Optimal. Leaf size=244 \[ -\frac {b \sec ^{-1}(a+b x)^2}{a}-\frac {\sec ^{-1}(a+b x)^2}{x}-\frac {2 i b \sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {2 i b \sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}-\frac {2 b \text {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {2 b \text {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}} \]
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Rubi [A]
time = 0.28, 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 = {5366, 4511,
4276, 3402, 2296, 2221, 2317, 2438} \begin {gather*} -\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} \end {gather*}
Antiderivative was successfully verified.
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Rule 2221
Rule 2296
Rule 2317
Rule 2438
Rule 3402
Rule 4276
Rule 4511
Rule 5366
Rubi steps
\begin {align*} \int \frac {\sec ^{-1}(a+b x)^2}{x^2} \, dx &=b \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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] Both result and optimal contain complex but leaf count is larger than twice
the leaf count of optimal. \(686\) vs. \(2(244)=488\).
time = 1.51, size = 686, normalized size = 2.81 \begin {gather*} -\frac {\frac {(a+b x) \sec ^{-1}(a+b x)^2}{x}+\frac {2 b \left (2 \sec ^{-1}(a+b x) \tanh ^{-1}\left (\frac {(-1+a) \cot \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt {-1+a^2}}\right )-2 \text {ArcCos}\left (\frac {1}{a}\right ) \tanh ^{-1}\left (\frac {(1+a) \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt {-1+a^2}}\right )+\left (\text {ArcCos}\left (\frac {1}{a}\right )-2 i \tanh ^{-1}\left (\frac {(-1+a) \cot \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt {-1+a^2}}\right )+2 i \tanh ^{-1}\left (\frac {(1+a) \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt {-1+a^2}}\right )\right ) \log \left (\frac {\sqrt {-1+a^2} e^{-\frac {1}{2} i \sec ^{-1}(a+b x)}}{\sqrt {2} \sqrt {a} \sqrt {-\frac {b x}{a+b x}}}\right )+\left (\text {ArcCos}\left (\frac {1}{a}\right )+2 i \left (\tanh ^{-1}\left (\frac {(-1+a) \cot \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt {-1+a^2}}\right )-\tanh ^{-1}\left (\frac {(1+a) \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt {-1+a^2}}\right )\right )\right ) \log \left (\frac {\sqrt {-1+a^2} e^{\frac {1}{2} i \sec ^{-1}(a+b x)}}{\sqrt {2} \sqrt {a} \sqrt {-\frac {b x}{a+b x}}}\right )-\left (\text {ArcCos}\left (\frac {1}{a}\right )-2 i \tanh ^{-1}\left (\frac {(1+a) \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt {-1+a^2}}\right )\right ) \log \left (\frac {(-1+a) \left (i+i a+\sqrt {-1+a^2}\right ) \left (-i+\tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )\right )}{a \left (-1+a+\sqrt {-1+a^2} \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )\right )}\right )-\left (\text {ArcCos}\left (\frac {1}{a}\right )+2 i \tanh ^{-1}\left (\frac {(1+a) \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt {-1+a^2}}\right )\right ) \log \left (\frac {(-1+a) \left (-i-i a+\sqrt {-1+a^2}\right ) \left (i+\tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )\right )}{a \left (-1+a+\sqrt {-1+a^2} \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )\right )}\right )+i \left (-\text {PolyLog}\left (2,\frac {\left (1-i \sqrt {-1+a^2}\right ) \left (1-a+\sqrt {-1+a^2} \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )\right )}{a \left (-1+a+\sqrt {-1+a^2} \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )\right )}\right )+\text {PolyLog}\left (2,\frac {\left (1+i \sqrt {-1+a^2}\right ) \left (1-a+\sqrt {-1+a^2} \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )\right )}{a \left (-1+a+\sqrt {-1+a^2} \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )\right )}\right )\right )\right )}{\sqrt {-1+a^2}}}{a} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.80, size = 336, normalized size = 1.38
method | result | size |
derivativedivides | \(b \left (-\frac {\left (b x +a \right ) \mathrm {arcsec}\left (b x +a \right )^{2}}{a b x}-\frac {2 i \sqrt {-a^{2}+1}\, \mathrm {arcsec}\left (b x +a \right ) \ln \left (\frac {-a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}+1}{1+\sqrt {-a^{2}+1}}\right )}{a \left (a^{2}-1\right )}+\frac {2 i \sqrt {-a^{2}+1}\, \mathrm {arcsec}\left (b x +a \right ) \ln \left (\frac {a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}-1}{-1+\sqrt {-a^{2}+1}}\right )}{a \left (a^{2}-1\right )}-\frac {2 \sqrt {-a^{2}+1}\, \dilog \left (\frac {-a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}+1}{1+\sqrt {-a^{2}+1}}\right )}{a \left (a^{2}-1\right )}+\frac {2 \sqrt {-a^{2}+1}\, \dilog \left (\frac {a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}-1}{-1+\sqrt {-a^{2}+1}}\right )}{a \left (a^{2}-1\right )}\right )\) | \(336\) |
default | \(b \left (-\frac {\left (b x +a \right ) \mathrm {arcsec}\left (b x +a \right )^{2}}{a b x}-\frac {2 i \sqrt {-a^{2}+1}\, \mathrm {arcsec}\left (b x +a \right ) \ln \left (\frac {-a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}+1}{1+\sqrt {-a^{2}+1}}\right )}{a \left (a^{2}-1\right )}+\frac {2 i \sqrt {-a^{2}+1}\, \mathrm {arcsec}\left (b x +a \right ) \ln \left (\frac {a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}-1}{-1+\sqrt {-a^{2}+1}}\right )}{a \left (a^{2}-1\right )}-\frac {2 \sqrt {-a^{2}+1}\, \dilog \left (\frac {-a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}+1}{1+\sqrt {-a^{2}+1}}\right )}{a \left (a^{2}-1\right )}+\frac {2 \sqrt {-a^{2}+1}\, \dilog \left (\frac {a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}-1}{-1+\sqrt {-a^{2}+1}}\right )}{a \left (a^{2}-1\right )}\right )\) | \(336\) |
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 [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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 \frac {\operatorname {asec}^{2}{\left (a + b x \right )}}{x^{2}}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \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 = -0.00 \begin {gather*} \int \frac {{\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}^2}{x^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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