Optimal. Leaf size=362 \[ -\frac {6 b \sec ^{-1}(a+b x) \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {6 b \sec ^{-1}(a+b x) \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a \sqrt {1-a^2}}-\frac {6 i b \text {Li}_3\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {6 i b \text {Li}_3\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a \sqrt {1-a^2}}-\frac {3 i b \sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {3 i b \sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a \sqrt {1-a^2}}-\frac {b \sec ^{-1}(a+b x)^3}{a}-\frac {\sec ^{-1}(a+b x)^3}{x} \]
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Rubi [A] time = 0.59, antiderivative size = 362, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5258, 4426, 4191, 3321, 2264, 2190, 2531, 2282, 6589} \[ -\frac {6 b \sec ^{-1}(a+b x) \text {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {6 b \sec ^{-1}(a+b x) \text {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a \sqrt {1-a^2}}-\frac {6 i b \text {PolyLog}\left (3,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {6 i b \text {PolyLog}\left (3,\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a \sqrt {1-a^2}}-\frac {3 i b \sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {3 i b \sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a \sqrt {1-a^2}}-\frac {b \sec ^{-1}(a+b x)^3}{a}-\frac {\sec ^{-1}(a+b x)^3}{x} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2264
Rule 2282
Rule 2531
Rule 3321
Rule 4191
Rule 4426
Rule 5258
Rule 6589
Rubi steps
\begin {align*} \int \frac {\sec ^{-1}(a+b x)^3}{x^2} \, dx &=b \operatorname {Subst}\left (\int \frac {x^3 \sec (x) \tan (x)}{(-a+\sec (x))^2} \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=-\frac {\sec ^{-1}(a+b x)^3}{x}+(3 b) \operatorname {Subst}\left (\int \frac {x^2}{-a+\sec (x)} \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=-\frac {\sec ^{-1}(a+b x)^3}{x}+(3 b) \operatorname {Subst}\left (\int \left (-\frac {x^2}{a}+\frac {x^2}{a (1-a \cos (x))}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=-\frac {b \sec ^{-1}(a+b x)^3}{a}-\frac {\sec ^{-1}(a+b x)^3}{x}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {x^2}{1-a \cos (x)} \, dx,x,\sec ^{-1}(a+b x)\right )}{a}\\ &=-\frac {b \sec ^{-1}(a+b x)^3}{a}-\frac {\sec ^{-1}(a+b x)^3}{x}+\frac {(6 b) \operatorname {Subst}\left (\int \frac {e^{i x} x^2}{-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)^3}{a}-\frac {\sec ^{-1}(a+b x)^3}{x}-\frac {(6 b) \operatorname {Subst}\left (\int \frac {e^{i x} x^2}{2-2 \sqrt {1-a^2}-2 a e^{i x}} \, dx,x,\sec ^{-1}(a+b x)\right )}{\sqrt {1-a^2}}+\frac {(6 b) \operatorname {Subst}\left (\int \frac {e^{i x} x^2}{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)^3}{a}-\frac {\sec ^{-1}(a+b x)^3}{x}-\frac {3 i b \sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {3 i b \sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {(6 i b) \operatorname {Subst}\left (\int x \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 {(6 i b) \operatorname {Subst}\left (\int x \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)^3}{a}-\frac {\sec ^{-1}(a+b x)^3}{x}-\frac {3 i b \sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {3 i b \sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}-\frac {6 b \sec ^{-1}(a+b x) \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 {6 b \sec ^{-1}(a+b x) \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 {(6 b) \operatorname {Subst}\left (\int \text {Li}_2\left (\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 {(6 b) \operatorname {Subst}\left (\int \text {Li}_2\left (\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)^3}{a}-\frac {\sec ^{-1}(a+b x)^3}{x}-\frac {3 i b \sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {3 i b \sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}-\frac {6 b \sec ^{-1}(a+b x) \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 {6 b \sec ^{-1}(a+b x) \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 {(6 i b) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {a x}{1-\sqrt {1-a^2}}\right )}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{a \sqrt {1-a^2}}+\frac {(6 i b) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {a x}{1+\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)^3}{a}-\frac {\sec ^{-1}(a+b x)^3}{x}-\frac {3 i b \sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {3 i b \sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}-\frac {6 b \sec ^{-1}(a+b x) \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 {6 b \sec ^{-1}(a+b x) \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 {6 i b \text {Li}_3\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {6 i b \text {Li}_3\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 [F] time = 180.00, size = 0, normalized size = 0.00 \[ \text {\$Aborted} \]
Verification is Not applicable to the result.
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fricas [F] time = 2.23, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arcsec}\left (b x + a\right )^{3}}{x^{2}}, x\right ) \]
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 \[ \int \frac {\operatorname {arcsec}\left (b x + a\right )^{3}}{x^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.44, size = 0, normalized size = 0.00 \[ \int \frac {\mathrm {arcsec}\left (b x +a \right )^{3}}{x^{2}}\, dx \]
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 \[ -\frac {4 \, \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )^{3} - 3 \, \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )^{2} - 3 \, x \int \frac {{\left (4 \, b x \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )^{2} - b x \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )^{2}\right )} \sqrt {b x + a + 1} \sqrt {b x + a - 1} - 4 \, {\left ({\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} - 1\right )} b x - a\right )} \log \left (b x + a\right )^{2} + {\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + {\left (a^{2} - 1\right )} b x - {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} - 1\right )} b x - a\right )} \log \left (b x + a\right )\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )\right )} \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )}{b^{3} x^{5} + 3 \, a b^{2} x^{4} + {\left (3 \, a^{2} - 1\right )} b x^{3} + {\left (a^{3} - a\right )} x^{2}}\,{d x}}{4 \, x} \]
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 \[ \int \frac {{\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}^3}{x^2} \,d x \]
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 \[ \int \frac {\operatorname {asec}^{3}{\left (a + b x \right )}}{x^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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