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

Optimal. Leaf size=106 \[ \frac{(d x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{d (m+1)}-\frac{b c \sqrt{1-c^2 x^2} (d x)^{m+2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},c^2 x^2\right )}{d^2 (m+1) (m+2) \sqrt{c x-1} \sqrt{c x+1}} \]

[Out]

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

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Rubi [A]  time = 0.053476, antiderivative size = 106, normalized size of antiderivative = 1., 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 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 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 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 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])

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*}

Mathematica [A]  time = 0.13697, size = 87, normalized size = 0.82 \[ \frac{x (d x)^m \left (-\frac{b c x \sqrt{1-c^2 x^2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},c^2 x^2\right )}{(m+2) \sqrt{c x-1} \sqrt{c x+1}}+a+b \cosh ^{-1}(c x)\right )}{m+1} \]

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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Maple [F]  time = 2.023, size = 0, normalized size = 0. \begin{align*} \int \left ( dx \right ) ^{m} \left ( a+b{\rm arccosh} \left (cx\right ) \right ) \, dx \end{align*}

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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \operatorname{arcosh}\left (c x\right ) + a\right )} \left (d x\right )^{m}, x\right ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d x\right )^{m} \left (a + b \operatorname{acosh}{\left (c x \right )}\right )\, dx \end{align*}

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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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out