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

Optimal. Leaf size=93 \[ \frac{2 a^2 \sqrt{a+b (c x)^n}}{n}-\frac{2 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+b (c x)^n}}{\sqrt{a}}\right )}{n}+\frac{2 a \left (a+b (c x)^n\right )^{3/2}}{3 n}+\frac{2 \left (a+b (c x)^n\right )^{5/2}}{5 n} \]

[Out]

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

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Rubi [A]  time = 0.0752527, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {367, 12, 266, 50, 63, 208} \[ \frac{2 a^2 \sqrt{a+b (c x)^n}}{n}-\frac{2 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+b (c x)^n}}{\sqrt{a}}\right )}{n}+\frac{2 a \left (a+b (c x)^n\right )^{3/2}}{3 n}+\frac{2 \left (a+b (c x)^n\right )^{5/2}}{5 n} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*(c*x)^n)^(5/2)/x,x]

[Out]

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

Rule 367

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.0780542, size = 77, normalized size = 0.83 \[ \frac{2 \sqrt{a+b (c x)^n} \left (23 a^2+11 a b (c x)^n+3 b^2 (c x)^{2 n}\right )-30 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+b (c x)^n}}{\sqrt{a}}\right )}{15 n} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*(c*x)^n)^(5/2)/x,x]

[Out]

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

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Maple [A]  time = 0.006, size = 70, normalized size = 0.8 \begin{align*}{\frac{1}{n} \left ({\frac{2}{5} \left ( a+b \left ( cx \right ) ^{n} \right ) ^{{\frac{5}{2}}}}+{\frac{2\,a}{3} \left ( a+b \left ( cx \right ) ^{n} \right ) ^{{\frac{3}{2}}}}+2\,\sqrt{a+b \left ( cx \right ) ^{n}}{a}^{2}-2\,{a}^{5/2}{\it Artanh} \left ({\frac{\sqrt{a+b \left ( cx \right ) ^{n}}}{\sqrt{a}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/n*(2/5*(a+b*(c*x)^n)^(5/2)+2/3*a*(a+b*(c*x)^n)^(3/2)+2*(a+b*(c*x)^n)^(1/2)*a^2-2*a^(5/2)*arctanh((a+b*(c*x)^
n)^(1/2)/a^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(((c*x)^n*b + a)^(5/2)/x, x)

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Fricas [A]  time = 1.84996, size = 393, normalized size = 4.23 \begin{align*} \left [\frac{15 \, a^{\frac{5}{2}} \log \left (\frac{\left (c x\right )^{n} b - 2 \, \sqrt{\left (c x\right )^{n} b + a} \sqrt{a} + 2 \, a}{\left (c x\right )^{n}}\right ) + 2 \,{\left (11 \, \left (c x\right )^{n} a b + 3 \, \left (c x\right )^{2 \, n} b^{2} + 23 \, a^{2}\right )} \sqrt{\left (c x\right )^{n} b + a}}{15 \, n}, \frac{2 \,{\left (15 \, \sqrt{-a} a^{2} \arctan \left (\frac{\sqrt{\left (c x\right )^{n} b + a} \sqrt{-a}}{a}\right ) +{\left (11 \, \left (c x\right )^{n} a b + 3 \, \left (c x\right )^{2 \, n} b^{2} + 23 \, a^{2}\right )} \sqrt{\left (c x\right )^{n} b + a}\right )}}{15 \, n}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 73.9411, size = 124, normalized size = 1.33 \begin{align*} - \begin{cases} - \left (a^{2} \sqrt{a + b} + 2 a b \sqrt{a + b} + b^{2} \sqrt{a + b}\right ) \log{\left (c x \right )} & \text{for}\: n = 0 \\- \frac{\frac{2 a^{3} \operatorname{atan}{\left (\frac{\sqrt{a + b \left (c x\right )^{n}}}{\sqrt{- a}} \right )}}{\sqrt{- a}} + 2 a^{2} \sqrt{a + b \left (c x\right )^{n}} + \frac{2 a \left (a + b \left (c x\right )^{n}\right )^{\frac{3}{2}}}{3} + \frac{2 \left (a + b \left (c x\right )^{n}\right )^{\frac{5}{2}}}{5}}{n} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Piecewise((-(a**2*sqrt(a + b) + 2*a*b*sqrt(a + b) + b**2*sqrt(a + b))*log(c*x), Eq(n, 0)), (-(2*a**3*atan(sqr
t(a + b*(c*x)**n)/sqrt(-a))/sqrt(-a) + 2*a**2*sqrt(a + b*(c*x)**n) + 2*a*(a + b*(c*x)**n)**(3/2)/3 + 2*(a + b*
(c*x)**n)**(5/2)/5)/n, True))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(((c*x)^n*b + a)^(5/2)/x, x)