3.1315 \(\int \frac {x^{3/2}}{\sqrt {1+x^5}} \, dx\)

Optimal. Leaf size=10 \[ \frac {2}{5} \sinh ^{-1}\left (x^{5/2}\right ) \]

[Out]

2/5*arcsinh(x^(5/2))

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Rubi [A]  time = 0.01, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {329, 275, 215} \[ \frac {2}{5} \sinh ^{-1}\left (x^{5/2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/Sqrt[1 + x^5],x]

[Out]

(2*ArcSinh[x^(5/2)])/5

Rule 215

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

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]

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]

Rubi steps

\begin {align*} \int \frac {x^{3/2}}{\sqrt {1+x^5}} \, dx &=2 \operatorname {Subst}\left (\int \frac {x^4}{\sqrt {1+x^{10}}} \, dx,x,\sqrt {x}\right )\\ &=\frac {2}{5} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^2}} \, dx,x,x^{5/2}\right )\\ &=\frac {2}{5} \sinh ^{-1}\left (x^{5/2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 10, normalized size = 1.00 \[ \frac {2}{5} \sinh ^{-1}\left (x^{5/2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)/Sqrt[1 + x^5],x]

[Out]

(2*ArcSinh[x^(5/2)])/5

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fricas [B]  time = 0.88, size = 22, normalized size = 2.20 \[ \frac {1}{5} \, \log \left (2 \, x^{5} + 2 \, \sqrt {x^{5} + 1} x^{\frac {5}{2}} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)/(x^5+1)^(1/2),x, algorithm="fricas")

[Out]

1/5*log(2*x^5 + 2*sqrt(x^5 + 1)*x^(5/2) + 1)

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giac [B]  time = 0.17, size = 16, normalized size = 1.60 \[ -\frac {2}{5} \, \log \left (-x^{\frac {5}{2}} + \sqrt {x^{5} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)/(x^5+1)^(1/2),x, algorithm="giac")

[Out]

-2/5*log(-x^(5/2) + sqrt(x^5 + 1))

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maple [A]  time = 0.13, size = 7, normalized size = 0.70 \[ \frac {2 \arcsinh \left (x^{\frac {5}{2}}\right )}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/(x^5+1)^(1/2),x)

[Out]

2/5*arcsinh(x^(5/2))

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maxima [B]  time = 0.93, size = 33, normalized size = 3.30 \[ \frac {1}{5} \, \log \left (\frac {\sqrt {x^{5} + 1}}{x^{\frac {5}{2}}} + 1\right ) - \frac {1}{5} \, \log \left (\frac {\sqrt {x^{5} + 1}}{x^{\frac {5}{2}}} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)/(x^5+1)^(1/2),x, algorithm="maxima")

[Out]

1/5*log(sqrt(x^5 + 1)/x^(5/2) + 1) - 1/5*log(sqrt(x^5 + 1)/x^(5/2) - 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.10 \[ \int \frac {x^{3/2}}{\sqrt {x^5+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/(x^5 + 1)^(1/2),x)

[Out]

int(x^(3/2)/(x^5 + 1)^(1/2), x)

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sympy [A]  time = 2.38, size = 8, normalized size = 0.80 \[ \frac {2 \operatorname {asinh}{\left (x^{\frac {5}{2}} \right )}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)/(x**5+1)**(1/2),x)

[Out]

2*asinh(x**(5/2))/5

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