Integrand size = 14, antiderivative size = 106 \[ \int (d x)^m (a+b \text {arccosh}(c x)) \, dx=\frac {(d x)^{1+m} (a+b \text {arccosh}(c x))}{d (1+m)}-\frac {b c (d x)^{2+m} \sqrt {1-c^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{d^2 (1+m) (2+m) \sqrt {-1+c x} \sqrt {1+c x}} \]
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Time = 0.05 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5883, 127, 372, 371} \[ \int (d x)^m (a+b \text {arccosh}(c x)) \, dx=\frac {(d x)^{m+1} (a+b \text {arccosh}(c x))}{d (m+1)}-\frac {b c \sqrt {1-c^2 x^2} (d x)^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},c^2 x^2\right )}{d^2 (m+1) (m+2) \sqrt {c x-1} \sqrt {c x+1}} \]
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Rule 127
Rule 371
Rule 372
Rule 5883
Rubi steps \begin{align*} \text {integral}& = \frac {(d x)^{1+m} (a+b \text {arccosh}(c x))}{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} (a+b \text {arccosh}(c x))}{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} (a+b \text {arccosh}(c x))}{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} (a+b \text {arccosh}(c x))}{d (1+m)}-\frac {b c (d x)^{2+m} \sqrt {1-c^2 x^2} \operatorname {Hypergeometric2F1}\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*}
Time = 0.11 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.82 \[ \int (d x)^m (a+b \text {arccosh}(c x)) \, dx=\frac {x (d x)^m \left (a+b \text {arccosh}(c x)-\frac {b c x \sqrt {1-c^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{(2+m) \sqrt {-1+c x} \sqrt {1+c x}}\right )}{1+m} \]
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\[\int \left (d x \right )^{m} \left (a +b \,\operatorname {arccosh}\left (c x \right )\right )d x\]
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\[ \int (d x)^m (a+b \text {arccosh}(c x)) \, dx=\int { {\left (b \operatorname {arcosh}\left (c x\right ) + a\right )} \left (d x\right )^{m} \,d x } \]
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\[ \int (d x)^m (a+b \text {arccosh}(c x)) \, dx=\int \left (d x\right )^{m} \left (a + b \operatorname {acosh}{\left (c x \right )}\right )\, dx \]
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\[ \int (d x)^m (a+b \text {arccosh}(c x)) \, dx=\int { {\left (b \operatorname {arcosh}\left (c x\right ) + a\right )} \left (d x\right )^{m} \,d x } \]
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\[ \int (d x)^m (a+b \text {arccosh}(c x)) \, dx=\int { {\left (b \operatorname {arcosh}\left (c x\right ) + a\right )} \left (d x\right )^{m} \,d x } \]
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Timed out. \[ \int (d x)^m (a+b \text {arccosh}(c x)) \, dx=\int \left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )\,{\left (d\,x\right )}^m \,d x \]
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