Optimal. Leaf size=180 \[ -\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{3 d e^4}+\frac {2 b \log \left (2-\frac {2}{c+d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{3 d e^4}-\frac {b^2 \text {Li}_2\left (\frac {2}{c+d x+1}-1\right )}{3 d e^4}-\frac {b^2}{3 d e^4 (c+d x)}+\frac {b^2 \tanh ^{-1}(c+d x)}{3 d e^4} \]
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Rubi [A] time = 0.27, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {6107, 12, 5916, 5982, 325, 206, 5988, 5932, 2447} \[ -\frac {b^2 \text {PolyLog}\left (2,\frac {2}{c+d x+1}-1\right )}{3 d e^4}-\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{3 d e^4}+\frac {2 b \log \left (2-\frac {2}{c+d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{3 d e^4}-\frac {b^2}{3 d e^4 (c+d x)}+\frac {b^2 \tanh ^{-1}(c+d x)}{3 d e^4} \]
Antiderivative was successfully verified.
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Rule 12
Rule 206
Rule 325
Rule 2447
Rule 5916
Rule 5932
Rule 5982
Rule 5988
Rule 6107
Rubi steps
\begin {align*} \int \frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{(c e+d e x)^4} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^2}{e^4 x^4} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^2}{x^4} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(x)}{x^3 \left (1-x^2\right )} \, dx,x,c+d x\right )}{3 d e^4}\\ &=-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(x)}{x^3} \, dx,x,c+d x\right )}{3 d e^4}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(x)}{x \left (1-x^2\right )} \, dx,x,c+d x\right )}{3 d e^4}\\ &=-\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{3 d e^4}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(x)}{x (1+x)} \, dx,x,c+d x\right )}{3 d e^4}+\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{x^2 \left (1-x^2\right )} \, dx,x,c+d x\right )}{3 d e^4}\\ &=-\frac {b^2}{3 d e^4 (c+d x)}-\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{3 d e^4}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}+\frac {2 b \left (a+b \tanh ^{-1}(c+d x)\right ) \log \left (2-\frac {2}{1+c+d x}\right )}{3 d e^4}+\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,c+d x\right )}{3 d e^4}-\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {\log \left (2-\frac {2}{1+x}\right )}{1-x^2} \, dx,x,c+d x\right )}{3 d e^4}\\ &=-\frac {b^2}{3 d e^4 (c+d x)}+\frac {b^2 \tanh ^{-1}(c+d x)}{3 d e^4}-\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{3 d e^4}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}+\frac {2 b \left (a+b \tanh ^{-1}(c+d x)\right ) \log \left (2-\frac {2}{1+c+d x}\right )}{3 d e^4}-\frac {b^2 \text {Li}_2\left (-1+\frac {2}{1+c+d x}\right )}{3 d e^4}\\ \end {align*}
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Mathematica [A] time = 0.54, size = 218, normalized size = 1.21 \[ -\frac {a^2-a b \left ((c+d x) \left (c^2+2 c d x+2 (c+d x)^2 \log \left (\frac {c+d x}{\sqrt {1-(c+d x)^2}}\right )+d^2 x^2-1\right )-2 \tanh ^{-1}(c+d x)\right )+b^2 \left ((c+d x)^3 \text {Li}_2\left (e^{-2 \tanh ^{-1}(c+d x)}\right )+(c+d x)^2+(c+d x)^2 \tanh ^{-1}(c+d x)^2+\left (1-(c+d x)^2\right ) \tanh ^{-1}(c+d x)^2+(c+d x) \tanh ^{-1}(c+d x) \left (-(c+d x)^2+(c+d x)^2 \left (-\tanh ^{-1}(c+d x)\right )-2 (c+d x)^2 \log \left (1-e^{-2 \tanh ^{-1}(c+d x)}\right )+1\right )\right )}{3 d e^4 (c+d x)^3} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \operatorname {artanh}\left (d x + c\right )^{2} + 2 \, a b \operatorname {artanh}\left (d x + c\right ) + a^{2}}{d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}, x\right ) \]
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 \[ \int \frac {{\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{4}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.08, size = 492, normalized size = 2.73 \[ -\frac {a^{2}}{3 d \,e^{4} \left (d x +c \right )^{3}}-\frac {b^{2} \arctanh \left (d x +c \right )^{2}}{3 d \,e^{4} \left (d x +c \right )^{3}}-\frac {b^{2} \arctanh \left (d x +c \right )}{3 d \,e^{4} \left (d x +c \right )^{2}}+\frac {2 b^{2} \ln \left (d x +c \right ) \arctanh \left (d x +c \right )}{3 d \,e^{4}}-\frac {b^{2} \arctanh \left (d x +c \right ) \ln \left (d x +c -1\right )}{3 d \,e^{4}}-\frac {b^{2} \arctanh \left (d x +c \right ) \ln \left (d x +c +1\right )}{3 d \,e^{4}}-\frac {b^{2}}{3 d \,e^{4} \left (d x +c \right )}-\frac {b^{2} \ln \left (d x +c -1\right )}{6 d \,e^{4}}+\frac {b^{2} \ln \left (d x +c +1\right )}{6 d \,e^{4}}-\frac {b^{2} \ln \left (d x +c -1\right )^{2}}{12 d \,e^{4}}+\frac {b^{2} \dilog \left (\frac {1}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{3 d \,e^{4}}+\frac {b^{2} \ln \left (d x +c -1\right ) \ln \left (\frac {1}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{6 d \,e^{4}}+\frac {b^{2} \ln \left (d x +c +1\right )^{2}}{12 d \,e^{4}}-\frac {b^{2} \ln \left (-\frac {d x}{2}-\frac {c}{2}+\frac {1}{2}\right ) \ln \left (d x +c +1\right )}{6 d \,e^{4}}+\frac {b^{2} \ln \left (-\frac {d x}{2}-\frac {c}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{6 d \,e^{4}}-\frac {b^{2} \dilog \left (d x +c \right )}{3 d \,e^{4}}-\frac {b^{2} \dilog \left (d x +c +1\right )}{3 d \,e^{4}}-\frac {b^{2} \ln \left (d x +c \right ) \ln \left (d x +c +1\right )}{3 d \,e^{4}}-\frac {2 a b \arctanh \left (d x +c \right )}{3 d \,e^{4} \left (d x +c \right )^{3}}-\frac {a b}{3 d \,e^{4} \left (d x +c \right )^{2}}+\frac {2 a b \ln \left (d x +c \right )}{3 d \,e^{4}}-\frac {a b \ln \left (d x +c -1\right )}{3 d \,e^{4}}-\frac {a b \ln \left (d x +c +1\right )}{3 d \,e^{4}} \]
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 \[ -\frac {1}{3} \, {\left (d {\left (\frac {1}{d^{4} e^{4} x^{2} + 2 \, c d^{3} e^{4} x + c^{2} d^{2} e^{4}} + \frac {\log \left (d x + c + 1\right )}{d^{2} e^{4}} - \frac {2 \, \log \left (d x + c\right )}{d^{2} e^{4}} + \frac {\log \left (d x + c - 1\right )}{d^{2} e^{4}}\right )} + \frac {2 \, \operatorname {artanh}\left (d x + c\right )}{d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}}\right )} a b - \frac {1}{12} \, b^{2} {\left (\frac {\log \left (-d x - c + 1\right )^{2}}{d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}} + 3 \, \int -\frac {3 \, {\left (d x + c - 1\right )} \log \left (d x + c + 1\right )^{2} + 2 \, {\left (d x - 3 \, {\left (d x + c - 1\right )} \log \left (d x + c + 1\right ) + c\right )} \log \left (-d x - c + 1\right )}{3 \, {\left (d^{5} e^{4} x^{5} + c^{5} e^{4} - c^{4} e^{4} + {\left (5 \, c d^{4} e^{4} - d^{4} e^{4}\right )} x^{4} + 2 \, {\left (5 \, c^{2} d^{3} e^{4} - 2 \, c d^{3} e^{4}\right )} x^{3} + 2 \, {\left (5 \, c^{3} d^{2} e^{4} - 3 \, c^{2} d^{2} e^{4}\right )} x^{2} + {\left (5 \, c^{4} d e^{4} - 4 \, c^{3} d e^{4}\right )} x\right )}}\,{d x}\right )} - \frac {a^{2}}{3 \, {\left (d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}\right )}} \]
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 \[ \int \frac {{\left (a+b\,\mathrm {atanh}\left (c+d\,x\right )\right )}^2}{{\left (c\,e+d\,e\,x\right )}^4} \,d x \]
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 \[ \frac {\int \frac {a^{2}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {b^{2} \operatorname {atanh}^{2}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {2 a b \operatorname {atanh}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx}{e^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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