\(\int \frac {x^4 \arctan (a x)}{(c+a^2 c x^2)^2} \, dx\) [183]

Optimal result

Integrand size = 20, antiderivative size = 96 \[ \int \frac {x^4 \arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {1}{4 a^5 c^2 \left (1+a^2 x^2\right )}+\frac {x \arctan (a x)}{a^4 c^2}+\frac {x \arctan (a x)}{2 a^4 c^2 \left (1+a^2 x^2\right )}-\frac {3 \arctan (a x)^2}{4 a^5 c^2}-\frac {\log \left (1+a^2 x^2\right )}{2 a^5 c^2} \]

[Out]

1/4/a^5/c^2/(a^2*x^2+1)+x*arctan(a*x)/a^4/c^2+1/2*x*arctan(a*x)/a^4/c^2/(a^2*x^2+1)-3/4*arctan(a*x)^2/a^5/c^2-
1/2*ln(a^2*x^2+1)/a^5/c^2

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {5084, 5036, 4930, 266, 5004, 5054} \[ \int \frac {x^4 \arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx=-\frac {3 \arctan (a x)^2}{4 a^5 c^2}+\frac {x \arctan (a x)}{a^4 c^2}+\frac {1}{4 a^5 c^2 \left (a^2 x^2+1\right )}-\frac {\log \left (a^2 x^2+1\right )}{2 a^5 c^2}+\frac {x \arctan (a x)}{2 a^4 c^2 \left (a^2 x^2+1\right )} \]

[In]

Int[(x^4*ArcTan[a*x])/(c + a^2*c*x^2)^2,x]

[Out]

1/(4*a^5*c^2*(1 + a^2*x^2)) + (x*ArcTan[a*x])/(a^4*c^2) + (x*ArcTan[a*x])/(2*a^4*c^2*(1 + a^2*x^2)) - (3*ArcTa
n[a*x]^2)/(4*a^5*c^2) - Log[1 + a^2*x^2]/(2*a^5*c^2)

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 4930

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x^n])^p, x] - Dist[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])

Rule 5004

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> 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]

Rule 5036

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2/
e, Int[(f*x)^(m - 2)*(a + b*ArcTan[c*x])^p, x], x] - Dist[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]

Rule 5054

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*(x_)^2*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(-b)*((d + e*x^2)^
(q + 1)/(4*c^3*d*(q + 1)^2)), x] + (-Dist[1/(2*c^2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x]), x]
, x] + Simp[x*(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])/(2*c^2*d*(q + 1))), x]) /; FreeQ[{a, b, c, d, e}, x] &&
EqQ[e, c^2*d] && LtQ[q, -1] && NeQ[q, -5/2]

Rule 5084

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[1/e, Int[
x^(m - 2)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p, x], x] - Dist[d/e, Int[x^(m - 2)*(d + e*x^2)^q*(a + b*Arc
Tan[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegersQ[p, 2*q] && LtQ[q, -1] && IGtQ[m
, 1] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {\int \frac {x^2 \arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx}{a^2}+\frac {\int \frac {x^2 \arctan (a x)}{c+a^2 c x^2} \, dx}{a^2 c} \\ & = \frac {1}{4 a^5 c^2 \left (1+a^2 x^2\right )}+\frac {x \arctan (a x)}{2 a^4 c^2 \left (1+a^2 x^2\right )}+\frac {\int \arctan (a x) \, dx}{a^4 c^2}-\frac {\int \frac {\arctan (a x)}{c+a^2 c x^2} \, dx}{2 a^4 c}-\frac {\int \frac {\arctan (a x)}{c+a^2 c x^2} \, dx}{a^4 c} \\ & = \frac {1}{4 a^5 c^2 \left (1+a^2 x^2\right )}+\frac {x \arctan (a x)}{a^4 c^2}+\frac {x \arctan (a x)}{2 a^4 c^2 \left (1+a^2 x^2\right )}-\frac {3 \arctan (a x)^2}{4 a^5 c^2}-\frac {\int \frac {x}{1+a^2 x^2} \, dx}{a^3 c^2} \\ & = \frac {1}{4 a^5 c^2 \left (1+a^2 x^2\right )}+\frac {x \arctan (a x)}{a^4 c^2}+\frac {x \arctan (a x)}{2 a^4 c^2 \left (1+a^2 x^2\right )}-\frac {3 \arctan (a x)^2}{4 a^5 c^2}-\frac {\log \left (1+a^2 x^2\right )}{2 a^5 c^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.82 \[ \int \frac {x^4 \arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {1+\left (6 a x+4 a^3 x^3\right ) \arctan (a x)-3 \left (1+a^2 x^2\right ) \arctan (a x)^2-2 \left (1+a^2 x^2\right ) \log \left (1+a^2 x^2\right )}{4 a^5 c^2 \left (1+a^2 x^2\right )} \]

[In]

Integrate[(x^4*ArcTan[a*x])/(c + a^2*c*x^2)^2,x]

[Out]

(1 + (6*a*x + 4*a^3*x^3)*ArcTan[a*x] - 3*(1 + a^2*x^2)*ArcTan[a*x]^2 - 2*(1 + a^2*x^2)*Log[1 + a^2*x^2])/(4*a^
5*c^2*(1 + a^2*x^2))

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.90

method	result	size
derivativedivides	\(\frac {\frac {\arctan \left (a x \right ) a x}{c^{2}}+\frac {a x \arctan \left (a x \right )}{2 c^{2} \left (a^{2} x^{2}+1\right )}-\frac {3 \arctan \left (a x \right )^{2}}{2 c^{2}}-\frac {-\frac {1}{2 \left (a^{2} x^{2}+1\right )}+\ln \left (a^{2} x^{2}+1\right )-\frac {3 \arctan \left (a x \right )^{2}}{2}}{2 c^{2}}}{a^{5}}\)	\(86\)
default	\(\frac {\frac {\arctan \left (a x \right ) a x}{c^{2}}+\frac {a x \arctan \left (a x \right )}{2 c^{2} \left (a^{2} x^{2}+1\right )}-\frac {3 \arctan \left (a x \right )^{2}}{2 c^{2}}-\frac {-\frac {1}{2 \left (a^{2} x^{2}+1\right )}+\ln \left (a^{2} x^{2}+1\right )-\frac {3 \arctan \left (a x \right )^{2}}{2}}{2 c^{2}}}{a^{5}}\)	\(86\)
parts	\(\frac {x \arctan \left (a x \right )}{a^{4} c^{2}}+\frac {x \arctan \left (a x \right )}{2 a^{4} c^{2} \left (a^{2} x^{2}+1\right )}-\frac {3 \arctan \left (a x \right )^{2}}{2 a^{5} c^{2}}-\frac {-\frac {3 \arctan \left (a x \right )^{2}}{4 a^{5}}+\frac {-\frac {1}{2 \left (a^{2} x^{2}+1\right )}+\ln \left (a^{2} x^{2}+1\right )}{2 a^{5}}}{c^{2}}\)	\(98\)
parallelrisch	\(\frac {4 \arctan \left (a x \right ) x^{3} a^{3}-3 x^{2} \arctan \left (a x \right )^{2} a^{2}-2 a^{2} \ln \left (a^{2} x^{2}+1\right ) x^{2}-a^{2} x^{2}+6 x \arctan \left (a x \right ) a -3 \arctan \left (a x \right )^{2}-2 \ln \left (a^{2} x^{2}+1\right )}{4 c^{2} \left (a^{2} x^{2}+1\right ) a^{5}}\)	\(101\)
risch	\(\frac {3 \ln \left (i a x +1\right )^{2}}{16 a^{5} c^{2}}-\frac {i \left (-3 i a^{2} x^{2} \ln \left (-i a x +1\right )+4 a^{3} x^{3}-3 i \ln \left (-i a x +1\right )+6 a x \right ) \ln \left (i a x +1\right )}{8 a^{5} c^{2} \left (a^{2} x^{2}+1\right )}+\frac {i \left (-3 i a^{2} x^{2} \ln \left (-i a x +1\right )^{2}-3 i \ln \left (-i a x +1\right )^{2}+8 a^{3} x^{3} \ln \left (-i a x +1\right )+12 a x \ln \left (-i a x +1\right )+8 i \ln \left (a^{2} x^{2}+1\right ) a^{2} x^{2}+8 i \ln \left (a^{2} x^{2}+1\right )-4 i\right )}{16 a^{5} \left (a x +i\right ) c^{2} \left (a x -i\right )}\)	\(209\)

[In]

int(x^4*arctan(a*x)/(a^2*c*x^2+c)^2,x,method=_RETURNVERBOSE)

[Out]

1/a^5*(1/c^2*arctan(a*x)*a*x+1/2*a*x*arctan(a*x)/c^2/(a^2*x^2+1)-3/2*arctan(a*x)^2/c^2-1/2/c^2*(-1/2/(a^2*x^2+
1)+ln(a^2*x^2+1)-3/2*arctan(a*x)^2))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.84 \[ \int \frac {x^4 \arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx=-\frac {3 \, {\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} - 2 \, {\left (2 \, a^{3} x^{3} + 3 \, a x\right )} \arctan \left (a x\right ) + 2 \, {\left (a^{2} x^{2} + 1\right )} \log \left (a^{2} x^{2} + 1\right ) - 1}{4 \, {\left (a^{7} c^{2} x^{2} + a^{5} c^{2}\right )}} \]

[In]

integrate(x^4*arctan(a*x)/(a^2*c*x^2+c)^2,x, algorithm="fricas")

[Out]

-1/4*(3*(a^2*x^2 + 1)*arctan(a*x)^2 - 2*(2*a^3*x^3 + 3*a*x)*arctan(a*x) + 2*(a^2*x^2 + 1)*log(a^2*x^2 + 1) - 1
)/(a^7*c^2*x^2 + a^5*c^2)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (90) = 180\).

Time = 0.55 (sec) , antiderivative size = 223, normalized size of antiderivative = 2.32 \[ \int \frac {x^4 \arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx=\begin {cases} \frac {4 a^{3} x^{3} \operatorname {atan}{\left (a x \right )}}{4 a^{7} c^{2} x^{2} + 4 a^{5} c^{2}} - \frac {2 a^{2} x^{2} \log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{4 a^{7} c^{2} x^{2} + 4 a^{5} c^{2}} - \frac {3 a^{2} x^{2} \operatorname {atan}^{2}{\left (a x \right )}}{4 a^{7} c^{2} x^{2} + 4 a^{5} c^{2}} + \frac {6 a x \operatorname {atan}{\left (a x \right )}}{4 a^{7} c^{2} x^{2} + 4 a^{5} c^{2}} - \frac {2 \log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{4 a^{7} c^{2} x^{2} + 4 a^{5} c^{2}} - \frac {3 \operatorname {atan}^{2}{\left (a x \right )}}{4 a^{7} c^{2} x^{2} + 4 a^{5} c^{2}} + \frac {1}{4 a^{7} c^{2} x^{2} + 4 a^{5} c^{2}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

[In]

integrate(x**4*atan(a*x)/(a**2*c*x**2+c)**2,x)

[Out]

Piecewise((4*a**3*x**3*atan(a*x)/(4*a**7*c**2*x**2 + 4*a**5*c**2) - 2*a**2*x**2*log(x**2 + a**(-2))/(4*a**7*c*
*2*x**2 + 4*a**5*c**2) - 3*a**2*x**2*atan(a*x)**2/(4*a**7*c**2*x**2 + 4*a**5*c**2) + 6*a*x*atan(a*x)/(4*a**7*c
**2*x**2 + 4*a**5*c**2) - 2*log(x**2 + a**(-2))/(4*a**7*c**2*x**2 + 4*a**5*c**2) - 3*atan(a*x)**2/(4*a**7*c**2
*x**2 + 4*a**5*c**2) + 1/(4*a**7*c**2*x**2 + 4*a**5*c**2), Ne(a, 0)), (0, True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.19 \[ \int \frac {x^4 \arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {1}{2} \, {\left (\frac {x}{a^{6} c^{2} x^{2} + a^{4} c^{2}} + \frac {2 \, x}{a^{4} c^{2}} - \frac {3 \, \arctan \left (a x\right )}{a^{5} c^{2}}\right )} \arctan \left (a x\right ) + \frac {{\left (3 \, {\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} - 2 \, {\left (a^{2} x^{2} + 1\right )} \log \left (a^{2} x^{2} + 1\right ) + 1\right )} a}{4 \, {\left (a^{8} c^{2} x^{2} + a^{6} c^{2}\right )}} \]

[In]

integrate(x^4*arctan(a*x)/(a^2*c*x^2+c)^2,x, algorithm="maxima")

[Out]

1/2*(x/(a^6*c^2*x^2 + a^4*c^2) + 2*x/(a^4*c^2) - 3*arctan(a*x)/(a^5*c^2))*arctan(a*x) + 1/4*(3*(a^2*x^2 + 1)*a
rctan(a*x)^2 - 2*(a^2*x^2 + 1)*log(a^2*x^2 + 1) + 1)*a/(a^8*c^2*x^2 + a^6*c^2)

Giac [F]

\[ \int \frac {x^4 \arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx=\int { \frac {x^{4} \arctan \left (a x\right )}{{\left (a^{2} c x^{2} + c\right )}^{2}} \,d x } \]

[In]

integrate(x^4*arctan(a*x)/(a^2*c*x^2+c)^2,x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.98 \[ \int \frac {x^4 \arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {1}{2\,a^2\,\left (2\,a^5\,c^2\,x^2+2\,a^3\,c^2\right )}-\frac {\ln \left (a^2\,x^2+1\right )}{2\,a^5\,c^2}+\frac {\mathrm {atan}\left (a\,x\right )\,\left (\frac {3\,x}{2\,a^6\,c^2}+\frac {x^3}{a^4\,c^2}\right )}{\frac {1}{a^2}+x^2}-\frac {3\,{\mathrm {atan}\left (a\,x\right )}^2}{4\,a^5\,c^2} \]

[In]

int((x^4*atan(a*x))/(c + a^2*c*x^2)^2,x)

[Out]

1/(2*a^2*(2*a^3*c^2 + 2*a^5*c^2*x^2)) - log(a^2*x^2 + 1)/(2*a^5*c^2) + (atan(a*x)*((3*x)/(2*a^6*c^2) + x^3/(a^
4*c^2)))/(1/a^2 + x^2) - (3*atan(a*x)^2)/(4*a^5*c^2)