Integrand size = 20, antiderivative size = 85 \[ \int \frac {\left (c+a^2 c x^2\right )^2 \arctan (a x)}{x^4} \, dx=-\frac {a c^2}{6 x^2}-\frac {c^2 \arctan (a x)}{3 x^3}-\frac {2 a^2 c^2 \arctan (a x)}{x}+a^4 c^2 x \arctan (a x)+\frac {5}{3} a^3 c^2 \log (x)-\frac {4}{3} a^3 c^2 \log \left (1+a^2 x^2\right ) \]
-1/6*a*c^2/x^2-1/3*c^2*arctan(a*x)/x^3-2*a^2*c^2*arctan(a*x)/x+a^4*c^2*x*a rctan(a*x)+5/3*a^3*c^2*ln(x)-4/3*a^3*c^2*ln(a^2*x^2+1)
Time = 0.04 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.80 \[ \int \frac {\left (c+a^2 c x^2\right )^2 \arctan (a x)}{x^4} \, dx=\frac {c^2 \left (2 \left (-1-6 a^2 x^2+3 a^4 x^4\right ) \arctan (a x)+a x \left (-1+10 a^2 x^2 \log (x)-8 a^2 x^2 \log \left (1+a^2 x^2\right )\right )\right )}{6 x^3} \]
(c^2*(2*(-1 - 6*a^2*x^2 + 3*a^4*x^4)*ArcTan[a*x] + a*x*(-1 + 10*a^2*x^2*Lo g[x] - 8*a^2*x^2*Log[1 + a^2*x^2])))/(6*x^3)
Time = 0.30 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {5483, 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.
\(\displaystyle \int \frac {\arctan (a x) \left (a^2 c x^2+c\right )^2}{x^4} \, dx\) |
\(\Big \downarrow \) 5483 |
\(\displaystyle \int \left (a^4 c^2 \arctan (a x)+\frac {2 a^2 c^2 \arctan (a x)}{x^2}+\frac {c^2 \arctan (a x)}{x^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle a^4 c^2 x \arctan (a x)+\frac {5}{3} a^3 c^2 \log (x)-\frac {2 a^2 c^2 \arctan (a x)}{x}-\frac {4}{3} a^3 c^2 \log \left (a^2 x^2+1\right )-\frac {c^2 \arctan (a x)}{3 x^3}-\frac {a c^2}{6 x^2}\) |
-1/6*(a*c^2)/x^2 - (c^2*ArcTan[a*x])/(3*x^3) - (2*a^2*c^2*ArcTan[a*x])/x + a^4*c^2*x*ArcTan[a*x] + (5*a^3*c^2*Log[x])/3 - (4*a^3*c^2*Log[1 + a^2*x^2 ])/3
3.2.64.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_)*((d_) + (e_. )*(x_)^2)^(q_), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*ArcTan[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2* d] && IGtQ[p, 0] && IGtQ[q, 1] && (EqQ[p, 1] || IntegerQ[m])
Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.88
method | result | size |
parts | \(a^{4} c^{2} x \arctan \left (a x \right )-\frac {2 a^{2} c^{2} \arctan \left (a x \right )}{x}-\frac {c^{2} \arctan \left (a x \right )}{3 x^{3}}-\frac {c^{2} a \left (4 a^{2} \ln \left (a^{2} x^{2}+1\right )+\frac {1}{2 x^{2}}-5 a^{2} \ln \left (x \right )\right )}{3}\) | \(75\) |
derivativedivides | \(a^{3} \left (a \,c^{2} x \arctan \left (a x \right )-\frac {2 c^{2} \arctan \left (a x \right )}{a x}-\frac {c^{2} \arctan \left (a x \right )}{3 a^{3} x^{3}}-\frac {c^{2} \left (4 \ln \left (a^{2} x^{2}+1\right )+\frac {1}{2 a^{2} x^{2}}-5 \ln \left (a x \right )\right )}{3}\right )\) | \(78\) |
default | \(a^{3} \left (a \,c^{2} x \arctan \left (a x \right )-\frac {2 c^{2} \arctan \left (a x \right )}{a x}-\frac {c^{2} \arctan \left (a x \right )}{3 a^{3} x^{3}}-\frac {c^{2} \left (4 \ln \left (a^{2} x^{2}+1\right )+\frac {1}{2 a^{2} x^{2}}-5 \ln \left (a x \right )\right )}{3}\right )\) | \(78\) |
parallelrisch | \(\frac {6 a^{4} c^{2} x^{4} \arctan \left (a x \right )+10 a^{3} c^{2} \ln \left (x \right ) x^{3}-8 a^{3} c^{2} \ln \left (a^{2} x^{2}+1\right ) x^{3}+a^{3} c^{2} x^{3}-12 a^{2} c^{2} x^{2} \arctan \left (a x \right )-a \,c^{2} x -2 c^{2} \arctan \left (a x \right )}{6 x^{3}}\) | \(97\) |
risch | \(-\frac {i c^{2} \left (3 a^{4} x^{4}-6 a^{2} x^{2}-1\right ) \ln \left (i a x +1\right )}{6 x^{3}}+\frac {i c^{2} \left (3 x^{4} \ln \left (-i a x +1\right ) a^{4}-10 i \ln \left (x \right ) a^{3} x^{3}+8 i \ln \left (-9 a^{2} x^{2}-9\right ) a^{3} x^{3}-6 a^{2} x^{2} \ln \left (-i a x +1\right )+i a x -\ln \left (-i a x +1\right )\right )}{6 x^{3}}\) | \(125\) |
meijerg | \(\frac {a^{3} c^{2} \left (\frac {4 a^{2} x^{2} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}-2 \ln \left (a^{2} x^{2}+1\right )\right )}{4}+\frac {a^{3} c^{2} \left (4 \ln \left (x \right )+4 \ln \left (a \right )-\frac {4 \arctan \left (\sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}-2 \ln \left (a^{2} x^{2}+1\right )\right )}{2}+\frac {a^{3} c^{2} \left (-\frac {2}{a^{2} x^{2}}+\frac {4}{9}-\frac {4 \ln \left (x \right )}{3}-\frac {4 \ln \left (a \right )}{3}+\frac {-\frac {4 a^{2} x^{2}}{9}+\frac {4}{3}}{a^{2} x^{2}}-\frac {4 \arctan \left (\sqrt {a^{2} x^{2}}\right )}{3 a^{2} x^{2} \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (a^{2} x^{2}+1\right )}{3}\right )}{4}\) | \(183\) |
a^4*c^2*x*arctan(a*x)-2*a^2*c^2*arctan(a*x)/x-1/3*c^2*arctan(a*x)/x^3-1/3* c^2*a*(4*a^2*ln(a^2*x^2+1)+1/2/x^2-5*a^2*ln(x))
Time = 0.27 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94 \[ \int \frac {\left (c+a^2 c x^2\right )^2 \arctan (a x)}{x^4} \, dx=-\frac {8 \, a^{3} c^{2} x^{3} \log \left (a^{2} x^{2} + 1\right ) - 10 \, a^{3} c^{2} x^{3} \log \left (x\right ) + a c^{2} x - 2 \, {\left (3 \, a^{4} c^{2} x^{4} - 6 \, a^{2} c^{2} x^{2} - c^{2}\right )} \arctan \left (a x\right )}{6 \, x^{3}} \]
-1/6*(8*a^3*c^2*x^3*log(a^2*x^2 + 1) - 10*a^3*c^2*x^3*log(x) + a*c^2*x - 2 *(3*a^4*c^2*x^4 - 6*a^2*c^2*x^2 - c^2)*arctan(a*x))/x^3
Time = 0.35 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.02 \[ \int \frac {\left (c+a^2 c x^2\right )^2 \arctan (a x)}{x^4} \, dx=\begin {cases} a^{4} c^{2} x \operatorname {atan}{\left (a x \right )} + \frac {5 a^{3} c^{2} \log {\left (x \right )}}{3} - \frac {4 a^{3} c^{2} \log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{3} - \frac {2 a^{2} c^{2} \operatorname {atan}{\left (a x \right )}}{x} - \frac {a c^{2}}{6 x^{2}} - \frac {c^{2} \operatorname {atan}{\left (a x \right )}}{3 x^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]
Piecewise((a**4*c**2*x*atan(a*x) + 5*a**3*c**2*log(x)/3 - 4*a**3*c**2*log( x**2 + a**(-2))/3 - 2*a**2*c**2*atan(a*x)/x - a*c**2/(6*x**2) - c**2*atan( a*x)/(3*x**3), Ne(a, 0)), (0, True))
Time = 0.20 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.89 \[ \int \frac {\left (c+a^2 c x^2\right )^2 \arctan (a x)}{x^4} \, dx=-\frac {1}{6} \, {\left (8 \, a^{2} c^{2} \log \left (a^{2} x^{2} + 1\right ) - 10 \, a^{2} c^{2} \log \left (x\right ) + \frac {c^{2}}{x^{2}}\right )} a + \frac {1}{3} \, {\left (3 \, a^{4} c^{2} x - \frac {6 \, a^{2} c^{2} x^{2} + c^{2}}{x^{3}}\right )} \arctan \left (a x\right ) \]
-1/6*(8*a^2*c^2*log(a^2*x^2 + 1) - 10*a^2*c^2*log(x) + c^2/x^2)*a + 1/3*(3 *a^4*c^2*x - (6*a^2*c^2*x^2 + c^2)/x^3)*arctan(a*x)
\[ \int \frac {\left (c+a^2 c x^2\right )^2 \arctan (a x)}{x^4} \, dx=\int { \frac {{\left (a^{2} c x^{2} + c\right )}^{2} \arctan \left (a x\right )}{x^{4}} \,d x } \]
Time = 0.54 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.92 \[ \int \frac {\left (c+a^2 c x^2\right )^2 \arctan (a x)}{x^4} \, dx=\frac {c^2\,\left (10\,a^3\,\ln \left (x\right )-8\,a^3\,\ln \left (a^2\,x^2+1\right )\right )}{6}-\frac {\frac {c^2\,\mathrm {atan}\left (a\,x\right )}{3}+\frac {a\,c^2\,x}{6}+2\,a^2\,c^2\,x^2\,\mathrm {atan}\left (a\,x\right )}{x^3}+a^4\,c^2\,x\,\mathrm {atan}\left (a\,x\right ) \]