3.26.9 \(\int \frac {1}{x (a+b x^n)^{3/2}} \, dx\) [2509]

Optimal. Leaf size=48 \[ \frac {2}{a n \sqrt {a+b x^n}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )}{a^{3/2} n} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {272, 53, 65, 214} \begin {gather*} \frac {2}{a n \sqrt {a+b x^n}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )}{a^{3/2} n} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^n)^(3/2)),x]

[Out]

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

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]]

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 48, normalized size = 1.00 \begin {gather*} \frac {2}{a n \sqrt {a+b x^n}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )}{a^{3/2} n} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x^n)^(3/2)),x]

[Out]

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

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Maple [A]
time = 0.23, size = 39, normalized size = 0.81

method	result	size
derivativedivides	\(\frac {-\frac {2 \arctanh \left (\frac {\sqrt {a +b \,x^{n}}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}}+\frac {2}{a \sqrt {a +b \,x^{n}}}}{n}\)	\(39\)
default	\(\frac {-\frac {2 \arctanh \left (\frac {\sqrt {a +b \,x^{n}}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}}+\frac {2}{a \sqrt {a +b \,x^{n}}}}{n}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 57, normalized size = 1.19 \begin {gather*} \frac {\log \left (\frac {\sqrt {b x^{n} + a} - \sqrt {a}}{\sqrt {b x^{n} + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}} n} + \frac {2}{\sqrt {b x^{n} + a} a n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((sqrt(b*x^n + a) - sqrt(a))/(sqrt(b*x^n + a) + sqrt(a)))/(a^(3/2)*n) + 2/(sqrt(b*x^n + a)*a*n)

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Fricas [A]
time = 0.42, size = 144, normalized size = 3.00 \begin {gather*} \left [\frac {{\left (\sqrt {a} b x^{n} + a^{\frac {3}{2}}\right )} \log \left (\frac {b x^{n} - 2 \, \sqrt {b x^{n} + a} \sqrt {a} + 2 \, a}{x^{n}}\right ) + 2 \, \sqrt {b x^{n} + a} a}{a^{2} b n x^{n} + a^{3} n}, \frac {2 \, {\left ({\left (\sqrt {-a} b x^{n} + \sqrt {-a} a\right )} \arctan \left (\frac {\sqrt {b x^{n} + a} \sqrt {-a}}{a}\right ) + \sqrt {b x^{n} + a} a\right )}}{a^{2} b n x^{n} + a^{3} n}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (39) = 78\).
time = 1.10, size = 184, normalized size = 3.83 \begin {gather*} \frac {2 a^{3} \sqrt {1 + \frac {b x^{n}}{a}}}{a^{\frac {9}{2}} n + a^{\frac {7}{2}} b n x^{n}} + \frac {a^{3} \log {\left (\frac {b x^{n}}{a} \right )}}{a^{\frac {9}{2}} n + a^{\frac {7}{2}} b n x^{n}} - \frac {2 a^{3} \log {\left (\sqrt {1 + \frac {b x^{n}}{a}} + 1 \right )}}{a^{\frac {9}{2}} n + a^{\frac {7}{2}} b n x^{n}} + \frac {a^{2} b x^{n} \log {\left (\frac {b x^{n}}{a} \right )}}{a^{\frac {9}{2}} n + a^{\frac {7}{2}} b n x^{n}} - \frac {2 a^{2} b x^{n} \log {\left (\sqrt {1 + \frac {b x^{n}}{a}} + 1 \right )}}{a^{\frac {9}{2}} n + a^{\frac {7}{2}} b n x^{n}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a**3*sqrt(1 + b*x**n/a)/(a**(9/2)*n + a**(7/2)*b*n*x**n) + a**3*log(b*x**n/a)/(a**(9/2)*n + a**(7/2)*b*n*x**
n) - 2*a**3*log(sqrt(1 + b*x**n/a) + 1)/(a**(9/2)*n + a**(7/2)*b*n*x**n) + a**2*b*x**n*log(b*x**n/a)/(a**(9/2)
*n + a**(7/2)*b*n*x**n) - 2*a**2*b*x**n*log(sqrt(1 + b*x**n/a) + 1)/(a**(9/2)*n + a**(7/2)*b*n*x**n)

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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*x^n)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((b*x^n + 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\,x^n\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b*x^n)^(3/2)),x)

[Out]

int(1/(x*(a + b*x^n)^(3/2)), x)

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