Optimal. Leaf size=279 \[ -\frac {x}{3 b^2}+\frac {2 a \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)}{b^3}-\frac {(a+b x) \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)}{3 b^3}+\frac {a^3 \text {sech}^{-1}(a+b x)^2}{3 b^3}+\frac {1}{3} x^3 \text {sech}^{-1}(a+b x)^2-\frac {2 \text {sech}^{-1}(a+b x) \text {ArcTan}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}-\frac {4 a^2 \text {sech}^{-1}(a+b x) \text {ArcTan}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}+\frac {2 a \log (a+b x)}{b^3}+\frac {i \text {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}+\frac {2 i a^2 \text {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}-\frac {i \text {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}-\frac {2 i a^2 \text {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{b^3} \]
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Rubi [A]
time = 0.17, antiderivative size = 279, normalized size of antiderivative = 1.00, number of steps
used = 17, number of rules used = 9, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6456, 5576,
4275, 4265, 2317, 2438, 4269, 3556, 4270} \begin {gather*} \frac {a^3 \text {sech}^{-1}(a+b x)^2}{3 b^3}-\frac {4 a^2 \text {sech}^{-1}(a+b x) \text {ArcTan}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}+\frac {2 i a^2 \text {Li}_2\left (-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}-\frac {2 i a^2 \text {Li}_2\left (i e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}-\frac {2 \text {sech}^{-1}(a+b x) \text {ArcTan}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}+\frac {i \text {Li}_2\left (-i e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}-\frac {i \text {Li}_2\left (i e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}+\frac {2 a \log (a+b x)}{b^3}+\frac {2 a \sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1) \text {sech}^{-1}(a+b x)}{b^3}-\frac {(a+b x) \sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1) \text {sech}^{-1}(a+b x)}{3 b^3}+\frac {1}{3} x^3 \text {sech}^{-1}(a+b x)^2-\frac {x}{3 b^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 2317
Rule 2438
Rule 3556
Rule 4265
Rule 4269
Rule 4270
Rule 4275
Rule 5576
Rule 6456
Rubi steps
\begin {align*} \int x^2 \text {sech}^{-1}(a+b x)^2 \, dx &=-\frac {\text {Subst}\left (\int x^2 \text {sech}(x) (-a+\text {sech}(x))^2 \tanh (x) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b^3}\\ &=\frac {1}{3} x^3 \text {sech}^{-1}(a+b x)^2-\frac {2 \text {Subst}\left (\int x (-a+\text {sech}(x))^3 \, dx,x,\text {sech}^{-1}(a+b x)\right )}{3 b^3}\\ &=\frac {1}{3} x^3 \text {sech}^{-1}(a+b x)^2-\frac {2 \text {Subst}\left (\int \left (-a^3 x+3 a^2 x \text {sech}(x)-3 a x \text {sech}^2(x)+x \text {sech}^3(x)\right ) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{3 b^3}\\ &=\frac {a^3 \text {sech}^{-1}(a+b x)^2}{3 b^3}+\frac {1}{3} x^3 \text {sech}^{-1}(a+b x)^2-\frac {2 \text {Subst}\left (\int x \text {sech}^3(x) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{3 b^3}+\frac {(2 a) \text {Subst}\left (\int x \text {sech}^2(x) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b^3}-\frac {\left (2 a^2\right ) \text {Subst}\left (\int x \text {sech}(x) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b^3}\\ &=-\frac {x}{3 b^2}+\frac {2 a \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)}{b^3}-\frac {(a+b x) \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)}{3 b^3}+\frac {a^3 \text {sech}^{-1}(a+b x)^2}{3 b^3}+\frac {1}{3} x^3 \text {sech}^{-1}(a+b x)^2-\frac {4 a^2 \text {sech}^{-1}(a+b x) \tan ^{-1}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}-\frac {\text {Subst}\left (\int x \text {sech}(x) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{3 b^3}-\frac {(2 a) \text {Subst}\left (\int \tanh (x) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b^3}+\frac {\left (2 i a^2\right ) \text {Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b^3}-\frac {\left (2 i a^2\right ) \text {Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b^3}\\ &=-\frac {x}{3 b^2}+\frac {2 a \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)}{b^3}-\frac {(a+b x) \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)}{3 b^3}+\frac {a^3 \text {sech}^{-1}(a+b x)^2}{3 b^3}+\frac {1}{3} x^3 \text {sech}^{-1}(a+b x)^2-\frac {2 \text {sech}^{-1}(a+b x) \tan ^{-1}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}-\frac {4 a^2 \text {sech}^{-1}(a+b x) \tan ^{-1}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}+\frac {2 a \log (a+b x)}{b^3}+\frac {i \text {Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{3 b^3}-\frac {i \text {Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{3 b^3}+\frac {\left (2 i a^2\right ) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}-\frac {\left (2 i a^2\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}\\ &=-\frac {x}{3 b^2}+\frac {2 a \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)}{b^3}-\frac {(a+b x) \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)}{3 b^3}+\frac {a^3 \text {sech}^{-1}(a+b x)^2}{3 b^3}+\frac {1}{3} x^3 \text {sech}^{-1}(a+b x)^2-\frac {2 \text {sech}^{-1}(a+b x) \tan ^{-1}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}-\frac {4 a^2 \text {sech}^{-1}(a+b x) \tan ^{-1}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}+\frac {2 a \log (a+b x)}{b^3}+\frac {2 i a^2 \text {Li}_2\left (-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}-\frac {2 i a^2 \text {Li}_2\left (i e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}+\frac {i \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}-\frac {i \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}\\ &=-\frac {x}{3 b^2}+\frac {2 a \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)}{b^3}-\frac {(a+b x) \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)}{3 b^3}+\frac {a^3 \text {sech}^{-1}(a+b x)^2}{3 b^3}+\frac {1}{3} x^3 \text {sech}^{-1}(a+b x)^2-\frac {2 \text {sech}^{-1}(a+b x) \tan ^{-1}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}-\frac {4 a^2 \text {sech}^{-1}(a+b x) \tan ^{-1}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}+\frac {2 a \log (a+b x)}{b^3}+\frac {i \text {Li}_2\left (-i e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}+\frac {2 i a^2 \text {Li}_2\left (-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}-\frac {i \text {Li}_2\left (i e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}-\frac {2 i a^2 \text {Li}_2\left (i e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}\\ \end {align*}
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Mathematica [A]
time = 1.39, size = 305, normalized size = 1.09 \begin {gather*} -\frac {2 (a+b x) \sqrt {-\frac {-1+a+b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)+6 a (a+b x)^2 \text {sech}^{-1}(a+b x)^2-2 (a+b x)^3 \text {sech}^{-1}(a+b x)^2+2 \left (a+b x-6 a \sqrt {-\frac {-1+a+b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)-3 a^2 (a+b x) \text {sech}^{-1}(a+b x)^2\right )+12 a \log \left (\frac {1}{a+b x}\right )-\left (1+6 a^2\right ) \left (\pi \log \left (1-i e^{\text {sech}^{-1}(a+b x)}\right )-2 i \text {sech}^{-1}(a+b x) \log \left (1-i e^{\text {sech}^{-1}(a+b x)}\right )-\pi \log \left (1+i e^{\text {sech}^{-1}(a+b x)}\right )+2 i \text {sech}^{-1}(a+b x) \log \left (1+i e^{\text {sech}^{-1}(a+b x)}\right )-\pi \log \left (\cot \left (\frac {1}{4} \left (\pi +2 i \text {sech}^{-1}(a+b x)\right )\right )\right )+2 i \text {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )-2 i \text {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )\right )}{6 b^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 1.29, size = 599, normalized size = 2.15
method | result | size |
derivativedivides | \(\frac {\mathrm {arcsech}\left (b x +a \right )^{2} a^{2} \left (b x +a \right )-\mathrm {arcsech}\left (b x +a \right )^{2} a \left (b x +a \right )^{2}+\frac {\mathrm {arcsech}\left (b x +a \right )^{2} \left (b x +a \right )^{3}}{3}+2 \,\mathrm {arcsech}\left (b x +a \right ) \sqrt {\frac {b x +a +1}{b x +a}}\, \sqrt {-\frac {b x +a -1}{b x +a}}\, a \left (b x +a \right )-\frac {\mathrm {arcsech}\left (b x +a \right ) \sqrt {\frac {b x +a +1}{b x +a}}\, \sqrt {-\frac {b x +a -1}{b x +a}}\, \left (b x +a \right )^{2}}{3}-2 a \,\mathrm {arcsech}\left (b x +a \right )-\frac {b x}{3}-\frac {a}{3}-\frac {i \mathrm {arcsech}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{3}+\frac {i \mathrm {arcsech}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{3}-\frac {i \dilog \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{3}+2 i a^{2} \dilog \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )-2 a \ln \left (1+\left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )^{2}\right )+4 a \ln \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )+\frac {i \dilog \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{3}-2 i a^{2} \mathrm {arcsech}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )-2 i a^{2} \dilog \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )+2 i a^{2} \mathrm {arcsech}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{b^{3}}\) | \(599\) |
default | \(\frac {\mathrm {arcsech}\left (b x +a \right )^{2} a^{2} \left (b x +a \right )-\mathrm {arcsech}\left (b x +a \right )^{2} a \left (b x +a \right )^{2}+\frac {\mathrm {arcsech}\left (b x +a \right )^{2} \left (b x +a \right )^{3}}{3}+2 \,\mathrm {arcsech}\left (b x +a \right ) \sqrt {\frac {b x +a +1}{b x +a}}\, \sqrt {-\frac {b x +a -1}{b x +a}}\, a \left (b x +a \right )-\frac {\mathrm {arcsech}\left (b x +a \right ) \sqrt {\frac {b x +a +1}{b x +a}}\, \sqrt {-\frac {b x +a -1}{b x +a}}\, \left (b x +a \right )^{2}}{3}-2 a \,\mathrm {arcsech}\left (b x +a \right )-\frac {b x}{3}-\frac {a}{3}-\frac {i \mathrm {arcsech}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{3}+\frac {i \mathrm {arcsech}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{3}-\frac {i \dilog \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{3}+2 i a^{2} \dilog \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )-2 a \ln \left (1+\left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )^{2}\right )+4 a \ln \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )+\frac {i \dilog \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{3}-2 i a^{2} \mathrm {arcsech}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )-2 i a^{2} \dilog \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )+2 i a^{2} \mathrm {arcsech}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{b^{3}}\) | \(599\) |
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 x^{2} \operatorname {asech}^{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 = -0.00 \begin {gather*} \int x^2\,{\mathrm {acosh}\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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