Optimal. Leaf size=154 \[ -\frac {a^2 \sec ^{-1}(a+b x)^2}{2 b^2}+\frac {2 i a \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {2 i a \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {\log (a+b x)}{b^2}-\frac {(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^2}-\frac {4 i a \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {1}{2} x^2 \sec ^{-1}(a+b x)^2 \]
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Rubi [A] time = 0.13, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {5258, 4426, 4190, 4181, 2279, 2391, 4184, 3475} \[ \frac {2 i a \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {2 i a \text {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {a^2 \sec ^{-1}(a+b x)^2}{2 b^2}+\frac {\log (a+b x)}{b^2}-\frac {(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^2}-\frac {4 i a \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {1}{2} x^2 \sec ^{-1}(a+b x)^2 \]
Antiderivative was successfully verified.
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Rule 2279
Rule 2391
Rule 3475
Rule 4181
Rule 4184
Rule 4190
Rule 4426
Rule 5258
Rubi steps
\begin {align*} \int x \sec ^{-1}(a+b x)^2 \, dx &=\frac {\operatorname {Subst}\left (\int x^2 \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)^2-\frac {\operatorname {Subst}\left (\int x (-a+\sec (x))^2 \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}\\ &=\frac {1}{2} x^2 \sec ^{-1}(a+b x)^2-\frac {\operatorname {Subst}\left (\int \left (a^2 x-2 a x \sec (x)+x \sec ^2(x)\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}\\ &=-\frac {a^2 \sec ^{-1}(a+b x)^2}{2 b^2}+\frac {1}{2} x^2 \sec ^{-1}(a+b x)^2-\frac {\operatorname {Subst}\left (\int x \sec ^2(x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}+\frac {(2 a) \operatorname {Subst}\left (\int x \sec (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}\\ &=-\frac {(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^2}-\frac {a^2 \sec ^{-1}(a+b x)^2}{2 b^2}+\frac {1}{2} x^2 \sec ^{-1}(a+b x)^2-\frac {4 i a \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {\operatorname {Subst}\left (\int \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}-\frac {(2 a) \operatorname {Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}+\frac {(2 a) \operatorname {Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}\\ &=-\frac {(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^2}-\frac {a^2 \sec ^{-1}(a+b x)^2}{2 b^2}+\frac {1}{2} x^2 \sec ^{-1}(a+b x)^2-\frac {4 i a \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {\log (a+b x)}{b^2}+\frac {(2 i a) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {(2 i a) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{b^2}\\ &=-\frac {(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^2}-\frac {a^2 \sec ^{-1}(a+b x)^2}{2 b^2}+\frac {1}{2} x^2 \sec ^{-1}(a+b x)^2-\frac {4 i a \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {\log (a+b x)}{b^2}+\frac {2 i a \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {2 i a \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}\\ \end {align*}
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Mathematica [A] time = 0.15, size = 142, normalized size = 0.92 \[ \frac {-\frac {1}{2} a^2 \sec ^{-1}(a+b x)^2+\frac {1}{2} b^2 x^2 \sec ^{-1}(a+b x)^2+2 i a \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )-2 i a \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )+\log (a+b x)-(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)-4 i a \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2} \]
Antiderivative was successfully verified.
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fricas [F] time = 1.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x \operatorname {arcsec}\left (b x + a\right )^{2}, x\right ) \]
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 \[ \int x \operatorname {arcsec}\left (b x + a\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.33, size = 264, normalized size = 1.71 \[ -\frac {a^{2} \mathrm {arcsec}\left (b x +a \right )^{2}}{2 b^{2}}-\frac {2 a \,\mathrm {arcsec}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b^{2}}+\frac {2 a \,\mathrm {arcsec}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b^{2}}+\frac {2 i a \dilog \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b^{2}}-\frac {2 i a \dilog \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b^{2}}+\frac {x^{2} \mathrm {arcsec}\left (b x +a \right )^{2}}{2}-\frac {\sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \mathrm {arcsec}\left (b x +a \right ) x}{b}-\frac {\sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \mathrm {arcsec}\left (b x +a \right ) a}{b^{2}}-\frac {\ln \left (\frac {1}{b x +a}\right )}{b^{2}} \]
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 \[ \frac {1}{2} \, x^{2} \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )^{2} - \frac {1}{8} \, x^{2} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )^{2} - \int \frac {2 \, \sqrt {b x + a + 1} \sqrt {b x + a - 1} b x^{2} \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) + 2 \, {\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} + {\left (3 \, a^{2} - 1\right )} b x^{2} + {\left (a^{3} - a\right )} x\right )} \log \left (b x + a\right )^{2} - {\left (b^{3} x^{4} + 2 \, a b^{2} x^{3} + {\left (a^{2} - 1\right )} b x^{2} + 2 \, {\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} + {\left (3 \, a^{2} - 1\right )} b x^{2} + {\left (a^{3} - a\right )} x\right )} \log \left (b x + a\right )\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}{2 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} - 1\right )} b x - a\right )}}\,{d x} \]
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 \[ \int x\,{\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}^2 \,d x \]
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 \[ \int x \operatorname {asec}^{2}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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