3.4.76 \(\int \frac {1}{x^4 \sqrt {1+x^3}} \, dx\)

Optimal. Leaf size=31 \[ \frac {1}{3} \tanh ^{-1}\left (\sqrt {x^3+1}\right )-\frac {\sqrt {x^3+1}}{3 x^3} \]

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Rubi [A]  time = 0.01, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {266, 51, 63, 207} \begin {gather*} \frac {1}{3} \tanh ^{-1}\left (\sqrt {x^3+1}\right )-\frac {\sqrt {x^3+1}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^4*Sqrt[1 + x^3]),x]

[Out]

-1/3*Sqrt[1 + x^3]/x^3 + ArcTanh[Sqrt[1 + x^3]]/3

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 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]]

Rubi steps

\begin {align*} \int \frac {1}{x^4 \sqrt {1+x^3}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {1+x}} \, dx,x,x^3\right )\\ &=-\frac {\sqrt {1+x^3}}{3 x^3}-\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1+x}} \, dx,x,x^3\right )\\ &=-\frac {\sqrt {1+x^3}}{3 x^3}-\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sqrt {1+x^3}\right )\\ &=-\frac {\sqrt {1+x^3}}{3 x^3}+\frac {1}{3} \tanh ^{-1}\left (\sqrt {1+x^3}\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 31, normalized size = 1.00 \begin {gather*} \frac {1}{3} \tanh ^{-1}\left (\sqrt {x^3+1}\right )-\frac {\sqrt {x^3+1}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*Sqrt[1 + x^3]),x]

[Out]

-1/3*Sqrt[1 + x^3]/x^3 + ArcTanh[Sqrt[1 + x^3]]/3

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IntegrateAlgebraic [A]  time = 0.04, size = 31, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {1+x^3}}{3 x^3}+\frac {1}{3} \tanh ^{-1}\left (\sqrt {1+x^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^4*Sqrt[1 + x^3]),x]

[Out]

-1/3*Sqrt[1 + x^3]/x^3 + ArcTanh[Sqrt[1 + x^3]]/3

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fricas [A]  time = 0.47, size = 44, normalized size = 1.42 \begin {gather*} \frac {x^{3} \log \left (\sqrt {x^{3} + 1} + 1\right ) - x^{3} \log \left (\sqrt {x^{3} + 1} - 1\right ) - 2 \, \sqrt {x^{3} + 1}}{6 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(x^3+1)^(1/2),x, algorithm="fricas")

[Out]

1/6*(x^3*log(sqrt(x^3 + 1) + 1) - x^3*log(sqrt(x^3 + 1) - 1) - 2*sqrt(x^3 + 1))/x^3

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giac [A]  time = 0.24, size = 38, normalized size = 1.23 \begin {gather*} -\frac {\sqrt {x^{3} + 1}}{3 \, x^{3}} + \frac {1}{6} \, \log \left (\sqrt {x^{3} + 1} + 1\right ) - \frac {1}{6} \, \log \left ({\left | \sqrt {x^{3} + 1} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(x^3+1)^(1/2),x, algorithm="giac")

[Out]

-1/3*sqrt(x^3 + 1)/x^3 + 1/6*log(sqrt(x^3 + 1) + 1) - 1/6*log(abs(sqrt(x^3 + 1) - 1))

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maple [A]  time = 0.23, size = 24, normalized size = 0.77

method	result	size
default	\(-\frac {\sqrt {x^{3}+1}}{3 x^{3}}+\frac {\arctanh \left (\sqrt {x^{3}+1}\right )}{3}\)	\(24\)
risch	\(-\frac {\sqrt {x^{3}+1}}{3 x^{3}}+\frac {\arctanh \left (\sqrt {x^{3}+1}\right )}{3}\)	\(24\)
elliptic	\(-\frac {\sqrt {x^{3}+1}}{3 x^{3}}+\frac {\arctanh \left (\sqrt {x^{3}+1}\right )}{3}\)	\(24\)
trager	\(-\frac {\sqrt {x^{3}+1}}{3 x^{3}}-\frac {\ln \left (-\frac {-x^{3}+2 \sqrt {x^{3}+1}-2}{x^{3}}\right )}{6}\)	\(38\)
meijerg	\(\frac {-\frac {\sqrt {\pi }}{x^{3}}-\frac {\left (1-2 \ln \relax (2)+3 \ln \relax (x )\right ) \sqrt {\pi }}{2}+\frac {\sqrt {\pi }\, \left (4 x^{3}+8\right )}{8 x^{3}}-\frac {\sqrt {\pi }\, \sqrt {x^{3}+1}}{x^{3}}+\ln \left (\frac {1}{2}+\frac {\sqrt {x^{3}+1}}{2}\right ) \sqrt {\pi }}{3 \sqrt {\pi }}\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(x^3+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/3*(x^3+1)^(1/2)/x^3+1/3*arctanh((x^3+1)^(1/2))

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maxima [A]  time = 0.35, size = 37, normalized size = 1.19 \begin {gather*} -\frac {\sqrt {x^{3} + 1}}{3 \, x^{3}} + \frac {1}{6} \, \log \left (\sqrt {x^{3} + 1} + 1\right ) - \frac {1}{6} \, \log \left (\sqrt {x^{3} + 1} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(x^3+1)^(1/2),x, algorithm="maxima")

[Out]

-1/3*sqrt(x^3 + 1)/x^3 + 1/6*log(sqrt(x^3 + 1) + 1) - 1/6*log(sqrt(x^3 + 1) - 1)

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mupad [B]  time = 0.16, size = 176, normalized size = 5.68 \begin {gather*} -\frac {\sqrt {x^3+1}}{3\,x^3}+\frac {\left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\sqrt {\frac {x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {\frac {1}{2}-x+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\Pi \left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2};\mathrm {asin}\left (\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )}{\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(x^3 + 1)^(1/2)),x)

[Out]

(((3^(1/2)*1i)/2 + 3/2)*((x + (3^(1/2)*1i)/2 - 1/2)/((3^(1/2)*1i)/2 - 3/2))^(1/2)*((x + 1)/((3^(1/2)*1i)/2 + 3
/2))^(1/2)*(((3^(1/2)*1i)/2 - x + 1/2)/((3^(1/2)*1i)/2 + 3/2))^(1/2)*ellipticPi((3^(1/2)*1i)/2 + 3/2, asin(((x
+ 1)/((3^(1/2)*1i)/2 + 3/2))^(1/2)), -((3^(1/2)*1i)/2 + 3/2)/((3^(1/2)*1i)/2 - 3/2)))/(x^3 - x*(((3^(1/2)*1i)
/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2) + 1) - ((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2))^(1/2) - (x^3 + 1)^(1/2)
/(3*x^3)

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sympy [A]  time = 1.46, size = 26, normalized size = 0.84 \begin {gather*} \frac {\operatorname {asinh}{\left (\frac {1}{x^{\frac {3}{2}}} \right )}}{3} - \frac {\sqrt {1 + \frac {1}{x^{3}}}}{3 x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(x**3+1)**(1/2),x)

[Out]

asinh(x**(-3/2))/3 - sqrt(1 + x**(-3))/(3*x**(3/2))

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