\(\int \arctan (a+b x) \, dx\) [47]

Optimal result

Integrand size = 6, antiderivative size = 33 \[ \int \arctan (a+b x) \, dx=\frac {(a+b x) \arctan (a+b x)}{b}-\frac {\log \left (1+(a+b x)^2\right )}{2 b} \]

[Out]

(b*x+a)*arctan(b*x+a)/b-1/2*ln(1+(b*x+a)^2)/b

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5147, 4930, 266} \[ \int \arctan (a+b x) \, dx=\frac {(a+b x) \arctan (a+b x)}{b}-\frac {\log \left ((a+b x)^2+1\right )}{2 b} \]

[In]

Int[ArcTan[a + b*x],x]

[Out]

((a + b*x)*ArcTan[a + b*x])/b - Log[1 + (a + b*x)^2]/(2*b)

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 4930

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x^n])^p, x] - Dist[b*c
*n*p, Int[x^n*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0
] && (EqQ[n, 1] || EqQ[p, 1])

Rule 5147

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcTan[x])^p, x],
x, c + d*x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}(\int \arctan (x) \, dx,x,a+b x)}{b} \\ & = \frac {(a+b x) \arctan (a+b x)}{b}-\frac {\text {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,a+b x\right )}{b} \\ & = \frac {(a+b x) \arctan (a+b x)}{b}-\frac {\log \left (1+(a+b x)^2\right )}{2 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \arctan (a+b x) \, dx=-\frac {-2 (a+b x) \arctan (a+b x)+\log \left (1+a^2+2 a b x+b^2 x^2\right )}{2 b} \]

[In]

Integrate[ArcTan[a + b*x],x]

[Out]

-1/2*(-2*(a + b*x)*ArcTan[a + b*x] + Log[1 + a^2 + 2*a*b*x + b^2*x^2])/b

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {\left (b x +a \right ) \arctan \left (b x +a \right )-\frac {\ln \left (1+\left (b x +a \right )^{2}\right )}{2}}{b}\)	\(30\)
default	\(\frac {\left (b x +a \right ) \arctan \left (b x +a \right )-\frac {\ln \left (1+\left (b x +a \right )^{2}\right )}{2}}{b}\)	\(30\)
parallelrisch	\(-\frac {-2 x \arctan \left (b x +a \right ) b^{2}-2 \arctan \left (b x +a \right ) a b +\ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) b}{2 b^{2}}\)	\(49\)
parts	\(x \arctan \left (b x +a \right )-b \left (\frac {\ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b^{2}}-\frac {a \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right )}{b^{2}}\right )\)	\(60\)
risch	\(-\frac {i x \ln \left (1+i \left (b x +a \right )\right )}{2}+\frac {i x \ln \left (1-i \left (b x +a \right )\right )}{2}+\frac {a \arctan \left (b x +a \right )}{b}-\frac {\ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b}\)	\(66\)

[In]

int(arctan(b*x+a),x,method=_RETURNVERBOSE)

[Out]

1/b*((b*x+a)*arctan(b*x+a)-1/2*ln(1+(b*x+a)^2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \arctan (a+b x) \, dx=\frac {2 \, {\left (b x + a\right )} \arctan \left (b x + a\right ) - \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{2 \, b} \]

[In]

integrate(arctan(b*x+a),x, algorithm="fricas")

[Out]

1/2*(2*(b*x + a)*arctan(b*x + a) - log(b^2*x^2 + 2*a*b*x + a^2 + 1))/b

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.39 \[ \int \arctan (a+b x) \, dx=\begin {cases} \frac {a \operatorname {atan}{\left (a + b x \right )}}{b} + x \operatorname {atan}{\left (a + b x \right )} - \frac {\log {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{2 b} & \text {for}\: b \neq 0 \\x \operatorname {atan}{\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(atan(b*x+a),x)

[Out]

Piecewise((a*atan(a + b*x)/b + x*atan(a + b*x) - log(a**2 + 2*a*b*x + b**2*x**2 + 1)/(2*b), Ne(b, 0)), (x*atan
(a), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \arctan (a+b x) \, dx=\frac {2 \, {\left (b x + a\right )} \arctan \left (b x + a\right ) - \log \left ({\left (b x + a\right )}^{2} + 1\right )}{2 \, b} \]

[In]

integrate(arctan(b*x+a),x, algorithm="maxima")

[Out]

1/2*(2*(b*x + a)*arctan(b*x + a) - log((b*x + a)^2 + 1))/b

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \arctan (a+b x) \, dx=\frac {2 \, {\left (b x + a\right )} \arctan \left (b x + a\right ) - \log \left ({\left (b x + a\right )}^{2} + 1\right )}{2 \, b} \]

[In]

integrate(arctan(b*x+a),x, algorithm="giac")

[Out]

1/2*(2*(b*x + a)*arctan(b*x + a) - log((b*x + a)^2 + 1))/b

Mupad [B] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.27 \[ \int \arctan (a+b x) \, dx=x\,\mathrm {atan}\left (a+b\,x\right )-\frac {\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )-2\,a\,\mathrm {atan}\left (a+b\,x\right )}{2\,b} \]

[In]

int(atan(a + b*x),x)

[Out]

x*atan(a + b*x) - (log(a^2 + b^2*x^2 + 2*a*b*x + 1) - 2*a*atan(a + b*x))/(2*b)