3.2.18 \(\int x^m \cosh ^{-1}(a x) \, dx\) [118]

Optimal. Leaf size=91 \[ \frac {x^{1+m} \cosh ^{-1}(a x)}{1+m}-\frac {a x^{2+m} \sqrt {1-a^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};a^2 x^2\right )}{\left (2+3 m+m^2\right ) \sqrt {-1+a x} \sqrt {1+a x}} \]

[Out]

x^(1+m)*arccosh(a*x)/(1+m)-a*x^(2+m)*hypergeom([1/2, 1+1/2*m],[2+1/2*m],a^2*x^2)*(-a^2*x^2+1)^(1/2)/(m^2+3*m+2
)/(a*x-1)^(1/2)/(a*x+1)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5883, 127, 372, 371} \begin {gather*} \frac {x^{m+1} \cosh ^{-1}(a x)}{m+1}-\frac {a \sqrt {1-a^2 x^2} x^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};a^2 x^2\right )}{\left (m^2+3 m+2\right ) \sqrt {a x-1} \sqrt {a x+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^m*ArcCosh[a*x],x]

[Out]

(x^(1 + m)*ArcCosh[a*x])/(1 + m) - (a*x^(2 + m)*Sqrt[1 - a^2*x^2]*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2,
a^2*x^2])/((2 + 3*m + m^2)*Sqrt[-1 + a*x]*Sqrt[1 + a*x])

Rule 127

Int[((f_.)*(x_))^(p_.)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Dist[(a + b*x)^Frac
Part[m]*((c + d*x)^FracPart[m]/(a*c + b*d*x^2)^FracPart[m]), Int[(a*c + b*d*x^2)^m*(f*x)^p, x], x] /; FreeQ[{a
, b, c, d, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m]

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/
(1 + b*(x^n/a))^FracPart[p]), Int[(c*x)^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 5883

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcC
osh[c*x])^n/(d*(m + 1))), x] - Dist[b*c*(n/(d*(m + 1))), Int[(d*x)^(m + 1)*((a + b*ArcCosh[c*x])^(n - 1)/(Sqrt
[1 + c*x]*Sqrt[-1 + c*x])), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rubi steps

\begin {align*} \int x^m \cosh ^{-1}(a x) \, dx &=\frac {x^{1+m} \cosh ^{-1}(a x)}{1+m}-\frac {a \int \frac {x^{1+m}}{\sqrt {-1+a x} \sqrt {1+a x}} \, dx}{1+m}\\ &=\frac {x^{1+m} \cosh ^{-1}(a x)}{1+m}-\frac {\left (a \sqrt {-1+a^2 x^2}\right ) \int \frac {x^{1+m}}{\sqrt {-1+a^2 x^2}} \, dx}{(1+m) \sqrt {-1+a x} \sqrt {1+a x}}\\ &=\frac {x^{1+m} \cosh ^{-1}(a x)}{1+m}-\frac {\left (a \sqrt {1-a^2 x^2}\right ) \int \frac {x^{1+m}}{\sqrt {1-a^2 x^2}} \, dx}{(1+m) \sqrt {-1+a x} \sqrt {1+a x}}\\ &=\frac {x^{1+m} \cosh ^{-1}(a x)}{1+m}-\frac {a x^{2+m} \sqrt {1-a^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};a^2 x^2\right )}{\left (2+3 m+m^2\right ) \sqrt {-1+a x} \sqrt {1+a x}}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 82, normalized size = 0.90 \begin {gather*} \frac {x^{1+m} \left (\cosh ^{-1}(a x)-\frac {a x \sqrt {1-a^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};a^2 x^2\right )}{(2+m) \sqrt {-1+a x} \sqrt {1+a x}}\right )}{1+m} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^m*ArcCosh[a*x],x]

[Out]

(x^(1 + m)*(ArcCosh[a*x] - (a*x*Sqrt[1 - a^2*x^2]*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, a^2*x^2])/((2 +
m)*Sqrt[-1 + a*x]*Sqrt[1 + a*x])))/(1 + m)

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Maple [F]
time = 4.65, size = 0, normalized size = 0.00 \[\int x^{m} \mathrm {arccosh}\left (a x \right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*arccosh(a*x),x)

[Out]

int(x^m*arccosh(a*x),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*arccosh(a*x),x, algorithm="maxima")

[Out]

-a^2*integrate(x^2*x^m/(a^2*(m + 1)*x^2 - m - 1), x) + a*integrate(x*x^m/(a^3*(m + 1)*x^3 - a*(m + 1)*x + (a^2
*(m + 1)*x^2 - m - 1)*sqrt(a*x + 1)*sqrt(a*x - 1)), x) + x*x^m*log(a*x + sqrt(a*x + 1)*sqrt(a*x - 1))/(m + 1)

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Fricas [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(x^m*arccosh(a*x),x, algorithm="fricas")

[Out]

integral(x^m*arccosh(a*x), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{m} \operatorname {acosh}{\left (a x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*acosh(a*x),x)

[Out]

Integral(x**m*acosh(a*x), x)

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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(x^m*arccosh(a*x),x, algorithm="giac")

[Out]

integrate(x^m*arccosh(a*x), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^m\,\mathrm {acosh}\left (a\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*acosh(a*x),x)

[Out]

int(x^m*acosh(a*x), x)

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