3.72 \(\int x^{-1+n} \cos ^{-1}(a+b x^n) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6715, 4804, 4620, 261} \[ \frac {\left (a+b x^n\right ) \cos ^{-1}\left (a+b x^n\right )}{b n}-\frac {\sqrt {1-\left (a+b x^n\right )^2}}{b n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 + n)*ArcCos[a + b*x^n],x]

[Out]

-(Sqrt[1 - (a + b*x^n)^2]/(b*n)) + ((a + b*x^n)*ArcCos[a + b*x^n])/(b*n)

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 4620

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcCos[c*x])^n, x] + Dist[b*c*n, Int[
(x*(a + b*ArcCos[c*x])^(n - 1))/Sqrt[1 - c^2*x^2], x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

Rule 4804

Int[((a_.) + ArcCos[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcCos[x])^n, x],
x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 43, normalized size = 0.90 \[ \frac {\left (a+b x^n\right ) \cos ^{-1}\left (a+b x^n\right )-\sqrt {1-\left (a+b x^n\right )^2}}{b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-Sqrt[1 - (a + b*x^n)^2] + (a + b*x^n)*ArcCos[a + b*x^n])/(b*n)

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fricas [A]  time = 0.44, size = 59, normalized size = 1.23 \[ \frac {b x^{n} \arccos \left (b x^{n} + a\right ) + a \arccos \left (b x^{n} + a\right ) - \sqrt {-b^{2} x^{2 \, n} - 2 \, a b x^{n} - a^{2} + 1}}{b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x^n*arccos(b*x^n + a) + a*arccos(b*x^n + a) - sqrt(-b^2*x^(2*n) - 2*a*b*x^n - a^2 + 1))/(b*n)

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giac [A]  time = 0.84, size = 41, normalized size = 0.85 \[ \frac {{\left (b x^{n} + a\right )} \arccos \left (b x^{n} + a\right ) - \sqrt {-{\left (b x^{n} + a\right )}^{2} + 1}}{b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*x^n + a)*arccos(b*x^n + a) - sqrt(-(b*x^n + a)^2 + 1))/(b*n)

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int x^{n -1} \arccos \left (a +b \,x^{n}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(n-1)*arccos(a+b*x^n),x)

[Out]

int(x^(n-1)*arccos(a+b*x^n),x)

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maxima [A]  time = 0.41, size = 41, normalized size = 0.85 \[ \frac {{\left (b x^{n} + a\right )} \arccos \left (b x^{n} + a\right ) - \sqrt {-{\left (b x^{n} + a\right )}^{2} + 1}}{b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*x^n + a)*arccos(b*x^n + a) - sqrt(-(b*x^n + a)^2 + 1))/(b*n)

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mupad [B]  time = 0.35, size = 111, normalized size = 2.31 \[ \frac {x^n\,\mathrm {acos}\left (a+b\,x^n\right )}{n}-\frac {\sqrt {1-b^2\,x^{2\,n}-2\,a\,b\,x^n-a^2}}{b\,n}-\frac {a\,\ln \left (\sqrt {1-b^2\,x^{2\,n}-2\,a\,b\,x^n-a^2}-\frac {a\,b+b^2\,x^n}{\sqrt {-b^2}}\right )}{n\,\sqrt {-b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(n - 1)*acos(a + b*x^n),x)

[Out]

(x^n*acos(a + b*x^n))/n - (1 - b^2*x^(2*n) - 2*a*b*x^n - a^2)^(1/2)/(b*n) - (a*log((1 - b^2*x^(2*n) - 2*a*b*x^
n - a^2)^(1/2) - (a*b + b^2*x^n)/(-b^2)^(1/2)))/(n*(-b^2)^(1/2))

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sympy [A]  time = 54.92, size = 76, normalized size = 1.58 \[ \begin {cases} \log {\relax (x )} \operatorname {acos}{\relax (a )} & \text {for}\: b = 0 \wedge n = 0 \\\log {\relax (x )} \operatorname {acos}{\left (a + b \right )} & \text {for}\: n = 0 \\\frac {x^{n} \operatorname {acos}{\relax (a )}}{n} & \text {for}\: b = 0 \\\frac {a \operatorname {acos}{\left (a + b x^{n} \right )}}{b n} + \frac {x^{n} \operatorname {acos}{\left (a + b x^{n} \right )}}{n} - \frac {\sqrt {- a^{2} - 2 a b x^{n} - b^{2} x^{2 n} + 1}}{b n} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1+n)*acos(a+b*x**n),x)

[Out]

Piecewise((log(x)*acos(a), Eq(b, 0) & Eq(n, 0)), (log(x)*acos(a + b), Eq(n, 0)), (x**n*acos(a)/n, Eq(b, 0)), (
a*acos(a + b*x**n)/(b*n) + x**n*acos(a + b*x**n)/n - sqrt(-a**2 - 2*a*b*x**n - b**2*x**(2*n) + 1)/(b*n), True)
)

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