Optimal. Leaf size=480 \[ \frac {2 a b d \log (e+f x)}{f^2-(d e-c f)^2}-\frac {a b d \log (-c-d x+1)}{f (-c f+d e+f)}+\frac {a b d \log (c+d x+1)}{f (-c f+d e-f)}-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}+\frac {b^2 d \text {Li}_2\left (-\frac {c+d x+1}{-c-d x+1}\right )}{2 f (-c f+d e+f)}+\frac {b^2 d \text {Li}_2\left (1-\frac {2}{c+d x+1}\right )}{2 f (-c f+d e-f)}-\frac {b^2 d \text {Li}_2\left (1-\frac {2}{c+d x+1}\right )}{(-c f+d e+f) (d e-(c+1) f)}+\frac {b^2 d \text {Li}_2\left (1-\frac {2 d (e+f x)}{(d e-c f+f) (c+d x+1)}\right )}{(-c f+d e+f) (d e-(c+1) f)}+\frac {b^2 d \log \left (\frac {2}{-c-d x+1}\right ) \coth ^{-1}(c+d x)}{f (-c f+d e+f)}-\frac {b^2 d \log \left (\frac {2}{c+d x+1}\right ) \coth ^{-1}(c+d x)}{f (-c f+d e-f)}+\frac {2 b^2 d \log \left (\frac {2}{c+d x+1}\right ) \coth ^{-1}(c+d x)}{(-c f+d e+f) (d e-(c+1) f)}-\frac {2 b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2 d (e+f x)}{(c+d x+1) (-c f+d e+f)}\right )}{(-c f+d e+f) (d e-(c+1) f)} \]
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Rubi [A] time = 1.74, antiderivative size = 485, normalized size of antiderivative = 1.01, number of steps used = 21, number of rules used = 19, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.950, Rules used = {6110, 1982, 705, 31, 632, 6741, 6122, 706, 633, 6688, 12, 6725, 72, 6742, 5919, 2402, 2315, 5921, 2447} \[ \frac {b^2 d \text {PolyLog}\left (2,-\frac {c+d x+1}{-c-d x+1}\right )}{2 f (-c f+d e+f)}+\frac {b^2 d \text {PolyLog}\left (2,1-\frac {2}{c+d x+1}\right )}{2 f (-c f+d e-f)}-\frac {b^2 d \text {PolyLog}\left (2,1-\frac {2}{c+d x+1}\right )}{(-c f+d e+f) (d e-(c+1) f)}+\frac {b^2 d \text {PolyLog}\left (2,1-\frac {2 d (e+f x)}{(c+d x+1) (-c f+d e+f)}\right )}{(-c f+d e+f) (d e-(c+1) f)}-\frac {a b d \log (-c-d x+1)}{f (-c f+d e+f)}+\frac {a b d \log (c+d x+1)}{f (-c f+d e-f)}-\frac {2 a b d \log (e+f x)}{(-c f+d e+f) (d e-(c+1) f)}-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}+\frac {b^2 d \log \left (\frac {2}{-c-d x+1}\right ) \coth ^{-1}(c+d x)}{f (-c f+d e+f)}-\frac {b^2 d \log \left (\frac {2}{c+d x+1}\right ) \coth ^{-1}(c+d x)}{f (-c f+d e-f)}+\frac {2 b^2 d \log \left (\frac {2}{c+d x+1}\right ) \coth ^{-1}(c+d x)}{(-c f+d e+f) (d e-(c+1) f)}-\frac {2 b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2 d (e+f x)}{(c+d x+1) (-c f+d e+f)}\right )}{(-c f+d e+f) (d e-(c+1) f)} \]
Antiderivative was successfully verified.
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Rule 12
Rule 31
Rule 72
Rule 632
Rule 633
Rule 705
Rule 706
Rule 1982
Rule 2315
Rule 2402
Rule 2447
Rule 5919
Rule 5921
Rule 6110
Rule 6122
Rule 6688
Rule 6725
Rule 6741
Rule 6742
Rubi steps
\begin {align*} \int \frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{(e+f x)^2} \, dx &=-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}+\frac {(2 b d) \int \frac {a+b \coth ^{-1}(c+d x)}{(e+f x) \left (1-(c+d x)^2\right )} \, dx}{f}\\ &=-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}+\frac {(2 b d) \int \frac {a+b \coth ^{-1}(c+d x)}{(e+f x) \left (1-c^2-2 c d x-d^2 x^2\right )} \, dx}{f}\\ &=-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {a+b \coth ^{-1}(x)}{\left (\frac {d e-c f}{d}+\frac {f x}{d}\right ) \left (1-x^2\right )} \, dx,x,c+d x\right )}{f}\\ &=-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {d \left (a+b \coth ^{-1}(x)\right )}{(d e-c f+f x) \left (1-x^2\right )} \, dx,x,c+d x\right )}{f}\\ &=-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}+\frac {(2 b d) \operatorname {Subst}\left (\int \frac {a+b \coth ^{-1}(x)}{(d e-c f+f x) \left (1-x^2\right )} \, dx,x,c+d x\right )}{f}\\ &=-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}+\frac {(2 b d) \operatorname {Subst}\left (\int \left (-\frac {a}{(-1+x) (1+x) (d e-c f+f x)}-\frac {b \coth ^{-1}(x)}{(-1+x) (1+x) (d e-c f+f x)}\right ) \, dx,x,c+d x\right )}{f}\\ &=-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}-\frac {(2 a b d) \operatorname {Subst}\left (\int \frac {1}{(-1+x) (1+x) (d e-c f+f x)} \, dx,x,c+d x\right )}{f}-\frac {\left (2 b^2 d\right ) \operatorname {Subst}\left (\int \frac {\coth ^{-1}(x)}{(-1+x) (1+x) (d e-c f+f x)} \, dx,x,c+d x\right )}{f}\\ &=-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}-\frac {(2 a b d) \operatorname {Subst}\left (\int \left (\frac {1}{2 (d e+f-c f) (-1+x)}+\frac {1}{2 (-d e+(1+c) f) (1+x)}+\frac {f^2}{(d e+(1-c) f) (d e-f-c f) (d e-c f+f x)}\right ) \, dx,x,c+d x\right )}{f}-\frac {\left (2 b^2 d\right ) \operatorname {Subst}\left (\int \left (\frac {\coth ^{-1}(x)}{2 (d e+f-c f) (-1+x)}+\frac {\coth ^{-1}(x)}{2 (-d e+(1+c) f) (1+x)}+\frac {f^2 \coth ^{-1}(x)}{(d e+(1-c) f) (d e-f-c f) (d e-c f+f x)}\right ) \, dx,x,c+d x\right )}{f}\\ &=-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}-\frac {a b d \log (1-c-d x)}{f (d e+f-c f)}+\frac {a b d \log (1+c+d x)}{f (d e-f-c f)}-\frac {2 a b d \log (e+f x)}{(d e+f-c f) (d e-(1+c) f)}+\frac {\left (b^2 d\right ) \operatorname {Subst}\left (\int \frac {\coth ^{-1}(x)}{1+x} \, dx,x,c+d x\right )}{f (d e-f-c f)}-\frac {\left (b^2 d\right ) \operatorname {Subst}\left (\int \frac {\coth ^{-1}(x)}{-1+x} \, dx,x,c+d x\right )}{f (d e+f-c f)}-\frac {\left (2 b^2 d f\right ) \operatorname {Subst}\left (\int \frac {\coth ^{-1}(x)}{d e-c f+f x} \, dx,x,c+d x\right )}{(d e+f-c f) (d e-(1+c) f)}\\ &=-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}+\frac {b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2}{1-c-d x}\right )}{f (d e+f-c f)}-\frac {a b d \log (1-c-d x)}{f (d e+f-c f)}-\frac {b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2}{1+c+d x}\right )}{f (d e-f-c f)}+\frac {2 b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac {a b d \log (1+c+d x)}{f (d e-f-c f)}-\frac {2 a b d \log (e+f x)}{(d e+f-c f) (d e-(1+c) f)}-\frac {2 b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac {\left (b^2 d\right ) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2}{1+x}\right )}{1-x^2} \, dx,x,c+d x\right )}{f (d e-f-c f)}-\frac {\left (b^2 d\right ) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{f (d e+f-c f)}-\frac {\left (2 b^2 d\right ) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2}{1+x}\right )}{1-x^2} \, dx,x,c+d x\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac {\left (2 b^2 d\right ) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2 (d e-c f+f x)}{(d e+f-c f) (1+x)}\right )}{1-x^2} \, dx,x,c+d x\right )}{(d e+f-c f) (d e-(1+c) f)}\\ &=-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}+\frac {b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2}{1-c-d x}\right )}{f (d e+f-c f)}-\frac {a b d \log (1-c-d x)}{f (d e+f-c f)}-\frac {b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2}{1+c+d x}\right )}{f (d e-f-c f)}+\frac {2 b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac {a b d \log (1+c+d x)}{f (d e-f-c f)}-\frac {2 a b d \log (e+f x)}{(d e+f-c f) (d e-(1+c) f)}-\frac {2 b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac {b^2 d \text {Li}_2\left (1-\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac {\left (b^2 d\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+c+d x}\right )}{f (d e-f-c f)}+\frac {\left (b^2 d\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c-d x}\right )}{f (d e+f-c f)}-\frac {\left (2 b^2 d\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}\\ &=-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}+\frac {b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2}{1-c-d x}\right )}{f (d e+f-c f)}-\frac {a b d \log (1-c-d x)}{f (d e+f-c f)}-\frac {b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2}{1+c+d x}\right )}{f (d e-f-c f)}+\frac {2 b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac {a b d \log (1+c+d x)}{f (d e-f-c f)}-\frac {2 a b d \log (e+f x)}{(d e+f-c f) (d e-(1+c) f)}-\frac {2 b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac {b^2 d \text {Li}_2\left (1-\frac {2}{1-c-d x}\right )}{2 f (d e+f-c f)}+\frac {b^2 d \text {Li}_2\left (1-\frac {2}{1+c+d x}\right )}{2 f (d e-f-c f)}-\frac {b^2 d \text {Li}_2\left (1-\frac {2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac {b^2 d \text {Li}_2\left (1-\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}\\ \end {align*}
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Mathematica [C] time = 8.90, size = 470, normalized size = 0.98 \[ \frac {-\frac {a^2}{f}+\frac {2 a b \left (\coth ^{-1}(c+d x) \left (c^2 (-f)+c d (e-f x)+d^2 e x+f\right )-d (e+f x) \log \left (-\frac {d (e+f x)}{(c+d x) \sqrt {1-\frac {1}{(c+d x)^2}}}\right )\right )}{(-c f+d e+f) (d e-(c+1) f)}+\frac {b^2 d \left (1-(c+d x)^2\right ) (e+f x) \left (\frac {f \left (-\text {Li}_2\left (\exp \left (-2 \left (\coth ^{-1}(c+d x)+\tanh ^{-1}\left (\frac {f}{d e-c f}\right )\right )\right )\right )-2 \tanh ^{-1}\left (\frac {f}{c f-d e}\right ) \log \left (1-\exp \left (-2 \left (\tanh ^{-1}\left (\frac {f}{d e-c f}\right )+\coth ^{-1}(c+d x)\right )\right )\right )+\coth ^{-1}(c+d x) \left (2 \log \left (1-\exp \left (-2 \left (\tanh ^{-1}\left (\frac {f}{d e-c f}\right )+\coth ^{-1}(c+d x)\right )\right )\right )+2 \tanh ^{-1}\left (\frac {f}{d e-c f}\right )+i \pi \right )+2 \tanh ^{-1}\left (\frac {f}{c f-d e}\right ) \log \left (i \sinh \left (\tanh ^{-1}\left (\frac {f}{d e-c f}\right )+\coth ^{-1}(c+d x)\right )\right )+i \pi \log \left (\frac {1}{\sqrt {1-\frac {1}{(c+d x)^2}}}\right )-i \pi \log \left (e^{2 \coth ^{-1}(c+d x)}+1\right )\right )}{\left (c^2-1\right ) f^2-2 c d e f+d^2 e^2}+\frac {\coth ^{-1}(c+d x)^2 e^{\tanh ^{-1}\left (\frac {f}{c f-d e}\right )}}{(c f-d e) \sqrt {1-\frac {f^2}{(d e-c f)^2}}}+\frac {\coth ^{-1}(c+d x)^2}{d e+d f x}\right )}{(c+d x)^2 \left (f-\frac {f}{(c+d x)^2}\right )}}{e+f x} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \operatorname {arcoth}\left (d x + c\right )^{2} + 2 \, a b \operatorname {arcoth}\left (d x + c\right ) + a^{2}}{f^{2} x^{2} + 2 \, e f x + e^{2}}, x\right ) \]
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 \[ \int \frac {{\left (b \operatorname {arcoth}\left (d x + c\right ) + a\right )}^{2}}{{\left (f x + e\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 783, normalized size = 1.63 \[ -\frac {d \,a^{2}}{\left (d f x +d e \right ) f}-\frac {d \,b^{2} \mathrm {arccoth}\left (d x +c \right )^{2}}{\left (d f x +d e \right ) f}-\frac {2 d \,b^{2} \mathrm {arccoth}\left (d x +c \right ) \ln \left (\left (d x +c \right ) f -c f +d e \right )}{\left (c f -d e -f \right ) \left (c f -d e +f \right )}+\frac {2 d \,b^{2} \mathrm {arccoth}\left (d x +c \right ) \ln \left (d x +c -1\right )}{f \left (2 c f -2 d e -2 f \right )}-\frac {2 d \,b^{2} \mathrm {arccoth}\left (d x +c \right ) \ln \left (d x +c +1\right )}{f \left (2 c f -2 d e +2 f \right )}+\frac {d \,b^{2} \ln \left (\left (d x +c \right ) f -c f +d e \right ) \ln \left (\frac {\left (d x +c \right ) f +f}{c f -d e +f}\right )}{\left (c f -d e -f \right ) \left (c f -d e +f \right )}+\frac {d \,b^{2} \dilog \left (\frac {\left (d x +c \right ) f +f}{c f -d e +f}\right )}{\left (c f -d e -f \right ) \left (c f -d e +f \right )}-\frac {d \,b^{2} \ln \left (\left (d x +c \right ) f -c f +d e \right ) \ln \left (\frac {\left (d x +c \right ) f -f}{c f -d e -f}\right )}{\left (c f -d e -f \right ) \left (c f -d e +f \right )}-\frac {d \,b^{2} \dilog \left (\frac {\left (d x +c \right ) f -f}{c f -d e -f}\right )}{\left (c f -d e -f \right ) \left (c f -d e +f \right )}+\frac {d \,b^{2} \ln \left (d x +c -1\right )^{2}}{4 f \left (c f -d e -f \right )}-\frac {d \,b^{2} \dilog \left (\frac {1}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{2 f \left (c f -d e -f \right )}-\frac {d \,b^{2} \ln \left (d x +c -1\right ) \ln \left (\frac {1}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{2 f \left (c f -d e -f \right )}+\frac {d \,b^{2} \ln \left (d x +c +1\right )^{2}}{4 f \left (c f -d e +f \right )}+\frac {d \,b^{2} \ln \left (-\frac {d x}{2}-\frac {c}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{2 f \left (c f -d e +f \right )}-\frac {d \,b^{2} \ln \left (-\frac {d x}{2}-\frac {c}{2}+\frac {1}{2}\right ) \ln \left (d x +c +1\right )}{2 f \left (c f -d e +f \right )}+\frac {d \,b^{2} \dilog \left (\frac {1}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{2 f \left (c f -d e +f \right )}-\frac {2 d a b \,\mathrm {arccoth}\left (d x +c \right )}{\left (d f x +d e \right ) f}-\frac {2 d a b \ln \left (\left (d x +c \right ) f -c f +d e \right )}{\left (c f -d e -f \right ) \left (c f -d e +f \right )}+\frac {2 d a b \ln \left (d x +c -1\right )}{f \left (2 c f -2 d e -2 f \right )}-\frac {2 d a b \ln \left (d x +c +1\right )}{f \left (2 c f -2 d e +2 f \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 \[ {\left (d {\left (\frac {\log \left (d x + c + 1\right )}{d e f - {\left (c + 1\right )} f^{2}} - \frac {\log \left (d x + c - 1\right )}{d e f - {\left (c - 1\right )} f^{2}} - \frac {2 \, \log \left (f x + e\right )}{d^{2} e^{2} - 2 \, c d e f + {\left (c^{2} - 1\right )} f^{2}}\right )} - \frac {2 \, \operatorname {arcoth}\left (d x + c\right )}{f^{2} x + e f}\right )} a b - \frac {1}{4} \, b^{2} {\left (\frac {\log \left (d x + c + 1\right )^{2}}{f^{2} x + e f} + \int -\frac {{\left (d f x + c f + f\right )} \log \left (d x + c - 1\right )^{2} + 2 \, {\left (d f x + d e - {\left (d f x + c f + f\right )} \log \left (d x + c - 1\right )\right )} \log \left (d x + c + 1\right )}{d f^{3} x^{3} + c e^{2} f + e^{2} f + {\left (2 \, d e f^{2} + c f^{3} + f^{3}\right )} x^{2} + {\left (d e^{2} f + 2 \, c e f^{2} + 2 \, e f^{2}\right )} x}\,{d x}\right )} - \frac {a^{2}}{f^{2} x + e f} \]
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 \[ \int \frac {{\left (a+b\,\mathrm {acoth}\left (c+d\,x\right )\right )}^2}{{\left (e+f\,x\right )}^2} \,d x \]
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 \[ \int \frac {\left (a + b \operatorname {acoth}{\left (c + d x \right )}\right )^{2}}{\left (e + f x\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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