Optimal. Leaf size=184 \[ -\frac {c^2 d (f x)^{m+3} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (m+3)}+\frac {d (f x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{f (m+1)}-\frac {b c d (3 m+7) \sqrt {1-c^2 x^2} (f x)^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{f^2 (m+1) (m+2) (m+3)^2 \sqrt {c x-1} \sqrt {c x+1}}+\frac {b c d \sqrt {c x-1} \sqrt {c x+1} (f x)^{m+2}}{f^2 (m+3)^2} \]
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Rubi [A] time = 0.26, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {14, 5731, 12, 460, 126, 365, 364} \[ -\frac {c^2 d (f x)^{m+3} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (m+3)}+\frac {d (f x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{f (m+1)}-\frac {b c d (3 m+7) \sqrt {1-c^2 x^2} (f x)^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{f^2 (m+1) (m+2) (m+3)^2 \sqrt {c x-1} \sqrt {c x+1}}+\frac {b c d \sqrt {c x-1} \sqrt {c x+1} (f x)^{m+2}}{f^2 (m+3)^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 14
Rule 126
Rule 364
Rule 365
Rule 460
Rule 5731
Rubi steps
\begin {align*} \int (f x)^m \left (d-c^2 d x^2\right ) \left (a+b \cosh ^{-1}(c x)\right ) \, dx &=\frac {d (f x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )}{f (1+m)}-\frac {c^2 d (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}-(b c) \int \frac {d (f x)^{1+m} \left (3+m-c^2 (1+m) x^2\right )}{f (1+m) (3+m) \sqrt {-1+c x} \sqrt {1+c x}} \, dx\\ &=\frac {d (f x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )}{f (1+m)}-\frac {c^2 d (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}-\frac {(b c d) \int \frac {(f x)^{1+m} \left (3+m-c^2 (1+m) x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{f \left (3+4 m+m^2\right )}\\ &=\frac {b c d (f x)^{2+m} \sqrt {-1+c x} \sqrt {1+c x}}{f^2 (3+m)^2}+\frac {d (f x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )}{f (1+m)}-\frac {c^2 d (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}-\frac {(b c d (7+3 m)) \int \frac {(f x)^{1+m}}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{f (1+m) (3+m)^2}\\ &=\frac {b c d (f x)^{2+m} \sqrt {-1+c x} \sqrt {1+c x}}{f^2 (3+m)^2}+\frac {d (f x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )}{f (1+m)}-\frac {c^2 d (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}-\frac {\left (b c d (7+3 m) \sqrt {-1+c^2 x^2}\right ) \int \frac {(f x)^{1+m}}{\sqrt {-1+c^2 x^2}} \, dx}{f (1+m) (3+m)^2 \sqrt {-1+c x} \sqrt {1+c x}}\\ &=\frac {b c d (f x)^{2+m} \sqrt {-1+c x} \sqrt {1+c x}}{f^2 (3+m)^2}+\frac {d (f x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )}{f (1+m)}-\frac {c^2 d (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}-\frac {\left (b c d (7+3 m) \sqrt {1-c^2 x^2}\right ) \int \frac {(f x)^{1+m}}{\sqrt {1-c^2 x^2}} \, dx}{f (1+m) (3+m)^2 \sqrt {-1+c x} \sqrt {1+c x}}\\ &=\frac {b c d (f x)^{2+m} \sqrt {-1+c x} \sqrt {1+c x}}{f^2 (3+m)^2}+\frac {d (f x)^{1+m} \left (a+b \cosh ^{-1}(c x)\right )}{f (1+m)}-\frac {c^2 d (f x)^{3+m} \left (a+b \cosh ^{-1}(c x)\right )}{f^3 (3+m)}-\frac {b c d (7+3 m) (f 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 )}{f^2 (1+m) (2+m) (3+m)^2 \sqrt {-1+c x} \sqrt {1+c x}}\\ \end {align*}
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Mathematica [A] time = 0.24, size = 191, normalized size = 1.04 \[ d x (f x)^m \left (-\frac {c^2 x^2 \left (a+b \cosh ^{-1}(c x)\right )}{m+3}+\frac {a+b \cosh ^{-1}(c x)}{m+1}-\frac {b c x \sqrt {1-c^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{\left (m^2+3 m+2\right ) \sqrt {c x-1} \sqrt {c x+1}}+\frac {b c^3 x^3 \sqrt {1-c^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {m+4}{2};\frac {m+6}{2};c^2 x^2\right )}{\left (m^2+7 m+12\right ) \sqrt {c x-1} \sqrt {c x+1}}\right ) \]
Antiderivative was successfully verified.
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fricas [F] time = 1.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a c^{2} d x^{2} - a d + {\left (b c^{2} d x^{2} - b d\right )} \operatorname {arcosh}\left (c x\right )\right )} \left (f x\right )^{m}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F(-2)] time = 180.00, size = 0, normalized size = 0.00 \[ \int \left (f x \right )^{m} \left (-c^{2} d \,x^{2}+d \right ) \left (a +b \,\mathrm {arccosh}\left (c x \right )\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {a c^{2} d f^{m} x^{3} x^{m}}{m + 3} - \frac {{\left (b c^{2} d f^{m} {\left (m + 1\right )} x^{3} - b d f^{m} {\left (m + 3\right )} x\right )} x^{m} \log \left (c x + \sqrt {c x + 1} \sqrt {c x - 1}\right )}{m^{2} + 4 \, m + 3} + \frac {\left (f x\right )^{m + 1} a d}{f {\left (m + 1\right )}} - \int \frac {{\left (b c^{3} d f^{m} {\left (m + 1\right )} x^{3} - b c d f^{m} {\left (m + 3\right )} x\right )} x^{m}}{{\left (m^{2} + 4 \, m + 3\right )} c^{3} x^{3} - {\left (m^{2} + 4 \, m + 3\right )} c x + {\left ({\left (m^{2} + 4 \, m + 3\right )} c^{2} x^{2} - m^{2} - 4 \, m - 3\right )} \sqrt {c x + 1} \sqrt {c x - 1}}\,{d x} + \int \frac {{\left (b c^{4} d f^{m} {\left (m + 1\right )} x^{4} - b c^{2} d f^{m} {\left (m + 3\right )} x^{2}\right )} x^{m}}{{\left (m^{2} + 4 \, m + 3\right )} c^{2} x^{2} - m^{2} - 4 \, m - 3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )\,\left (d-c^2\,d\,x^2\right )\,{\left (f\,x\right )}^m \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ - d \left (\int \left (- a \left (f x\right )^{m}\right )\, dx + \int \left (- b \left (f x\right )^{m} \operatorname {acosh}{\left (c x \right )}\right )\, dx + \int a c^{2} x^{2} \left (f x\right )^{m}\, dx + \int b c^{2} x^{2} \left (f x\right )^{m} \operatorname {acosh}{\left (c x \right )}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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