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

Optimal. Leaf size=23 \[ \frac {2}{3} \sqrt {a} \sinh ^{-1}\left (\frac {(a x)^{3/2}}{a^{3/2}}\right ) \]

[Out]

2/3*arcsinh((a*x)^(3/2)/a^(3/2))*a^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {329, 275, 215} \[ \frac {2}{3} \sqrt {a} \sinh ^{-1}\left (\frac {(a x)^{3/2}}{a^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*x]/Sqrt[1 + x^3],x]

[Out]

(2*Sqrt[a]*ArcSinh[(a*x)^(3/2)/a^(3/2)])/3

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

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Mathematica [A]  time = 0.01, size = 22, normalized size = 0.96 \[ \frac {2 \sqrt {a x} \sinh ^{-1}\left (x^{3/2}\right )}{3 \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*x]/Sqrt[1 + x^3],x]

[Out]

(2*Sqrt[a*x]*ArcSinh[x^(3/2)])/(3*Sqrt[x])

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fricas [B]  time = 0.50, size = 85, normalized size = 3.70 \[ \left [\frac {1}{6} \, \sqrt {a} \log \left (-8 \, a x^{6} - 8 \, a x^{3} - 4 \, {\left (2 \, x^{4} + x\right )} \sqrt {x^{3} + 1} \sqrt {a x} \sqrt {a} - a\right ), -\frac {1}{3} \, \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {x^{3} + 1} \sqrt {a x} \sqrt {-a} x}{2 \, a x^{3} + a}\right )\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*sqrt(a)*log(-8*a*x^6 - 8*a*x^3 - 4*(2*x^4 + x)*sqrt(x^3 + 1)*sqrt(a*x)*sqrt(a) - a), -1/3*sqrt(-a)*arctan
(2*sqrt(x^3 + 1)*sqrt(a*x)*sqrt(-a)*x/(2*a*x^3 + a))]

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giac [B]  time = 0.19, size = 35, normalized size = 1.52 \[ -\frac {2 \, a^{\frac {5}{2}} \log \left (-\sqrt {a x} a^{\frac {3}{2}} x + \sqrt {a^{4} x^{3} + a^{4}}\right )}{3 \, {\left | a \right |}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*a^(5/2)*log(-sqrt(a*x)*a^(3/2)*x + sqrt(a^4*x^3 + a^4))/abs(a)^2

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maple [C]  time = 0.14, size = 321, normalized size = 13.96 \[ -\frac {4 \sqrt {a x}\, \sqrt {x^{3}+1}\, \left (1+i \sqrt {3}\right ) \sqrt {\frac {\left (3+i \sqrt {3}\right ) x}{\left (1+i \sqrt {3}\right ) \left (x +1\right )}}\, \left (x +1\right )^{2} \sqrt {\frac {2 x +i \sqrt {3}-1}{\left (i \sqrt {3}-1\right ) \left (x +1\right )}}\, \sqrt {\frac {-2 x +i \sqrt {3}+1}{\left (1+i \sqrt {3}\right ) \left (x +1\right )}}\, \left (\EllipticF \left (\sqrt {\frac {\left (3+i \sqrt {3}\right ) x}{\left (1+i \sqrt {3}\right ) \left (x +1\right )}}, \sqrt {\frac {\left (-3+i \sqrt {3}\right ) \left (1+i \sqrt {3}\right )}{\left (i \sqrt {3}-1\right ) \left (3+i \sqrt {3}\right )}}\right )-\EllipticPi \left (\sqrt {\frac {\left (3+i \sqrt {3}\right ) x}{\left (1+i \sqrt {3}\right ) \left (x +1\right )}}, \frac {1+i \sqrt {3}}{3+i \sqrt {3}}, \sqrt {\frac {\left (-3+i \sqrt {3}\right ) \left (1+i \sqrt {3}\right )}{\left (i \sqrt {3}-1\right ) \left (3+i \sqrt {3}\right )}}\right )\right ) a}{\sqrt {\left (x^{3}+1\right ) a x}\, \left (3+i \sqrt {3}\right ) \sqrt {-\left (x +1\right ) \left (2 x +i \sqrt {3}-1\right ) \left (-2 x +i \sqrt {3}+1\right ) a x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*(a*x)^(1/2)*(x^3+1)^(1/2)*a*(1+I*3^(1/2))*((3+I*3^(1/2))/(1+I*3^(1/2))/(x+1)*x)^(1/2)*(x+1)^2*((2*x+I*3^(1/
2)-1)/(I*3^(1/2)-1)/(x+1))^(1/2)*((-2*x+I*3^(1/2)+1)/(1+I*3^(1/2))/(x+1))^(1/2)*(EllipticF(((3+I*3^(1/2))/(1+I
*3^(1/2))/(x+1)*x)^(1/2),((-3+I*3^(1/2))*(1+I*3^(1/2))/(I*3^(1/2)-1)/(3+I*3^(1/2)))^(1/2))-EllipticPi(((3+I*3^
(1/2))/(1+I*3^(1/2))/(x+1)*x)^(1/2),(1+I*3^(1/2))/(3+I*3^(1/2)),((-3+I*3^(1/2))*(1+I*3^(1/2))/(I*3^(1/2)-1)/(3
+I*3^(1/2)))^(1/2)))/((x^3+1)*a*x)^(1/2)/(3+I*3^(1/2))/(-(x+1)*(2*x+I*3^(1/2)-1)*(-2*x+I*3^(1/2)+1)*a*x)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a x}}{\sqrt {x^{3} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a*x)/sqrt(x^3 + 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \[ \int \frac {\sqrt {a\,x}}{\sqrt {x^3+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x)^(1/2)/(x^3 + 1)^(1/2),x)

[Out]

int((a*x)^(1/2)/(x^3 + 1)^(1/2), x)

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sympy [A]  time = 1.15, size = 14, normalized size = 0.61 \[ \frac {2 \sqrt {a} \operatorname {asinh}{\left (x^{\frac {3}{2}} \right )}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(a)*asinh(x**(3/2))/3

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