3.15.20 \(\int \frac {1+x^2+2 x^4}{\sqrt [4]{1+x^2} (2+3 x^2+x^4)} \, dx\)

Optimal. Leaf size=101 \[ \frac {4 x}{\sqrt [4]{x^2+1}}+\frac {7}{4} \tan ^{-1}\left (\frac {\sqrt [4]{x^2+1}-x}{\sqrt [4]{x^2+1}}\right )-\frac {7}{4} \tan ^{-1}\left (\frac {\sqrt [4]{x^2+1}+x}{\sqrt [4]{x^2+1}}\right )-\frac {7}{4} \tanh ^{-1}\left (\frac {2 x \sqrt [4]{x^2+1}}{x^2+2 \sqrt {x^2+1}}\right ) \]

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Rubi [A]  time = 0.27, antiderivative size = 73, normalized size of antiderivative = 0.72, number of steps used = 9, number of rules used = 6, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {1688, 6725, 196, 285, 403, 397} \begin {gather*} \frac {4 x}{\sqrt [4]{x^2+1}}+\frac {7}{2} \tan ^{-1}\left (\frac {\sqrt {x^2+1}+1}{x \sqrt [4]{x^2+1}}\right )+\frac {7}{2} \tanh ^{-1}\left (\frac {1-\sqrt {x^2+1}}{x \sqrt [4]{x^2+1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x^2 + 2*x^4)/((1 + x^2)^(1/4)*(2 + 3*x^2 + x^4)),x]

[Out]

(4*x)/(1 + x^2)^(1/4) + (7*ArcTan[(1 + Sqrt[1 + x^2])/(x*(1 + x^2)^(1/4))])/2 + (7*ArcTanh[(1 - Sqrt[1 + x^2])
/(x*(1 + x^2)^(1/4))])/2

Rule 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 285

Int[((c_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[(2*c*(c*x)^(m - 1))/(b*(2*m - 3)*(a + b*x
^2)^(1/4)), x] - Dist[(2*a*c^2*(m - 1))/(b*(2*m - 3)), Int[(c*x)^(m - 2)/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a
, b, c}, x] && PosQ[b/a] && IntegerQ[2*m] && GtQ[m, 3/2]

Rule 397

Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> With[{q = Rt[b^2/a, 4]}, -Simp[(b*ArcT
an[(b + q^2*Sqrt[a + b*x^2])/(q^3*x*(a + b*x^2)^(1/4))])/(2*a*d*q), x] - Simp[(b*ArcTanh[(b - q^2*Sqrt[a + b*x
^2])/(q^3*x*(a + b*x^2)^(1/4))])/(2*a*d*q), x]] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && PosQ[b^2/a
]

Rule 403

Int[((a_) + (b_.)*(x_)^2)^(p_)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/(b*c - a*d), Int[(a + b*x^2)^p, x],
x] - Dist[d/(b*c - a*d), Int[(a + b*x^2)^(p + 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*
d, 0] && LtQ[p, -1] && EqQ[Denominator[p], 4] && (EqQ[p, -5/4] || EqQ[p, -7/4])

Rule 1688

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

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {1+x^2+2 x^4}{\sqrt [4]{1+x^2} \left (2+3 x^2+x^4\right )} \, dx &=\int \frac {1+x^2+2 x^4}{\left (1+x^2\right )^{5/4} \left (2+x^2\right )} \, dx\\ &=\int \left (-\frac {3}{\left (1+x^2\right )^{5/4}}+\frac {2 x^2}{\left (1+x^2\right )^{5/4}}+\frac {7}{\left (1+x^2\right )^{5/4} \left (2+x^2\right )}\right ) \, dx\\ &=2 \int \frac {x^2}{\left (1+x^2\right )^{5/4}} \, dx-3 \int \frac {1}{\left (1+x^2\right )^{5/4}} \, dx+7 \int \frac {1}{\left (1+x^2\right )^{5/4} \left (2+x^2\right )} \, dx\\ &=\frac {4 x}{\sqrt [4]{1+x^2}}-6 E\left (\left .\frac {1}{2} \tan ^{-1}(x)\right |2\right )-4 \int \frac {1}{\left (1+x^2\right )^{5/4}} \, dx+7 \int \frac {1}{\left (1+x^2\right )^{5/4}} \, dx-7 \int \frac {1}{\sqrt [4]{1+x^2} \left (2+x^2\right )} \, dx\\ &=\frac {4 x}{\sqrt [4]{1+x^2}}+\frac {7}{2} \tan ^{-1}\left (\frac {1+\sqrt {1+x^2}}{x \sqrt [4]{1+x^2}}\right )+\frac {7}{2} \tanh ^{-1}\left (\frac {1-\sqrt {1+x^2}}{x \sqrt [4]{1+x^2}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.20, size = 127, normalized size = 1.26 \begin {gather*} \frac {2 x \left (\frac {21 F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};-x^2,-\frac {x^2}{2}\right )}{\left (x^2+2\right ) \left (x^2 \left (2 F_1\left (\frac {3}{2};\frac {1}{4},2;\frac {5}{2};-x^2,-\frac {x^2}{2}\right )+F_1\left (\frac {3}{2};\frac {5}{4},1;\frac {5}{2};-x^2,-\frac {x^2}{2}\right )\right )-6 F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};-x^2,-\frac {x^2}{2}\right )\right )}+2\right )}{\sqrt [4]{x^2+1}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(1 + x^2 + 2*x^4)/((1 + x^2)^(1/4)*(2 + 3*x^2 + x^4)),x]

[Out]

(2*x*(2 + (21*AppellF1[1/2, 1/4, 1, 3/2, -x^2, -1/2*x^2])/((2 + x^2)*(-6*AppellF1[1/2, 1/4, 1, 3/2, -x^2, -1/2
*x^2] + x^2*(2*AppellF1[3/2, 1/4, 2, 5/2, -x^2, -1/2*x^2] + AppellF1[3/2, 5/4, 1, 5/2, -x^2, -1/2*x^2])))))/(1
+ x^2)^(1/4)

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IntegrateAlgebraic [A]  time = 0.39, size = 101, normalized size = 1.00 \begin {gather*} \frac {4 x}{\sqrt [4]{1+x^2}}+\frac {7}{4} \tan ^{-1}\left (\frac {-x+\sqrt [4]{1+x^2}}{\sqrt [4]{1+x^2}}\right )-\frac {7}{4} \tan ^{-1}\left (\frac {x+\sqrt [4]{1+x^2}}{\sqrt [4]{1+x^2}}\right )-\frac {7}{4} \tanh ^{-1}\left (\frac {2 x \sqrt [4]{1+x^2}}{x^2+2 \sqrt {1+x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 + x^2 + 2*x^4)/((1 + x^2)^(1/4)*(2 + 3*x^2 + x^4)),x]

[Out]

(4*x)/(1 + x^2)^(1/4) + (7*ArcTan[(-x + (1 + x^2)^(1/4))/(1 + x^2)^(1/4)])/4 - (7*ArcTan[(x + (1 + x^2)^(1/4))
/(1 + x^2)^(1/4)])/4 - (7*ArcTanh[(2*x*(1 + x^2)^(1/4))/(x^2 + 2*Sqrt[1 + x^2])])/4

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fricas [A]  time = 0.46, size = 137, normalized size = 1.36 \begin {gather*} \frac {14 \, {\left (x^{2} + 1\right )} \arctan \left (\frac {x + 2 \, {\left (x^{2} + 1\right )}^{\frac {1}{4}}}{x}\right ) + 14 \, {\left (x^{2} + 1\right )} \arctan \left (-\frac {x - 2 \, {\left (x^{2} + 1\right )}^{\frac {1}{4}}}{x}\right ) - 7 \, {\left (x^{2} + 1\right )} \log \left (\frac {x^{2} + 2 \, {\left (x^{2} + 1\right )}^{\frac {1}{4}} x + 2 \, \sqrt {x^{2} + 1}}{x^{2}}\right ) + 7 \, {\left (x^{2} + 1\right )} \log \left (\frac {x^{2} - 2 \, {\left (x^{2} + 1\right )}^{\frac {1}{4}} x + 2 \, \sqrt {x^{2} + 1}}{x^{2}}\right ) + 32 \, {\left (x^{2} + 1\right )}^{\frac {3}{4}} x}{8 \, {\left (x^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^4+x^2+1)/(x^2+1)^(1/4)/(x^4+3*x^2+2),x, algorithm="fricas")

[Out]

1/8*(14*(x^2 + 1)*arctan((x + 2*(x^2 + 1)^(1/4))/x) + 14*(x^2 + 1)*arctan(-(x - 2*(x^2 + 1)^(1/4))/x) - 7*(x^2
+ 1)*log((x^2 + 2*(x^2 + 1)^(1/4)*x + 2*sqrt(x^2 + 1))/x^2) + 7*(x^2 + 1)*log((x^2 - 2*(x^2 + 1)^(1/4)*x + 2*
sqrt(x^2 + 1))/x^2) + 32*(x^2 + 1)^(3/4)*x)/(x^2 + 1)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, x^{4} + x^{2} + 1}{{\left (x^{4} + 3 \, x^{2} + 2\right )} {\left (x^{2} + 1\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^4+x^2+1)/(x^2+1)^(1/4)/(x^4+3*x^2+2),x, algorithm="giac")

[Out]

integrate((2*x^4 + x^2 + 1)/((x^4 + 3*x^2 + 2)*(x^2 + 1)^(1/4)), x)

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maple [C]  time = 3.05, size = 291, normalized size = 2.88

method	result	size
trager	\(\frac {4 x}{\left (x^{2}+1\right )^{\frac {1}{4}}}+\frac {7 \ln \left (\frac {8 \left (x^{2}+1\right )^{\frac {3}{4}} \RootOf \left (32 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )-2 \left (x^{2}+1\right )^{\frac {3}{4}}+x \sqrt {x^{2}+1}+8 \left (x^{2}+1\right )^{\frac {1}{4}} \RootOf \left (32 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )-8 \RootOf \left (32 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right ) x +x}{x^{2}+2}\right )}{2}-14 \ln \left (\frac {8 \left (x^{2}+1\right )^{\frac {3}{4}} \RootOf \left (32 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )-2 \left (x^{2}+1\right )^{\frac {3}{4}}+x \sqrt {x^{2}+1}+8 \left (x^{2}+1\right )^{\frac {1}{4}} \RootOf \left (32 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )-8 \RootOf \left (32 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right ) x +x}{x^{2}+2}\right ) \RootOf \left (32 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )+14 \RootOf \left (32 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right ) \ln \left (-\frac {8 \left (x^{2}+1\right )^{\frac {3}{4}} \RootOf \left (32 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )-x \sqrt {x^{2}+1}+8 \left (x^{2}+1\right )^{\frac {1}{4}} \RootOf \left (32 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )-8 \RootOf \left (32 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right ) x -2 \left (x^{2}+1\right )^{\frac {1}{4}}+x}{x^{2}+2}\right )\)	\(291\)
risch	\(\frac {4 x}{\left (x^{2}+1\right )^{\frac {1}{4}}}+7 \RootOf \left (8 \textit {\_Z}^{2}+4 \textit {\_Z} +1\right ) \ln \left (-\frac {4 \RootOf \left (8 \textit {\_Z}^{2}+4 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {3}{4}}-x \sqrt {x^{2}+1}+4 \RootOf \left (8 \textit {\_Z}^{2}+4 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {1}{4}}+4 \RootOf \left (8 \textit {\_Z}^{2}+4 \textit {\_Z} +1\right ) x +2 \left (x^{2}+1\right )^{\frac {1}{4}}+x}{x^{2}+2}\right )-\frac {7 \ln \left (\frac {4 \RootOf \left (8 \textit {\_Z}^{2}+4 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {3}{4}}+2 \left (x^{2}+1\right )^{\frac {3}{4}}+x \sqrt {x^{2}+1}+4 \RootOf \left (8 \textit {\_Z}^{2}+4 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {1}{4}}+4 \RootOf \left (8 \textit {\_Z}^{2}+4 \textit {\_Z} +1\right ) x +x}{x^{2}+2}\right )}{2}-7 \ln \left (\frac {4 \RootOf \left (8 \textit {\_Z}^{2}+4 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {3}{4}}+2 \left (x^{2}+1\right )^{\frac {3}{4}}+x \sqrt {x^{2}+1}+4 \RootOf \left (8 \textit {\_Z}^{2}+4 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {1}{4}}+4 \RootOf \left (8 \textit {\_Z}^{2}+4 \textit {\_Z} +1\right ) x +x}{x^{2}+2}\right ) \RootOf \left (8 \textit {\_Z}^{2}+4 \textit {\_Z} +1\right )\)	\(291\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^4+x^2+1)/(x^2+1)^(1/4)/(x^4+3*x^2+2),x,method=_RETURNVERBOSE)

[Out]

4*x/(x^2+1)^(1/4)+7/2*ln((8*(x^2+1)^(3/4)*RootOf(32*_Z^2-8*_Z+1)-2*(x^2+1)^(3/4)+x*(x^2+1)^(1/2)+8*(x^2+1)^(1/
4)*RootOf(32*_Z^2-8*_Z+1)-8*RootOf(32*_Z^2-8*_Z+1)*x+x)/(x^2+2))-14*ln((8*(x^2+1)^(3/4)*RootOf(32*_Z^2-8*_Z+1)
-2*(x^2+1)^(3/4)+x*(x^2+1)^(1/2)+8*(x^2+1)^(1/4)*RootOf(32*_Z^2-8*_Z+1)-8*RootOf(32*_Z^2-8*_Z+1)*x+x)/(x^2+2))
*RootOf(32*_Z^2-8*_Z+1)+14*RootOf(32*_Z^2-8*_Z+1)*ln(-(8*(x^2+1)^(3/4)*RootOf(32*_Z^2-8*_Z+1)-x*(x^2+1)^(1/2)+
8*(x^2+1)^(1/4)*RootOf(32*_Z^2-8*_Z+1)-8*RootOf(32*_Z^2-8*_Z+1)*x-2*(x^2+1)^(1/4)+x)/(x^2+2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, x^{4} + x^{2} + 1}{{\left (x^{4} + 3 \, x^{2} + 2\right )} {\left (x^{2} + 1\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^4+x^2+1)/(x^2+1)^(1/4)/(x^4+3*x^2+2),x, algorithm="maxima")

[Out]

integrate((2*x^4 + x^2 + 1)/((x^4 + 3*x^2 + 2)*(x^2 + 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+x^2+1}{{\left (x^2+1\right )}^{1/4}\,\left (x^4+3\,x^2+2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 + 2*x^4 + 1)/((x^2 + 1)^(1/4)*(3*x^2 + x^4 + 2)),x)

[Out]

int((x^2 + 2*x^4 + 1)/((x^2 + 1)^(1/4)*(3*x^2 + x^4 + 2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**4+x**2+1)/(x**2+1)**(1/4)/(x**4+3*x**2+2),x)

[Out]

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

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