Integrand size = 15, antiderivative size = 229 \[ \int (e+f x)^2 \cot ^{-1}(\coth (a+b x)) \, dx=\frac {(e+f x)^3 \cot ^{-1}(\coth (a+b x))}{3 f}-\frac {(e+f x)^3 \arctan \left (e^{2 a+2 b x}\right )}{3 f}+\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{2 a+2 b x}\right )}{4 b}-\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{2 a+2 b x}\right )}{4 b}-\frac {i f (e+f x) \operatorname {PolyLog}\left (3,-i e^{2 a+2 b x}\right )}{4 b^2}+\frac {i f (e+f x) \operatorname {PolyLog}\left (3,i e^{2 a+2 b x}\right )}{4 b^2}+\frac {i f^2 \operatorname {PolyLog}\left (4,-i e^{2 a+2 b x}\right )}{8 b^3}-\frac {i f^2 \operatorname {PolyLog}\left (4,i e^{2 a+2 b x}\right )}{8 b^3} \]
1/3*(f*x+e)^3*arccot(coth(b*x+a))/f-1/3*(f*x+e)^3*arctan(exp(2*b*x+2*a))/f +1/4*I*(f*x+e)^2*polylog(2,-I*exp(2*b*x+2*a))/b-1/4*I*(f*x+e)^2*polylog(2, I*exp(2*b*x+2*a))/b-1/4*I*f*(f*x+e)*polylog(3,-I*exp(2*b*x+2*a))/b^2+1/4*I *f*(f*x+e)*polylog(3,I*exp(2*b*x+2*a))/b^2+1/8*I*f^2*polylog(4,-I*exp(2*b* x+2*a))/b^3-1/8*I*f^2*polylog(4,I*exp(2*b*x+2*a))/b^3
Time = 0.13 (sec) , antiderivative size = 375, normalized size of antiderivative = 1.64 \[ \int (e+f x)^2 \cot ^{-1}(\coth (a+b x)) \, dx=\frac {1}{3} x \left (3 e^2+3 e f x+f^2 x^2\right ) \cot ^{-1}(\coth (a+b x))-\frac {i \left (12 b^3 e^2 x \log \left (1-i e^{2 (a+b x)}\right )+12 b^3 e f x^2 \log \left (1-i e^{2 (a+b x)}\right )+4 b^3 f^2 x^3 \log \left (1-i e^{2 (a+b x)}\right )-12 b^3 e^2 x \log \left (1+i e^{2 (a+b x)}\right )-12 b^3 e f x^2 \log \left (1+i e^{2 (a+b x)}\right )-4 b^3 f^2 x^3 \log \left (1+i e^{2 (a+b x)}\right )-6 b^2 (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{2 (a+b x)}\right )+6 b^2 (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{2 (a+b x)}\right )+6 b e f \operatorname {PolyLog}\left (3,-i e^{2 (a+b x)}\right )+6 b f^2 x \operatorname {PolyLog}\left (3,-i e^{2 (a+b x)}\right )-6 b e f \operatorname {PolyLog}\left (3,i e^{2 (a+b x)}\right )-6 b f^2 x \operatorname {PolyLog}\left (3,i e^{2 (a+b x)}\right )-3 f^2 \operatorname {PolyLog}\left (4,-i e^{2 (a+b x)}\right )+3 f^2 \operatorname {PolyLog}\left (4,i e^{2 (a+b x)}\right )\right )}{24 b^3} \]
(x*(3*e^2 + 3*e*f*x + f^2*x^2)*ArcCot[Coth[a + b*x]])/3 - ((I/24)*(12*b^3* e^2*x*Log[1 - I*E^(2*(a + b*x))] + 12*b^3*e*f*x^2*Log[1 - I*E^(2*(a + b*x) )] + 4*b^3*f^2*x^3*Log[1 - I*E^(2*(a + b*x))] - 12*b^3*e^2*x*Log[1 + I*E^( 2*(a + b*x))] - 12*b^3*e*f*x^2*Log[1 + I*E^(2*(a + b*x))] - 4*b^3*f^2*x^3* Log[1 + I*E^(2*(a + b*x))] - 6*b^2*(e + f*x)^2*PolyLog[2, (-I)*E^(2*(a + b *x))] + 6*b^2*(e + f*x)^2*PolyLog[2, I*E^(2*(a + b*x))] + 6*b*e*f*PolyLog[ 3, (-I)*E^(2*(a + b*x))] + 6*b*f^2*x*PolyLog[3, (-I)*E^(2*(a + b*x))] - 6* b*e*f*PolyLog[3, I*E^(2*(a + b*x))] - 6*b*f^2*x*PolyLog[3, I*E^(2*(a + b*x ))] - 3*f^2*PolyLog[4, (-I)*E^(2*(a + b*x))] + 3*f^2*PolyLog[4, I*E^(2*(a + b*x))]))/b^3
Time = 0.83 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {5709, 3042, 4668, 3011, 7163, 2720, 7143}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int (e+f x)^2 \cot ^{-1}(\coth (a+b x)) \, dx\) |
\(\Big \downarrow \) 5709 |
\(\displaystyle \frac {(e+f x)^3 \cot ^{-1}(\coth (a+b x))}{3 f}-\frac {b \int (e+f x)^3 \text {sech}(2 a+2 b x)dx}{3 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(e+f x)^3 \cot ^{-1}(\coth (a+b x))}{3 f}-\frac {b \int (e+f x)^3 \csc \left (2 i a+2 i b x+\frac {\pi }{2}\right )dx}{3 f}\) |
\(\Big \downarrow \) 4668 |
\(\displaystyle \frac {(e+f x)^3 \cot ^{-1}(\coth (a+b x))}{3 f}-\frac {b \left (-\frac {3 i f \int (e+f x)^2 \log \left (1-i e^{2 a+2 b x}\right )dx}{2 b}+\frac {3 i f \int (e+f x)^2 \log \left (1+i e^{2 a+2 b x}\right )dx}{2 b}+\frac {(e+f x)^3 \arctan \left (e^{2 a+2 b x}\right )}{b}\right )}{3 f}\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle \frac {(e+f x)^3 \cot ^{-1}(\coth (a+b x))}{3 f}-\frac {b \left (\frac {3 i f \left (\frac {f \int (e+f x) \operatorname {PolyLog}\left (2,-i e^{2 a+2 b x}\right )dx}{b}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{2 a+2 b x}\right )}{2 b}\right )}{2 b}-\frac {3 i f \left (\frac {f \int (e+f x) \operatorname {PolyLog}\left (2,i e^{2 a+2 b x}\right )dx}{b}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,i e^{2 a+2 b x}\right )}{2 b}\right )}{2 b}+\frac {(e+f x)^3 \arctan \left (e^{2 a+2 b x}\right )}{b}\right )}{3 f}\) |
\(\Big \downarrow \) 7163 |
\(\displaystyle \frac {(e+f x)^3 \cot ^{-1}(\coth (a+b x))}{3 f}-\frac {b \left (\frac {3 i f \left (\frac {f \left (\frac {(e+f x) \operatorname {PolyLog}\left (3,-i e^{2 a+2 b x}\right )}{2 b}-\frac {f \int \operatorname {PolyLog}\left (3,-i e^{2 a+2 b x}\right )dx}{2 b}\right )}{b}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{2 a+2 b x}\right )}{2 b}\right )}{2 b}-\frac {3 i f \left (\frac {f \left (\frac {(e+f x) \operatorname {PolyLog}\left (3,i e^{2 a+2 b x}\right )}{2 b}-\frac {f \int \operatorname {PolyLog}\left (3,i e^{2 a+2 b x}\right )dx}{2 b}\right )}{b}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,i e^{2 a+2 b x}\right )}{2 b}\right )}{2 b}+\frac {(e+f x)^3 \arctan \left (e^{2 a+2 b x}\right )}{b}\right )}{3 f}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {(e+f x)^3 \cot ^{-1}(\coth (a+b x))}{3 f}-\frac {b \left (\frac {3 i f \left (\frac {f \left (\frac {(e+f x) \operatorname {PolyLog}\left (3,-i e^{2 a+2 b x}\right )}{2 b}-\frac {f \int e^{-2 a-2 b x} \operatorname {PolyLog}\left (3,-i e^{2 a+2 b x}\right )de^{2 a+2 b x}}{4 b^2}\right )}{b}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{2 a+2 b x}\right )}{2 b}\right )}{2 b}-\frac {3 i f \left (\frac {f \left (\frac {(e+f x) \operatorname {PolyLog}\left (3,i e^{2 a+2 b x}\right )}{2 b}-\frac {f \int e^{-2 a-2 b x} \operatorname {PolyLog}\left (3,i e^{2 a+2 b x}\right )de^{2 a+2 b x}}{4 b^2}\right )}{b}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,i e^{2 a+2 b x}\right )}{2 b}\right )}{2 b}+\frac {(e+f x)^3 \arctan \left (e^{2 a+2 b x}\right )}{b}\right )}{3 f}\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {(e+f x)^3 \cot ^{-1}(\coth (a+b x))}{3 f}-\frac {b \left (\frac {(e+f x)^3 \arctan \left (e^{2 a+2 b x}\right )}{b}+\frac {3 i f \left (\frac {f \left (\frac {(e+f x) \operatorname {PolyLog}\left (3,-i e^{2 a+2 b x}\right )}{2 b}-\frac {f \operatorname {PolyLog}\left (4,-i e^{2 a+2 b x}\right )}{4 b^2}\right )}{b}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{2 a+2 b x}\right )}{2 b}\right )}{2 b}-\frac {3 i f \left (\frac {f \left (\frac {(e+f x) \operatorname {PolyLog}\left (3,i e^{2 a+2 b x}\right )}{2 b}-\frac {f \operatorname {PolyLog}\left (4,i e^{2 a+2 b x}\right )}{4 b^2}\right )}{b}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,i e^{2 a+2 b x}\right )}{2 b}\right )}{2 b}\right )}{3 f}\) |
((e + f*x)^3*ArcCot[Coth[a + b*x]])/(3*f) - (b*(((e + f*x)^3*ArcTan[E^(2*a + 2*b*x)])/b + (((3*I)/2)*f*(-1/2*((e + f*x)^2*PolyLog[2, (-I)*E^(2*a + 2 *b*x)])/b + (f*(((e + f*x)*PolyLog[3, (-I)*E^(2*a + 2*b*x)])/(2*b) - (f*Po lyLog[4, (-I)*E^(2*a + 2*b*x)])/(4*b^2)))/b))/b - (((3*I)/2)*f*(-1/2*((e + f*x)^2*PolyLog[2, I*E^(2*a + 2*b*x)])/b + (f*(((e + f*x)*PolyLog[3, I*E^( 2*a + 2*b*x)])/(2*b) - (f*PolyLog[4, I*E^(2*a + 2*b*x)])/(4*b^2)))/b))/b)) /(3*f)
3.3.1.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) *(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F])) Int[(f + g*x)^( m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e , f, g, n}, x] && GtQ[m, 0]
Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_ ))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^((-I)*e + f*fz*x)/E^( I*k*Pi)]/(f*fz*I)), x] + (-Simp[d*(m/(f*fz*I)) Int[(c + d*x)^(m - 1)*Log[ 1 - E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x] + Simp[d*(m/(f*fz*I)) Int[(c + d*x)^(m - 1)*Log[1 + E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c , d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]
Int[ArcCot[Coth[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcCot[Coth[a + b*x]]/(f*(m + 1))), x] - Simp[b/ (f*(m + 1)) Int[(e + f*x)^(m + 1)*Sech[2*a + 2*b*x], x], x] /; FreeQ[{a, b, e, f}, x] && IGtQ[m, 0]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d , e, n, p}, x] && EqQ[b*d, a*e]
Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_. )*(x_))))^(p_.)], x_Symbol] :> Simp[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a + b*x)))^p]/(b*c*p*Log[F])), x] - Simp[f*(m/(b*c*p*Log[F])) Int[(e + f*x) ^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c , d, e, f, n, p}, x] && GtQ[m, 0]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 21.01 (sec) , antiderivative size = 2668, normalized size of antiderivative = 11.65
-1/2*I*f^2/b^3*a^2*dilog(((-I)^(1/2)-exp(b*x+a))/(-I)^(1/2))-1/2*I*f^2/b^3 *a^2*dilog(((-I)^(1/2)+exp(b*x+a))/(-I)^(1/2))+1/4*I*f/b^2*e*polylog(3,I*e xp(2*b*x+2*a))+1/6*I*f^2/b^3*a^3*ln(exp(2*b*x+2*a)+I)+1/3*I*f^2/b^3*ln(1-I *exp(2*b*x+2*a))*a^3-1/4*I*f^2/b*polylog(2,I*exp(2*b*x+2*a))*x^2+1/2*I/b*a *e^2*ln(exp(2*b*x+2*a)+I)-1/2*I/b*e^2*ln(((-I)^(1/2)-exp(b*x+a))/(-I)^(1/2 ))*a-1/2*I/b*e^2*ln(((-I)^(1/2)+exp(b*x+a))/(-I)^(1/2))*a+1/8*I*f^2*polylo g(4,-I*exp(2*b*x+2*a))/b^3-1/8*I*f^2*polylog(4,I*exp(2*b*x+2*a))/b^3-1/6*I *f^2/b^3*a^3*ln(-exp(2*b*x+2*a)+I)+1/2*I*f*e*ln(1+I*exp(2*b*x+2*a))*x^2-1/ 3*I*f^2/b^3*ln(1+I*exp(2*b*x+2*a))*a^3+1/4*I*f^2/b*polylog(2,-I*exp(2*b*x+ 2*a))*x^2-1/4*I*f^2/b^3*polylog(2,-I*exp(2*b*x+2*a))*a^2-1/4*I*f^2/b^2*pol ylog(3,-I*exp(2*b*x+2*a))*x+1/2*I*f^2/b^3*a^3*ln(1+exp(b*x+a)*(-1)^(3/4))+ 1/2*I*f^2/b^3*a^3*ln(1-exp(b*x+a)*(-1)^(3/4))+1/2*I*f^2/b^3*a^2*dilog(1+ex p(b*x+a)*(-1)^(3/4))+1/2*I*f^2/b^3*a^2*dilog(1-exp(b*x+a)*(-1)^(3/4))-1/4* I*f/b^2*e*polylog(3,-I*exp(2*b*x+2*a))+1/2*I/b*e^2*ln(1+exp(b*x+a)*(-1)^(3 /4))*a+1/2*I/b*e^2*ln(1-exp(b*x+a)*(-1)^(3/4))*a-1/2*I/b*a*e^2*ln(-exp(2*b *x+2*a)+I)-1/2*I*f*ln(exp(2*b*x+2*a)-I)*x^2*e-1/2*I*f*e*ln(1-I*exp(2*b*x+2 *a))*x^2+1/4*I*f^2/b^3*polylog(2,I*exp(2*b*x+2*a))*a^2+1/4*I*f^2/b^2*polyl og(3,I*exp(2*b*x+2*a))*x-1/2*I*f^2/b^3*a^3*ln(((-I)^(1/2)-exp(b*x+a))/(-I) ^(1/2))-1/2*I*f^2/b^3*a^3*ln(((-I)^(1/2)+exp(b*x+a))/(-I)^(1/2))+1/2*I*f/b ^2*e*ln(1+I*exp(2*b*x+2*a))*a^2+1/2*I*f/b*e*polylog(2,-I*exp(2*b*x+2*a)...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1002 vs. \(2 (180) = 360\).
Time = 0.33 (sec) , antiderivative size = 1002, normalized size of antiderivative = 4.38 \[ \int (e+f x)^2 \cot ^{-1}(\coth (a+b x)) \, dx=\text {Too large to display} \]
1/6*(-6*I*f^2*polylog(4, 1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a))) - 6*I*f^2*polylog(4, -1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a))) + 6*I*f ^2*polylog(4, 1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a))) + 6*I*f^2*po lylog(4, -1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a))) + 2*(b^3*f^2*x^3 + 3*b^3*e*f*x^2 + 3*b^3*e^2*x)*arctan(sinh(b*x + a)/cosh(b*x + a)) - 3*(I *b^2*f^2*x^2 + 2*I*b^2*e*f*x + I*b^2*e^2)*dilog(1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a))) - 3*(I*b^2*f^2*x^2 + 2*I*b^2*e*f*x + I*b^2*e^2)*dilog (-1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a))) - 3*(-I*b^2*f^2*x^2 - 2*I *b^2*e*f*x - I*b^2*e^2)*dilog(1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a ))) - 3*(-I*b^2*f^2*x^2 - 2*I*b^2*e*f*x - I*b^2*e^2)*dilog(-1/2*sqrt(-4*I) *(cosh(b*x + a) + sinh(b*x + a))) + (-I*b^3*f^2*x^3 - 3*I*b^3*e*f*x^2 - 3* I*b^3*e^2*x - 3*I*a*b^2*e^2 + 3*I*a^2*b*e*f - I*a^3*f^2)*log(1/2*sqrt(4*I) *(cosh(b*x + a) + sinh(b*x + a)) + 1) + (-I*b^3*f^2*x^3 - 3*I*b^3*e*f*x^2 - 3*I*b^3*e^2*x - 3*I*a*b^2*e^2 + 3*I*a^2*b*e*f - I*a^3*f^2)*log(-1/2*sqrt (4*I)*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (I*b^3*f^2*x^3 + 3*I*b^3*e*f* x^2 + 3*I*b^3*e^2*x + 3*I*a*b^2*e^2 - 3*I*a^2*b*e*f + I*a^3*f^2)*log(1/2*s qrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (I*b^3*f^2*x^3 + 3*I*b^3* e*f*x^2 + 3*I*b^3*e^2*x + 3*I*a*b^2*e^2 - 3*I*a^2*b*e*f + I*a^3*f^2)*log(- 1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (3*I*a*b^2*e^2 - 3*I *a^2*b*e*f + I*a^3*f^2)*log(I*sqrt(4*I) + 2*cosh(b*x + a) + 2*sinh(b*x ...
\[ \int (e+f x)^2 \cot ^{-1}(\coth (a+b x)) \, dx=\int \left (e + f x\right )^{2} \operatorname {acot}{\left (\coth {\left (a + b x \right )} \right )}\, dx \]
\[ \int (e+f x)^2 \cot ^{-1}(\coth (a+b x)) \, dx=\int { {\left (f x + e\right )}^{2} \operatorname {arccot}\left (\coth \left (b x + a\right )\right ) \,d x } \]
1/3*(f^2*x^3 + 3*e*f*x^2 + 3*e^2*x)*arctan((e^(2*b*x + 2*a) - 1)/(e^(2*b*x + 2*a) + 1)) - integrate(2/3*(b*f^2*x^3*e^(2*a) + 3*b*e*f*x^2*e^(2*a) + 3 *b*e^2*x*e^(2*a))*e^(2*b*x)/(e^(4*b*x + 4*a) + 1), x)
\[ \int (e+f x)^2 \cot ^{-1}(\coth (a+b x)) \, dx=\int { {\left (f x + e\right )}^{2} \operatorname {arccot}\left (\coth \left (b x + a\right )\right ) \,d x } \]
Timed out. \[ \int (e+f x)^2 \cot ^{-1}(\coth (a+b x)) \, dx=\int \mathrm {acot}\left (\mathrm {coth}\left (a+b\,x\right )\right )\,{\left (e+f\,x\right )}^2 \,d x \]