3.2.78 \(\int \frac {1}{x \sqrt {a x^2+b x^3}} \, dx\)

Optimal. Leaf size=54 \[ \frac {b \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{3/2}}-\frac {\sqrt {a x^2+b x^3}}{a x^2} \]

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Rubi [A]  time = 0.05, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2025, 2008, 206} \begin {gather*} \frac {b \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{3/2}}-\frac {\sqrt {a x^2+b x^3}}{a x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*Sqrt[a*x^2 + b*x^3]),x]

[Out]

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

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 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]

Rule 2025

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 66, normalized size = 1.22 \begin {gather*} \frac {2 b x (a+b x) \left (\frac {\tanh ^{-1}\left (\sqrt {\frac {b x}{a}+1}\right )}{2 \sqrt {\frac {b x}{a}+1}}-\frac {a}{2 b x}\right )}{a^2 \sqrt {x^2 (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*Sqrt[a*x^2 + b*x^3]),x]

[Out]

(2*b*x*(a + b*x)*(-1/2*a/(b*x) + ArcTanh[Sqrt[1 + (b*x)/a]]/(2*Sqrt[1 + (b*x)/a])))/(a^2*Sqrt[x^2*(a + b*x)])

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IntegrateAlgebraic [A]  time = 0.06, size = 54, normalized size = 1.00 \begin {gather*} \frac {b \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{3/2}}-\frac {\sqrt {a x^2+b x^3}}{a x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x*Sqrt[a*x^2 + b*x^3]),x]

[Out]

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

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fricas [A]  time = 0.40, size = 127, normalized size = 2.35 \begin {gather*} \left [\frac {\sqrt {a} b x^{2} \log \left (\frac {b x^{2} + 2 \, a x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) - 2 \, \sqrt {b x^{3} + a x^{2}} a}{2 \, a^{2} x^{2}}, -\frac {\sqrt {-a} b x^{2} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{a x}\right ) + \sqrt {b x^{3} + a x^{2}} a}{a^{2} x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Warning, choosing root of [1,0,%%%{-4,[1
,0,0]%%%}+%%%{-2,[0,1,1]%%%},0,%%%{1,[0,2,2]%%%}] at parameters values [62.4600259969,-13,46]-1/a*sqrt(a*(1/x)
^2+b/x)-2*b/4/a/sqrt(a)*ln(abs(2*sqrt(a)*(sqrt(a*(1/x)^2+b/x)-sqrt(a)/x)-b))

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maple [A]  time = 0.05, size = 55, normalized size = 1.02 \begin {gather*} -\frac {\sqrt {b x +a}\, \left (-a b x \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+\sqrt {b x +a}\, a^{\frac {3}{2}}\right )}{\sqrt {b \,x^{3}+a \,x^{2}}\, a^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*x+a)^(1/2)*((b*x+a)^(1/2)*a^(3/2)-arctanh((b*x+a)^(1/2)/a^(1/2))*x*a*b)/(b*x^3+a*x^2)^(1/2)/a^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {b x^{3} + a x^{2}} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{x\,\sqrt {b\,x^3+a\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a*x^2 + b*x^3)^(1/2)),x)

[Out]

int(1/(x*(a*x^2 + b*x^3)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \sqrt {x^{2} \left (a + b x\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x*sqrt(x**2*(a + b*x))), x)

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