Optimal. Leaf size=94 \[ \frac {(a+b x) \sec ^{-1}(a+b x)^2}{b}+\frac {4 i \sec ^{-1}(a+b x) \text {ArcTan}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b}-\frac {2 i \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {2 i \text {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b} \]
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Rubi [A]
time = 0.05, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 6, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5360, 5324,
3842, 4266, 2317, 2438} \begin {gather*} \frac {4 i \sec ^{-1}(a+b x) \text {ArcTan}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b}-\frac {2 i \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {2 i \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {(a+b x) \sec ^{-1}(a+b x)^2}{b} \end {gather*}
Antiderivative was successfully verified.
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Rule 2317
Rule 2438
Rule 3842
Rule 4266
Rule 5324
Rule 5360
Rubi steps
\begin {align*} \int \sec ^{-1}(a+b x)^2 \, dx &=\frac {\text {Subst}\left (\int \sec ^{-1}(x)^2 \, dx,x,a+b x\right )}{b}\\ &=\frac {\text {Subst}\left (\int x^2 \sec (x) \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b}\\ &=\frac {(a+b x) \sec ^{-1}(a+b x)^2}{b}-\frac {2 \text {Subst}\left (\int x \sec (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b}\\ &=\frac {(a+b x) \sec ^{-1}(a+b x)^2}{b}+\frac {4 i \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {2 \text {Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b}-\frac {2 \text {Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b}\\ &=\frac {(a+b x) \sec ^{-1}(a+b x)^2}{b}+\frac {4 i \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b}-\frac {(2 i) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {(2 i) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{b}\\ &=\frac {(a+b x) \sec ^{-1}(a+b x)^2}{b}+\frac {4 i \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b}-\frac {2 i \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {2 i \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b}\\ \end {align*}
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Mathematica [A]
time = 0.08, size = 111, normalized size = 1.18 \begin {gather*} \frac {\sec ^{-1}(a+b x) \left ((a+b x) \sec ^{-1}(a+b x)-2 \log \left (1-i e^{i \sec ^{-1}(a+b x)}\right )+2 \log \left (1+i e^{i \sec ^{-1}(a+b x)}\right )\right )-2 i \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )+2 i \text {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.26, size = 162, normalized size = 1.72
method | result | size |
derivativedivides | \(\frac {\mathrm {arcsec}\left (b x +a \right )^{2} \left (b x +a \right )+2 \,\mathrm {arcsec}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-2 \,\mathrm {arcsec}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-2 i \dilog \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+2 i \dilog \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b}\) | \(162\) |
default | \(\frac {\mathrm {arcsec}\left (b x +a \right )^{2} \left (b x +a \right )+2 \,\mathrm {arcsec}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-2 \,\mathrm {arcsec}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-2 i \dilog \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+2 i \dilog \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b}\) | \(162\) |
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 \operatorname {asec}^{2}{\left (a + b x \right )}\, 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 = -0.01 \begin {gather*} \int {\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}^2 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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