Integrand size = 11, antiderivative size = 198 \[ \int \arctan (c+d \tan (a+b x)) \, dx=x \arctan (c+d \tan (a+b x))+\frac {1}{2} i x \log \left (1+\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )-\frac {1}{2} i x \log \left (1+\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )+\frac {\operatorname {PolyLog}\left (2,-\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )}{4 b}-\frac {\operatorname {PolyLog}\left (2,-\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )}{4 b} \]
x*arctan(c+d*tan(b*x+a))+1/2*I*x*ln(1+(1+I*c+d)*exp(2*I*a+2*I*b*x)/(1+I*c- d))-1/2*I*x*ln(1+(c+I*(1-d))*exp(2*I*a+2*I*b*x)/(c+I*(1+d)))+1/4*polylog(2 ,-(1+I*c+d)*exp(2*I*a+2*I*b*x)/(1+I*c-d))/b-1/4*polylog(2,-(c+I*(1-d))*exp (2*I*a+2*I*b*x)/(c+I*(1+d)))/b
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(549\) vs. \(2(198)=396\).
Time = 5.61 (sec) , antiderivative size = 549, normalized size of antiderivative = 2.77 \[ \int \arctan (c+d \tan (a+b x)) \, dx=x \arctan (c+d \tan (a+b x))-\frac {x \left (4 a \sqrt {-d^2} \arctan (c+d \tan (a+b x))-i d \log (1-i \tan (a+b x)) \log \left (\frac {-c d+\sqrt {-d^2}-d^2 \tan (a+b x)}{-c d+i d^2+\sqrt {-d^2}}\right )+i d \log (1+i \tan (a+b x)) \log \left (\frac {c d-\sqrt {-d^2}+d^2 \tan (a+b x)}{c d+i d^2-\sqrt {-d^2}}\right )+i d \log (1-i \tan (a+b x)) \log \left (\frac {c d+\sqrt {-d^2}+d^2 \tan (a+b x)}{c d-i d^2+\sqrt {-d^2}}\right )-i d \log (1+i \tan (a+b x)) \log \left (\frac {c d+\sqrt {-d^2}+d^2 \tan (a+b x)}{c d+i d^2+\sqrt {-d^2}}\right )-i d \operatorname {PolyLog}\left (2,\frac {d^2 (1-i \tan (a+b x))}{i c d+d^2-i \sqrt {-d^2}}\right )+i d \operatorname {PolyLog}\left (2,\frac {d^2 (1-i \tan (a+b x))}{i c d+d^2+i \sqrt {-d^2}}\right )+i d \operatorname {PolyLog}\left (2,\frac {d^2 (1+i \tan (a+b x))}{-i c d+d^2+i \sqrt {-d^2}}\right )-i d \operatorname {PolyLog}\left (2,\frac {d^2 (1+i \tan (a+b x))}{d^2-i \left (c d+\sqrt {-d^2}\right )}\right )\right )}{2 \sqrt {-d^2} (2 a-i \log (1-i \tan (a+b x))+i \log (1+i \tan (a+b x)))} \]
x*ArcTan[c + d*Tan[a + b*x]] - (x*(4*a*Sqrt[-d^2]*ArcTan[c + d*Tan[a + b*x ]] - I*d*Log[1 - I*Tan[a + b*x]]*Log[(-(c*d) + Sqrt[-d^2] - d^2*Tan[a + b* x])/(-(c*d) + I*d^2 + Sqrt[-d^2])] + I*d*Log[1 + I*Tan[a + b*x]]*Log[(c*d - Sqrt[-d^2] + d^2*Tan[a + b*x])/(c*d + I*d^2 - Sqrt[-d^2])] + I*d*Log[1 - I*Tan[a + b*x]]*Log[(c*d + Sqrt[-d^2] + d^2*Tan[a + b*x])/(c*d - I*d^2 + Sqrt[-d^2])] - I*d*Log[1 + I*Tan[a + b*x]]*Log[(c*d + Sqrt[-d^2] + d^2*Tan [a + b*x])/(c*d + I*d^2 + Sqrt[-d^2])] - I*d*PolyLog[2, (d^2*(1 - I*Tan[a + b*x]))/(I*c*d + d^2 - I*Sqrt[-d^2])] + I*d*PolyLog[2, (d^2*(1 - I*Tan[a + b*x]))/(I*c*d + d^2 + I*Sqrt[-d^2])] + I*d*PolyLog[2, (d^2*(1 + I*Tan[a + b*x]))/((-I)*c*d + d^2 + I*Sqrt[-d^2])] - I*d*PolyLog[2, (d^2*(1 + I*Tan [a + b*x]))/(d^2 - I*(c*d + Sqrt[-d^2]))]))/(2*Sqrt[-d^2]*(2*a - I*Log[1 - I*Tan[a + b*x]] + I*Log[1 + I*Tan[a + b*x]]))
Time = 0.67 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.40, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {5690, 2620, 2715, 2838}
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.
| \(\displaystyle \int \arctan (d \tan (a+b x)+c) \, dx\) |
| \(\Big \downarrow \) 5690 |
| \(\displaystyle b (-i c-d+1) \int \frac {e^{2 i a+2 i b x} x}{-i c+(-i c-d+1) e^{2 i a+2 i b x}+d+1}dx-b (i c+d+1) \int \frac {e^{2 i a+2 i b x} x}{i c+(i c+d+1) e^{2 i a+2 i b x}-d+1}dx+x \arctan (d \tan (a+b x)+c)\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle -b (i c+d+1) \left (\frac {\int \log \left (\frac {e^{2 i a+2 i b x} (i c+d+1)}{i c-d+1}+1\right )dx}{2 b (c-i (d+1))}-\frac {x \log \left (1+\frac {(i c+d+1) e^{2 i a+2 i b x}}{i c-d+1}\right )}{2 b (c-i (d+1))}\right )+b (-i c-d+1) \left (\frac {x \log \left (1+\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (d+1)}\right )}{2 b (c+i (1-d))}-\frac {\int \log \left (\frac {e^{2 i a+2 i b x} (c+i (1-d))}{c+i (d+1)}+1\right )dx}{2 b (c+i (1-d))}\right )+x \arctan (d \tan (a+b x)+c)\) |
| \(\Big \downarrow \) 2715 |
| \(\displaystyle -b (i c+d+1) \left (-\frac {i \int e^{-2 i a-2 i b x} \log \left (\frac {e^{2 i a+2 i b x} (i c+d+1)}{i c-d+1}+1\right )de^{2 i a+2 i b x}}{4 b^2 (c-i (d+1))}-\frac {x \log \left (1+\frac {(i c+d+1) e^{2 i a+2 i b x}}{i c-d+1}\right )}{2 b (c-i (d+1))}\right )+b (-i c-d+1) \left (\frac {i \int e^{-2 i a-2 i b x} \log \left (\frac {e^{2 i a+2 i b x} (c+i (1-d))}{c+i (d+1)}+1\right )de^{2 i a+2 i b x}}{4 b^2 (c+i (1-d))}+\frac {x \log \left (1+\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (d+1)}\right )}{2 b (c+i (1-d))}\right )+x \arctan (d \tan (a+b x)+c)\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle x \arctan (d \tan (a+b x)+c)-b (i c+d+1) \left (\frac {i \operatorname {PolyLog}\left (2,-\frac {(i c+d+1) e^{2 i a+2 i b x}}{i c-d+1}\right )}{4 b^2 (c-i (d+1))}-\frac {x \log \left (1+\frac {(i c+d+1) e^{2 i a+2 i b x}}{i c-d+1}\right )}{2 b (c-i (d+1))}\right )+b (-i c-d+1) \left (\frac {x \log \left (1+\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (d+1)}\right )}{2 b (c+i (1-d))}-\frac {i \operatorname {PolyLog}\left (2,-\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (d+1)}\right )}{4 b^2 (c+i (1-d))}\right )\) |
x*ArcTan[c + d*Tan[a + b*x]] - b*(1 + I*c + d)*(-1/2*(x*Log[1 + ((1 + I*c + d)*E^((2*I)*a + (2*I)*b*x))/(1 + I*c - d)])/(b*(c - I*(1 + d))) + ((I/4) *PolyLog[2, -(((1 + I*c + d)*E^((2*I)*a + (2*I)*b*x))/(1 + I*c - d))])/(b^ 2*(c - I*(1 + d)))) + b*(1 - I*c - d)*((x*Log[1 + ((c + I*(1 - d))*E^((2*I )*a + (2*I)*b*x))/(c + I*(1 + d))])/(2*b*(c + I*(1 - d))) - ((I/4)*PolyLog [2, -(((c + I*(1 - d))*E^((2*I)*a + (2*I)*b*x))/(c + I*(1 + d)))])/(b^2*(c + I*(1 - d))))
3.1.50.3.1 Defintions of rubi rules used
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[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[1/(d*e*n*Log[F]) Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) ))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
Int[ArcTan[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]], x_Symbol] :> Simp[x*ArcT an[c + d*Tan[a + b*x]], x] + (Simp[b*(1 - I*c - d) Int[x*(E^(2*I*a + 2*I* b*x)/(1 - I*c + d + (1 - I*c - d)*E^(2*I*a + 2*I*b*x))), x], x] - Simp[b*(1 + I*c + d) Int[x*(E^(2*I*a + 2*I*b*x)/(1 + I*c - d + (1 + I*c + d)*E^(2* I*a + 2*I*b*x))), x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[(c + I*d)^2, -1]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1000 vs. \(2 (168 ) = 336\).
Time = 2.41 (sec) , antiderivative size = 1001, normalized size of antiderivative = 5.06
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(1001\) |
| default | \(\text {Expression too large to display}\) | \(1001\) |
| risch | \(\text {Expression too large to display}\) | \(4973\) |
1/b/d*(d*arctan(tan(b*x+a))*arctan(c+d*tan(b*x+a))-d^2*(1/2*I/d*arctan(-(c +d*tan(b*x+a))/d+c/d)*ln(1-(c-I*d-I)*(1+I*((c+d*tan(b*x+a))/d-c/d))^2/(((c +d*tan(b*x+a))/d-c/d)^2+1)/(-I*d+I-c))-1/2/d*arctan(-(c+d*tan(b*x+a))/d+c/ d)^2-1/4/d*polylog(2,(c-I*d-I)*(1+I*((c+d*tan(b*x+a))/d-c/d))^2/(((c+d*tan (b*x+a))/d-c/d)^2+1)/(-I*d+I-c))+1/2/(I+c+I*d)*ln(1-(c-I*d+I)*(1+I*((c+d*t an(b*x+a))/d-c/d))^2/(((c+d*tan(b*x+a))/d-c/d)^2+1)/(-I*d-I-c))*arctan(-(c +d*tan(b*x+a))/d+c/d)+1/2/d/(I+c+I*d)*ln(1-(c-I*d+I)*(1+I*((c+d*tan(b*x+a) )/d-c/d))^2/(((c+d*tan(b*x+a))/d-c/d)^2+1)/(-I*d-I-c))*arctan(-(c+d*tan(b* x+a))/d+c/d)-1/2*I/d/(I+c+I*d)*ln(1-(c-I*d+I)*(1+I*((c+d*tan(b*x+a))/d-c/d ))^2/(((c+d*tan(b*x+a))/d-c/d)^2+1)/(-I*d-I-c))*c*arctan(-(c+d*tan(b*x+a)) /d+c/d)+1/2*I/(I+c+I*d)*arctan(-(c+d*tan(b*x+a))/d+c/d)^2+1/4*I/(I+c+I*d)* polylog(2,(c-I*d+I)*(1+I*((c+d*tan(b*x+a))/d-c/d))^2/(((c+d*tan(b*x+a))/d- c/d)^2+1)/(-I*d-I-c))+1/2*I/d/(I+c+I*d)*arctan(-(c+d*tan(b*x+a))/d+c/d)^2+ 1/2/d/(I+c+I*d)*c*arctan(-(c+d*tan(b*x+a))/d+c/d)^2+1/4*I/d/(I+c+I*d)*poly log(2,(c-I*d+I)*(1+I*((c+d*tan(b*x+a))/d-c/d))^2/(((c+d*tan(b*x+a))/d-c/d) ^2+1)/(-I*d-I-c))+1/4/d/(I+c+I*d)*polylog(2,(c-I*d+I)*(1+I*((c+d*tan(b*x+a ))/d-c/d))^2/(((c+d*tan(b*x+a))/d-c/d)^2+1)/(-I*d-I-c))*c))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1101 vs. \(2 (141) = 282\).
Time = 0.34 (sec) , antiderivative size = 1101, normalized size of antiderivative = 5.56 \[ \int \arctan (c+d \tan (a+b x)) \, dx=\text {Too large to display} \]
1/8*(8*b*x*arctan(d*tan(b*x + a) + c) - 2*(I*b*x + I*a)*log(-2*((I*c*d - d ^2 + d)*tan(b*x + a)^2 - c^2 - I*c*d + (I*c^2 - 2*c*d - I*d^2 + I)*tan(b*x + a) + d - 1)/((c^2 + d^2 - 2*d + 1)*tan(b*x + a)^2 + c^2 + d^2 - 2*d + 1 )) - 2*(-I*b*x - I*a)*log(-2*((I*c*d - d^2 - d)*tan(b*x + a)^2 - c^2 - I*c *d + (I*c^2 - 2*c*d - I*d^2 + I)*tan(b*x + a) - d - 1)/((c^2 + d^2 + 2*d + 1)*tan(b*x + a)^2 + c^2 + d^2 + 2*d + 1)) - 2*(-I*b*x - I*a)*log(-2*((-I* c*d - d^2 + d)*tan(b*x + a)^2 - c^2 + I*c*d + (-I*c^2 - 2*c*d + I*d^2 - I) *tan(b*x + a) + d - 1)/((c^2 + d^2 - 2*d + 1)*tan(b*x + a)^2 + c^2 + d^2 - 2*d + 1)) - 2*(I*b*x + I*a)*log(-2*((-I*c*d - d^2 - d)*tan(b*x + a)^2 - c ^2 + I*c*d + (-I*c^2 - 2*c*d + I*d^2 - I)*tan(b*x + a) - d - 1)/((c^2 + d^ 2 + 2*d + 1)*tan(b*x + a)^2 + c^2 + d^2 + 2*d + 1)) + 2*I*a*log(((I*c*d + d^2 + d)*tan(b*x + a)^2 - c^2 + I*c*d + (I*c^2 + I*d^2 + 2*I*d + I)*tan(b* x + a) - d - 1)/(tan(b*x + a)^2 + 1)) - 2*I*a*log(((I*c*d + d^2 - d)*tan(b *x + a)^2 - c^2 + I*c*d + (I*c^2 + I*d^2 - 2*I*d + I)*tan(b*x + a) + d - 1 )/(tan(b*x + a)^2 + 1)) + 2*I*a*log(((I*c*d - d^2 + d)*tan(b*x + a)^2 + c^ 2 + I*c*d + (I*c^2 + I*d^2 - 2*I*d + I)*tan(b*x + a) - d + 1)/(tan(b*x + a )^2 + 1)) - 2*I*a*log(((I*c*d - d^2 - d)*tan(b*x + a)^2 + c^2 + I*c*d + (I *c^2 + I*d^2 + 2*I*d + I)*tan(b*x + a) + d + 1)/(tan(b*x + a)^2 + 1)) + di log(2*((I*c*d - d^2 + d)*tan(b*x + a)^2 - c^2 - I*c*d + (I*c^2 - 2*c*d - I *d^2 + I)*tan(b*x + a) + d - 1)/((c^2 + d^2 - 2*d + 1)*tan(b*x + a)^2 +...
\[ \int \arctan (c+d \tan (a+b x)) \, dx=\int \operatorname {atan}{\left (c + d \tan {\left (a + b x \right )} \right )}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 433 vs. \(2 (141) = 282\).
Time = 0.34 (sec) , antiderivative size = 433, normalized size of antiderivative = 2.19 \[ \int \arctan (c+d \tan (a+b x)) \, dx=\frac {d {\left (\frac {8 \, {\left (b x + a\right )} \arctan \left (\frac {d^{2} \tan \left (b x + a\right ) + c d}{d}\right )}{d} - \frac {4 \, {\left (b x + a\right )} \arctan \left (\frac {c d + {\left (d^{2} + d\right )} \tan \left (b x + a\right )}{c^{2} + d^{2} + 2 \, d + 1}, \frac {c d \tan \left (b x + a\right ) + c^{2} + d + 1}{c^{2} + d^{2} + 2 \, d + 1}\right ) - 4 \, {\left (b x + a\right )} \arctan \left (\frac {c d + {\left (d^{2} - d\right )} \tan \left (b x + a\right )}{c^{2} + d^{2} - 2 \, d + 1}, \frac {c d \tan \left (b x + a\right ) + c^{2} - d + 1}{c^{2} + d^{2} - 2 \, d + 1}\right ) + \log \left (\tan \left (b x + a\right )^{2} + 1\right ) \log \left (\frac {d^{2} \tan \left (b x + a\right )^{2} + 2 \, c d \tan \left (b x + a\right ) + c^{2} + 1}{c^{2} + d^{2} + 2 \, d + 1}\right ) - \log \left (\tan \left (b x + a\right )^{2} + 1\right ) \log \left (\frac {d^{2} \tan \left (b x + a\right )^{2} + 2 \, c d \tan \left (b x + a\right ) + c^{2} + 1}{c^{2} + d^{2} - 2 \, d + 1}\right ) + 2 \, {\rm Li}_2\left (-\frac {i \, d \tan \left (b x + a\right ) - d}{i \, c + d + 1}\right ) - 2 \, {\rm Li}_2\left (-\frac {i \, d \tan \left (b x + a\right ) - d}{i \, c + d - 1}\right ) + 2 \, {\rm Li}_2\left (\frac {i \, d \tan \left (b x + a\right ) + d}{-i \, c + d + 1}\right ) - 2 \, {\rm Li}_2\left (\frac {i \, d \tan \left (b x + a\right ) + d}{-i \, c + d - 1}\right )}{d}\right )} + 8 \, {\left (b x + a\right )} \arctan \left (d \tan \left (b x + a\right ) + c\right ) - 8 \, {\left (b x + a\right )} \arctan \left (\frac {d^{2} \tan \left (b x + a\right ) + c d}{d}\right )}{8 \, b} \]
1/8*(d*(8*(b*x + a)*arctan((d^2*tan(b*x + a) + c*d)/d)/d - (4*(b*x + a)*ar ctan2((c*d + (d^2 + d)*tan(b*x + a))/(c^2 + d^2 + 2*d + 1), (c*d*tan(b*x + a) + c^2 + d + 1)/(c^2 + d^2 + 2*d + 1)) - 4*(b*x + a)*arctan2((c*d + (d^ 2 - d)*tan(b*x + a))/(c^2 + d^2 - 2*d + 1), (c*d*tan(b*x + a) + c^2 - d + 1)/(c^2 + d^2 - 2*d + 1)) + log(tan(b*x + a)^2 + 1)*log((d^2*tan(b*x + a)^ 2 + 2*c*d*tan(b*x + a) + c^2 + 1)/(c^2 + d^2 + 2*d + 1)) - log(tan(b*x + a )^2 + 1)*log((d^2*tan(b*x + a)^2 + 2*c*d*tan(b*x + a) + c^2 + 1)/(c^2 + d^ 2 - 2*d + 1)) + 2*dilog(-(I*d*tan(b*x + a) - d)/(I*c + d + 1)) - 2*dilog(- (I*d*tan(b*x + a) - d)/(I*c + d - 1)) + 2*dilog((I*d*tan(b*x + a) + d)/(-I *c + d + 1)) - 2*dilog((I*d*tan(b*x + a) + d)/(-I*c + d - 1)))/d) + 8*(b*x + a)*arctan(d*tan(b*x + a) + c) - 8*(b*x + a)*arctan((d^2*tan(b*x + a) + c*d)/d))/b
\[ \int \arctan (c+d \tan (a+b x)) \, dx=\int { \arctan \left (d \tan \left (b x + a\right ) + c\right ) \,d x } \]
Timed out. \[ \int \arctan (c+d \tan (a+b x)) \, dx=\int \mathrm {atan}\left (c+d\,\mathrm {tan}\left (a+b\,x\right )\right ) \,d x \]