3.5.7 \(\int \frac {1}{\sqrt {\frac {a+b x^3}{x}}} \, dx\) [407]

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

[Out]

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

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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 = {2004, 2033, 212} \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 212

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

Rule 2004

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

Rule 2033

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} \text {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 = 0.01, size = 63, normalized size = 1.97 \begin {gather*} \frac {2 \sqrt {a+b x^3} \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {b} x^{3/2}}\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[a + b*x^3]/(Sqrt[b]*x^(3/2))])/(3*Sqrt[b]*Sqrt[x]*Sqrt[(a + b*x^3)/x])

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.41, size = 477, normalized size = 14.91

method	result	size
default	\(-\frac {4 \left (b \,x^{3}+a \right ) \left (-1+i \sqrt {3}\right ) \sqrt {-\frac {\left (i \sqrt {3}-3\right ) x b}{\left (-1+i \sqrt {3}\right ) \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}}\, \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )^{2} \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}+2 b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}}{\left (1+i \sqrt {3}\right ) \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}}\, \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}-2 b x -\left (-a \,b^{2}\right )^{\frac {1}{3}}}{\left (-1+i \sqrt {3}\right ) \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}}\, \left (\EllipticF \left (\sqrt {-\frac {\left (i \sqrt {3}-3\right ) x b}{\left (-1+i \sqrt {3}\right ) \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}}, \sqrt {\frac {\left (i \sqrt {3}+3\right ) \left (-1+i \sqrt {3}\right )}{\left (1+i \sqrt {3}\right ) \left (i \sqrt {3}-3\right )}}\right )-\EllipticPi \left (\sqrt {-\frac {\left (i \sqrt {3}-3\right ) x b}{\left (-1+i \sqrt {3}\right ) \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}}, \frac {-1+i \sqrt {3}}{i \sqrt {3}-3}, \sqrt {\frac {\left (i \sqrt {3}+3\right ) \left (-1+i \sqrt {3}\right )}{\left (1+i \sqrt {3}\right ) \left (i \sqrt {3}-3\right )}}\right )\right )}{b^{2} \sqrt {\frac {b \,x^{3}+a}{x}}\, \sqrt {x \left (b \,x^{3}+a \right )}\, \left (i \sqrt {3}-3\right ) \sqrt {\frac {x \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) \left (i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}+2 b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) \left (i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}-2 b x -\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}{b^{2}}}}\)	\(477\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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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Fricas [A]
time = 1.71, 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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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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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,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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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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