\(\int \frac {(c+a^2 c x^2)^3 \arctan (a x)}{x^4} \, dx\) [172]

Optimal result

Integrand size = 20, antiderivative size = 116 \[ \int \frac {\left (c+a^2 c x^2\right )^3 \arctan (a x)}{x^4} \, dx=-\frac {a c^3}{6 x^2}-\frac {1}{6} a^5 c^3 x^2-\frac {c^3 \arctan (a x)}{3 x^3}-\frac {3 a^2 c^3 \arctan (a x)}{x}+3 a^4 c^3 x \arctan (a x)+\frac {1}{3} a^6 c^3 x^3 \arctan (a x)+\frac {8}{3} a^3 c^3 \log (x)-\frac {8}{3} a^3 c^3 \log \left (1+a^2 x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5068, 4930, 266, 4946, 272, 46, 36, 29, 31, 45} \[ \int \frac {\left (c+a^2 c x^2\right )^3 \arctan (a x)}{x^4} \, dx=\frac {1}{3} a^6 c^3 x^3 \arctan (a x)-\frac {1}{6} a^5 c^3 x^2+3 a^4 c^3 x \arctan (a x)+\frac {8}{3} a^3 c^3 \log (x)-\frac {3 a^2 c^3 \arctan (a x)}{x}-\frac {8}{3} a^3 c^3 \log \left (a^2 x^2+1\right )-\frac {c^3 \arctan (a x)}{3 x^3}-\frac {a c^3}{6 x^2} \]

[In]

Int[((c + a^2*c*x^2)^3*ArcTan[a*x])/x^4,x]

[Out]

-1/6*(a*c^3)/x^2 - (a^5*c^3*x^2)/6 - (c^3*ArcTan[a*x])/(3*x^3) - (3*a^2*c^3*ArcTan[a*x])/x + 3*a^4*c^3*x*ArcTa
n[a*x] + (a^6*c^3*x^3*ArcTan[a*x])/3 + (8*a^3*c^3*Log[x])/3 - (8*a^3*c^3*Log[1 + a^2*x^2])/3

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 46

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] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 4946

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] - Dist[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]

Rule 5068

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Int[Ex
pandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*ArcTan[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e,
c^2*d] && IGtQ[p, 0] && IGtQ[q, 1] && (EqQ[p, 1] || IntegerQ[m])

Rubi steps \begin{align*} \text {integral}& = \int \left (3 a^4 c^3 \arctan (a x)+\frac {c^3 \arctan (a x)}{x^4}+\frac {3 a^2 c^3 \arctan (a x)}{x^2}+a^6 c^3 x^2 \arctan (a x)\right ) \, dx \\ & = c^3 \int \frac {\arctan (a x)}{x^4} \, dx+\left (3 a^2 c^3\right ) \int \frac {\arctan (a x)}{x^2} \, dx+\left (3 a^4 c^3\right ) \int \arctan (a x) \, dx+\left (a^6 c^3\right ) \int x^2 \arctan (a x) \, dx \\ & = -\frac {c^3 \arctan (a x)}{3 x^3}-\frac {3 a^2 c^3 \arctan (a x)}{x}+3 a^4 c^3 x \arctan (a x)+\frac {1}{3} a^6 c^3 x^3 \arctan (a x)+\frac {1}{3} \left (a c^3\right ) \int \frac {1}{x^3 \left (1+a^2 x^2\right )} \, dx+\left (3 a^3 c^3\right ) \int \frac {1}{x \left (1+a^2 x^2\right )} \, dx-\left (3 a^5 c^3\right ) \int \frac {x}{1+a^2 x^2} \, dx-\frac {1}{3} \left (a^7 c^3\right ) \int \frac {x^3}{1+a^2 x^2} \, dx \\ & = -\frac {c^3 \arctan (a x)}{3 x^3}-\frac {3 a^2 c^3 \arctan (a x)}{x}+3 a^4 c^3 x \arctan (a x)+\frac {1}{3} a^6 c^3 x^3 \arctan (a x)-\frac {3}{2} a^3 c^3 \log \left (1+a^2 x^2\right )+\frac {1}{6} \left (a c^3\right ) \text {Subst}\left (\int \frac {1}{x^2 \left (1+a^2 x\right )} \, dx,x,x^2\right )+\frac {1}{2} \left (3 a^3 c^3\right ) \text {Subst}\left (\int \frac {1}{x \left (1+a^2 x\right )} \, dx,x,x^2\right )-\frac {1}{6} \left (a^7 c^3\right ) \text {Subst}\left (\int \frac {x}{1+a^2 x} \, dx,x,x^2\right ) \\ & = -\frac {c^3 \arctan (a x)}{3 x^3}-\frac {3 a^2 c^3 \arctan (a x)}{x}+3 a^4 c^3 x \arctan (a x)+\frac {1}{3} a^6 c^3 x^3 \arctan (a x)-\frac {3}{2} a^3 c^3 \log \left (1+a^2 x^2\right )+\frac {1}{6} \left (a c^3\right ) \text {Subst}\left (\int \left (\frac {1}{x^2}-\frac {a^2}{x}+\frac {a^4}{1+a^2 x}\right ) \, dx,x,x^2\right )+\frac {1}{2} \left (3 a^3 c^3\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{2} \left (3 a^5 c^3\right ) \text {Subst}\left (\int \frac {1}{1+a^2 x} \, dx,x,x^2\right )-\frac {1}{6} \left (a^7 c^3\right ) \text {Subst}\left (\int \left (\frac {1}{a^2}-\frac {1}{a^2 \left (1+a^2 x\right )}\right ) \, dx,x,x^2\right ) \\ & = -\frac {a c^3}{6 x^2}-\frac {1}{6} a^5 c^3 x^2-\frac {c^3 \arctan (a x)}{3 x^3}-\frac {3 a^2 c^3 \arctan (a x)}{x}+3 a^4 c^3 x \arctan (a x)+\frac {1}{3} a^6 c^3 x^3 \arctan (a x)+\frac {8}{3} a^3 c^3 \log (x)-\frac {8}{3} a^3 c^3 \log \left (1+a^2 x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.72 \[ \int \frac {\left (c+a^2 c x^2\right )^3 \arctan (a x)}{x^4} \, dx=\frac {c^3 \left (2 \left (-1-9 a^2 x^2+9 a^4 x^4+a^6 x^6\right ) \arctan (a x)-a x \left (1+a^4 x^4-16 a^2 x^2 \log (x)+16 a^2 x^2 \log \left (1+a^2 x^2\right )\right )\right )}{6 x^3} \]

[In]

Integrate[((c + a^2*c*x^2)^3*ArcTan[a*x])/x^4,x]

[Out]

(c^3*(2*(-1 - 9*a^2*x^2 + 9*a^4*x^4 + a^6*x^6)*ArcTan[a*x] - a*x*(1 + a^4*x^4 - 16*a^2*x^2*Log[x] + 16*a^2*x^2
*Log[1 + a^2*x^2])))/(6*x^3)

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.85

method	result	size
parts	\(\frac {a^{6} c^{3} x^{3} \arctan \left (a x \right )}{3}+3 a^{4} c^{3} x \arctan \left (a x \right )-\frac {3 a^{2} c^{3} \arctan \left (a x \right )}{x}-\frac {c^{3} \arctan \left (a x \right )}{3 x^{3}}-\frac {c^{3} a \left (\frac {a^{4} x^{2}}{2}+8 a^{2} \ln \left (a^{2} x^{2}+1\right )+\frac {1}{2 x^{2}}-8 a^{2} \ln \left (x \right )\right )}{3}\)	\(99\)
derivativedivides	\(a^{3} \left (\frac {a^{3} c^{3} x^{3} \arctan \left (a x \right )}{3}+3 a \,c^{3} x \arctan \left (a x \right )-\frac {c^{3} \arctan \left (a x \right )}{3 a^{3} x^{3}}-\frac {3 c^{3} \arctan \left (a x \right )}{a x}-\frac {c^{3} \left (8 \ln \left (a^{2} x^{2}+1\right )+\frac {a^{2} x^{2}}{2}+\frac {1}{2 a^{2} x^{2}}-8 \ln \left (a x \right )\right )}{3}\right )\)	\(102\)
default	\(a^{3} \left (\frac {a^{3} c^{3} x^{3} \arctan \left (a x \right )}{3}+3 a \,c^{3} x \arctan \left (a x \right )-\frac {c^{3} \arctan \left (a x \right )}{3 a^{3} x^{3}}-\frac {3 c^{3} \arctan \left (a x \right )}{a x}-\frac {c^{3} \left (8 \ln \left (a^{2} x^{2}+1\right )+\frac {a^{2} x^{2}}{2}+\frac {1}{2 a^{2} x^{2}}-8 \ln \left (a x \right )\right )}{3}\right )\)	\(102\)
parallelrisch	\(\frac {2 a^{6} c^{3} x^{6} \arctan \left (a x \right )-a^{5} c^{3} x^{5}+18 a^{4} c^{3} x^{4} \arctan \left (a x \right )+16 a^{3} c^{3} \ln \left (x \right ) x^{3}-16 a^{3} c^{3} \ln \left (a^{2} x^{2}+1\right ) x^{3}+a^{3} c^{3} x^{3}-18 a^{2} c^{3} x^{2} \arctan \left (a x \right )-a \,c^{3} x -2 c^{3} \arctan \left (a x \right )}{6 x^{3}}\)	\(123\)
risch	\(-\frac {i c^{3} \left (a^{6} x^{6}+9 a^{4} x^{4}-9 a^{2} x^{2}-1\right ) \ln \left (i a x +1\right )}{6 x^{3}}+\frac {i c^{3} \left (a^{6} x^{6} \ln \left (-i a x +1\right )+i a^{5} x^{5}+9 x^{4} \ln \left (-i a x +1\right ) a^{4}-16 i \ln \left (x \right ) a^{3} x^{3}+16 i \ln \left (2 a^{2} x^{2}+2\right ) a^{3} x^{3}-9 a^{2} x^{2} \ln \left (-i a x +1\right )+i a x -\ln \left (-i a x +1\right )\right )}{6 x^{3}}\)	\(156\)
meijerg	\(\frac {a^{3} c^{3} \left (-\frac {2 a^{2} x^{2}}{3}+\frac {4 a^{4} x^{4} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{3 \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (a^{2} x^{2}+1\right )}{3}\right )}{4}+\frac {3 a^{3} c^{3} \left (\frac {4 a^{2} x^{2} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}-2 \ln \left (a^{2} x^{2}+1\right )\right )}{4}+\frac {3 a^{3} c^{3} \left (4 \ln \left (x \right )+4 \ln \left (a \right )-\frac {4 \arctan \left (\sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}-2 \ln \left (a^{2} x^{2}+1\right )\right )}{4}+\frac {a^{3} c^{3} \left (-\frac {2}{a^{2} x^{2}}+\frac {4}{9}-\frac {4 \ln \left (x \right )}{3}-\frac {4 \ln \left (a \right )}{3}+\frac {-\frac {4 a^{2} x^{2}}{9}+\frac {4}{3}}{a^{2} x^{2}}-\frac {4 \arctan \left (\sqrt {a^{2} x^{2}}\right )}{3 a^{2} x^{2} \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (a^{2} x^{2}+1\right )}{3}\right )}{4}\)	\(239\)

[In]

int((a^2*c*x^2+c)^3*arctan(a*x)/x^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.86 \[ \int \frac {\left (c+a^2 c x^2\right )^3 \arctan (a x)}{x^4} \, dx=-\frac {a^{5} c^{3} x^{5} + 16 \, a^{3} c^{3} x^{3} \log \left (a^{2} x^{2} + 1\right ) - 16 \, a^{3} c^{3} x^{3} \log \left (x\right ) + a c^{3} x - 2 \, {\left (a^{6} c^{3} x^{6} + 9 \, a^{4} c^{3} x^{4} - 9 \, a^{2} c^{3} x^{2} - c^{3}\right )} \arctan \left (a x\right )}{6 \, x^{3}} \]

[In]

integrate((a^2*c*x^2+c)^3*arctan(a*x)/x^4,x, algorithm="fricas")

[Out]

-1/6*(a^5*c^3*x^5 + 16*a^3*c^3*x^3*log(a^2*x^2 + 1) - 16*a^3*c^3*x^3*log(x) + a*c^3*x - 2*(a^6*c^3*x^6 + 9*a^4
*c^3*x^4 - 9*a^2*c^3*x^2 - c^3)*arctan(a*x))/x^3

Sympy [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.01 \[ \int \frac {\left (c+a^2 c x^2\right )^3 \arctan (a x)}{x^4} \, dx=\begin {cases} \frac {a^{6} c^{3} x^{3} \operatorname {atan}{\left (a x \right )}}{3} - \frac {a^{5} c^{3} x^{2}}{6} + 3 a^{4} c^{3} x \operatorname {atan}{\left (a x \right )} + \frac {8 a^{3} c^{3} \log {\left (x \right )}}{3} - \frac {8 a^{3} c^{3} \log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{3} - \frac {3 a^{2} c^{3} \operatorname {atan}{\left (a x \right )}}{x} - \frac {a c^{3}}{6 x^{2}} - \frac {c^{3} \operatorname {atan}{\left (a x \right )}}{3 x^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

[In]

integrate((a**2*c*x**2+c)**3*atan(a*x)/x**4,x)

[Out]

Piecewise((a**6*c**3*x**3*atan(a*x)/3 - a**5*c**3*x**2/6 + 3*a**4*c**3*x*atan(a*x) + 8*a**3*c**3*log(x)/3 - 8*
a**3*c**3*log(x**2 + a**(-2))/3 - 3*a**2*c**3*atan(a*x)/x - a*c**3/(6*x**2) - c**3*atan(a*x)/(3*x**3), Ne(a, 0
)), (0, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.83 \[ \int \frac {\left (c+a^2 c x^2\right )^3 \arctan (a x)}{x^4} \, dx=-\frac {1}{6} \, {\left (a^{4} c^{3} x^{2} + 16 \, a^{2} c^{3} \log \left (a^{2} x^{2} + 1\right ) - 16 \, a^{2} c^{3} \log \left (x\right ) + \frac {c^{3}}{x^{2}}\right )} a + \frac {1}{3} \, {\left (a^{6} c^{3} x^{3} + 9 \, a^{4} c^{3} x - \frac {9 \, a^{2} c^{3} x^{2} + c^{3}}{x^{3}}\right )} \arctan \left (a x\right ) \]

[In]

integrate((a^2*c*x^2+c)^3*arctan(a*x)/x^4,x, algorithm="maxima")

[Out]

-1/6*(a^4*c^3*x^2 + 16*a^2*c^3*log(a^2*x^2 + 1) - 16*a^2*c^3*log(x) + c^3/x^2)*a + 1/3*(a^6*c^3*x^3 + 9*a^4*c^
3*x - (9*a^2*c^3*x^2 + c^3)/x^3)*arctan(a*x)

Giac [F]

\[ \int \frac {\left (c+a^2 c x^2\right )^3 \arctan (a x)}{x^4} \, dx=\int { \frac {{\left (a^{2} c x^{2} + c\right )}^{3} \arctan \left (a x\right )}{x^{4}} \,d x } \]

[In]

integrate((a^2*c*x^2+c)^3*arctan(a*x)/x^4,x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.61 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.84 \[ \int \frac {\left (c+a^2 c x^2\right )^3 \arctan (a x)}{x^4} \, dx=-\frac {c^3\,\left (2\,\mathrm {atan}\left (a\,x\right )+a\,x-a^3\,x^3+a^5\,x^5+18\,a^2\,x^2\,\mathrm {atan}\left (a\,x\right )-18\,a^4\,x^4\,\mathrm {atan}\left (a\,x\right )-2\,a^6\,x^6\,\mathrm {atan}\left (a\,x\right )+16\,a^3\,x^3\,\ln \left (a^2\,x^2+1\right )-16\,a^3\,x^3\,\ln \left (x\right )\right )}{6\,x^3} \]

[In]

int((atan(a*x)*(c + a^2*c*x^2)^3)/x^4,x)

[Out]

-(c^3*(2*atan(a*x) + a*x - a^3*x^3 + a^5*x^5 + 18*a^2*x^2*atan(a*x) - 18*a^4*x^4*atan(a*x) - 2*a^6*x^6*atan(a*
x) + 16*a^3*x^3*log(a^2*x^2 + 1) - 16*a^3*x^3*log(x)))/(6*x^3)