RETURNVERBOSE)

[Out]

1/3*x/(x^4+1)^(1/4)-5/72*RootOf(_Z^2+RootOf(_Z^4-54)^2)*ln(-(-2*(x^4+1)^(1/2)*RootOf(_Z^4-54)^2*RootOf(_Z^2+Ro
otOf(_Z^4-54)^2)*x^2-6*(x^4+1)^(1/4)*RootOf(_Z^4-54)^2*x^3+15*RootOf(_Z^2+RootOf(_Z^4-54)^2)*x^4+36*(x^4+1)^(3
/4)*x+6*RootOf(_Z^2+RootOf(_Z^4-54)^2))/(x^4-2))-5/72*RootOf(_Z^4-54)*ln(-(2*(x^4+1)^(1/2)*RootOf(_Z^4-54)^3*x
^2+6*(x^4+1)^(1/4)*RootOf(_Z^4-54)^2*x^3+15*RootOf(_Z^4-54)*x^4+36*(x^4+1)^(3/4)*x+6*RootOf(_Z^4-54))/(x^4-2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, x^{4} + 1}{{\left (x^{8} - x^{4} - 2\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((2*x^4+1)/(x^4+1)^(1/4)/(x^8-x^4-2),x, algorithm="maxima")

[Out]

integrate((2*x^4 + 1)/((x^8 - x^4 - 2)*(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 {2\,x^4+1}{{\left (x^4+1\right )}^{1/4}\,\left (-x^8+x^4+2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x^4 + 1)/((x^4 + 1)^(1/4)*(x^4 - x^8 + 2)),x)

[Out]

int(-(2*x^4 + 1)/((x^4 + 1)^(1/4)*(x^4 - x^8 + 2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 x^{4} + 1}{\left (x^{4} - 2\right ) \left (x^{4} + 1\right )^{\frac {5}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**4+1)/(x**4+1)**(1/4)/(x**8-x**4-2),x)

[Out]

Integral((2*x**4 + 1)/((x**4 - 2)*(x**4 + 1)**(5/4)), x)

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