\(\int \frac {(1+2 x^4) \sqrt {1+2 x^8}}{x} \, dx\) [1042]

Optimal result

Integrand size = 22, antiderivative size = 78 \[ \int \frac {\left (1+2 x^4\right ) \sqrt {1+2 x^8}}{x} \, dx=\frac {1}{4} \left (1+x^4\right ) \sqrt {1+2 x^8}-\frac {1}{2} \text {arctanh}\left (\sqrt {2} x^4+\sqrt {1+2 x^8}\right )+\frac {\log \left (\sqrt {2} x^4+\sqrt {1+2 x^8}\right )}{4 \sqrt {2}} \]

[Out]

1/4*(x^4+1)*(2*x^8+1)^(1/2)-1/2*arctanh(2^(1/2)*x^4+(2*x^8+1)^(1/2))+1/8*ln(2^(1/2)*x^4+(2*x^8+1)^(1/2))*2^(1/
2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.72, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1489, 829, 858, 221, 272, 65, 213} \[ \int \frac {\left (1+2 x^4\right ) \sqrt {1+2 x^8}}{x} \, dx=\frac {\text {arcsinh}\left (\sqrt {2} x^4\right )}{4 \sqrt {2}}-\frac {1}{4} \text {arctanh}\left (\sqrt {2 x^8+1}\right )+\frac {1}{4} \sqrt {2 x^8+1} \left (x^4+1\right ) \]

[In]

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

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 213

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

Rule 221

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 272

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

Rule 829

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m
+ 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*((a + c*x^2)^p/(c*e^2*(m + 2*p + 1)*(m +
2*p + 2))), x] + Dist[2*(p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2))), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*
m, 2*p])

Rule 858

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 1489

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, c, d, e, m, n, p, q}, x
] && EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \text {Subst}\left (\int \frac {(1+2 x) \sqrt {1+2 x^2}}{x} \, dx,x,x^4\right ) \\ & = \frac {1}{4} \left (1+x^4\right ) \sqrt {1+2 x^8}+\frac {1}{16} \text {Subst}\left (\int \frac {4+4 x}{x \sqrt {1+2 x^2}} \, dx,x,x^4\right ) \\ & = \frac {1}{4} \left (1+x^4\right ) \sqrt {1+2 x^8}+\frac {1}{4} \text {Subst}\left (\int \frac {1}{\sqrt {1+2 x^2}} \, dx,x,x^4\right )+\frac {1}{4} \text {Subst}\left (\int \frac {1}{x \sqrt {1+2 x^2}} \, dx,x,x^4\right ) \\ & = \frac {1}{4} \left (1+x^4\right ) \sqrt {1+2 x^8}+\frac {\text {arcsinh}\left (\sqrt {2} x^4\right )}{4 \sqrt {2}}+\frac {1}{8} \text {Subst}\left (\int \frac {1}{x \sqrt {1+2 x}} \, dx,x,x^8\right ) \\ & = \frac {1}{4} \left (1+x^4\right ) \sqrt {1+2 x^8}+\frac {\text {arcsinh}\left (\sqrt {2} x^4\right )}{4 \sqrt {2}}+\frac {1}{8} \text {Subst}\left (\int \frac {1}{-\frac {1}{2}+\frac {x^2}{2}} \, dx,x,\sqrt {1+2 x^8}\right ) \\ & = \frac {1}{4} \left (1+x^4\right ) \sqrt {1+2 x^8}+\frac {\text {arcsinh}\left (\sqrt {2} x^4\right )}{4 \sqrt {2}}-\frac {1}{4} \text {arctanh}\left (\sqrt {1+2 x^8}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.04 \[ \int \frac {\left (1+2 x^4\right ) \sqrt {1+2 x^8}}{x} \, dx=\frac {1}{4} \left (1+x^4\right ) \sqrt {1+2 x^8}+\frac {1}{2} \text {arctanh}\left (\sqrt {2} x^4-\sqrt {1+2 x^8}\right )-\frac {\log \left (-\sqrt {2} x^4+\sqrt {1+2 x^8}\right )}{4 \sqrt {2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.84 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.67

method	result	size
pseudoelliptic	\(\frac {\sqrt {2 x^{8}+1}}{4}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {2 x^{8}+1}}\right )}{4}+\frac {x^{4} \sqrt {2 x^{8}+1}}{4}+\frac {\sqrt {2}\, \operatorname {arcsinh}\left (\sqrt {2}\, x^{4}\right )}{8}\)	\(52\)
trager	\(\left (\frac {x^{4}}{4}+\frac {1}{4}\right ) \sqrt {2 x^{8}+1}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x^{4}-\sqrt {2 x^{8}+1}\right )}{8}-\frac {\ln \left (\frac {\sqrt {2 x^{8}+1}+1}{x^{4}}\right )}{4}\)	\(68\)
meijerg	\(-\frac {4 \sqrt {\pi }-4 \sqrt {\pi }\, \sqrt {2 x^{8}+1}+4 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {2 x^{8}+1}}{2}\right )-2 \left (2-\ln \left (2\right )+8 \ln \left (x \right )\right ) \sqrt {\pi }}{16 \sqrt {\pi }}-\frac {\sqrt {2}\, \left (-2 \sqrt {\pi }\, \sqrt {2}\, x^{4} \sqrt {2 x^{8}+1}-2 \sqrt {\pi }\, \operatorname {arcsinh}\left (\sqrt {2}\, x^{4}\right )\right )}{16 \sqrt {\pi }}\)	\(103\)

[In]

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

[Out]

1/4*(2*x^8+1)^(1/2)-1/4*arctanh(1/(2*x^8+1)^(1/2))+1/4*x^4*(2*x^8+1)^(1/2)+1/8*2^(1/2)*arcsinh(2^(1/2)*x^4)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.78 \[ \int \frac {\left (1+2 x^4\right ) \sqrt {1+2 x^8}}{x} \, dx=\frac {1}{4} \, \sqrt {2 \, x^{8} + 1} {\left (x^{4} + 1\right )} + \frac {1}{8} \, \sqrt {2} \log \left (-\sqrt {2} x^{4} - \sqrt {2 \, x^{8} + 1}\right ) + \frac {1}{4} \, \log \left (\frac {\sqrt {2 \, x^{8} + 1} - 1}{x^{4}}\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 16.29 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.94 \[ \int \frac {\left (1+2 x^4\right ) \sqrt {1+2 x^8}}{x} \, dx=\frac {x^{4} \sqrt {2 x^{8} + 1}}{4} + \frac {\sqrt {2 x^{8} + 1}}{4} + \frac {\log {\left (\sqrt {2 x^{8} + 1} - 1 \right )}}{8} - \frac {\log {\left (\sqrt {2 x^{8} + 1} + 1 \right )}}{8} + \frac {\sqrt {2} \operatorname {asinh}{\left (\sqrt {2} x^{4} \right )}}{8} \]

[In]

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

[Out]

x**4*sqrt(2*x**8 + 1)/4 + sqrt(2*x**8 + 1)/4 + log(sqrt(2*x**8 + 1) - 1)/8 - log(sqrt(2*x**8 + 1) + 1)/8 + sqr
t(2)*asinh(sqrt(2)*x**4)/8

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.46 \[ \int \frac {\left (1+2 x^4\right ) \sqrt {1+2 x^8}}{x} \, dx=-\frac {1}{16} \, \sqrt {2} \log \left (-\frac {\sqrt {2} - \frac {\sqrt {2 \, x^{8} + 1}}{x^{4}}}{\sqrt {2} + \frac {\sqrt {2 \, x^{8} + 1}}{x^{4}}}\right ) + \frac {1}{4} \, \sqrt {2 \, x^{8} + 1} + \frac {\sqrt {2 \, x^{8} + 1}}{4 \, x^{4} {\left (\frac {2 \, x^{8} + 1}{x^{8}} - 2\right )}} - \frac {1}{8} \, \log \left (\sqrt {2 \, x^{8} + 1} + 1\right ) + \frac {1}{8} \, \log \left (\sqrt {2 \, x^{8} + 1} - 1\right ) \]

[In]

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

[Out]

-1/16*sqrt(2)*log(-(sqrt(2) - sqrt(2*x^8 + 1)/x^4)/(sqrt(2) + sqrt(2*x^8 + 1)/x^4)) + 1/4*sqrt(2*x^8 + 1) + 1/
4*sqrt(2*x^8 + 1)/(x^4*((2*x^8 + 1)/x^8 - 2)) - 1/8*log(sqrt(2*x^8 + 1) + 1) + 1/8*log(sqrt(2*x^8 + 1) - 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.10 \[ \int \frac {\left (1+2 x^4\right ) \sqrt {1+2 x^8}}{x} \, dx=\frac {1}{4} \, \sqrt {2 \, x^{8} + 1} {\left (x^{4} + 1\right )} - \frac {1}{8} \, \sqrt {2} \log \left (-\sqrt {2} x^{4} + \sqrt {2 \, x^{8} + 1}\right ) + \frac {1}{4} \, \log \left (\sqrt {2} x^{4} - \sqrt {2 \, x^{8} + 1} + 1\right ) - \frac {1}{4} \, \log \left (-\sqrt {2} x^{4} + \sqrt {2 \, x^{8} + 1} + 1\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.33 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.60 \[ \int \frac {\left (1+2 x^4\right ) \sqrt {1+2 x^8}}{x} \, dx=\frac {\sqrt {2}\,\mathrm {asinh}\left (\sqrt {2}\,x^4\right )}{8}-\frac {\mathrm {atanh}\left (\sqrt {2}\,\sqrt {x^8+\frac {1}{2}}\right )}{4}+\frac {\sqrt {2}\,\sqrt {x^8+\frac {1}{2}}\,\left (\frac {x^4}{2}+\frac {1}{2}\right )}{2} \]

[In]

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

[Out]

(2^(1/2)*asinh(2^(1/2)*x^4))/8 - atanh(2^(1/2)*(x^8 + 1/2)^(1/2))/4 + (2^(1/2)*(x^8 + 1/2)^(1/2)*(x^4/2 + 1/2)
)/2