3.2657 \(\int x^{-1-4 n} \sqrt {a+b x^n} \, dx\)

Optimal. Leaf size=142 \[ \frac {5 b^4 \tanh ^{-1}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )}{64 a^{7/2} n}-\frac {5 b^3 x^{-n} \sqrt {a+b x^n}}{64 a^3 n}+\frac {5 b^2 x^{-2 n} \sqrt {a+b x^n}}{96 a^2 n}-\frac {x^{-4 n} \sqrt {a+b x^n}}{4 n}-\frac {b x^{-3 n} \sqrt {a+b x^n}}{24 a n} \]

[Out]

5/64*b^4*arctanh((a+b*x^n)^(1/2)/a^(1/2))/a^(7/2)/n-1/4*(a+b*x^n)^(1/2)/n/(x^(4*n))-1/24*b*(a+b*x^n)^(1/2)/a/n
/(x^(3*n))+5/96*b^2*(a+b*x^n)^(1/2)/a^2/n/(x^(2*n))-5/64*b^3*(a+b*x^n)^(1/2)/a^3/n/(x^n)

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Rubi [A]  time = 0.07, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {266, 47, 51, 63, 208} \[ \frac {5 b^2 x^{-2 n} \sqrt {a+b x^n}}{96 a^2 n}-\frac {5 b^3 x^{-n} \sqrt {a+b x^n}}{64 a^3 n}+\frac {5 b^4 \tanh ^{-1}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )}{64 a^{7/2} n}-\frac {x^{-4 n} \sqrt {a+b x^n}}{4 n}-\frac {b x^{-3 n} \sqrt {a+b x^n}}{24 a n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 - 4*n)*Sqrt[a + b*x^n],x]

[Out]

-Sqrt[a + b*x^n]/(4*n*x^(4*n)) - (b*Sqrt[a + b*x^n])/(24*a*n*x^(3*n)) + (5*b^2*Sqrt[a + b*x^n])/(96*a^2*n*x^(2
*n)) - (5*b^3*Sqrt[a + b*x^n])/(64*a^3*n*x^n) + (5*b^4*ArcTanh[Sqrt[a + b*x^n]/Sqrt[a]])/(64*a^(7/2)*n)

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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 x^{-1-4 n} \sqrt {a+b x^n} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x^5} \, dx,x,x^n\right )}{n}\\ &=-\frac {x^{-4 n} \sqrt {a+b x^n}}{4 n}+\frac {b \operatorname {Subst}\left (\int \frac {1}{x^4 \sqrt {a+b x}} \, dx,x,x^n\right )}{8 n}\\ &=-\frac {x^{-4 n} \sqrt {a+b x^n}}{4 n}-\frac {b x^{-3 n} \sqrt {a+b x^n}}{24 a n}-\frac {\left (5 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt {a+b x}} \, dx,x,x^n\right )}{48 a n}\\ &=-\frac {x^{-4 n} \sqrt {a+b x^n}}{4 n}-\frac {b x^{-3 n} \sqrt {a+b x^n}}{24 a n}+\frac {5 b^2 x^{-2 n} \sqrt {a+b x^n}}{96 a^2 n}+\frac {\left (5 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,x^n\right )}{64 a^2 n}\\ &=-\frac {x^{-4 n} \sqrt {a+b x^n}}{4 n}-\frac {b x^{-3 n} \sqrt {a+b x^n}}{24 a n}+\frac {5 b^2 x^{-2 n} \sqrt {a+b x^n}}{96 a^2 n}-\frac {5 b^3 x^{-n} \sqrt {a+b x^n}}{64 a^3 n}-\frac {\left (5 b^4\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^n\right )}{128 a^3 n}\\ &=-\frac {x^{-4 n} \sqrt {a+b x^n}}{4 n}-\frac {b x^{-3 n} \sqrt {a+b x^n}}{24 a n}+\frac {5 b^2 x^{-2 n} \sqrt {a+b x^n}}{96 a^2 n}-\frac {5 b^3 x^{-n} \sqrt {a+b x^n}}{64 a^3 n}-\frac {\left (5 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^n}\right )}{64 a^3 n}\\ &=-\frac {x^{-4 n} \sqrt {a+b x^n}}{4 n}-\frac {b x^{-3 n} \sqrt {a+b x^n}}{24 a n}+\frac {5 b^2 x^{-2 n} \sqrt {a+b x^n}}{96 a^2 n}-\frac {5 b^3 x^{-n} \sqrt {a+b x^n}}{64 a^3 n}+\frac {5 b^4 \tanh ^{-1}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )}{64 a^{7/2} n}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 42, normalized size = 0.30 \[ -\frac {2 b^4 \left (a+b x^n\right )^{3/2} \, _2F_1\left (\frac {3}{2},5;\frac {5}{2};\frac {b x^n}{a}+1\right )}{3 a^5 n} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 - 4*n)*Sqrt[a + b*x^n],x]

[Out]

(-2*b^4*(a + b*x^n)^(3/2)*Hypergeometric2F1[3/2, 5, 5/2, 1 + (b*x^n)/a])/(3*a^5*n)

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fricas [A]  time = 0.59, size = 209, normalized size = 1.47 \[ \left [\frac {15 \, \sqrt {a} b^{4} x^{4 \, n} \log \left (\frac {b x^{n} + 2 \, \sqrt {b x^{n} + a} \sqrt {a} + 2 \, a}{x^{n}}\right ) - 2 \, {\left (15 \, a b^{3} x^{3 \, n} - 10 \, a^{2} b^{2} x^{2 \, n} + 8 \, a^{3} b x^{n} + 48 \, a^{4}\right )} \sqrt {b x^{n} + a}}{384 \, a^{4} n x^{4 \, n}}, -\frac {15 \, \sqrt {-a} b^{4} x^{4 \, n} \arctan \left (\frac {\sqrt {b x^{n} + a} \sqrt {-a}}{a}\right ) + {\left (15 \, a b^{3} x^{3 \, n} - 10 \, a^{2} b^{2} x^{2 \, n} + 8 \, a^{3} b x^{n} + 48 \, a^{4}\right )} \sqrt {b x^{n} + a}}{192 \, a^{4} n x^{4 \, n}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-4*n)*(a+b*x^n)^(1/2),x, algorithm="fricas")

[Out]

[1/384*(15*sqrt(a)*b^4*x^(4*n)*log((b*x^n + 2*sqrt(b*x^n + a)*sqrt(a) + 2*a)/x^n) - 2*(15*a*b^3*x^(3*n) - 10*a
^2*b^2*x^(2*n) + 8*a^3*b*x^n + 48*a^4)*sqrt(b*x^n + a))/(a^4*n*x^(4*n)), -1/192*(15*sqrt(-a)*b^4*x^(4*n)*arcta
n(sqrt(b*x^n + a)*sqrt(-a)/a) + (15*a*b^3*x^(3*n) - 10*a^2*b^2*x^(2*n) + 8*a^3*b*x^n + 48*a^4)*sqrt(b*x^n + a)
)/(a^4*n*x^(4*n))]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b x^{n} + a} x^{-4 \, n - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-4*n)*(a+b*x^n)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*x^n + a)*x^(-4*n - 1), x)

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maple [F]  time = 0.21, size = 0, normalized size = 0.00 \[ \int \sqrt {b \,x^{n}+a}\, x^{-4 n -1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-1-4*n)*(b*x^n+a)^(1/2),x)

[Out]

int(x^(-1-4*n)*(b*x^n+a)^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b x^{n} + a} x^{-4 \, n - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-4*n)*(a+b*x^n)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*x^n + a)*x^(-4*n - 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+b\,x^n}}{x^{4\,n+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^n)^(1/2)/x^(4*n + 1),x)

[Out]

int((a + b*x^n)^(1/2)/x^(4*n + 1), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1-4*n)*(a+b*x**n)**(1/2),x)

[Out]

Timed out

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