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

Optimal. Leaf size=94 \[ \frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )}{8 \sqrt{b}}-\frac{5}{8} a (a+2 b x) \sqrt{a x+b x^2}+\frac{2 \left (a x+b x^2\right )^{5/2}}{x^2}-\frac{5}{3} b \left (a x+b x^2\right )^{3/2} \]

[Out]

(-5*a*(a + 2*b*x)*Sqrt[a*x + b*x^2])/8 - (5*b*(a*x + b*x^2)^(3/2))/3 + (2*(a*x + b*x^2)^(5/2))/x^2 + (5*a^3*Ar
cTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/(8*Sqrt[b])

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Rubi [A]  time = 0.0391494, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {662, 664, 612, 620, 206} \[ \frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )}{8 \sqrt{b}}-\frac{5}{8} a (a+2 b x) \sqrt{a x+b x^2}+\frac{2 \left (a x+b x^2\right )^{5/2}}{x^2}-\frac{5}{3} b \left (a x+b x^2\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(a*x + b*x^2)^(5/2)/x^3,x]

[Out]

(-5*a*(a + 2*b*x)*Sqrt[a*x + b*x^2])/8 - (5*b*(a*x + b*x^2)^(3/2))/3 + (2*(a*x + b*x^2)^(5/2))/x^2 + (5*a^3*Ar
cTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/(8*Sqrt[b])

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 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*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^3} \, dx &=\frac{2 \left (a x+b x^2\right )^{5/2}}{x^2}-(5 b) \int \frac{\left (a x+b x^2\right )^{3/2}}{x} \, dx\\ &=-\frac{5}{3} b \left (a x+b x^2\right )^{3/2}+\frac{2 \left (a x+b x^2\right )^{5/2}}{x^2}-\frac{1}{2} (5 a b) \int \sqrt{a x+b x^2} \, dx\\ &=-\frac{5}{8} a (a+2 b x) \sqrt{a x+b x^2}-\frac{5}{3} b \left (a x+b x^2\right )^{3/2}+\frac{2 \left (a x+b x^2\right )^{5/2}}{x^2}+\frac{1}{16} \left (5 a^3\right ) \int \frac{1}{\sqrt{a x+b x^2}} \, dx\\ &=-\frac{5}{8} a (a+2 b x) \sqrt{a x+b x^2}-\frac{5}{3} b \left (a x+b x^2\right )^{3/2}+\frac{2 \left (a x+b x^2\right )^{5/2}}{x^2}+\frac{1}{8} \left (5 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a x+b x^2}}\right )\\ &=-\frac{5}{8} a (a+2 b x) \sqrt{a x+b x^2}-\frac{5}{3} b \left (a x+b x^2\right )^{3/2}+\frac{2 \left (a x+b x^2\right )^{5/2}}{x^2}+\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )}{8 \sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.124836, size = 80, normalized size = 0.85 \[ \frac{1}{24} \sqrt{x (a+b x)} \left (\frac{15 a^{5/2} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{b} \sqrt{x} \sqrt{\frac{b x}{a}+1}}+33 a^2+26 a b x+8 b^2 x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x + b*x^2)^(5/2)/x^3,x]

[Out]

(Sqrt[x*(a + b*x)]*(33*a^2 + 26*a*b*x + 8*b^2*x^2 + (15*a^(5/2)*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(Sqrt[b]*S
qrt[x]*Sqrt[1 + (b*x)/a])))/24

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Maple [B]  time = 0.05, size = 158, normalized size = 1.7 \begin{align*} 2\,{\frac{ \left ( b{x}^{2}+ax \right ) ^{7/2}}{a{x}^{3}}}-{\frac{16\,b}{3\,{a}^{2}{x}^{2}} \left ( b{x}^{2}+ax \right ) ^{{\frac{7}{2}}}}+{\frac{16\,{b}^{2}}{3\,{a}^{2}} \left ( b{x}^{2}+ax \right ) ^{{\frac{5}{2}}}}+{\frac{10\,{b}^{2}x}{3\,a} \left ( b{x}^{2}+ax \right ) ^{{\frac{3}{2}}}}+{\frac{5\,b}{3} \left ( b{x}^{2}+ax \right ) ^{{\frac{3}{2}}}}-{\frac{5\,abx}{4}\sqrt{b{x}^{2}+ax}}-{\frac{5\,{a}^{2}}{8}\sqrt{b{x}^{2}+ax}}+{\frac{5\,{a}^{3}}{16}\ln \left ({ \left ({\frac{a}{2}}+bx \right ){\frac{1}{\sqrt{b}}}}+\sqrt{b{x}^{2}+ax} \right ){\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.92348, size = 346, normalized size = 3.68 \begin{align*} \left [\frac{15 \, a^{3} \sqrt{b} \log \left (2 \, b x + a + 2 \, \sqrt{b x^{2} + a x} \sqrt{b}\right ) + 2 \,{\left (8 \, b^{3} x^{2} + 26 \, a b^{2} x + 33 \, a^{2} b\right )} \sqrt{b x^{2} + a x}}{48 \, b}, -\frac{15 \, a^{3} \sqrt{-b} \arctan \left (\frac{\sqrt{b x^{2} + a x} \sqrt{-b}}{b x}\right ) -{\left (8 \, b^{3} x^{2} + 26 \, a b^{2} x + 33 \, a^{2} b\right )} \sqrt{b x^{2} + a x}}{24 \, b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(15*a^3*sqrt(b)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) + 2*(8*b^3*x^2 + 26*a*b^2*x + 33*a^2*b)*sqr
t(b*x^2 + a*x))/b, -1/24*(15*a^3*sqrt(-b)*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x)) - (8*b^3*x^2 + 26*a*b^2*x +
33*a^2*b)*sqrt(b*x^2 + a*x))/b]

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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^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.24978, size = 97, normalized size = 1.03 \begin{align*} -\frac{5 \, a^{3} \log \left ({\left | -2 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )} \sqrt{b} - a \right |}\right )}{16 \, \sqrt{b}} + \frac{1}{24} \, \sqrt{b x^{2} + a x}{\left (33 \, a^{2} + 2 \,{\left (4 \, b^{2} x + 13 \, a b\right )} x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/16*a^3*log(abs(-2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) - a))/sqrt(b) + 1/24*sqrt(b*x^2 + a*x)*(33*a^2 +
2*(4*b^2*x + 13*a*b)*x)