Optimal. Leaf size=84 \[ \frac {c^2 \cos \left (\frac {2 a}{b}\right ) \text {Ci}\left (\frac {2 a}{b}+2 \sec ^{-1}(c x)\right )}{b^2}+\frac {c^2 \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sec ^{-1}(c x)\right )}{b^2}-\frac {c^2 \sin \left (2 \sec ^{-1}(c x)\right )}{2 b \left (a+b \sec ^{-1}(c x)\right )} \]
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Rubi [A] time = 0.15, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5222, 4406, 12, 3297, 3303, 3299, 3302} \[ \frac {c^2 \cos \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 a}{b}+2 \sec ^{-1}(c x)\right )}{b^2}+\frac {c^2 \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sec ^{-1}(c x)\right )}{b^2}-\frac {c^2 \sin \left (2 \sec ^{-1}(c x)\right )}{2 b \left (a+b \sec ^{-1}(c x)\right )} \]
Antiderivative was successfully verified.
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Rule 12
Rule 3297
Rule 3299
Rule 3302
Rule 3303
Rule 4406
Rule 5222
Rubi steps
\begin {align*} \int \frac {1}{x^3 \left (a+b \sec ^{-1}(c x)\right )^2} \, dx &=c^2 \operatorname {Subst}\left (\int \frac {\cos (x) \sin (x)}{(a+b x)^2} \, dx,x,\sec ^{-1}(c x)\right )\\ &=c^2 \operatorname {Subst}\left (\int \frac {\sin (2 x)}{2 (a+b x)^2} \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac {1}{2} c^2 \operatorname {Subst}\left (\int \frac {\sin (2 x)}{(a+b x)^2} \, dx,x,\sec ^{-1}(c x)\right )\\ &=-\frac {c^2 \sin \left (2 \sec ^{-1}(c x)\right )}{2 b \left (a+b \sec ^{-1}(c x)\right )}+\frac {c^2 \operatorname {Subst}\left (\int \frac {\cos (2 x)}{a+b x} \, dx,x,\sec ^{-1}(c x)\right )}{b}\\ &=-\frac {c^2 \sin \left (2 \sec ^{-1}(c x)\right )}{2 b \left (a+b \sec ^{-1}(c x)\right )}+\frac {\left (c^2 \cos \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sec ^{-1}(c x)\right )}{b}+\frac {\left (c^2 \sin \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sec ^{-1}(c x)\right )}{b}\\ &=\frac {c^2 \cos \left (\frac {2 a}{b}\right ) \text {Ci}\left (\frac {2 a}{b}+2 \sec ^{-1}(c x)\right )}{b^2}-\frac {c^2 \sin \left (2 \sec ^{-1}(c x)\right )}{2 b \left (a+b \sec ^{-1}(c x)\right )}+\frac {c^2 \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sec ^{-1}(c x)\right )}{b^2}\\ \end {align*}
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Mathematica [A] time = 0.37, size = 80, normalized size = 0.95 \[ \frac {c \left (-\frac {b \sqrt {1-\frac {1}{c^2 x^2}}}{a x+b x \sec ^{-1}(c x)}+c \cos \left (\frac {2 a}{b}\right ) \text {Ci}\left (2 \left (\frac {a}{b}+\sec ^{-1}(c x)\right )\right )+c \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\sec ^{-1}(c x)\right )\right )\right )}{b^2} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{b^{2} x^{3} \operatorname {arcsec}\left (c x\right )^{2} + 2 \, a b x^{3} \operatorname {arcsec}\left (c x\right ) + a^{2} x^{3}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.16, size = 357, normalized size = 4.25 \[ {\left (\frac {2 \, b c \arccos \left (\frac {1}{c x}\right ) \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (\frac {1}{c x}\right )\right )}{b^{3} \arccos \left (\frac {1}{c x}\right ) + a b^{2}} + \frac {2 \, b c \arccos \left (\frac {1}{c x}\right ) \cos \left (\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (\frac {1}{c x}\right )\right )}{b^{3} \arccos \left (\frac {1}{c x}\right ) + a b^{2}} + \frac {2 \, a c \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (\frac {1}{c x}\right )\right )}{b^{3} \arccos \left (\frac {1}{c x}\right ) + a b^{2}} + \frac {2 \, a c \cos \left (\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (\frac {1}{c x}\right )\right )}{b^{3} \arccos \left (\frac {1}{c x}\right ) + a b^{2}} - \frac {b c \arccos \left (\frac {1}{c x}\right ) \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (\frac {1}{c x}\right )\right )}{b^{3} \arccos \left (\frac {1}{c x}\right ) + a b^{2}} - \frac {a c \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (\frac {1}{c x}\right )\right )}{b^{3} \arccos \left (\frac {1}{c x}\right ) + a b^{2}} - \frac {b \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{{\left (b^{3} \arccos \left (\frac {1}{c x}\right ) + a b^{2}\right )} x}\right )} c \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.10, size = 77, normalized size = 0.92 \[ c^{2} \left (-\frac {\sin \left (2 \,\mathrm {arcsec}\left (c x \right )\right )}{2 \left (a +b \,\mathrm {arcsec}\left (c x \right )\right ) b}+\frac {\Si \left (\frac {2 a}{b}+2 \,\mathrm {arcsec}\left (c x \right )\right ) \sin \left (\frac {2 a}{b}\right )+\Ci \left (\frac {2 a}{b}+2 \,\mathrm {arcsec}\left (c x \right )\right ) \cos \left (\frac {2 a}{b}\right )}{b^{2}}\right ) \]
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 {4 \, \sqrt {c x + 1} \sqrt {c x - 1} {\left (b \arctan \left (\sqrt {c x + 1} \sqrt {c x - 1}\right ) + a\right )} + 4 \, {\left (4 \, b^{3} x^{2} \arctan \left (\sqrt {c x + 1} \sqrt {c x - 1}\right )^{2} + b^{3} x^{2} \log \left (c^{2} x^{2}\right )^{2} + 8 \, b^{3} x^{2} \log \relax (c) \log \relax (x) + 4 \, b^{3} x^{2} \log \relax (x)^{2} + 8 \, a b^{2} x^{2} \arctan \left (\sqrt {c x + 1} \sqrt {c x - 1}\right ) + 4 \, {\left (b^{3} \log \relax (c)^{2} + a^{2} b\right )} x^{2} - 4 \, {\left (b^{3} x^{2} \log \relax (c) + b^{3} x^{2} \log \relax (x)\right )} \log \left (c^{2} x^{2}\right )\right )} \int \frac {{\left (a c^{2} x^{2} + {\left (b c^{2} x^{2} - 2 \, b\right )} \arctan \left (\sqrt {c x + 1} \sqrt {c x - 1}\right ) - 2 \, a\right )} \sqrt {c x + 1} \sqrt {c x - 1}}{4 \, {\left (b^{3} c^{2} \log \relax (c)^{2} + a^{2} b c^{2}\right )} x^{5} - 4 \, {\left (b^{3} \log \relax (c)^{2} + a^{2} b\right )} x^{3} + 4 \, {\left (b^{3} c^{2} x^{5} - b^{3} x^{3}\right )} \arctan \left (\sqrt {c x + 1} \sqrt {c x - 1}\right )^{2} + {\left (b^{3} c^{2} x^{5} - b^{3} x^{3}\right )} \log \left (c^{2} x^{2}\right )^{2} + 4 \, {\left (b^{3} c^{2} x^{5} - b^{3} x^{3}\right )} \log \relax (x)^{2} + 8 \, {\left (a b^{2} c^{2} x^{5} - a b^{2} x^{3}\right )} \arctan \left (\sqrt {c x + 1} \sqrt {c x - 1}\right ) - 4 \, {\left (b^{3} c^{2} x^{5} \log \relax (c) - b^{3} x^{3} \log \relax (c) + {\left (b^{3} c^{2} x^{5} - b^{3} x^{3}\right )} \log \relax (x)\right )} \log \left (c^{2} x^{2}\right ) + 8 \, {\left (b^{3} c^{2} x^{5} \log \relax (c) - b^{3} x^{3} \log \relax (c)\right )} \log \relax (x)}\,{d x}}{4 \, b^{3} x^{2} \arctan \left (\sqrt {c x + 1} \sqrt {c x - 1}\right )^{2} + b^{3} x^{2} \log \left (c^{2} x^{2}\right )^{2} + 8 \, b^{3} x^{2} \log \relax (c) \log \relax (x) + 4 \, b^{3} x^{2} \log \relax (x)^{2} + 8 \, a b^{2} x^{2} \arctan \left (\sqrt {c x + 1} \sqrt {c x - 1}\right ) + 4 \, {\left (b^{3} \log \relax (c)^{2} + a^{2} b\right )} x^{2} - 4 \, {\left (b^{3} x^{2} \log \relax (c) + b^{3} x^{2} \log \relax (x)\right )} \log \left (c^{2} x^{2}\right )} \]
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 \[ \int \frac {1}{x^3\,{\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}^2} \,d x \]
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 \[ \int \frac {1}{x^{3} \left (a + b \operatorname {asec}{\left (c x \right )}\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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