Integrand size = 18, antiderivative size = 40 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^2} \, dx=-\frac {c \arctan (a x)}{x}+a^2 c x \arctan (a x)+a c \log (x)-a c \log \left (1+a^2 x^2\right ) \]
Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^2} \, dx=-\frac {c \arctan (a x)}{x}+a^2 c x \arctan (a x)+a c \log (x)-a c \log \left (1+a^2 x^2\right ) \]
Time = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.60, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {5485, 5345, 240, 5361, 243, 47, 14, 16}
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.
\(\displaystyle \int \frac {\arctan (a x) \left (a^2 c x^2+c\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 5485 |
\(\displaystyle a^2 c \int \arctan (a x)dx+c \int \frac {\arctan (a x)}{x^2}dx\) |
\(\Big \downarrow \) 5345 |
\(\displaystyle a^2 c \left (x \arctan (a x)-a \int \frac {x}{a^2 x^2+1}dx\right )+c \int \frac {\arctan (a x)}{x^2}dx\) |
\(\Big \downarrow \) 240 |
\(\displaystyle c \int \frac {\arctan (a x)}{x^2}dx+a^2 c \left (x \arctan (a x)-\frac {\log \left (a^2 x^2+1\right )}{2 a}\right )\) |
\(\Big \downarrow \) 5361 |
\(\displaystyle c \left (a \int \frac {1}{x \left (a^2 x^2+1\right )}dx-\frac {\arctan (a x)}{x}\right )+a^2 c \left (x \arctan (a x)-\frac {\log \left (a^2 x^2+1\right )}{2 a}\right )\) |
\(\Big \downarrow \) 243 |
\(\displaystyle c \left (\frac {1}{2} a \int \frac {1}{x^2 \left (a^2 x^2+1\right )}dx^2-\frac {\arctan (a x)}{x}\right )+a^2 c \left (x \arctan (a x)-\frac {\log \left (a^2 x^2+1\right )}{2 a}\right )\) |
\(\Big \downarrow \) 47 |
\(\displaystyle c \left (\frac {1}{2} a \left (\int \frac {1}{x^2}dx^2-a^2 \int \frac {1}{a^2 x^2+1}dx^2\right )-\frac {\arctan (a x)}{x}\right )+a^2 c \left (x \arctan (a x)-\frac {\log \left (a^2 x^2+1\right )}{2 a}\right )\) |
\(\Big \downarrow \) 14 |
\(\displaystyle c \left (\frac {1}{2} a \left (\log \left (x^2\right )-a^2 \int \frac {1}{a^2 x^2+1}dx^2\right )-\frac {\arctan (a x)}{x}\right )+a^2 c \left (x \arctan (a x)-\frac {\log \left (a^2 x^2+1\right )}{2 a}\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle a^2 c \left (x \arctan (a x)-\frac {\log \left (a^2 x^2+1\right )}{2 a}\right )+c \left (\frac {1}{2} a \left (\log \left (x^2\right )-\log \left (a^2 x^2+1\right )\right )-\frac {\arctan (a x)}{x}\right )\) |
c*(-(ArcTan[a*x]/x) + (a*(Log[x^2] - Log[1 + a^2*x^2]))/2) + a^2*c*(x*ArcT an[a*x] - Log[1 + a^2*x^2]/(2*a))
3.2.54.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, 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_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x ^2, x]]/(2*b), x] /; FreeQ[{a, b}, 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_Symbol] :> Simp[x*(a + b*ArcTan[c*x^n])^p, x] - Simp[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])
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_.)*((f_.)*(x_))^(m_)*((d_) + (e_. )*(x_)^2)^(q_.), x_Symbol] :> Simp[d Int[(f*x)^m*(d + e*x^2)^(q - 1)*(a + b*ArcTan[c*x])^p, x], x] + Simp[c^2*(d/f^2) Int[(f*x)^(m + 2)*(d + e*x^2 )^(q - 1)*(a + b*ArcTan[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && GtQ[q, 0] && IGtQ[p, 0] && (RationalQ[m] || (EqQ[p, 1] && IntegerQ[q]))
Time = 0.12 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.02
method | result | size |
parts | \(a^{2} c x \arctan \left (a x \right )-\frac {c \arctan \left (a x \right )}{x}-c a \left (\ln \left (a^{2} x^{2}+1\right )-\ln \left (x \right )\right )\) | \(41\) |
derivativedivides | \(a \left (a c x \arctan \left (a x \right )-\frac {c \arctan \left (a x \right )}{a x}-c \left (-\ln \left (a x \right )+\ln \left (a^{2} x^{2}+1\right )\right )\right )\) | \(45\) |
default | \(a \left (a c x \arctan \left (a x \right )-\frac {c \arctan \left (a x \right )}{a x}-c \left (-\ln \left (a x \right )+\ln \left (a^{2} x^{2}+1\right )\right )\right )\) | \(45\) |
parallelrisch | \(\frac {a^{2} c \,x^{2} \arctan \left (a x \right )+c a \ln \left (x \right ) x -c a \ln \left (a^{2} x^{2}+1\right ) x -c \arctan \left (a x \right )}{x}\) | \(46\) |
risch | \(-\frac {i c \left (a^{2} x^{2}-1\right ) \ln \left (i a x +1\right )}{2 x}+\frac {i c \left (a^{2} x^{2} \ln \left (-i a x +1\right )-2 i a \ln \left (x \right ) x +2 i a \ln \left (-2 a^{2} x^{2}-2\right ) x -\ln \left (-i a x +1\right )\right )}{2 x}\) | \(82\) |
meijerg | \(\frac {a c \left (\frac {4 a^{2} x^{2} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}-2 \ln \left (a^{2} x^{2}+1\right )\right )}{4}+\frac {a c \left (4 \ln \left (x \right )+4 \ln \left (a \right )-\frac {4 \arctan \left (\sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}-2 \ln \left (a^{2} x^{2}+1\right )\right )}{4}\) | \(92\) |
Time = 0.25 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.12 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^2} \, dx=-\frac {a c x \log \left (a^{2} x^{2} + 1\right ) - a c x \log \left (x\right ) - {\left (a^{2} c x^{2} - c\right )} \arctan \left (a x\right )}{x} \]
Time = 0.27 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.02 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^2} \, dx=\begin {cases} a^{2} c x \operatorname {atan}{\left (a x \right )} + a c \log {\left (x \right )} - a c \log {\left (x^{2} + \frac {1}{a^{2}} \right )} - \frac {c \operatorname {atan}{\left (a x \right )}}{x} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]
Piecewise((a**2*c*x*atan(a*x) + a*c*log(x) - a*c*log(x**2 + a**(-2)) - c*a tan(a*x)/x, Ne(a, 0)), (0, True))
Time = 0.23 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^2} \, dx=-{\left (c \log \left (a^{2} x^{2} + 1\right ) - c \log \left (x\right )\right )} a + {\left (a^{2} c x - \frac {c}{x}\right )} \arctan \left (a x\right ) \]
\[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^2} \, dx=\int { \frac {{\left (a^{2} c x^{2} + c\right )} \arctan \left (a x\right )}{x^{2}} \,d x } \]
Time = 0.18 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.05 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^2} \, dx=a^2\,c\,x\,\mathrm {atan}\left (a\,x\right )-\frac {c\,\mathrm {atan}\left (a\,x\right )}{x}-c\,\left (a\,\ln \left (a^2\,x^2+1\right )-a\,\ln \left (x\right )\right ) \]