3.663 \(\int \frac{1}{x (a+b (c x)^n)^{5/2}} \, dx\)

Optimal. Leaf size=75 \[ \frac{2}{a^2 n \sqrt{a+b (c x)^n}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+b (c x)^n}}{\sqrt{a}}\right )}{a^{5/2} n}+\frac{2}{3 a n \left (a+b (c x)^n\right )^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0642267, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {367, 12, 266, 51, 63, 208} \[ \frac{2}{a^2 n \sqrt{a+b (c x)^n}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+b (c x)^n}}{\sqrt{a}}\right )}{a^{5/2} n}+\frac{2}{3 a n \left (a+b (c x)^n\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*(c*x)^n)^(5/2)),x]

[Out]

2/(3*a*n*(a + b*(c*x)^n)^(3/2)) + 2/(a^2*n*Sqrt[a + b*(c*x)^n]) - (2*ArcTanh[Sqrt[a + b*(c*x)^n]/Sqrt[a]])/(a^
(5/2)*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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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{1}{x \left (a+b (c x)^n\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{c}{x \left (a+b x^n\right )^{5/2}} \, dx,x,c x\right )}{c}\\ &=\operatorname{Subst}\left (\int \frac{1}{x \left (a+b x^n\right )^{5/2}} \, dx,x,c x\right )\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{x (a+b x)^{5/2}} \, dx,x,(c x)^n\right )}{n}\\ &=\frac{2}{3 a n \left (a+b (c x)^n\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{1}{x (a+b x)^{3/2}} \, dx,x,(c x)^n\right )}{a n}\\ &=\frac{2}{3 a n \left (a+b (c x)^n\right )^{3/2}}+\frac{2}{a^2 n \sqrt{a+b (c x)^n}}+\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,(c x)^n\right )}{a^2 n}\\ &=\frac{2}{3 a n \left (a+b (c x)^n\right )^{3/2}}+\frac{2}{a^2 n \sqrt{a+b (c x)^n}}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b (c x)^n}\right )}{a^2 b n}\\ &=\frac{2}{3 a n \left (a+b (c x)^n\right )^{3/2}}+\frac{2}{a^2 n \sqrt{a+b (c x)^n}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+b (c x)^n}}{\sqrt{a}}\right )}{a^{5/2} n}\\ \end{align*}

Mathematica [C]  time = 0.0339142, size = 43, normalized size = 0.57 \[ \frac{2 \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{b (c x)^n}{a}+1\right )}{3 a n \left (a+b (c x)^n\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*(c*x)^n)^(5/2)),x]

[Out]

(2*Hypergeometric2F1[-3/2, 1, -1/2, 1 + (b*(c*x)^n)/a])/(3*a*n*(a + b*(c*x)^n)^(3/2))

________________________________________________________________________________________

Maple [A]  time = 0.007, size = 59, normalized size = 0.8 \begin{align*}{\frac{1}{n} \left ( 2\,{\frac{1}{\sqrt{a+b \left ( cx \right ) ^{n}}{a}^{2}}}+{\frac{2}{3\,a} \left ( a+b \left ( cx \right ) ^{n} \right ) ^{-{\frac{3}{2}}}}-2\,{\frac{1}{{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(1/x/(a+b*(c*x)^n)^(5/2),x)

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\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(1/x/(a+b*(c*x)^n)^(5/2),x, algorithm="maxima")

[Out]

integrate(1/(((c*x)^n*b + a)^(5/2)*x), x)

________________________________________________________________________________________

Fricas [B]  time = 1.97957, size = 603, normalized size = 8.04 \begin{align*} \left [\frac{3 \,{\left (2 \, \left (c x\right )^{n} a^{\frac{3}{2}} b + \left (c x\right )^{2 \, n} \sqrt{a} b^{2} + a^{\frac{5}{2}}\right )} \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 (3 \, \left (c x\right )^{n} a b + 4 \, a^{2}\right )} \sqrt{\left (c x\right )^{n} b + a}}{3 \,{\left (2 \, \left (c x\right )^{n} a^{4} b n + \left (c x\right )^{2 \, n} a^{3} b^{2} n + a^{5} n\right )}}, \frac{2 \,{\left (3 \,{\left (2 \, \left (c x\right )^{n} \sqrt{-a} a b + \left (c x\right )^{2 \, n} \sqrt{-a} b^{2} + \sqrt{-a} a^{2}\right )} \arctan \left (\frac{\sqrt{\left (c x\right )^{n} b + a} \sqrt{-a}}{a}\right ) +{\left (3 \, \left (c x\right )^{n} a b + 4 \, a^{2}\right )} \sqrt{\left (c x\right )^{n} b + a}\right )}}{3 \,{\left (2 \, \left (c x\right )^{n} a^{4} b n + \left (c x\right )^{2 \, n} a^{3} b^{2} n + a^{5} n\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a+b*(c*x)^n)^(5/2),x, algorithm="fricas")

[Out]

[1/3*(3*(2*(c*x)^n*a^(3/2)*b + (c*x)^(2*n)*sqrt(a)*b^2 + a^(5/2))*log(((c*x)^n*b - 2*sqrt((c*x)^n*b + a)*sqrt(
a) + 2*a)/(c*x)^n) + 2*(3*(c*x)^n*a*b + 4*a^2)*sqrt((c*x)^n*b + a))/(2*(c*x)^n*a^4*b*n + (c*x)^(2*n)*a^3*b^2*n
+ a^5*n), 2/3*(3*(2*(c*x)^n*sqrt(-a)*a*b + (c*x)^(2*n)*sqrt(-a)*b^2 + sqrt(-a)*a^2)*arctan(sqrt((c*x)^n*b + a
)*sqrt(-a)/a) + (3*(c*x)^n*a*b + 4*a^2)*sqrt((c*x)^n*b + a))/(2*(c*x)^n*a^4*b*n + (c*x)^(2*n)*a^3*b^2*n + a^5*
n)]

________________________________________________________________________________________

Sympy [A]  time = 11.8503, size = 70, normalized size = 0.93 \begin{align*} \frac{2}{3 a n \left (a + b \left (c x\right )^{n}\right )^{\frac{3}{2}}} + \frac{2}{a^{2} n \sqrt{a + b \left (c x\right )^{n}}} + \frac{2 \operatorname{atan}{\left (\frac{\sqrt{a + b \left (c x\right )^{n}}}{\sqrt{- a}} \right )}}{a^{2} n \sqrt{- a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a+b*(c*x)**n)**(5/2),x)

[Out]

2/(3*a*n*(a + b*(c*x)**n)**(3/2)) + 2/(a**2*n*sqrt(a + b*(c*x)**n)) + 2*atan(sqrt(a + b*(c*x)**n)/sqrt(-a))/(a
**2*n*sqrt(-a))

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\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(1/x/(a+b*(c*x)^n)^(5/2),x, algorithm="giac")

[Out]

integrate(1/(((c*x)^n*b + a)^(5/2)*x), x)