Integrand size = 20, antiderivative size = 80 \[ \int \frac {x^4 \arctan (a x)}{c+a^2 c x^2} \, dx=-\frac {x^2}{6 a^3 c}-\frac {x \arctan (a x)}{a^4 c}+\frac {x^3 \arctan (a x)}{3 a^2 c}+\frac {\arctan (a x)^2}{2 a^5 c}+\frac {2 \log \left (1+a^2 x^2\right )}{3 a^5 c} \] Output:
-1/6*x^2/a^3/c-x*arctan(a*x)/a^4/c+1/3*x^3*arctan(a*x)/a^2/c+1/2*arctan(a* x)^2/a^5/c+2/3*ln(a^2*x^2+1)/a^5/c
Time = 0.06 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.70 \[ \int \frac {x^4 \arctan (a x)}{c+a^2 c x^2} \, dx=\frac {-a^2 x^2+2 a x \left (-3+a^2 x^2\right ) \arctan (a x)+3 \arctan (a x)^2+4 \log \left (1+a^2 x^2\right )}{6 a^5 c} \] Input:
Integrate[(x^4*ArcTan[a*x])/(c + a^2*c*x^2),x]
Output:
(-(a^2*x^2) + 2*a*x*(-3 + a^2*x^2)*ArcTan[a*x] + 3*ArcTan[a*x]^2 + 4*Log[1 + a^2*x^2])/(6*a^5*c)
Time = 0.64 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.22, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5451, 27, 5361, 243, 49, 2009, 5451, 5345, 240, 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.
\(\displaystyle \int \frac {x^4 \arctan (a x)}{a^2 c x^2+c} \, dx\) |
\(\Big \downarrow \) 5451 |
\(\displaystyle \frac {\int x^2 \arctan (a x)dx}{a^2 c}-\frac {\int \frac {x^2 \arctan (a x)}{c \left (a^2 x^2+1\right )}dx}{a^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int x^2 \arctan (a x)dx}{a^2 c}-\frac {\int \frac {x^2 \arctan (a x)}{a^2 x^2+1}dx}{a^2 c}\) |
\(\Big \downarrow \) 5361 |
\(\displaystyle \frac {\frac {1}{3} x^3 \arctan (a x)-\frac {1}{3} a \int \frac {x^3}{a^2 x^2+1}dx}{a^2 c}-\frac {\int \frac {x^2 \arctan (a x)}{a^2 x^2+1}dx}{a^2 c}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\frac {1}{3} x^3 \arctan (a x)-\frac {1}{6} a \int \frac {x^2}{a^2 x^2+1}dx^2}{a^2 c}-\frac {\int \frac {x^2 \arctan (a x)}{a^2 x^2+1}dx}{a^2 c}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\frac {1}{3} x^3 \arctan (a x)-\frac {1}{6} a \int \left (\frac {1}{a^2}-\frac {1}{a^2 \left (a^2 x^2+1\right )}\right )dx^2}{a^2 c}-\frac {\int \frac {x^2 \arctan (a x)}{a^2 x^2+1}dx}{a^2 c}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{3} x^3 \arctan (a x)-\frac {1}{6} a \left (\frac {x^2}{a^2}-\frac {\log \left (a^2 x^2+1\right )}{a^4}\right )}{a^2 c}-\frac {\int \frac {x^2 \arctan (a x)}{a^2 x^2+1}dx}{a^2 c}\) |
\(\Big \downarrow \) 5451 |
\(\displaystyle \frac {\frac {1}{3} x^3 \arctan (a x)-\frac {1}{6} a \left (\frac {x^2}{a^2}-\frac {\log \left (a^2 x^2+1\right )}{a^4}\right )}{a^2 c}-\frac {\frac {\int \arctan (a x)dx}{a^2}-\frac {\int \frac {\arctan (a x)}{a^2 x^2+1}dx}{a^2}}{a^2 c}\) |
\(\Big \downarrow \) 5345 |
\(\displaystyle \frac {\frac {1}{3} x^3 \arctan (a x)-\frac {1}{6} a \left (\frac {x^2}{a^2}-\frac {\log \left (a^2 x^2+1\right )}{a^4}\right )}{a^2 c}-\frac {\frac {x \arctan (a x)-a \int \frac {x}{a^2 x^2+1}dx}{a^2}-\frac {\int \frac {\arctan (a x)}{a^2 x^2+1}dx}{a^2}}{a^2 c}\) |
\(\Big \downarrow \) 240 |
\(\displaystyle \frac {\frac {1}{3} x^3 \arctan (a x)-\frac {1}{6} a \left (\frac {x^2}{a^2}-\frac {\log \left (a^2 x^2+1\right )}{a^4}\right )}{a^2 c}-\frac {\frac {x \arctan (a x)-\frac {\log \left (a^2 x^2+1\right )}{2 a}}{a^2}-\frac {\int \frac {\arctan (a x)}{a^2 x^2+1}dx}{a^2}}{a^2 c}\) |
\(\Big \downarrow \) 5419 |
\(\displaystyle \frac {\frac {1}{3} x^3 \arctan (a x)-\frac {1}{6} a \left (\frac {x^2}{a^2}-\frac {\log \left (a^2 x^2+1\right )}{a^4}\right )}{a^2 c}-\frac {\frac {x \arctan (a x)-\frac {\log \left (a^2 x^2+1\right )}{2 a}}{a^2}-\frac {\arctan (a x)^2}{2 a^3}}{a^2 c}\) |
Input:
Int[(x^4*ArcTan[a*x])/(c + a^2*c*x^2),x]
Output:
((x^3*ArcTan[a*x])/3 - (a*(x^2/a^2 - Log[1 + a^2*x^2]/a^4))/6)/(a^2*c) - ( -1/2*ArcTan[a*x]^2/a^3 + (x*ArcTan[a*x] - Log[1 + a^2*x^2]/(2*a))/a^2)/(a^ 2*c)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_)/((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_.)/((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 = 0.46 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.72
method | result | size |
parallelrisch | \(\frac {2 \arctan \left (a x \right ) x^{3} a^{3}-a^{2} x^{2}-6 \arctan \left (a x \right ) a x +3 \arctan \left (a x \right )^{2}+4 \ln \left (a^{2} x^{2}+1\right )}{6 c \,a^{5}}\) | \(58\) |
derivativedivides | \(\frac {\frac {\arctan \left (a x \right ) a^{3} x^{3}}{3 c}-\frac {\arctan \left (a x \right ) a x}{c}+\frac {\arctan \left (a x \right )^{2}}{c}-\frac {\frac {a^{2} x^{2}}{2}-2 \ln \left (a^{2} x^{2}+1\right )+\frac {3 \arctan \left (a x \right )^{2}}{2}}{3 c}}{a^{5}}\) | \(76\) |
default | \(\frac {\frac {\arctan \left (a x \right ) a^{3} x^{3}}{3 c}-\frac {\arctan \left (a x \right ) a x}{c}+\frac {\arctan \left (a x \right )^{2}}{c}-\frac {\frac {a^{2} x^{2}}{2}-2 \ln \left (a^{2} x^{2}+1\right )+\frac {3 \arctan \left (a x \right )^{2}}{2}}{3 c}}{a^{5}}\) | \(76\) |
parts | \(\frac {x^{3} \arctan \left (a x \right )}{3 a^{2} c}-\frac {x \arctan \left (a x \right )}{a^{4} c}+\frac {\arctan \left (a x \right )^{2}}{a^{5} c}-\frac {\frac {\arctan \left (a x \right )^{2}}{2 a^{5}}+\frac {\frac {a^{2} x^{2}}{2}-2 \ln \left (a^{2} x^{2}+1\right )}{3 a^{5}}}{c}\) | \(86\) |
risch | \(-\frac {\ln \left (i a x +1\right )^{2}}{8 a^{5} c}-\frac {i \left (2 a^{3} x^{3}+3 i \ln \left (-i a x +1\right )-6 a x \right ) \ln \left (i a x +1\right )}{12 c \,a^{5}}+\frac {i x^{3} \ln \left (-i a x +1\right )}{6 a^{2} c}-\frac {i x \ln \left (-i a x +1\right )}{2 a^{4} c}-\frac {x^{2}}{6 a^{3} c}-\frac {\ln \left (-i a x +1\right )^{2}}{8 a^{5} c}+\frac {2 \ln \left (-a^{2} x^{2}-1\right )}{3 a^{5} c}\) | \(147\) |
Input:
int(x^4*arctan(a*x)/(a^2*c*x^2+c),x,method=_RETURNVERBOSE)
Output:
1/6*(2*arctan(a*x)*x^3*a^3-a^2*x^2-6*arctan(a*x)*a*x+3*arctan(a*x)^2+4*ln( a^2*x^2+1))/c/a^5
Time = 0.11 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.68 \[ \int \frac {x^4 \arctan (a x)}{c+a^2 c x^2} \, dx=-\frac {a^{2} x^{2} - 2 \, {\left (a^{3} x^{3} - 3 \, a x\right )} \arctan \left (a x\right ) - 3 \, \arctan \left (a x\right )^{2} - 4 \, \log \left (a^{2} x^{2} + 1\right )}{6 \, a^{5} c} \] Input:
integrate(x^4*arctan(a*x)/(a^2*c*x^2+c),x, algorithm="fricas")
Output:
-1/6*(a^2*x^2 - 2*(a^3*x^3 - 3*a*x)*arctan(a*x) - 3*arctan(a*x)^2 - 4*log( a^2*x^2 + 1))/(a^5*c)
Time = 0.37 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.88 \[ \int \frac {x^4 \arctan (a x)}{c+a^2 c x^2} \, dx=\begin {cases} \frac {x^{3} \operatorname {atan}{\left (a x \right )}}{3 a^{2} c} - \frac {x^{2}}{6 a^{3} c} - \frac {x \operatorname {atan}{\left (a x \right )}}{a^{4} c} + \frac {2 \log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{3 a^{5} c} + \frac {\operatorname {atan}^{2}{\left (a x \right )}}{2 a^{5} c} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \] Input:
integrate(x**4*atan(a*x)/(a**2*c*x**2+c),x)
Output:
Piecewise((x**3*atan(a*x)/(3*a**2*c) - x**2/(6*a**3*c) - x*atan(a*x)/(a**4 *c) + 2*log(x**2 + a**(-2))/(3*a**5*c) + atan(a*x)**2/(2*a**5*c), Ne(a, 0) ), (0, True))
Time = 0.11 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.92 \[ \int \frac {x^4 \arctan (a x)}{c+a^2 c x^2} \, dx=\frac {1}{3} \, {\left (\frac {a^{2} x^{3} - 3 \, x}{a^{4} c} + \frac {3 \, \arctan \left (a x\right )}{a^{5} c}\right )} \arctan \left (a x\right ) - \frac {a^{2} x^{2} + 3 \, \arctan \left (a x\right )^{2} - 4 \, \log \left (a^{2} x^{2} + 1\right )}{6 \, a^{5} c} \] Input:
integrate(x^4*arctan(a*x)/(a^2*c*x^2+c),x, algorithm="maxima")
Output:
1/3*((a^2*x^3 - 3*x)/(a^4*c) + 3*arctan(a*x)/(a^5*c))*arctan(a*x) - 1/6*(a ^2*x^2 + 3*arctan(a*x)^2 - 4*log(a^2*x^2 + 1))/(a^5*c)
Time = 0.11 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.71 \[ \int \frac {x^4 \arctan (a x)}{c+a^2 c x^2} \, dx=\frac {2 \, a^{3} x^{3} \arctan \left (a x\right ) - a^{2} x^{2} - 6 \, a x \arctan \left (a x\right ) + 3 \, \arctan \left (a x\right )^{2} + 4 \, \log \left (a^{2} x^{2} + 1\right )}{6 \, a^{5} c} \] Input:
integrate(x^4*arctan(a*x)/(a^2*c*x^2+c),x, algorithm="giac")
Output:
1/6*(2*a^3*x^3*arctan(a*x) - a^2*x^2 - 6*a*x*arctan(a*x) + 3*arctan(a*x)^2 + 4*log(a^2*x^2 + 1))/(a^5*c)
Time = 0.18 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.91 \[ \int \frac {x^4 \arctan (a x)}{c+a^2 c x^2} \, dx=\frac {2\,\ln \left (a^2\,x^2+1\right )}{3\,a^5\,c}-a^2\,\mathrm {atan}\left (a\,x\right )\,\left (\frac {x}{a^6\,c}-\frac {x^3}{3\,a^4\,c}\right )-\frac {x^2}{6\,a^3\,c}+\frac {{\mathrm {atan}\left (a\,x\right )}^2}{2\,a^5\,c} \] Input:
int((x^4*atan(a*x))/(c + a^2*c*x^2),x)
Output:
(2*log(a^2*x^2 + 1))/(3*a^5*c) - a^2*atan(a*x)*(x/(a^6*c) - x^3/(3*a^4*c)) - x^2/(6*a^3*c) + atan(a*x)^2/(2*a^5*c)
Time = 0.21 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.71 \[ \int \frac {x^4 \arctan (a x)}{c+a^2 c x^2} \, dx=\frac {3 \mathit {atan} \left (a x \right )^{2}+2 \mathit {atan} \left (a x \right ) a^{3} x^{3}-6 \mathit {atan} \left (a x \right ) a x +4 \,\mathrm {log}\left (a^{2} x^{2}+1\right )-a^{2} x^{2}}{6 a^{5} c} \] Input:
int(x^4*atan(a*x)/(a^2*c*x^2+c),x)
Output:
(3*atan(a*x)**2 + 2*atan(a*x)*a**3*x**3 - 6*atan(a*x)*a*x + 4*log(a**2*x** 2 + 1) - a**2*x**2)/(6*a**5*c)