3.8.82 \(\int \frac {(-1+x^2) \sqrt {1+x^4}}{x^5} \, dx\)

Optimal. Leaf size=60 \[ \frac {\sqrt {x^4+1} \left (1-2 x^2\right )}{4 x^4}+\frac {1}{2} \log \left (\sqrt {x^4+1}+x^2\right )+\frac {1}{2} \tanh ^{-1}\left (\sqrt {x^4+1}+x^2\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 46, normalized size of antiderivative = 0.77, number of steps used = 7, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {1252, 811, 844, 215, 266, 63, 207} \begin {gather*} \frac {1}{4} \tanh ^{-1}\left (\sqrt {x^4+1}\right )+\frac {1}{2} \sinh ^{-1}\left (x^2\right )+\frac {\sqrt {x^4+1} \left (1-2 x^2\right )}{4 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-1 + x^2)*Sqrt[1 + x^4])/x^5,x]

[Out]

((1 - 2*x^2)*Sqrt[1 + x^4])/(4*x^4) + ArcSinh[x^2]/2 + ArcTanh[Sqrt[1 + x^4]]/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 811

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((d + e*x)^
(m + 1)*(a + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 + a*e^2) - 2*c*d^2*p*(e*f - d*g) - e*(g*(m + 1)*(c*d^2 + a*e
^2) + 2*c*d*p*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2 + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2
+ a*e^2)), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) - c*(2*c*d*(d*g*(2*p + 1
) - e*f*(m + 2*p + 2)) - 2*a*e^2*g*(m + 1))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2
, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

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 1252

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m
- 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 41, normalized size = 0.68 \begin {gather*} \frac {1}{4} \left (\tanh ^{-1}\left (\sqrt {x^4+1}\right )+2 \sinh ^{-1}\left (x^2\right )+\frac {\sqrt {x^4+1} \left (1-2 x^2\right )}{x^4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-1 + x^2)*Sqrt[1 + x^4])/x^5,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.17, size = 64, normalized size = 1.07 \begin {gather*} \frac {\left (1-2 x^2\right ) \sqrt {1+x^4}}{4 x^4}-\frac {1}{2} \tanh ^{-1}\left (x^2-\sqrt {1+x^4}\right )-\frac {1}{2} \log \left (-x^2+\sqrt {1+x^4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-1 + x^2)*Sqrt[1 + x^4])/x^5,x]

[Out]

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

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fricas [A]  time = 0.45, size = 85, normalized size = 1.42 \begin {gather*} \frac {x^{4} \log \left (-x^{2} + \sqrt {x^{4} + 1} + 1\right ) - 2 \, x^{4} \log \left (-x^{2} + \sqrt {x^{4} + 1}\right ) - x^{4} \log \left (-x^{2} + \sqrt {x^{4} + 1} - 1\right ) - 2 \, x^{4} - \sqrt {x^{4} + 1} {\left (2 \, x^{2} - 1\right )}}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-1)*(x^4+1)^(1/2)/x^5,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.71, size = 118, normalized size = 1.97 \begin {gather*} -\frac {{\left (x^{2} - \sqrt {x^{4} + 1}\right )}^{3} - 2 \, {\left (x^{2} - \sqrt {x^{4} + 1}\right )}^{2} + x^{2} - \sqrt {x^{4} + 1} + 2}{2 \, {\left ({\left (x^{2} - \sqrt {x^{4} + 1}\right )}^{2} - 1\right )}^{2}} - \frac {1}{4} \, \log \left (x^{2} - \sqrt {x^{4} + 1} + 1\right ) + \frac {1}{4} \, \log \left (-x^{2} + \sqrt {x^{4} + 1} + 1\right ) - \frac {1}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-1)*(x^4+1)^(1/2)/x^5,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.40, size = 42, normalized size = 0.70

method	result	size
elliptic	\(\frac {\arcsinh \left (x^{2}\right )}{2}+\frac {\sqrt {x^{4}+1}}{4 x^{4}}+\frac {\arctanh \left (\frac {1}{\sqrt {x^{4}+1}}\right )}{4}-\frac {\sqrt {x^{4}+1}}{2 x^{2}}\)	\(42\)
risch	\(-\frac {2 x^{6}-x^{4}+2 x^{2}-1}{4 x^{4} \sqrt {x^{4}+1}}+\frac {\arcsinh \left (x^{2}\right )}{2}+\frac {\arctanh \left (\frac {1}{\sqrt {x^{4}+1}}\right )}{4}\)	\(47\)
default	\(\frac {\left (x^{4}+1\right )^{\frac {3}{2}}}{4 x^{4}}-\frac {\sqrt {x^{4}+1}}{4}+\frac {\arctanh \left (\frac {1}{\sqrt {x^{4}+1}}\right )}{4}-\frac {\left (x^{4}+1\right )^{\frac {3}{2}}}{2 x^{2}}+\frac {x^{2} \sqrt {x^{4}+1}}{2}+\frac {\arcsinh \left (x^{2}\right )}{2}\)	\(63\)
trager	\(-\frac {\left (2 x^{2}-1\right ) \sqrt {x^{4}+1}}{4 x^{4}}-\frac {\ln \left (-\frac {-2 x^{6}+2 x^{4} \sqrt {x^{4}+1}-2 x^{4}+2 x^{2} \sqrt {x^{4}+1}-2 x^{2}+\sqrt {x^{4}+1}-1}{x^{2}}\right )}{4}\)	\(77\)
meijerg	\(\frac {-\frac {\sqrt {\pi }\, \left (4 x^{4}+8\right )}{4 x^{4}}+\frac {2 \sqrt {\pi }\, \sqrt {x^{4}+1}}{x^{4}}+2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {x^{4}+1}}{2}\right )-\left (-2 \ln \relax (2)-1+4 \ln \relax (x )\right ) \sqrt {\pi }+\frac {2 \sqrt {\pi }}{x^{4}}}{8 \sqrt {\pi }}-\frac {\frac {4 \sqrt {\pi }\, \sqrt {x^{4}+1}}{x^{2}}-4 \sqrt {\pi }\, \arcsinh \left (x^{2}\right )}{8 \sqrt {\pi }}\)	\(108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2-1)*(x^4+1)^(1/2)/x^5,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.42, size = 81, normalized size = 1.35 \begin {gather*} -\frac {\sqrt {x^{4} + 1}}{2 \, x^{2}} + \frac {\sqrt {x^{4} + 1}}{4 \, x^{4}} + \frac {1}{8} \, \log \left (\sqrt {x^{4} + 1} + 1\right ) - \frac {1}{8} \, \log \left (\sqrt {x^{4} + 1} - 1\right ) + \frac {1}{4} \, \log \left (\frac {\sqrt {x^{4} + 1}}{x^{2}} + 1\right ) - \frac {1}{4} \, \log \left (\frac {\sqrt {x^{4} + 1}}{x^{2}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-1)*(x^4+1)^(1/2)/x^5,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.97, size = 45, normalized size = 0.75 \begin {gather*} \frac {\mathrm {asinh}\left (x^2\right )}{2}-\frac {\sqrt {x^4+1}}{2\,x^2}+\frac {\sqrt {x^4+1}}{4\,x^4}-\frac {\mathrm {atan}\left (\sqrt {x^4+1}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2 - 1)*(x^4 + 1)^(1/2))/x^5,x)

[Out]

asinh(x^2)/2 - (atan((x^4 + 1)^(1/2)*1i)*1i)/4 - (x^4 + 1)^(1/2)/(2*x^2) + (x^4 + 1)^(1/2)/(4*x^4)

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sympy [A]  time = 4.70, size = 58, normalized size = 0.97 \begin {gather*} - \frac {x^{2}}{2 \sqrt {x^{4} + 1}} + \frac {\operatorname {asinh}{\left (\frac {1}{x^{2}} \right )}}{4} + \frac {\operatorname {asinh}{\left (x^{2} \right )}}{2} + \frac {\sqrt {1 + \frac {1}{x^{4}}}}{4 x^{2}} - \frac {1}{2 x^{2} \sqrt {x^{4} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2-1)*(x**4+1)**(1/2)/x**5,x)

[Out]

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

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