Optimal. Leaf size=244 \[ -\frac {2 b \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {2 b \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a \sqrt {1-a^2}}-\frac {2 i b \sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {2 i b \sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a \sqrt {1-a^2}}-\frac {b \sec ^{-1}(a+b x)^2}{a}-\frac {\sec ^{-1}(a+b x)^2}{x} \]
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Rubi [A] time = 0.39, antiderivative size = 244, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {5258, 4426, 4191, 3321, 2264, 2190, 2279, 2391} \[ -\frac {2 b \text {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {2 b \text {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a \sqrt {1-a^2}}-\frac {2 i b \sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {2 i b \sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a \sqrt {1-a^2}}-\frac {b \sec ^{-1}(a+b x)^2}{a}-\frac {\sec ^{-1}(a+b x)^2}{x} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2264
Rule 2279
Rule 2391
Rule 3321
Rule 4191
Rule 4426
Rule 5258
Rubi steps
\begin {align*} \int \frac {\sec ^{-1}(a+b x)^2}{x^2} \, dx &=b \operatorname {Subst}\left (\int \frac {x^2 \sec (x) \tan (x)}{(-a+\sec (x))^2} \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=-\frac {\sec ^{-1}(a+b x)^2}{x}+(2 b) \operatorname {Subst}\left (\int \frac {x}{-a+\sec (x)} \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=-\frac {\sec ^{-1}(a+b x)^2}{x}+(2 b) \operatorname {Subst}\left (\int \left (-\frac {x}{a}+\frac {x}{a (1-a \cos (x))}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=-\frac {b \sec ^{-1}(a+b x)^2}{a}-\frac {\sec ^{-1}(a+b x)^2}{x}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {x}{1-a \cos (x)} \, dx,x,\sec ^{-1}(a+b x)\right )}{a}\\ &=-\frac {b \sec ^{-1}(a+b x)^2}{a}-\frac {\sec ^{-1}(a+b x)^2}{x}+\frac {(4 b) \operatorname {Subst}\left (\int \frac {e^{i x} x}{-a+2 e^{i x}-a e^{2 i x}} \, dx,x,\sec ^{-1}(a+b x)\right )}{a}\\ &=-\frac {b \sec ^{-1}(a+b x)^2}{a}-\frac {\sec ^{-1}(a+b x)^2}{x}-\frac {(4 b) \operatorname {Subst}\left (\int \frac {e^{i x} x}{2-2 \sqrt {1-a^2}-2 a e^{i x}} \, dx,x,\sec ^{-1}(a+b x)\right )}{\sqrt {1-a^2}}+\frac {(4 b) \operatorname {Subst}\left (\int \frac {e^{i x} x}{2+2 \sqrt {1-a^2}-2 a e^{i x}} \, dx,x,\sec ^{-1}(a+b x)\right )}{\sqrt {1-a^2}}\\ &=-\frac {b \sec ^{-1}(a+b x)^2}{a}-\frac {\sec ^{-1}(a+b x)^2}{x}-\frac {2 i b \sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {2 i b \sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {(2 i b) \operatorname {Subst}\left (\int \log \left (1-\frac {2 a e^{i x}}{2-2 \sqrt {1-a^2}}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{a \sqrt {1-a^2}}-\frac {(2 i b) \operatorname {Subst}\left (\int \log \left (1-\frac {2 a e^{i x}}{2+2 \sqrt {1-a^2}}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{a \sqrt {1-a^2}}\\ &=-\frac {b \sec ^{-1}(a+b x)^2}{a}-\frac {\sec ^{-1}(a+b x)^2}{x}-\frac {2 i b \sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {2 i b \sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {2 a x}{2-2 \sqrt {1-a^2}}\right )}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{a \sqrt {1-a^2}}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {2 a x}{2+2 \sqrt {1-a^2}}\right )}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{a \sqrt {1-a^2}}\\ &=-\frac {b \sec ^{-1}(a+b x)^2}{a}-\frac {\sec ^{-1}(a+b x)^2}{x}-\frac {2 i b \sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {2 i b \sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}-\frac {2 b \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {2 b \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}\\ \end {align*}
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Mathematica [B] time = 2.29, size = 686, normalized size = 2.81 \[ -\frac {\frac {(a+b x) \sec ^{-1}(a+b x)^2}{x}+\frac {2 b \left (i \left (\text {Li}_2\left (\frac {\left (i \sqrt {a^2-1}+1\right ) \left (-a+\sqrt {a^2-1} \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )+1\right )}{a \left (a+\sqrt {a^2-1} \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )-1\right )}\right )-\text {Li}_2\left (\frac {\left (1-i \sqrt {a^2-1}\right ) \left (-a+\sqrt {a^2-1} \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )+1\right )}{a \left (a+\sqrt {a^2-1} \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )-1\right )}\right )\right )+2 \sec ^{-1}(a+b x) \tanh ^{-1}\left (\frac {(a-1) \cot \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt {a^2-1}}\right )-2 \cos ^{-1}\left (\frac {1}{a}\right ) \tanh ^{-1}\left (\frac {(a+1) \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt {a^2-1}}\right )-\log \left (\frac {(a-1) \left (\sqrt {a^2-1}+i a+i\right ) \left (\tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )-i\right )}{a \left (\sqrt {a^2-1} \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )+a-1\right )}\right ) \left (\cos ^{-1}\left (\frac {1}{a}\right )-2 i \tanh ^{-1}\left (\frac {(a+1) \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt {a^2-1}}\right )\right )-\log \left (\frac {(a-1) \left (\sqrt {a^2-1}-i a-i\right ) \left (\tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )+i\right )}{a \left (\sqrt {a^2-1} \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )+a-1\right )}\right ) \left (\cos ^{-1}\left (\frac {1}{a}\right )+2 i \tanh ^{-1}\left (\frac {(a+1) \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt {a^2-1}}\right )\right )+\log \left (\frac {\sqrt {a^2-1} e^{-\frac {1}{2} i \sec ^{-1}(a+b x)}}{\sqrt {2} \sqrt {a} \sqrt {-\frac {b x}{a+b x}}}\right ) \left (-2 i \tanh ^{-1}\left (\frac {(a-1) \cot \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt {a^2-1}}\right )+2 i \tanh ^{-1}\left (\frac {(a+1) \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt {a^2-1}}\right )+\cos ^{-1}\left (\frac {1}{a}\right )\right )+\log \left (\frac {\sqrt {a^2-1} e^{\frac {1}{2} i \sec ^{-1}(a+b x)}}{\sqrt {2} \sqrt {a} \sqrt {-\frac {b x}{a+b x}}}\right ) \left (\cos ^{-1}\left (\frac {1}{a}\right )+2 i \left (\tanh ^{-1}\left (\frac {(a-1) \cot \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt {a^2-1}}\right )-\tanh ^{-1}\left (\frac {(a+1) \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt {a^2-1}}\right )\right )\right )\right )}{\sqrt {a^2-1}}}{a} \]
Antiderivative was successfully verified.
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fricas [F] time = 2.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arcsec}\left (b x + a\right )^{2}}{x^{2}}, x\right ) \]
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 \[ \int \frac {\operatorname {arcsec}\left (b x + a\right )^{2}}{x^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.84, size = 341, normalized size = 1.40 \[ -\frac {b \mathrm {arcsec}\left (b x +a \right )^{2}}{a}-\frac {\mathrm {arcsec}\left (b x +a \right )^{2}}{x}-\frac {2 i b \sqrt {-a^{2}+1}\, \mathrm {arcsec}\left (b x +a \right ) \ln \left (\frac {-a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}+1}{1+\sqrt {-a^{2}+1}}\right )}{a \left (a^{2}-1\right )}+\frac {2 i b \sqrt {-a^{2}+1}\, \mathrm {arcsec}\left (b x +a \right ) \ln \left (\frac {a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}-1}{-1+\sqrt {-a^{2}+1}}\right )}{a \left (a^{2}-1\right )}-\frac {2 b \sqrt {-a^{2}+1}\, \dilog \left (\frac {-a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}+1}{1+\sqrt {-a^{2}+1}}\right )}{a \left (a^{2}-1\right )}+\frac {2 b \sqrt {-a^{2}+1}\, \dilog \left (\frac {a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}-1}{-1+\sqrt {-a^{2}+1}}\right )}{a \left (a^{2}-1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
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 \[ \int \frac {{\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}^2}{x^2} \,d x \]
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 \[ \int \frac {\operatorname {asec}^{2}{\left (a + b x \right )}}{x^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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