3.3.57 \(\int \frac {1}{x \sqrt {a x^2+b x^3}} \, dx\) [257]

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

[Out]

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

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Rubi [A]
time = 0.04, 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 = {2050, 2033, 212} \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 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 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]

Rule 2050

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 \text {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 = 0.05, size = 60, normalized size = 1.11 \begin {gather*} \frac {-\sqrt {a} (a+b x)+b x \sqrt {a+b x} \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2} \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]

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

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Maple [A]
time = 0.37, size = 55, normalized size = 1.02

method	result	size
default	\(-\frac {\sqrt {b x +a}\, \left (\sqrt {b x +a}\, a^{\frac {3}{2}}-\arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a b x \right )}{\sqrt {b \,x^{3}+a \,x^{2}}\, a^{\frac {5}{2}}}\)	\(55\)
risch	\(-\frac {b x +a}{a \sqrt {x^{2} \left (b x +a \right )}}+\frac {b \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {b x +a}\, x}{a^{\frac {3}{2}} \sqrt {x^{2} \left (b x +a \right )}}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*x+a)^(1/2)*((b*x+a)^(1/2)*a^(3/2)-arctanh((b*x+a)^(1/2)/a^(1/2))*a*b*x)/(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*} \text {Failed to integrate} \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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Fricas [A]
time = 1.99, 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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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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Giac [A]
time = 1.31, size = 51, normalized size = 0.94 \begin {gather*} -\frac {\frac {b^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {\sqrt {b x + a} b}{a x}}{b \mathrm {sgn}\left (x\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="giac")

[Out]

-(b^2*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a) + sqrt(b*x + a)*b/(a*x))/(b*sgn(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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