Optimal. Leaf size=476 \[ \frac {2 b^2 d (c+d x) \text {Li}_2\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{f^2 \left (a^2-b^2\right )^2}+\frac {2 b^2 d (c+d x) \log \left (\frac {(a+b) e^{2 e+2 f x}}{a-b}+1\right )}{f^2 \left (a^2-b^2\right )^2}+\frac {2 b^2 (c+d x)^2 \log \left (\frac {(a+b) e^{2 e+2 f x}}{a-b}+1\right )}{f \left (a^2-b^2\right )^2}-\frac {2 b^2 (c+d x)^2}{f \left (a^2-b^2\right )^2}+\frac {b^2 d^2 \text {Li}_2\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{f^3 \left (a^2-b^2\right )^2}-\frac {b^2 d^2 \text {Li}_3\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{f^3 \left (a^2-b^2\right )^2}+\frac {2 b^2 (c+d x)^2}{f (a-b) (a+b)^2 \left ((a+b) e^{2 e+2 f x}+a-b\right )}-\frac {2 b d (c+d x) \text {Li}_2\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{f^2 (a-b)^2 (a+b)}-\frac {2 b (c+d x)^2 \log \left (\frac {(a+b) e^{2 e+2 f x}}{a-b}+1\right )}{f (a-b)^2 (a+b)}+\frac {(c+d x)^3}{3 d (a-b)^2}+\frac {b d^2 \text {Li}_3\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{f^3 (a-b)^2 (a+b)} \]
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Rubi [A] time = 1.65, antiderivative size = 476, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 11, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {3734, 2190, 2531, 2282, 6589, 2254, 2185, 2184, 2191, 2279, 2391} \[ \frac {2 b^2 d (c+d x) \text {PolyLog}\left (2,-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{f^2 \left (a^2-b^2\right )^2}+\frac {b^2 d^2 \text {PolyLog}\left (2,-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{f^3 \left (a^2-b^2\right )^2}-\frac {b^2 d^2 \text {PolyLog}\left (3,-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{f^3 \left (a^2-b^2\right )^2}-\frac {2 b d (c+d x) \text {PolyLog}\left (2,-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{f^2 (a-b)^2 (a+b)}+\frac {b d^2 \text {PolyLog}\left (3,-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{f^3 (a-b)^2 (a+b)}+\frac {2 b^2 d (c+d x) \log \left (\frac {(a+b) e^{2 e+2 f x}}{a-b}+1\right )}{f^2 \left (a^2-b^2\right )^2}+\frac {2 b^2 (c+d x)^2 \log \left (\frac {(a+b) e^{2 e+2 f x}}{a-b}+1\right )}{f \left (a^2-b^2\right )^2}-\frac {2 b^2 (c+d x)^2}{f \left (a^2-b^2\right )^2}+\frac {2 b^2 (c+d x)^2}{f (a-b) (a+b)^2 \left ((a+b) e^{2 e+2 f x}+a-b\right )}-\frac {2 b (c+d x)^2 \log \left (\frac {(a+b) e^{2 e+2 f x}}{a-b}+1\right )}{f (a-b)^2 (a+b)}+\frac {(c+d x)^3}{3 d (a-b)^2} \]
Antiderivative was successfully verified.
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Rule 2184
Rule 2185
Rule 2190
Rule 2191
Rule 2254
Rule 2279
Rule 2282
Rule 2391
Rule 2531
Rule 3734
Rule 6589
Rubi steps
\begin {align*} \int \frac {(c+d x)^2}{(a+b \tanh (e+f x))^2} \, dx &=\int \left (\frac {(c+d x)^2}{(a-b)^2}+\frac {4 b e^{2 e+2 f x} (c+d x)^2}{(a-b)^2 \left (-a \left (1-\frac {b}{a}\right )-a \left (1+\frac {b}{a}\right ) e^{2 e+2 f x}\right )}+\frac {4 b^2 e^{4 e+4 f x} (c+d x)^2}{(a-b)^2 \left (a \left (1-\frac {b}{a}\right )+a \left (1+\frac {b}{a}\right ) e^{2 e+2 f x}\right )^2}\right ) \, dx\\ &=\frac {(c+d x)^3}{3 (a-b)^2 d}+\frac {(4 b) \int \frac {e^{2 e+2 f x} (c+d x)^2}{-a \left (1-\frac {b}{a}\right )-a \left (1+\frac {b}{a}\right ) e^{2 e+2 f x}} \, dx}{(a-b)^2}+\frac {\left (4 b^2\right ) \int \frac {e^{4 e+4 f x} (c+d x)^2}{\left (a \left (1-\frac {b}{a}\right )+a \left (1+\frac {b}{a}\right ) e^{2 e+2 f x}\right )^2} \, dx}{(a-b)^2}\\ &=\frac {(c+d x)^3}{3 (a-b)^2 d}-\frac {2 b (c+d x)^2 \log \left (1+\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{(a-b)^2 (a+b) f}+\frac {\left (4 b^2\right ) \int \left (\frac {(c+d x)^2}{(a+b)^2}+\frac {(a-b)^2 (c+d x)^2}{(a+b)^2 \left (a-b+(a+b) e^{2 e+2 f x}\right )^2}+\frac {2 (-a+b) (c+d x)^2}{(a+b)^2 \left (a-b+(a+b) e^{2 e+2 f x}\right )}\right ) \, dx}{(a-b)^2}+\frac {(4 b d) \int (c+d x) \log \left (1+\frac {\left (1+\frac {b}{a}\right ) e^{2 e+2 f x}}{1-\frac {b}{a}}\right ) \, dx}{(a-b)^2 (a+b) f}\\ &=\frac {(c+d x)^3}{3 (a-b)^2 d}+\frac {4 b^2 (c+d x)^3}{3 \left (a^2-b^2\right )^2 d}-\frac {2 b (c+d x)^2 \log \left (1+\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{(a-b)^2 (a+b) f}-\frac {2 b d (c+d x) \text {Li}_2\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{(a-b)^2 (a+b) f^2}+\frac {\left (4 b^2\right ) \int \frac {(c+d x)^2}{\left (a-b+(a+b) e^{2 e+2 f x}\right )^2} \, dx}{(a+b)^2}-\frac {\left (8 b^2\right ) \int \frac {(c+d x)^2}{a-b+(a+b) e^{2 e+2 f x}} \, dx}{(a-b) (a+b)^2}+\frac {\left (2 b d^2\right ) \int \text {Li}_2\left (-\frac {\left (1+\frac {b}{a}\right ) e^{2 e+2 f x}}{1-\frac {b}{a}}\right ) \, dx}{(a-b)^2 (a+b) f^2}\\ &=\frac {(c+d x)^3}{3 (a-b)^2 d}-\frac {4 b^2 (c+d x)^3}{3 \left (a^2-b^2\right )^2 d}-\frac {2 b (c+d x)^2 \log \left (1+\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{(a-b)^2 (a+b) f}-\frac {2 b d (c+d x) \text {Li}_2\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{(a-b)^2 (a+b) f^2}+\frac {\left (4 b^2\right ) \int \frac {(c+d x)^2}{a-b+(a+b) e^{2 e+2 f x}} \, dx}{(a-b) (a+b)^2}+\frac {\left (8 b^2\right ) \int \frac {e^{2 e+2 f x} (c+d x)^2}{a-b+(a+b) e^{2 e+2 f x}} \, dx}{(a-b)^2 (a+b)}-\frac {\left (4 b^2\right ) \int \frac {e^{2 e+2 f x} (c+d x)^2}{\left (a-b+(a+b) e^{2 e+2 f x}\right )^2} \, dx}{a^2-b^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+2 f x}\right )}{(a-b)^2 (a+b) f^3}\\ &=\frac {2 b^2 (c+d x)^2}{(a-b) (a+b)^2 \left (a-b+(a+b) e^{2 e+2 f x}\right ) f}+\frac {(c+d x)^3}{3 (a-b)^2 d}-\frac {2 b (c+d x)^2 \log \left (1+\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{(a-b)^2 (a+b) f}+\frac {4 b^2 (c+d x)^2 \log \left (1+\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{\left (a^2-b^2\right )^2 f}-\frac {2 b d (c+d x) \text {Li}_2\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{(a-b)^2 (a+b) f^2}+\frac {b d^2 \text {Li}_3\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{(a-b)^2 (a+b) f^3}-\frac {\left (4 b^2\right ) \int \frac {e^{2 e+2 f x} (c+d x)^2}{a-b+(a+b) e^{2 e+2 f x}} \, dx}{(a-b)^2 (a+b)}-\frac {\left (4 b^2 d\right ) \int \frac {c+d x}{a-b+(a+b) e^{2 e+2 f x}} \, dx}{(a-b) (a+b)^2 f}-\frac {\left (8 b^2 d\right ) \int (c+d x) \log \left (1+\frac {(a+b) e^{2 e+2 f x}}{a-b}\right ) \, dx}{\left (a^2-b^2\right )^2 f}\\ &=-\frac {2 b^2 (c+d x)^2}{\left (a^2-b^2\right )^2 f}+\frac {2 b^2 (c+d x)^2}{(a-b) (a+b)^2 \left (a-b+(a+b) e^{2 e+2 f x}\right ) f}+\frac {(c+d x)^3}{3 (a-b)^2 d}-\frac {2 b (c+d x)^2 \log \left (1+\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{(a-b)^2 (a+b) f}+\frac {2 b^2 (c+d x)^2 \log \left (1+\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{\left (a^2-b^2\right )^2 f}-\frac {2 b d (c+d x) \text {Li}_2\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{(a-b)^2 (a+b) f^2}+\frac {4 b^2 d (c+d x) \text {Li}_2\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{\left (a^2-b^2\right )^2 f^2}+\frac {b d^2 \text {Li}_3\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{(a-b)^2 (a+b) f^3}-\frac {\left (4 b^2 d^2\right ) \int \text {Li}_2\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right ) \, dx}{\left (a^2-b^2\right )^2 f^2}+\frac {\left (4 b^2 d\right ) \int \frac {e^{2 e+2 f x} (c+d x)}{a-b+(a+b) e^{2 e+2 f x}} \, dx}{(a-b)^2 (a+b) f}+\frac {\left (4 b^2 d\right ) \int (c+d x) \log \left (1+\frac {(a+b) e^{2 e+2 f x}}{a-b}\right ) \, dx}{\left (a^2-b^2\right )^2 f}\\ &=-\frac {2 b^2 (c+d x)^2}{\left (a^2-b^2\right )^2 f}+\frac {2 b^2 (c+d x)^2}{(a-b) (a+b)^2 \left (a-b+(a+b) e^{2 e+2 f x}\right ) f}+\frac {(c+d x)^3}{3 (a-b)^2 d}+\frac {2 b^2 d (c+d x) \log \left (1+\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{\left (a^2-b^2\right )^2 f^2}-\frac {2 b (c+d x)^2 \log \left (1+\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{(a-b)^2 (a+b) f}+\frac {2 b^2 (c+d x)^2 \log \left (1+\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{\left (a^2-b^2\right )^2 f}-\frac {2 b d (c+d x) \text {Li}_2\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{(a-b)^2 (a+b) f^2}+\frac {2 b^2 d (c+d x) \text {Li}_2\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{\left (a^2-b^2\right )^2 f^2}+\frac {b d^2 \text {Li}_3\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{(a-b)^2 (a+b) f^3}-\frac {\left (2 b^2 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+2 f x}\right )}{\left (a^2-b^2\right )^2 f^3}-\frac {\left (2 b^2 d^2\right ) \int \log \left (1+\frac {(a+b) e^{2 e+2 f x}}{a-b}\right ) \, dx}{\left (a^2-b^2\right )^2 f^2}+\frac {\left (2 b^2 d^2\right ) \int \text {Li}_2\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right ) \, dx}{\left (a^2-b^2\right )^2 f^2}\\ &=-\frac {2 b^2 (c+d x)^2}{\left (a^2-b^2\right )^2 f}+\frac {2 b^2 (c+d x)^2}{(a-b) (a+b)^2 \left (a-b+(a+b) e^{2 e+2 f x}\right ) f}+\frac {(c+d x)^3}{3 (a-b)^2 d}+\frac {2 b^2 d (c+d x) \log \left (1+\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{\left (a^2-b^2\right )^2 f^2}-\frac {2 b (c+d x)^2 \log \left (1+\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{(a-b)^2 (a+b) f}+\frac {2 b^2 (c+d x)^2 \log \left (1+\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{\left (a^2-b^2\right )^2 f}-\frac {2 b d (c+d x) \text {Li}_2\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{(a-b)^2 (a+b) f^2}+\frac {2 b^2 d (c+d x) \text {Li}_2\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{\left (a^2-b^2\right )^2 f^2}+\frac {b d^2 \text {Li}_3\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{(a-b)^2 (a+b) f^3}-\frac {2 b^2 d^2 \text {Li}_3\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{\left (a^2-b^2\right )^2 f^3}-\frac {\left (b^2 d^2\right ) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {(a+b) x}{a-b}\right )}{x} \, dx,x,e^{2 e+2 f x}\right )}{\left (a^2-b^2\right )^2 f^3}+\frac {\left (b^2 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+2 f x}\right )}{\left (a^2-b^2\right )^2 f^3}\\ &=-\frac {2 b^2 (c+d x)^2}{\left (a^2-b^2\right )^2 f}+\frac {2 b^2 (c+d x)^2}{(a-b) (a+b)^2 \left (a-b+(a+b) e^{2 e+2 f x}\right ) f}+\frac {(c+d x)^3}{3 (a-b)^2 d}+\frac {2 b^2 d (c+d x) \log \left (1+\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{\left (a^2-b^2\right )^2 f^2}-\frac {2 b (c+d x)^2 \log \left (1+\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{(a-b)^2 (a+b) f}+\frac {2 b^2 (c+d x)^2 \log \left (1+\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{\left (a^2-b^2\right )^2 f}+\frac {b^2 d^2 \text {Li}_2\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{\left (a^2-b^2\right )^2 f^3}-\frac {2 b d (c+d x) \text {Li}_2\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{(a-b)^2 (a+b) f^2}+\frac {2 b^2 d (c+d x) \text {Li}_2\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{\left (a^2-b^2\right )^2 f^2}+\frac {b d^2 \text {Li}_3\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{(a-b)^2 (a+b) f^3}-\frac {b^2 d^2 \text {Li}_3\left (-\frac {(a+b) e^{2 e+2 f x}}{a-b}\right )}{\left (a^2-b^2\right )^2 f^3}\\ \end {align*}
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Mathematica [A] time = 9.39, size = 506, normalized size = 1.06 \[ \frac {\frac {f^2 (a-b) (a+b) \left (f x \left (a^2-b^2\right ) \left (3 c^2+3 c d x+d^2 x^2\right ) \cosh (2 e+f x)+f x \left (a^2+b^2\right ) \left (3 c^2+3 c d x+d^2 x^2\right ) \cosh (f x)+2 b \sinh (f x) \left (a f x \left (3 c^2+3 c d x+d^2 x^2\right )-3 b (c+d x)^2\right )\right )}{(a \cosh (e)+b \sinh (e)) (a \cosh (e+f x)+b \sinh (e+f x))}-\frac {12 b d f^2 x^2 (a-b) (2 a c f-b d)}{a \left (e^{2 e}+1\right )+b \left (e^{2 e}-1\right )}-\frac {24 b c f^2 x (a-b) (a c f-b d)}{a \left (e^{2 e}+1\right )+b \left (e^{2 e}-1\right )}-6 b d (b d-2 a c f) \text {Li}_2\left (\frac {(b-a) e^{-2 (e+f x)}}{a+b}\right )+12 b d f x (b d-2 a c f) \log \left (\frac {(a-b) e^{-2 (e+f x)}}{a+b}+1\right )+12 b c f (a c f-b d) \left (2 f x-\log \left ((a+b) e^{2 (e+f x)}+a-b\right )\right )-\frac {8 a b d^2 f^3 x^3 (a-b)}{a \left (e^{2 e}+1\right )+b \left (e^{2 e}-1\right )}-12 a b d^2 f^2 x^2 \log \left (\frac {(a-b) e^{-2 (e+f x)}}{a+b}+1\right )+6 a b d^2 \left (2 f x \text {Li}_2\left (\frac {(b-a) e^{-2 (e+f x)}}{a+b}\right )+\text {Li}_3\left (\frac {(b-a) e^{-2 (e+f x)}}{a+b}\right )\right )}{6 f^3 (a-b)^2 (a+b)^2} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.56, size = 3693, normalized size = 7.76 \[ \text {result too large to display} \]
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 \[ \int \frac {{\left (d x + c\right )}^{2}}{{\left (b \tanh \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.85, size = 1352, normalized size = 2.84 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.92, size = 753, normalized size = 1.58 \[ -\frac {4 \, b^{2} c d f x}{a^{4} f^{2} - 2 \, a^{2} b^{2} f^{2} + b^{4} 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 )} a b d^{2}}{a^{4} f^{3} - 2 \, a^{2} b^{2} f^{3} + b^{4} f^{3}} + \frac {2 \, b^{2} c d \log \left ({\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )} + a - b\right )}{a^{4} f^{2} - 2 \, a^{2} b^{2} f^{2} + b^{4} f^{2}} - c^{2} {\left (\frac {2 \, a b \log \left (-{\left (a - b\right )} e^{\left (-2 \, f x - 2 \, e\right )} - a - b\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} f} + \frac {2 \, b^{2}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4} + {\left (a^{4} - 2 \, a^{3} b + 2 \, a b^{3} - b^{4}\right )} e^{\left (-2 \, f x - 2 \, e\right )}\right )} f} - \frac {f x + e}{{\left (a^{2} + 2 \, a b + b^{2}\right )} f}\right )} - \frac {{\left (2 \, a b c d f - b^{2} d^{2}\right )} {\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 )}}{a^{4} f^{3} - 2 \, a^{2} b^{2} f^{3} + b^{4} f^{3}} + \frac {2 \, {\left (2 \, a b d^{2} f^{3} x^{3} + 3 \, {\left (2 \, a b c d f - b^{2} d^{2}\right )} f^{2} x^{2}\right )}}{3 \, {\left (a^{4} f^{3} - 2 \, a^{2} b^{2} f^{3} + b^{4} f^{3}\right )}} + \frac {12 \, b^{2} c d x + {\left (a^{2} d^{2} f - 2 \, a b d^{2} f + b^{2} d^{2} f\right )} x^{3} + 3 \, {\left (a^{2} c d f - 2 \, a b c d f + {\left (c d f + 2 \, d^{2}\right )} b^{2}\right )} x^{2} + {\left ({\left (a^{2} d^{2} f e^{\left (2 \, e\right )} - b^{2} d^{2} f e^{\left (2 \, e\right )}\right )} x^{3} + 3 \, {\left (a^{2} c d f e^{\left (2 \, e\right )} - b^{2} c d f e^{\left (2 \, e\right )}\right )} x^{2}\right )} e^{\left (2 \, f x\right )}}{3 \, {\left (a^{4} f - 2 \, a^{2} b^{2} f + b^{4} f + {\left (a^{4} f e^{\left (2 \, e\right )} + 2 \, a^{3} b f e^{\left (2 \, e\right )} - 2 \, a b^{3} f e^{\left (2 \, e\right )} - b^{4} f e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c+d\,x\right )}^2}{{\left (a+b\,\mathrm {tanh}\left (e+f\,x\right )\right )}^2} \,d x \]
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 \[ \int \frac {\left (c + d x\right )^{2}}{\left (a + b \tanh {\left (e + f x \right )}\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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