Integrand size = 18, antiderivative size = 133 \[ \int x \text {arctanh}(1-i d+d \tan (a+b x)) \, dx=\frac {1}{6} i b x^3+\frac {1}{2} x^2 \text {arctanh}(1-i d+d \tan (a+b x))-\frac {1}{4} x^2 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+\frac {i x \operatorname {PolyLog}\left (2,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )}{4 b}-\frac {\operatorname {PolyLog}\left (3,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )}{8 b^2} \] Output:
1/6*I*b*x^3+1/2*x^2*arctanh(1-I*d+d*tan(b*x+a))-1/4*x^2*ln(1+(1-I*d)*exp(2 *I*a+2*I*b*x))+1/4*I*x*polylog(2,-(1-I*d)*exp(2*I*a+2*I*b*x))/b-1/8*polylo g(3,-(1-I*d)*exp(2*I*a+2*I*b*x))/b^2
Time = 0.17 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.89 \[ \int x \text {arctanh}(1-i d+d \tan (a+b x)) \, dx=\frac {1}{2} x^2 \text {arctanh}(1-i d+d \tan (a+b x))-\frac {2 b^2 x^2 \log \left (1+\frac {i e^{-2 i (a+b x)}}{i+d}\right )+2 i b x \operatorname {PolyLog}\left (2,-\frac {i e^{-2 i (a+b x)}}{i+d}\right )+\operatorname {PolyLog}\left (3,-\frac {i e^{-2 i (a+b x)}}{i+d}\right )}{8 b^2} \] Input:
Integrate[x*ArcTanh[1 - I*d + d*Tan[a + b*x]],x]
Output:
(x^2*ArcTanh[1 - I*d + d*Tan[a + b*x]])/2 - (2*b^2*x^2*Log[1 + I/((I + d)* E^((2*I)*(a + b*x)))] + (2*I)*b*x*PolyLog[2, (-I)/((I + d)*E^((2*I)*(a + b *x)))] + PolyLog[3, (-I)/((I + d)*E^((2*I)*(a + b*x)))])/(8*b^2)
Time = 0.70 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.29, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6817, 2615, 2620, 3011, 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.
\(\displaystyle \int x \text {arctanh}(d \tan (a+b x)-i d+1) \, dx\) |
\(\Big \downarrow \) 6817 |
\(\displaystyle \frac {1}{2} i b \int \frac {x^2}{e^{2 i a+2 i b x} (1-i d)+1}dx+\frac {1}{2} x^2 \text {arctanh}(d \tan (a+b x)-i d+1)\) |
\(\Big \downarrow \) 2615 |
\(\displaystyle \frac {1}{2} i b \left (\frac {x^3}{3}-(1-i d) \int \frac {e^{2 i a+2 i b x} x^2}{e^{2 i a+2 i b x} (1-i d)+1}dx\right )+\frac {1}{2} x^2 \text {arctanh}(d \tan (a+b x)-i d+1)\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle \frac {1}{2} i b \left (\frac {x^3}{3}-(1-i d) \left (\frac {x^2 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )}{2 b (d+i)}-\frac {\int x \log \left (e^{2 i a+2 i b x} (1-i d)+1\right )dx}{b (d+i)}\right )\right )+\frac {1}{2} x^2 \text {arctanh}(d \tan (a+b x)-i d+1)\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle \frac {1}{2} i b \left (\frac {x^3}{3}-(1-i d) \left (\frac {x^2 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )}{2 b (d+i)}-\frac {\frac {i x \operatorname {PolyLog}\left (2,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )}{2 b}-\frac {i \int \operatorname {PolyLog}\left (2,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )dx}{2 b}}{b (d+i)}\right )\right )+\frac {1}{2} x^2 \text {arctanh}(d \tan (a+b x)-i d+1)\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {1}{2} i b \left (\frac {x^3}{3}-(1-i d) \left (\frac {x^2 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )}{2 b (d+i)}-\frac {\frac {i x \operatorname {PolyLog}\left (2,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )}{2 b}-\frac {\int e^{-2 i a-2 i b x} \operatorname {PolyLog}\left (2,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )de^{2 i a+2 i b x}}{4 b^2}}{b (d+i)}\right )\right )+\frac {1}{2} x^2 \text {arctanh}(d \tan (a+b x)-i d+1)\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {1}{2} x^2 \text {arctanh}(d \tan (a+b x)-i d+1)+\frac {1}{2} i b \left (\frac {x^3}{3}-(1-i d) \left (\frac {x^2 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )}{2 b (d+i)}-\frac {\frac {i x \operatorname {PolyLog}\left (2,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )}{2 b}-\frac {\operatorname {PolyLog}\left (3,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )}{4 b^2}}{b (d+i)}\right )\right )\) |
Input:
Int[x*ArcTanh[1 - I*d + d*Tan[a + b*x]],x]
Output:
(x^2*ArcTanh[1 - I*d + d*Tan[a + b*x]])/2 + (I/2)*b*(x^3/3 - (1 - I*d)*((x ^2*Log[1 + (1 - I*d)*E^((2*I)*a + (2*I)*b*x)])/(2*b*(I + d)) - (((I/2)*x*P olyLog[2, -((1 - I*d)*E^((2*I)*a + (2*I)*b*x))])/b - PolyLog[3, -((1 - I*d )*E^((2*I)*a + (2*I)*b*x))]/(4*b^2))/(b*(I + d))))
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_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_ .), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcTanh[c + d*Tan[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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 3.32 (sec) , antiderivative size = 2187, normalized size of antiderivative = 16.44
Input:
int(x*arctanh(1-I*d+d*tan(b*x+a)),x,method=_RETURNVERBOSE)
Output:
1/2/b^2*d*a^2/(I+d)*ln(1+I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))-1/4/b^2*d/(I+d )*ln(1-I*(I+d)*exp(2*I*(b*x+a)))*a^2-1/4*I/b^2/(I+d)*ln(1-I*(I+d)*exp(2*I* (b*x+a)))*a^2+1/2*I/b^2*a^2/(I+d)*ln(1-I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))+ 1/2*I/b^2*a^2/(I+d)*ln(1+I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))-1/4*I/b^2*a^2/ (I+d)*ln(I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d+I)+1/6*I*b*x^3-1/4/b^2*a^2* d/(I+d)*ln(I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d+I)+1/2/b^2*d*a^2/(I+d)*ln (1-I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))+1/4*I/b*d/(I+d)*polylog(2,I*(I+d)*ex p(2*I*(b*x+a)))*x+1/4*I/b^2*d/(I+d)*polylog(2,I*(I+d)*exp(2*I*(b*x+a)))*a+ 1/2*I/b*a/(I+d)*ln(1-I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))*x+1/2*I/b*a/(I+d)* ln(1+I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))*x+1/2/b*d*a/(I+d)*ln(1-I*exp(I*(b* x+a))*(-I*(I+d))^(1/2))*x+1/2/b*d*a/(I+d)*ln(1+I*exp(I*(b*x+a))*(-I*(I+d)) ^(1/2))*x-1/2/b*d/(I+d)*ln(1-I*(I+d)*exp(2*I*(b*x+a)))*a*x-1/2*I/b^2*d*a/( I+d)*dilog(1-I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))-1/2*I/b^2*d*a/(I+d)*dilog( 1+I*exp(I*(b*x+a))*(-I*(I+d))^(1/2))-1/2*I/b/(I+d)*ln(1-I*(I+d)*exp(2*I*(b *x+a)))*a*x-1/8*(2*I*Pi+2*ln(d)+I*Pi*csgn(I/(exp(2*I*(b*x+a))+1))*csgn(I/( exp(2*I*(b*x+a))+1)*exp(2*I*(b*x+a)))^2+I*Pi*csgn(1/(exp(2*I*(b*x+a))+1)*e xp(2*I*(b*x+a))*d)^3+I*Pi*csgn(I*(I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d+I) /(exp(2*I*(b*x+a))+1))*csgn((I*exp(2*I*(b*x+a))+exp(2*I*(b*x+a))*d+I)/(exp (2*I*(b*x+a))+1))-I*Pi*csgn(I/(exp(2*I*(b*x+a))+1))*csgn(I*(I*exp(2*I*(b*x +a))+exp(2*I*(b*x+a))*d+I)/(exp(2*I*(b*x+a))+1))^2-I*Pi*csgn(1/(exp(2*I...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 293 vs. \(2 (93) = 186\).
Time = 0.10 (sec) , antiderivative size = 293, normalized size of antiderivative = 2.20 \[ \int x \text {arctanh}(1-i d+d \tan (a+b x)) \, dx=\frac {2 i \, b^{3} x^{3} + 3 \, b^{2} x^{2} \log \left (-\frac {{\left ({\left (d + i\right )} e^{\left (2 i \, b x + 2 i \, a\right )} + i\right )} e^{\left (-2 i \, b x - 2 i \, a\right )}}{d}\right ) + 2 i \, a^{3} + 6 i \, b x {\rm Li}_2\left (\frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) + 6 i \, b x {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) - 3 \, a^{2} \log \left (\frac {2 \, {\left (d + i\right )} e^{\left (i \, b x + i \, a\right )} + i \, \sqrt {4 i \, d - 4}}{2 \, {\left (d + i\right )}}\right ) - 3 \, a^{2} \log \left (\frac {2 \, {\left (d + i\right )} e^{\left (i \, b x + i \, a\right )} - i \, \sqrt {4 i \, d - 4}}{2 \, {\left (d + i\right )}}\right ) - 3 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (\frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) - 3 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (-\frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) - 6 \, {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) - 6 \, {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right )}{12 \, b^{2}} \] Input:
integrate(x*arctanh(1-I*d+d*tan(b*x+a)),x, algorithm="fricas")
Output:
1/12*(2*I*b^3*x^3 + 3*b^2*x^2*log(-((d + I)*e^(2*I*b*x + 2*I*a) + I)*e^(-2 *I*b*x - 2*I*a)/d) + 2*I*a^3 + 6*I*b*x*dilog(1/2*sqrt(4*I*d - 4)*e^(I*b*x + I*a)) + 6*I*b*x*dilog(-1/2*sqrt(4*I*d - 4)*e^(I*b*x + I*a)) - 3*a^2*log( 1/2*(2*(d + I)*e^(I*b*x + I*a) + I*sqrt(4*I*d - 4))/(d + I)) - 3*a^2*log(1 /2*(2*(d + I)*e^(I*b*x + I*a) - I*sqrt(4*I*d - 4))/(d + I)) - 3*(b^2*x^2 - a^2)*log(1/2*sqrt(4*I*d - 4)*e^(I*b*x + I*a) + 1) - 3*(b^2*x^2 - a^2)*log (-1/2*sqrt(4*I*d - 4)*e^(I*b*x + I*a) + 1) - 6*polylog(3, 1/2*sqrt(4*I*d - 4)*e^(I*b*x + I*a)) - 6*polylog(3, -1/2*sqrt(4*I*d - 4)*e^(I*b*x + I*a))) /b^2
\[ \int x \text {arctanh}(1-i d+d \tan (a+b x)) \, dx=\int x \operatorname {atanh}{\left (d \tan {\left (a + b x \right )} - i d + 1 \right )}\, dx \] Input:
integrate(x*atanh(1-I*d+d*tan(b*x+a)),x)
Output:
Integral(x*atanh(d*tan(a + b*x) - I*d + 1), x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 248 vs. \(2 (93) = 186\).
Time = 0.05 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.86 \[ \int x \text {arctanh}(1-i d+d \tan (a+b x)) \, dx=\frac {\frac {12 \, {\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a\right )} \operatorname {artanh}\left (d \tan \left (b x + a\right ) - i \, d + 1\right )}{b} - \frac {-4 i \, {\left (b x + a\right )}^{3} + 12 i \, {\left (b x + a\right )}^{2} a - 6 i \, b x {\rm Li}_2\left ({\left (i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}\right ) - 6 \, {\left (-i \, {\left (b x + a\right )}^{2} + 2 i \, {\left (b x + a\right )} a\right )} \arctan \left (-d \cos \left (2 \, b x + 2 \, a\right ) + \sin \left (2 \, b x + 2 \, a\right ), d \sin \left (2 \, b x + 2 \, a\right ) + \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \, {\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a\right )} \log \left ({\left (d^{2} + 1\right )} \cos \left (2 \, b x + 2 \, a\right )^{2} + {\left (d^{2} + 1\right )} \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, d \sin \left (2 \, b x + 2 \, a\right ) + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \, {\rm Li}_{3}({\left (i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )})}{b}}{24 \, b} \] Input:
integrate(x*arctanh(1-I*d+d*tan(b*x+a)),x, algorithm="maxima")
Output:
1/24*(12*((b*x + a)^2 - 2*(b*x + a)*a)*arctanh(d*tan(b*x + a) - I*d + 1)/b - (-4*I*(b*x + a)^3 + 12*I*(b*x + a)^2*a - 6*I*b*x*dilog((I*d - 1)*e^(2*I *b*x + 2*I*a)) - 6*(-I*(b*x + a)^2 + 2*I*(b*x + a)*a)*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 *((b*x + a)^2 - 2*(b*x + a)*a)*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* polylog(3, (I*d - 1)*e^(2*I*b*x + 2*I*a)))/b)/b
\[ \int x \text {arctanh}(1-i d+d \tan (a+b x)) \, dx=\int { x \operatorname {artanh}\left (d \tan \left (b x + a\right ) - i \, d + 1\right ) \,d x } \] Input:
integrate(x*arctanh(1-I*d+d*tan(b*x+a)),x, algorithm="giac")
Output:
integrate(x*arctanh(d*tan(b*x + a) - I*d + 1), x)
Timed out. \[ \int x \text {arctanh}(1-i d+d \tan (a+b x)) \, dx=\int x\,\mathrm {atanh}\left (d\,\mathrm {tan}\left (a+b\,x\right )+1-d\,1{}\mathrm {i}\right ) \,d x \] Input:
int(x*atanh(d*tan(a + b*x) - d*1i + 1),x)
Output:
int(x*atanh(d*tan(a + b*x) - d*1i + 1), x)
\[ \int x \text {arctanh}(1-i d+d \tan (a+b x)) \, dx=\int \mathit {atanh} \left (\tan \left (b x +a \right ) d -d i +1\right ) x d x \] Input:
int(x*atanh(1-I*d+d*tan(b*x+a)),x)
Output:
int(atanh(tan(a + b*x)*d - d*i + 1)*x,x)