Optimal. Leaf size=69 \[ \frac {i \text {Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )}{2 d}+\frac {i \sec ^{-1}(a+b x)^2}{2 d}-\frac {\sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{d} \]
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Rubi [A] time = 0.09, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {5256, 12, 5218, 4626, 3719, 2190, 2279, 2391} \[ \frac {i \text {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )}{2 d}+\frac {i \sec ^{-1}(a+b x)^2}{2 d}-\frac {\sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 12
Rule 2190
Rule 2279
Rule 2391
Rule 3719
Rule 4626
Rule 5218
Rule 5256
Rubi steps
\begin {align*} \int \frac {\sec ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {b \sec ^{-1}(x)}{d x} \, dx,x,a+b x\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\sec ^{-1}(x)}{x} \, dx,x,a+b x\right )}{d}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {\cos ^{-1}(x)}{x} \, dx,x,\frac {1}{a+b x}\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int x \tan (x) \, dx,x,\cos ^{-1}\left (\frac {1}{a+b x}\right )\right )}{d}\\ &=\frac {i \cos ^{-1}\left (\frac {1}{a+b x}\right )^2}{2 d}-\frac {(2 i) \operatorname {Subst}\left (\int \frac {e^{2 i x} x}{1+e^{2 i x}} \, dx,x,\cos ^{-1}\left (\frac {1}{a+b x}\right )\right )}{d}\\ &=\frac {i \cos ^{-1}\left (\frac {1}{a+b x}\right )^2}{2 d}-\frac {\cos ^{-1}\left (\frac {1}{a+b x}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac {1}{a+b x}\right )}\right )}{d}+\frac {\operatorname {Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\cos ^{-1}\left (\frac {1}{a+b x}\right )\right )}{d}\\ &=\frac {i \cos ^{-1}\left (\frac {1}{a+b x}\right )^2}{2 d}-\frac {\cos ^{-1}\left (\frac {1}{a+b x}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac {1}{a+b x}\right )}\right )}{d}-\frac {i \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \cos ^{-1}\left (\frac {1}{a+b x}\right )}\right )}{2 d}\\ &=\frac {i \cos ^{-1}\left (\frac {1}{a+b x}\right )^2}{2 d}-\frac {\cos ^{-1}\left (\frac {1}{a+b x}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac {1}{a+b x}\right )}\right )}{d}+\frac {i \text {Li}_2\left (-e^{2 i \cos ^{-1}\left (\frac {1}{a+b x}\right )}\right )}{2 d}\\ \end {align*}
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Mathematica [A] time = 0.05, size = 59, normalized size = 0.86 \[ \frac {i \left (\text {Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )+\sec ^{-1}(a+b x) \left (\sec ^{-1}(a+b x)+2 i \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )\right )\right )}{2 d} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {arcsec}\left (b x + a\right )}{b d x + a d}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.59, size = 115, normalized size = 1.67 \[ -\frac {1}{4} \, b^{2} {\left (\frac {2 \, {\left (b x + a\right )}^{2} \arccos \left (\frac {1}{{\left ({\left (b x + a\right )} {\left (\frac {a}{b x + a} - 1\right )} - a\right )} {\left (\frac {a}{b x + a} - 1\right )} + a}\right )}{b^{3} d} - \frac {{\left (b x + a\right )} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )} - \frac {1}{{\left (b x + a\right )} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )}}}{b^{3} d}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.35, size = 92, normalized size = 1.33 \[ \frac {i \mathrm {arcsec}\left (b x +a \right )^{2}}{2 d}-\frac {\mathrm {arcsec}\left (b x +a \right ) \ln \left (1+\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )^{2}\right )}{d}+\frac {i \polylog \left (2, -\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )^{2}\right )}{2 d} \]
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 \[ -\frac {-i \, {\left (b {\left (\frac {\log \left (b x + a + 1\right ) \log \left (b x + a\right ) - \log \left (b x + a\right )^{2} + \log \left (b x + a\right ) \log \left (b x + a - 1\right )}{b^{2} d} - \frac {\log \left (b x + a + 1\right ) \log \left (b x + a\right ) + {\rm Li}_2\left (-b x - a\right )}{b^{2} d} - \frac {\log \left (b x + a\right ) \log \left (-b x - a + 1\right ) + {\rm Li}_2\left (b x + a\right )}{b^{2} d}\right )} - {\left (\frac {\log \left (b x + a + 1\right )}{b d} - \frac {2 \, \log \left (b x + a\right )}{b d} + \frac {\log \left (b x + a - 1\right )}{b d}\right )} \log \left (b x + a\right )\right )} b d + 2 \, b \int \frac {\log \left (b x + a\right )}{\sqrt {b x + a + 1} \sqrt {b x + a - 1} b x + \sqrt {b x + a + 1} \sqrt {b x + a - 1} a}\,{d x} - 2 \, \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) \log \left (b x + a\right ) + i \, \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right ) \log \left (b x + a\right ) - i \, \log \left (b x + a + 1\right ) \log \left (b x + a\right ) - i \, \log \left (b x + a\right )^{2} - i \, \log \left (b x + a\right ) \log \left (-b x - a + 1\right ) - i \, {\rm Li}_2\left (b x + a\right ) - i \, {\rm Li}_2\left (-b x - a\right )}{2 \, d} \]
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 \[ \int \frac {\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}{d\,x+\frac {a\,d}{b}} \,d x \]
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 \[ \frac {b \int \frac {\operatorname {asec}{\left (a + b x \right )}}{a + b x}\, dx}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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