3.165 \(\int (d x)^m (a+b \cosh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=106 \[ \frac {(d x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{d (m+1)}-\frac {b c \sqrt {1-c^2 x^2} (d x)^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{d^2 (m+1) (m+2) \sqrt {c x-1} \sqrt {c x+1}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 106, 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 = {5662, 126, 365, 364} \[ \frac {(d x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{d (m+1)}-\frac {b c \sqrt {1-c^2 x^2} (d x)^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{d^2 (m+1) (m+2) \sqrt {c x-1} \sqrt {c x+1}} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^m*(a + b*ArcCosh[c*x]),x]

[Out]

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

Rule 126

Int[((f_.)*(x_))^(p_.)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Dist[((a + b*x)^Fra
cPart[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[m - n, 0]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 365

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 87, normalized size = 0.82 \[ \frac {x (d x)^m \left (a-\frac {b c x \sqrt {1-c^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{(m+2) \sqrt {c x-1} \sqrt {c x+1}}+b \cosh ^{-1}(c x)\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^m*(a + b*ArcCosh[c*x]),x]

[Out]

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

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fricas [F]  time = 1.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \operatorname {arcosh}\left (c x\right ) + a\right )} \left (d x\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(a+b*arccosh(c*x)),x, algorithm="fricas")

[Out]

integral((b*arccosh(c*x) + a)*(d*x)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(a+b*arccosh(c*x)),x, algorithm="giac")

[Out]

integrate((b*arccosh(c*x) + a)*(d*x)^m, x)

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maple [F]  time = 2.96, size = 0, normalized size = 0.00 \[ \int \left (d x \right )^{m} \left (a +b \,\mathrm {arccosh}\left (c x \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m*(a+b*arccosh(c*x)),x)

[Out]

int((d*x)^m*(a+b*arccosh(c*x)),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(a+b*arccosh(c*x)),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(c*x))*(d*x)^m,x)

[Out]

int((a + b*acosh(c*x))*(d*x)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**m*(a+b*acosh(c*x)),x)

[Out]

Integral((d*x)**m*(a + b*acosh(c*x)), x)

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