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

Optimal. Leaf size=89 \[ 5 b^2 \sqrt{a x+b x^2}+5 a b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )-\frac{2 \left (a x+b x^2\right )^{5/2}}{3 x^4}-\frac{10 b \left (a x+b x^2\right )^{3/2}}{3 x^2} \]

[Out]

5*b^2*Sqrt[a*x + b*x^2] - (10*b*(a*x + b*x^2)^(3/2))/(3*x^2) - (2*(a*x + b*x^2)^(5/2))/(3*x^4) + 5*a*b^(3/2)*A
rcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]]

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Rubi [A]  time = 0.0405703, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {662, 664, 620, 206} \[ 5 b^2 \sqrt{a x+b x^2}+5 a b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )-\frac{2 \left (a x+b x^2\right )^{5/2}}{3 x^4}-\frac{10 b \left (a x+b x^2\right )^{3/2}}{3 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a*x + b*x^2)^(5/2)/x^5,x]

[Out]

5*b^2*Sqrt[a*x + b*x^2] - (10*b*(a*x + b*x^2)^(3/2))/(3*x^2) - (2*(a*x + b*x^2)^(5/2))/(3*x^4) + 5*a*b^(3/2)*A
rcTanh[(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 664

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 + 2*p + 1)), x] - Dist[(p*(2*c*d - b*e))/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*
(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] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*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^5} \, dx &=-\frac{2 \left (a x+b x^2\right )^{5/2}}{3 x^4}+\frac{1}{3} (5 b) \int \frac{\left (a x+b x^2\right )^{3/2}}{x^3} \, dx\\ &=-\frac{10 b \left (a x+b x^2\right )^{3/2}}{3 x^2}-\frac{2 \left (a x+b x^2\right )^{5/2}}{3 x^4}+\left (5 b^2\right ) \int \frac{\sqrt{a x+b x^2}}{x} \, dx\\ &=5 b^2 \sqrt{a x+b x^2}-\frac{10 b \left (a x+b x^2\right )^{3/2}}{3 x^2}-\frac{2 \left (a x+b x^2\right )^{5/2}}{3 x^4}+\frac{1}{2} \left (5 a b^2\right ) \int \frac{1}{\sqrt{a x+b x^2}} \, dx\\ &=5 b^2 \sqrt{a x+b x^2}-\frac{10 b \left (a x+b x^2\right )^{3/2}}{3 x^2}-\frac{2 \left (a x+b x^2\right )^{5/2}}{3 x^4}+\left (5 a b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a x+b x^2}}\right )\\ &=5 b^2 \sqrt{a x+b x^2}-\frac{10 b \left (a x+b x^2\right )^{3/2}}{3 x^2}-\frac{2 \left (a x+b x^2\right )^{5/2}}{3 x^4}+5 a b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0128478, size = 50, normalized size = 0.56 \[ -\frac{2 a^2 \sqrt{x (a+b x)} \, _2F_1\left (-\frac{5}{2},-\frac{3}{2};-\frac{1}{2};-\frac{b x}{a}\right )}{3 x^2 \sqrt{\frac{b x}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x + b*x^2)^(5/2)/x^5,x]

[Out]

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

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Maple [B]  time = 0.05, size = 209, normalized size = 2.4 \begin{align*} -{\frac{2}{3\,a{x}^{5}} \left ( b{x}^{2}+ax \right ) ^{{\frac{7}{2}}}}-{\frac{8\,b}{3\,{a}^{2}{x}^{4}} \left ( b{x}^{2}+ax \right ) ^{{\frac{7}{2}}}}+16\,{\frac{{b}^{2} \left ( b{x}^{2}+ax \right ) ^{7/2}}{{a}^{3}{x}^{3}}}-{\frac{128\,{b}^{3}}{3\,{a}^{4}{x}^{2}} \left ( b{x}^{2}+ax \right ) ^{{\frac{7}{2}}}}+{\frac{128\,{b}^{4}}{3\,{a}^{4}} \left ( b{x}^{2}+ax \right ) ^{{\frac{5}{2}}}}+{\frac{80\,{b}^{4}x}{3\,{a}^{3}} \left ( b{x}^{2}+ax \right ) ^{{\frac{3}{2}}}}+{\frac{40\,{b}^{3}}{3\,{a}^{2}} \left ( b{x}^{2}+ax \right ) ^{{\frac{3}{2}}}}-10\,{\frac{{b}^{3}\sqrt{b{x}^{2}+ax}x}{a}}-5\,{b}^{2}\sqrt{b{x}^{2}+ax}+{\frac{5\,a}{2}{b}^{{\frac{3}{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^5,x)

[Out]

-2/3/a/x^5*(b*x^2+a*x)^(7/2)-8/3*b/a^2/x^4*(b*x^2+a*x)^(7/2)+16*b^2/a^3/x^3*(b*x^2+a*x)^(7/2)-128/3*b^3/a^4/x^
2*(b*x^2+a*x)^(7/2)+128/3*b^4/a^4*(b*x^2+a*x)^(5/2)+80/3*b^4/a^3*(b*x^2+a*x)^(3/2)*x+40/3*b^3/a^2*(b*x^2+a*x)^
(3/2)-10*b^3/a*(b*x^2+a*x)^(1/2)*x-5*b^2*(b*x^2+a*x)^(1/2)+5/2*b^(3/2)*a*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^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.9333, size = 343, normalized size = 3.85 \begin{align*} \left [\frac{15 \, a b^{\frac{3}{2}} x^{2} \log \left (2 \, b x + a + 2 \, \sqrt{b x^{2} + a x} \sqrt{b}\right ) + 2 \,{\left (3 \, b^{2} x^{2} - 14 \, a b x - 2 \, a^{2}\right )} \sqrt{b x^{2} + a x}}{6 \, x^{2}}, -\frac{15 \, a \sqrt{-b} b x^{2} \arctan \left (\frac{\sqrt{b x^{2} + a x} \sqrt{-b}}{b x}\right ) -{\left (3 \, b^{2} x^{2} - 14 \, a b x - 2 \, a^{2}\right )} \sqrt{b x^{2} + a x}}{3 \, x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(15*a*b^(3/2)*x^2*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) + 2*(3*b^2*x^2 - 14*a*b*x - 2*a^2)*sqrt(b*
x^2 + a*x))/x^2, -1/3*(15*a*sqrt(-b)*b*x^2*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x)) - (3*b^2*x^2 - 14*a*b*x -
2*a^2)*sqrt(b*x^2 + a*x))/x^2]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.21262, size = 180, normalized size = 2.02 \begin{align*} -\frac{5}{2} \, a b^{\frac{3}{2}} \log \left ({\left | -2 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )} \sqrt{b} - a \right |}\right ) + \sqrt{b x^{2} + a x} b^{2} + \frac{2 \,{\left (9 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )}^{2} a^{2} b + 3 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )} a^{3} \sqrt{b} + a^{4}\right )}}{3 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/2*a*b^(3/2)*log(abs(-2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) - a)) + sqrt(b*x^2 + a*x)*b^2 + 2/3*(9*(sqrt
(b)*x - sqrt(b*x^2 + a*x))^2*a^2*b + 3*(sqrt(b)*x - sqrt(b*x^2 + a*x))*a^3*sqrt(b) + a^4)/(sqrt(b)*x - sqrt(b*
x^2 + a*x))^3