Optimal. Leaf size=156 \[ \frac {b d (c+d x) \text {Li}_2\left (\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{f^2 \left (a^2-b^2\right )}-\frac {b (c+d x)^2 \log \left (1-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{f \left (a^2-b^2\right )}+\frac {b d^2 \text {Li}_3\left (\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 f^3 \left (a^2-b^2\right )}+\frac {(c+d x)^3}{3 d (a+b)} \]
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Rubi [A] time = 0.28, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3731, 2190, 2531, 2282, 6589} \[ \frac {b d (c+d x) \text {PolyLog}\left (2,\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{f^2 \left (a^2-b^2\right )}+\frac {b d^2 \text {PolyLog}\left (3,\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 f^3 \left (a^2-b^2\right )}-\frac {b (c+d x)^2 \log \left (1-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{f \left (a^2-b^2\right )}+\frac {(c+d x)^3}{3 d (a+b)} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2282
Rule 2531
Rule 3731
Rule 6589
Rubi steps
\begin {align*} \int \frac {(c+d x)^2}{a+b \coth (e+f x)} \, dx &=\frac {(c+d x)^3}{3 (a+b) d}-(2 b) \int \frac {e^{-2 (e+f x)} (c+d x)^2}{(a+b)^2+\left (-a^2+b^2\right ) e^{-2 (e+f x)}} \, dx\\ &=\frac {(c+d x)^3}{3 (a+b) d}-\frac {b (c+d x)^2 \log \left (1-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f}+\frac {(2 b d) \int (c+d x) \log \left (1+\frac {\left (-a^2+b^2\right ) e^{-2 (e+f x)}}{(a+b)^2}\right ) \, dx}{\left (a^2-b^2\right ) f}\\ &=\frac {(c+d x)^3}{3 (a+b) d}-\frac {b (c+d x)^2 \log \left (1-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f}+\frac {b d (c+d x) \text {Li}_2\left (\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f^2}-\frac {\left (b d^2\right ) \int \text {Li}_2\left (-\frac {\left (-a^2+b^2\right ) e^{-2 (e+f x)}}{(a+b)^2}\right ) \, dx}{\left (a^2-b^2\right ) f^2}\\ &=\frac {(c+d x)^3}{3 (a+b) d}-\frac {b (c+d x)^2 \log \left (1-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f}+\frac {b d (c+d x) \text {Li}_2\left (\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f^2}+\frac {\left (b d^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {(a-b) x}{a+b}\right )}{x} \, dx,x,e^{-2 (e+f x)}\right )}{2 \left (a^2-b^2\right ) f^3}\\ &=\frac {(c+d x)^3}{3 (a+b) d}-\frac {b (c+d x)^2 \log \left (1-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f}+\frac {b d (c+d x) \text {Li}_2\left (\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f^2}+\frac {b d^2 \text {Li}_3\left (\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 \left (a^2-b^2\right ) f^3}\\ \end {align*}
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Mathematica [A] time = 3.69, size = 191, normalized size = 1.22 \[ \frac {x \sinh (e) \left (3 c^2+3 c d x+d^2 x^2\right )}{3 (a \sinh (e)+b \cosh (e))}+\frac {b \left (\frac {3 d \left (2 f (c+d x) \text {Li}_2\left (\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )+d \text {Li}_3\left (\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )\right )}{f^3 (a-b)}-\frac {6 (c+d x)^2 \log \left (\frac {(b-a) e^{-2 (e+f x)}}{a+b}+1\right )}{f (a-b)}+\frac {4 (c+d x)^3}{d \left (a \left (e^{2 e}-1\right )+b \left (e^{2 e}+1\right )\right )}\right )}{6 (a+b)} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.44, size = 492, normalized size = 3.15 \[ \frac {{\left (a + b\right )} d^{2} f^{3} x^{3} + 3 \, {\left (a + b\right )} c d f^{3} x^{2} + 3 \, {\left (a + b\right )} c^{2} f^{3} x + 6 \, b d^{2} {\rm polylog}\left (3, \sqrt {\frac {a + b}{a - b}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )}\right ) + 6 \, b d^{2} {\rm polylog}\left (3, -\sqrt {\frac {a + b}{a - b}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )}\right ) - 6 \, {\left (b d^{2} f x + b c d f\right )} {\rm Li}_2\left (\sqrt {\frac {a + b}{a - b}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )}\right ) - 6 \, {\left (b d^{2} f x + b c d f\right )} {\rm Li}_2\left (-\sqrt {\frac {a + b}{a - b}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )}\right ) - 3 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (2 \, {\left (a + b\right )} \cosh \left (f x + e\right ) + 2 \, {\left (a + b\right )} \sinh \left (f x + e\right ) + 2 \, {\left (a - b\right )} \sqrt {\frac {a + b}{a - b}}\right ) - 3 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (2 \, {\left (a + b\right )} \cosh \left (f x + e\right ) + 2 \, {\left (a + b\right )} \sinh \left (f x + e\right ) - 2 \, {\left (a - b\right )} \sqrt {\frac {a + b}{a - b}}\right ) - 3 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (\sqrt {\frac {a + b}{a - b}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )} + 1\right ) - 3 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (-\sqrt {\frac {a + b}{a - b}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )} + 1\right )}{3 \, {\left (a^{2} - b^{2}\right )} f^{3}} \]
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 \[ \int \frac {{\left (d x + c\right )}^{2}}{b \coth \left (f x + e\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.85, size = 722, normalized size = 4.63 \[ \frac {d^{2} x^{3}}{3 a +3 b}+\frac {c d \,x^{2}}{a +b}+\frac {c^{2} x}{a +b}+\frac {2 b \,c^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f \left (a +b \right ) \left (a -b \right )}-\frac {b \,c^{2} \ln \left (a \,{\mathrm e}^{2 f x +2 e}+b \,{\mathrm e}^{2 f x +2 e}-a +b \right )}{f \left (a +b \right ) \left (a -b \right )}+\frac {2 b \,d^{2} e^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3} \left (a +b \right ) \left (a -b \right )}-\frac {b \,d^{2} e^{2} \ln \left (a \,{\mathrm e}^{2 f x +2 e}+b \,{\mathrm e}^{2 f x +2 e}-a +b \right )}{f^{3} \left (a +b \right ) \left (a -b \right )}+\frac {2 b \,d^{2} x^{3}}{3 \left (a +b \right ) \left (a -b \right )}-\frac {2 b \,d^{2} e^{2} x}{f^{2} \left (a +b \right ) \left (a -b \right )}-\frac {4 b \,d^{2} e^{3}}{3 f^{3} \left (a +b \right ) \left (a -b \right )}-\frac {b \,d^{2} \ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{a -b}\right ) x^{2}}{f \left (a +b \right ) \left (a -b \right )}+\frac {b \,d^{2} \ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{a -b}\right ) e^{2}}{f^{3} \left (a +b \right ) \left (a -b \right )}-\frac {b \,d^{2} \polylog \left (2, \frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{a -b}\right ) x}{f^{2} \left (a +b \right ) \left (a -b \right )}+\frac {b \,d^{2} \polylog \left (3, \frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{a -b}\right )}{2 f^{3} \left (a +b \right ) \left (a -b \right )}-\frac {4 b c d e \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2} \left (a +b \right ) \left (a -b \right )}+\frac {2 b c d e \ln \left (a \,{\mathrm e}^{2 f x +2 e}+b \,{\mathrm e}^{2 f x +2 e}-a +b \right )}{f^{2} \left (a +b \right ) \left (a -b \right )}-\frac {2 b c d \ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{a -b}\right ) x}{f \left (a +b \right ) \left (a -b \right )}-\frac {2 b c d \ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{a -b}\right ) e}{f^{2} \left (a +b \right ) \left (a -b \right )}+\frac {2 b d c \,x^{2}}{\left (a +b \right ) \left (a -b \right )}+\frac {4 b d c e x}{f \left (a +b \right ) \left (a -b \right )}+\frac {2 b d c \,e^{2}}{f^{2} \left (a +b \right ) \left (a -b \right )}-\frac {b c d \polylog \left (2, \frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{a -b}\right )}{f^{2} \left (a +b \right ) \left (a -b \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.49, size = 332, normalized size = 2.13 \[ -\frac {{\left (2 \, f x \log \left (-\frac {{\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a - b} + 1\right ) + {\rm Li}_2\left (\frac {{\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a - b}\right )\right )} b c d}{a^{2} f^{2} - b^{2} f^{2}} - \frac {{\left (2 \, f^{2} x^{2} \log \left (-\frac {{\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a - b} + 1\right ) + 2 \, f x {\rm Li}_2\left (\frac {{\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a - b}\right ) - {\rm Li}_{3}(\frac {{\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a - b})\right )} b d^{2}}{2 \, {\left (a^{2} f^{3} - b^{2} f^{3}\right )}} - c^{2} {\left (\frac {b \log \left (-{\left (a - b\right )} e^{\left (-2 \, f x - 2 \, e\right )} + a + b\right )}{{\left (a^{2} - b^{2}\right )} f} - \frac {f x + e}{{\left (a + b\right )} f}\right )} + \frac {2 \, {\left (b d^{2} f^{3} x^{3} + 3 \, b c d f^{3} x^{2}\right )}}{3 \, {\left (a^{2} f^{3} - b^{2} f^{3}\right )}} + \frac {d^{2} x^{3} + 3 \, c d x^{2}}{3 \, {\left (a + b\right )}} \]
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 \[ \int \frac {{\left (c+d\,x\right )}^2}{a+b\,\mathrm {coth}\left (e+f\,x\right )} \,d x \]
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 \[ \int \frac {\left (c + d x\right )^{2}}{a + b \coth {\left (e + f x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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