3.2.65 \(\int (d x)^m (a+b \cosh ^{-1}(c x)) \, dx\) [165]

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

[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 = 0.04, 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 = {5883, 127, 372, 371} \begin {gather*} \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}} \end {gather*}

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 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 (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 = 0.10, size = 87, normalized size = 0.82 \begin {gather*} \frac {x (d x)^m \left (a+b \cosh ^{-1}(c x)-\frac {b c x \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 )}{(2+m) \sqrt {-1+c x} \sqrt {1+c x}}\right )}{1+m} \end {gather*}

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 = 9.59, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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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((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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d x\right )^{m} \left (a + b \operatorname {acosh}{\left (c x \right )}\right )\, dx \end {gather*}

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

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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