\(\int \frac {\arctan (a+b x)}{x^4} \, dx\) [8]

Optimal result

Integrand size = 10, antiderivative size = 129 \[ \int \frac {\arctan (a+b x)}{x^4} \, dx=-\frac {b}{6 \left (1+a^2\right ) x^2}+\frac {2 a b^2}{3 \left (1+a^2\right )^2 x}+\frac {a \left (3-a^2\right ) b^3 \arctan (a+b x)}{3 \left (1+a^2\right )^3}-\frac {\arctan (a+b x)}{3 x^3}-\frac {\left (1-3 a^2\right ) b^3 \log (x)}{3 \left (1+a^2\right )^3}+\frac {\left (1-3 a^2\right ) b^3 \log \left (1+(a+b x)^2\right )}{6 \left (1+a^2\right )^3} \] Output:

-1/6*b/(a^2+1)/x^2+2/3*a*b^2/(a^2+1)^2/x+1/3*a*(-a^2+3)*b^3*arctan(b*x+a)/ 
(a^2+1)^3-1/3*arctan(b*x+a)/x^3-1/3*(-3*a^2+1)*b^3*ln(x)/(a^2+1)^3+1/6*(-3 
*a^2+1)*b^3*ln(1+(b*x+a)^2)/(a^2+1)^3
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.99 \[ \int \frac {\arctan (a+b x)}{x^4} \, dx=\frac {-2 \left (1+a^2\right )^3 \arctan (a+b x)+2 \left (-1+3 a^2\right ) b^3 x^3 \log (x)+i (i+a)^3 b^3 x^3 \log (i-a-b x)-(-i+a) b x \left ((i+a) \left (1+a^2-4 a b x\right )+i (-i+a)^2 b^2 x^2 \log (i+a+b x)\right )}{6 \left (1+a^2\right )^3 x^3} \] Input:

Integrate[ArcTan[a + b*x]/x^4,x]
 

Output:

(-2*(1 + a^2)^3*ArcTan[a + b*x] + 2*(-1 + 3*a^2)*b^3*x^3*Log[x] + I*(I + a 
)^3*b^3*x^3*Log[I - a - b*x] - (-I + a)*b*x*((I + a)*(1 + a^2 - 4*a*b*x) + 
 I*(-I + a)^2*b^2*x^2*Log[I + a + b*x]))/(6*(1 + a^2)^3*x^3)
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5568, 896, 25, 480, 657, 2009}

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.

\(\displaystyle \int \frac {\arctan (a+b x)}{x^4} \, dx\)

\(\Big \downarrow \) 5568

\(\displaystyle \frac {1}{3} b \int \frac {1}{x^3 \left ((a+b x)^2+1\right )}dx-\frac {\arctan (a+b x)}{3 x^3}\)

\(\Big \downarrow \) 896

\(\displaystyle \frac {1}{3} b^3 \int \frac {1}{b^3 x^3 \left ((a+b x)^2+1\right )}d(a+b x)-\frac {\arctan (a+b x)}{3 x^3}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {1}{3} b^3 \int -\frac {1}{b^3 x^3 \left ((a+b x)^2+1\right )}d(a+b x)-\frac {\arctan (a+b x)}{3 x^3}\)

\(\Big \downarrow \) 480

\(\displaystyle \frac {1}{3} b^3 \left (-\frac {\int \frac {2 a+b x}{b^2 x^2 \left ((a+b x)^2+1\right )}d(a+b x)}{a^2+1}-\frac {1}{2 \left (a^2+1\right ) b^2 x^2}\right )-\frac {\arctan (a+b x)}{3 x^3}\)

\(\Big \downarrow \) 657

\(\displaystyle \frac {1}{3} b^3 \left (-\frac {\int \left (\frac {2 a}{\left (a^2+1\right ) b^2 x^2}-\frac {3 a^2-1}{\left (a^2+1\right )^2 b x}+\frac {-a \left (3-a^2\right )-\left (1-3 a^2\right ) (a+b x)}{\left (a^2+1\right )^2 \left ((a+b x)^2+1\right )}\right )d(a+b x)}{a^2+1}-\frac {1}{2 \left (a^2+1\right ) b^2 x^2}\right )-\frac {\arctan (a+b x)}{3 x^3}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} b^3 \left (-\frac {-\frac {\left (3-a^2\right ) a \arctan (a+b x)}{\left (a^2+1\right )^2}-\frac {2 a}{\left (a^2+1\right ) b x}+\frac {\left (1-3 a^2\right ) \log (-b x)}{\left (a^2+1\right )^2}-\frac {\left (1-3 a^2\right ) \log \left ((a+b x)^2+1\right )}{2 \left (a^2+1\right )^2}}{a^2+1}-\frac {1}{2 \left (a^2+1\right ) b^2 x^2}\right )-\frac {\arctan (a+b x)}{3 x^3}\)

Input:

Int[ArcTan[a + b*x]/x^4,x]
 

Output:

-1/3*ArcTan[a + b*x]/x^3 + (b^3*(-1/2*1/((1 + a^2)*b^2*x^2) - ((-2*a)/((1 
+ a^2)*b*x) - (a*(3 - a^2)*ArcTan[a + b*x])/(1 + a^2)^2 + ((1 - 3*a^2)*Log 
[-(b*x)])/(1 + a^2)^2 - ((1 - 3*a^2)*Log[1 + (a + b*x)^2])/(2*(1 + a^2)^2) 
)/(1 + a^2)))/3
 

Defintions of rubi rules used

rule 25

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

rule 480

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

rule 657

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( 
x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 
2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
 

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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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 = 0.14 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(b^{3} \left (-\frac {\arctan \left (b x +a \right )}{3 b^{3} x^{3}}-\frac {\left (-3 a^{2}+1\right ) \ln \left (-b x \right )}{3 \left (a^{2}+1\right )^{3}}-\frac {1}{6 \left (a^{2}+1\right ) b^{2} x^{2}}+\frac {2 a}{3 \left (a^{2}+1\right )^{2} b x}-\frac {\frac {\left (3 a^{2}-1\right ) \ln \left (1+\left (b x +a \right )^{2}\right )}{2}+\left (a^{3}-3 a \right ) \arctan \left (b x +a \right )}{3 \left (a^{2}+1\right )^{3}}\right )\)	\(115\)
default	\(b^{3} \left (-\frac {\arctan \left (b x +a \right )}{3 b^{3} x^{3}}-\frac {\left (-3 a^{2}+1\right ) \ln \left (-b x \right )}{3 \left (a^{2}+1\right )^{3}}-\frac {1}{6 \left (a^{2}+1\right ) b^{2} x^{2}}+\frac {2 a}{3 \left (a^{2}+1\right )^{2} b x}-\frac {\frac {\left (3 a^{2}-1\right ) \ln \left (1+\left (b x +a \right )^{2}\right )}{2}+\left (a^{3}-3 a \right ) \arctan \left (b x +a \right )}{3 \left (a^{2}+1\right )^{3}}\right )\)	\(115\)
parts	\(-\frac {\arctan \left (b x +a \right )}{3 x^{3}}+\frac {b \left (-\frac {b^{3} \left (\frac {\left (3 a^{2} b -b \right ) \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b^{2}}+\frac {\left (4 a^{3}-4 a -\frac {\left (3 a^{2} b -b \right ) a}{b}\right ) \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right )}{b}\right )}{\left (a^{2}+1\right )^{3}}-\frac {1}{2 \left (a^{2}+1\right ) x^{2}}+\frac {b^{2} \left (3 a^{2}-1\right ) \ln \left (x \right )}{\left (a^{2}+1\right )^{3}}+\frac {2 a b}{\left (a^{2}+1\right )^{2} x}\right )}{3}\)	\(155\)
parallelrisch	\(\frac {-2 x^{3} \arctan \left (b x +a \right ) a^{3} b^{3}+6 \ln \left (x \right ) x^{3} a^{2} b^{3}-3 \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) x^{3} a^{2} b^{3}+6 x^{3} \arctan \left (b x +a \right ) a \,b^{3}-7 x^{3} a^{2} b^{3}-2 b^{3} \ln \left (x \right ) x^{3}+b^{3} \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) x^{3}+4 a^{3} b^{2} x^{2}-2 \arctan \left (b x +a \right ) a^{6}+b^{3} x^{3}-a^{4} b x +4 a \,b^{2} x^{2}-6 \arctan \left (b x +a \right ) a^{4}-2 a^{2} b x -6 \arctan \left (b x +a \right ) a^{2}-b x -2 \arctan \left (b x +a \right )}{6 x^{3} \left (a^{6}+3 a^{4}+3 a^{2}+1\right )}\)	\(230\)
risch	\(\text {Expression too large to display}\)	\(1295\)

Input:

int(arctan(b*x+a)/x^4,x,method=_RETURNVERBOSE)
 

Output:

b^3*(-1/3*arctan(b*x+a)/b^3/x^3-1/3*(-3*a^2+1)/(a^2+1)^3*ln(-b*x)-1/6/(a^2 
+1)/b^2/x^2+2/3/(a^2+1)^2*a/b/x-1/3/(a^2+1)^3*(1/2*(3*a^2-1)*ln(1+(b*x+a)^ 
2)+(a^3-3*a)*arctan(b*x+a)))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.05 \[ \int \frac {\arctan (a+b x)}{x^4} \, dx=-\frac {{\left (3 \, a^{2} - 1\right )} b^{3} x^{3} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) - 2 \, {\left (3 \, a^{2} - 1\right )} b^{3} x^{3} \log \left (x\right ) - 4 \, {\left (a^{3} + a\right )} b^{2} x^{2} + {\left (a^{4} + 2 \, a^{2} + 1\right )} b x + 2 \, {\left ({\left (a^{3} - 3 \, a\right )} b^{3} x^{3} + a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1\right )} \arctan \left (b x + a\right )}{6 \, {\left (a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1\right )} x^{3}} \] Input:

integrate(arctan(b*x+a)/x^4,x, algorithm="fricas")
 

Output:

-1/6*((3*a^2 - 1)*b^3*x^3*log(b^2*x^2 + 2*a*b*x + a^2 + 1) - 2*(3*a^2 - 1) 
*b^3*x^3*log(x) - 4*(a^3 + a)*b^2*x^2 + (a^4 + 2*a^2 + 1)*b*x + 2*((a^3 - 
3*a)*b^3*x^3 + a^6 + 3*a^4 + 3*a^2 + 1)*arctan(b*x + a))/((a^6 + 3*a^4 + 3 
*a^2 + 1)*x^3)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.47 (sec) , antiderivative size = 760, normalized size of antiderivative = 5.89 \[ \int \frac {\arctan (a+b x)}{x^4} \, dx =\text {Too large to display} \] Input:

integrate(atan(b*x+a)/x**4,x)
 

Output:

Piecewise((I*b**3*atan(b*x - I)/24 + I*b**2/(24*x) - b/(24*x**2) - atan(b* 
x - I)/(3*x**3) - I/(18*x**3), Eq(a, -I)), (-I*b**3*atan(b*x + I)/24 - I*b 
**2/(24*x) - b/(24*x**2) - atan(b*x + I)/(3*x**3) + I/(18*x**3), Eq(a, I)) 
, (-2*a**6*atan(a + b*x)/(6*a**6*x**3 + 18*a**4*x**3 + 18*a**2*x**3 + 6*x* 
*3) - a**4*b*x/(6*a**6*x**3 + 18*a**4*x**3 + 18*a**2*x**3 + 6*x**3) - 6*a* 
*4*atan(a + b*x)/(6*a**6*x**3 + 18*a**4*x**3 + 18*a**2*x**3 + 6*x**3) - 2* 
a**3*b**3*x**3*atan(a + b*x)/(6*a**6*x**3 + 18*a**4*x**3 + 18*a**2*x**3 + 
6*x**3) + 4*a**3*b**2*x**2/(6*a**6*x**3 + 18*a**4*x**3 + 18*a**2*x**3 + 6* 
x**3) + 6*a**2*b**3*x**3*log(x)/(6*a**6*x**3 + 18*a**4*x**3 + 18*a**2*x**3 
 + 6*x**3) - 3*a**2*b**3*x**3*log(a**2 + 2*a*b*x + b**2*x**2 + 1)/(6*a**6* 
x**3 + 18*a**4*x**3 + 18*a**2*x**3 + 6*x**3) - 2*a**2*b*x/(6*a**6*x**3 + 1 
8*a**4*x**3 + 18*a**2*x**3 + 6*x**3) - 6*a**2*atan(a + b*x)/(6*a**6*x**3 + 
 18*a**4*x**3 + 18*a**2*x**3 + 6*x**3) + 6*a*b**3*x**3*atan(a + b*x)/(6*a* 
*6*x**3 + 18*a**4*x**3 + 18*a**2*x**3 + 6*x**3) + 4*a*b**2*x**2/(6*a**6*x* 
*3 + 18*a**4*x**3 + 18*a**2*x**3 + 6*x**3) - 2*b**3*x**3*log(x)/(6*a**6*x* 
*3 + 18*a**4*x**3 + 18*a**2*x**3 + 6*x**3) + b**3*x**3*log(a**2 + 2*a*b*x 
+ b**2*x**2 + 1)/(6*a**6*x**3 + 18*a**4*x**3 + 18*a**2*x**3 + 6*x**3) - b* 
x/(6*a**6*x**3 + 18*a**4*x**3 + 18*a**2*x**3 + 6*x**3) - 2*atan(a + b*x)/( 
6*a**6*x**3 + 18*a**4*x**3 + 18*a**2*x**3 + 6*x**3), True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.28 \[ \int \frac {\arctan (a+b x)}{x^4} \, dx=-\frac {1}{6} \, {\left (\frac {2 \, {\left (a^{3} - 3 \, a\right )} b^{2} \arctan \left (\frac {b^{2} x + a b}{b}\right )}{a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1} + \frac {{\left (3 \, a^{2} - 1\right )} b^{2} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1} - \frac {2 \, {\left (3 \, a^{2} - 1\right )} b^{2} \log \left (x\right )}{a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1} - \frac {4 \, a b x - a^{2} - 1}{{\left (a^{4} + 2 \, a^{2} + 1\right )} x^{2}}\right )} b - \frac {\arctan \left (b x + a\right )}{3 \, x^{3}} \] Input:

integrate(arctan(b*x+a)/x^4,x, algorithm="maxima")
 

Output:

-1/6*(2*(a^3 - 3*a)*b^2*arctan((b^2*x + a*b)/b)/(a^6 + 3*a^4 + 3*a^2 + 1) 
+ (3*a^2 - 1)*b^2*log(b^2*x^2 + 2*a*b*x + a^2 + 1)/(a^6 + 3*a^4 + 3*a^2 + 
1) - 2*(3*a^2 - 1)*b^2*log(x)/(a^6 + 3*a^4 + 3*a^2 + 1) - (4*a*b*x - a^2 - 
 1)/((a^4 + 2*a^2 + 1)*x^2))*b - 1/3*arctan(b*x + a)/x^3
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.37 \[ \int \frac {\arctan (a+b x)}{x^4} \, dx=-\frac {1}{6} \, b {\left (\frac {{\left (3 \, a^{2} b^{2} - b^{2}\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1} - \frac {2 \, {\left (3 \, a^{2} b^{2} - b^{2}\right )} \log \left ({\left | x \right |}\right )}{a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1} + \frac {2 \, {\left (a^{3} b^{3} - 3 \, a b^{3}\right )} \arctan \left (b x + a\right )}{{\left (a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1\right )} b} + \frac {a^{4} + 2 \, a^{2} - 4 \, {\left (a^{3} b + a b\right )} x + 1}{{\left (a^{2} + 1\right )}^{3} x^{2}}\right )} - \frac {\arctan \left (b x + a\right )}{3 \, x^{3}} \] Input:

integrate(arctan(b*x+a)/x^4,x, algorithm="giac")
 

Output:

-1/6*b*((3*a^2*b^2 - b^2)*log(b^2*x^2 + 2*a*b*x + a^2 + 1)/(a^6 + 3*a^4 + 
3*a^2 + 1) - 2*(3*a^2*b^2 - b^2)*log(abs(x))/(a^6 + 3*a^4 + 3*a^2 + 1) + 2 
*(a^3*b^3 - 3*a*b^3)*arctan(b*x + a)/((a^6 + 3*a^4 + 3*a^2 + 1)*b) + (a^4 
+ 2*a^2 - 4*(a^3*b + a*b)*x + 1)/((a^2 + 1)^3*x^2)) - 1/3*arctan(b*x + a)/ 
x^3
 

Mupad [B] (verification not implemented)

Time = 1.46 (sec) , antiderivative size = 288, normalized size of antiderivative = 2.23 \[ \int \frac {\arctan (a+b x)}{x^4} \, dx=-\frac {\frac {b\,x}{6}+\mathrm {atan}\left (a+b\,x\right )\,\left (\frac {a^2}{3}+\frac {1}{3}\right )+\frac {b^2\,x^2\,\mathrm {atan}\left (a+b\,x\right )}{3}+\frac {x^3\,\left (b^3-7\,a^2\,b^3\right )}{6\,\left (a^4+2\,a^2+1\right )}-\frac {a\,b^2\,x^2}{3\,\left (a^2+1\right )}-\frac {2\,a\,b^4\,x^4}{3\,{\left (a^2+1\right )}^2}+\frac {2\,a\,b\,x\,\mathrm {atan}\left (a+b\,x\right )}{3}}{a^2\,x^3+2\,a\,b\,x^4+b^2\,x^5+x^3}-\frac {\ln \left (x\right )\,\left (\frac {b^3}{3}-a^2\,b^3\right )}{a^6+3\,a^4+3\,a^2+1}-\frac {b^3\,\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )\,\left (3\,a^2-1\right )}{6\,\left (a^6+3\,a^4+3\,a^2+1\right )}-\frac {a\,\mathrm {atan}\left (\frac {2\,x\,b^2+2\,a\,b}{2\,\sqrt {b^2\,\left (a^2+1\right )-a^2\,b^2}}\right )\,\left (a^2-3\right )\,{\left (b^2\right )}^{3/2}}{3\,\left (a^6+3\,a^4+3\,a^2+1\right )} \] Input:

int(atan(a + b*x)/x^4,x)
 

Output:

- ((b*x)/6 + atan(a + b*x)*(a^2/3 + 1/3) + (b^2*x^2*atan(a + b*x))/3 + (x^ 
3*(b^3 - 7*a^2*b^3))/(6*(2*a^2 + a^4 + 1)) - (a*b^2*x^2)/(3*(a^2 + 1)) - ( 
2*a*b^4*x^4)/(3*(a^2 + 1)^2) + (2*a*b*x*atan(a + b*x))/3)/(x^3 + a^2*x^3 + 
 b^2*x^5 + 2*a*b*x^4) - (log(x)*(b^3/3 - a^2*b^3))/(3*a^2 + 3*a^4 + a^6 + 
1) - (b^3*log(a^2 + b^2*x^2 + 2*a*b*x + 1)*(3*a^2 - 1))/(6*(3*a^2 + 3*a^4 
+ a^6 + 1)) - (a*atan((2*a*b + 2*b^2*x)/(2*(b^2*(a^2 + 1) - a^2*b^2)^(1/2) 
))*(a^2 - 3)*(b^2)^(3/2))/(3*(3*a^2 + 3*a^4 + a^6 + 1))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.64 \[ \int \frac {\arctan (a+b x)}{x^4} \, dx=\frac {-2 \mathit {atan} \left (b x +a \right ) a^{6}-6 \mathit {atan} \left (b x +a \right ) a^{4}-2 \mathit {atan} \left (b x +a \right ) a^{3} b^{3} x^{3}-6 \mathit {atan} \left (b x +a \right ) a^{2}+6 \mathit {atan} \left (b x +a \right ) a \,b^{3} x^{3}-2 \mathit {atan} \left (b x +a \right )-3 \,\mathrm {log}\left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{2} b^{3} x^{3}+\mathrm {log}\left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) b^{3} x^{3}+6 \,\mathrm {log}\left (x \right ) a^{2} b^{3} x^{3}-2 \,\mathrm {log}\left (x \right ) b^{3} x^{3}-a^{4} b x +4 a^{3} b^{2} x^{2}-2 a^{2} b x +4 a \,b^{2} x^{2}-b x}{6 x^{3} \left (a^{6}+3 a^{4}+3 a^{2}+1\right )} \] Input:

int(atan(b*x+a)/x^4,x)
 

Output:

( - 2*atan(a + b*x)*a**6 - 6*atan(a + b*x)*a**4 - 2*atan(a + b*x)*a**3*b** 
3*x**3 - 6*atan(a + b*x)*a**2 + 6*atan(a + b*x)*a*b**3*x**3 - 2*atan(a + b 
*x) - 3*log(a**2 + 2*a*b*x + b**2*x**2 + 1)*a**2*b**3*x**3 + log(a**2 + 2* 
a*b*x + b**2*x**2 + 1)*b**3*x**3 + 6*log(x)*a**2*b**3*x**3 - 2*log(x)*b**3 
*x**3 - a**4*b*x + 4*a**3*b**2*x**2 - 2*a**2*b*x + 4*a*b**2*x**2 - b*x)/(6 
*x**3*(a**6 + 3*a**4 + 3*a**2 + 1))