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

Optimal. Leaf size=52 \[ \frac {\sqrt {x^3} \tan ^{-1}\left (\sqrt {x}\right )}{x^{3/2}}-\frac {\sqrt {x^3} \tanh ^{-1}\left (\sqrt {x}\right )}{x^{3/2}}+\tan ^{-1}\left (\sqrt {x}\right )+\tanh ^{-1}\left (\sqrt {x}\right ) \]

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Rubi [A]  time = 0.13, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6729, 1593, 6725, 329, 212, 206, 203, 15, 298} \begin {gather*} \frac {\sqrt {x^3} \tan ^{-1}\left (\sqrt {x}\right )}{x^{3/2}}-\frac {\sqrt {x^3} \tanh ^{-1}\left (\sqrt {x}\right )}{x^{3/2}}+\tan ^{-1}\left (\sqrt {x}\right )+\tanh ^{-1}\left (\sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[x] + Sqrt[x^3])^(-1),x]

[Out]

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

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 203

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

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 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 329

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

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 6729

Int[(u_.)/((a_.)*(x_)^(m_.) + (b_.)*Sqrt[(c_.)*(x_)^(n_)]), x_Symbol] :> Int[(u*(a*x^m - b*Sqrt[c*x^n]))/(a^2*
x^(2*m) - b^2*c*x^n), x] /; FreeQ[{a, b, c, m, n}, x]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 49, normalized size = 0.94 \begin {gather*} \frac {\left (x^{3/2}+\sqrt {x^3}\right ) \tan ^{-1}\left (\sqrt {x}\right )+\left (x^{3/2}-\sqrt {x^3}\right ) \tanh ^{-1}\left (\sqrt {x}\right )}{x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[x] + Sqrt[x^3])^(-1),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 2.58, size = 39, normalized size = 0.75 \begin {gather*} \tan ^{-1}\left (\frac {\sqrt {x^3}}{x}\right )-\tanh ^{-1}\left (\frac {\sqrt {x^3}}{x}\right )+\tan ^{-1}\left (\sqrt {x}\right )+\tanh ^{-1}\left (\sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(Sqrt[x] + Sqrt[x^3])^(-1),x]

[Out]

ArcTan[Sqrt[x]] + ArcTan[Sqrt[x^3]/x] + ArcTanh[Sqrt[x]] - ArcTanh[Sqrt[x^3]/x]

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fricas [A]  time = 1.41, size = 6, normalized size = 0.12 \begin {gather*} 2 \, \arctan \left (\sqrt {x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(sqrt(x))

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giac [A]  time = 0.32, size = 6, normalized size = 0.12 \begin {gather*} 2 \, \arctan \left (\sqrt {x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(sqrt(x))

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maple [A]  time = 0.01, size = 30, normalized size = 0.58 \begin {gather*} \frac {2 \arctan \left (\sqrt {\frac {\sqrt {x^{3}}}{x^{\frac {3}{2}}}}\, \sqrt {x}\right )}{\sqrt {\frac {\sqrt {x^{3}}}{x^{\frac {3}{2}}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(1/2)+(x^3)^(1/2)),x)

[Out]

2/((x^3)^(1/2)/x^(3/2))^(1/2)*arctan(x^(1/2)*((x^3)^(1/2)/x^(3/2))^(1/2))

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maxima [A]  time = 0.99, size = 6, normalized size = 0.12 \begin {gather*} 2 \, \arctan \left (\sqrt {x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(sqrt(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{\sqrt {x^3}+\sqrt {x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((x^3)^(1/2) + x^(1/2)),x)

[Out]

int(1/((x^3)^(1/2) + x^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {x} + \sqrt {x^{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(x) + sqrt(x**3)), x)

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