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

Optimal. Leaf size=114 \[ -\frac {1}{8} \tan ^{-1}\left (\sqrt {\sqrt {x^2+1}+x}\right )-\frac {1}{8} \tanh ^{-1}\left (\sqrt {\sqrt {x^2+1}+x}\right )+\frac {384 x^8+456 x^6+48 x^4-57 x^2+\sqrt {x^2+1} \left (384 x^7+264 x^5-36 x^3-30 x\right )-8}{24 x^3 \left (\sqrt {x^2+1}+x\right )^{9/2}} \]

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Rubi [A]  time = 0.36, antiderivative size = 141, normalized size of antiderivative = 1.24, number of steps used = 13, number of rules used = 11, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.458, Rules used = {6742, 2117, 14, 2119, 457, 288, 290, 329, 212, 206, 203} \begin {gather*} \sqrt {\sqrt {x^2+1}+x}+\frac {1}{24 x \sqrt {\sqrt {x^2+1}+x}}+\frac {1}{12 x^2 \left (\sqrt {x^2+1}+x\right )^{3/2}}-\frac {1}{3 \left (\sqrt {x^2+1}+x\right )^{3/2}}-\frac {1}{8} \tan ^{-1}\left (\sqrt {\sqrt {x^2+1}+x}\right )-\frac {1}{8} \tanh ^{-1}\left (\sqrt {\sqrt {x^2+1}+x}\right )-\frac {1}{3 x^3 \sqrt {\sqrt {x^2+1}+x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*1/(x + Sqrt[1 + x^2])^(3/2) + 1/(12*x^2*(x + Sqrt[1 + x^2])^(3/2)) - 1/(3*x^3*Sqrt[x + Sqrt[1 + x^2]]) +
1/(24*x*Sqrt[x + Sqrt[1 + x^2]]) + Sqrt[x + Sqrt[1 + x^2]] - ArcTan[Sqrt[x + Sqrt[1 + x^2]]]/8 - ArcTanh[Sqrt[
x + Sqrt[1 + x^2]]]/8

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 288

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

Rule 290

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

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

Rule 2117

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Dist[1/(2*
e), Subst[Int[((g + h*x^n)^p*(d^2 + a*f^2 - 2*d*x + x^2))/(d - x)^2, x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /
; FreeQ[{a, c, d, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]

Rule 2119

Int[((g_.) + (h_.)*(x_))^(m_.)*((e_.)*(x_) + (f_.)*Sqrt[(a_.) + (c_.)*(x_)^2])^(n_.), x_Symbol] :> Dist[1/(2^(
m + 1)*e^(m + 1)), Subst[Int[x^(n - m - 2)*(a*f^2 + x^2)*(-(a*f^2*h) + 2*e*g*x + h*x^2)^m, x], x, e*x + f*Sqrt
[a + c*x^2]], x] /; FreeQ[{a, c, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[m]

Rule 6742

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

Rubi steps

\begin {align*} \int \frac {1+x^4}{x^4 \sqrt {x+\sqrt {1+x^2}}} \, dx &=\int \left (\frac {1}{\sqrt {x+\sqrt {1+x^2}}}+\frac {1}{x^4 \sqrt {x+\sqrt {1+x^2}}}\right ) \, dx\\ &=\int \frac {1}{\sqrt {x+\sqrt {1+x^2}}} \, dx+\int \frac {1}{x^4 \sqrt {x+\sqrt {1+x^2}}} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1+x^2}{x^{5/2}} \, dx,x,x+\sqrt {1+x^2}\right )+8 \operatorname {Subst}\left (\int \frac {x^{3/2} \left (1+x^2\right )}{\left (-1+x^2\right )^4} \, dx,x,x+\sqrt {1+x^2}\right )\\ &=-\frac {1}{3 x^3 \sqrt {x+\sqrt {1+x^2}}}+\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {1}{x^{5/2}}+\frac {1}{\sqrt {x}}\right ) \, dx,x,x+\sqrt {1+x^2}\right )-\frac {4}{3} \operatorname {Subst}\left (\int \frac {x^{3/2}}{\left (-1+x^2\right )^3} \, dx,x,x+\sqrt {1+x^2}\right )\\ &=-\frac {1}{3 \left (x+\sqrt {1+x^2}\right )^{3/2}}+\frac {1}{12 x^2 \left (x+\sqrt {1+x^2}\right )^{3/2}}-\frac {1}{3 x^3 \sqrt {x+\sqrt {1+x^2}}}+\sqrt {x+\sqrt {1+x^2}}-\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (-1+x^2\right )^2} \, dx,x,x+\sqrt {1+x^2}\right )\\ &=-\frac {1}{3 \left (x+\sqrt {1+x^2}\right )^{3/2}}+\frac {1}{12 x^2 \left (x+\sqrt {1+x^2}\right )^{3/2}}-\frac {1}{3 x^3 \sqrt {x+\sqrt {1+x^2}}}+\frac {1}{24 x \sqrt {x+\sqrt {1+x^2}}}+\sqrt {x+\sqrt {1+x^2}}+\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (-1+x^2\right )} \, dx,x,x+\sqrt {1+x^2}\right )\\ &=-\frac {1}{3 \left (x+\sqrt {1+x^2}\right )^{3/2}}+\frac {1}{12 x^2 \left (x+\sqrt {1+x^2}\right )^{3/2}}-\frac {1}{3 x^3 \sqrt {x+\sqrt {1+x^2}}}+\frac {1}{24 x \sqrt {x+\sqrt {1+x^2}}}+\sqrt {x+\sqrt {1+x^2}}+\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{-1+x^4} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right )\\ &=-\frac {1}{3 \left (x+\sqrt {1+x^2}\right )^{3/2}}+\frac {1}{12 x^2 \left (x+\sqrt {1+x^2}\right )^{3/2}}-\frac {1}{3 x^3 \sqrt {x+\sqrt {1+x^2}}}+\frac {1}{24 x \sqrt {x+\sqrt {1+x^2}}}+\sqrt {x+\sqrt {1+x^2}}-\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right )-\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right )\\ &=-\frac {1}{3 \left (x+\sqrt {1+x^2}\right )^{3/2}}+\frac {1}{12 x^2 \left (x+\sqrt {1+x^2}\right )^{3/2}}-\frac {1}{3 x^3 \sqrt {x+\sqrt {1+x^2}}}+\frac {1}{24 x \sqrt {x+\sqrt {1+x^2}}}+\sqrt {x+\sqrt {1+x^2}}-\frac {1}{8} \tan ^{-1}\left (\sqrt {x+\sqrt {1+x^2}}\right )-\frac {1}{8} \tanh ^{-1}\left (\sqrt {x+\sqrt {1+x^2}}\right )\\ \end {align*}

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Mathematica [C]  time = 24.91, size = 9486, normalized size = 83.21 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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IntegrateAlgebraic [A]  time = 0.20, size = 114, normalized size = 1.00 \begin {gather*} \frac {-8-57 x^2+48 x^4+456 x^6+384 x^8+\sqrt {1+x^2} \left (-30 x-36 x^3+264 x^5+384 x^7\right )}{24 x^3 \left (x+\sqrt {1+x^2}\right )^{9/2}}-\frac {1}{8} \tan ^{-1}\left (\sqrt {x+\sqrt {1+x^2}}\right )-\frac {1}{8} \tanh ^{-1}\left (\sqrt {x+\sqrt {1+x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8 - 57*x^2 + 48*x^4 + 456*x^6 + 384*x^8 + Sqrt[1 + x^2]*(-30*x - 36*x^3 + 264*x^5 + 384*x^7))/(24*x^3*(x + S
qrt[1 + x^2])^(9/2)) - ArcTan[Sqrt[x + Sqrt[1 + x^2]]]/8 - ArcTanh[Sqrt[x + Sqrt[1 + x^2]]]/8

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fricas [A]  time = 0.46, size = 109, normalized size = 0.96 \begin {gather*} -\frac {6 \, x^{3} \arctan \left (\sqrt {x + \sqrt {x^{2} + 1}}\right ) + 3 \, x^{3} \log \left (\sqrt {x + \sqrt {x^{2} + 1}} + 1\right ) - 3 \, x^{3} \log \left (\sqrt {x + \sqrt {x^{2} + 1}} - 1\right ) + 2 \, {\left (16 \, x^{5} - 19 \, x^{3} - {\left (16 \, x^{4} - 3 \, x^{2} - 8\right )} \sqrt {x^{2} + 1} - 10 \, x\right )} \sqrt {x + \sqrt {x^{2} + 1}}}{48 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(6*x^3*arctan(sqrt(x + sqrt(x^2 + 1))) + 3*x^3*log(sqrt(x + sqrt(x^2 + 1)) + 1) - 3*x^3*log(sqrt(x + sqr
t(x^2 + 1)) - 1) + 2*(16*x^5 - 19*x^3 - (16*x^4 - 3*x^2 - 8)*sqrt(x^2 + 1) - 10*x)*sqrt(x + sqrt(x^2 + 1)))/x^
3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.07, size = 84, normalized size = 0.74

method	result	size
meijerg	\(-\frac {\sqrt {2}\, \hypergeom \left (\left [\frac {1}{4}, \frac {3}{4}, \frac {7}{4}\right ], \left [\frac {3}{2}, \frac {11}{4}\right ], -\frac {1}{x^{2}}\right )}{7 x^{\frac {7}{2}}}-\frac {-\frac {32 \sqrt {\pi }\, \sqrt {2}\, \cosh \left (\frac {3 \arcsinh \left (\frac {1}{x}\right )}{2}\right )}{3 x^{\frac {3}{2}}}-\frac {8 \sqrt {\pi }\, \sqrt {2}\, x^{\frac {3}{2}} \left (-\frac {4}{3 x^{4}}-\frac {2}{3 x^{2}}+\frac {2}{3}\right ) \sinh \left (\frac {3 \arcsinh \left (\frac {1}{x}\right )}{2}\right )}{\sqrt {1+\frac {1}{x^{2}}}}}{8 \sqrt {\pi }}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7*2^(1/2)/x^(7/2)*hypergeom([1/4,3/4,7/4],[3/2,11/4],-1/x^2)-1/8/Pi^(1/2)*(-32/3*Pi^(1/2)*2^(1/2)/x^(3/2)*c
osh(3/2*arcsinh(1/x))-8*Pi^(1/2)*2^(1/2)*x^(3/2)*(-4/3/x^4-2/3/x^2+2/3)*sinh(3/2*arcsinh(1/x))/(1+1/x^2)^(1/2)
)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 8.12, size = 90, normalized size = 0.79 \begin {gather*} \frac {4 x}{3 \sqrt {x + \sqrt {x^{2} + 1}}} + \frac {2 \sqrt {x^{2} + 1}}{3 \sqrt {x + \sqrt {x^{2} + 1}}} - \frac {\Gamma \left (\frac {1}{4}\right ) \Gamma \left (\frac {3}{4}\right ) \Gamma \left (\frac {7}{4}\right ) {{}_{3}F_{2}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4}, \frac {7}{4} \\ \frac {3}{2}, \frac {11}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{2}}} \right )}}{4 \pi x^{\frac {7}{2}} \Gamma \left (\frac {11}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*x/(3*sqrt(x + sqrt(x**2 + 1))) + 2*sqrt(x**2 + 1)/(3*sqrt(x + sqrt(x**2 + 1))) - gamma(1/4)*gamma(3/4)*gamma
(7/4)*hyper((1/4, 3/4, 7/4), (3/2, 11/4), exp_polar(I*pi)/x**2)/(4*pi*x**(7/2)*gamma(11/4))

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