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

Optimal. Leaf size=115 \[ -\frac{5 b^2 \sqrt{a x^2+b x^3}}{8 a^3 x^2}+\frac{5 b^3 \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right )}{8 a^{7/2}}+\frac{5 b \sqrt{a x^2+b x^3}}{12 a^2 x^3}-\frac{\sqrt{a x^2+b x^3}}{3 a x^4} \]

[Out]

-Sqrt[a*x^2 + b*x^3]/(3*a*x^4) + (5*b*Sqrt[a*x^2 + b*x^3])/(12*a^2*x^3) - (5*b^2*Sqrt[a*x^2 + b*x^3])/(8*a^3*x
^2) + (5*b^3*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]])/(8*a^(7/2))

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Rubi [A]  time = 0.135388, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2025, 2008, 206} \[ -\frac{5 b^2 \sqrt{a x^2+b x^3}}{8 a^3 x^2}+\frac{5 b^3 \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right )}{8 a^{7/2}}+\frac{5 b \sqrt{a x^2+b x^3}}{12 a^2 x^3}-\frac{\sqrt{a x^2+b x^3}}{3 a x^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*Sqrt[a*x^2 + b*x^3]),x]

[Out]

-Sqrt[a*x^2 + b*x^3]/(3*a*x^4) + (5*b*Sqrt[a*x^2 + b*x^3])/(12*a^2*x^3) - (5*b^2*Sqrt[a*x^2 + b*x^3])/(8*a^3*x
^2) + (5*b^3*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]])/(8*a^(7/2))

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]

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 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])

Rubi steps

\begin{align*} \int \frac{1}{x^3 \sqrt{a x^2+b x^3}} \, dx &=-\frac{\sqrt{a x^2+b x^3}}{3 a x^4}-\frac{(5 b) \int \frac{1}{x^2 \sqrt{a x^2+b x^3}} \, dx}{6 a}\\ &=-\frac{\sqrt{a x^2+b x^3}}{3 a x^4}+\frac{5 b \sqrt{a x^2+b x^3}}{12 a^2 x^3}+\frac{\left (5 b^2\right ) \int \frac{1}{x \sqrt{a x^2+b x^3}} \, dx}{8 a^2}\\ &=-\frac{\sqrt{a x^2+b x^3}}{3 a x^4}+\frac{5 b \sqrt{a x^2+b x^3}}{12 a^2 x^3}-\frac{5 b^2 \sqrt{a x^2+b x^3}}{8 a^3 x^2}-\frac{\left (5 b^3\right ) \int \frac{1}{\sqrt{a x^2+b x^3}} \, dx}{16 a^3}\\ &=-\frac{\sqrt{a x^2+b x^3}}{3 a x^4}+\frac{5 b \sqrt{a x^2+b x^3}}{12 a^2 x^3}-\frac{5 b^2 \sqrt{a x^2+b x^3}}{8 a^3 x^2}+\frac{\left (5 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-a x^2} \, dx,x,\frac{x}{\sqrt{a x^2+b x^3}}\right )}{8 a^3}\\ &=-\frac{\sqrt{a x^2+b x^3}}{3 a x^4}+\frac{5 b \sqrt{a x^2+b x^3}}{12 a^2 x^3}-\frac{5 b^2 \sqrt{a x^2+b x^3}}{8 a^3 x^2}+\frac{5 b^3 \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right )}{8 a^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0102403, size = 40, normalized size = 0.35 \[ \frac{2 b^3 \sqrt{x^2 (a+b x)} \, _2F_1\left (\frac{1}{2},4;\frac{3}{2};\frac{b x}{a}+1\right )}{a^4 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b^3*Sqrt[x^2*(a + b*x)]*Hypergeometric2F1[1/2, 4, 3/2, 1 + (b*x)/a])/(a^4*x)

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Maple [A]  time = 0.006, size = 95, normalized size = 0.8 \begin{align*} -{\frac{1}{24\,{x}^{2}}\sqrt{bx+a} \left ( 15\,{a}^{3/2}\sqrt{bx+a}{x}^{2}{b}^{2}-15\,{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ){x}^{3}a{b}^{3}-10\,{a}^{5/2}\sqrt{bx+a}xb+8\,{a}^{7/2}\sqrt{bx+a} \right ){\frac{1}{\sqrt{b{x}^{3}+a{x}^{2}}}}{a}^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24/x^2*(b*x+a)^(1/2)*(15*a^(3/2)*(b*x+a)^(1/2)*x^2*b^2-15*arctanh((b*x+a)^(1/2)/a^(1/2))*x^3*a*b^3-10*a^(5/
2)*(b*x+a)^(1/2)*x*b+8*a^(7/2)*(b*x+a)^(1/2))/(b*x^3+a*x^2)^(1/2)/a^(9/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.809012, size = 402, normalized size = 3.5 \begin{align*} \left [\frac{15 \, \sqrt{a} b^{3} x^{4} \log \left (\frac{b x^{2} + 2 \, a x + 2 \, \sqrt{b x^{3} + a x^{2}} \sqrt{a}}{x^{2}}\right ) - 2 \,{\left (15 \, a b^{2} x^{2} - 10 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt{b x^{3} + a x^{2}}}{48 \, a^{4} x^{4}}, -\frac{15 \, \sqrt{-a} b^{3} x^{4} \arctan \left (\frac{\sqrt{b x^{3} + a x^{2}} \sqrt{-a}}{a x}\right ) +{\left (15 \, a b^{2} x^{2} - 10 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt{b x^{3} + a x^{2}}}{24 \, a^{4} x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(15*sqrt(a)*b^3*x^4*log((b*x^2 + 2*a*x + 2*sqrt(b*x^3 + a*x^2)*sqrt(a))/x^2) - 2*(15*a*b^2*x^2 - 10*a^2*
b*x + 8*a^3)*sqrt(b*x^3 + a*x^2))/(a^4*x^4), -1/24*(15*sqrt(-a)*b^3*x^4*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(a
*x)) + (15*a*b^2*x^2 - 10*a^2*b*x + 8*a^3)*sqrt(b*x^3 + a*x^2))/(a^4*x^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \sqrt{x^{2} \left (a + b x\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError