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

Optimal. Leaf size=101 \[ \frac {2 \sqrt {x^4+1} x^4+\left (2 x^4+3\right ) x^2}{8 x \sqrt {\sqrt {x^4+1}+x^2}}-\frac {11 \tanh ^{-1}\left (\frac {\sqrt {2} x \sqrt {\sqrt {x^4+1}+x^2}}{\sqrt {x^4+1}+x^2+1}\right )}{4 \sqrt {2}} \]

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Rubi [C]  time = 0.50, antiderivative size = 161, normalized size of antiderivative = 1.59, number of steps used = 11, number of rules used = 7, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {6742, 2132, 206, 2133, 321, 216, 215} \begin {gather*} \left (\frac {3}{16}-\frac {3 i}{16}\right ) \sqrt {1-i x^2} x+\left (\frac {3}{16}+\frac {3 i}{16}\right ) \sqrt {1+i x^2} x-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {\sqrt {x^4+1}+x^2}}\right )}{\sqrt {2}}+\left (\frac {1}{8}+\frac {i}{8}\right ) \sqrt {1-i x^2} x^3+\left (\frac {1}{8}-\frac {i}{8}\right ) \sqrt {1+i x^2} x^3+\frac {3 i \sin ^{-1}\left (\sqrt [4]{-1} x\right )}{8 \sqrt {2}}-\frac {3 \sinh ^{-1}\left (\sqrt [4]{-1} x\right )}{8 \sqrt {2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(3/16 - (3*I)/16)*x*Sqrt[1 - I*x^2] + (1/8 + I/8)*x^3*Sqrt[1 - I*x^2] + (3/16 + (3*I)/16)*x*Sqrt[1 + I*x^2] +
(1/8 - I/8)*x^3*Sqrt[1 + I*x^2] + (((3*I)/8)*ArcSin[(-1)^(1/4)*x])/Sqrt[2] - (3*ArcSinh[(-1)^(1/4)*x])/(8*Sqrt
[2]) - ArcTanh[(Sqrt[2]*x)/Sqrt[x^2 + Sqrt[1 + x^4]]]/Sqrt[2]

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 215

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

Rule 216

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

Rule 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2132

Int[Sqrt[(c_.)*(x_)^2 + (d_.)*Sqrt[(a_) + (b_.)*(x_)^4]]/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Dist[d, Subst
[Int[1/(1 - 2*c*x^2), x], x, x/Sqrt[c*x^2 + d*Sqrt[a + b*x^4]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[c^2 - b*d
^2, 0]

Rule 2133

Int[(((c_.) + (d_.)*(x_))^(m_.)*Sqrt[(b_.)*(x_)^2 + Sqrt[(a_) + (e_.)*(x_)^4]])/Sqrt[(a_) + (e_.)*(x_)^4], x_S
ymbol] :> Dist[(1 - I)/2, Int[(c + d*x)^m/Sqrt[Sqrt[a] - I*b*x^2], x], x] + Dist[(1 + I)/2, Int[(c + d*x)^m/Sq
rt[Sqrt[a] + I*b*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && EqQ[e, b^2] && GtQ[a, 0]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {\left (-1+x^4\right ) \sqrt {x^2+\sqrt {1+x^4}}}{\sqrt {1+x^4}} \, dx &=\int \left (-\frac {\sqrt {x^2+\sqrt {1+x^4}}}{\sqrt {1+x^4}}+\frac {x^4 \sqrt {x^2+\sqrt {1+x^4}}}{\sqrt {1+x^4}}\right ) \, dx\\ &=-\int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{\sqrt {1+x^4}} \, dx+\int \frac {x^4 \sqrt {x^2+\sqrt {1+x^4}}}{\sqrt {1+x^4}} \, dx\\ &=\left (\frac {1}{2}-\frac {i}{2}\right ) \int \frac {x^4}{\sqrt {1-i x^2}} \, dx+\left (\frac {1}{2}+\frac {i}{2}\right ) \int \frac {x^4}{\sqrt {1+i x^2}} \, dx-\operatorname {Subst}\left (\int \frac {1}{1-2 x^2} \, dx,x,\frac {x}{\sqrt {x^2+\sqrt {1+x^4}}}\right )\\ &=\left (\frac {1}{8}+\frac {i}{8}\right ) x^3 \sqrt {1-i x^2}+\left (\frac {1}{8}-\frac {i}{8}\right ) x^3 \sqrt {1+i x^2}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^2+\sqrt {1+x^4}}}\right )}{\sqrt {2}}+\left (-\frac {3}{8}-\frac {3 i}{8}\right ) \int \frac {x^2}{\sqrt {1-i x^2}} \, dx+\left (-\frac {3}{8}+\frac {3 i}{8}\right ) \int \frac {x^2}{\sqrt {1+i x^2}} \, dx\\ &=\left (\frac {3}{16}-\frac {3 i}{16}\right ) x \sqrt {1-i x^2}+\left (\frac {1}{8}+\frac {i}{8}\right ) x^3 \sqrt {1-i x^2}+\left (\frac {3}{16}+\frac {3 i}{16}\right ) x \sqrt {1+i x^2}+\left (\frac {1}{8}-\frac {i}{8}\right ) x^3 \sqrt {1+i x^2}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^2+\sqrt {1+x^4}}}\right )}{\sqrt {2}}+\left (-\frac {3}{16}-\frac {3 i}{16}\right ) \int \frac {1}{\sqrt {1+i x^2}} \, dx+\left (-\frac {3}{16}+\frac {3 i}{16}\right ) \int \frac {1}{\sqrt {1-i x^2}} \, dx\\ &=\left (\frac {3}{16}-\frac {3 i}{16}\right ) x \sqrt {1-i x^2}+\left (\frac {1}{8}+\frac {i}{8}\right ) x^3 \sqrt {1-i x^2}+\left (\frac {3}{16}+\frac {3 i}{16}\right ) x \sqrt {1+i x^2}+\left (\frac {1}{8}-\frac {i}{8}\right ) x^3 \sqrt {1+i x^2}+\frac {3 i \sin ^{-1}\left (\sqrt [4]{-1} x\right )}{8 \sqrt {2}}-\frac {3 \sinh ^{-1}\left (\sqrt [4]{-1} x\right )}{8 \sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^2+\sqrt {1+x^4}}}\right )}{\sqrt {2}}\\ \end {align*}

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Mathematica [C]  time = 2.56, size = 286, normalized size = 2.83 \begin {gather*} \frac {x \left (x^4+\sqrt {x^4+1} x^2+1\right ) \left (\frac {\left (2 x^4+2 \sqrt {x^4+1} x^2+1\right )^2 \, _2F_1\left (-\frac {3}{2},-\frac {1}{2};\frac {1}{2};2 x^4+2 \sqrt {x^4+1} x^2+1\right )}{\sqrt {-x^2 \left (\sqrt {x^4+1}+x^2\right )} \left (16 x^{12}+28 x^8+13 x^4+16 \sqrt {x^4+1} x^{10}+20 \sqrt {x^4+1} x^6+5 \sqrt {x^4+1} x^2+1\right )}+\frac {2 \left (\log \left (1-\frac {\sqrt {x^2 \left (\sqrt {x^4+1}+x^2\right )}}{\sqrt {2} x^2}\right )-\log \left (\frac {\sqrt {x^2 \left (\sqrt {x^4+1}+x^2\right )}}{\sqrt {2} x^2}+1\right )\right )}{\sqrt {x^4+1} \sqrt {x^2 \left (\sqrt {x^4+1}+x^2\right )}}\right )}{4 \sqrt {2} \sqrt {\sqrt {x^4+1}+x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(1 + x^4 + x^2*Sqrt[1 + x^4])*(((1 + 2*x^4 + 2*x^2*Sqrt[1 + x^4])^2*Hypergeometric2F1[-3/2, -1/2, 1/2, 1 +
2*x^4 + 2*x^2*Sqrt[1 + x^4]])/(Sqrt[-(x^2*(x^2 + Sqrt[1 + x^4]))]*(1 + 13*x^4 + 28*x^8 + 16*x^12 + 5*x^2*Sqrt[
1 + x^4] + 20*x^6*Sqrt[1 + x^4] + 16*x^10*Sqrt[1 + x^4])) + (2*(Log[1 - Sqrt[x^2*(x^2 + Sqrt[1 + x^4])]/(Sqrt[
2]*x^2)] - Log[1 + Sqrt[x^2*(x^2 + Sqrt[1 + x^4])]/(Sqrt[2]*x^2)]))/(Sqrt[1 + x^4]*Sqrt[x^2*(x^2 + Sqrt[1 + x^
4])])))/(4*Sqrt[2]*Sqrt[x^2 + Sqrt[1 + x^4]])

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IntegrateAlgebraic [A]  time = 0.31, size = 114, normalized size = 1.13 \begin {gather*} \frac {2 x^4 \sqrt {1+x^4}+x^2 \left (3+2 x^4\right )}{8 x \sqrt {x^2+\sqrt {1+x^4}}}-\frac {11 \tanh ^{-1}\left (\frac {-\frac {1}{\sqrt {2}}+\frac {x^2}{\sqrt {2}}+\frac {\sqrt {1+x^4}}{\sqrt {2}}}{x \sqrt {x^2+\sqrt {1+x^4}}}\right )}{4 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^4*Sqrt[1 + x^4] + x^2*(3 + 2*x^4))/(8*x*Sqrt[x^2 + Sqrt[1 + x^4]]) - (11*ArcTanh[(-(1/Sqrt[2]) + x^2/Sqrt
[2] + Sqrt[1 + x^4]/Sqrt[2])/(x*Sqrt[x^2 + Sqrt[1 + x^4]])])/(4*Sqrt[2])

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fricas [A]  time = 1.03, size = 90, normalized size = 0.89 \begin {gather*} -\frac {1}{8} \, {\left (x^{3} - 3 \, \sqrt {x^{4} + 1} x\right )} \sqrt {x^{2} + \sqrt {x^{4} + 1}} + \frac {11}{32} \, \sqrt {2} \log \left (4 \, x^{4} + 4 \, \sqrt {x^{4} + 1} x^{2} - 2 \, {\left (\sqrt {2} x^{3} + \sqrt {2} \sqrt {x^{4} + 1} x\right )} \sqrt {x^{2} + \sqrt {x^{4} + 1}} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(x^3 - 3*sqrt(x^4 + 1)*x)*sqrt(x^2 + sqrt(x^4 + 1)) + 11/32*sqrt(2)*log(4*x^4 + 4*sqrt(x^4 + 1)*x^2 - 2*(
sqrt(2)*x^3 + sqrt(2)*sqrt(x^4 + 1)*x)*sqrt(x^2 + sqrt(x^4 + 1)) + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (x^{4}-1\right ) \sqrt {x^{2}+\sqrt {x^{4}+1}}}{\sqrt {x^{4}+1}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (x^4-1\right )\,\sqrt {\sqrt {x^4+1}+x^2}}{\sqrt {x^4+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 4.52, size = 32, normalized size = 0.32 \begin {gather*} - \frac {{G_{3, 3}^{2, 2}\left (\begin {matrix} 1, 1 & \frac {1}{2} \\\frac {1}{4}, \frac {3}{4} & 0 \end {matrix} \middle | {x^{4}} \right )}}{4 \sqrt {\pi }} + \frac {{G_{3, 3}^{2, 2}\left (\begin {matrix} 2, 1 & \frac {3}{2} \\\frac {5}{4}, \frac {7}{4} & 0 \end {matrix} \middle | {x^{4}} \right )}}{4 \sqrt {\pi }} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-meijerg(((1, 1), (1/2,)), ((1/4, 3/4), (0,)), x**4)/(4*sqrt(pi)) + meijerg(((2, 1), (3/2,)), ((5/4, 7/4), (0,
)), x**4)/(4*sqrt(pi))

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