Integrand size = 19, antiderivative size = 124 \[ \int x \arctan (c+(-1+i c) \tan (a+b x)) \, dx=\frac {b x^3}{6}+\frac {1}{2} x^2 \arctan (c-(1-i c) \tan (a+b x))+\frac {1}{4} i x^2 \log \left (1+i c e^{2 i a+2 i b x}\right )+\frac {x \operatorname {PolyLog}\left (2,-i c e^{2 i a+2 i b x}\right )}{4 b}+\frac {i \operatorname {PolyLog}\left (3,-i c e^{2 i a+2 i b x}\right )}{8 b^2} \]
1/6*b*x^3+1/2*x^2*arctan(c-(1-I*c)*tan(b*x+a))+1/4*I*x^2*ln(1+I*c*exp(2*I* a+2*I*b*x))+1/4*x*polylog(2,-I*c*exp(2*I*a+2*I*b*x))/b+1/8*I*polylog(3,-I* c*exp(2*I*a+2*I*b*x))/b^2
Time = 0.21 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.90 \[ \int x \arctan (c+(-1+i c) \tan (a+b x)) \, dx=\frac {1}{2} x^2 \arctan (c+i (i+c) \tan (a+b x))+\frac {i \left (2 b^2 x^2 \log \left (1-\frac {i e^{-2 i (a+b x)}}{c}\right )+2 i b x \operatorname {PolyLog}\left (2,\frac {i e^{-2 i (a+b x)}}{c}\right )+\operatorname {PolyLog}\left (3,\frac {i e^{-2 i (a+b x)}}{c}\right )\right )}{8 b^2} \]
(x^2*ArcTan[c + I*(I + c)*Tan[a + b*x]])/2 + ((I/8)*(2*b^2*x^2*Log[1 - I/( c*E^((2*I)*(a + b*x)))] + (2*I)*b*x*PolyLog[2, I/(c*E^((2*I)*(a + b*x)))] + PolyLog[3, I/(c*E^((2*I)*(a + b*x)))]))/b^2
Time = 0.65 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.26, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {5694, 25, 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.
\(\displaystyle \int x \arctan (c+(-1+i c) \tan (a+b x)) \, dx\) |
\(\Big \downarrow \) 5694 |
\(\displaystyle \frac {1}{2} x^2 \arctan (c-(1-i c) \tan (a+b x))-\frac {1}{2} i b \int -\frac {x^2}{i-c e^{2 i a+2 i b x}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} i b \int \frac {x^2}{i-c e^{2 i a+2 i b x}}dx+\frac {1}{2} x^2 \arctan (c-(1-i c) \tan (a+b x))\) |
\(\Big \downarrow \) 2615 |
\(\displaystyle \frac {1}{2} i b \left (-i c \int \frac {e^{2 i a+2 i b x} x^2}{i-c e^{2 i a+2 i b x}}dx-\frac {i x^3}{3}\right )+\frac {1}{2} x^2 \arctan (c-(1-i c) \tan (a+b x))\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle \frac {1}{2} i b \left (-i c \left (\frac {i x^2 \log \left (1+i c e^{2 i a+2 i b x}\right )}{2 b c}-\frac {i \int x \log \left (i e^{2 i a+2 i b x} c+1\right )dx}{b c}\right )-\frac {i x^3}{3}\right )+\frac {1}{2} x^2 \arctan (c-(1-i c) \tan (a+b x))\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle \frac {1}{2} i b \left (-i c \left (\frac {i x^2 \log \left (1+i c e^{2 i a+2 i b x}\right )}{2 b c}-\frac {i \left (\frac {i x \operatorname {PolyLog}\left (2,-i c e^{2 i a+2 i b x}\right )}{2 b}-\frac {i \int \operatorname {PolyLog}\left (2,-i c e^{2 i a+2 i b x}\right )dx}{2 b}\right )}{b c}\right )-\frac {i x^3}{3}\right )+\frac {1}{2} x^2 \arctan (c-(1-i c) \tan (a+b x))\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {1}{2} i b \left (-i c \left (\frac {i x^2 \log \left (1+i c e^{2 i a+2 i b x}\right )}{2 b c}-\frac {i \left (\frac {i x \operatorname {PolyLog}\left (2,-i c e^{2 i a+2 i b x}\right )}{2 b}-\frac {\int e^{-2 i a-2 i b x} \operatorname {PolyLog}\left (2,-i c e^{2 i a+2 i b x}\right )de^{2 i a+2 i b x}}{4 b^2}\right )}{b c}\right )-\frac {i x^3}{3}\right )+\frac {1}{2} x^2 \arctan (c-(1-i c) \tan (a+b x))\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {1}{2} x^2 \arctan (c-(1-i c) \tan (a+b x))+\frac {1}{2} i b \left (-i c \left (\frac {i x^2 \log \left (1+i c e^{2 i a+2 i b x}\right )}{2 b c}-\frac {i \left (\frac {i x \operatorname {PolyLog}\left (2,-i c e^{2 i a+2 i b x}\right )}{2 b}-\frac {\operatorname {PolyLog}\left (3,-i c e^{2 i a+2 i b x}\right )}{4 b^2}\right )}{b c}\right )-\frac {i x^3}{3}\right )\) |
(x^2*ArcTan[c - (1 - I*c)*Tan[a + b*x]])/2 + (I/2)*b*((-1/3*I)*x^3 - I*c*( ((I/2)*x^2*Log[1 + I*c*E^((2*I)*a + (2*I)*b*x)])/(b*c) - (I*(((I/2)*x*Poly Log[2, (-I)*c*E^((2*I)*a + (2*I)*b*x)])/b - PolyLog[3, (-I)*c*E^((2*I)*a + (2*I)*b*x)]/(4*b^2)))/(b*c)))
3.1.57.3.1 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[ArcTan[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_. ), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcTan[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] && E qQ[(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 = 0.98 (sec) , antiderivative size = 1453, normalized size of antiderivative = 11.72
1/2*I/b*ln(I*exp(2*I*(b*x+a))*c+1)*a*x+1/8*I/b^2*polylog(3,-I*exp(2*I*(b*x +a))*c)-1/2*I/b*a*ln(1+I*exp(I*(b*x+a))*(I*c)^(1/2))*x+1/8*I*(-I*Pi*csgn(I *exp(2*I*(b*x+a)))^3-2*I*Pi+I*Pi*csgn(I/(exp(2*I*(b*x+a))+1)*(I+c))*csgn(I *exp(2*I*(b*x+a))*(I+c)/(exp(2*I*(b*x+a))+1))^2+2*ln(I+c)-I*Pi*csgn(I*exp( I*(b*x+a)))^2*csgn(I*exp(2*I*(b*x+a)))+I*Pi*csgn(I*(exp(2*I*(b*x+a))*c-I)/ (exp(2*I*(b*x+a))+1))^3-I*Pi*csgn(I*exp(2*I*(b*x+a))*(I+c)/(exp(2*I*(b*x+a ))+1))^3+I*Pi*csgn(I*(exp(2*I*(b*x+a))*c-I)/(exp(2*I*(b*x+a))+1))*csgn((ex p(2*I*(b*x+a))*c-I)/(exp(2*I*(b*x+a))+1))-I*Pi*csgn(exp(2*I*(b*x+a))*(I+c) /(exp(2*I*(b*x+a))+1))^3-I*Pi*csgn(I/(exp(2*I*(b*x+a))+1)*(I+c))^3-I*Pi*cs gn(I*exp(2*I*(b*x+a))*(I+c)/(exp(2*I*(b*x+a))+1))*csgn(exp(2*I*(b*x+a))*(I +c)/(exp(2*I*(b*x+a))+1))+I*Pi*csgn((exp(2*I*(b*x+a))*c-I)/(exp(2*I*(b*x+a ))+1))^2-I*Pi*csgn(I/(exp(2*I*(b*x+a))+1))*csgn(I*(I+c))*csgn(I/(exp(2*I*( b*x+a))+1)*(I+c))-I*Pi*csgn(I*exp(2*I*(b*x+a)))*csgn(I/(exp(2*I*(b*x+a))+1 )*(I+c))*csgn(I*exp(2*I*(b*x+a))*(I+c)/(exp(2*I*(b*x+a))+1))+I*Pi*csgn(I/( exp(2*I*(b*x+a))+1))*csgn(I*(exp(2*I*(b*x+a))*c-I))*csgn(I*(exp(2*I*(b*x+a ))*c-I)/(exp(2*I*(b*x+a))+1))-I*Pi*csgn(I/(exp(2*I*(b*x+a))+1))*csgn(I*(ex p(2*I*(b*x+a))*c-I)/(exp(2*I*(b*x+a))+1))^2-I*Pi*csgn(I*(exp(2*I*(b*x+a))* c-I))*csgn(I*(exp(2*I*(b*x+a))*c-I)/(exp(2*I*(b*x+a))+1))^2+I*Pi*csgn(I/(e xp(2*I*(b*x+a))+1))*csgn(I/(exp(2*I*(b*x+a))+1)*(I+c))^2+I*Pi*csgn(I*(I+c) )*csgn(I/(exp(2*I*(b*x+a))+1)*(I+c))^2+I*Pi*csgn(I*exp(2*I*(b*x+a)))*cs...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 271 vs. \(2 (86) = 172\).
Time = 0.27 (sec) , antiderivative size = 271, normalized size of antiderivative = 2.19 \[ \int x \arctan (c+(-1+i c) \tan (a+b x)) \, dx=\frac {2 \, b^{3} x^{3} + 3 i \, b^{2} x^{2} \log \left (-\frac {{\left (c + i\right )} e^{\left (2 i \, b x + 2 i \, a\right )}}{c e^{\left (2 i \, b x + 2 i \, a\right )} - i}\right ) + 2 \, a^{3} + 6 \, b x {\rm Li}_2\left (\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + 6 \, b x {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + 3 i \, a^{2} \log \left (\frac {2 \, c e^{\left (i \, b x + i \, a\right )} + i \, \sqrt {-4 i \, c}}{2 \, c}\right ) + 3 i \, a^{2} \log \left (\frac {2 \, c e^{\left (i \, b x + i \, a\right )} - i \, \sqrt {-4 i \, c}}{2 \, c}\right ) - 3 \, {\left (-i \, b^{2} x^{2} + i \, a^{2}\right )} \log \left (\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) - 3 \, {\left (-i \, b^{2} x^{2} + i \, a^{2}\right )} \log \left (-\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) + 6 i \, {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + 6 i \, {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (i \, b x + i \, a\right )}\right )}{12 \, b^{2}} \]
1/12*(2*b^3*x^3 + 3*I*b^2*x^2*log(-(c + I)*e^(2*I*b*x + 2*I*a)/(c*e^(2*I*b *x + 2*I*a) - I)) + 2*a^3 + 6*b*x*dilog(1/2*sqrt(-4*I*c)*e^(I*b*x + I*a)) + 6*b*x*dilog(-1/2*sqrt(-4*I*c)*e^(I*b*x + I*a)) + 3*I*a^2*log(1/2*(2*c*e^ (I*b*x + I*a) + I*sqrt(-4*I*c))/c) + 3*I*a^2*log(1/2*(2*c*e^(I*b*x + I*a) - I*sqrt(-4*I*c))/c) - 3*(-I*b^2*x^2 + I*a^2)*log(1/2*sqrt(-4*I*c)*e^(I*b* x + I*a) + 1) - 3*(-I*b^2*x^2 + I*a^2)*log(-1/2*sqrt(-4*I*c)*e^(I*b*x + I* a) + 1) + 6*I*polylog(3, 1/2*sqrt(-4*I*c)*e^(I*b*x + I*a)) + 6*I*polylog(3 , -1/2*sqrt(-4*I*c)*e^(I*b*x + I*a)))/b^2
Exception generated. \[ \int x \arctan (c+(-1+i c) \tan (a+b x)) \, dx=\text {Exception raised: CoercionFailed} \]
Exception raised: CoercionFailed >> Cannot convert _t0**2 + exp(2*I*a) of type <class 'sympy.core.add.Add'> to QQ_I[x,b,_t0,exp(I*a)]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (86) = 172\).
Time = 0.22 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.76 \[ \int x \arctan (c+(-1+i c) \tan (a+b x)) \, dx=\frac {\frac {6 \, {\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a\right )} \arctan \left ({\left (i \, c - 1\right )} \tan \left (b x + a\right ) + c\right )}{b} + \frac {{\left (-4 i \, {\left (b x + a\right )}^{3} + 12 i \, {\left (b x + a\right )}^{2} a - 6 i \, b x {\rm Li}_2\left (-i \, c e^{\left (2 i \, b x + 2 i \, a\right )}\right ) - 6 \, {\left (-i \, {\left (b x + a\right )}^{2} + 2 i \, {\left (b x + a\right )} a\right )} \arctan \left (c \cos \left (2 \, b x + 2 \, a\right ), -c \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \, {\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a\right )} \log \left (c^{2} \cos \left (2 \, b x + 2 \, a\right )^{2} + c^{2} \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, c \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \, {\rm Li}_{3}(-i \, c e^{\left (2 i \, b x + 2 i \, a\right )})\right )} {\left (i \, c - 1\right )}}{b {\left (c + i\right )}}}{12 \, b} \]
1/12*(6*((b*x + a)^2 - 2*(b*x + a)*a)*arctan((I*c - 1)*tan(b*x + a) + c)/b + (-4*I*(b*x + a)^3 + 12*I*(b*x + a)^2*a - 6*I*b*x*dilog(-I*c*e^(2*I*b*x + 2*I*a)) - 6*(-I*(b*x + a)^2 + 2*I*(b*x + a)*a)*arctan2(c*cos(2*b*x + 2*a ), -c*sin(2*b*x + 2*a) + 1) + 3*((b*x + a)^2 - 2*(b*x + a)*a)*log(c^2*cos( 2*b*x + 2*a)^2 + c^2*sin(2*b*x + 2*a)^2 - 2*c*sin(2*b*x + 2*a) + 1) + 3*po lylog(3, -I*c*e^(2*I*b*x + 2*I*a)))*(I*c - 1)/(b*(c + I)))/b
\[ \int x \arctan (c+(-1+i c) \tan (a+b x)) \, dx=\int { x \arctan \left ({\left (i \, c - 1\right )} \tan \left (b x + a\right ) + c\right ) \,d x } \]
Timed out. \[ \int x \arctan (c+(-1+i c) \tan (a+b x)) \, dx=\int x\,\mathrm {atan}\left (c+\mathrm {tan}\left (a+b\,x\right )\,\left (-1+c\,1{}\mathrm {i}\right )\right ) \,d x \]