\(\int \frac {\arctan (a+b x)}{x^2} \, dx\) [6]

Optimal result

Integrand size = 10, antiderivative size = 62 \[ \int \frac {\arctan (a+b x)}{x^2} \, dx=-\frac {a b \arctan (a+b x)}{1+a^2}-\frac {\arctan (a+b x)}{x}+\frac {b \log (x)}{1+a^2}-\frac {b \log \left (1+(a+b x)^2\right )}{2 \left (1+a^2\right )} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.08 \[ \int \frac {\arctan (a+b x)}{x^2} \, dx=-\frac {\arctan (a+b x)}{x}+\frac {b (2 \log (x)+i (i+a) \log (i-a-b x)+(-1-i a) \log (i+a+b x))}{2 \left (1+a^2\right )} \] Input:

Integrate[ArcTan[a + b*x]/x^2,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {5568, 896, 25, 479, 452, 216, 240}

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.

\(\displaystyle \int \frac {\arctan (a+b x)}{x^2} \, dx\)

\(\Big \downarrow \) 5568

\(\displaystyle b \int \frac {1}{x \left ((a+b x)^2+1\right )}dx-\frac {\arctan (a+b x)}{x}\)

\(\Big \downarrow \) 896

\(\displaystyle b \int \frac {1}{b x \left ((a+b x)^2+1\right )}d(a+b x)-\frac {\arctan (a+b x)}{x}\)

\(\Big \downarrow \) 25

\(\displaystyle -b \int -\frac {1}{b x \left ((a+b x)^2+1\right )}d(a+b x)-\frac {\arctan (a+b x)}{x}\)

\(\Big \downarrow \) 479

\(\displaystyle b \left (\frac {\log (-b x)}{a^2+1}-\frac {\int \frac {2 a+b x}{(a+b x)^2+1}d(a+b x)}{a^2+1}\right )-\frac {\arctan (a+b x)}{x}\)

\(\Big \downarrow \) 452

\(\displaystyle b \left (\frac {\log (-b x)}{a^2+1}-\frac {a \int \frac {1}{(a+b x)^2+1}d(a+b x)+\int \frac {a+b x}{(a+b x)^2+1}d(a+b x)}{a^2+1}\right )-\frac {\arctan (a+b x)}{x}\)

\(\Big \downarrow \) 216

\(\displaystyle b \left (\frac {\log (-b x)}{a^2+1}-\frac {\int \frac {a+b x}{(a+b x)^2+1}d(a+b x)+a \arctan (a+b x)}{a^2+1}\right )-\frac {\arctan (a+b x)}{x}\)

\(\Big \downarrow \) 240

\(\displaystyle b \left (\frac {\log (-b x)}{a^2+1}-\frac {a \arctan (a+b x)+\frac {1}{2} \log \left ((a+b x)^2+1\right )}{a^2+1}\right )-\frac {\arctan (a+b x)}{x}\)

Input:

Int[ArcTan[a + b*x]/x^2,x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 240

Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x 
^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
 

rule 452

Int[((c_) + (d_.)*(x_))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[c   Int[1/ 
(a + b*x^2), x], x] + Simp[d   Int[x/(a + b*x^2), x], x] /; FreeQ[{a, b, c, 
 d}, x] && NeQ[b*c^2 + a*d^2, 0]
 

rule 479

Int[1/(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)), x_Symbol] :> Simp[d*(Log 
[RemoveContent[c + d*x, x]]/(b*c^2 + a*d^2)), x] + Simp[b/(b*c^2 + a*d^2) 
 Int[(c - d*x)/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d}, x]
 

rule 896

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coeff 
icient[v, x, 0], d = Coefficient[v, x, 1]}, Simp[1/d^(m + 1)   Subst[Int[Si 
mplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]] /; 
 FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]
 

rule 5568

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m 
_), x_Symbol] :> Simp[(e + f*x)^(m + 1)*((a + b*ArcTan[c + d*x])^p/(f*(m + 
1))), x] - Simp[b*d*(p/(f*(m + 1)))   Int[(e + f*x)^(m + 1)*((a + b*ArcTan[ 
c + d*x])^(p - 1)/(1 + (c + d*x)^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x 
] && IGtQ[p, 0] && ILtQ[m, -1]
 

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.98

method	result	size
derivativedivides	\(b \left (-\frac {\arctan \left (b x +a \right )}{b x}+\frac {\ln \left (-b x \right )}{a^{2}+1}-\frac {\frac {\ln \left (1+\left (b x +a \right )^{2}\right )}{2}+a \arctan \left (b x +a \right )}{a^{2}+1}\right )\)	\(61\)
default	\(b \left (-\frac {\arctan \left (b x +a \right )}{b x}+\frac {\ln \left (-b x \right )}{a^{2}+1}-\frac {\frac {\ln \left (1+\left (b x +a \right )^{2}\right )}{2}+a \arctan \left (b x +a \right )}{a^{2}+1}\right )\)	\(61\)
parts	\(-\frac {\arctan \left (b x +a \right )}{x}+b \left (-\frac {b \left (\frac {\ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b}+\frac {a \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right )}{b}\right )}{a^{2}+1}+\frac {\ln \left (x \right )}{a^{2}+1}\right )\)	\(82\)
parallelrisch	\(\frac {-2 x \arctan \left (b x +a \right ) a^{2} b^{2}+2 b^{2} \ln \left (x \right ) a x -b^{2} \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a x -2 \arctan \left (b x +a \right ) a^{3} b -2 \arctan \left (b x +a \right ) b a}{2 x a b \left (a^{2}+1\right )}\)	\(91\)
risch	\(\frac {i \ln \left (1+i \left (b x +a \right )\right )}{2 x}-\frac {i \left (a^{2} \ln \left (1-i \left (b x +a \right )\right )+\ln \left (1-i \left (b x +a \right )\right )-i \ln \left (\left (-a^{2} b +3 i a b \right ) x -3 a +2 i a^{2}-a^{3}\right ) b x -i \ln \left (\left (-a^{2} b -3 i a b \right ) x -3 a -2 i a^{2}-a^{3}\right ) b x +2 i \ln \left (x \right ) b x +\ln \left (\left (-a^{2} b +3 i a b \right ) x -3 a +2 i a^{2}-a^{3}\right ) a b x -\ln \left (\left (-a^{2} b -3 i a b \right ) x -3 a -2 i a^{2}-a^{3}\right ) a b x \right )}{2 x \left (i+a \right ) \left (a -i\right )}\)	\(210\)

Input:

int(arctan(b*x+a)/x^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.92 \[ \int \frac {\arctan (a+b x)}{x^2} \, dx=-\frac {b x \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) - 2 \, b x \log \left (x\right ) + 2 \, {\left (a b x + a^{2} + 1\right )} \arctan \left (b x + a\right )}{2 \, {\left (a^{2} + 1\right )} x} \] Input:

integrate(arctan(b*x+a)/x^2,x, algorithm="fricas")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.79 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.71 \[ \int \frac {\arctan (a+b x)}{x^2} \, dx=\begin {cases} - \frac {i b \operatorname {atan}{\left (b x - i \right )}}{2} - \frac {\operatorname {atan}{\left (b x - i \right )}}{x} - \frac {i}{2 x} & \text {for}\: a = - i \\\frac {i b \operatorname {atan}{\left (b x + i \right )}}{2} - \frac {\operatorname {atan}{\left (b x + i \right )}}{x} + \frac {i}{2 x} & \text {for}\: a = i \\- \frac {2 a^{2} \operatorname {atan}{\left (a + b x \right )}}{2 a^{2} x + 2 x} - \frac {2 a b x \operatorname {atan}{\left (a + b x \right )}}{2 a^{2} x + 2 x} + \frac {2 b x \log {\left (x \right )}}{2 a^{2} x + 2 x} - \frac {b x \log {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{2 a^{2} x + 2 x} - \frac {2 \operatorname {atan}{\left (a + b x \right )}}{2 a^{2} x + 2 x} & \text {otherwise} \end {cases} \] Input:

integrate(atan(b*x+a)/x**2,x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.24 \[ \int \frac {\arctan (a+b x)}{x^2} \, dx=-\frac {1}{2} \, b {\left (\frac {2 \, a \arctan \left (\frac {b^{2} x + a b}{b}\right )}{a^{2} + 1} + \frac {\log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{a^{2} + 1} - \frac {2 \, \log \left (x\right )}{a^{2} + 1}\right )} - \frac {\arctan \left (b x + a\right )}{x} \] Input:

integrate(arctan(b*x+a)/x^2,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.13 \[ \int \frac {\arctan (a+b x)}{x^2} \, dx=-\frac {1}{2} \, b {\left (\frac {2 \, a \arctan \left (b x + a\right )}{a^{2} + 1} + \frac {\log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{a^{2} + 1} - \frac {2 \, \log \left ({\left | x \right |}\right )}{a^{2} + 1}\right )} - \frac {\arctan \left (b x + a\right )}{x} \] Input:

integrate(arctan(b*x+a)/x^2,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 1.43 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.02 \[ \int \frac {\arctan (a+b x)}{x^2} \, dx=-\frac {\mathrm {atan}\left (a+b\,x\right )}{x}-\frac {\frac {b\,x\,\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}{2}-b\,x\,\ln \left (x\right )+a\,b\,x\,\mathrm {atan}\left (a+b\,x\right )}{x\,\left (a^2+1\right )} \] Input:

int(atan(a + b*x)/x^2,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.15 \[ \int \frac {\arctan (a+b x)}{x^2} \, dx=\frac {-2 \mathit {atan} \left (b x +a \right ) a^{2}-2 \mathit {atan} \left (b x +a \right ) a b x -2 \mathit {atan} \left (b x +a \right )-\mathrm {log}\left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) b x +2 \,\mathrm {log}\left (x \right ) b x}{2 x \left (a^{2}+1\right )} \] Input:

int(atan(b*x+a)/x^2,x)
 

Output:

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