Integrand size = 18, antiderivative size = 66 \[ \int x^2 \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=-\frac {c x^2}{15 a}-\frac {1}{20} a c x^4+\frac {1}{3} c x^3 \arctan (a x)+\frac {1}{5} a^2 c x^5 \arctan (a x)+\frac {c \log \left (1+a^2 x^2\right )}{15 a^3} \]
-1/15*c*x^2/a-1/20*a*c*x^4+1/3*c*x^3*arctan(a*x)+1/5*a^2*c*x^5*arctan(a*x) +1/15*c*ln(a^2*x^2+1)/a^3
Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00 \[ \int x^2 \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=-\frac {c x^2}{15 a}-\frac {1}{20} a c x^4+\frac {1}{3} c x^3 \arctan (a x)+\frac {1}{5} a^2 c x^5 \arctan (a x)+\frac {c \log \left (1+a^2 x^2\right )}{15 a^3} \]
-1/15*(c*x^2)/a - (a*c*x^4)/20 + (c*x^3*ArcTan[a*x])/3 + (a^2*c*x^5*ArcTan [a*x])/5 + (c*Log[1 + a^2*x^2])/(15*a^3)
Time = 0.32 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.48, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {5485, 5361, 243, 49, 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.
\(\displaystyle \int x^2 \arctan (a x) \left (a^2 c x^2+c\right ) \, dx\) |
\(\Big \downarrow \) 5485 |
\(\displaystyle a^2 c \int x^4 \arctan (a x)dx+c \int x^2 \arctan (a x)dx\) |
\(\Big \downarrow \) 5361 |
\(\displaystyle a^2 c \left (\frac {1}{5} x^5 \arctan (a x)-\frac {1}{5} a \int \frac {x^5}{a^2 x^2+1}dx\right )+c \left (\frac {1}{3} x^3 \arctan (a x)-\frac {1}{3} a \int \frac {x^3}{a^2 x^2+1}dx\right )\) |
\(\Big \downarrow \) 243 |
\(\displaystyle c \left (\frac {1}{3} x^3 \arctan (a x)-\frac {1}{6} a \int \frac {x^2}{a^2 x^2+1}dx^2\right )+a^2 c \left (\frac {1}{5} x^5 \arctan (a x)-\frac {1}{10} a \int \frac {x^4}{a^2 x^2+1}dx^2\right )\) |
\(\Big \downarrow \) 49 |
\(\displaystyle c \left (\frac {1}{3} x^3 \arctan (a x)-\frac {1}{6} a \int \left (\frac {1}{a^2}-\frac {1}{a^2 \left (a^2 x^2+1\right )}\right )dx^2\right )+a^2 c \left (\frac {1}{5} x^5 \arctan (a x)-\frac {1}{10} a \int \left (\frac {x^2}{a^2}+\frac {1}{a^4 \left (a^2 x^2+1\right )}-\frac {1}{a^4}\right )dx^2\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle c \left (\frac {1}{3} x^3 \arctan (a x)-\frac {1}{6} a \left (\frac {x^2}{a^2}-\frac {\log \left (a^2 x^2+1\right )}{a^4}\right )\right )+a^2 c \left (\frac {1}{5} x^5 \arctan (a x)-\frac {1}{10} a \left (-\frac {x^2}{a^4}+\frac {x^4}{2 a^2}+\frac {\log \left (a^2 x^2+1\right )}{a^6}\right )\right )\) |
a^2*c*((x^5*ArcTan[a*x])/5 - (a*(-(x^2/a^4) + x^4/(2*a^2) + Log[1 + a^2*x^ 2]/a^6))/10) + c*((x^3*ArcTan[a*x])/3 - (a*(x^2/a^2 - Log[1 + a^2*x^2]/a^4 ))/6)
3.2.50.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
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_.)*((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 = 0.18 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.95
method | result | size |
derivativedivides | \(\frac {\frac {c \arctan \left (a x \right ) a^{5} x^{5}}{5}+\frac {c \arctan \left (a x \right ) a^{3} x^{3}}{3}-\frac {c \left (\frac {3 a^{4} x^{4}}{4}+a^{2} x^{2}-\ln \left (a^{2} x^{2}+1\right )\right )}{15}}{a^{3}}\) | \(63\) |
default | \(\frac {\frac {c \arctan \left (a x \right ) a^{5} x^{5}}{5}+\frac {c \arctan \left (a x \right ) a^{3} x^{3}}{3}-\frac {c \left (\frac {3 a^{4} x^{4}}{4}+a^{2} x^{2}-\ln \left (a^{2} x^{2}+1\right )\right )}{15}}{a^{3}}\) | \(63\) |
parallelrisch | \(\frac {12 c \arctan \left (a x \right ) a^{5} x^{5}-3 a^{4} c \,x^{4}+20 c \arctan \left (a x \right ) a^{3} x^{3}-4 a^{2} c \,x^{2}+4 c \ln \left (a^{2} x^{2}+1\right )}{60 a^{3}}\) | \(64\) |
parts | \(\frac {a^{2} c \,x^{5} \arctan \left (a x \right )}{5}+\frac {c \,x^{3} \arctan \left (a x \right )}{3}-\frac {c a \left (\frac {\frac {3}{2} a^{2} x^{4}+2 x^{2}}{2 a^{2}}-\frac {\ln \left (a^{2} x^{2}+1\right )}{a^{4}}\right )}{15}\) | \(64\) |
risch | \(-\frac {i c \,x^{3} \left (3 a^{2} x^{2}+5\right ) \ln \left (i a x +1\right )}{30}+\frac {i c \,a^{2} x^{5} \ln \left (-i a x +1\right )}{10}-\frac {a c \,x^{4}}{20}+\frac {i c \,x^{3} \ln \left (-i a x +1\right )}{6}-\frac {c \,x^{2}}{15 a}+\frac {c \ln \left (-a^{2} x^{2}-1\right )}{15 a^{3}}-\frac {c}{45 a^{3}}\) | \(99\) |
meijerg | \(\frac {c \left (\frac {a^{2} x^{2} \left (-3 a^{2} x^{2}+6\right )}{15}+\frac {4 a^{6} x^{6} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{5 \sqrt {a^{2} x^{2}}}-\frac {2 \ln \left (a^{2} x^{2}+1\right )}{5}\right )}{4 a^{3}}+\frac {c \left (-\frac {2 a^{2} x^{2}}{3}+\frac {4 a^{4} x^{4} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{3 \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (a^{2} x^{2}+1\right )}{3}\right )}{4 a^{3}}\) | \(120\) |
1/a^3*(1/5*c*arctan(a*x)*a^5*x^5+1/3*c*arctan(a*x)*a^3*x^3-1/15*c*(3/4*a^4 *x^4+a^2*x^2-ln(a^2*x^2+1)))
Time = 0.25 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.94 \[ \int x^2 \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=-\frac {3 \, a^{4} c x^{4} + 4 \, a^{2} c x^{2} - 4 \, {\left (3 \, a^{5} c x^{5} + 5 \, a^{3} c x^{3}\right )} \arctan \left (a x\right ) - 4 \, c \log \left (a^{2} x^{2} + 1\right )}{60 \, a^{3}} \]
-1/60*(3*a^4*c*x^4 + 4*a^2*c*x^2 - 4*(3*a^5*c*x^5 + 5*a^3*c*x^3)*arctan(a* x) - 4*c*log(a^2*x^2 + 1))/a^3
Time = 0.29 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.92 \[ \int x^2 \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=\begin {cases} \frac {a^{2} c x^{5} \operatorname {atan}{\left (a x \right )}}{5} - \frac {a c x^{4}}{20} + \frac {c x^{3} \operatorname {atan}{\left (a x \right )}}{3} - \frac {c x^{2}}{15 a} + \frac {c \log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{15 a^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]
Piecewise((a**2*c*x**5*atan(a*x)/5 - a*c*x**4/20 + c*x**3*atan(a*x)/3 - c* x**2/(15*a) + c*log(x**2 + a**(-2))/(15*a**3), Ne(a, 0)), (0, True))
Time = 0.19 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.95 \[ \int x^2 \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=-\frac {1}{60} \, a {\left (\frac {3 \, a^{2} c x^{4} + 4 \, c x^{2}}{a^{2}} - \frac {4 \, c \log \left (a^{2} x^{2} + 1\right )}{a^{4}}\right )} + \frac {1}{15} \, {\left (3 \, a^{2} c x^{5} + 5 \, c x^{3}\right )} \arctan \left (a x\right ) \]
-1/60*a*((3*a^2*c*x^4 + 4*c*x^2)/a^2 - 4*c*log(a^2*x^2 + 1)/a^4) + 1/15*(3 *a^2*c*x^5 + 5*c*x^3)*arctan(a*x)
\[ \int x^2 \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=\int { {\left (a^{2} c x^{2} + c\right )} x^{2} \arctan \left (a x\right ) \,d x } \]
Time = 0.25 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.88 \[ \int x^2 \left (c+a^2 c x^2\right ) \arctan (a x) \, dx=\frac {\frac {c\,\ln \left (a^2\,x^2+1\right )}{15}-\frac {a^2\,c\,x^2}{15}}{a^3}+\frac {c\,x^3\,\mathrm {atan}\left (a\,x\right )}{3}-\frac {a\,c\,x^4}{20}+\frac {a^2\,c\,x^5\,\mathrm {atan}\left (a\,x\right )}{5} \]