Integrand size = 15, antiderivative size = 351 \[ \int x^2 \cot ^{-1}(c+d \coth (a+b x)) \, dx=\frac {1}{3} x^3 \cot ^{-1}(c+d \coth (a+b x))-\frac {1}{6} i x^3 \log \left (1-\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )+\frac {1}{6} i x^3 \log \left (1-\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )-\frac {i x^2 \operatorname {PolyLog}\left (2,\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{4 b}+\frac {i x^2 \operatorname {PolyLog}\left (2,\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{4 b}+\frac {i x \operatorname {PolyLog}\left (3,\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{4 b^2}-\frac {i x \operatorname {PolyLog}\left (3,\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{4 b^2}-\frac {i \operatorname {PolyLog}\left (4,\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{8 b^3}+\frac {i \operatorname {PolyLog}\left (4,\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{8 b^3} \]
[Out]
Time = 0.33 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5310, 2221, 2611, 6744, 2320, 6724} \[ \int x^2 \cot ^{-1}(c+d \coth (a+b x)) \, dx=-\frac {i \operatorname {PolyLog}\left (4,\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{8 b^3}+\frac {i \operatorname {PolyLog}\left (4,\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{8 b^3}+\frac {i x \operatorname {PolyLog}\left (3,\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{4 b^2}-\frac {i x \operatorname {PolyLog}\left (3,\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{4 b^2}-\frac {i x^2 \operatorname {PolyLog}\left (2,\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )}{4 b}+\frac {i x^2 \operatorname {PolyLog}\left (2,\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )}{4 b}-\frac {1}{6} i x^3 \log \left (1-\frac {(-c-d+i) e^{2 a+2 b x}}{-c+d+i}\right )+\frac {1}{6} i x^3 \log \left (1-\frac {(c+d+i) e^{2 a+2 b x}}{c-d+i}\right )+\frac {1}{3} x^3 \cot ^{-1}(d \coth (a+b x)+c) \]
[In]
[Out]
Rule 2221
Rule 2320
Rule 2611
Rule 5310
Rule 6724
Rule 6744
Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} x^3 \cot ^{-1}(c+d \coth (a+b x))+\frac {1}{3} (b (1-i (c+d))) \int \frac {e^{2 a+2 b x} x^3}{i+c-d+(-i-c-d) e^{2 a+2 b x}} \, dx-\frac {1}{3} (b (1+i (c+d))) \int \frac {e^{2 a+2 b x} x^3}{i-c+d+(-i+c+d) e^{2 a+2 b x}} \, dx \\ & = \frac {1}{3} x^3 \cot ^{-1}(c+d \coth (a+b x))-\frac {1}{6} i x^3 \log \left (1-\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )+\frac {1}{6} i x^3 \log \left (1-\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )-\frac {1}{2} i \int x^2 \log \left (1+\frac {(-i-c-d) e^{2 a+2 b x}}{i+c-d}\right ) \, dx+\frac {1}{2} i \int x^2 \log \left (1+\frac {(-i+c+d) e^{2 a+2 b x}}{i-c+d}\right ) \, dx \\ & = \frac {1}{3} x^3 \cot ^{-1}(c+d \coth (a+b x))-\frac {1}{6} i x^3 \log \left (1-\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )+\frac {1}{6} i x^3 \log \left (1-\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )-\frac {i x^2 \operatorname {PolyLog}\left (2,\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{4 b}+\frac {i x^2 \operatorname {PolyLog}\left (2,\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{4 b}-\frac {i \int x \operatorname {PolyLog}\left (2,-\frac {(-i-c-d) e^{2 a+2 b x}}{i+c-d}\right ) \, dx}{2 b}+\frac {i \int x \operatorname {PolyLog}\left (2,-\frac {(-i+c+d) e^{2 a+2 b x}}{i-c+d}\right ) \, dx}{2 b} \\ & = \frac {1}{3} x^3 \cot ^{-1}(c+d \coth (a+b x))-\frac {1}{6} i x^3 \log \left (1-\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )+\frac {1}{6} i x^3 \log \left (1-\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )-\frac {i x^2 \operatorname {PolyLog}\left (2,\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{4 b}+\frac {i x^2 \operatorname {PolyLog}\left (2,\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{4 b}+\frac {i x \operatorname {PolyLog}\left (3,\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{4 b^2}-\frac {i x \operatorname {PolyLog}\left (3,\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{4 b^2}+\frac {i \int \operatorname {PolyLog}\left (3,-\frac {(-i-c-d) e^{2 a+2 b x}}{i+c-d}\right ) \, dx}{4 b^2}-\frac {i \int \operatorname {PolyLog}\left (3,-\frac {(-i+c+d) e^{2 a+2 b x}}{i-c+d}\right ) \, dx}{4 b^2} \\ & = \frac {1}{3} x^3 \cot ^{-1}(c+d \coth (a+b x))-\frac {1}{6} i x^3 \log \left (1-\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )+\frac {1}{6} i x^3 \log \left (1-\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )-\frac {i x^2 \operatorname {PolyLog}\left (2,\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{4 b}+\frac {i x^2 \operatorname {PolyLog}\left (2,\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{4 b}+\frac {i x \operatorname {PolyLog}\left (3,\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{4 b^2}-\frac {i x \operatorname {PolyLog}\left (3,\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{4 b^2}-\frac {i \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (3,\frac {(-i+c+d) x}{-i+c-d}\right )}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^3}+\frac {i \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (3,\frac {(i+c+d) x}{i+c-d}\right )}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^3} \\ & = \frac {1}{3} x^3 \cot ^{-1}(c+d \coth (a+b x))-\frac {1}{6} i x^3 \log \left (1-\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )+\frac {1}{6} i x^3 \log \left (1-\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )-\frac {i x^2 \operatorname {PolyLog}\left (2,\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{4 b}+\frac {i x^2 \operatorname {PolyLog}\left (2,\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{4 b}+\frac {i x \operatorname {PolyLog}\left (3,\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{4 b^2}-\frac {i x \operatorname {PolyLog}\left (3,\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{4 b^2}-\frac {i \operatorname {PolyLog}\left (4,\frac {(i-c-d) e^{2 a+2 b x}}{i-c+d}\right )}{8 b^3}+\frac {i \operatorname {PolyLog}\left (4,\frac {(i+c+d) e^{2 a+2 b x}}{i+c-d}\right )}{8 b^3} \\ \end{align*}
Time = 0.36 (sec) , antiderivative size = 442, normalized size of antiderivative = 1.26 \[ \int x^2 \cot ^{-1}(c+d \coth (a+b x)) \, dx=\frac {1}{3} x^3 \cot ^{-1}(c+d \coth (a+b x))-\frac {d \left (4 b^3 x^3 \log \left (1-\frac {\left (1+(c+d)^2\right ) e^{2 (a+b x)}}{1+c^2-d^2+2 \sqrt {-d^2}}\right )-4 b^3 x^3 \log \left (1+\frac {\left (1+(c+d)^2\right ) e^{2 (a+b x)}}{-1-c^2+d^2+2 \sqrt {-d^2}}\right )+6 b^2 x^2 \operatorname {PolyLog}\left (2,\frac {\left (1+c^2+2 c d+d^2\right ) e^{2 (a+b x)}}{1+c^2-d^2+2 \sqrt {-d^2}}\right )-6 b^2 x^2 \operatorname {PolyLog}\left (2,-\frac {\left (1+(c+d)^2\right ) e^{2 (a+b x)}}{-1-c^2+d^2+2 \sqrt {-d^2}}\right )-6 b x \operatorname {PolyLog}\left (3,\frac {\left (1+c^2+2 c d+d^2\right ) e^{2 (a+b x)}}{1+c^2-d^2+2 \sqrt {-d^2}}\right )+6 b x \operatorname {PolyLog}\left (3,-\frac {\left (1+(c+d)^2\right ) e^{2 (a+b x)}}{-1-c^2+d^2+2 \sqrt {-d^2}}\right )-3 \operatorname {PolyLog}\left (4,\frac {\left (1+c^2+2 c d+d^2\right ) e^{2 (a+b x)}}{1+c^2-d^2-2 \sqrt {-d^2}}\right )+3 \operatorname {PolyLog}\left (4,\frac {\left (1+c^2+2 c d+d^2\right ) e^{2 (a+b x)}}{1+c^2-d^2+2 \sqrt {-d^2}}\right )\right )}{24 b^3 \sqrt {-d^2}} \]
[In]
[Out]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 39.01 (sec) , antiderivative size = 6844, normalized size of antiderivative = 19.50
[In]
[Out]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1269 vs. \(2 (259) = 518\).
Time = 0.36 (sec) , antiderivative size = 1269, normalized size of antiderivative = 3.62 \[ \int x^2 \cot ^{-1}(c+d \coth (a+b x)) \, dx=\text {Too large to display} \]
[In]
[Out]
Timed out. \[ \int x^2 \cot ^{-1}(c+d \coth (a+b x)) \, dx=\text {Timed out} \]
[In]
[Out]
\[ \int x^2 \cot ^{-1}(c+d \coth (a+b x)) \, dx=\int { x^{2} \operatorname {arccot}\left (d \coth \left (b x + a\right ) + c\right ) \,d x } \]
[In]
[Out]
\[ \int x^2 \cot ^{-1}(c+d \coth (a+b x)) \, dx=\int { x^{2} \operatorname {arccot}\left (d \coth \left (b x + a\right ) + c\right ) \,d x } \]
[In]
[Out]
Timed out. \[ \int x^2 \cot ^{-1}(c+d \coth (a+b x)) \, dx=\int x^2\,\mathrm {acot}\left (c+d\,\mathrm {coth}\left (a+b\,x\right )\right ) \,d x \]
[In]
[Out]