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

Optimal. Leaf size=62 \[ -\frac{3 b^2 \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{4 \sqrt{a}}-\frac{(a+b x)^{3/2}}{2 x^2}-\frac{3 b \sqrt{a+b x}}{4 x} \]

[Out]

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

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Rubi [A]  time = 0.0160989, antiderivative size = 62, normalized size of antiderivative = 1., 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 = {47, 63, 208} \[ -\frac{3 b^2 \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{4 \sqrt{a}}-\frac{(a+b x)^{3/2}}{2 x^2}-\frac{3 b \sqrt{a+b x}}{4 x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(3/2)/x^3,x]

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && 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{(a+b x)^{3/2}}{x^3} \, dx &=-\frac{(a+b x)^{3/2}}{2 x^2}+\frac{1}{4} (3 b) \int \frac{\sqrt{a+b x}}{x^2} \, dx\\ &=-\frac{3 b \sqrt{a+b x}}{4 x}-\frac{(a+b x)^{3/2}}{2 x^2}+\frac{1}{8} \left (3 b^2\right ) \int \frac{1}{x \sqrt{a+b x}} \, dx\\ &=-\frac{3 b \sqrt{a+b x}}{4 x}-\frac{(a+b x)^{3/2}}{2 x^2}+\frac{1}{4} (3 b) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x}\right )\\ &=-\frac{3 b \sqrt{a+b x}}{4 x}-\frac{(a+b x)^{3/2}}{2 x^2}-\frac{3 b^2 \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{4 \sqrt{a}}\\ \end{align*}

Mathematica [A]  time = 0.0531582, size = 68, normalized size = 1.1 \[ -\frac{2 a^2+3 b^2 x^2 \sqrt{\frac{b x}{a}+1} \tanh ^{-1}\left (\sqrt{\frac{b x}{a}+1}\right )+7 a b x+5 b^2 x^2}{4 x^2 \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(3/2)/x^3,x]

[Out]

-(2*a^2 + 7*a*b*x + 5*b^2*x^2 + 3*b^2*x^2*Sqrt[1 + (b*x)/a]*ArcTanh[Sqrt[1 + (b*x)/a]])/(4*x^2*Sqrt[a + b*x])

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Maple [A]  time = 0.008, size = 51, normalized size = 0.8 \begin{align*} 2\,{b}^{2} \left ({\frac{-5/8\, \left ( bx+a \right ) ^{3/2}+3/8\,a\sqrt{bx+a}}{{b}^{2}{x}^{2}}}-3/8\,{\frac{1}{\sqrt{a}}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^2*((-5/8*(b*x+a)^(3/2)+3/8*a*(b*x+a)^(1/2))/b^2/x^2-3/8*arctanh((b*x+a)^(1/2)/a^(1/2))/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((b*x+a)^(3/2)/x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.55413, size = 296, normalized size = 4.77 \begin{align*} \left [\frac{3 \, \sqrt{a} b^{2} x^{2} \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) - 2 \,{\left (5 \, a b x + 2 \, a^{2}\right )} \sqrt{b x + a}}{8 \, a x^{2}}, \frac{3 \, \sqrt{-a} b^{2} x^{2} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) -{\left (5 \, a b x + 2 \, a^{2}\right )} \sqrt{b x + a}}{4 \, a x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 4.1397, size = 76, normalized size = 1.23 \begin{align*} - \frac{a \sqrt{b} \sqrt{\frac{a}{b x} + 1}}{2 x^{\frac{3}{2}}} - \frac{5 b^{\frac{3}{2}} \sqrt{\frac{a}{b x} + 1}}{4 \sqrt{x}} - \frac{3 b^{2} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} \sqrt{x}} \right )}}{4 \sqrt{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*sqrt(b)*sqrt(a/(b*x) + 1)/(2*x**(3/2)) - 5*b**(3/2)*sqrt(a/(b*x) + 1)/(4*sqrt(x)) - 3*b**2*asinh(sqrt(a)/(s
qrt(b)*sqrt(x)))/(4*sqrt(a))

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Giac [A]  time = 1.23718, size = 86, normalized size = 1.39 \begin{align*} \frac{\frac{3 \, b^{3} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - \frac{5 \,{\left (b x + a\right )}^{\frac{3}{2}} b^{3} - 3 \, \sqrt{b x + a} a b^{3}}{b^{2} x^{2}}}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(3*b^3*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) - (5*(b*x + a)^(3/2)*b^3 - 3*sqrt(b*x + a)*a*b^3)/(b^2*x^2)
)/b