3.2514 \(\int \frac {1}{x (a+b x^n)^{5/2}} \, dx\)

Optimal. Leaf size=69 \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )}{a^{5/2} n}+\frac {2}{a^2 n \sqrt {a+b x^n}}+\frac {2}{3 a n \left (a+b x^n\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {266, 51, 63, 208} \[ \frac {2}{a^2 n \sqrt {a+b x^n}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )}{a^{5/2} n}+\frac {2}{3 a n \left (a+b x^n\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^n)^(5/2)),x]

[Out]

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

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]]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 39, normalized size = 0.57 \[ \frac {2 \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {b x^n}{a}+1\right )}{3 a n \left (a+b x^n\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x^n)^(5/2)),x]

[Out]

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

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fricas [A]  time = 1.44, size = 230, normalized size = 3.33 \[ \left [\frac {3 \, {\left (\sqrt {a} b^{2} x^{2 \, n} + 2 \, a^{\frac {3}{2}} b x^{n} + a^{\frac {5}{2}}\right )} \log \left (\frac {b x^{n} - 2 \, \sqrt {b x^{n} + a} \sqrt {a} + 2 \, a}{x^{n}}\right ) + 2 \, {\left (3 \, a b x^{n} + 4 \, a^{2}\right )} \sqrt {b x^{n} + a}}{3 \, {\left (a^{3} b^{2} n x^{2 \, n} + 2 \, a^{4} b n x^{n} + a^{5} n\right )}}, \frac {2 \, {\left (3 \, {\left (\sqrt {-a} b^{2} x^{2 \, n} + 2 \, \sqrt {-a} a b x^{n} + \sqrt {-a} a^{2}\right )} \arctan \left (\frac {\sqrt {b x^{n} + a} \sqrt {-a}}{a}\right ) + {\left (3 \, a b x^{n} + 4 \, a^{2}\right )} \sqrt {b x^{n} + a}\right )}}{3 \, {\left (a^{3} b^{2} n x^{2 \, n} + 2 \, a^{4} b n x^{n} + a^{5} n\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{n} + a\right )}^{\frac {5}{2}} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((b*x^n + a)^(5/2)*x), x)

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maple [A]  time = 0.01, size = 53, normalized size = 0.77 \[ \frac {\frac {2}{3 \left (b \,x^{n}+a \right )^{\frac {3}{2}} a}-\frac {2 \arctanh \left (\frac {\sqrt {b \,x^{n}+a}}{\sqrt {a}}\right )}{a^{\frac {5}{2}}}+\frac {2}{\sqrt {b \,x^{n}+a}\, a^{2}}}{n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.18, size = 67, normalized size = 0.97 \[ \frac {\log \left (\frac {\sqrt {b x^{n} + a} - \sqrt {a}}{\sqrt {b x^{n} + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}} n} + \frac {2 \, {\left (3 \, b x^{n} + 4 \, a\right )}}{3 \, {\left (b x^{n} + a\right )}^{\frac {3}{2}} a^{2} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((sqrt(b*x^n + a) - sqrt(a))/(sqrt(b*x^n + a) + sqrt(a)))/(a^(5/2)*n) + 2/3*(3*b*x^n + 4*a)/((b*x^n + a)^(3
/2)*a^2*n)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x\,{\left (a+b\,x^n\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b*x^n)^(5/2)),x)

[Out]

int(1/(x*(a + b*x^n)^(5/2)), x)

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sympy [B]  time = 5.32, size = 860, normalized size = 12.46 \[ \frac {8 a^{7} \sqrt {1 + \frac {b x^{n}}{a}}}{3 a^{\frac {19}{2}} n + 9 a^{\frac {17}{2}} b n x^{n} + 9 a^{\frac {15}{2}} b^{2} n x^{2 n} + 3 a^{\frac {13}{2}} b^{3} n x^{3 n}} + \frac {3 a^{7} \log {\left (\frac {b x^{n}}{a} \right )}}{3 a^{\frac {19}{2}} n + 9 a^{\frac {17}{2}} b n x^{n} + 9 a^{\frac {15}{2}} b^{2} n x^{2 n} + 3 a^{\frac {13}{2}} b^{3} n x^{3 n}} - \frac {6 a^{7} \log {\left (\sqrt {1 + \frac {b x^{n}}{a}} + 1 \right )}}{3 a^{\frac {19}{2}} n + 9 a^{\frac {17}{2}} b n x^{n} + 9 a^{\frac {15}{2}} b^{2} n x^{2 n} + 3 a^{\frac {13}{2}} b^{3} n x^{3 n}} + \frac {14 a^{6} b x^{n} \sqrt {1 + \frac {b x^{n}}{a}}}{3 a^{\frac {19}{2}} n + 9 a^{\frac {17}{2}} b n x^{n} + 9 a^{\frac {15}{2}} b^{2} n x^{2 n} + 3 a^{\frac {13}{2}} b^{3} n x^{3 n}} + \frac {9 a^{6} b x^{n} \log {\left (\frac {b x^{n}}{a} \right )}}{3 a^{\frac {19}{2}} n + 9 a^{\frac {17}{2}} b n x^{n} + 9 a^{\frac {15}{2}} b^{2} n x^{2 n} + 3 a^{\frac {13}{2}} b^{3} n x^{3 n}} - \frac {18 a^{6} b x^{n} \log {\left (\sqrt {1 + \frac {b x^{n}}{a}} + 1 \right )}}{3 a^{\frac {19}{2}} n + 9 a^{\frac {17}{2}} b n x^{n} + 9 a^{\frac {15}{2}} b^{2} n x^{2 n} + 3 a^{\frac {13}{2}} b^{3} n x^{3 n}} + \frac {6 a^{5} b^{2} x^{2 n} \sqrt {1 + \frac {b x^{n}}{a}}}{3 a^{\frac {19}{2}} n + 9 a^{\frac {17}{2}} b n x^{n} + 9 a^{\frac {15}{2}} b^{2} n x^{2 n} + 3 a^{\frac {13}{2}} b^{3} n x^{3 n}} + \frac {9 a^{5} b^{2} x^{2 n} \log {\left (\frac {b x^{n}}{a} \right )}}{3 a^{\frac {19}{2}} n + 9 a^{\frac {17}{2}} b n x^{n} + 9 a^{\frac {15}{2}} b^{2} n x^{2 n} + 3 a^{\frac {13}{2}} b^{3} n x^{3 n}} - \frac {18 a^{5} b^{2} x^{2 n} \log {\left (\sqrt {1 + \frac {b x^{n}}{a}} + 1 \right )}}{3 a^{\frac {19}{2}} n + 9 a^{\frac {17}{2}} b n x^{n} + 9 a^{\frac {15}{2}} b^{2} n x^{2 n} + 3 a^{\frac {13}{2}} b^{3} n x^{3 n}} + \frac {3 a^{4} b^{3} x^{3 n} \log {\left (\frac {b x^{n}}{a} \right )}}{3 a^{\frac {19}{2}} n + 9 a^{\frac {17}{2}} b n x^{n} + 9 a^{\frac {15}{2}} b^{2} n x^{2 n} + 3 a^{\frac {13}{2}} b^{3} n x^{3 n}} - \frac {6 a^{4} b^{3} x^{3 n} \log {\left (\sqrt {1 + \frac {b x^{n}}{a}} + 1 \right )}}{3 a^{\frac {19}{2}} n + 9 a^{\frac {17}{2}} b n x^{n} + 9 a^{\frac {15}{2}} b^{2} n x^{2 n} + 3 a^{\frac {13}{2}} b^{3} n x^{3 n}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*a**7*sqrt(1 + b*x**n/a)/(3*a**(19/2)*n + 9*a**(17/2)*b*n*x**n + 9*a**(15/2)*b**2*n*x**(2*n) + 3*a**(13/2)*b*
*3*n*x**(3*n)) + 3*a**7*log(b*x**n/a)/(3*a**(19/2)*n + 9*a**(17/2)*b*n*x**n + 9*a**(15/2)*b**2*n*x**(2*n) + 3*
a**(13/2)*b**3*n*x**(3*n)) - 6*a**7*log(sqrt(1 + b*x**n/a) + 1)/(3*a**(19/2)*n + 9*a**(17/2)*b*n*x**n + 9*a**(
15/2)*b**2*n*x**(2*n) + 3*a**(13/2)*b**3*n*x**(3*n)) + 14*a**6*b*x**n*sqrt(1 + b*x**n/a)/(3*a**(19/2)*n + 9*a*
*(17/2)*b*n*x**n + 9*a**(15/2)*b**2*n*x**(2*n) + 3*a**(13/2)*b**3*n*x**(3*n)) + 9*a**6*b*x**n*log(b*x**n/a)/(3
*a**(19/2)*n + 9*a**(17/2)*b*n*x**n + 9*a**(15/2)*b**2*n*x**(2*n) + 3*a**(13/2)*b**3*n*x**(3*n)) - 18*a**6*b*x
**n*log(sqrt(1 + b*x**n/a) + 1)/(3*a**(19/2)*n + 9*a**(17/2)*b*n*x**n + 9*a**(15/2)*b**2*n*x**(2*n) + 3*a**(13
/2)*b**3*n*x**(3*n)) + 6*a**5*b**2*x**(2*n)*sqrt(1 + b*x**n/a)/(3*a**(19/2)*n + 9*a**(17/2)*b*n*x**n + 9*a**(1
5/2)*b**2*n*x**(2*n) + 3*a**(13/2)*b**3*n*x**(3*n)) + 9*a**5*b**2*x**(2*n)*log(b*x**n/a)/(3*a**(19/2)*n + 9*a*
*(17/2)*b*n*x**n + 9*a**(15/2)*b**2*n*x**(2*n) + 3*a**(13/2)*b**3*n*x**(3*n)) - 18*a**5*b**2*x**(2*n)*log(sqrt
(1 + b*x**n/a) + 1)/(3*a**(19/2)*n + 9*a**(17/2)*b*n*x**n + 9*a**(15/2)*b**2*n*x**(2*n) + 3*a**(13/2)*b**3*n*x
**(3*n)) + 3*a**4*b**3*x**(3*n)*log(b*x**n/a)/(3*a**(19/2)*n + 9*a**(17/2)*b*n*x**n + 9*a**(15/2)*b**2*n*x**(2
*n) + 3*a**(13/2)*b**3*n*x**(3*n)) - 6*a**4*b**3*x**(3*n)*log(sqrt(1 + b*x**n/a) + 1)/(3*a**(19/2)*n + 9*a**(1
7/2)*b*n*x**n + 9*a**(15/2)*b**2*n*x**(2*n) + 3*a**(13/2)*b**3*n*x**(3*n))

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