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

Optimal. Leaf size=90 \[ \frac {\sqrt {1-2 x^4} \left (-2 x^2-1\right )}{4 x^4}+\frac {i \log \left (\sqrt {1-2 x^4}+i \sqrt {2} x^2\right )}{\sqrt {2}}+i \tan ^{-1}\left (\sqrt {2} x^2-i \sqrt {1-2 x^4}\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 59, normalized size of antiderivative = 0.66, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1252, 811, 844, 216, 266, 63, 206} \begin {gather*} \frac {1}{2} \tanh ^{-1}\left (\sqrt {1-2 x^4}\right )-\frac {\sin ^{-1}\left (\sqrt {2} x^2\right )}{\sqrt {2}}-\frac {\sqrt {1-2 x^4} \left (2 x^2+1\right )}{4 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.05, size = 59, normalized size = 0.66 \begin {gather*} \frac {1}{4} \left (2 \tanh ^{-1}\left (\sqrt {1-2 x^4}\right )-2 \sqrt {2} \sin ^{-1}\left (\sqrt {2} x^2\right )-\frac {\sqrt {1-2 x^4} \left (2 x^2+1\right )}{x^4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.18, size = 84, normalized size = 0.93 \begin {gather*} \frac {\left (-1-2 x^2\right ) \sqrt {1-2 x^4}}{4 x^4}-\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} x^2}{-1+\sqrt {1-2 x^4}}\right )+\frac {\log \left (x^2\right )}{2}-\frac {1}{2} \log \left (-1+\sqrt {1-2 x^4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1 - 2*x^2)*Sqrt[1 - 2*x^4])/(4*x^4) - Sqrt[2]*ArcTan[(Sqrt[2]*x^2)/(-1 + Sqrt[1 - 2*x^4])] + Log[x^2]/2 - L
og[-1 + Sqrt[1 - 2*x^4]]/2

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fricas [A]  time = 0.94, size = 78, normalized size = 0.87 \begin {gather*} \frac {4 \, \sqrt {2} x^{4} \arctan \left (\frac {\sqrt {2} \sqrt {-2 \, x^{4} + 1} - \sqrt {2}}{2 \, x^{2}}\right ) - 2 \, x^{4} \log \left (\frac {\sqrt {-2 \, x^{4} + 1} - 1}{x^{2}}\right ) - \sqrt {-2 \, 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)*(-2*x^4+1)^(1/2)/x^5,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.25, size = 148, normalized size = 1.64 \begin {gather*} \frac {x^{4} {\left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {-2 \, x^{4} + 1} - \sqrt {2}\right )}}{x^{2}} - 1\right )}}{2 \, {\left (\sqrt {2} \sqrt {-2 \, x^{4} + 1} - \sqrt {2}\right )}^{2}} - \frac {1}{2} \, \sqrt {2} \arcsin \left (\sqrt {2} x^{2}\right ) - \frac {\sqrt {2} {\left (\sqrt {2} \sqrt {-2 \, x^{4} + 1} - \sqrt {2}\right )}}{8 \, x^{2}} + \frac {{\left (\sqrt {2} \sqrt {-2 \, x^{4} + 1} - \sqrt {2}\right )}^{2}}{32 \, x^{4}} - \frac {1}{2} \, \log \left (-\frac {\sqrt {2} \sqrt {-2 \, x^{4} + 1} - \sqrt {2}}{2 \, x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^4*(sqrt(2)*(sqrt(2)*sqrt(-2*x^4 + 1) - sqrt(2))/x^2 - 1)/(sqrt(2)*sqrt(-2*x^4 + 1) - sqrt(2))^2 - 1/2*sq
rt(2)*arcsin(sqrt(2)*x^2) - 1/8*sqrt(2)*(sqrt(2)*sqrt(-2*x^4 + 1) - sqrt(2))/x^2 + 1/32*(sqrt(2)*sqrt(-2*x^4 +
1) - sqrt(2))^2/x^4 - 1/2*log(-1/2*(sqrt(2)*sqrt(-2*x^4 + 1) - sqrt(2))/x^2)

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maple [A]  time = 0.53, size = 55, normalized size = 0.61

method	result	size
elliptic	\(-\frac {\sqrt {2}\, \arcsin \left (\sqrt {2}\, x^{2}\right )}{2}-\frac {\sqrt {-2 x^{4}+1}}{4 x^{4}}+\frac {\arctanh \left (\frac {1}{\sqrt {-2 x^{4}+1}}\right )}{2}-\frac {\sqrt {-2 x^{4}+1}}{2 x^{2}}\)	\(55\)
risch	\(\frac {4 x^{6}+2 x^{4}-2 x^{2}-1}{4 x^{4} \sqrt {-2 x^{4}+1}}-\frac {\sqrt {2}\, \arcsin \left (\sqrt {2}\, x^{2}\right )}{2}+\frac {\arctanh \left (\frac {1}{\sqrt {-2 x^{4}+1}}\right )}{2}\)	\(58\)
trager	\(-\frac {\left (2 x^{2}+1\right ) \sqrt {-2 x^{4}+1}}{4 x^{4}}+\frac {\RootOf \left (\textit {\_Z}^{2}+2\right ) \ln \left (-\RootOf \left (\textit {\_Z}^{2}+2\right ) x^{2}-\sqrt {-2 x^{4}+1}\right )}{2}-\frac {\ln \left (\frac {\sqrt {-2 x^{4}+1}-1}{x^{2}}\right )}{2}\)	\(73\)
default	\(-\frac {\left (-2 x^{4}+1\right )^{\frac {3}{2}}}{4 x^{4}}-\frac {\sqrt {-2 x^{4}+1}}{2}+\frac {\arctanh \left (\frac {1}{\sqrt {-2 x^{4}+1}}\right )}{2}-\frac {\left (-2 x^{4}+1\right )^{\frac {3}{2}}}{2 x^{2}}-x^{2} \sqrt {-2 x^{4}+1}-\frac {\sqrt {2}\, \arcsin \left (\sqrt {2}\, x^{2}\right )}{2}\)	\(80\)
meijerg	\(\frac {\frac {\sqrt {\pi }\, \left (-8 x^{4}+8\right )}{8 x^{4}}-\frac {\sqrt {\pi }\, \sqrt {-2 x^{4}+1}}{x^{4}}+2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-2 x^{4}+1}}{2}\right )-\left (-\ln \relax (2)-1+4 \ln \relax (x )+i \pi \right ) \sqrt {\pi }-\frac {\sqrt {\pi }}{x^{4}}}{4 \sqrt {\pi }}-\frac {i \sqrt {2}\, \left (-\frac {2 i \sqrt {\pi }\, \sqrt {2}\, \sqrt {-2 x^{4}+1}}{x^{2}}-4 i \sqrt {\pi }\, \arcsin \left (\sqrt {2}\, x^{2}\right )\right )}{8 \sqrt {\pi }}\)	\(131\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.45, size = 80, normalized size = 0.89 \begin {gather*} \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-2 \, x^{4} + 1}}{2 \, x^{2}}\right ) - \frac {\sqrt {-2 \, x^{4} + 1}}{2 \, x^{2}} - \frac {\sqrt {-2 \, x^{4} + 1}}{4 \, x^{4}} + \frac {1}{4} \, \log \left (\sqrt {-2 \, x^{4} + 1} + 1\right ) - \frac {1}{4} \, \log \left (\sqrt {-2 \, x^{4} + 1} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.11, size = 71, normalized size = 0.79 \begin {gather*} -\frac {\ln \left (\sqrt {\frac {1}{2\,x^4}-1}-\sqrt {\frac {1}{2\,x^4}}\right )}{2}-\frac {\sqrt {2}\,\mathrm {asin}\left (\sqrt {2}\,x^2\right )}{2}-\frac {\sqrt {2}\,\sqrt {\frac {1}{2}-x^4}}{2\,x^2}-\frac {\sqrt {2}\,\sqrt {\frac {1}{2}-x^4}}{4\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- log((1/(2*x^4) - 1)^(1/2) - (1/(2*x^4))^(1/2))/2 - (2^(1/2)*asin(2^(1/2)*x^2))/2 - (2^(1/2)*(1/2 - x^4)^(1/2
))/(2*x^2) - (2^(1/2)*(1/2 - x^4)^(1/2))/(4*x^4)

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sympy [A]  time = 5.12, size = 230, normalized size = 2.56 \begin {gather*} \begin {cases} - \frac {i x^{2}}{\sqrt {2 x^{4} - 1}} + \frac {\sqrt {2} i \operatorname {acosh}{\left (\sqrt {2} x^{2} \right )}}{2} + \frac {i}{2 x^{2} \sqrt {2 x^{4} - 1}} & \text {for}\: 2 \left |{x^{4}}\right | > 1 \\\frac {x^{2}}{\sqrt {1 - 2 x^{4}}} - \frac {\sqrt {2} \operatorname {asin}{\left (\sqrt {2} x^{2} \right )}}{2} - \frac {1}{2 x^{2} \sqrt {1 - 2 x^{4}}} & \text {otherwise} \end {cases} + \begin {cases} \frac {\operatorname {acosh}{\left (\frac {\sqrt {2}}{2 x^{2}} \right )}}{2} + \frac {\sqrt {2}}{4 x^{2} \sqrt {-1 + \frac {1}{2 x^{4}}}} - \frac {\sqrt {2}}{8 x^{6} \sqrt {-1 + \frac {1}{2 x^{4}}}} & \text {for}\: \frac {1}{2 \left |{x^{4}}\right |} > 1 \\- \frac {i \operatorname {asin}{\left (\frac {\sqrt {2}}{2 x^{2}} \right )}}{2} - \frac {\sqrt {2} i}{4 x^{2} \sqrt {1 - \frac {1}{2 x^{4}}}} + \frac {\sqrt {2} i}{8 x^{6} \sqrt {1 - \frac {1}{2 x^{4}}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-I*x**2/sqrt(2*x**4 - 1) + sqrt(2)*I*acosh(sqrt(2)*x**2)/2 + I/(2*x**2*sqrt(2*x**4 - 1)), 2*Abs(x**
4) > 1), (x**2/sqrt(1 - 2*x**4) - sqrt(2)*asin(sqrt(2)*x**2)/2 - 1/(2*x**2*sqrt(1 - 2*x**4)), True)) + Piecewi
se((acosh(sqrt(2)/(2*x**2))/2 + sqrt(2)/(4*x**2*sqrt(-1 + 1/(2*x**4))) - sqrt(2)/(8*x**6*sqrt(-1 + 1/(2*x**4))
), 1/(2*Abs(x**4)) > 1), (-I*asin(sqrt(2)/(2*x**2))/2 - sqrt(2)*I/(4*x**2*sqrt(1 - 1/(2*x**4))) + sqrt(2)*I/(8
*x**6*sqrt(1 - 1/(2*x**4))), True))

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