3.239 \(\int \frac{\sqrt{a x^2+b x^3}}{x^5} \, dx\)

Optimal. Leaf size=112 \[ \frac{b^2 \sqrt{a x^2+b x^3}}{8 a^2 x^2}-\frac{b^3 \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right )}{8 a^{5/2}}-\frac{b \sqrt{a x^2+b x^3}}{12 a x^3}-\frac{\sqrt{a x^2+b x^3}}{3 x^4} \]

[Out]

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

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Rubi [A]  time = 0.137877, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2020, 2025, 2008, 206} \[ \frac{b^2 \sqrt{a x^2+b x^3}}{8 a^2 x^2}-\frac{b^3 \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right )}{8 a^{5/2}}-\frac{b \sqrt{a x^2+b x^3}}{12 a x^3}-\frac{\sqrt{a x^2+b x^3}}{3 x^4} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*x^2 + b*x^3]/x^5,x]

[Out]

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

Rule 2020

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

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

Mathematica [C]  time = 0.0124094, size = 42, normalized size = 0.38 \[ \frac{2 b^3 \left (x^2 (a+b x)\right )^{3/2} \, _2F_1\left (\frac{3}{2},4;\frac{5}{2};\frac{b x}{a}+1\right )}{3 a^4 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*x^2 + b*x^3]/x^5,x]

[Out]

(2*b^3*(x^2*(a + b*x))^(3/2)*Hypergeometric2F1[3/2, 4, 5/2, 1 + (b*x)/a])/(3*a^4*x^3)

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Maple [A]  time = 0.012, size = 89, normalized size = 0.8 \begin{align*} -{\frac{1}{24\,{x}^{4}}\sqrt{b{x}^{3}+a{x}^{2}} \left ( 3\,{a}^{9/2}\sqrt{bx+a}+8\,{a}^{7/2} \left ( bx+a \right ) ^{3/2}-3\,{a}^{5/2} \left ( bx+a \right ) ^{5/2}+3\,{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ){a}^{2}{b}^{3}{x}^{3} \right ){a}^{-{\frac{9}{2}}}{\frac{1}{\sqrt{bx+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a*x^2)^(1/2)/x^5,x)

[Out]

-1/24*(b*x^3+a*x^2)^(1/2)*(3*a^(9/2)*(b*x+a)^(1/2)+8*a^(7/2)*(b*x+a)^(3/2)-3*a^(5/2)*(b*x+a)^(5/2)+3*arctanh((
b*x+a)^(1/2)/a^(1/2))*a^2*b^3*x^3)/x^4/(b*x+a)^(1/2)/a^(9/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b x^{3} + a x^{2}}}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*x^3 + a*x^2)/x^5, x)

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Fricas [A]  time = 0.933463, size = 393, normalized size = 3.51 \begin{align*} \left [\frac{3 \, \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 (3 \, a b^{2} x^{2} - 2 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt{b x^{3} + a x^{2}}}{48 \, a^{3} x^{4}}, \frac{3 \, \sqrt{-a} b^{3} x^{4} \arctan \left (\frac{\sqrt{b x^{3} + a x^{2}} \sqrt{-a}}{a x}\right ) +{\left (3 \, a b^{2} x^{2} - 2 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt{b x^{3} + a x^{2}}}{24 \, a^{3} x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a*x^2)^(1/2)/x^5,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} \left (a + b x\right )}}{x^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a*x**2)**(1/2)/x**5,x)

[Out]

Integral(sqrt(x**2*(a + b*x))/x**5, x)

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Giac [A]  time = 1.18835, size = 116, normalized size = 1.04 \begin{align*} \frac{{\left (\frac{3 \, b^{4} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2}} + \frac{3 \,{\left (b x + a\right )}^{\frac{5}{2}} b^{4} - 8 \,{\left (b x + a\right )}^{\frac{3}{2}} a b^{4} - 3 \, \sqrt{b x + a} a^{2} b^{4}}{a^{2} b^{3} x^{3}}\right )} \mathrm{sgn}\left (x\right )}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a*x^2)^(1/2)/x^5,x, algorithm="giac")

[Out]

1/24*(3*b^4*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^2) + (3*(b*x + a)^(5/2)*b^4 - 8*(b*x + a)^(3/2)*a*b^4 -
3*sqrt(b*x + a)*a^2*b^4)/(a^2*b^3*x^3))*sgn(x)/b