3.248 \(\int \frac{(a x^2+b x^3)^{3/2}}{x^6} \, dx\)

Optimal. Leaf size=81 \[ -\frac{3 b^2 \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right )}{4 \sqrt{a}}-\frac{3 b \sqrt{a x^2+b x^3}}{4 x^2}-\frac{\left (a x^2+b x^3\right )^{3/2}}{2 x^5} \]

[Out]

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

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Rubi [A]  time = 0.0921833, antiderivative size = 81, 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 = {2020, 2008, 206} \[ -\frac{3 b^2 \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right )}{4 \sqrt{a}}-\frac{3 b \sqrt{a x^2+b x^3}}{4 x^2}-\frac{\left (a x^2+b x^3\right )^{3/2}}{2 x^5} \]

Antiderivative was successfully verified.

[In]

Int[(a*x^2 + b*x^3)^(3/2)/x^6,x]

[Out]

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

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

Mathematica [A]  time = 0.0463637, size = 72, normalized size = 0.89 \[ -\frac{2 a^2+3 b^2 x^2 \sqrt{\frac{b x}{a}+1} \tanh ^{-1}\left (\sqrt{\frac{b x}{a}+1}\right )+7 a b x+5 b^2 x^2}{4 x \sqrt{x^2 (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x^2 + b*x^3)^(3/2)/x^6,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*Sqrt[x^2*(a + b*x
)])

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Maple [A]  time = 0.01, size = 74, normalized size = 0.9 \begin{align*}{\frac{1}{4\,{x}^{5}} \left ( b{x}^{3}+a{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( -3\,{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ){x}^{2}{b}^{2}+3\,{a}^{3/2}\sqrt{bx+a}-5\,\sqrt{a} \left ( bx+a \right ) ^{3/2} \right ){\frac{1}{\sqrt{a}}} \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^6,x)

[Out]

1/4*(b*x^3+a*x^2)^(3/2)*(-3*arctanh((b*x+a)^(1/2)/a^(1/2))*x^2*b^2+3*a^(3/2)*(b*x+a)^(1/2)-5*a^(1/2)*(b*x+a)^(
3/2))/x^5/(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^{6}}\,{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^6,x, algorithm="maxima")

[Out]

integrate((b*x^3 + a*x^2)^(3/2)/x^6, x)

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Fricas [A]  time = 0.928418, size = 342, normalized size = 4.22 \begin{align*} \left [\frac{3 \, \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 (5 \, a b x + 2 \, a^{2}\right )}}{8 \, a x^{3}}, \frac{3 \, \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 (5 \, a b x + 2 \, a^{2}\right )}}{4 \, a x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a*x^2)^(3/2)/x^6,x, algorithm="fricas")

[Out]

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

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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^{6}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a*x**2)**(3/2)/x**6,x)

[Out]

Integral((x**2*(a + b*x))**(3/2)/x**6, x)

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Giac [A]  time = 1.27062, size = 95, normalized size = 1.17 \begin{align*} \frac{\frac{3 \, b^{3} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right ) \mathrm{sgn}\left (x\right )}{\sqrt{-a}} - \frac{5 \,{\left (b x + a\right )}^{\frac{3}{2}} b^{3} \mathrm{sgn}\left (x\right ) - 3 \, \sqrt{b x + a} a b^{3} \mathrm{sgn}\left (x\right )}{b^{2} x^{2}}}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a*x^2)^(3/2)/x^6,x, algorithm="giac")

[Out]

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