Optimal. Leaf size=206 \[ -\frac {b e (f x)^{1+m} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{c^2 f (2+m) (3+m)}+\frac {d (f x)^{1+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f (1+m)}+\frac {e (f x)^{3+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (3+m)}+\frac {b \left (e (1+m)^2+c^2 d (2+m) (3+m)\right ) (f x)^{1+m} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};c^2 x^2\right )}{c^2 f (1+m)^2 (2+m) (3+m)} \]
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Rubi [A]
time = 0.13, antiderivative size = 192, normalized size of antiderivative = 0.93, number of steps
used = 4, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {14, 6436, 12,
470, 371} \begin {gather*} \frac {d (f x)^{m+1} \left (a+b \text {sech}^{-1}(c x)\right )}{f (m+1)}+\frac {e (f x)^{m+3} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (m+3)}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} (f x)^{m+1} \left (\frac {e}{c^2 (m+2) (m+3)}+\frac {d}{(m+1)^2}\right ) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};c^2 x^2\right )}{f}-\frac {b e \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} (f x)^{m+1}}{c^2 f (m+2) (m+3)} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 14
Rule 371
Rule 470
Rule 6436
Rubi steps
\begin {align*} \int (f x)^m \left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right ) \, dx &=\frac {d (f x)^{1+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f (1+m)}+\frac {e (f x)^{3+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (3+m)}+\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {(f x)^m \left (d (3+m)+e (1+m) x^2\right )}{(1+m) (3+m) \sqrt {1-c^2 x^2}} \, dx\\ &=\frac {d (f x)^{1+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f (1+m)}+\frac {e (f x)^{3+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (3+m)}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {(f x)^m \left (d (3+m)+e (1+m) x^2\right )}{\sqrt {1-c^2 x^2}} \, dx}{3+4 m+m^2}\\ &=-\frac {b e (f x)^{1+m} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{c^2 f (2+m) (3+m)}+\frac {d (f x)^{1+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f (1+m)}+\frac {e (f x)^{3+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (3+m)}+\frac {\left (b \left (\frac {e (1+m)^2}{c^2 (2+m)}+d (3+m)\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {(f x)^m}{\sqrt {1-c^2 x^2}} \, dx}{3+4 m+m^2}\\ &=-\frac {b e (f x)^{1+m} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{c^2 f (2+m) (3+m)}+\frac {d (f x)^{1+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f (1+m)}+\frac {e (f x)^{3+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (3+m)}+\frac {b \left (\frac {e (1+m)^2}{c^2 (2+m)}+d (3+m)\right ) (f x)^{1+m} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};c^2 x^2\right )}{f (1+m) \left (3+4 m+m^2\right )}\\ \end {align*}
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Mathematica [F]
time = 0.07, size = 0, normalized size = 0.00 \begin {gather*} \int (f x)^m \left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right ) \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [F]
time = 0.70, size = 0, normalized size = 0.00 \[\int \left (f x \right )^{m} \left (e \,x^{2}+d \right ) \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )\, dx\]
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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (f x\right )^{m} \left (a + b \operatorname {asech}{\left (c x \right )}\right ) \left (d + e x^{2}\right )\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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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.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (f\,x\right )}^m\,\left (e\,x^2+d\right )\,\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right ) \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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