Optimal. Leaf size=119 \[ -\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^3 (c+d x)}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3}+\frac {b^2 \log (c+d x)}{d e^3}-\frac {b^2 \log \left (1-(c+d x)^2\right )}{2 d e^3} \]
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Rubi [A] time = 0.17, antiderivative size = 119, 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, 266, 36, 31, 29, 5948} \[ -\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^3 (c+d x)}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3}+\frac {b^2 \log (c+d x)}{d e^3}-\frac {b^2 \log \left (1-(c+d x)^2\right )}{2 d e^3} \]
Antiderivative was successfully verified.
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Rule 12
Rule 29
Rule 31
Rule 36
Rule 266
Rule 5916
Rule 5948
Rule 5982
Rule 6107
Rubi steps
\begin {align*} \int \frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{(c e+d e x)^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^2}{e^3 x^3} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^2}{x^3} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}+\frac {b \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(x)}{x^2 \left (1-x^2\right )} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}+\frac {b \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(x)}{x^2} \, dx,x,c+d x\right )}{d e^3}+\frac {b \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(x)}{1-x^2} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^3 (c+d x)}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}+\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{x \left (1-x^2\right )} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^3 (c+d x)}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}+\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{(1-x) x} \, dx,x,(c+d x)^2\right )}{2 d e^3}\\ &=-\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^3 (c+d x)}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}+\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,(c+d x)^2\right )}{2 d e^3}+\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,(c+d x)^2\right )}{2 d e^3}\\ &=-\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^3 (c+d x)}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}+\frac {b^2 \log (c+d x)}{d e^3}-\frac {b^2 \log \left (1-(c+d x)^2\right )}{2 d e^3}\\ \end {align*}
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Mathematica [A] time = 0.17, size = 136, normalized size = 1.14 \[ \frac {-\frac {a^2}{(c+d x)^2}-\frac {2 a b}{c+d x}-b (a+b) \log (-c-d x+1)+b (a-b) \log (c+d x+1)-\frac {2 b \tanh ^{-1}(c+d x) (a+b (c+d x))}{(c+d x)^2}+\frac {b^2 \left (c^2+2 c d x+d^2 x^2-1\right ) \tanh ^{-1}(c+d x)^2}{(c+d x)^2}+2 b^2 \log (c+d x)}{2 d e^3} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.49, size = 270, normalized size = 2.27 \[ -\frac {8 \, a b d x + 8 \, a b c - {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} - b^{2}\right )} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )^{2} + 4 \, a^{2} - 4 \, {\left ({\left (a b - b^{2}\right )} d^{2} x^{2} + 2 \, {\left (a b - b^{2}\right )} c d x + {\left (a b - b^{2}\right )} c^{2}\right )} \log \left (d x + c + 1\right ) - 8 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \log \left (d x + c\right ) + 4 \, {\left ({\left (a b + b^{2}\right )} d^{2} x^{2} + 2 \, {\left (a b + b^{2}\right )} c d x + {\left (a b + b^{2}\right )} c^{2}\right )} \log \left (d x + c - 1\right ) + 4 \, {\left (b^{2} d x + b^{2} c + a b\right )} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{8 \, {\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.18, size = 356, normalized size = 2.99 \[ \frac {{\left (\frac {{\left (d x + c + 1\right )} b^{2} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )^{2}}{d x + c - 1} + \frac {2 \, {\left (d x + c + 1\right )}^{2} b^{2} \log \left (-\frac {d x + c + 1}{d x + c - 1} - 1\right )}{{\left (d x + c - 1\right )}^{2}} + \frac {4 \, {\left (d x + c + 1\right )} b^{2} \log \left (-\frac {d x + c + 1}{d x + c - 1} - 1\right )}{d x + c - 1} + 2 \, b^{2} \log \left (-\frac {d x + c + 1}{d x + c - 1} - 1\right ) + \frac {4 \, {\left (d x + c + 1\right )} a b \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{d x + c - 1} - \frac {2 \, {\left (d x + c + 1\right )}^{2} b^{2} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{{\left (d x + c - 1\right )}^{2}} - \frac {2 \, {\left (d x + c + 1\right )} b^{2} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{d x + c - 1} + \frac {4 \, {\left (d x + c + 1\right )} a^{2}}{d x + c - 1} + \frac {4 \, {\left (d x + c + 1\right )} a b}{d x + c - 1} + 4 \, a b\right )} {\left ({\left (c + 1\right )} d - {\left (c - 1\right )} d\right )}}{4 \, {\left (\frac {{\left (d x + c + 1\right )}^{2} d^{2} e^{3}}{{\left (d x + c - 1\right )}^{2}} + \frac {2 \, {\left (d x + c + 1\right )} d^{2} e^{3}}{d x + c - 1} + d^{2} e^{3}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.08, size = 371, normalized size = 3.12 \[ -\frac {a^{2}}{2 d \,e^{3} \left (d x +c \right )^{2}}-\frac {b^{2} \arctanh \left (d x +c \right )^{2}}{2 d \,e^{3} \left (d x +c \right )^{2}}-\frac {b^{2} \arctanh \left (d x +c \right )}{d \,e^{3} \left (d x +c \right )}-\frac {b^{2} \arctanh \left (d x +c \right ) \ln \left (d x +c -1\right )}{2 d \,e^{3}}+\frac {b^{2} \arctanh \left (d x +c \right ) \ln \left (d x +c +1\right )}{2 d \,e^{3}}-\frac {b^{2} \ln \left (d x +c -1\right )^{2}}{8 d \,e^{3}}+\frac {b^{2} \ln \left (d x +c -1\right ) \ln \left (\frac {1}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,e^{3}}+\frac {b^{2} \ln \left (d x +c \right )}{d \,e^{3}}-\frac {b^{2} \ln \left (d x +c -1\right )}{2 d \,e^{3}}-\frac {b^{2} \ln \left (d x +c +1\right )}{2 d \,e^{3}}-\frac {b^{2} \ln \left (d x +c +1\right )^{2}}{8 d \,e^{3}}+\frac {b^{2} \ln \left (-\frac {d x}{2}-\frac {c}{2}+\frac {1}{2}\right ) \ln \left (d x +c +1\right )}{4 d \,e^{3}}-\frac {b^{2} \ln \left (-\frac {d x}{2}-\frac {c}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,e^{3}}-\frac {a b \arctanh \left (d x +c \right )}{d \,e^{3} \left (d x +c \right )^{2}}-\frac {a b}{d \,e^{3} \left (d x +c \right )}-\frac {a b \ln \left (d x +c -1\right )}{2 d \,e^{3}}+\frac {a b \ln \left (d x +c +1\right )}{2 d \,e^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.36, size = 329, normalized size = 2.76 \[ -\frac {1}{2} \, {\left (d {\left (\frac {2}{d^{3} e^{3} x + c d^{2} e^{3}} - \frac {\log \left (d x + c + 1\right )}{d^{2} e^{3}} + \frac {\log \left (d x + c - 1\right )}{d^{2} e^{3}}\right )} + \frac {2 \, \operatorname {artanh}\left (d x + c\right )}{d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}}\right )} a b - \frac {1}{8} \, {\left (d^{2} {\left (\frac {\log \left (d x + c + 1\right )^{2} - 2 \, \log \left (d x + c + 1\right ) \log \left (d x + c - 1\right ) + \log \left (d x + c - 1\right )^{2} + 4 \, \log \left (d x + c - 1\right )}{d^{3} e^{3}} + \frac {4 \, \log \left (d x + c + 1\right )}{d^{3} e^{3}} - \frac {8 \, \log \left (d x + c\right )}{d^{3} e^{3}}\right )} + 4 \, d {\left (\frac {2}{d^{3} e^{3} x + c d^{2} e^{3}} - \frac {\log \left (d x + c + 1\right )}{d^{2} e^{3}} + \frac {\log \left (d x + c - 1\right )}{d^{2} e^{3}}\right )} \operatorname {artanh}\left (d x + c\right )\right )} b^{2} - \frac {b^{2} \operatorname {artanh}\left (d x + c\right )^{2}}{2 \, {\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} - \frac {a^{2}}{2 \, {\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.42, size = 776, normalized size = 6.52 \[ {\ln \left (1-d\,x-c\right )}^2\,\left (\frac {b^2}{8\,d\,e^3}-\frac {b^2}{2\,d\,\left (4\,c^2\,e^3+8\,c\,d\,e^3\,x+4\,d^2\,e^3\,x^2\right )}\right )+{\ln \left (c+d\,x+1\right )}^2\,\left (\frac {b^2}{8\,d\,e^3}-\frac {b^2}{8\,d^2\,e^3\,\left (2\,c\,x+d\,x^2+\frac {c^2}{d}\right )}\right )+\ln \left (1-d\,x-c\right )\,\left (\ln \left (c+d\,x+1\right )\,\left (\frac {b^2}{2\,d\,\left (2\,c^2\,e^3+4\,c\,d\,e^3\,x+2\,d^2\,e^3\,x^2\right )}-\frac {b^2\,\left (c^2+2\,c\,d\,x+d^2\,x^2\right )}{2\,d\,\left (2\,c^2\,e^3+4\,c\,d\,e^3\,x+2\,d^2\,e^3\,x^2\right )}\right )+\frac {b^2}{2\,d\,\left (4\,c^2\,e^3+8\,c\,d\,e^3\,x+4\,d^2\,e^3\,x^2\right )}+\frac {b\,\left (4\,a-b\right )}{2\,d\,\left (4\,c^2\,e^3+8\,c\,d\,e^3\,x+4\,d^2\,e^3\,x^2\right )}-\frac {b^2\,\left (x\,\left (4\,c\,d-d+d\,\left (2\,c-1\right )\right )-c+c^2+c\,\left (2\,c-1\right )+3\,d^2\,x^2+1\right )}{2\,d\,\left (4\,c^2\,e^3+8\,c\,d\,e^3\,x+4\,d^2\,e^3\,x^2\right )}+\frac {b^2\,\left (x\,\left (2\,d\,e^3+d\,\left (4\,c\,e^3+2\,e^3\right )+8\,c\,d\,e^3\right )+2\,c\,e^3+2\,e^3+c\,\left (4\,c\,e^3+2\,e^3\right )+2\,c^2\,e^3+6\,d^2\,e^3\,x^2\right )}{4\,d\,e^3\,\left (4\,c^2\,e^3+8\,c\,d\,e^3\,x+4\,d^2\,e^3\,x^2\right )}\right )-\frac {\frac {a^2+2\,b\,c\,a}{2\,d}+a\,b\,x}{c^2\,e^3+2\,c\,d\,e^3\,x+d^2\,e^3\,x^2}-\frac {\ln \left (c+d\,x+1\right )\,\left (x\,\left (\frac {2\,b^2\,c+b^2}{4\,d\,e^3}+\frac {b^2\,c}{4\,d\,e^3}-\frac {b^2\,\left (3\,c-1\right )}{4\,d\,e^3}\right )+\frac {b^2\,c^2+b^2\,c+b^2+4\,a\,b}{8\,d^2\,e^3}-\frac {b^2\,\left (\frac {c^2-c+1}{2\,d}+\frac {c\,\left (2\,c-1\right )}{2\,d}\right )}{4\,d\,e^3}+\frac {c\,\left (2\,b^2\,c+b^2\right )}{8\,d^2\,e^3}\right )}{2\,c\,x+d\,x^2+\frac {c^2}{d}}+\frac {b^2\,\ln \left (c+d\,x\right )}{d\,e^3}-\frac {\ln \left (c+d\,x-1\right )\,\left (b^2+a\,b\right )}{2\,d\,e^3}+\frac {\ln \left (c+d\,x+1\right )\,\left (a\,b-b^2\right )}{2\,d\,e^3} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 4.31, size = 1102, normalized size = 9.26 \[ \begin {cases} - \frac {a^{2}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {2 a b c^{2} \operatorname {atanh}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {4 a b c d x \operatorname {atanh}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {2 a b c}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {2 a b d^{2} x^{2} \operatorname {atanh}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {2 a b d x}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {2 a b \operatorname {atanh}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {2 b^{2} c^{2} \log {\left (\frac {c}{d} + x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {2 b^{2} c^{2} \log {\left (\frac {c}{d} + x + \frac {1}{d} \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {b^{2} c^{2} \operatorname {atanh}^{2}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {2 b^{2} c^{2} \operatorname {atanh}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {4 b^{2} c d x \log {\left (\frac {c}{d} + x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {4 b^{2} c d x \log {\left (\frac {c}{d} + x + \frac {1}{d} \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {2 b^{2} c d x \operatorname {atanh}^{2}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {4 b^{2} c d x \operatorname {atanh}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {2 b^{2} c \operatorname {atanh}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {2 b^{2} d^{2} x^{2} \log {\left (\frac {c}{d} + x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {2 b^{2} d^{2} x^{2} \log {\left (\frac {c}{d} + x + \frac {1}{d} \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {b^{2} d^{2} x^{2} \operatorname {atanh}^{2}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {2 b^{2} d^{2} x^{2} \operatorname {atanh}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {2 b^{2} d x \operatorname {atanh}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {b^{2} \operatorname {atanh}^{2}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} & \text {for}\: d \neq 0 \\\frac {x \left (a + b \operatorname {atanh}{\relax (c )}\right )^{2}}{c^{3} e^{3}} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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