3.2499 \(\int \frac{(a+b x^n)^{5/2}}{x} \, dx\)

Optimal. Leaf size=85 \[ \frac{2 a^2 \sqrt{a+b x^n}}{n}-\frac{2 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+b x^n}}{\sqrt{a}}\right )}{n}+\frac{2 a \left (a+b x^n\right )^{3/2}}{3 n}+\frac{2 \left (a+b x^n\right )^{5/2}}{5 n} \]

[Out]

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

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Rubi [A]  time = 0.0414165, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {266, 50, 63, 208} \[ \frac{2 a^2 \sqrt{a+b x^n}}{n}-\frac{2 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+b x^n}}{\sqrt{a}}\right )}{n}+\frac{2 a \left (a+b x^n\right )^{3/2}}{3 n}+\frac{2 \left (a+b x^n\right )^{5/2}}{5 n} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^n)^(5/2)/x,x]

[Out]

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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

Mathematica [A]  time = 0.0330789, size = 69, normalized size = 0.81 \[ \frac{2 \sqrt{a+b x^n} \left (23 a^2+11 a b x^n+3 b^2 x^{2 n}\right )-30 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+b x^n}}{\sqrt{a}}\right )}{15 n} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^n)^(5/2)/x,x]

[Out]

(2*Sqrt[a + b*x^n]*(23*a^2 + 11*a*b*x^n + 3*b^2*x^(2*n)) - 30*a^(5/2)*ArcTanh[Sqrt[a + b*x^n]/Sqrt[a]])/(15*n)

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Maple [A]  time = 0.002, size = 62, normalized size = 0.7 \begin{align*}{\frac{1}{n} \left ({\frac{2}{5} \left ( a+b{x}^{n} \right ) ^{{\frac{5}{2}}}}+{\frac{2\,a}{3} \left ( a+b{x}^{n} \right ) ^{{\frac{3}{2}}}}+2\,{a}^{2}\sqrt{a+b{x}^{n}}-2\,{a}^{5/2}{\it Artanh} \left ({\frac{\sqrt{a+b{x}^{n}}}{\sqrt{a}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.08146, size = 339, normalized size = 3.99 \begin{align*} \left [\frac{15 \, a^{\frac{5}{2}} \log \left (\frac{b x^{n} - 2 \, \sqrt{b x^{n} + a} \sqrt{a} + 2 \, a}{x^{n}}\right ) + 2 \,{\left (3 \, b^{2} x^{2 \, n} + 11 \, a b x^{n} + 23 \, a^{2}\right )} \sqrt{b x^{n} + a}}{15 \, n}, \frac{2 \,{\left (15 \, \sqrt{-a} a^{2} \arctan \left (\frac{\sqrt{b x^{n} + a} \sqrt{-a}}{a}\right ) +{\left (3 \, b^{2} x^{2 \, n} + 11 \, a b x^{n} + 23 \, a^{2}\right )} \sqrt{b x^{n} + a}\right )}}{15 \, n}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/15*(15*a^(5/2)*log((b*x^n - 2*sqrt(b*x^n + a)*sqrt(a) + 2*a)/x^n) + 2*(3*b^2*x^(2*n) + 11*a*b*x^n + 23*a^2)
*sqrt(b*x^n + a))/n, 2/15*(15*sqrt(-a)*a^2*arctan(sqrt(b*x^n + a)*sqrt(-a)/a) + (3*b^2*x^(2*n) + 11*a*b*x^n +
23*a^2)*sqrt(b*x^n + a))/n]

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Sympy [A]  time = 19.1493, size = 117, normalized size = 1.38 \begin{align*} \frac{46 a^{\frac{5}{2}} \sqrt{1 + \frac{b x^{n}}{a}}}{15 n} + \frac{a^{\frac{5}{2}} \log{\left (\frac{b x^{n}}{a} \right )}}{n} - \frac{2 a^{\frac{5}{2}} \log{\left (\sqrt{1 + \frac{b x^{n}}{a}} + 1 \right )}}{n} + \frac{22 a^{\frac{3}{2}} b x^{n} \sqrt{1 + \frac{b x^{n}}{a}}}{15 n} + \frac{2 \sqrt{a} b^{2} x^{2 n} \sqrt{1 + \frac{b x^{n}}{a}}}{5 n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{n} + a\right )}^{\frac{5}{2}}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^n + a)^(5/2)/x, x)