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

Optimal. Leaf size=32 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {\frac {a}{x}+b x^2}}\right )}{3 \sqrt {b}} \]

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Rubi [A]  time = 0.01, antiderivative size = 32, 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 = {1979, 2008, 206} \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {\frac {a}{x}+b x^2}}\right )}{3 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[(a + b*x^3)/x],x]

[Out]

(2*ArcTanh[(Sqrt[b]*x)/Sqrt[a/x + b*x^2]])/(3*Sqrt[b])

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 1979

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && GeneralizedBinomialQ[u, x] &&  !Gene
ralizedBinomialMatchQ[u, x]

Rule 2008

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[2/(2 - n), Subst[Int[1/(1 - a*x^2), x], x, x/Sq
rt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, n}, x] && NeQ[n, 2]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 63, normalized size = 1.97 \begin {gather*} \frac {2 \sqrt {a+b x^3} \tanh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a+b x^3}}\right )}{3 \sqrt {b} \sqrt {x} \sqrt {\frac {a+b x^3}{x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[(a + b*x^3)/x],x]

[Out]

(2*Sqrt[a + b*x^3]*ArcTanh[(Sqrt[b]*x^(3/2))/Sqrt[a + b*x^3]])/(3*Sqrt[b]*Sqrt[x]*Sqrt[(a + b*x^3)/x])

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IntegrateAlgebraic [A]  time = 0.36, size = 34, normalized size = 1.06 \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {\sqrt {\frac {a+b x^3}{x}}}{\sqrt {b} x}\right )}{3 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/Sqrt[(a + b*x^3)/x],x]

[Out]

(2*ArcTanh[Sqrt[(a + b*x^3)/x]/(Sqrt[b]*x)])/(3*Sqrt[b])

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fricas [A]  time = 0.57, size = 102, normalized size = 3.19 \begin {gather*} \left [\frac {\log \left (-8 \, b^{2} x^{6} - 8 \, a b x^{3} - a^{2} - 4 \, {\left (2 \, b x^{5} + a x^{2}\right )} \sqrt {b} \sqrt {\frac {b x^{3} + a}{x}}\right )}{6 \, \sqrt {b}}, -\frac {\sqrt {-b} \arctan \left (\frac {2 \, \sqrt {-b} x^{2} \sqrt {\frac {b x^{3} + a}{x}}}{2 \, b x^{3} + a}\right )}{3 \, b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*log(-8*b^2*x^6 - 8*a*b*x^3 - a^2 - 4*(2*b*x^5 + a*x^2)*sqrt(b)*sqrt((b*x^3 + a)/x))/sqrt(b), -1/3*sqrt(-b
)*arctan(2*sqrt(-b)*x^2*sqrt((b*x^3 + a)/x)/(2*b*x^3 + a))/b]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x)]Limit: Max order reached or unable to make series expansion Error: Bad Argument Value

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maple [C]  time = 0.57, size = 477, normalized size = 14.91 \begin {gather*} -\frac {4 \left (b \,x^{3}+a \right ) \left (i \sqrt {3}-1\right ) \sqrt {-\frac {\left (i \sqrt {3}-3\right ) b x}{\left (i \sqrt {3}-1\right ) \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}}\, \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )^{2} \sqrt {\frac {2 b x +i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}+\left (-a \,b^{2}\right )^{\frac {1}{3}}}{\left (1+i \sqrt {3}\right ) \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}}\, \sqrt {\frac {-2 b x +i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}-\left (-a \,b^{2}\right )^{\frac {1}{3}}}{\left (i \sqrt {3}-1\right ) \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}}\, \left (\EllipticF \left (\sqrt {-\frac {\left (i \sqrt {3}-3\right ) b x}{\left (i \sqrt {3}-1\right ) \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}}, \sqrt {\frac {\left (i \sqrt {3}+3\right ) \left (i \sqrt {3}-1\right )}{\left (1+i \sqrt {3}\right ) \left (i \sqrt {3}-3\right )}}\right )-\EllipticPi \left (\sqrt {-\frac {\left (i \sqrt {3}-3\right ) b x}{\left (i \sqrt {3}-1\right ) \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}}, \frac {i \sqrt {3}-1}{i \sqrt {3}-3}, \sqrt {\frac {\left (i \sqrt {3}+3\right ) \left (i \sqrt {3}-1\right )}{\left (1+i \sqrt {3}\right ) \left (i \sqrt {3}-3\right )}}\right )\right )}{\sqrt {\frac {b \,x^{3}+a}{x}}\, \sqrt {\left (b \,x^{3}+a \right ) x}\, \left (i \sqrt {3}-3\right ) \sqrt {\frac {\left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) \left (2 b x +i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}+\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) \left (-2 b x +i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}-\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) x}{b^{2}}}\, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt((b*x^3 + a)/x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {1}{\sqrt {\frac {b\,x^3+a}{x}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x^3)/x)^(1/2),x)

[Out]

int(1/((a + b*x^3)/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 {\frac {a + b x^{3}}{x}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt((a + b*x**3)/x), x)

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