3.1.19 \(\int \frac {\sqrt {1+x^3}}{x} \, dx\) [19]

Optimal. Leaf size=28 \[ \frac {2 \sqrt {1+x^3}}{3}-\frac {2}{3} \tanh ^{-1}\left (\sqrt {1+x^3}\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 28, 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 = {272, 52, 65, 213} \begin {gather*} \frac {2 \sqrt {x^3+1}}{3}-\frac {2}{3} \tanh ^{-1}\left (\sqrt {x^3+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + x^3]/x,x]

[Out]

(2*Sqrt[1 + x^3])/3 - (2*ArcTanh[Sqrt[1 + x^3]])/3

Rule 52

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

Rule 65

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 213

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

Rule 272

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

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Mathematica [A]
time = 0.02, size = 28, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {1+x^3}}{3}-\frac {2}{3} \tanh ^{-1}\left (\sqrt {1+x^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + x^3]/x,x]

[Out]

(2*Sqrt[1 + x^3])/3 - (2*ArcTanh[Sqrt[1 + x^3]])/3

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Maple [A]
time = 0.12, size = 21, normalized size = 0.75

method	result	size
default	\(-\frac {2 \arctanh \left (\sqrt {x^{3}+1}\right )}{3}+\frac {2 \sqrt {x^{3}+1}}{3}\)	\(21\)
elliptic	\(-\frac {2 \arctanh \left (\sqrt {x^{3}+1}\right )}{3}+\frac {2 \sqrt {x^{3}+1}}{3}\)	\(21\)
trager	\(\frac {2 \sqrt {x^{3}+1}}{3}-\frac {\ln \left (-\frac {x^{3}+2 \sqrt {x^{3}+1}+2}{x^{3}}\right )}{3}\)	\(33\)
meijerg	\(-\frac {-2 \left (2-2 \ln \left (2\right )+3 \ln \left (x \right )\right ) \sqrt {\pi }+4 \sqrt {\pi }-4 \sqrt {\pi }\, \sqrt {x^{3}+1}+4 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {x^{3}+1}}{2}\right )}{6 \sqrt {\pi }}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 4.98, size = 34, normalized size = 1.21 \begin {gather*} \frac {2}{3} \, \sqrt {x^{3} + 1} - \frac {1}{3} \, \log \left (\sqrt {x^{3} + 1} + 1\right ) + \frac {1}{3} \, \log \left (\sqrt {x^{3} + 1} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.46, size = 34, normalized size = 1.21 \begin {gather*} \frac {2}{3} \, \sqrt {x^{3} + 1} - \frac {1}{3} \, \log \left (\sqrt {x^{3} + 1} + 1\right ) + \frac {1}{3} \, \log \left (\sqrt {x^{3} + 1} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.56, size = 48, normalized size = 1.71 \begin {gather*} \frac {2 x^{\frac {3}{2}}}{3 \sqrt {1 + \frac {1}{x^{3}}}} - \frac {2 \operatorname {asinh}{\left (\frac {1}{x^{\frac {3}{2}}} \right )}}{3} + \frac {2}{3 x^{\frac {3}{2}} \sqrt {1 + \frac {1}{x^{3}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.47, size = 35, normalized size = 1.25 \begin {gather*} \frac {2}{3} \, \sqrt {x^{3} + 1} - \frac {1}{3} \, \log \left (\sqrt {x^{3} + 1} + 1\right ) + \frac {1}{3} \, \log \left ({\left | \sqrt {x^{3} + 1} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.13, size = 174, normalized size = 6.21 \begin {gather*} \frac {2\,\sqrt {x^3+1}}{3}-\frac {2\,\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((x^3 + 1)^(1/2)/x,x)

[Out]

(2*(x^3 + 1)^(1/2))/3 - (2*((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)

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