Optimal. Leaf size=247 \[ \frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {\left (e-\sqrt {e^2-c^2 d^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}+\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {\left (\sqrt {e^2-c^2 d^2}+e\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}-\frac {\log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{e}-\frac {i b \text {Li}_2\left (-\frac {\left (e-\sqrt {e^2-c^2 d^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}-\frac {i b \text {Li}_2\left (-\frac {\left (e+\sqrt {e^2-c^2 d^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}+\frac {i b \text {Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )}{2 e} \]
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Rubi [A] time = 0.37, antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5224, 2518} \[ -\frac {i b \text {PolyLog}\left (2,-\frac {\left (e-\sqrt {e^2-c^2 d^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}-\frac {i b \text {PolyLog}\left (2,-\frac {\left (\sqrt {e^2-c^2 d^2}+e\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}+\frac {i b \text {PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )}{2 e}+\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {\left (e-\sqrt {e^2-c^2 d^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}+\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {\left (\sqrt {e^2-c^2 d^2}+e\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}-\frac {\log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{e} \]
Antiderivative was successfully verified.
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Rule 2518
Rule 5224
Rubi steps
\begin {align*} \int \frac {a+b \sec ^{-1}(c x)}{d+e x} \, dx &=\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {\left (e-\sqrt {-c^2 d^2+e^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}+\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {\left (e+\sqrt {-c^2 d^2+e^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )}{e}-\frac {b \int \frac {\log \left (1+\frac {\left (e-\sqrt {-c^2 d^2+e^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{\sqrt {1-\frac {1}{c^2 x^2}} x^2} \, dx}{c e}-\frac {b \int \frac {\log \left (1+\frac {\left (e+\sqrt {-c^2 d^2+e^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{\sqrt {1-\frac {1}{c^2 x^2}} x^2} \, dx}{c e}+\frac {b \int \frac {\log \left (1+e^{2 i \sec ^{-1}(c x)}\right )}{\sqrt {1-\frac {1}{c^2 x^2}} x^2} \, dx}{c e}\\ &=\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {\left (e-\sqrt {-c^2 d^2+e^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}+\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {\left (e+\sqrt {-c^2 d^2+e^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )}{e}-\frac {i b \text {Li}_2\left (-\frac {\left (e-\sqrt {-c^2 d^2+e^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}-\frac {i b \text {Li}_2\left (-\frac {\left (e+\sqrt {-c^2 d^2+e^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}+\frac {i b \text {Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )}{2 e}\\ \end {align*}
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Mathematica [A] time = 0.62, size = 333, normalized size = 1.35 \[ \frac {a \log (d+e x)}{e}+\frac {b \left (-i \left (\text {Li}_2\left (\frac {\left (\sqrt {e^2-c^2 d^2}-e\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )+\text {Li}_2\left (-\frac {\left (e+\sqrt {e^2-c^2 d^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )\right )+\log \left (1+\frac {\left (e-\sqrt {e^2-c^2 d^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right ) \left (2 \sin ^{-1}\left (\frac {\sqrt {\frac {e}{c d}+1}}{\sqrt {2}}\right )+\sec ^{-1}(c x)\right )+\log \left (1+\frac {\left (\sqrt {e^2-c^2 d^2}+e\right ) e^{i \sec ^{-1}(c x)}}{c d}\right ) \left (\sec ^{-1}(c x)-2 \sin ^{-1}\left (\frac {\sqrt {\frac {e}{c d}+1}}{\sqrt {2}}\right )\right )+4 i \sin ^{-1}\left (\frac {\sqrt {\frac {e}{c d}+1}}{\sqrt {2}}\right ) \tan ^{-1}\left (\frac {(e-c d) \tan \left (\frac {1}{2} \sec ^{-1}(c x)\right )}{\sqrt {e^2-c^2 d^2}}\right )+\frac {1}{2} i \text {Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )-\sec ^{-1}(c x) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )\right )}{e} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {arcsec}\left (c x\right ) + a}{e x + d}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.82, size = 456, normalized size = 1.85 \[ \frac {a \ln \left (c e x +d c \right )}{e}-\frac {b \,\mathrm {arcsec}\left (c x \right ) \ln \left (1+i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{e}-\frac {b \,\mathrm {arcsec}\left (c x \right ) \ln \left (1-i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{e}+\frac {i b \dilog \left (1+i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{e}+\frac {i b \dilog \left (1-i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{e}+\frac {b \,\mathrm {arcsec}\left (c x \right ) \ln \left (\frac {-d c \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )+\sqrt {-c^{2} d^{2}+e^{2}}-e}{-e +\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{e}+\frac {b \,\mathrm {arcsec}\left (c x \right ) \ln \left (\frac {d c \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )+\sqrt {-c^{2} d^{2}+e^{2}}+e}{e +\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{e}-\frac {i b \dilog \left (\frac {-d c \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )+\sqrt {-c^{2} d^{2}+e^{2}}-e}{-e +\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{e}-\frac {i b \dilog \left (\frac {d c \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )+\sqrt {-c^{2} d^{2}+e^{2}}+e}{e +\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{e} \]
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 \[ b \int \frac {\arctan \left (\sqrt {c x + 1} \sqrt {c x - 1}\right )}{e x + d}\,{d x} + \frac {a \log \left (e x + d\right )}{e} \]
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 \[ \int \frac {a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )}{d+e\,x} \,d x \]
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 \[ \int \frac {a + b \operatorname {asec}{\left (c x \right )}}{d + e x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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