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

Optimal. Leaf size=57 \[ -\frac{3 b}{a^2 \sqrt{a+b x}}+\frac{3 b \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{a^{5/2}}-\frac{1}{a x \sqrt{a+b x}} \]

[Out]

(-3*b)/(a^2*Sqrt[a + b*x]) - 1/(a*x*Sqrt[a + b*x]) + (3*b*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/a^(5/2)

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Rubi [A]  time = 0.0165216, antiderivative size = 59, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {51, 63, 208} \[ -\frac{3 \sqrt{a+b x}}{a^2 x}+\frac{3 b \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{a^{5/2}}+\frac{2}{a x \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x)^(3/2)),x]

[Out]

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

Mathematica [C]  time = 0.0156457, size = 31, normalized size = 0.54 \[ -\frac{2 b \, _2F_1\left (-\frac{1}{2},2;\frac{1}{2};\frac{b x}{a}+1\right )}{a^2 \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a + b*x)^(3/2)),x]

[Out]

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

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Maple [A]  time = 0.012, size = 55, normalized size = 1. \begin{align*} 2\,b \left ( -{\frac{1}{{a}^{2}} \left ( 1/2\,{\frac{\sqrt{bx+a}}{bx}}-3/2\,{\frac{1}{\sqrt{a}}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) } \right ) }-{\frac{1}{{a}^{2}\sqrt{bx+a}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.63446, size = 346, normalized size = 6.07 \begin{align*} \left [\frac{3 \,{\left (b^{2} x^{2} + a b x\right )} \sqrt{a} \log \left (\frac{b x + 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) - 2 \,{\left (3 \, a b x + a^{2}\right )} \sqrt{b x + a}}{2 \,{\left (a^{3} b x^{2} + a^{4} x\right )}}, -\frac{3 \,{\left (b^{2} x^{2} + a b x\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (3 \, a b x + a^{2}\right )} \sqrt{b x + a}}{a^{3} b x^{2} + a^{4} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 4.14375, size = 73, normalized size = 1.28 \begin{align*} - \frac{1}{a \sqrt{b} x^{\frac{3}{2}} \sqrt{\frac{a}{b x} + 1}} - \frac{3 \sqrt{b}}{a^{2} \sqrt{x} \sqrt{\frac{a}{b x} + 1}} + \frac{3 b \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} \sqrt{x}} \right )}}{a^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(a*sqrt(b)*x**(3/2)*sqrt(a/(b*x) + 1)) - 3*sqrt(b)/(a**2*sqrt(x)*sqrt(a/(b*x) + 1)) + 3*b*asinh(sqrt(a)/(sq
rt(b)*sqrt(x)))/a**(5/2)

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Giac [A]  time = 1.18841, size = 86, normalized size = 1.51 \begin{align*} -\frac{3 \, b \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2}} - \frac{3 \,{\left (b x + a\right )} b - 2 \, a b}{{\left ({\left (b x + a\right )}^{\frac{3}{2}} - \sqrt{b x + a} a\right )} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*b*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^2) - (3*(b*x + a)*b - 2*a*b)/(((b*x + a)^(3/2) - sqrt(b*x + a)
*a)*a^2)