Optimal. Leaf size=78 \[ -\frac {a^2 \sec ^{-1}(a+b x)}{2 b^2}-\frac {(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{2 b^2}+\frac {a \tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )}{b^2}+\frac {1}{2} x^2 \sec ^{-1}(a+b x) \]
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Rubi [A] time = 0.05, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5258, 4426, 3773, 3770, 3767, 8} \[ -\frac {a^2 \sec ^{-1}(a+b x)}{2 b^2}-\frac {(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{2 b^2}+\frac {a \tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )}{b^2}+\frac {1}{2} x^2 \sec ^{-1}(a+b x) \]
Antiderivative was successfully verified.
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Rule 8
Rule 3767
Rule 3770
Rule 3773
Rule 4426
Rule 5258
Rubi steps
\begin {align*} \int x \sec ^{-1}(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int x \sec (x) (-a+\sec (x)) \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}\\ &=\frac {1}{2} x^2 \sec ^{-1}(a+b x)-\frac {\operatorname {Subst}\left (\int (-a+\sec (x))^2 \, dx,x,\sec ^{-1}(a+b x)\right )}{2 b^2}\\ &=-\frac {a^2 \sec ^{-1}(a+b x)}{2 b^2}+\frac {1}{2} x^2 \sec ^{-1}(a+b x)-\frac {\operatorname {Subst}\left (\int \sec ^2(x) \, dx,x,\sec ^{-1}(a+b x)\right )}{2 b^2}+\frac {a \operatorname {Subst}\left (\int \sec (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}\\ &=-\frac {a^2 \sec ^{-1}(a+b x)}{2 b^2}+\frac {1}{2} x^2 \sec ^{-1}(a+b x)+\frac {a \tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )}{b^2}+\frac {\operatorname {Subst}\left (\int 1 \, dx,x,-(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}\right )}{2 b^2}\\ &=-\frac {(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{2 b^2}-\frac {a^2 \sec ^{-1}(a+b x)}{2 b^2}+\frac {1}{2} x^2 \sec ^{-1}(a+b x)+\frac {a \tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )}{b^2}\\ \end {align*}
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Mathematica [A] time = 0.12, size = 110, normalized size = 1.41 \[ \frac {-(a+b x) \sqrt {\frac {a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}}+2 a \log \left ((a+b x) \left (\sqrt {\frac {a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}}+1\right )\right )+a^2 \sin ^{-1}\left (\frac {1}{a+b x}\right )+b^2 x^2 \sec ^{-1}(a+b x)}{2 b^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 1.44, size = 104, normalized size = 1.33 \[ \frac {b^{2} x^{2} \operatorname {arcsec}\left (b x + a\right ) - 2 \, a^{2} \arctan \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - 2 \, a \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}}{2 \, b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.16, size = 133, normalized size = 1.71 \[ -\frac {1}{4} \, b {\left (\frac {2 \, {\left (b x + a\right )}^{2} {\left (\frac {2 \, a}{b x + a} - 1\right )} \arccos \left (-\frac {1}{{\left (b x + a\right )} {\left (\frac {a}{b x + a} - 1\right )} - a}\right )}{b^{3}} + \frac {{\left (b x + a\right )} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )} + 4 \, a \log \left (-{\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )} {\left | b x + a \right |}\right ) - \frac {1}{{\left (b x + a\right )} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )}}}{b^{3}}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 126, normalized size = 1.62 \[ \frac {x^{2} \mathrm {arcsec}\left (b x +a \right )}{2}-\frac {a^{2} \mathrm {arcsec}\left (b x +a \right )}{2 b^{2}}+\frac {\sqrt {-1+\left (b x +a \right )^{2}}\, a \ln \left (b x +a +\sqrt {-1+\left (b x +a \right )^{2}}\right )}{b^{2} \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )}-\frac {-1+\left (b x +a \right )^{2}}{2 b^{2} \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )} \]
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 \[ \frac {1}{2} \, x^{2} \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) - \int \frac {{\left (b^{2} x^{3} + a b x^{2}\right )} e^{\left (\frac {1}{2} \, \log \left (b x + a + 1\right ) + \frac {1}{2} \, \log \left (b x + a - 1\right )\right )}}{2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + {\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )} e^{\left (\log \left (b x + a + 1\right ) + \log \left (b x + a - 1\right )\right )} - 1\right )}}\,{d x} \]
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 \[ \int x\,\mathrm {acos}\left (\frac {1}{a+b\,x}\right ) \,d x \]
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 \[ \int x \operatorname {asec}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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