Optimal. Leaf size=176 \[ -\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (c x+1)}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (c x+1)^2}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{9 c (c x+1)^3}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{24 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c (c x+1)^3}-\frac {11 b^2}{144 c (c x+1)}-\frac {5 b^2}{144 c (c x+1)^2}-\frac {b^2}{54 c (c x+1)^3}+\frac {11 b^2 \tanh ^{-1}(c x)}{144 c} \]
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Rubi [A] time = 0.22, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5928, 5926, 627, 44, 207, 5948} \[ -\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (c x+1)}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (c x+1)^2}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{9 c (c x+1)^3}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{24 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c (c x+1)^3}-\frac {11 b^2}{144 c (c x+1)}-\frac {5 b^2}{144 c (c x+1)^2}-\frac {b^2}{54 c (c x+1)^3}+\frac {11 b^2 \tanh ^{-1}(c x)}{144 c} \]
Antiderivative was successfully verified.
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Rule 44
Rule 207
Rule 627
Rule 5926
Rule 5928
Rule 5948
Rubi steps
\begin {align*} \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{(1+c x)^4} \, dx &=-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c (1+c x)^3}+\frac {1}{3} (2 b) \int \left (\frac {a+b \tanh ^{-1}(c x)}{2 (1+c x)^4}+\frac {a+b \tanh ^{-1}(c x)}{4 (1+c x)^3}+\frac {a+b \tanh ^{-1}(c x)}{8 (1+c x)^2}-\frac {a+b \tanh ^{-1}(c x)}{8 \left (-1+c^2 x^2\right )}\right ) \, dx\\ &=-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c (1+c x)^3}+\frac {1}{12} b \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx-\frac {1}{12} b \int \frac {a+b \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx+\frac {1}{6} b \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^3} \, dx+\frac {1}{3} b \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^4} \, dx\\ &=-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{9 c (1+c x)^3}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (1+c x)^2}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (1+c x)}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{24 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c (1+c x)^3}+\frac {1}{12} b^2 \int \frac {1}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx+\frac {1}{12} b^2 \int \frac {1}{(1+c x) \left (1-c^2 x^2\right )} \, dx+\frac {1}{9} b^2 \int \frac {1}{(1+c x)^3 \left (1-c^2 x^2\right )} \, dx\\ &=-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{9 c (1+c x)^3}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (1+c x)^2}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (1+c x)}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{24 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c (1+c x)^3}+\frac {1}{12} b^2 \int \frac {1}{(1-c x) (1+c x)^3} \, dx+\frac {1}{12} b^2 \int \frac {1}{(1-c x) (1+c x)^2} \, dx+\frac {1}{9} b^2 \int \frac {1}{(1-c x) (1+c x)^4} \, dx\\ &=-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{9 c (1+c x)^3}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (1+c x)^2}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (1+c x)}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{24 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c (1+c x)^3}+\frac {1}{12} b^2 \int \left (\frac {1}{2 (1+c x)^2}-\frac {1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx+\frac {1}{12} b^2 \int \left (\frac {1}{2 (1+c x)^3}+\frac {1}{4 (1+c x)^2}-\frac {1}{4 \left (-1+c^2 x^2\right )}\right ) \, dx+\frac {1}{9} b^2 \int \left (\frac {1}{2 (1+c x)^4}+\frac {1}{4 (1+c x)^3}+\frac {1}{8 (1+c x)^2}-\frac {1}{8 \left (-1+c^2 x^2\right )}\right ) \, dx\\ &=-\frac {b^2}{54 c (1+c x)^3}-\frac {5 b^2}{144 c (1+c x)^2}-\frac {11 b^2}{144 c (1+c x)}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{9 c (1+c x)^3}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (1+c x)^2}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (1+c x)}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{24 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c (1+c x)^3}-\frac {1}{72} b^2 \int \frac {1}{-1+c^2 x^2} \, dx-\frac {1}{48} b^2 \int \frac {1}{-1+c^2 x^2} \, dx-\frac {1}{24} b^2 \int \frac {1}{-1+c^2 x^2} \, dx\\ &=-\frac {b^2}{54 c (1+c x)^3}-\frac {5 b^2}{144 c (1+c x)^2}-\frac {11 b^2}{144 c (1+c x)}+\frac {11 b^2 \tanh ^{-1}(c x)}{144 c}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{9 c (1+c x)^3}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (1+c x)^2}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (1+c x)}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{24 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c (1+c x)^3}\\ \end {align*}
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Mathematica [A] time = 0.18, size = 168, normalized size = 0.95 \[ -\frac {16 \left (18 a^2+6 a b+b^2\right )+24 b \tanh ^{-1}(c x) \left (24 a+b \left (3 c^2 x^2+9 c x+10\right )\right )+6 b (12 a+11 b) (c x+1)^2+6 b (12 a+5 b) (c x+1)+3 b (12 a+11 b) (c x+1)^3 \log (1-c x)-3 b (12 a+11 b) (c x+1)^3 \log (c x+1)-36 b^2 \left (c^3 x^3+3 c^2 x^2+3 c x-7\right ) \tanh ^{-1}(c x)^2}{864 c (c x+1)^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.62, size = 203, normalized size = 1.15 \[ -\frac {6 \, {\left (12 \, a b + 11 \, b^{2}\right )} c^{2} x^{2} + 54 \, {\left (4 \, a b + 3 \, b^{2}\right )} c x - 9 \, {\left (b^{2} c^{3} x^{3} + 3 \, b^{2} c^{2} x^{2} + 3 \, b^{2} c x - 7 \, b^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )^{2} + 288 \, a^{2} + 240 \, a b + 112 \, b^{2} - 3 \, {\left ({\left (12 \, a b + 11 \, b^{2}\right )} c^{3} x^{3} + 3 \, {\left (12 \, a b + 7 \, b^{2}\right )} c^{2} x^{2} + 3 \, {\left (12 \, a b - b^{2}\right )} c x - 84 \, a b - 29 \, b^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{864 \, {\left (c^{4} x^{3} + 3 \, c^{3} x^{2} + 3 \, c^{2} x + c\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.47, size = 333, normalized size = 1.89 \[ \frac {1}{1728} \, c {\left (\frac {18 \, {\left (\frac {3 \, {\left (c x + 1\right )}^{2} b^{2}}{{\left (c x - 1\right )}^{2}} - \frac {3 \, {\left (c x + 1\right )} b^{2}}{c x - 1} + b^{2}\right )} {\left (c x - 1\right )}^{3} \log \left (-\frac {c x + 1}{c x - 1}\right )^{2}}{{\left (c x + 1\right )}^{3} c^{2}} + \frac {6 \, {\left (\frac {36 \, {\left (c x + 1\right )}^{2} a b}{{\left (c x - 1\right )}^{2}} - \frac {36 \, {\left (c x + 1\right )} a b}{c x - 1} + 12 \, a b + \frac {18 \, {\left (c x + 1\right )}^{2} b^{2}}{{\left (c x - 1\right )}^{2}} - \frac {9 \, {\left (c x + 1\right )} b^{2}}{c x - 1} + 2 \, b^{2}\right )} {\left (c x - 1\right )}^{3} \log \left (-\frac {c x + 1}{c x - 1}\right )}{{\left (c x + 1\right )}^{3} c^{2}} + \frac {{\left (\frac {216 \, {\left (c x + 1\right )}^{2} a^{2}}{{\left (c x - 1\right )}^{2}} - \frac {216 \, {\left (c x + 1\right )} a^{2}}{c x - 1} + 72 \, a^{2} + \frac {216 \, {\left (c x + 1\right )}^{2} a b}{{\left (c x - 1\right )}^{2}} - \frac {108 \, {\left (c x + 1\right )} a b}{c x - 1} + 24 \, a b + \frac {108 \, {\left (c x + 1\right )}^{2} b^{2}}{{\left (c x - 1\right )}^{2}} - \frac {27 \, {\left (c x + 1\right )} b^{2}}{c x - 1} + 4 \, b^{2}\right )} {\left (c x - 1\right )}^{3}}{{\left (c x + 1\right )}^{3} c^{2}}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.07, size = 386, normalized size = 2.19 \[ -\frac {a^{2}}{3 c \left (c x +1\right )^{3}}-\frac {b^{2} \arctanh \left (c x \right )^{2}}{3 c \left (c x +1\right )^{3}}-\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{24 c}-\frac {b^{2} \arctanh \left (c x \right )}{9 c \left (c x +1\right )^{3}}-\frac {b^{2} \arctanh \left (c x \right )}{12 c \left (c x +1\right )^{2}}-\frac {b^{2} \arctanh \left (c x \right )}{12 c \left (c x +1\right )}+\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )}{24 c}-\frac {b^{2} \ln \left (c x -1\right )^{2}}{96 c}+\frac {b^{2} \ln \left (c x -1\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{48 c}-\frac {b^{2} \ln \left (c x +1\right )^{2}}{96 c}+\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{48 c}-\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{48 c}-\frac {11 b^{2} \ln \left (c x -1\right )}{288 c}-\frac {b^{2}}{54 c \left (c x +1\right )^{3}}-\frac {5 b^{2}}{144 c \left (c x +1\right )^{2}}-\frac {11 b^{2}}{144 c \left (c x +1\right )}+\frac {11 b^{2} \ln \left (c x +1\right )}{288 c}-\frac {2 a b \arctanh \left (c x \right )}{3 c \left (c x +1\right )^{3}}-\frac {a b \ln \left (c x -1\right )}{24 c}-\frac {a b}{9 c \left (c x +1\right )^{3}}-\frac {a b}{12 c \left (c x +1\right )^{2}}-\frac {a b}{12 c \left (c x +1\right )}+\frac {a b \ln \left (c x +1\right )}{24 c} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.35, size = 445, normalized size = 2.53 \[ -\frac {1}{72} \, {\left (c {\left (\frac {2 \, {\left (3 \, c^{2} x^{2} + 9 \, c x + 10\right )}}{c^{5} x^{3} + 3 \, c^{4} x^{2} + 3 \, c^{3} x + c^{2}} - \frac {3 \, \log \left (c x + 1\right )}{c^{2}} + \frac {3 \, \log \left (c x - 1\right )}{c^{2}}\right )} + \frac {48 \, \operatorname {artanh}\left (c x\right )}{c^{4} x^{3} + 3 \, c^{3} x^{2} + 3 \, c^{2} x + c}\right )} a b - \frac {1}{864} \, {\left (12 \, c {\left (\frac {2 \, {\left (3 \, c^{2} x^{2} + 9 \, c x + 10\right )}}{c^{5} x^{3} + 3 \, c^{4} x^{2} + 3 \, c^{3} x + c^{2}} - \frac {3 \, \log \left (c x + 1\right )}{c^{2}} + \frac {3 \, \log \left (c x - 1\right )}{c^{2}}\right )} \operatorname {artanh}\left (c x\right ) + \frac {{\left (66 \, c^{2} x^{2} + 9 \, {\left (c^{3} x^{3} + 3 \, c^{2} x^{2} + 3 \, c x + 1\right )} \log \left (c x + 1\right )^{2} + 9 \, {\left (c^{3} x^{3} + 3 \, c^{2} x^{2} + 3 \, c x + 1\right )} \log \left (c x - 1\right )^{2} + 162 \, c x - 3 \, {\left (11 \, c^{3} x^{3} + 33 \, c^{2} x^{2} + 33 \, c x + 6 \, {\left (c^{3} x^{3} + 3 \, c^{2} x^{2} + 3 \, c x + 1\right )} \log \left (c x - 1\right ) + 11\right )} \log \left (c x + 1\right ) + 33 \, {\left (c^{3} x^{3} + 3 \, c^{2} x^{2} + 3 \, c x + 1\right )} \log \left (c x - 1\right ) + 112\right )} c^{2}}{c^{6} x^{3} + 3 \, c^{5} x^{2} + 3 \, c^{4} x + c^{3}}\right )} b^{2} - \frac {b^{2} \operatorname {artanh}\left (c x\right )^{2}}{3 \, {\left (c^{4} x^{3} + 3 \, c^{3} x^{2} + 3 \, c^{2} x + c\right )}} - \frac {a^{2}}{3 \, {\left (c^{4} x^{3} + 3 \, c^{3} x^{2} + 3 \, c^{2} x + c\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.30, size = 498, normalized size = 2.83 \[ \ln \left (1-c\,x\right )\,\left (\ln \left (c\,x+1\right )\,\left (\frac {b^2}{3\,c\,\left (2\,c^3\,x^3+6\,c^2\,x^2+6\,c\,x+2\right )}-\frac {b^2\,\left (c^3\,x^3+3\,c^2\,x^2+3\,c\,x+1\right )}{24\,c\,\left (2\,c^3\,x^3+6\,c^2\,x^2+6\,c\,x+2\right )}\right )+\frac {b^2}{3\,c\,\left (6\,c^3\,x^3+18\,c^2\,x^2+18\,c\,x+6\right )}+\frac {b\,\left (6\,a-b\right )}{3\,c\,\left (6\,c^3\,x^3+18\,c^2\,x^2+18\,c\,x+6\right )}+\frac {b^2\,\left (11\,c^3\,x^3+45\,c^2\,x^2+69\,c\,x+51\right )}{48\,c\,\left (6\,c^3\,x^3+18\,c^2\,x^2+18\,c\,x+6\right )}\right )-\frac {x\,\left (27\,b^2+36\,a\,b\right )+x^2\,\left (11\,c\,b^2+12\,a\,c\,b\right )+\frac {8\,\left (18\,a^2+15\,a\,b+7\,b^2\right )}{3\,c}}{144\,c^3\,x^3+432\,c^2\,x^2+432\,c\,x+144}+{\ln \left (c\,x+1\right )}^2\,\left (\frac {b^2}{96\,c}-\frac {b^2}{12\,c^2\,\left (3\,x+3\,c\,x^2+\frac {1}{c}+c^2\,x^3\right )}\right )+{\ln \left (1-c\,x\right )}^2\,\left (\frac {b^2}{96\,c}-\frac {b^2}{3\,c\,\left (4\,c^3\,x^3+12\,c^2\,x^2+12\,c\,x+4\right )}\right )-\frac {\ln \left (c\,x+1\right )\,\left (\frac {7\,b^2}{96\,c^2}+\frac {5\,b^2\,x^2}{32}+\frac {23\,b^2\,x}{96\,c}+\frac {11\,b^2\,c\,x^3}{288}+\frac {b\,\left (16\,a+5\,b\right )}{48\,c^2}\right )}{3\,x+3\,c\,x^2+\frac {1}{c}+c^2\,x^3}-\frac {b\,\mathrm {atan}\left (c\,x\,1{}\mathrm {i}\right )\,\left (6\,a+11\,b\right )\,1{}\mathrm {i}}{72\,c} \]
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 \[ \int \frac {\left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{2}}{\left (c x + 1\right )^{4}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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