Integrand size = 12, antiderivative size = 76 \[ \int x (a+b \arctan (c x))^2 \, dx=-\frac {a b x}{c}-\frac {b^2 x \arctan (c x)}{c}+\frac {(a+b \arctan (c x))^2}{2 c^2}+\frac {1}{2} x^2 (a+b \arctan (c x))^2+\frac {b^2 \log \left (1+c^2 x^2\right )}{2 c^2} \] Output:
-a*b*x/c-b^2*x*arctan(c*x)/c+1/2*(a+b*arctan(c*x))^2/c^2+1/2*x^2*(a+b*arct an(c*x))^2+1/2*b^2*ln(c^2*x^2+1)/c^2
Time = 0.09 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.99 \[ \int x (a+b \arctan (c x))^2 \, dx=\frac {a c x (-2 b+a c x)+2 b \left (a-b c x+a c^2 x^2\right ) \arctan (c x)+b^2 \left (1+c^2 x^2\right ) \arctan (c x)^2+b^2 \log \left (1+c^2 x^2\right )}{2 c^2} \] Input:
Integrate[x*(a + b*ArcTan[c*x])^2,x]
Output:
(a*c*x*(-2*b + a*c*x) + 2*b*(a - b*c*x + a*c^2*x^2)*ArcTan[c*x] + b^2*(1 + c^2*x^2)*ArcTan[c*x]^2 + b^2*Log[1 + c^2*x^2])/(2*c^2)
Time = 0.41 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {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.
\(\displaystyle \int x (a+b \arctan (c x))^2 \, dx\) |
\(\Big \downarrow \) 5361 |
\(\displaystyle \frac {1}{2} x^2 (a+b \arctan (c x))^2-b c \int \frac {x^2 (a+b \arctan (c x))}{c^2 x^2+1}dx\) |
\(\Big \downarrow \) 5451 |
\(\displaystyle \frac {1}{2} x^2 (a+b \arctan (c x))^2-b c \left (\frac {\int (a+b \arctan (c x))dx}{c^2}-\frac {\int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx}{c^2}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} x^2 (a+b \arctan (c x))^2-b c \left (\frac {a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}}{c^2}-\frac {\int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx}{c^2}\right )\) |
\(\Big \downarrow \) 5419 |
\(\displaystyle \frac {1}{2} x^2 (a+b \arctan (c x))^2-b c \left (\frac {a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}}{c^2}-\frac {(a+b \arctan (c x))^2}{2 b c^3}\right )\) |
Input:
Int[x*(a + b*ArcTan[c*x])^2,x]
Output:
(x^2*(a + b*ArcTan[c*x])^2)/2 - b*c*(-1/2*(a + b*ArcTan[c*x])^2/(b*c^3) + (a*x + b*x*ArcTan[c*x] - (b*Log[1 + c^2*x^2])/(2*c))/c^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[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 = 0.14 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.16
method | result | size |
parts | \(\frac {a^{2} x^{2}}{2}+\frac {b^{2} \left (\frac {c^{2} x^{2} \arctan \left (c x \right )^{2}}{2}+\frac {\arctan \left (c x \right )^{2}}{2}-c x \arctan \left (c x \right )+\frac {\ln \left (c^{2} x^{2}+1\right )}{2}\right )}{c^{2}}+a b \arctan \left (c x \right ) x^{2}-\frac {a b x}{c}+\frac {a b \arctan \left (c x \right )}{c^{2}}\) | \(88\) |
derivativedivides | \(\frac {\frac {x^{2} c^{2} a^{2}}{2}+b^{2} \left (\frac {c^{2} x^{2} \arctan \left (c x \right )^{2}}{2}+\frac {\arctan \left (c x \right )^{2}}{2}-c x \arctan \left (c x \right )+\frac {\ln \left (c^{2} x^{2}+1\right )}{2}\right )+2 a b \left (\frac {c^{2} x^{2} \arctan \left (c x \right )}{2}-\frac {c x}{2}+\frac {\arctan \left (c x \right )}{2}\right )}{c^{2}}\) | \(91\) |
default | \(\frac {\frac {x^{2} c^{2} a^{2}}{2}+b^{2} \left (\frac {c^{2} x^{2} \arctan \left (c x \right )^{2}}{2}+\frac {\arctan \left (c x \right )^{2}}{2}-c x \arctan \left (c x \right )+\frac {\ln \left (c^{2} x^{2}+1\right )}{2}\right )+2 a b \left (\frac {c^{2} x^{2} \arctan \left (c x \right )}{2}-\frac {c x}{2}+\frac {\arctan \left (c x \right )}{2}\right )}{c^{2}}\) | \(91\) |
parallelrisch | \(\frac {b^{2} \arctan \left (c x \right )^{2} x^{2} c^{2}+2 a b \arctan \left (c x \right ) x^{2} c^{2}+x^{2} c^{2} a^{2}-2 b^{2} \arctan \left (c x \right ) x c -2 a b c x +b^{2} \arctan \left (c x \right )^{2}+b^{2} \ln \left (c^{2} x^{2}+1\right )+2 a b \arctan \left (c x \right )-a^{2}}{2 c^{2}}\) | \(101\) |
risch | \(-\frac {b^{2} \left (c^{2} x^{2}+1\right ) \ln \left (i c x +1\right )^{2}}{8 c^{2}}-\frac {i b \left (2 a \,c^{2} x^{2}+i b \,c^{2} x^{2} \ln \left (-i c x +1\right )-2 b c x +i b \ln \left (-i c x +1\right )\right ) \ln \left (i c x +1\right )}{4 c^{2}}+\frac {i a b \,x^{2} \ln \left (-i c x +1\right )}{2}-\frac {b^{2} x^{2} \ln \left (-i c x +1\right )^{2}}{8}-\frac {i b^{2} x \ln \left (-i c x +1\right )}{2 c}+\frac {a^{2} x^{2}}{2}-\frac {b^{2} \ln \left (-i c x +1\right )^{2}}{8 c^{2}}-\frac {a b x}{c}+\frac {a b \arctan \left (c x \right )}{c^{2}}+\frac {b^{2} \ln \left (c^{2} x^{2}+1\right )}{2 c^{2}}\) | \(203\) |
Input:
int(x*(a+b*arctan(c*x))^2,x,method=_RETURNVERBOSE)
Output:
1/2*a^2*x^2+b^2/c^2*(1/2*c^2*x^2*arctan(c*x)^2+1/2*arctan(c*x)^2-c*x*arcta n(c*x)+1/2*ln(c^2*x^2+1))+a*b*arctan(c*x)*x^2-a*b*x/c+1/c^2*a*b*arctan(c*x )
Time = 0.09 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.09 \[ \int x (a+b \arctan (c x))^2 \, dx=\frac {a^{2} c^{2} x^{2} - 2 \, a b c x + {\left (b^{2} c^{2} x^{2} + b^{2}\right )} \arctan \left (c x\right )^{2} + b^{2} \log \left (c^{2} x^{2} + 1\right ) + 2 \, {\left (a b c^{2} x^{2} - b^{2} c x + a b\right )} \arctan \left (c x\right )}{2 \, c^{2}} \] Input:
integrate(x*(a+b*arctan(c*x))^2,x, algorithm="fricas")
Output:
1/2*(a^2*c^2*x^2 - 2*a*b*c*x + (b^2*c^2*x^2 + b^2)*arctan(c*x)^2 + b^2*log (c^2*x^2 + 1) + 2*(a*b*c^2*x^2 - b^2*c*x + a*b)*arctan(c*x))/c^2
Time = 0.29 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.41 \[ \int x (a+b \arctan (c x))^2 \, dx=\begin {cases} \frac {a^{2} x^{2}}{2} + a b x^{2} \operatorname {atan}{\left (c x \right )} - \frac {a b x}{c} + \frac {a b \operatorname {atan}{\left (c x \right )}}{c^{2}} + \frac {b^{2} x^{2} \operatorname {atan}^{2}{\left (c x \right )}}{2} - \frac {b^{2} x \operatorname {atan}{\left (c x \right )}}{c} + \frac {b^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c^{2}} + \frac {b^{2} \operatorname {atan}^{2}{\left (c x \right )}}{2 c^{2}} & \text {for}\: c \neq 0 \\\frac {a^{2} x^{2}}{2} & \text {otherwise} \end {cases} \] Input:
integrate(x*(a+b*atan(c*x))**2,x)
Output:
Piecewise((a**2*x**2/2 + a*b*x**2*atan(c*x) - a*b*x/c + a*b*atan(c*x)/c**2 + b**2*x**2*atan(c*x)**2/2 - b**2*x*atan(c*x)/c + b**2*log(x**2 + c**(-2) )/(2*c**2) + b**2*atan(c*x)**2/(2*c**2), Ne(c, 0)), (a**2*x**2/2, True))
Time = 0.14 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.37 \[ \int x (a+b \arctan (c x))^2 \, dx=\frac {1}{2} \, b^{2} x^{2} \arctan \left (c x\right )^{2} + \frac {1}{2} \, a^{2} x^{2} + {\left (x^{2} \arctan \left (c x\right ) - c {\left (\frac {x}{c^{2}} - \frac {\arctan \left (c x\right )}{c^{3}}\right )}\right )} a b - \frac {1}{2} \, {\left (2 \, c {\left (\frac {x}{c^{2}} - \frac {\arctan \left (c x\right )}{c^{3}}\right )} \arctan \left (c x\right ) + \frac {\arctan \left (c x\right )^{2} - \log \left (c^{2} x^{2} + 1\right )}{c^{2}}\right )} b^{2} \] Input:
integrate(x*(a+b*arctan(c*x))^2,x, algorithm="maxima")
Output:
1/2*b^2*x^2*arctan(c*x)^2 + 1/2*a^2*x^2 + (x^2*arctan(c*x) - c*(x/c^2 - ar ctan(c*x)/c^3))*a*b - 1/2*(2*c*(x/c^2 - arctan(c*x)/c^3)*arctan(c*x) + (ar ctan(c*x)^2 - log(c^2*x^2 + 1))/c^2)*b^2
Time = 0.14 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.37 \[ \int x (a+b \arctan (c x))^2 \, dx=\frac {b^{2} c^{2} x^{2} \arctan \left (c x\right )^{2} + 2 \, a b c^{2} x^{2} \arctan \left (c x\right ) + a^{2} c^{2} x^{2} - 2 \, b^{2} c x \arctan \left (c x\right ) - 2 \, \pi a b \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (x\right ) - 2 \, a b c x + b^{2} \arctan \left (c x\right )^{2} + 2 \, a b \arctan \left (c x\right ) + b^{2} \log \left (c^{2} x^{2} + 1\right )}{2 \, c^{2}} \] Input:
integrate(x*(a+b*arctan(c*x))^2,x, algorithm="giac")
Output:
1/2*(b^2*c^2*x^2*arctan(c*x)^2 + 2*a*b*c^2*x^2*arctan(c*x) + a^2*c^2*x^2 - 2*b^2*c*x*arctan(c*x) - 2*pi*a*b*sgn(c)*sgn(x) - 2*a*b*c*x + b^2*arctan(c *x)^2 + 2*a*b*arctan(c*x) + b^2*log(c^2*x^2 + 1))/c^2
Time = 0.22 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.16 \[ \int x (a+b \arctan (c x))^2 \, dx=\frac {\frac {b^2\,{\mathrm {atan}\left (c\,x\right )}^2}{2}+\frac {b^2\,\ln \left (c^2\,x^2+1\right )}{2}-c\,\left (x\,\mathrm {atan}\left (c\,x\right )\,b^2+a\,x\,b\right )+a\,b\,\mathrm {atan}\left (c\,x\right )}{c^2}+\frac {a^2\,x^2}{2}+\frac {b^2\,x^2\,{\mathrm {atan}\left (c\,x\right )}^2}{2}+a\,b\,x^2\,\mathrm {atan}\left (c\,x\right ) \] Input:
int(x*(a + b*atan(c*x))^2,x)
Output:
((b^2*atan(c*x)^2)/2 + (b^2*log(c^2*x^2 + 1))/2 - c*(b^2*x*atan(c*x) + a*b *x) + a*b*atan(c*x))/c^2 + (a^2*x^2)/2 + (b^2*x^2*atan(c*x)^2)/2 + a*b*x^2 *atan(c*x)
Time = 0.18 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.25 \[ \int x (a+b \arctan (c x))^2 \, dx=\frac {\mathit {atan} \left (c x \right )^{2} b^{2} c^{2} x^{2}+\mathit {atan} \left (c x \right )^{2} b^{2}+2 \mathit {atan} \left (c x \right ) a b \,c^{2} x^{2}+2 \mathit {atan} \left (c x \right ) a b -2 \mathit {atan} \left (c x \right ) b^{2} c x +\mathrm {log}\left (c^{2} x^{2}+1\right ) b^{2}+a^{2} c^{2} x^{2}-2 a b c x}{2 c^{2}} \] Input:
int(x*(a+b*atan(c*x))^2,x)
Output:
(atan(c*x)**2*b**2*c**2*x**2 + atan(c*x)**2*b**2 + 2*atan(c*x)*a*b*c**2*x* *2 + 2*atan(c*x)*a*b - 2*atan(c*x)*b**2*c*x + log(c**2*x**2 + 1)*b**2 + a* *2*c**2*x**2 - 2*a*b*c*x)/(2*c**2)