3.2499 \(\int \frac {(a+b x^n)^{5/2}}{x} \, dx\)

Optimal. Leaf size=85 \[ -\frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )}{n}+\frac {2 a^2 \sqrt {a+b x^n}}{n}+\frac {2 a \left (a+b x^n\right )^{3/2}}{3 n}+\frac {2 \left (a+b x^n\right )^{5/2}}{5 n} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {266, 50, 63, 208} \[ \frac {2 a^2 \sqrt {a+b x^n}}{n}-\frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )}{n}+\frac {2 a \left (a+b x^n\right )^{3/2}}{3 n}+\frac {2 \left (a+b x^n\right )^{5/2}}{5 n} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^n)^(5/2)/x,x]

[Out]

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

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

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Mathematica [A]  time = 0.05, size = 69, normalized size = 0.81 \[ \frac {2 \sqrt {a+b x^n} \left (23 a^2+11 a b x^n+3 b^2 x^{2 n}\right )-30 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )}{15 n} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^n)^(5/2)/x,x]

[Out]

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

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fricas [A]  time = 1.14, size = 144, normalized size = 1.69 \[ \left [\frac {15 \, a^{\frac {5}{2}} \log \left (\frac {b x^{n} - 2 \, \sqrt {b x^{n} + a} \sqrt {a} + 2 \, a}{x^{n}}\right ) + 2 \, {\left (3 \, b^{2} x^{2 \, n} + 11 \, a b x^{n} + 23 \, a^{2}\right )} \sqrt {b x^{n} + a}}{15 \, n}, \frac {2 \, {\left (15 \, \sqrt {-a} a^{2} \arctan \left (\frac {\sqrt {b x^{n} + a} \sqrt {-a}}{a}\right ) + {\left (3 \, b^{2} x^{2 \, n} + 11 \, a b x^{n} + 23 \, a^{2}\right )} \sqrt {b x^{n} + a}\right )}}{15 \, n}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^n + a)^(5/2)/x, x)

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maple [A]  time = 0.00, size = 62, normalized size = 0.73 \[ \frac {-2 a^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {b \,x^{n}+a}}{\sqrt {a}}\right )+2 \sqrt {b \,x^{n}+a}\, a^{2}+\frac {2 \left (b \,x^{n}+a \right )^{\frac {3}{2}} a}{3}+\frac {2 \left (b \,x^{n}+a \right )^{\frac {5}{2}}}{5}}{n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.31, size = 83, normalized size = 0.98 \[ \frac {a^{\frac {5}{2}} \log \left (\frac {\sqrt {b x^{n} + a} - \sqrt {a}}{\sqrt {b x^{n} + a} + \sqrt {a}}\right )}{n} + \frac {2 \, {\left (3 \, {\left (b x^{n} + a\right )}^{\frac {5}{2}} + 5 \, {\left (b x^{n} + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x^{n} + a} a^{2}\right )}}{15 \, n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^(5/2)*log((sqrt(b*x^n + a) - sqrt(a))/(sqrt(b*x^n + a) + sqrt(a)))/n + 2/15*(3*(b*x^n + a)^(5/2) + 5*(b*x^n
+ a)^(3/2)*a + 15*sqrt(b*x^n + a)*a^2)/n

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^n)^(5/2)/x,x)

[Out]

int((a + b*x^n)^(5/2)/x, x)

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sympy [A]  time = 7.98, size = 117, normalized size = 1.38 \[ \frac {46 a^{\frac {5}{2}} \sqrt {1 + \frac {b x^{n}}{a}}}{15 n} + \frac {a^{\frac {5}{2}} \log {\left (\frac {b x^{n}}{a} \right )}}{n} - \frac {2 a^{\frac {5}{2}} \log {\left (\sqrt {1 + \frac {b x^{n}}{a}} + 1 \right )}}{n} + \frac {22 a^{\frac {3}{2}} b x^{n} \sqrt {1 + \frac {b x^{n}}{a}}}{15 n} + \frac {2 \sqrt {a} b^{2} x^{2 n} \sqrt {1 + \frac {b x^{n}}{a}}}{5 n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

46*a**(5/2)*sqrt(1 + b*x**n/a)/(15*n) + a**(5/2)*log(b*x**n/a)/n - 2*a**(5/2)*log(sqrt(1 + b*x**n/a) + 1)/n +
22*a**(3/2)*b*x**n*sqrt(1 + b*x**n/a)/(15*n) + 2*sqrt(a)*b**2*x**(2*n)*sqrt(1 + b*x**n/a)/(5*n)

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