3.238 \(\int \frac{\sqrt{a x^2+b x^3}}{x^4} \, dx\)

Optimal. Leaf size=84 \[ \frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right )}{4 a^{3/2}}-\frac{b \sqrt{a x^2+b x^3}}{4 a x^2}-\frac{\sqrt{a x^2+b x^3}}{2 x^3} \]

[Out]

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

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Rubi [A]  time = 0.0915473, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2020, 2025, 2008, 206} \[ \frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right )}{4 a^{3/2}}-\frac{b \sqrt{a x^2+b x^3}}{4 a x^2}-\frac{\sqrt{a x^2+b x^3}}{2 x^3} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*x^2 + b*x^3]/x^4,x]

[Out]

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

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

Mathematica [C]  time = 0.0118765, size = 42, normalized size = 0.5 \[ -\frac{2 b^2 \left (x^2 (a+b x)\right )^{3/2} \, _2F_1\left (\frac{3}{2},3;\frac{5}{2};\frac{b x}{a}+1\right )}{3 a^3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*x^2 + b*x^3]/x^4,x]

[Out]

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

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Maple [A]  time = 0.01, size = 73, normalized size = 0.9 \begin{align*} -{\frac{1}{4\,{x}^{3}}\sqrt{b{x}^{3}+a{x}^{2}} \left ({a}^{{\frac{3}{2}}} \left ( bx+a \right ) ^{{\frac{3}{2}}}-{\it Artanh} \left ({\sqrt{bx+a}{\frac{1}{\sqrt{a}}}} \right ) a{b}^{2}{x}^{2}+{a}^{{\frac{5}{2}}}\sqrt{bx+a} \right ){a}^{-{\frac{5}{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^4,x)

[Out]

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

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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^{4}}\,{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^4,x, algorithm="maxima")

[Out]

integrate(sqrt(b*x^3 + a*x^2)/x^4, x)

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Fricas [A]  time = 0.865725, size = 338, normalized size = 4.02 \begin{align*} \left [\frac{\sqrt{a} b^{2} x^{3} \log \left (\frac{b x^{2} + 2 \, a x + 2 \, \sqrt{b x^{3} + a x^{2}} \sqrt{a}}{x^{2}}\right ) - 2 \, \sqrt{b x^{3} + a x^{2}}{\left (a b x + 2 \, a^{2}\right )}}{8 \, a^{2} x^{3}}, -\frac{\sqrt{-a} b^{2} x^{3} \arctan \left (\frac{\sqrt{b x^{3} + a x^{2}} \sqrt{-a}}{a x}\right ) + \sqrt{b x^{3} + a x^{2}}{\left (a b x + 2 \, a^{2}\right )}}{4 \, a^{2} x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a*x^2)^(1/2)/x^4,x, algorithm="fricas")

[Out]

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

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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^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a*x**2)**(1/2)/x**4,x)

[Out]

Integral(sqrt(x**2*(a + b*x))/x**4, x)

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Giac [A]  time = 1.21381, size = 92, normalized size = 1.1 \begin{align*} -\frac{{\left (\frac{b^{3} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a} + \frac{{\left (b x + a\right )}^{\frac{3}{2}} b^{3} + \sqrt{b x + a} a b^{3}}{a b^{2} x^{2}}\right )} \mathrm{sgn}\left (x\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a*x^2)^(1/2)/x^4,x, algorithm="giac")

[Out]

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