3.117 \(\int x^m \cosh ^{-1}(a x)^2 \, dx\)

Optimal. Leaf size=154 \[ -\frac {2 a^2 x^{m+3} \, _3F_2\left (1,\frac {m}{2}+\frac {3}{2},\frac {m}{2}+\frac {3}{2};\frac {m}{2}+2,\frac {m}{2}+\frac {5}{2};a^2 x^2\right )}{m^3+6 m^2+11 m+6}-\frac {2 a \sqrt {1-a x} x^{m+2} \cosh ^{-1}(a x) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};a^2 x^2\right )}{\left (m^2+3 m+2\right ) \sqrt {a x-1}}+\frac {x^{m+1} \cosh ^{-1}(a x)^2}{m+1} \]

[Out]

x^(1+m)*arccosh(a*x)^2/(1+m)-2*a^2*x^(3+m)*HypergeometricPFQ([1, 3/2+1/2*m, 3/2+1/2*m],[2+1/2*m, 5/2+1/2*m],a^
2*x^2)/(m^3+6*m^2+11*m+6)-2*a*x^(2+m)*arccosh(a*x)*hypergeom([1/2, 1+1/2*m],[2+1/2*m],a^2*x^2)*(-a*x+1)^(1/2)/
(m^2+3*m+2)/(a*x-1)^(1/2)

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Rubi [A]  time = 0.24, antiderivative size = 167, normalized size of antiderivative = 1.08, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5662, 5763} \[ -\frac {2 a^2 x^{m+3} \, _3F_2\left (1,\frac {m}{2}+\frac {3}{2},\frac {m}{2}+\frac {3}{2};\frac {m}{2}+2,\frac {m}{2}+\frac {5}{2};a^2 x^2\right )}{m^3+6 m^2+11 m+6}-\frac {2 a \sqrt {1-a^2 x^2} x^{m+2} \cosh ^{-1}(a x) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};a^2 x^2\right )}{\left (m^2+3 m+2\right ) \sqrt {a x-1} \sqrt {a x+1}}+\frac {x^{m+1} \cosh ^{-1}(a x)^2}{m+1} \]

Antiderivative was successfully verified.

[In]

Int[x^m*ArcCosh[a*x]^2,x]

[Out]

(x^(1 + m)*ArcCosh[a*x]^2)/(1 + m) - (2*a*x^(2 + m)*Sqrt[1 - a^2*x^2]*ArcCosh[a*x]*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]) - (2*a^2*x^(3 + m)*HypergeometricPF
Q[{1, 3/2 + m/2, 3/2 + m/2}, {2 + m/2, 5/2 + m/2}, a^2*x^2])/(6 + 11*m + 6*m^2 + m^3)

Rule 5662

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))/(Sqr
t[-1 + c*x]*Sqrt[1 + c*x]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 5763

Int[(((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_))/(Sqrt[(d1_) + (e1_.)*(x_)]*Sqrt[(d2_) + (e2_.)*(x_
)]), x_Symbol] :> Simp[((f*x)^(m + 1)*Sqrt[1 - c^2*x^2]*(a + b*ArcCosh[c*x])*Hypergeometric2F1[1/2, (1 + m)/2,
(3 + m)/2, c^2*x^2])/(f*(m + 1)*Sqrt[d1 + e1*x]*Sqrt[d2 + e2*x]), x] + Simp[(b*c*(f*x)^(m + 2)*Hypergeometric
PFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2])/(Sqrt[-(d1*d2)]*f^2*(m + 1)*(m + 2)), x] /; FreeQ[{
a, b, c, d1, e1, d2, e2, f, m}, x] && EqQ[e1 - c*d1, 0] && EqQ[e2 + c*d2, 0] && GtQ[d1, 0] && LtQ[d2, 0] &&  !
IntegerQ[m]

Rubi steps

\begin {align*} \int x^m \cosh ^{-1}(a x)^2 \, dx &=\frac {x^{1+m} \cosh ^{-1}(a x)^2}{1+m}-\frac {(2 a) \int \frac {x^{1+m} \cosh ^{-1}(a x)}{\sqrt {-1+a x} \sqrt {1+a x}} \, dx}{1+m}\\ &=\frac {x^{1+m} \cosh ^{-1}(a x)^2}{1+m}-\frac {2 a x^{2+m} \sqrt {1-a^2 x^2} \cosh ^{-1}(a x) \, _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}}-\frac {2 a^2 x^{3+m} \, _3F_2\left (1,\frac {3}{2}+\frac {m}{2},\frac {3}{2}+\frac {m}{2};2+\frac {m}{2},\frac {5}{2}+\frac {m}{2};a^2 x^2\right )}{6+11 m+6 m^2+m^3}\\ \end {align*}

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Mathematica [A]  time = 0.38, size = 143, normalized size = 0.93 \[ \frac {x^{m+1} \left (\cosh ^{-1}(a x)^2-\frac {2 a x \left (\frac {a x \, _3F_2\left (1,\frac {m}{2}+\frac {3}{2},\frac {m}{2}+\frac {3}{2};\frac {m}{2}+2,\frac {m}{2}+\frac {5}{2};a^2 x^2\right )}{m+3}+\frac {\sqrt {1-a^2 x^2} \cosh ^{-1}(a x) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};a^2 x^2\right )}{\sqrt {a x-1} \sqrt {a x+1}}\right )}{m+2}\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*ArcCosh[a*x]^2,x]

[Out]

(x^(1 + m)*(ArcCosh[a*x]^2 - (2*a*x*((Sqrt[1 - a^2*x^2]*ArcCosh[a*x]*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)
/2, a^2*x^2])/(Sqrt[-1 + a*x]*Sqrt[1 + a*x]) + (a*x*HypergeometricPFQ[{1, 3/2 + m/2, 3/2 + m/2}, {2 + m/2, 5/2
+ m/2}, a^2*x^2])/(3 + m)))/(2 + m)))/(1 + m)

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fricas [F]  time = 0.74, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{m} \operatorname {arcosh}\left (a x\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*arccosh(a*x)^2,x, algorithm="fricas")

[Out]

integral(x^m*arccosh(a*x)^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{m} \operatorname {arcosh}\left (a x\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*arccosh(a*x)^2,x, algorithm="giac")

[Out]

integrate(x^m*arccosh(a*x)^2, x)

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maple [F]  time = 1.14, size = 0, normalized size = 0.00 \[ \int x^{m} \mathrm {arccosh}\left (a x \right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*arccosh(a*x)^2,x)

[Out]

int(x^m*arccosh(a*x)^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {x x^{m} \log \left (a x + \sqrt {a x + 1} \sqrt {a x - 1}\right )^{2}}{m + 1} - \int \frac {2 \, {\left (\sqrt {a x + 1} \sqrt {a x - 1} a^{2} x^{2} x^{m} + {\left (a^{3} x^{3} - a x\right )} x^{m}\right )} \log \left (a x + \sqrt {a x + 1} \sqrt {a x - 1}\right )}{a^{3} {\left (m + 1\right )} x^{3} - a {\left (m + 1\right )} x + {\left (a^{2} {\left (m + 1\right )} x^{2} - m - 1\right )} \sqrt {a x + 1} \sqrt {a x - 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*arccosh(a*x)^2,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^m\,{\mathrm {acosh}\left (a\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*acosh(a*x)^2,x)

[Out]

int(x^m*acosh(a*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{m} \operatorname {acosh}^{2}{\left (a x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*acosh(a*x)**2,x)

[Out]

Integral(x**m*acosh(a*x)**2, x)

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