Integrand size = 20, antiderivative size = 168 \[ \int x^2 \text {arctanh}(1+i d+d \cot (a+b x)) \, dx=\frac {1}{12} i b x^4+\frac {1}{3} x^3 \text {arctanh}(1+i d+d \cot (a+b x))-\frac {1}{6} x^3 \log \left (1-(1+i d) e^{2 i a+2 i b x}\right )+\frac {i x^2 \operatorname {PolyLog}\left (2,(1+i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac {x \operatorname {PolyLog}\left (3,(1+i d) e^{2 i a+2 i b x}\right )}{4 b^2}-\frac {i \operatorname {PolyLog}\left (4,(1+i d) e^{2 i a+2 i b x}\right )}{8 b^3} \]
1/12*I*b*x^4+1/3*x^3*arctanh(1+I*d+d*cot(b*x+a))-1/6*x^3*ln(1-(1+I*d)*exp( 2*I*a+2*I*b*x))+1/4*I*x^2*polylog(2,(1+I*d)*exp(2*I*a+2*I*b*x))/b-1/4*x*po lylog(3,(1+I*d)*exp(2*I*a+2*I*b*x))/b^2-1/8*I*polylog(4,(1+I*d)*exp(2*I*a+ 2*I*b*x))/b^3
Time = 0.28 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.92 \[ \int x^2 \text {arctanh}(1+i d+d \cot (a+b x)) \, dx=\frac {1}{3} x^3 \text {arctanh}(1+i d+d \cot (a+b x))-\frac {4 b^3 x^3 \log \left (1+\frac {i e^{-2 i (a+b x)}}{-i+d}\right )+6 i b^2 x^2 \operatorname {PolyLog}\left (2,-\frac {i e^{-2 i (a+b x)}}{-i+d}\right )+6 b x \operatorname {PolyLog}\left (3,-\frac {i e^{-2 i (a+b x)}}{-i+d}\right )-3 i \operatorname {PolyLog}\left (4,-\frac {i e^{-2 i (a+b x)}}{-i+d}\right )}{24 b^3} \]
(x^3*ArcTanh[1 + I*d + d*Cot[a + b*x]])/3 - (4*b^3*x^3*Log[1 + I/((-I + d) *E^((2*I)*(a + b*x)))] + (6*I)*b^2*x^2*PolyLog[2, (-I)/((-I + d)*E^((2*I)* (a + b*x)))] + 6*b*x*PolyLog[3, (-I)/((-I + d)*E^((2*I)*(a + b*x)))] - (3* I)*PolyLog[4, (-I)/((-I + d)*E^((2*I)*(a + b*x)))])/(24*b^3)
Time = 0.91 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.30, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {6819, 2615, 2620, 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.
\(\displaystyle \int x^2 \text {arctanh}(d \cot (a+b x)+i d+1) \, dx\) |
\(\Big \downarrow \) 6819 |
\(\displaystyle \frac {1}{3} i b \int \frac {x^3}{1-(i d+1) e^{2 i a+2 i b x}}dx+\frac {1}{3} x^3 \text {arctanh}(d \cot (a+b x)+i d+1)\) |
\(\Big \downarrow \) 2615 |
\(\displaystyle \frac {1}{3} i b \left (\frac {x^4}{4}+(1+i d) \int \frac {e^{2 i a+2 i b x} x^3}{1-(i d+1) e^{2 i a+2 i b x}}dx\right )+\frac {1}{3} x^3 \text {arctanh}(d \cot (a+b x)+i d+1)\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle \frac {1}{3} i b \left (\frac {x^4}{4}+(1+i d) \left (\frac {3 \int x^2 \log \left (1-(i d+1) e^{2 i a+2 i b x}\right )dx}{2 b (-d+i)}-\frac {x^3 \log \left (1-(1+i d) e^{2 i a+2 i b x}\right )}{2 b (-d+i)}\right )\right )+\frac {1}{3} x^3 \text {arctanh}(d \cot (a+b x)+i d+1)\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle \frac {1}{3} i b \left (\frac {x^4}{4}+(1+i d) \left (\frac {3 \left (\frac {i x^2 \operatorname {PolyLog}\left (2,(i d+1) e^{2 i a+2 i b x}\right )}{2 b}-\frac {i \int x \operatorname {PolyLog}\left (2,(i d+1) e^{2 i a+2 i b x}\right )dx}{b}\right )}{2 b (-d+i)}-\frac {x^3 \log \left (1-(1+i d) e^{2 i a+2 i b x}\right )}{2 b (-d+i)}\right )\right )+\frac {1}{3} x^3 \text {arctanh}(d \cot (a+b x)+i d+1)\) |
\(\Big \downarrow \) 7163 |
\(\displaystyle \frac {1}{3} i b \left (\frac {x^4}{4}+(1+i d) \left (\frac {3 \left (\frac {i x^2 \operatorname {PolyLog}\left (2,(i d+1) e^{2 i a+2 i b x}\right )}{2 b}-\frac {i \left (\frac {i \int \operatorname {PolyLog}\left (3,(i d+1) e^{2 i a+2 i b x}\right )dx}{2 b}-\frac {i x \operatorname {PolyLog}\left (3,(i d+1) e^{2 i a+2 i b x}\right )}{2 b}\right )}{b}\right )}{2 b (-d+i)}-\frac {x^3 \log \left (1-(1+i d) e^{2 i a+2 i b x}\right )}{2 b (-d+i)}\right )\right )+\frac {1}{3} x^3 \text {arctanh}(d \cot (a+b x)+i d+1)\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {1}{3} i b \left (\frac {x^4}{4}+(1+i d) \left (\frac {3 \left (\frac {i x^2 \operatorname {PolyLog}\left (2,(i d+1) e^{2 i a+2 i b x}\right )}{2 b}-\frac {i \left (\frac {\int e^{-2 i a-2 i b x} \operatorname {PolyLog}\left (3,(i d+1) e^{2 i a+2 i b x}\right )de^{2 i a+2 i b x}}{4 b^2}-\frac {i x \operatorname {PolyLog}\left (3,(i d+1) e^{2 i a+2 i b x}\right )}{2 b}\right )}{b}\right )}{2 b (-d+i)}-\frac {x^3 \log \left (1-(1+i d) e^{2 i a+2 i b x}\right )}{2 b (-d+i)}\right )\right )+\frac {1}{3} x^3 \text {arctanh}(d \cot (a+b x)+i d+1)\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {1}{3} x^3 \text {arctanh}(d \cot (a+b x)+i d+1)+\frac {1}{3} i b \left (\frac {x^4}{4}+(1+i d) \left (\frac {3 \left (\frac {i x^2 \operatorname {PolyLog}\left (2,(i d+1) e^{2 i a+2 i b x}\right )}{2 b}-\frac {i \left (\frac {\operatorname {PolyLog}\left (4,(i d+1) e^{2 i a+2 i b x}\right )}{4 b^2}-\frac {i x \operatorname {PolyLog}\left (3,(i d+1) e^{2 i a+2 i b x}\right )}{2 b}\right )}{b}\right )}{2 b (-d+i)}-\frac {x^3 \log \left (1-(1+i d) e^{2 i a+2 i b x}\right )}{2 b (-d+i)}\right )\right )\) |
(x^3*ArcTanh[1 + I*d + d*Cot[a + b*x]])/3 + (I/3)*b*(x^4/4 + (1 + I*d)*(-1 /2*(x^3*Log[1 - (1 + I*d)*E^((2*I)*a + (2*I)*b*x)])/(b*(I - d)) + (3*(((I/ 2)*x^2*PolyLog[2, (1 + I*d)*E^((2*I)*a + (2*I)*b*x)])/b - (I*(((-1/2*I)*x* PolyLog[3, (1 + I*d)*E^((2*I)*a + (2*I)*b*x)])/b + PolyLog[4, (1 + I*d)*E^ ((2*I)*a + (2*I)*b*x)]/(4*b^2)))/b))/(2*b*(I - d))))
3.4.38.3.1 Defintions of rubi rules used
Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x _))))^(n_.)), x_Symbol] :> Simp[(c + d*x)^(m + 1)/(a*d*(m + 1)), x] - Simp[ b/a Int[(c + d*x)^m*((F^(g*(e + f*x)))^n/(a + b*(F^(g*(e + f*x)))^n)), x] , x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ ((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp [((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si mp[d*(m/(b*f*g*n*Log[F])) Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x )))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
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[ArcTanh[(c_.) + Cot[(a_.) + (b_.)*(x_)]*(d_.)]*((e_.) + (f_.)*(x_))^(m_ .), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcTanh[c + d*Cot[a + b*x]]/(f*(m + 1))), x] + Simp[I*(b/(f*(m + 1))) Int[(e + f*x)^(m + 1)/(c - I*d - c*E^ (2*I*a + 2*I*b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && EqQ[(c - I*d)^2, 1]
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 = 2.03 (sec) , antiderivative size = 2387, normalized size of antiderivative = 14.21
-1/2*I/b^3*a^3/(I-d)*ln(1+I*exp(I*(b*x+a))*(I*(I-d))^(1/2))+1/6*I/b^3*a^3/ (I-d)*ln(I*exp(2*I*(b*x+a))-exp(2*I*(b*x+a))*d-I)+1/12*I*b*x^4+1/2/b^3*d*a ^3/(I-d)*ln(1-I*exp(I*(b*x+a))*(I*(I-d))^(1/2))+1/2/b^3*d*a^3/(I-d)*ln(1+I *exp(I*(b*x+a))*(I*(I-d))^(1/2))+1/4/b^2*d/(I-d)*polylog(3,-I*(I-d)*exp(2* I*(b*x+a)))*x-1/6/b^3*a^3*d/(I-d)*ln(I*exp(2*I*(b*x+a))-exp(2*I*(b*x+a))*d -I)-1/3/b^3*d/(I-d)*ln(1+I*(I-d)*exp(2*I*(b*x+a)))*a^3+1/8*I/b^3*d/(I-d)*p olylog(4,-I*(I-d)*exp(2*I*(b*x+a)))+1/3*I/b^3/(I-d)*ln(1+I*(I-d)*exp(2*I*( b*x+a)))*a^3-1/4*I/b^2/(I-d)*polylog(3,-I*(I-d)*exp(2*I*(b*x+a)))*x-1/2*I/ b^3*a^3/(I-d)*ln(1-I*exp(I*(b*x+a))*(I*(I-d))^(1/2))+1/6*d/(I-d)*ln(1+I*(I -d)*exp(2*I*(b*x+a)))*x^3-1/4/b/(I-d)*polylog(2,-I*(I-d)*exp(2*I*(b*x+a))) *x^2+1/4/b^3/(I-d)*polylog(2,-I*(I-d)*exp(2*I*(b*x+a)))*a^2-1/2/b^3*a^2/(I -d)*dilog(1-I*exp(I*(b*x+a))*(I*(I-d))^(1/2))-1/2/b^3*a^2/(I-d)*dilog(1+I* exp(I*(b*x+a))*(I*(I-d))^(1/2))-1/6*I/(I-d)*ln(1+I*(I-d)*exp(2*I*(b*x+a))) *x^3-1/12*(-I*Pi*csgn(I*d/(exp(2*I*(b*x+a))-1)*exp(2*I*(b*x+a)))*csgn(1/(e xp(2*I*(b*x+a))-1)*exp(2*I*(b*x+a))*d)+I*Pi*csgn(I*(exp(2*I*(b*x+a))*d-I*e xp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))-1))*csgn((exp(2*I*(b*x+a))*d-I*exp(2* I*(b*x+a))+I)/(exp(2*I*(b*x+a))-1))+I*Pi*csgn(I*d/(exp(2*I*(b*x+a))-1)*exp (2*I*(b*x+a)))*csgn(1/(exp(2*I*(b*x+a))-1)*exp(2*I*(b*x+a))*d)^2+I*Pi*csgn (I*exp(2*I*(b*x+a)))*csgn(I/(exp(2*I*(b*x+a))-1)*exp(2*I*(b*x+a)))^2-2*I*P i-I*Pi*csgn((exp(2*I*(b*x+a))*d-I*exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))...
Time = 0.27 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.07 \[ \int x^2 \text {arctanh}(1+i d+d \cot (a+b x)) \, dx=\frac {2 i \, b^{4} x^{4} + 4 \, b^{3} x^{3} \log \left (-\frac {{\left ({\left (d - i\right )} e^{\left (2 i \, b x + 2 i \, a\right )} + i\right )} e^{\left (-2 i \, b x - 2 i \, a\right )}}{d}\right ) + 6 i \, b^{2} x^{2} {\rm Li}_2\left (-{\left (-i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}\right ) - 2 i \, a^{4} + 4 \, a^{3} \log \left (\frac {{\left (d - i\right )} e^{\left (2 i \, b x + 2 i \, a\right )} + i}{d - i}\right ) - 6 \, b x {\rm polylog}\left (3, {\left (i \, d + 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}\right ) - 4 \, {\left (b^{3} x^{3} + a^{3}\right )} \log \left ({\left (-i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )} + 1\right ) - 3 i \, {\rm polylog}\left (4, {\left (i \, d + 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}\right )}{24 \, b^{3}} \]
1/24*(2*I*b^4*x^4 + 4*b^3*x^3*log(-((d - I)*e^(2*I*b*x + 2*I*a) + I)*e^(-2 *I*b*x - 2*I*a)/d) + 6*I*b^2*x^2*dilog(-(-I*d - 1)*e^(2*I*b*x + 2*I*a)) - 2*I*a^4 + 4*a^3*log(((d - I)*e^(2*I*b*x + 2*I*a) + I)/(d - I)) - 6*b*x*pol ylog(3, (I*d + 1)*e^(2*I*b*x + 2*I*a)) - 4*(b^3*x^3 + a^3)*log((-I*d - 1)* e^(2*I*b*x + 2*I*a) + 1) - 3*I*polylog(4, (I*d + 1)*e^(2*I*b*x + 2*I*a)))/ b^3
\[ \int x^2 \text {arctanh}(1+i d+d \cot (a+b x)) \, dx=\int x^{2} \operatorname {atanh}{\left (d \cot {\left (a + b x \right )} + i d + 1 \right )}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 344 vs. \(2 (119) = 238\).
Time = 0.22 (sec) , antiderivative size = 344, normalized size of antiderivative = 2.05 \[ \int x^2 \text {arctanh}(1+i d+d \cot (a+b x)) \, dx=\frac {\frac {12 \, {\left ({\left (b x + a\right )}^{3} - 3 \, {\left (b x + a\right )}^{2} a + 3 \, {\left (b x + a\right )} a^{2}\right )} \operatorname {artanh}\left (d \cot \left (b x + a\right ) + i \, d + 1\right )}{b^{2}} - \frac {-3 i \, {\left (b x + a\right )}^{4} + 12 i \, {\left (b x + a\right )}^{3} a - 18 i \, {\left (b x + a\right )}^{2} a^{2} - 2 \, {\left (4 i \, {\left (b x + a\right )}^{3} - 9 i \, {\left (b x + a\right )}^{2} a + 9 i \, {\left (b x + a\right )} a^{2}\right )} \arctan \left (d \cos \left (2 \, b x + 2 \, a\right ) + \sin \left (2 \, b x + 2 \, a\right ), d \sin \left (2 \, b x + 2 \, a\right ) - \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - 3 \, {\left (4 i \, {\left (b x + a\right )}^{2} - 6 i \, {\left (b x + a\right )} a + 3 i \, a^{2}\right )} {\rm Li}_2\left ({\left (i \, d + 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + {\left (4 \, {\left (b x + a\right )}^{3} - 9 \, {\left (b x + a\right )}^{2} a + 9 \, {\left (b x + a\right )} a^{2}\right )} \log \left ({\left (d^{2} + 1\right )} \cos \left (2 \, b x + 2 \, a\right )^{2} + {\left (d^{2} + 1\right )} \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, d \sin \left (2 \, b x + 2 \, a\right ) - 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \, {\left (4 \, b x + a\right )} {\rm Li}_{3}({\left (i \, d + 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}) + 6 i \, {\rm Li}_{4}({\left (i \, d + 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )})}{b^{2}}}{36 \, b} \]
1/36*(12*((b*x + a)^3 - 3*(b*x + a)^2*a + 3*(b*x + a)*a^2)*arctanh(d*cot(b *x + a) + I*d + 1)/b^2 - (-3*I*(b*x + a)^4 + 12*I*(b*x + a)^3*a - 18*I*(b* x + a)^2*a^2 - 2*(4*I*(b*x + a)^3 - 9*I*(b*x + a)^2*a + 9*I*(b*x + a)*a^2) *arctan2(d*cos(2*b*x + 2*a) + sin(2*b*x + 2*a), d*sin(2*b*x + 2*a) - cos(2 *b*x + 2*a) + 1) - 3*(4*I*(b*x + a)^2 - 6*I*(b*x + a)*a + 3*I*a^2)*dilog(( I*d + 1)*e^(2*I*b*x + 2*I*a)) + (4*(b*x + a)^3 - 9*(b*x + a)^2*a + 9*(b*x + a)*a^2)*log((d^2 + 1)*cos(2*b*x + 2*a)^2 + (d^2 + 1)*sin(2*b*x + 2*a)^2 + 2*d*sin(2*b*x + 2*a) - 2*cos(2*b*x + 2*a) + 1) + 3*(4*b*x + a)*polylog(3 , (I*d + 1)*e^(2*I*b*x + 2*I*a)) + 6*I*polylog(4, (I*d + 1)*e^(2*I*b*x + 2 *I*a)))/b^2)/b
\[ \int x^2 \text {arctanh}(1+i d+d \cot (a+b x)) \, dx=\int { x^{2} \operatorname {artanh}\left (d \cot \left (b x + a\right ) + i \, d + 1\right ) \,d x } \]
Timed out. \[ \int x^2 \text {arctanh}(1+i d+d \cot (a+b x)) \, dx=\int x^2\,\mathrm {atanh}\left (d\,\mathrm {cot}\left (a+b\,x\right )+1+d\,1{}\mathrm {i}\right ) \,d x \]