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

Optimal. Leaf size=101 \[ -\frac{5 a^4 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )}{64 b^{3/2}}+\frac{5 a^2 (a+2 b x) \sqrt{a x+b x^2}}{64 b}+\frac{5}{24} a \left (a x+b x^2\right )^{3/2}+\frac{\left (a x+b x^2\right )^{5/2}}{4 x} \]

[Out]

(5*a^2*(a + 2*b*x)*Sqrt[a*x + b*x^2])/(64*b) + (5*a*(a*x + b*x^2)^(3/2))/24 + (a*x + b*x^2)^(5/2)/(4*x) - (5*a
^4*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/(64*b^(3/2))

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Rubi [A]  time = 0.041171, antiderivative size = 101, 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 = {664, 612, 620, 206} \[ -\frac{5 a^4 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )}{64 b^{3/2}}+\frac{5 a^2 (a+2 b x) \sqrt{a x+b x^2}}{64 b}+\frac{5}{24} a \left (a x+b x^2\right )^{3/2}+\frac{\left (a x+b x^2\right )^{5/2}}{4 x} \]

Antiderivative was successfully verified.

[In]

Int[(a*x + b*x^2)^(5/2)/x^2,x]

[Out]

(5*a^2*(a + 2*b*x)*Sqrt[a*x + b*x^2])/(64*b) + (5*a*(a*x + b*x^2)^(3/2))/24 + (a*x + b*x^2)^(5/2)/(4*x) - (5*a
^4*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/(64*b^(3/2))

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^2} \, dx &=\frac{\left (a x+b x^2\right )^{5/2}}{4 x}+\frac{1}{8} (5 a) \int \frac{\left (a x+b x^2\right )^{3/2}}{x} \, dx\\ &=\frac{5}{24} a \left (a x+b x^2\right )^{3/2}+\frac{\left (a x+b x^2\right )^{5/2}}{4 x}+\frac{1}{16} \left (5 a^2\right ) \int \sqrt{a x+b x^2} \, dx\\ &=\frac{5 a^2 (a+2 b x) \sqrt{a x+b x^2}}{64 b}+\frac{5}{24} a \left (a x+b x^2\right )^{3/2}+\frac{\left (a x+b x^2\right )^{5/2}}{4 x}-\frac{\left (5 a^4\right ) \int \frac{1}{\sqrt{a x+b x^2}} \, dx}{128 b}\\ &=\frac{5 a^2 (a+2 b x) \sqrt{a x+b x^2}}{64 b}+\frac{5}{24} a \left (a x+b x^2\right )^{3/2}+\frac{\left (a x+b x^2\right )^{5/2}}{4 x}-\frac{\left (5 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a x+b x^2}}\right )}{64 b}\\ &=\frac{5 a^2 (a+2 b x) \sqrt{a x+b x^2}}{64 b}+\frac{5}{24} a \left (a x+b x^2\right )^{3/2}+\frac{\left (a x+b x^2\right )^{5/2}}{4 x}-\frac{5 a^4 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )}{64 b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.131444, size = 98, normalized size = 0.97 \[ \frac{\sqrt{x (a+b x)} \left (\sqrt{b} \left (118 a^2 b x+15 a^3+136 a b^2 x^2+48 b^3 x^3\right )-\frac{15 a^{7/2} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{x} \sqrt{\frac{b x}{a}+1}}\right )}{192 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x + b*x^2)^(5/2)/x^2,x]

[Out]

(Sqrt[x*(a + b*x)]*(Sqrt[b]*(15*a^3 + 118*a^2*b*x + 136*a*b^2*x^2 + 48*b^3*x^3) - (15*a^(7/2)*ArcSinh[(Sqrt[b]
*Sqrt[x])/Sqrt[a]])/(Sqrt[x]*Sqrt[1 + (b*x)/a])))/(192*b^(3/2))

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Maple [A]  time = 0.046, size = 135, normalized size = 1.3 \begin{align*}{\frac{2}{3\,a{x}^{2}} \left ( b{x}^{2}+ax \right ) ^{{\frac{7}{2}}}}-{\frac{2\,b}{3\,a} \left ( b{x}^{2}+ax \right ) ^{{\frac{5}{2}}}}-{\frac{5\,bx}{12} \left ( b{x}^{2}+ax \right ) ^{{\frac{3}{2}}}}-{\frac{5\,a}{24} \left ( b{x}^{2}+ax \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{a}^{2}x}{32}\sqrt{b{x}^{2}+ax}}+{\frac{5\,{a}^{3}}{64\,b}\sqrt{b{x}^{2}+ax}}-{\frac{5\,{a}^{4}}{128}\ln \left ({ \left ({\frac{a}{2}}+bx \right ){\frac{1}{\sqrt{b}}}}+\sqrt{b{x}^{2}+ax} \right ){b}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/a/x^2*(b*x^2+a*x)^(7/2)-2/3*b/a*(b*x^2+a*x)^(5/2)-5/12*b*(b*x^2+a*x)^(3/2)*x-5/24*a*(b*x^2+a*x)^(3/2)+5/32
*a^2*(b*x^2+a*x)^(1/2)*x+5/64/b*a^3*(b*x^2+a*x)^(1/2)-5/128/b^(3/2)*a^4*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^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.11876, size = 406, normalized size = 4.02 \begin{align*} \left [\frac{15 \, a^{4} \sqrt{b} \log \left (2 \, b x + a - 2 \, \sqrt{b x^{2} + a x} \sqrt{b}\right ) + 2 \,{\left (48 \, b^{4} x^{3} + 136 \, a b^{3} x^{2} + 118 \, a^{2} b^{2} x + 15 \, a^{3} b\right )} \sqrt{b x^{2} + a x}}{384 \, b^{2}}, \frac{15 \, a^{4} \sqrt{-b} \arctan \left (\frac{\sqrt{b x^{2} + a x} \sqrt{-b}}{b x}\right ) +{\left (48 \, b^{4} x^{3} + 136 \, a b^{3} x^{2} + 118 \, a^{2} b^{2} x + 15 \, a^{3} b\right )} \sqrt{b x^{2} + a x}}{192 \, b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/384*(15*a^4*sqrt(b)*log(2*b*x + a - 2*sqrt(b*x^2 + a*x)*sqrt(b)) + 2*(48*b^4*x^3 + 136*a*b^3*x^2 + 118*a^2*
b^2*x + 15*a^3*b)*sqrt(b*x^2 + a*x))/b^2, 1/192*(15*a^4*sqrt(-b)*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x)) + (4
8*b^4*x^3 + 136*a*b^3*x^2 + 118*a^2*b^2*x + 15*a^3*b)*sqrt(b*x^2 + a*x))/b^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^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.18341, size = 113, normalized size = 1.12 \begin{align*} \frac{5 \, a^{4} \log \left ({\left | -2 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )} \sqrt{b} - a \right |}\right )}{128 \, b^{\frac{3}{2}}} + \frac{1}{192} \, \sqrt{b x^{2} + a x}{\left (\frac{15 \, a^{3}}{b} + 2 \,{\left (59 \, a^{2} + 4 \,{\left (6 \, b^{2} x + 17 \, a b\right )} x\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^2,x, algorithm="giac")

[Out]

5/128*a^4*log(abs(-2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) - a))/b^(3/2) + 1/192*sqrt(b*x^2 + a*x)*(15*a^3/b
+ 2*(59*a^2 + 4*(6*b^2*x + 17*a*b)*x)*x)