\(\int \frac {\sqrt {1+x^3}}{x^4} \, dx\) [378]

Optimal result

Integrand size = 13, antiderivative size = 31 \[ \int \frac {\sqrt {1+x^3}}{x^4} \, dx=-\frac {\sqrt {1+x^3}}{3 x^3}-\frac {1}{3} \text {arctanh}\left (\sqrt {1+x^3}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {272, 43, 65, 213} \[ \int \frac {\sqrt {1+x^3}}{x^4} \, dx=-\frac {1}{3} \text {arctanh}\left (\sqrt {x^3+1}\right )-\frac {\sqrt {x^3+1}}{3 x^3} \]

[In]

Int[Sqrt[1 + x^3]/x^4,x]

[Out]

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

Rule 43

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

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*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {\sqrt {1+x}}{x^2} \, dx,x,x^3\right ) \\ & = -\frac {\sqrt {1+x^3}}{3 x^3}+\frac {1}{6} \text {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} \text {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} \text {arctanh}\left (\sqrt {1+x^3}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {1+x^3}}{x^4} \, dx=-\frac {\sqrt {1+x^3}}{3 x^3}-\frac {1}{3} \text {arctanh}\left (\sqrt {1+x^3}\right ) \]

[In]

Integrate[Sqrt[1 + x^3]/x^4,x]

[Out]

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

Maple [A] (verified)

Time = 2.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77

method	result	size
default	\(-\frac {\sqrt {x^{3}+1}}{3 x^{3}}-\frac {\operatorname {arctanh}\left (\sqrt {x^{3}+1}\right )}{3}\)	\(24\)
risch	\(-\frac {\sqrt {x^{3}+1}}{3 x^{3}}-\frac {\operatorname {arctanh}\left (\sqrt {x^{3}+1}\right )}{3}\)	\(24\)
elliptic	\(-\frac {\sqrt {x^{3}+1}}{3 x^{3}}-\frac {\operatorname {arctanh}\left (\sqrt {x^{3}+1}\right )}{3}\)	\(24\)
trager	\(-\frac {\sqrt {x^{3}+1}}{3 x^{3}}+\frac {\ln \left (-\frac {-x^{3}+2 \sqrt {x^{3}+1}-2}{x^{3}}\right )}{6}\)	\(38\)
pseudoelliptic	\(\frac {\ln \left (\sqrt {x^{3}+1}-1\right ) x^{3}-\ln \left (\sqrt {x^{3}+1}+1\right ) x^{3}-2 \sqrt {x^{3}+1}}{6 x^{3}}\)	\(45\)
meijerg	\(-\frac {-\frac {\sqrt {\pi }\, \left (4 x^{3}+8\right )}{4 x^{3}}+\frac {2 \sqrt {\pi }\, \sqrt {x^{3}+1}}{x^{3}}+2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {x^{3}+1}}{2}\right )-\left (-2 \ln \left (2\right )-1+3 \ln \left (x \right )\right ) \sqrt {\pi }+\frac {2 \sqrt {\pi }}{x^{3}}}{6 \sqrt {\pi }}\)	\(77\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {\sqrt {1+x^3}}{x^4} \, dx=-\frac {x^{3} \log \left (\sqrt {x^{3} + 1} + 1\right ) - x^{3} \log \left (\sqrt {x^{3} + 1} - 1\right ) + 2 \, \sqrt {x^{3} + 1}}{6 \, x^{3}} \]

[In]

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

Sympy [A] (verification not implemented)

Time = 0.81 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {\sqrt {1+x^3}}{x^4} \, dx=- \frac {\operatorname {asinh}{\left (\frac {1}{x^{\frac {3}{2}}} \right )}}{3} - \frac {\sqrt {1 + \frac {1}{x^{3}}}}{3 x^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19 \[ \int \frac {\sqrt {1+x^3}}{x^4} \, dx=-\frac {\sqrt {x^{3} + 1}}{3 \, x^{3}} - \frac {1}{6} \, \log \left (\sqrt {x^{3} + 1} + 1\right ) + \frac {1}{6} \, \log \left (\sqrt {x^{3} + 1} - 1\right ) \]

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {\sqrt {1+x^3}}{x^4} \, dx=-\frac {\sqrt {x^{3} + 1}}{3 \, x^{3}} - \frac {1}{6} \, \log \left (\sqrt {x^{3} + 1} + 1\right ) + \frac {1}{6} \, \log \left ({\left | \sqrt {x^{3} + 1} - 1 \right |}\right ) \]

[In]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 177, normalized size of antiderivative = 5.71 \[ \int \frac {\sqrt {1+x^3}}{x^4} \, dx=-\frac {\sqrt {x^3+1}}{3\,x^3}-\frac {\left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\sqrt {\frac {x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {\frac {1}{2}-x+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\Pi \left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2};\mathrm {asin}\left (\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )}{\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \]

[In]

int((x^3 + 1)^(1/2)/x^4,x)

[Out]

- (x^3 + 1)^(1/2)/(3*x^3) - (((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)