Optimal. Leaf size=116 \[ \frac{a^3 \sec ^{-1}(a+b x)}{3 b^3}-\frac{\left (6 a^2+1\right ) \tanh ^{-1}\left (\sqrt{1-\frac{1}{(a+b x)^2}}\right )}{6 b^3}+\frac{5 a (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{6 b^3}-\frac{x (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{6 b^2}+\frac{1}{3} x^3 \sec ^{-1}(a+b x) \]
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Rubi [A] time = 0.089765, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {5258, 4426, 3782, 3770, 3767, 8} \[ \frac{a^3 \sec ^{-1}(a+b x)}{3 b^3}-\frac{\left (6 a^2+1\right ) \tanh ^{-1}\left (\sqrt{1-\frac{1}{(a+b x)^2}}\right )}{6 b^3}+\frac{5 a (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{6 b^3}-\frac{x (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{6 b^2}+\frac{1}{3} x^3 \sec ^{-1}(a+b x) \]
Antiderivative was successfully verified.
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Rule 5258
Rule 4426
Rule 3782
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int x^2 \sec ^{-1}(a+b x) \, dx &=\frac{\operatorname{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{\operatorname{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{\operatorname{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) \operatorname{Subst}\left (\int \sec ^2(x) \, dx,x,\sec ^{-1}(a+b x)\right )}{6 b^3}-\frac{\left (1+6 a^2\right ) \operatorname{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) \operatorname{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*}
Mathematica [A] time = 0.174164, size = 131, normalized size = 1.13 \[ \frac{\left (5 a^2+4 a b x-b^2 x^2\right ) \sqrt{\frac{a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}}-\left (6 a^2+1\right ) \log \left ((a+b x) \left (\sqrt{\frac{a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}}+1\right )\right )-2 a^3 \sin ^{-1}\left (\frac{1}{a+b x}\right )+2 b^3 x^3 \sec ^{-1}(a+b x)}{6 b^3} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.212, size = 273, normalized size = 2.4 \begin{align*}{\frac{{x}^{3}{\rm arcsec} \left (bx+a\right )}{3}}-{\frac{{a}^{3}}{3\,{b}^{3} \left ( bx+a \right ) }\sqrt{-1+ \left ( bx+a \right ) ^{2}}\arctan \left ({\frac{1}{\sqrt{-1+ \left ( bx+a \right ) ^{2}}}} \right ){\frac{1}{\sqrt{{\frac{-1+ \left ( bx+a \right ) ^{2}}{ \left ( bx+a \right ) ^{2}}}}}}}-{\frac{ \left ( -1+ \left ( bx+a \right ) ^{2} \right ) x}{6\,{b}^{2} \left ( bx+a \right ) }{\frac{1}{\sqrt{{\frac{-1+ \left ( bx+a \right ) ^{2}}{ \left ( bx+a \right ) ^{2}}}}}}}-{\frac{{a}^{2}}{{b}^{3} \left ( bx+a \right ) }\sqrt{-1+ \left ( bx+a \right ) ^{2}}\ln \left ( bx+a+\sqrt{-1+ \left ( bx+a \right ) ^{2}} \right ){\frac{1}{\sqrt{{\frac{-1+ \left ( bx+a \right ) ^{2}}{ \left ( bx+a \right ) ^{2}}}}}}}+{\frac{ \left ( -5+5\, \left ( bx+a \right ) ^{2} \right ) a}{6\,{b}^{3} \left ( bx+a \right ) }{\frac{1}{\sqrt{{\frac{-1+ \left ( bx+a \right ) ^{2}}{ \left ( bx+a \right ) ^{2}}}}}}}-{\frac{1}{6\,{b}^{3} \left ( bx+a \right ) }\sqrt{-1+ \left ( bx+a \right ) ^{2}}\ln \left ( bx+a+\sqrt{-1+ \left ( bx+a \right ) ^{2}} \right ){\frac{1}{\sqrt{{\frac{-1+ \left ( bx+a \right ) ^{2}}{ \left ( bx+a \right ) ^{2}}}}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{3} \, x^{3} \arctan \left (\sqrt{b x + a + 1} \sqrt{b x + a - 1}\right ) - \int \frac{{\left (b^{2} x^{4} + a b x^{3}\right )} e^{\left (\frac{1}{2} \, \log \left (b x + a + 1\right ) + \frac{1}{2} \, \log \left (b x + a - 1\right )\right )}}{3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2} +{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )} e^{\left (\log \left (b x + a + 1\right ) + \log \left (b x + a - 1\right )\right )} - 1\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.53533, size = 284, normalized size = 2.45 \begin{align*} \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{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{asec}{\left (a + b x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{arcsec}\left (b x + a\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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