Integrand size = 16, antiderivative size = 90 \[ \int x^3 \left (a+b \arctan \left (c x^2\right )\right )^2 \, dx=-\frac {a b x^2}{2 c}-\frac {b^2 x^2 \arctan \left (c x^2\right )}{2 c}+\frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{4 c^2}+\frac {1}{4} x^4 \left (a+b \arctan \left (c x^2\right )\right )^2+\frac {b^2 \log \left (1+c^2 x^4\right )}{4 c^2} \]
-1/2*a*b*x^2/c-1/2*b^2*x^2*arctan(c*x^2)/c+1/4*(a+b*arctan(c*x^2))^2/c^2+1 /4*x^4*(a+b*arctan(c*x^2))^2+1/4*b^2*ln(c^2*x^4+1)/c^2
Time = 0.08 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.94 \[ \int x^3 \left (a+b \arctan \left (c x^2\right )\right )^2 \, dx=\frac {a c x^2 \left (-2 b+a c x^2\right )+2 b \left (a-b c x^2+a c^2 x^4\right ) \arctan \left (c x^2\right )+b^2 \left (1+c^2 x^4\right ) \arctan \left (c x^2\right )^2+b^2 \log \left (1+c^2 x^4\right )}{4 c^2} \]
(a*c*x^2*(-2*b + a*c*x^2) + 2*b*(a - b*c*x^2 + a*c^2*x^4)*ArcTan[c*x^2] + b^2*(1 + c^2*x^4)*ArcTan[c*x^2]^2 + b^2*Log[1 + c^2*x^4])/(4*c^2)
Time = 0.45 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5363, 5361, 5451, 2009, 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.
\(\displaystyle \int x^3 \left (a+b \arctan \left (c x^2\right )\right )^2 \, dx\) |
\(\Big \downarrow \) 5363 |
\(\displaystyle \frac {1}{2} \int x^2 \left (a+b \arctan \left (c x^2\right )\right )^2dx^2\) |
\(\Big \downarrow \) 5361 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \left (a+b \arctan \left (c x^2\right )\right )^2-b c \int \frac {x^4 \left (a+b \arctan \left (c x^2\right )\right )}{c^2 x^4+1}dx^2\right )\) |
\(\Big \downarrow \) 5451 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \left (a+b \arctan \left (c x^2\right )\right )^2-b c \left (\frac {\int \left (a+b \arctan \left (c x^2\right )\right )dx^2}{c^2}-\frac {\int \frac {a+b \arctan \left (c x^2\right )}{c^2 x^4+1}dx^2}{c^2}\right )\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \left (a+b \arctan \left (c x^2\right )\right )^2-b c \left (\frac {a x^2+b x^2 \arctan \left (c x^2\right )-\frac {b \log \left (c^2 x^4+1\right )}{2 c}}{c^2}-\frac {\int \frac {a+b \arctan \left (c x^2\right )}{c^2 x^4+1}dx^2}{c^2}\right )\right )\) |
\(\Big \downarrow \) 5419 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \left (a+b \arctan \left (c x^2\right )\right )^2-b c \left (\frac {a x^2+b x^2 \arctan \left (c x^2\right )-\frac {b \log \left (c^2 x^4+1\right )}{2 c}}{c^2}-\frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{2 b c^3}\right )\right )\) |
((x^4*(a + b*ArcTan[c*x^2])^2)/2 - b*c*(-1/2*(a + b*ArcTan[c*x^2])^2/(b*c^ 3) + (a*x^2 + b*x^2*ArcTan[c*x^2] - (b*Log[1 + c^2*x^4])/(2*c))/c^2))/2
3.1.76.3.1 Defintions of rubi rules used
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_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*ArcTan[c*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 1] && IntegerQ[Simplif y[(m + 1)/n]]
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[f^2/e Int[(f*x)^(m - 2)*(a + b*ArcTan[c*x] )^p, x], x] - Simp[d*(f^2/e) 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] && GtQ[m, 1]
Time = 1.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.22
method | result | size |
parallelrisch | \(\frac {b^{2} x^{4} \arctan \left (c \,x^{2}\right )^{2} c^{2}+2 a b \,x^{4} \arctan \left (c \,x^{2}\right ) c^{2}+a^{2} c^{2} x^{4}-2 b^{2} \arctan \left (c \,x^{2}\right ) x^{2} c -2 a b c \,x^{2}+b^{2} \arctan \left (c \,x^{2}\right )^{2}+b^{2} \ln \left (c^{2} x^{4}+1\right )+2 a b \arctan \left (c \,x^{2}\right )}{4 c^{2}}\) | \(110\) |
default | \(\frac {a^{2} x^{4}}{4}+\frac {b^{2} x^{4} \arctan \left (c \,x^{2}\right )^{2}}{4}-\frac {b^{2} x^{2} \arctan \left (c \,x^{2}\right )}{2 c}+\frac {b^{2} \arctan \left (c \,x^{2}\right )^{2}}{4 c^{2}}+\frac {b^{2} \ln \left (c^{2} x^{4}+1\right )}{4 c^{2}}+\frac {a b \,x^{4} \arctan \left (c \,x^{2}\right )}{2}-\frac {a b \,x^{2}}{2 c}+\frac {a b \arctan \left (c \,x^{2}\right )}{2 c^{2}}\) | \(113\) |
parts | \(\frac {a^{2} x^{4}}{4}+\frac {b^{2} x^{4} \arctan \left (c \,x^{2}\right )^{2}}{4}-\frac {b^{2} x^{2} \arctan \left (c \,x^{2}\right )}{2 c}+\frac {b^{2} \arctan \left (c \,x^{2}\right )^{2}}{4 c^{2}}+\frac {b^{2} \ln \left (c^{2} x^{4}+1\right )}{4 c^{2}}+\frac {a b \,x^{4} \arctan \left (c \,x^{2}\right )}{2}-\frac {a b \,x^{2}}{2 c}+\frac {a b \arctan \left (c \,x^{2}\right )}{2 c^{2}}\) | \(113\) |
risch | \(-\frac {b^{2} \left (c^{2} x^{4}+1\right ) \ln \left (i c \,x^{2}+1\right )^{2}}{16 c^{2}}-\frac {i b \left (4 a^{2} c^{2} x^{4}+2 i x^{4} b \ln \left (-i c \,x^{2}+1\right ) a \,c^{2}-4 a b c \,x^{2}+b^{2}+2 i b \ln \left (-i c \,x^{2}+1\right ) a \right ) \ln \left (i c \,x^{2}+1\right )}{16 a \,c^{2}}+\frac {i a b \,x^{4} \ln \left (-i c \,x^{2}+1\right )}{4}-\frac {b^{2} x^{4} \ln \left (-i c \,x^{2}+1\right )^{2}}{16}+\frac {a^{2} x^{4}}{4}+\frac {i b^{3} \ln \left (c^{2} x^{4}+1\right )}{32 a \,c^{2}}-\frac {a b \,x^{2}}{2 c}+\frac {a b \arctan \left (c \,x^{2}\right )}{2 c^{2}}-\frac {i b^{2} x^{2} \ln \left (-i c \,x^{2}+1\right )}{4 c}-\frac {b^{3} \arctan \left (c \,x^{2}\right )}{16 a \,c^{2}}-\frac {b^{2} \ln \left (-i c \,x^{2}+1\right )^{2}}{16 c^{2}}+\frac {b^{2} \ln \left (c^{2} x^{4}+1\right )}{4 c^{2}}+\frac {b^{2}}{4 c^{2}}\) | \(286\) |
1/4*(b^2*x^4*arctan(c*x^2)^2*c^2+2*a*b*x^4*arctan(c*x^2)*c^2+a^2*c^2*x^4-2 *b^2*arctan(c*x^2)*x^2*c-2*a*b*c*x^2+b^2*arctan(c*x^2)^2+b^2*ln(c^2*x^4+1) +2*a*b*arctan(c*x^2))/c^2
Time = 0.26 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.11 \[ \int x^3 \left (a+b \arctan \left (c x^2\right )\right )^2 \, dx=\frac {a^{2} c^{2} x^{4} - 2 \, a b c x^{2} + {\left (b^{2} c^{2} x^{4} + b^{2}\right )} \arctan \left (c x^{2}\right )^{2} - 2 \, a b \arctan \left (\frac {1}{c x^{2}}\right ) + b^{2} \log \left (c^{2} x^{4} + 1\right ) + 2 \, {\left (a b c^{2} x^{4} - b^{2} c x^{2}\right )} \arctan \left (c x^{2}\right )}{4 \, c^{2}} \]
1/4*(a^2*c^2*x^4 - 2*a*b*c*x^2 + (b^2*c^2*x^4 + b^2)*arctan(c*x^2)^2 - 2*a *b*arctan(1/(c*x^2)) + b^2*log(c^2*x^4 + 1) + 2*(a*b*c^2*x^4 - b^2*c*x^2)* arctan(c*x^2))/c^2
Time = 16.83 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.72 \[ \int x^3 \left (a+b \arctan \left (c x^2\right )\right )^2 \, dx=\begin {cases} \frac {a^{2} x^{4}}{4} + \frac {a b x^{4} \operatorname {atan}{\left (c x^{2} \right )}}{2} - \frac {a b x^{2}}{2 c} + \frac {a b \operatorname {atan}{\left (c x^{2} \right )}}{2 c^{2}} + \frac {b^{2} x^{4} \operatorname {atan}^{2}{\left (c x^{2} \right )}}{4} - \frac {b^{2} x^{2} \operatorname {atan}{\left (c x^{2} \right )}}{2 c} + \frac {b^{2} \sqrt {- \frac {1}{c^{2}}} \operatorname {atan}{\left (c x^{2} \right )}}{2 c} + \frac {b^{2} \log {\left (x^{2} + \sqrt {- \frac {1}{c^{2}}} \right )}}{2 c^{2}} + \frac {b^{2} \operatorname {atan}^{2}{\left (c x^{2} \right )}}{4 c^{2}} & \text {for}\: c \neq 0 \\\frac {a^{2} x^{4}}{4} & \text {otherwise} \end {cases} \]
Piecewise((a**2*x**4/4 + a*b*x**4*atan(c*x**2)/2 - a*b*x**2/(2*c) + a*b*at an(c*x**2)/(2*c**2) + b**2*x**4*atan(c*x**2)**2/4 - b**2*x**2*atan(c*x**2) /(2*c) + b**2*sqrt(-1/c**2)*atan(c*x**2)/(2*c) + b**2*log(x**2 + sqrt(-1/c **2))/(2*c**2) + b**2*atan(c*x**2)**2/(4*c**2), Ne(c, 0)), (a**2*x**4/4, T rue))
Time = 0.33 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.40 \[ \int x^3 \left (a+b \arctan \left (c x^2\right )\right )^2 \, dx=\frac {1}{4} \, b^{2} x^{4} \arctan \left (c x^{2}\right )^{2} + \frac {1}{4} \, a^{2} x^{4} + \frac {1}{2} \, {\left (x^{4} \arctan \left (c x^{2}\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\arctan \left (c x^{2}\right )}{c^{3}}\right )}\right )} a b - \frac {1}{4} \, {\left (2 \, c {\left (\frac {x^{2}}{c^{2}} - \frac {\arctan \left (c x^{2}\right )}{c^{3}}\right )} \arctan \left (c x^{2}\right ) + \frac {\arctan \left (c x^{2}\right )^{2} - \log \left (4 \, c^{5} x^{4} + 4 \, c^{3}\right )}{c^{2}}\right )} b^{2} \]
1/4*b^2*x^4*arctan(c*x^2)^2 + 1/4*a^2*x^4 + 1/2*(x^4*arctan(c*x^2) - c*(x^ 2/c^2 - arctan(c*x^2)/c^3))*a*b - 1/4*(2*c*(x^2/c^2 - arctan(c*x^2)/c^3)*a rctan(c*x^2) + (arctan(c*x^2)^2 - log(4*c^5*x^4 + 4*c^3))/c^2)*b^2
Time = 0.27 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.11 \[ \int x^3 \left (a+b \arctan \left (c x^2\right )\right )^2 \, dx=\frac {a^{2} c x^{4} + \frac {2 \, {\left (c^{2} x^{4} \arctan \left (c x^{2}\right ) - c x^{2} + \arctan \left (c x^{2}\right )\right )} a b}{c} + \frac {{\left (c^{2} x^{4} \arctan \left (c x^{2}\right )^{2} - 2 \, c x^{2} \arctan \left (c x^{2}\right ) + \arctan \left (c x^{2}\right )^{2} + \log \left (c^{2} x^{4} + 1\right )\right )} b^{2}}{c}}{4 \, c} \]
1/4*(a^2*c*x^4 + 2*(c^2*x^4*arctan(c*x^2) - c*x^2 + arctan(c*x^2))*a*b/c + (c^2*x^4*arctan(c*x^2)^2 - 2*c*x^2*arctan(c*x^2) + arctan(c*x^2)^2 + log( c^2*x^4 + 1))*b^2/c)/c
Time = 0.80 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.24 \[ \int x^3 \left (a+b \arctan \left (c x^2\right )\right )^2 \, dx=\frac {a^2\,x^4}{4}+\frac {b^2\,{\mathrm {atan}\left (c\,x^2\right )}^2}{4\,c^2}+\frac {b^2\,x^4\,{\mathrm {atan}\left (c\,x^2\right )}^2}{4}+\frac {b^2\,\ln \left (c^2\,x^4+1\right )}{4\,c^2}-\frac {b^2\,x^2\,\mathrm {atan}\left (c\,x^2\right )}{2\,c}-\frac {a\,b\,x^2}{2\,c}+\frac {a\,b\,\mathrm {atan}\left (c\,x^2\right )}{2\,c^2}+\frac {a\,b\,x^4\,\mathrm {atan}\left (c\,x^2\right )}{2} \]