Optimal. Leaf size=455 \[ -\frac {3 b c d (f x)^{2+m} \sqrt {d-c^2 d x^2}}{f^2 (2+m)^2 (4+m) \sqrt {-1+c x} \sqrt {1+c x}}-\frac {b c d (f x)^{2+m} \sqrt {d-c^2 d x^2}}{f^2 (2+m) (4+m) \sqrt {-1+c x} \sqrt {1+c x}}+\frac {b c^3 d (f x)^{4+m} \sqrt {d-c^2 d x^2}}{f^4 (4+m)^2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {3 d (f x)^{1+m} \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{f \left (8+6 m+m^2\right )}+\frac {(f x)^{1+m} \left (d-c^2 d x^2\right )^{3/2} \left (a+b \cosh ^{-1}(c x)\right )}{f (4+m)}+\frac {3 d (f x)^{1+m} \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right ) \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};c^2 x^2\right )}{f (4+m) \left (2+3 m+m^2\right ) \sqrt {1-c x} \sqrt {1+c x}}-\frac {3 b c d (f x)^{2+m} \sqrt {d-c^2 d x^2} \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};c^2 x^2\right )}{f^2 (1+m) (2+m)^2 (4+m) \sqrt {-1+c x} \sqrt {1+c x}} \]
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Rubi [A]
time = 0.35, antiderivative size = 455, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {5930, 5926,
5949, 32, 74, 14} \begin {gather*} -\frac {3 b c d \sqrt {d-c^2 d x^2} (f x)^{m+2} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;c^2 x^2\right )}{f^2 (m+1) (m+2)^2 (m+4) \sqrt {c x-1} \sqrt {c x+1}}+\frac {3 d \sqrt {d-c^2 d x^2} (f x)^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};c^2 x^2\right ) \left (a+b \cosh ^{-1}(c x)\right )}{f (m+4) \left (m^2+3 m+2\right ) \sqrt {1-c x} \sqrt {c x+1}}+\frac {3 d \sqrt {d-c^2 d x^2} (f x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{f \left (m^2+6 m+8\right )}+\frac {\left (d-c^2 d x^2\right )^{3/2} (f x)^{m+1} \left (a+b \cosh ^{-1}(c x)\right )}{f (m+4)}-\frac {b c d \sqrt {d-c^2 d x^2} (f x)^{m+2}}{f^2 (m+2) (m+4) \sqrt {c x-1} \sqrt {c x+1}}-\frac {3 b c d \sqrt {d-c^2 d x^2} (f x)^{m+2}}{f^2 (m+2)^2 (m+4) \sqrt {c x-1} \sqrt {c x+1}}+\frac {b c^3 d \sqrt {d-c^2 d x^2} (f x)^{m+4}}{f^4 (m+4)^2 \sqrt {c x-1} \sqrt {c x+1}} \end {gather*}
Antiderivative was successfully verified.
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Rule 14
Rule 32
Rule 74
Rule 5926
Rule 5930
Rule 5949
Rubi steps
\begin {align*} \int (f x)^m \left (d-c^2 d x^2\right )^{3/2} \left (a+b \cosh ^{-1}(c x)\right ) \, dx &=-\frac {\left (d \sqrt {d-c^2 d x^2}\right ) \int (f x)^m (-1+c x)^{3/2} (1+c x)^{3/2} \left (a+b \cosh ^{-1}(c x)\right ) \, dx}{\sqrt {-1+c x} \sqrt {1+c x}}\\ &=\frac {d (f x)^{1+m} (1-c x) (1+c x) \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{f (4+m)}+\frac {\left (3 d \sqrt {d-c^2 d x^2}\right ) \int (f x)^m \sqrt {-1+c x} \sqrt {1+c x} \left (a+b \cosh ^{-1}(c x)\right ) \, dx}{(4+m) \sqrt {-1+c x} \sqrt {1+c x}}+\frac {\left (b c d \sqrt {d-c^2 d x^2}\right ) \int (f x)^{1+m} \left (-1+c^2 x^2\right ) \, dx}{f (4+m) \sqrt {-1+c x} \sqrt {1+c x}}\\ &=\frac {3 d (f x)^{1+m} \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{f \left (8+6 m+m^2\right )}+\frac {d (f x)^{1+m} (1-c x) (1+c x) \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{f (4+m)}+\frac {\left (b c d \sqrt {d-c^2 d x^2}\right ) \int \left (-(f x)^{1+m}+\frac {c^2 (f x)^{3+m}}{f^2}\right ) \, dx}{f (4+m) \sqrt {-1+c x} \sqrt {1+c x}}-\frac {\left (3 d \sqrt {d-c^2 d x^2}\right ) \int \frac {(f x)^m \left (a+b \cosh ^{-1}(c x)\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{(2+m) (4+m) \sqrt {-1+c x} \sqrt {1+c x}}-\frac {\left (3 b c d \sqrt {d-c^2 d x^2}\right ) \int (f x)^{1+m} \, dx}{f (2+m) (4+m) \sqrt {-1+c x} \sqrt {1+c x}}\\ &=-\frac {3 b c d (f x)^{2+m} \sqrt {d-c^2 d x^2}}{f^2 (2+m)^2 (4+m) \sqrt {-1+c x} \sqrt {1+c x}}-\frac {b c d (f x)^{2+m} \sqrt {d-c^2 d x^2}}{f^2 (2+m) (4+m) \sqrt {-1+c x} \sqrt {1+c x}}+\frac {b c^3 d (f x)^{4+m} \sqrt {d-c^2 d x^2}}{f^4 (4+m)^2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {3 d (f x)^{1+m} \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{f \left (8+6 m+m^2\right )}+\frac {d (f x)^{1+m} (1-c x) (1+c x) \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{f (4+m)}+\frac {3 d (f x)^{1+m} \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right ) \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};c^2 x^2\right )}{f (1+m) (2+m) (4+m) (1-c x) (1+c x)}-\frac {3 b c d (f x)^{2+m} \sqrt {d-c^2 d x^2} \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};c^2 x^2\right )}{f^2 (1+m) (2+m)^2 (4+m) \sqrt {-1+c x} \sqrt {1+c x}}\\ \end {align*}
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Mathematica [A]
time = 0.59, size = 274, normalized size = 0.60 \begin {gather*} -\frac {d x (f x)^m \sqrt {d-c^2 d x^2} \left (\frac {3 b c x}{(2+m)^2}+b c x \left (\frac {1}{2+m}-\frac {c^2 x^2}{4+m}\right )-\frac {3 \sqrt {-1+c x} \sqrt {1+c x} \left (a+b \cosh ^{-1}(c x)\right )}{2+m}+(-1+c x)^{3/2} (1+c x)^{3/2} \left (a+b \cosh ^{-1}(c x)\right )+\frac {3 \sqrt {1-c^2 x^2} \left (a+b \cosh ^{-1}(c x)\right ) \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};c^2 x^2\right )}{(1+m) (2+m) \sqrt {-1+c x} \sqrt {1+c x}}+\frac {3 b c x \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};c^2 x^2\right )}{(1+m) (2+m)^2}\right )}{(4+m) \sqrt {-1+c x} \sqrt {1+c x}} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \left (f x \right )^{m} \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \,\mathrm {arccosh}\left (c x \right )\right )\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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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.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
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 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^{3/2}\,{\left (f\,x\right )}^m \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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