3.6.5 \(\int x \sqrt {-1+x^4} \, dx\)

Optimal. Leaf size=39 \[ \frac {1}{4} x^2 \sqrt {x^4-1}-\frac {1}{2} \tanh ^{-1}\left (\frac {\sqrt {x^4-1}}{x^2+1}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 35, normalized size of antiderivative = 0.90, number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {275, 195, 217, 206} \begin {gather*} \frac {1}{4} x^2 \sqrt {x^4-1}-\frac {1}{4} \tanh ^{-1}\left (\frac {x^2}{\sqrt {x^4-1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[-1 + x^4],x]

[Out]

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 275

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 42, normalized size = 1.08 \begin {gather*} \frac {\left (x^4-1\right ) \left (\sin ^{-1}\left (x^2\right )+\sqrt {1-x^4} x^2\right )}{4 \sqrt {-\left (x^4-1\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[-1 + x^4],x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.14, size = 39, normalized size = 1.00 \begin {gather*} \frac {1}{4} x^2 \sqrt {-1+x^4}-\frac {1}{2} \tanh ^{-1}\left (\frac {\sqrt {-1+x^4}}{1+x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x*Sqrt[-1 + x^4],x]

[Out]

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

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fricas [A]  time = 0.43, size = 29, normalized size = 0.74 \begin {gather*} \frac {1}{4} \, \sqrt {x^{4} - 1} x^{2} + \frac {1}{4} \, \log \left (-x^{2} + \sqrt {x^{4} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.30, size = 29, normalized size = 0.74 \begin {gather*} \frac {1}{4} \, \sqrt {x^{4} - 1} x^{2} + \frac {1}{4} \, \log \left (x^{2} - \sqrt {x^{4} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.15, size = 28, normalized size = 0.72

method	result	size
default	\(\frac {x^{2} \sqrt {x^{4}-1}}{4}-\frac {\ln \left (x^{2}+\sqrt {x^{4}-1}\right )}{4}\)	\(28\)
risch	\(\frac {x^{2} \sqrt {x^{4}-1}}{4}-\frac {\ln \left (x^{2}+\sqrt {x^{4}-1}\right )}{4}\)	\(28\)
elliptic	\(\frac {x^{2} \sqrt {x^{4}-1}}{4}-\frac {\ln \left (x^{2}+\sqrt {x^{4}-1}\right )}{4}\)	\(28\)
trager	\(\frac {x^{2} \sqrt {x^{4}-1}}{4}+\frac {\ln \left (x^{2}-\sqrt {x^{4}-1}\right )}{4}\)	\(30\)
meijerg	\(\frac {i \sqrt {\mathrm {signum}\left (x^{4}-1\right )}\, \left (-2 i \sqrt {\pi }\, x^{2} \sqrt {-x^{4}+1}-2 i \sqrt {\pi }\, \arcsin \left (x^{2}\right )\right )}{8 \sqrt {-\mathrm {signum}\left (x^{4}-1\right )}\, \sqrt {\pi }}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 58, normalized size = 1.49 \begin {gather*} -\frac {\sqrt {x^{4} - 1}}{4 \, x^{2} {\left (\frac {x^{4} - 1}{x^{4}} - 1\right )}} - \frac {1}{8} \, \log \left (\frac {\sqrt {x^{4} - 1}}{x^{2}} + 1\right ) + \frac {1}{8} \, \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*(x^4-1)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int x\,\sqrt {x^4-1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(x^4 - 1)^(1/2),x)

[Out]

int(x*(x^4 - 1)^(1/2), x)

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sympy [A]  time = 1.28, size = 60, normalized size = 1.54 \begin {gather*} \begin {cases} \frac {x^{6}}{4 \sqrt {x^{4} - 1}} - \frac {x^{2}}{4 \sqrt {x^{4} - 1}} - \frac {\operatorname {acosh}{\left (x^{2} \right )}}{4} & \text {for}\: \left |{x^{4}}\right | > 1 \\\frac {i x^{2} \sqrt {1 - x^{4}}}{4} + \frac {i \operatorname {asin}{\left (x^{2} \right )}}{4} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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