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

Optimal. Leaf size=87 \[ -\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{4 a^{5/2}}+\frac {3 b \sqrt {a x^2+b x^3}}{4 a^2 x^2}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3} \]

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Rubi [A]  time = 0.09, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, 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 {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{4 a^{5/2}}+\frac {3 b \sqrt {a x^2+b x^3}}{4 a^2 x^2}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*Sqrt[a*x^2 + b*x^3]),x]

[Out]

-Sqrt[a*x^2 + b*x^3]/(2*a*x^3) + (3*b*Sqrt[a*x^2 + b*x^3])/(4*a^2*x^2) - (3*b^2*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2
+ b*x^3]])/(4*a^(5/2))

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^2 \sqrt {a x^2+b x^3}} \, dx &=-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}-\frac {(3 b) \int \frac {1}{x \sqrt {a x^2+b x^3}} \, dx}{4 a}\\ &=-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}+\frac {3 b \sqrt {a x^2+b x^3}}{4 a^2 x^2}+\frac {\left (3 b^2\right ) \int \frac {1}{\sqrt {a x^2+b x^3}} \, dx}{8 a^2}\\ &=-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}+\frac {3 b \sqrt {a x^2+b x^3}}{4 a^2 x^2}-\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {x}{\sqrt {a x^2+b x^3}}\right )}{4 a^2}\\ &=-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}+\frac {3 b \sqrt {a x^2+b x^3}}{4 a^2 x^2}-\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{4 a^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 40, normalized size = 0.46 \begin {gather*} -\frac {2 b^2 \sqrt {x^2 (a+b x)} \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};\frac {b x}{a}+1\right )}{a^3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^2*Sqrt[x^2*(a + b*x)]*Hypergeometric2F1[1/2, 3, 3/2, 1 + (b*x)/a])/(a^3*x)

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IntegrateAlgebraic [A]  time = 0.07, size = 69, normalized size = 0.79 \begin {gather*} \frac {(3 b x-2 a) \sqrt {a x^2+b x^3}}{4 a^2 x^3}-\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{4 a^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*a + 3*b*x)*Sqrt[a*x^2 + b*x^3])/(4*a^2*x^3) - (3*b^2*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]])/(4*a^(5/2)
)

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fricas [A]  time = 0.41, size = 153, normalized size = 1.76 \begin {gather*} \left [\frac {3 \, \sqrt {a} b^{2} x^{3} \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}} {\left (3 \, a b x - 2 \, a^{2}\right )}}{8 \, a^{3} x^{3}}, \frac {3 \, \sqrt {-a} b^{2} x^{3} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{a x}\right ) + \sqrt {b x^{3} + a x^{2}} {\left (3 \, a b x - 2 \, a^{2}\right )}}{4 \, a^{3} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^2/(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]2*(-4*a/16/a^2/x+
6*b/16/a^2)*sqrt(a*(1/x)^2+b/x)+6*b^2/16/a^2/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 = 77, normalized size = 0.89 \begin {gather*} -\frac {\sqrt {b x +a}\, \left (3 a \,b^{2} x^{2} \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )-3 \sqrt {b x +a}\, a^{\frac {3}{2}} b x +2 \sqrt {b x +a}\, a^{\frac {5}{2}}\right )}{4 \sqrt {b \,x^{3}+a \,x^{2}}\, a^{\frac {7}{2}} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b*x+a)^(1/2)*(2*(b*x+a)^(1/2)*a^(5/2)-3*(b*x+a)^(1/2)*a^(3/2)*x*b+3*arctanh((b*x+a)^(1/2)/a^(1/2))*a*x^2
*b^2)/x/(b*x^3+a*x^2)^(1/2)/a^(7/2)

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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^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.41, size = 44, normalized size = 0.51 \begin {gather*} -\frac {2\,\sqrt {\frac {a}{b\,x}+1}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {5}{2};\ \frac {7}{2};\ -\frac {a}{b\,x}\right )}{5\,x\,\sqrt {b\,x^3+a\,x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a*x^2 + b*x^3)^(1/2)),x)

[Out]

-(2*(a/(b*x) + 1)^(1/2)*hypergeom([1/2, 5/2], 7/2, -a/(b*x)))/(5*x*(a*x^2 + b*x^3)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{2} \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**2/(b*x**3+a*x**2)**(1/2),x)

[Out]

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

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