3.30 \(\int \frac{(a x+b x^2)^{5/2}}{x^4} \, dx\)

Optimal. Leaf size=92 \[ \frac{15}{4} a^2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )-\frac{2 \left (a x+b x^2\right )^{5/2}}{x^3}+\frac{5 b \left (a x+b x^2\right )^{3/2}}{2 x}+\frac{15}{4} a b \sqrt{a x+b x^2} \]

[Out]

(15*a*b*Sqrt[a*x + b*x^2])/4 + (5*b*(a*x + b*x^2)^(3/2))/(2*x) - (2*(a*x + b*x^2)^(5/2))/x^3 + (15*a^2*Sqrt[b]
*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/4

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Rubi [A]  time = 0.04064, antiderivative size = 92, 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} \[ \frac{15}{4} a^2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )-\frac{2 \left (a x+b x^2\right )^{5/2}}{x^3}+\frac{5 b \left (a x+b x^2\right )^{3/2}}{2 x}+\frac{15}{4} a b \sqrt{a x+b x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a*x + b*x^2)^(5/2)/x^4,x]

[Out]

(15*a*b*Sqrt[a*x + b*x^2])/4 + (5*b*(a*x + b*x^2)^(3/2))/(2*x) - (2*(a*x + b*x^2)^(5/2))/x^3 + (15*a^2*Sqrt[b]
*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/4

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^4} \, dx &=-\frac{2 \left (a x+b x^2\right )^{5/2}}{x^3}+(5 b) \int \frac{\left (a x+b x^2\right )^{3/2}}{x^2} \, dx\\ &=\frac{5 b \left (a x+b x^2\right )^{3/2}}{2 x}-\frac{2 \left (a x+b x^2\right )^{5/2}}{x^3}+\frac{1}{4} (15 a b) \int \frac{\sqrt{a x+b x^2}}{x} \, dx\\ &=\frac{15}{4} a b \sqrt{a x+b x^2}+\frac{5 b \left (a x+b x^2\right )^{3/2}}{2 x}-\frac{2 \left (a x+b x^2\right )^{5/2}}{x^3}+\frac{1}{8} \left (15 a^2 b\right ) \int \frac{1}{\sqrt{a x+b x^2}} \, dx\\ &=\frac{15}{4} a b \sqrt{a x+b x^2}+\frac{5 b \left (a x+b x^2\right )^{3/2}}{2 x}-\frac{2 \left (a x+b x^2\right )^{5/2}}{x^3}+\frac{1}{4} \left (15 a^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a x+b x^2}}\right )\\ &=\frac{15}{4} a b \sqrt{a x+b x^2}+\frac{5 b \left (a x+b x^2\right )^{3/2}}{2 x}-\frac{2 \left (a x+b x^2\right )^{5/2}}{x^3}+\frac{15}{4} a^2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0124159, size = 48, normalized size = 0.52 \[ -\frac{2 a^2 \sqrt{x (a+b x)} \, _2F_1\left (-\frac{5}{2},-\frac{1}{2};\frac{1}{2};-\frac{b x}{a}\right )}{x \sqrt{\frac{b x}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x + b*x^2)^(5/2)/x^4,x]

[Out]

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

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Maple [B]  time = 0.048, size = 185, normalized size = 2. \begin{align*} -2\,{\frac{ \left ( b{x}^{2}+ax \right ) ^{7/2}}{a{x}^{4}}}+12\,{\frac{b \left ( b{x}^{2}+ax \right ) ^{7/2}}{{a}^{2}{x}^{3}}}-32\,{\frac{{b}^{2} \left ( b{x}^{2}+ax \right ) ^{7/2}}{{x}^{2}{a}^{3}}}+32\,{\frac{{b}^{3} \left ( b{x}^{2}+ax \right ) ^{5/2}}{{a}^{3}}}+20\,{\frac{{b}^{3} \left ( b{x}^{2}+ax \right ) ^{3/2}x}{{a}^{2}}}+10\,{\frac{{b}^{2} \left ( b{x}^{2}+ax \right ) ^{3/2}}{a}}-{\frac{15\,{b}^{2}x}{2}\sqrt{b{x}^{2}+ax}}-{\frac{15\,ab}{4}\sqrt{b{x}^{2}+ax}}+{\frac{15\,{a}^{2}}{8}\sqrt{b}\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^4,x)

[Out]

-2/a/x^4*(b*x^2+a*x)^(7/2)+12*b/a^2/x^3*(b*x^2+a*x)^(7/2)-32*b^2/a^3/x^2*(b*x^2+a*x)^(7/2)+32*b^3/a^3*(b*x^2+a
*x)^(5/2)+20*b^3/a^2*(b*x^2+a*x)^(3/2)*x+10*b^2/a*(b*x^2+a*x)^(3/2)-15/2*b^2*(b*x^2+a*x)^(1/2)*x-15/4*a*b*(b*x
^2+a*x)^(1/2)+15/8*b^(1/2)*a^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^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.08301, size = 332, normalized size = 3.61 \begin{align*} \left [\frac{15 \, a^{2} \sqrt{b} x \log \left (2 \, b x + a + 2 \, \sqrt{b x^{2} + a x} \sqrt{b}\right ) + 2 \,{\left (2 \, b^{2} x^{2} + 9 \, a b x - 8 \, a^{2}\right )} \sqrt{b x^{2} + a x}}{8 \, x}, -\frac{15 \, a^{2} \sqrt{-b} x \arctan \left (\frac{\sqrt{b x^{2} + a x} \sqrt{-b}}{b x}\right ) -{\left (2 \, b^{2} x^{2} + 9 \, a b x - 8 \, a^{2}\right )} \sqrt{b x^{2} + a x}}{4 \, x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a*x)^(5/2)/x^4,x, algorithm="fricas")

[Out]

[1/8*(15*a^2*sqrt(b)*x*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) + 2*(2*b^2*x^2 + 9*a*b*x - 8*a^2)*sqrt(b*x
^2 + a*x))/x, -1/4*(15*a^2*sqrt(-b)*x*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x)) - (2*b^2*x^2 + 9*a*b*x - 8*a^2)
*sqrt(b*x^2 + a*x))/x]

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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^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a*x)**(5/2)/x**4,x)

[Out]

Integral((x*(a + b*x))**(5/2)/x**4, x)

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Giac [A]  time = 1.26531, size = 120, normalized size = 1.3 \begin{align*} -\frac{15}{8} \, a^{2} \sqrt{b} \log \left ({\left | -2 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )} \sqrt{b} - a \right |}\right ) + \frac{2 \, a^{3}}{\sqrt{b} x - \sqrt{b x^{2} + a x}} + \frac{1}{4} \,{\left (2 \, b^{2} x + 9 \, a b\right )} \sqrt{b x^{2} + a x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a*x)^(5/2)/x^4,x, algorithm="giac")

[Out]

-15/8*a^2*sqrt(b)*log(abs(-2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) - a)) + 2*a^3/(sqrt(b)*x - sqrt(b*x^2 + a
*x)) + 1/4*(2*b^2*x + 9*a*b)*sqrt(b*x^2 + a*x)