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

Optimal. Leaf size=74 \[ -2 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right )+\frac{2 a \sqrt{a x^2+b x^3}}{x}+\frac{2 \left (a x^2+b x^3\right )^{3/2}}{3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.0958373, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2021, 2008, 206} \[ -2 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right )+\frac{2 a \sqrt{a x^2+b x^3}}{x}+\frac{2 \left (a x^2+b x^3\right )^{3/2}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a*x^2 + b*x^3)^(3/2)/x^4,x]

[Out]

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

Rule 2021

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 + n*p + 1)), x] + Dist[(a*(n - j)*p)/(c^j*(m + n*p + 1)), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*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{\left (a x^2+b x^3\right )^{3/2}}{x^4} \, dx &=\frac{2 \left (a x^2+b x^3\right )^{3/2}}{3 x^3}+a \int \frac{\sqrt{a x^2+b x^3}}{x^2} \, dx\\ &=\frac{2 a \sqrt{a x^2+b x^3}}{x}+\frac{2 \left (a x^2+b x^3\right )^{3/2}}{3 x^3}+a^2 \int \frac{1}{\sqrt{a x^2+b x^3}} \, dx\\ &=\frac{2 a \sqrt{a x^2+b x^3}}{x}+\frac{2 \left (a x^2+b x^3\right )^{3/2}}{3 x^3}-\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-a x^2} \, dx,x,\frac{x}{\sqrt{a x^2+b x^3}}\right )\\ &=\frac{2 a \sqrt{a x^2+b x^3}}{x}+\frac{2 \left (a x^2+b x^3\right )^{3/2}}{3 x^3}-2 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0426705, size = 68, normalized size = 0.92 \[ \frac{2 x \sqrt{a+b x} \left (\sqrt{a+b x} (4 a+b x)-3 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )\right )}{3 \sqrt{x^2 (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x^2 + b*x^3)^(3/2)/x^4,x]

[Out]

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

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Maple [A]  time = 0.007, size = 63, normalized size = 0.9 \begin{align*} -{\frac{2}{3\,{x}^{3}} \left ( b{x}^{3}+a{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( 3\,{a}^{3/2}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) - \left ( bx+a \right ) ^{{\frac{3}{2}}}-3\,a\sqrt{bx+a} \right ) \left ( bx+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a*x^2)^(3/2)/x^4,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{3} + a x^{2}\right )}^{\frac{3}{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)^(3/2)/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.810448, size = 296, normalized size = 4. \begin{align*} \left [\frac{3 \, a^{\frac{3}{2}} x \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 (b x + 4 \, a\right )}}{3 \, x}, \frac{2 \,{\left (3 \, \sqrt{-a} a x \arctan \left (\frac{\sqrt{b x^{3} + a x^{2}} \sqrt{-a}}{a x}\right ) + \sqrt{b x^{3} + a x^{2}}{\left (b x + 4 \, a\right )}\right )}}{3 \, x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (x^{2} \left (a + b x\right )\right )^{\frac{3}{2}}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.19226, size = 115, normalized size = 1.55 \begin{align*} \frac{2 \, a^{2} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right ) \mathrm{sgn}\left (x\right )}{\sqrt{-a}} + \frac{2}{3} \,{\left (b x + a\right )}^{\frac{3}{2}} \mathrm{sgn}\left (x\right ) + 2 \, \sqrt{b x + a} a \mathrm{sgn}\left (x\right ) - \frac{2 \,{\left (3 \, a^{2} \arctan \left (\frac{\sqrt{a}}{\sqrt{-a}}\right ) + 4 \, \sqrt{-a} a^{\frac{3}{2}}\right )} \mathrm{sgn}\left (x\right )}{3 \, \sqrt{-a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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