3.32 \(\int \frac{(a x+b x^2)^{5/2}}{x^6} \, dx\)

Optimal. Leaf size=91 \[ -\frac{2 b^2 \sqrt{a x+b x^2}}{x}+2 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )-\frac{2 b \left (a x+b x^2\right )^{3/2}}{3 x^3}-\frac{2 \left (a x+b x^2\right )^{5/2}}{5 x^5} \]

[Out]

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

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Rubi [A]  time = 0.0406987, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {662, 620, 206} \[ -\frac{2 b^2 \sqrt{a x+b x^2}}{x}+2 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )-\frac{2 b \left (a x+b x^2\right )^{3/2}}{3 x^3}-\frac{2 \left (a x+b x^2\right )^{5/2}}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Int[(a*x + b*x^2)^(5/2)/x^6,x]

[Out]

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

Rule 662

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2
)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[
p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

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

Mathematica [C]  time = 0.013242, size = 50, normalized size = 0.55 \[ -\frac{2 a^2 \sqrt{x (a+b x)} \, _2F_1\left (-\frac{5}{2},-\frac{5}{2};-\frac{3}{2};-\frac{b x}{a}\right )}{5 x^3 \sqrt{\frac{b x}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x + b*x^2)^(5/2)/x^6,x]

[Out]

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

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Maple [B]  time = 0.048, size = 232, normalized size = 2.6 \begin{align*} -{\frac{2}{5\,a{x}^{6}} \left ( b{x}^{2}+ax \right ) ^{{\frac{7}{2}}}}-{\frac{4\,b}{15\,{a}^{2}{x}^{5}} \left ( b{x}^{2}+ax \right ) ^{{\frac{7}{2}}}}-{\frac{16\,{b}^{2}}{15\,{a}^{3}{x}^{4}} \left ( b{x}^{2}+ax \right ) ^{{\frac{7}{2}}}}+{\frac{32\,{b}^{3}}{5\,{a}^{4}{x}^{3}} \left ( b{x}^{2}+ax \right ) ^{{\frac{7}{2}}}}-{\frac{256\,{b}^{4}}{15\,{a}^{5}{x}^{2}} \left ( b{x}^{2}+ax \right ) ^{{\frac{7}{2}}}}+{\frac{256\,{b}^{5}}{15\,{a}^{5}} \left ( b{x}^{2}+ax \right ) ^{{\frac{5}{2}}}}+{\frac{32\,{b}^{5}x}{3\,{a}^{4}} \left ( b{x}^{2}+ax \right ) ^{{\frac{3}{2}}}}+{\frac{16\,{b}^{4}}{3\,{a}^{3}} \left ( b{x}^{2}+ax \right ) ^{{\frac{3}{2}}}}-4\,{\frac{{b}^{4}\sqrt{b{x}^{2}+ax}x}{{a}^{2}}}-2\,{\frac{{b}^{3}\sqrt{b{x}^{2}+ax}}{a}}+{b}^{{\frac{5}{2}}}\ln \left ({ \left ({\frac{a}{2}}+bx \right ){\frac{1}{\sqrt{b}}}}+\sqrt{b{x}^{2}+ax} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a*x)^(5/2)/x^6,x)

[Out]

-2/5/a/x^6*(b*x^2+a*x)^(7/2)-4/15*b/a^2/x^5*(b*x^2+a*x)^(7/2)-16/15*b^2/a^3/x^4*(b*x^2+a*x)^(7/2)+32/5*b^3/a^4
/x^3*(b*x^2+a*x)^(7/2)-256/15*b^4/a^5/x^2*(b*x^2+a*x)^(7/2)+256/15*b^5/a^5*(b*x^2+a*x)^(5/2)+32/3*b^5/a^4*(b*x
^2+a*x)^(3/2)*x+16/3*b^4/a^3*(b*x^2+a*x)^(3/2)-4*b^4/a^2*(b*x^2+a*x)^(1/2)*x-2*b^3/a*(b*x^2+a*x)^(1/2)+b^(5/2)
*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a*x)^(5/2)/x^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.03784, size = 346, normalized size = 3.8 \begin{align*} \left [\frac{15 \, b^{\frac{5}{2}} x^{3} \log \left (2 \, b x + a + 2 \, \sqrt{b x^{2} + a x} \sqrt{b}\right ) - 2 \,{\left (23 \, b^{2} x^{2} + 11 \, a b x + 3 \, a^{2}\right )} \sqrt{b x^{2} + a x}}{15 \, x^{3}}, -\frac{2 \,{\left (15 \, \sqrt{-b} b^{2} x^{3} \arctan \left (\frac{\sqrt{b x^{2} + a x} \sqrt{-b}}{b x}\right ) +{\left (23 \, b^{2} x^{2} + 11 \, a b x + 3 \, a^{2}\right )} \sqrt{b x^{2} + a x}\right )}}{15 \, x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a*x)^(5/2)/x^6,x, algorithm="fricas")

[Out]

[1/15*(15*b^(5/2)*x^3*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) - 2*(23*b^2*x^2 + 11*a*b*x + 3*a^2)*sqrt(b*
x^2 + a*x))/x^3, -2/15*(15*sqrt(-b)*b^2*x^3*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x)) + (23*b^2*x^2 + 11*a*b*x
+ 3*a^2)*sqrt(b*x^2 + a*x))/x^3]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (x \left (a + b x\right )\right )^{\frac{5}{2}}}{x^{6}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a*x)**(5/2)/x**6,x)

[Out]

Integral((x*(a + b*x))**(5/2)/x**6, x)

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Giac [B]  time = 1.29556, size = 236, normalized size = 2.59 \begin{align*} -b^{\frac{5}{2}} \log \left ({\left | -2 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )} \sqrt{b} - a \right |}\right ) + \frac{2 \,{\left (45 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )}^{4} a b^{2} + 45 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )}^{3} a^{2} b^{\frac{3}{2}} + 35 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )}^{2} a^{3} b + 15 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )} a^{4} \sqrt{b} + 3 \, a^{5}\right )}}{15 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a*x)^(5/2)/x^6,x, algorithm="giac")

[Out]

-b^(5/2)*log(abs(-2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) - a)) + 2/15*(45*(sqrt(b)*x - sqrt(b*x^2 + a*x))^4
*a*b^2 + 45*(sqrt(b)*x - sqrt(b*x^2 + a*x))^3*a^2*b^(3/2) + 35*(sqrt(b)*x - sqrt(b*x^2 + a*x))^2*a^3*b + 15*(s
qrt(b)*x - sqrt(b*x^2 + a*x))*a^4*sqrt(b) + 3*a^5)/(sqrt(b)*x - sqrt(b*x^2 + a*x))^5