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

Optimal. Leaf size=110 \[ -\frac{15 b^2 \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right )}{4 a^{7/2}}+\frac{15 b \sqrt{a x^2+b x^3}}{4 a^3 x^2}-\frac{5 \sqrt{a x^2+b x^3}}{2 a^2 x^3}+\frac{2}{a x \sqrt{a x^2+b x^3}} \]

[Out]

2/(a*x*Sqrt[a*x^2 + b*x^3]) - (5*Sqrt[a*x^2 + b*x^3])/(2*a^2*x^3) + (15*b*Sqrt[a*x^2 + b*x^3])/(4*a^3*x^2) - (
15*b^2*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]])/(4*a^(7/2))

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Rubi [A]  time = 0.105175, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {2006, 2025, 2008, 206} \[ -\frac{15 b^2 \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right )}{4 a^{7/2}}+\frac{15 b \sqrt{a x^2+b x^3}}{4 a^3 x^2}-\frac{5 \sqrt{a x^2+b x^3}}{2 a^2 x^3}+\frac{2}{a x \sqrt{a x^2+b x^3}} \]

Antiderivative was successfully verified.

[In]

Int[(a*x^2 + b*x^3)^(-3/2),x]

[Out]

2/(a*x*Sqrt[a*x^2 + b*x^3]) - (5*Sqrt[a*x^2 + b*x^3])/(2*a^2*x^3) + (15*b*Sqrt[a*x^2 + b*x^3])/(4*a^3*x^2) - (
15*b^2*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]])/(4*a^(7/2))

Rule 2006

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

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}{\left (a x^2+b x^3\right )^{3/2}} \, dx &=\frac{2}{a x \sqrt{a x^2+b x^3}}+\frac{5 \int \frac{1}{x^2 \sqrt{a x^2+b x^3}} \, dx}{a}\\ &=\frac{2}{a x \sqrt{a x^2+b x^3}}-\frac{5 \sqrt{a x^2+b x^3}}{2 a^2 x^3}-\frac{(15 b) \int \frac{1}{x \sqrt{a x^2+b x^3}} \, dx}{4 a^2}\\ &=\frac{2}{a x \sqrt{a x^2+b x^3}}-\frac{5 \sqrt{a x^2+b x^3}}{2 a^2 x^3}+\frac{15 b \sqrt{a x^2+b x^3}}{4 a^3 x^2}+\frac{\left (15 b^2\right ) \int \frac{1}{\sqrt{a x^2+b x^3}} \, dx}{8 a^3}\\ &=\frac{2}{a x \sqrt{a x^2+b x^3}}-\frac{5 \sqrt{a x^2+b x^3}}{2 a^2 x^3}+\frac{15 b \sqrt{a x^2+b x^3}}{4 a^3 x^2}-\frac{\left (15 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^3}\\ &=\frac{2}{a x \sqrt{a x^2+b x^3}}-\frac{5 \sqrt{a x^2+b x^3}}{2 a^2 x^3}+\frac{15 b \sqrt{a x^2+b x^3}}{4 a^3 x^2}-\frac{15 b^2 \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right )}{4 a^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0072385, size = 38, normalized size = 0.35 \[ \frac{2 b^2 x \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};\frac{b x}{a}+1\right )}{a^3 \sqrt{x^2 (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x^2 + b*x^3)^(-3/2),x]

[Out]

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

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Maple [A]  time = 0.012, size = 76, normalized size = 0.7 \begin{align*} -{\frac{x \left ( bx+a \right ) }{4} \left ( 15\,{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) \sqrt{bx+a}{x}^{2}{b}^{2}-5\,{a}^{3/2}xb-15\,{x}^{2}{b}^{2}\sqrt{a}+2\,{a}^{5/2} \right ) \left ( b{x}^{3}+a{x}^{2} \right ) ^{-{\frac{3}{2}}}{a}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x*(b*x+a)*(15*arctanh((b*x+a)^(1/2)/a^(1/2))*(b*x+a)^(1/2)*x^2*b^2-5*a^(3/2)*x*b-15*x^2*b^2*a^(1/2)+2*a^(
5/2))/(b*x^3+a*x^2)^(3/2)/a^(7/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a*x^2)^(-3/2), x)

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Fricas [A]  time = 0.901782, size = 466, normalized size = 4.24 \begin{align*} \left [\frac{15 \,{\left (b^{3} x^{4} + a b^{2} x^{3}\right )} \sqrt{a} \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} + 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt{b x^{3} + a x^{2}}}{8 \,{\left (a^{4} b x^{4} + a^{5} x^{3}\right )}}, \frac{15 \,{\left (b^{3} x^{4} + a b^{2} x^{3}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{b x^{3} + a x^{2}} \sqrt{-a}}{a x}\right ) +{\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt{b x^{3} + a x^{2}}}{4 \,{\left (a^{4} b x^{4} + a^{5} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(15*(b^3*x^4 + a*b^2*x^3)*sqrt(a)*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 + 5*a^2*b*x - 2*a^3)*sqrt(b*x^3 + a*x^2))/(a^4*b*x^4 + a^5*x^3), 1/4*(15*(b^3*x^4 + a*b^2*x^3)*sqrt(-a)*ar
ctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(a*x)) + (15*a*b^2*x^2 + 5*a^2*b*x - 2*a^3)*sqrt(b*x^3 + a*x^2))/(a^4*b*x^4
+ a^5*x^3)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x