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

Optimal. Leaf size=78 \[ \frac {1}{4} \sqrt {2 x^8+1} \left (x^4+1\right )+\frac {\log \left (\sqrt {2 x^8+1}+\sqrt {2} x^4\right )}{4 \sqrt {2}}-\frac {1}{2} \tanh ^{-1}\left (\sqrt {2 x^8+1}+\sqrt {2} x^4\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 56, normalized size of antiderivative = 0.72, number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1475, 815, 844, 215, 266, 63, 207} \begin {gather*} -\frac {1}{4} \tanh ^{-1}\left (\sqrt {2 x^8+1}\right )+\frac {\sinh ^{-1}\left (\sqrt {2} x^4\right )}{4 \sqrt {2}}+\frac {1}{4} \sqrt {2 x^8+1} \left (x^4+1\right ) \end {gather*}

Antiderivative was successfully verified.

[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 63

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 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[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 266

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 815

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 844

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 1475

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

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Mathematica [A]  time = 0.04, size = 53, normalized size = 0.68 \begin {gather*} \frac {1}{8} \left (-2 \tanh ^{-1}\left (\sqrt {2 x^8+1}\right )+\sqrt {2} \sinh ^{-1}\left (\sqrt {2} x^4\right )+2 \sqrt {2 x^8+1} \left (x^4+1\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.16, size = 81, normalized size = 1.04 \begin {gather*} \frac {1}{4} \left (1+x^4\right ) \sqrt {1+2 x^8}+\frac {1}{2} \tanh ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((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])

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fricas [A]  time = 0.48, size = 61, normalized size = 0.78 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.32, size = 86, normalized size = 1.10 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [C]  time = 0.55, size = 68, normalized size = 0.87

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 114, normalized size = 1.46 \begin {gather*} -\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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 1.24, size = 47, normalized size = 0.60 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 31.28, size = 65, normalized size = 0.83 \begin {gather*} \frac {x^{4} \sqrt {2 x^{8} + 1}}{4} + \frac {\sqrt {2 x^{8} + 1}}{4} + \frac {\log {\left (x^{8} \right )}}{8} - \frac {\log {\left (\sqrt {2 x^{8} + 1} + 1 \right )}}{4} + \frac {\sqrt {2} \operatorname {asinh}{\left (\sqrt {2} x^{4} \right )}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

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

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