Integrand size = 11, antiderivative size = 37 \[ \int x^m \arctan (\tan (a+b x)) \, dx=-\frac {b x^{2+m}}{2+3 m+m^2}+\frac {x^{1+m} \arctan (\tan (a+b x))}{1+m} \] Output:
-b*x^(2+m)/(m^2+3*m+2)+x^(1+m)*arctan(tan(b*x+a))/(1+m)
Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.92 \[ \int x^m \arctan (\tan (a+b x)) \, dx=x^m \left (\frac {b x^2}{2+m}+\frac {x (-b x+\arctan (\tan (a+b x)))}{1+m}\right ) \] Input:
Integrate[x^m*ArcTan[Tan[a + b*x]],x]
Output:
x^m*((b*x^2)/(2 + m) + (x*(-(b*x) + ArcTan[Tan[a + b*x]]))/(1 + m))
Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2599, 15}
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.
\(\displaystyle \int x^m \arctan (\tan (a+b x)) \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {x^{m+1} \arctan (\tan (a+b x))}{m+1}-\frac {b \int x^{m+1}dx}{m+1}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {x^{m+1} \arctan (\tan (a+b x))}{m+1}-\frac {b x^{m+2}}{(m+1) (m+2)}\) |
Input:
Int[x^m*ArcTan[Tan[a + b*x]],x]
Output:
-((b*x^(2 + m))/((1 + m)*(2 + m))) + (x^(1 + m)*ArcTan[Tan[a + b*x]])/(1 + m)
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Sim plify[D[v, x]]}, Simp[u^(m + 1)*(v^n/(a*(m + 1))), x] - Simp[b*(n/(a*(m + 1 ))) Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m, n} , x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0 ] && !(ILtQ[m + n, -2] && (FractionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ [n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] && !IntegerQ[m]) || (ILt Q[m, 0] && !IntegerQ[n]))
Time = 0.44 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.11
method | result | size |
default | \(\frac {b \,x^{2} {\mathrm e}^{m \ln \left (x \right )}}{2+m}+\frac {\left (\arctan \left (\tan \left (b x +a \right )\right )-b x \right ) x \,{\mathrm e}^{m \ln \left (x \right )}}{1+m}\) | \(41\) |
parallelrisch | \(-\frac {b \,x^{m} x^{2}-2 \arctan \left (\tan \left (b x +a \right )\right ) x \,x^{m}-x \,x^{m} \arctan \left (\tan \left (b x +a \right )\right ) m}{\left (1+m \right ) \left (2+m \right )}\) | \(49\) |
risch | \(-\frac {i x \,x^{m} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{1+m}-\frac {x \left (4 b x +2 \pi \operatorname {csgn}\left (i {\mathrm e}^{2 i \left (b x +a \right )}\right )^{3}+2 \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 i \left (b x +a \right )}}{{\mathrm e}^{2 i \left (b x +a \right )}+1}\right )^{3}+\pi m \operatorname {csgn}\left (i {\mathrm e}^{2 i \left (b x +a \right )}\right )^{3}+\pi m \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 i \left (b x +a \right )}}{{\mathrm e}^{2 i \left (b x +a \right )}+1}\right )^{3}-2 \pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 i \left (b x +a \right )}+1}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 i \left (b x +a \right )}}{{\mathrm e}^{2 i \left (b x +a \right )}+1}\right )^{2}+2 \pi \operatorname {csgn}\left (i {\mathrm e}^{i \left (b x +a \right )}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 i \left (b x +a \right )}\right )-4 \pi \,\operatorname {csgn}\left (i {\mathrm e}^{i \left (b x +a \right )}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 i \left (b x +a \right )}\right )^{2}-2 \pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 i \left (b x +a \right )}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 i \left (b x +a \right )}}{{\mathrm e}^{2 i \left (b x +a \right )}+1}\right )^{2}+\pi m \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 i \left (b x +a \right )}+1}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 i \left (b x +a \right )}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 i \left (b x +a \right )}}{{\mathrm e}^{2 i \left (b x +a \right )}+1}\right )-\pi m \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 i \left (b x +a \right )}+1}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 i \left (b x +a \right )}}{{\mathrm e}^{2 i \left (b x +a \right )}+1}\right )^{2}+\pi m \operatorname {csgn}\left (i {\mathrm e}^{i \left (b x +a \right )}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 i \left (b x +a \right )}\right )-2 \pi m \,\operatorname {csgn}\left (i {\mathrm e}^{i \left (b x +a \right )}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 i \left (b x +a \right )}\right )^{2}-\pi m \,\operatorname {csgn}\left (i {\mathrm e}^{2 i \left (b x +a \right )}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 i \left (b x +a \right )}}{{\mathrm e}^{2 i \left (b x +a \right )}+1}\right )^{2}+2 \pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 i \left (b x +a \right )}+1}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 i \left (b x +a \right )}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 i \left (b x +a \right )}}{{\mathrm e}^{2 i \left (b x +a \right )}+1}\right )\right ) x^{m}}{4 \left (1+m \right ) \left (2+m \right )}\) | \(572\) |
Input:
int(x^m*arctan(tan(b*x+a)),x,method=_RETURNVERBOSE)
Output:
b/(2+m)*x^2*exp(m*ln(x))+(arctan(tan(b*x+a))-b*x)/(1+m)*x*exp(m*ln(x))
Time = 0.09 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.89 \[ \int x^m \arctan (\tan (a+b x)) \, dx=\frac {{\left ({\left (b m + b\right )} x^{2} + {\left (a m + 2 \, a\right )} x\right )} x^{m}}{m^{2} + 3 \, m + 2} \] Input:
integrate(x^m*arctan(tan(b*x+a)),x, algorithm="fricas")
Output:
((b*m + b)*x^2 + (a*m + 2*a)*x)*x^m/(m^2 + 3*m + 2)
Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (31) = 62\).
Time = 0.81 (sec) , antiderivative size = 158, normalized size of antiderivative = 4.27 \[ \int x^m \arctan (\tan (a+b x)) \, dx=\begin {cases} b \log {\left (x \right )} - \frac {\operatorname {atan}{\left (\tan {\left (a + b x \right )} \right )} + \pi \left \lfloor {\frac {a + b x - \frac {\pi }{2}}{\pi }}\right \rfloor }{x} & \text {for}\: m = -2 \\- b x \log {\left (x \right )} + b x + \left (\operatorname {atan}{\left (\tan {\left (a + b x \right )} \right )} + 2 \pi \left \lfloor {\frac {a + b x - \frac {\pi }{2}}{\pi }}\right \rfloor \right ) \log {\left (x \right )} & \text {for}\: m = -1 \\- \frac {b x^{2} x^{m}}{m^{2} + 3 m + 2} + \frac {m x x^{m} \left (\operatorname {atan}{\left (\tan {\left (a + b x \right )} \right )} + \pi \left \lfloor {\frac {a + b x - \frac {\pi }{2}}{\pi }}\right \rfloor \right )}{m^{2} + 3 m + 2} + \frac {2 x x^{m} \left (\operatorname {atan}{\left (\tan {\left (a + b x \right )} \right )} + \pi \left \lfloor {\frac {a + b x - \frac {\pi }{2}}{\pi }}\right \rfloor \right )}{m^{2} + 3 m + 2} & \text {otherwise} \end {cases} \] Input:
integrate(x**m*atan(tan(b*x+a)),x)
Output:
Piecewise((b*log(x) - (atan(tan(a + b*x)) + pi*floor((a + b*x - pi/2)/pi)) /x, Eq(m, -2)), (-b*x*log(x) + b*x + (atan(tan(a + b*x)) + 2*pi*floor((a + b*x - pi/2)/pi))*log(x), Eq(m, -1)), (-b*x**2*x**m/(m**2 + 3*m + 2) + m*x *x**m*(atan(tan(a + b*x)) + pi*floor((a + b*x - pi/2)/pi))/(m**2 + 3*m + 2 ) + 2*x*x**m*(atan(tan(a + b*x)) + pi*floor((a + b*x - pi/2)/pi))/(m**2 + 3*m + 2), True))
Time = 0.05 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.03 \[ \int x^m \arctan (\tan (a+b x)) \, dx=-\frac {b x^{2} x^{m}}{{\left (m + 2\right )} {\left (m + 1\right )}} + \frac {x^{m + 1} \arctan \left (\tan \left (b x + a\right )\right )}{m + 1} \] Input:
integrate(x^m*arctan(tan(b*x+a)),x, algorithm="maxima")
Output:
-b*x^2*x^m/((m + 2)*(m + 1)) + x^(m + 1)*arctan(tan(b*x + a))/(m + 1)
\[ \int x^m \arctan (\tan (a+b x)) \, dx=\int { x^{m} \arctan \left (\tan \left (b x + a\right )\right ) \,d x } \] Input:
integrate(x^m*arctan(tan(b*x+a)),x, algorithm="giac")
Output:
sage0*x
Timed out. \[ \int x^m \arctan (\tan (a+b x)) \, dx=\int x^m\,\mathrm {atan}\left (\mathrm {tan}\left (a+b\,x\right )\right ) \,d x \] Input:
int(x^m*atan(tan(a + b*x)),x)
Output:
int(x^m*atan(tan(a + b*x)), x)
Time = 0.18 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.03 \[ \int x^m \arctan (\tan (a+b x)) \, dx=\frac {x^{m} x \left (\mathit {atan} \left (\tan \left (b x +a \right )\right ) m +2 \mathit {atan} \left (\tan \left (b x +a \right )\right )-b x \right )}{m^{2}+3 m +2} \] Input:
int(x^m*atan(tan(b*x+a)),x)
Output:
(x**m*x*(atan(tan(a + b*x))*m + 2*atan(tan(a + b*x)) - b*x))/(m**2 + 3*m + 2)