Integrand size = 20, antiderivative size = 52 \[ \int \frac {\arctan (a x)}{x^2 \left (c+a^2 c x^2\right )} \, dx=-\frac {\arctan (a x)}{c x}-\frac {a \arctan (a x)^2}{2 c}+\frac {a \log (x)}{c}-\frac {a \log \left (1+a^2 x^2\right )}{2 c} \] Output:
-arctan(a*x)/c/x-1/2*a*arctan(a*x)^2/c+a*ln(x)/c-1/2*a*ln(a^2*x^2+1)/c
Time = 0.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00 \[ \int \frac {\arctan (a x)}{x^2 \left (c+a^2 c x^2\right )} \, dx=-\frac {\arctan (a x)}{c x}-\frac {a \arctan (a x)^2}{2 c}+\frac {a \log (x)}{c}-\frac {a \log \left (1+a^2 x^2\right )}{2 c} \] Input:
Integrate[ArcTan[a*x]/(x^2*(c + a^2*c*x^2)),x]
Output:
-(ArcTan[a*x]/(c*x)) - (a*ArcTan[a*x]^2)/(2*c) + (a*Log[x])/c - (a*Log[1 + a^2*x^2])/(2*c)
Time = 0.38 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5453, 27, 5361, 243, 47, 14, 16, 5419}
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.
\(\displaystyle \int \frac {\arctan (a x)}{x^2 \left (a^2 c x^2+c\right )} \, dx\) |
\(\Big \downarrow \) 5453 |
\(\displaystyle \frac {\int \frac {\arctan (a x)}{x^2}dx}{c}-a^2 \int \frac {\arctan (a x)}{c \left (a^2 x^2+1\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\arctan (a x)}{x^2}dx}{c}-\frac {a^2 \int \frac {\arctan (a x)}{a^2 x^2+1}dx}{c}\) |
\(\Big \downarrow \) 5361 |
\(\displaystyle \frac {a \int \frac {1}{x \left (a^2 x^2+1\right )}dx-\frac {\arctan (a x)}{x}}{c}-\frac {a^2 \int \frac {\arctan (a x)}{a^2 x^2+1}dx}{c}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\frac {1}{2} a \int \frac {1}{x^2 \left (a^2 x^2+1\right )}dx^2-\frac {\arctan (a x)}{x}}{c}-\frac {a^2 \int \frac {\arctan (a x)}{a^2 x^2+1}dx}{c}\) |
\(\Big \downarrow \) 47 |
\(\displaystyle \frac {\frac {1}{2} a \left (\int \frac {1}{x^2}dx^2-a^2 \int \frac {1}{a^2 x^2+1}dx^2\right )-\frac {\arctan (a x)}{x}}{c}-\frac {a^2 \int \frac {\arctan (a x)}{a^2 x^2+1}dx}{c}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {\frac {1}{2} a \left (\log \left (x^2\right )-a^2 \int \frac {1}{a^2 x^2+1}dx^2\right )-\frac {\arctan (a x)}{x}}{c}-\frac {a^2 \int \frac {\arctan (a x)}{a^2 x^2+1}dx}{c}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {\frac {1}{2} a \left (\log \left (x^2\right )-\log \left (a^2 x^2+1\right )\right )-\frac {\arctan (a x)}{x}}{c}-\frac {a^2 \int \frac {\arctan (a x)}{a^2 x^2+1}dx}{c}\) |
\(\Big \downarrow \) 5419 |
\(\displaystyle \frac {\frac {1}{2} a \left (\log \left (x^2\right )-\log \left (a^2 x^2+1\right )\right )-\frac {\arctan (a x)}{x}}{c}-\frac {a \arctan (a x)^2}{2 c}\) |
Input:
Int[ArcTan[a*x]/(x^2*(c + a^2*c*x^2)),x]
Output:
-1/2*(a*ArcTan[a*x]^2)/c + (-(ArcTan[a*x]/x) + (a*(Log[x^2] - Log[1 + a^2* x^2]))/2)/c
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[b/(b*c - a*d) Int[1/(a + b*x), x], x] - Simp[d/(b*c - a*d) Int[1/(c + d*x), x ], x] /; FreeQ[{a, b, c, d}, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m + 1)) Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] & & IntegerQ[m])) && NeQ[m, -1]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbo l] :> Simp[(a + b*ArcTan[c*x])^(p + 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && NeQ[p, -1]
Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e _.)*(x_)^2), x_Symbol] :> Simp[1/d Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Simp[e/(d*f^2) Int[(f*x)^(m + 2)*((a + b*ArcTan[c*x])^p/(d + e*x^2) ), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]
Time = 0.43 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.88
method | result | size |
parallelrisch | \(\frac {-a \arctan \left (a x \right )^{2} x +2 a x \ln \left (x \right )-a \ln \left (a^{2} x^{2}+1\right ) x -2 \arctan \left (a x \right )}{2 c x}\) | \(46\) |
parts | \(-\frac {a \arctan \left (a x \right )^{2}}{c}-\frac {\arctan \left (a x \right )}{c x}-\frac {-\frac {a \arctan \left (a x \right )^{2}}{2}-a \left (-\frac {\ln \left (a^{2} x^{2}+1\right )}{2}+\ln \left (a x \right )\right )}{c}\) | \(61\) |
derivativedivides | \(a \left (-\frac {\arctan \left (a x \right )^{2}}{c}-\frac {\arctan \left (a x \right )}{c a x}-\frac {\frac {\ln \left (a^{2} x^{2}+1\right )}{2}-\ln \left (a x \right )-\frac {\arctan \left (a x \right )^{2}}{2}}{c}\right )\) | \(62\) |
default | \(a \left (-\frac {\arctan \left (a x \right )^{2}}{c}-\frac {\arctan \left (a x \right )}{c a x}-\frac {\frac {\ln \left (a^{2} x^{2}+1\right )}{2}-\ln \left (a x \right )-\frac {\arctan \left (a x \right )^{2}}{2}}{c}\right )\) | \(62\) |
risch | \(\frac {a \ln \left (i a x +1\right )^{2}}{8 c}-\frac {\left (a x \ln \left (-i a x +1\right )-2 i\right ) \ln \left (i a x +1\right )}{4 c x}-\frac {-a \ln \left (-i a x +1\right )^{2} x -8 a x \ln \left (x \right )+4 a x \ln \left (-3 a^{2} x^{2}-3\right )+4 i \ln \left (-i a x +1\right )}{8 c x}\) | \(103\) |
Input:
int(arctan(a*x)/x^2/(a^2*c*x^2+c),x,method=_RETURNVERBOSE)
Output:
1/2*(-a*arctan(a*x)^2*x+2*a*x*ln(x)-a*ln(a^2*x^2+1)*x-2*arctan(a*x))/c/x
Time = 0.09 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.83 \[ \int \frac {\arctan (a x)}{x^2 \left (c+a^2 c x^2\right )} \, dx=-\frac {a x \arctan \left (a x\right )^{2} + a x \log \left (a^{2} x^{2} + 1\right ) - 2 \, a x \log \left (x\right ) + 2 \, \arctan \left (a x\right )}{2 \, c x} \] Input:
integrate(arctan(a*x)/x^2/(a^2*c*x^2+c),x, algorithm="fricas")
Output:
-1/2*(a*x*arctan(a*x)^2 + a*x*log(a^2*x^2 + 1) - 2*a*x*log(x) + 2*arctan(a *x))/(c*x)
Time = 0.40 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.81 \[ \int \frac {\arctan (a x)}{x^2 \left (c+a^2 c x^2\right )} \, dx=\begin {cases} \frac {a \log {\left (x \right )}}{c} - \frac {a \log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{2 c} - \frac {a \operatorname {atan}^{2}{\left (a x \right )}}{2 c} - \frac {\operatorname {atan}{\left (a x \right )}}{c x} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \] Input:
integrate(atan(a*x)/x**2/(a**2*c*x**2+c),x)
Output:
Piecewise((a*log(x)/c - a*log(x**2 + a**(-2))/(2*c) - a*atan(a*x)**2/(2*c) - atan(a*x)/(c*x), Ne(a, 0)), (0, True))
Time = 0.12 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.02 \[ \int \frac {\arctan (a x)}{x^2 \left (c+a^2 c x^2\right )} \, dx=-{\left (\frac {a \arctan \left (a x\right )}{c} + \frac {1}{c x}\right )} \arctan \left (a x\right ) + \frac {{\left (\arctan \left (a x\right )^{2} - \log \left (a^{2} x^{2} + 1\right ) + 2 \, \log \left (x\right )\right )} a}{2 \, c} \] Input:
integrate(arctan(a*x)/x^2/(a^2*c*x^2+c),x, algorithm="maxima")
Output:
-(a*arctan(a*x)/c + 1/(c*x))*arctan(a*x) + 1/2*(arctan(a*x)^2 - log(a^2*x^ 2 + 1) + 2*log(x))*a/c
\[ \int \frac {\arctan (a x)}{x^2 \left (c+a^2 c x^2\right )} \, dx=\int { \frac {\arctan \left (a x\right )}{{\left (a^{2} c x^{2} + c\right )} x^{2}} \,d x } \] Input:
integrate(arctan(a*x)/x^2/(a^2*c*x^2+c),x, algorithm="giac")
Output:
integrate(arctan(a*x)/((a^2*c*x^2 + c)*x^2), x)
Time = 0.72 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.92 \[ \int \frac {\arctan (a x)}{x^2 \left (c+a^2 c x^2\right )} \, dx=\frac {a\,\ln \left (x\right )}{c}-\frac {a\,\ln \left (a^2\,x^2+1\right )}{2\,c}-\frac {a\,{\mathrm {atan}\left (a\,x\right )}^2}{2\,c}-\frac {\mathrm {atan}\left (a\,x\right )}{c\,x} \] Input:
int(atan(a*x)/(x^2*(c + a^2*c*x^2)),x)
Output:
(a*log(x))/c - (a*log(a^2*x^2 + 1))/(2*c) - (a*atan(a*x)^2)/(2*c) - atan(a *x)/(c*x)
Time = 0.21 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.87 \[ \int \frac {\arctan (a x)}{x^2 \left (c+a^2 c x^2\right )} \, dx=\frac {-\mathit {atan} \left (a x \right )^{2} a x -2 \mathit {atan} \left (a x \right )-\mathrm {log}\left (a^{2} x^{2}+1\right ) a x +2 \,\mathrm {log}\left (x \right ) a x}{2 c x} \] Input:
int(atan(a*x)/x^2/(a^2*c*x^2+c),x)
Output:
( - atan(a*x)**2*a*x - 2*atan(a*x) - log(a**2*x**2 + 1)*a*x + 2*log(x)*a*x )/(2*c*x)