3.7.62 \(\int \frac {1}{x (a+b (c x)^n)^{3/2}} \, dx\) [662]

Optimal. Leaf size=52 \[ \frac {2}{a n \sqrt {a+b (c x)^n}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b (c x)^n}}{\sqrt {a}}\right )}{a^{3/2} n} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {374, 12, 272, 53, 65, 214} \begin {gather*} \frac {2}{a n \sqrt {a+b (c x)^n}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b (c x)^n}}{\sqrt {a}}\right )}{a^{3/2} n} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*(c*x)^n)^(3/2)),x]

[Out]

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

Rule 12

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

Rule 53

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 65

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 214

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

Rule 272

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 374

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]

Rubi steps

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

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Mathematica [A]
time = 0.13, size = 52, normalized size = 1.00 \begin {gather*} \frac {2}{a n \sqrt {a+b (c x)^n}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b (c x)^n}}{\sqrt {a}}\right )}{a^{3/2} n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.45, size = 43, normalized size = 0.83

method	result	size
derivativedivides	\(\frac {\frac {2}{a \sqrt {a +b \left (c x \right )^{n}}}-\frac {2 \arctanh \left (\frac {\sqrt {a +b \left (c x \right )^{n}}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}}}{n}\)	\(43\)
default	\(\frac {\frac {2}{a \sqrt {a +b \left (c x \right )^{n}}}-\frac {2 \arctanh \left (\frac {\sqrt {a +b \left (c x \right )^{n}}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}}}{n}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(a+b*(c*x)^n)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a+b*(c*x)^n)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/(((c*x)^n*b + a)^(3/2)*x), x)

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Fricas [A]
time = 0.36, size = 164, normalized size = 3.15 \begin {gather*} \left [\frac {{\left (\left (c x\right )^{n} \sqrt {a} b + a^{\frac {3}{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 \, \sqrt {\left (c x\right )^{n} b + a} a}{\left (c x\right )^{n} a^{2} b n + a^{3} n}, \frac {2 \, {\left ({\left (\left (c x\right )^{n} \sqrt {-a} b + \sqrt {-a} a\right )} \arctan \left (\frac {\sqrt {\left (c x\right )^{n} b + a} \sqrt {-a}}{a}\right ) + \sqrt {\left (c x\right )^{n} b + a} a\right )}}{\left (c x\right )^{n} a^{2} b n + a^{3} n}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a+b*(c*x)^n)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 5.55, size = 48, normalized size = 0.92 \begin {gather*} \frac {2}{a 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 n \sqrt {- a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a+b*(c*x)**n)**(3/2),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a+b*(c*x)^n)^(3/2),x, algorithm="giac")

[Out]

integrate(1/(((c*x)^n*b + a)^(3/2)*x), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{x\,{\left (a+b\,{\left (c\,x\right )}^n\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b*(c*x)^n)^(3/2)),x)

[Out]

int(1/(x*(a + b*(c*x)^n)^(3/2)), x)

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