Optimal. Leaf size=116 \[ \frac {5 a (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{6 b^3}-\frac {x (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{6 b^2}+\frac {a^3 \sec ^{-1}(a+b x)}{3 b^3}+\frac {1}{3} x^3 \sec ^{-1}(a+b x)-\frac {\left (1+6 a^2\right ) \tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )}{6 b^3} \]
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Rubi [A]
time = 0.07, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5366, 4511,
3867, 3855, 3852, 8} \begin {gather*} \frac {a^3 \sec ^{-1}(a+b x)}{3 b^3}-\frac {\left (6 a^2+1\right ) \tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )}{6 b^3}+\frac {5 a (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{6 b^3}-\frac {x (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{6 b^2}+\frac {1}{3} x^3 \sec ^{-1}(a+b x) \end {gather*}
Antiderivative was successfully verified.
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Rule 8
Rule 3852
Rule 3855
Rule 3867
Rule 4511
Rule 5366
Rubi steps
\begin {align*} \int x^2 \sec ^{-1}(a+b x) \, dx &=\frac {\text {Subst}\left (\int x \sec (x) (-a+\sec (x))^2 \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {1}{3} x^3 \sec ^{-1}(a+b x)-\frac {\text {Subst}\left (\int (-a+\sec (x))^3 \, dx,x,\sec ^{-1}(a+b x)\right )}{3 b^3}\\ &=-\frac {x (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{6 b^2}+\frac {1}{3} x^3 \sec ^{-1}(a+b x)-\frac {\text {Subst}\left (\int \left (-2 a^3+\left (1+6 a^2\right ) \sec (x)-5 a \sec ^2(x)\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{6 b^3}\\ &=-\frac {x (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{6 b^2}+\frac {a^3 \sec ^{-1}(a+b x)}{3 b^3}+\frac {1}{3} x^3 \sec ^{-1}(a+b x)+\frac {(5 a) \text {Subst}\left (\int \sec ^2(x) \, dx,x,\sec ^{-1}(a+b x)\right )}{6 b^3}-\frac {\left (1+6 a^2\right ) \text {Subst}\left (\int \sec (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{6 b^3}\\ &=-\frac {x (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{6 b^2}+\frac {a^3 \sec ^{-1}(a+b x)}{3 b^3}+\frac {1}{3} x^3 \sec ^{-1}(a+b x)-\frac {\left (1+6 a^2\right ) \tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )}{6 b^3}-\frac {(5 a) \text {Subst}\left (\int 1 \, dx,x,-(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}\right )}{6 b^3}\\ &=\frac {5 a (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{6 b^3}-\frac {x (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{6 b^2}+\frac {a^3 \sec ^{-1}(a+b x)}{3 b^3}+\frac {1}{3} x^3 \sec ^{-1}(a+b x)-\frac {\left (1+6 a^2\right ) \tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )}{6 b^3}\\ \end {align*}
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Mathematica [A]
time = 0.12, size = 131, normalized size = 1.13 \begin {gather*} \frac {\left (5 a^2+4 a b x-b^2 x^2\right ) \sqrt {\frac {-1+a^2+2 a b x+b^2 x^2}{(a+b x)^2}}+2 b^3 x^3 \sec ^{-1}(a+b x)-2 a^3 \text {ArcSin}\left (\frac {1}{a+b x}\right )-\left (1+6 a^2\right ) \log \left ((a+b x) \left (1+\sqrt {\frac {-1+a^2+2 a b x+b^2 x^2}{(a+b x)^2}}\right )\right )}{6 b^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.15, size = 190, normalized size = 1.64
method | result | size |
derivativedivides | \(\frac {-\frac {\mathrm {arcsec}\left (b x +a \right ) a^{3}}{3}+\mathrm {arcsec}\left (b x +a \right ) a^{2} \left (b x +a \right )-\mathrm {arcsec}\left (b x +a \right ) a \left (b x +a \right )^{2}+\frac {\mathrm {arcsec}\left (b x +a \right ) \left (b x +a \right )^{3}}{3}-\frac {\sqrt {\left (b x +a \right )^{2}-1}\, \left (2 a^{3} \arctan \left (\frac {1}{\sqrt {\left (b x +a \right )^{2}-1}}\right )+6 a^{2} \ln \left (b x +a +\sqrt {\left (b x +a \right )^{2}-1}\right )-6 a \sqrt {\left (b x +a \right )^{2}-1}+\left (b x +a \right ) \sqrt {\left (b x +a \right )^{2}-1}+\ln \left (b x +a +\sqrt {\left (b x +a \right )^{2}-1}\right )\right )}{6 \sqrt {\frac {\left (b x +a \right )^{2}-1}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )}}{b^{3}}\) | \(190\) |
default | \(\frac {-\frac {\mathrm {arcsec}\left (b x +a \right ) a^{3}}{3}+\mathrm {arcsec}\left (b x +a \right ) a^{2} \left (b x +a \right )-\mathrm {arcsec}\left (b x +a \right ) a \left (b x +a \right )^{2}+\frac {\mathrm {arcsec}\left (b x +a \right ) \left (b x +a \right )^{3}}{3}-\frac {\sqrt {\left (b x +a \right )^{2}-1}\, \left (2 a^{3} \arctan \left (\frac {1}{\sqrt {\left (b x +a \right )^{2}-1}}\right )+6 a^{2} \ln \left (b x +a +\sqrt {\left (b x +a \right )^{2}-1}\right )-6 a \sqrt {\left (b x +a \right )^{2}-1}+\left (b x +a \right ) \sqrt {\left (b x +a \right )^{2}-1}+\ln \left (b x +a +\sqrt {\left (b x +a \right )^{2}-1}\right )\right )}{6 \sqrt {\frac {\left (b x +a \right )^{2}-1}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )}}{b^{3}}\) | \(190\) |
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 [A]
time = 3.18, size = 117, normalized size = 1.01 \begin {gather*} \frac {2 \, b^{3} x^{3} \operatorname {arcsec}\left (b x + a\right ) + 4 \, a^{3} \arctan \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) + {\left (6 \, a^{2} + 1\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1} {\left (b x - 5 \, a\right )}}{6 \, b^{3}} \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 x^{2} \operatorname {asec}{\left (a + b x \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 204 vs.
\(2 (100) = 200\).
time = 0.45, size = 204, normalized size = 1.76 \begin {gather*} -\frac {1}{24} \, b {\left (\frac {8 \, {\left (b x + a\right )}^{3} {\left (\frac {3 \, a}{b x + a} - \frac {3 \, a^{2}}{{\left (b x + a\right )}^{2}} - 1\right )} \arccos \left (-\frac {1}{{\left (b x + a\right )} {\left (\frac {a}{b x + a} - 1\right )} - a}\right )}{b^{4}} - \frac {{\left (b x + a\right )}^{2} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )}^{2} + 12 \, {\left (b x + a\right )} a {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )} + 4 \, {\left (6 \, a^{2} + 1\right )} \log \left (-{\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )} {\left | b x + a \right |}\right ) - \frac {12 \, {\left (b x + a\right )} a {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )} + 1}{{\left (b x + a\right )}^{2} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )}^{2}}}{b^{4}}\right )} \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 = -0.01 \begin {gather*} \int x^2\,\mathrm {acos}\left (\frac {1}{a+b\,x}\right ) \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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