Integrand size = 17, antiderivative size = 105 \[ \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {1}{16 a c^3 \left (1+a^2 x^2\right )^2}+\frac {3}{16 a c^3 \left (1+a^2 x^2\right )}+\frac {x \arctan (a x)}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac {3 x \arctan (a x)}{8 c^3 \left (1+a^2 x^2\right )}+\frac {3 \arctan (a x)^2}{16 a c^3} \] Output:
1/16/a/c^3/(a^2*x^2+1)^2+3/16/a/c^3/(a^2*x^2+1)+1/4*x*arctan(a*x)/c^3/(a^2 *x^2+1)^2+3/8*x*arctan(a*x)/c^3/(a^2*x^2+1)+3/16*arctan(a*x)^2/a/c^3
Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.65 \[ \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {4+3 a^2 x^2+2 a x \left (5+3 a^2 x^2\right ) \arctan (a x)+3 \left (1+a^2 x^2\right )^2 \arctan (a x)^2}{16 a c^3 \left (1+a^2 x^2\right )^2} \] Input:
Integrate[ArcTan[a*x]/(c + a^2*c*x^2)^3,x]
Output:
(4 + 3*a^2*x^2 + 2*a*x*(5 + 3*a^2*x^2)*ArcTan[a*x] + 3*(1 + a^2*x^2)^2*Arc Tan[a*x]^2)/(16*a*c^3*(1 + a^2*x^2)^2)
Time = 0.31 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {5431, 27, 5427, 241}
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.
\(\displaystyle \int \frac {\arctan (a x)}{\left (a^2 c x^2+c\right )^3} \, dx\) |
\(\Big \downarrow \) 5431 |
\(\displaystyle \frac {3 \int \frac {\arctan (a x)}{c^2 \left (a^2 x^2+1\right )^2}dx}{4 c}+\frac {x \arctan (a x)}{4 c^3 \left (a^2 x^2+1\right )^2}+\frac {1}{16 a c^3 \left (a^2 x^2+1\right )^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3 \int \frac {\arctan (a x)}{\left (a^2 x^2+1\right )^2}dx}{4 c^3}+\frac {x \arctan (a x)}{4 c^3 \left (a^2 x^2+1\right )^2}+\frac {1}{16 a c^3 \left (a^2 x^2+1\right )^2}\) |
\(\Big \downarrow \) 5427 |
\(\displaystyle \frac {3 \left (-\frac {1}{2} a \int \frac {x}{\left (a^2 x^2+1\right )^2}dx+\frac {x \arctan (a x)}{2 \left (a^2 x^2+1\right )}+\frac {\arctan (a x)^2}{4 a}\right )}{4 c^3}+\frac {x \arctan (a x)}{4 c^3 \left (a^2 x^2+1\right )^2}+\frac {1}{16 a c^3 \left (a^2 x^2+1\right )^2}\) |
\(\Big \downarrow \) 241 |
\(\displaystyle \frac {x \arctan (a x)}{4 c^3 \left (a^2 x^2+1\right )^2}+\frac {3 \left (\frac {x \arctan (a x)}{2 \left (a^2 x^2+1\right )}+\frac {1}{4 a \left (a^2 x^2+1\right )}+\frac {\arctan (a x)^2}{4 a}\right )}{4 c^3}+\frac {1}{16 a c^3 \left (a^2 x^2+1\right )^2}\) |
Input:
Int[ArcTan[a*x]/(c + a^2*c*x^2)^3,x]
Output:
1/(16*a*c^3*(1 + a^2*x^2)^2) + (x*ArcTan[a*x])/(4*c^3*(1 + a^2*x^2)^2) + ( 3*(1/(4*a*(1 + a^2*x^2)) + (x*ArcTan[a*x])/(2*(1 + a^2*x^2)) + ArcTan[a*x] ^2/(4*a)))/(4*c^3)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x^2)^(p + 1)/ (2*b*(p + 1)), x] /; FreeQ[{a, b, p}, x] && NeQ[p, -1]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2)^2, x_Sym bol] :> Simp[x*((a + b*ArcTan[c*x])^p/(2*d*(d + e*x^2))), x] + (Simp[(a + b *ArcTan[c*x])^(p + 1)/(2*b*c*d^2*(p + 1)), x] - Simp[b*c*(p/2) Int[x*((a + b*ArcTan[c*x])^(p - 1)/(d + e*x^2)^2), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[p, 0]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol ] :> Simp[b*((d + e*x^2)^(q + 1)/(4*c*d*(q + 1)^2)), x] + (-Simp[x*(d + e*x ^2)^(q + 1)*((a + b*ArcTan[c*x])/(2*d*(q + 1))), x] + Simp[(2*q + 3)/(2*d*( q + 1)) Int[(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x]), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && LtQ[q, -1] && NeQ[q, -3/2]
Time = 0.60 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.89
method | result | size |
parallelrisch | \(\frac {3 a^{4} \arctan \left (a x \right )^{2} x^{4}-4 a^{4} x^{4}+6 \arctan \left (a x \right ) x^{3} a^{3}+6 \arctan \left (a x \right )^{2} x^{2} a^{2}-5 a^{2} x^{2}+10 \arctan \left (a x \right ) a x +3 \arctan \left (a x \right )^{2}}{16 c^{3} \left (a^{2} x^{2}+1\right )^{2} a}\) | \(93\) |
derivativedivides | \(\frac {\frac {a x \arctan \left (a x \right )}{4 c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {3 a x \arctan \left (a x \right )}{8 c^{3} \left (a^{2} x^{2}+1\right )}+\frac {3 \arctan \left (a x \right )^{2}}{8 c^{3}}-\frac {-\frac {3}{2 \left (a^{2} x^{2}+1\right )}-\frac {1}{2 \left (a^{2} x^{2}+1\right )^{2}}+\frac {3 \arctan \left (a x \right )^{2}}{2}}{8 c^{3}}}{a}\) | \(101\) |
default | \(\frac {\frac {a x \arctan \left (a x \right )}{4 c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {3 a x \arctan \left (a x \right )}{8 c^{3} \left (a^{2} x^{2}+1\right )}+\frac {3 \arctan \left (a x \right )^{2}}{8 c^{3}}-\frac {-\frac {3}{2 \left (a^{2} x^{2}+1\right )}-\frac {1}{2 \left (a^{2} x^{2}+1\right )^{2}}+\frac {3 \arctan \left (a x \right )^{2}}{2}}{8 c^{3}}}{a}\) | \(101\) |
parts | \(\frac {x \arctan \left (a x \right )}{4 c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {3 x \arctan \left (a x \right )}{8 c^{3} \left (a^{2} x^{2}+1\right )}+\frac {3 \arctan \left (a x \right )^{2}}{8 a \,c^{3}}-\frac {\frac {-\frac {3}{2 \left (a^{2} x^{2}+1\right )}-\frac {1}{2 \left (a^{2} x^{2}+1\right )^{2}}}{a}+\frac {3 \arctan \left (a x \right )^{2}}{2 a}}{8 c^{3}}\) | \(106\) |
risch | \(-\frac {3 \ln \left (i a x +1\right )^{2}}{64 a \,c^{3}}+\frac {\left (3 x^{4} \ln \left (-i a x +1\right ) a^{4}+6 a^{2} x^{2} \ln \left (-i a x +1\right )-6 i a^{3} x^{3}+3 \ln \left (-i a x +1\right )-10 i a x \right ) \ln \left (i a x +1\right )}{32 a \left (a^{2} x^{2}+1\right )^{2} c^{3}}-\frac {3 a^{4} x^{4} \ln \left (-i a x +1\right )^{2}+6 a^{2} x^{2} \ln \left (-i a x +1\right )^{2}-12 i a^{3} x^{3} \ln \left (-i a x +1\right )-12 a^{2} x^{2}+3 \ln \left (-i a x +1\right )^{2}-20 i a x \ln \left (-i a x +1\right )-16}{64 \left (a x +i\right )^{2} a \,c^{3} \left (a x -i\right )^{2}}\) | \(216\) |
Input:
int(arctan(a*x)/(a^2*c*x^2+c)^3,x,method=_RETURNVERBOSE)
Output:
1/16*(3*a^4*arctan(a*x)^2*x^4-4*a^4*x^4+6*arctan(a*x)*x^3*a^3+6*arctan(a*x )^2*x^2*a^2-5*a^2*x^2+10*arctan(a*x)*a*x+3*arctan(a*x)^2)/c^3/(a^2*x^2+1)^ 2/a
Time = 0.10 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.81 \[ \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {3 \, a^{2} x^{2} + 3 \, {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} + 2 \, {\left (3 \, a^{3} x^{3} + 5 \, a x\right )} \arctan \left (a x\right ) + 4}{16 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{3} c^{3} x^{2} + a c^{3}\right )}} \] Input:
integrate(arctan(a*x)/(a^2*c*x^2+c)^3,x, algorithm="fricas")
Output:
1/16*(3*a^2*x^2 + 3*(a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x)^2 + 2*(3*a^3*x^3 + 5*a*x)*arctan(a*x) + 4)/(a^5*c^3*x^4 + 2*a^3*c^3*x^2 + a*c^3)
Exception generated. \[ \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\text {Exception raised: RecursionError} \] Input:
integrate(atan(a*x)/(a**2*c*x**2+c)**3,x)
Output:
Exception raised: RecursionError >> maximum recursion depth exceeded in co mparison
Time = 0.12 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.23 \[ \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {1}{8} \, {\left (\frac {3 \, a^{2} x^{3} + 5 \, x}{a^{4} c^{3} x^{4} + 2 \, a^{2} c^{3} x^{2} + c^{3}} + \frac {3 \, \arctan \left (a x\right )}{a c^{3}}\right )} \arctan \left (a x\right ) + \frac {{\left (3 \, a^{2} x^{2} - 3 \, {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} + 4\right )} a}{16 \, {\left (a^{6} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{2} c^{3}\right )}} \] Input:
integrate(arctan(a*x)/(a^2*c*x^2+c)^3,x, algorithm="maxima")
Output:
1/8*((3*a^2*x^3 + 5*x)/(a^4*c^3*x^4 + 2*a^2*c^3*x^2 + c^3) + 3*arctan(a*x) /(a*c^3))*arctan(a*x) + 1/16*(3*a^2*x^2 - 3*(a^4*x^4 + 2*a^2*x^2 + 1)*arct an(a*x)^2 + 4)*a/(a^6*c^3*x^4 + 2*a^4*c^3*x^2 + a^2*c^3)
\[ \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\int { \frac {\arctan \left (a x\right )}{{\left (a^{2} c x^{2} + c\right )}^{3}} \,d x } \] Input:
integrate(arctan(a*x)/(a^2*c*x^2+c)^3,x, algorithm="giac")
Output:
integrate(arctan(a*x)/(a^2*c*x^2 + c)^3, x)
Time = 0.73 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.81 \[ \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {3\,a^4\,x^4\,{\mathrm {atan}\left (a\,x\right )}^2+6\,a^3\,x^3\,\mathrm {atan}\left (a\,x\right )+6\,a^2\,x^2\,{\mathrm {atan}\left (a\,x\right )}^2+3\,a^2\,x^2+10\,a\,x\,\mathrm {atan}\left (a\,x\right )+3\,{\mathrm {atan}\left (a\,x\right )}^2+4}{16\,a\,c^3\,{\left (a^2\,x^2+1\right )}^2} \] Input:
int(atan(a*x)/(c + a^2*c*x^2)^3,x)
Output:
(3*a^2*x^2 + 3*atan(a*x)^2 + 6*a^3*x^3*atan(a*x) + 10*a*x*atan(a*x) + 6*a^ 2*x^2*atan(a*x)^2 + 3*a^4*x^4*atan(a*x)^2 + 4)/(16*a*c^3*(a^2*x^2 + 1)^2)
Time = 0.20 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.89 \[ \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {6 \mathit {atan} \left (a x \right )^{2} a^{4} x^{4}+12 \mathit {atan} \left (a x \right )^{2} a^{2} x^{2}+6 \mathit {atan} \left (a x \right )^{2}+12 \mathit {atan} \left (a x \right ) a^{3} x^{3}+20 \mathit {atan} \left (a x \right ) a x -3 a^{4} x^{4}+5}{32 a \,c^{3} \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right )} \] Input:
int(atan(a*x)/(a^2*c*x^2+c)^3,x)
Output:
(6*atan(a*x)**2*a**4*x**4 + 12*atan(a*x)**2*a**2*x**2 + 6*atan(a*x)**2 + 1 2*atan(a*x)*a**3*x**3 + 20*atan(a*x)*a*x - 3*a**4*x**4 + 5)/(32*a*c**3*(a* *4*x**4 + 2*a**2*x**2 + 1))