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

Optimal. Leaf size=73 \[ -\frac{\left (a x^2+b x^3\right )^{3/2}}{x^4}+\frac{3 b \sqrt{a x^2+b x^3}}{x}-3 \sqrt{a} b \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0931158, antiderivative size = 73, 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, 2021, 2008, 206} \[ -\frac{\left (a x^2+b x^3\right )^{3/2}}{x^4}+\frac{3 b \sqrt{a x^2+b x^3}}{x}-3 \sqrt{a} b \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a*x^2 + b*x^3)^(3/2)/x^5,x]

[Out]

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

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

Mathematica [C]  time = 0.0134288, size = 40, normalized size = 0.55 \[ \frac{2 b \left (x^2 (a+b x)\right )^{5/2} \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};\frac{b x}{a}+1\right )}{5 a^2 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x^2 + b*x^3)^(3/2)/x^5,x]

[Out]

(2*b*(x^2*(a + b*x))^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, 1 + (b*x)/a])/(5*a^2*x^5)

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Maple [A]  time = 0.011, size = 72, normalized size = 1. \begin{align*} -{\frac{1}{{x}^{4}} \left ( b{x}^{3}+a{x}^{2} \right ) ^{{\frac{3}{2}}} \left ({a}^{{\frac{3}{2}}}\sqrt{bx+a}-2\,\sqrt{bx+a}bx\sqrt{a}+3\,{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) xab \right ) \left ( bx+a \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^{5}}\,{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^5,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.898792, size = 304, normalized size = 4.16 \begin{align*} \left [\frac{3 \, \sqrt{a} b x^{2} \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 (2 \, b x - a\right )}}{2 \, x^{2}}, \frac{3 \, \sqrt{-a} b x^{2} \arctan \left (\frac{\sqrt{b x^{3} + a x^{2}} \sqrt{-a}}{a x}\right ) + \sqrt{b x^{3} + a x^{2}}{\left (2 \, b x - a\right )}}{x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.45642, size = 84, normalized size = 1.15 \begin{align*} \frac{\frac{3 \, a b^{2} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right ) \mathrm{sgn}\left (x\right )}{\sqrt{-a}} + 2 \, \sqrt{b x + a} b^{2} \mathrm{sgn}\left (x\right ) - \frac{\sqrt{b x + a} a b \mathrm{sgn}\left (x\right )}{x}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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