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

Optimal. Leaf size=65 \[ 2 a^2 \sqrt{a+b x}-2 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )+\frac{2}{3} a (a+b x)^{3/2}+\frac{2}{5} (a+b x)^{5/2} \]

[Out]

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

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Rubi [A]  time = 0.0208395, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {50, 63, 208} \[ 2 a^2 \sqrt{a+b x}-2 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )+\frac{2}{3} a (a+b x)^{3/2}+\frac{2}{5} (a+b x)^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(5/2)/x,x]

[Out]

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.0887927, size = 58, normalized size = 0.89 \[ -2 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )+\frac{2}{3} a \sqrt{a+b x} (4 a+b x)+\frac{2}{5} (a+b x)^{5/2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(5/2)/x,x]

[Out]

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

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Maple [A]  time = 0.003, size = 50, normalized size = 0.8 \begin{align*}{\frac{2\,a}{3} \left ( bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{2}{5} \left ( bx+a \right ) ^{{\frac{5}{2}}}}-2\,{a}^{5/2}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) +2\,{a}^{2}\sqrt{bx+a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*a*(b*x+a)^(3/2)+2/5*(b*x+a)^(5/2)-2*a^(5/2)*arctanh((b*x+a)^(1/2)/a^(1/2))+2*a^2*(b*x+a)^(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+a)^(5/2)/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.57475, size = 288, normalized size = 4.43 \begin{align*} \left [a^{\frac{5}{2}} \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + \frac{2}{15} \,{\left (3 \, b^{2} x^{2} + 11 \, a b x + 23 \, a^{2}\right )} \sqrt{b x + a}, 2 \, \sqrt{-a} a^{2} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) + \frac{2}{15} \,{\left (3 \, b^{2} x^{2} + 11 \, a b x + 23 \, a^{2}\right )} \sqrt{b x + a}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 4.68259, size = 97, normalized size = 1.49 \begin{align*} \frac{46 a^{\frac{5}{2}} \sqrt{1 + \frac{b x}{a}}}{15} + a^{\frac{5}{2}} \log{\left (\frac{b x}{a} \right )} - 2 a^{\frac{5}{2}} \log{\left (\sqrt{1 + \frac{b x}{a}} + 1 \right )} + \frac{22 a^{\frac{3}{2}} b x \sqrt{1 + \frac{b x}{a}}}{15} + \frac{2 \sqrt{a} b^{2} x^{2} \sqrt{1 + \frac{b x}{a}}}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

46*a**(5/2)*sqrt(1 + b*x/a)/15 + a**(5/2)*log(b*x/a) - 2*a**(5/2)*log(sqrt(1 + b*x/a) + 1) + 22*a**(3/2)*b*x*s
qrt(1 + b*x/a)/15 + 2*sqrt(a)*b**2*x**2*sqrt(1 + b*x/a)/5

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Giac [A]  time = 1.18573, size = 76, normalized size = 1.17 \begin{align*} \frac{2 \, a^{3} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} + \frac{2}{5} \,{\left (b x + a\right )}^{\frac{5}{2}} + \frac{2}{3} \,{\left (b x + a\right )}^{\frac{3}{2}} a + 2 \, \sqrt{b x + a} a^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a^3*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) + 2/5*(b*x + a)^(5/2) + 2/3*(b*x + a)^(3/2)*a + 2*sqrt(b*x + a)*
a^2