\(\int \frac {(c+a^2 c x^2) \arctan (a x)}{x^4} \, dx\) [156]

Optimal result

Integrand size = 18, antiderivative size = 63 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^4} \, dx=-\frac {a c}{6 x^2}-\frac {c \arctan (a x)}{3 x^3}-\frac {a^2 c \arctan (a x)}{x}+\frac {2}{3} a^3 c \log (x)-\frac {1}{3} a^3 c \log \left (1+a^2 x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {5070, 4946, 272, 46, 36, 29, 31} \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^4} \, dx=\frac {2}{3} a^3 c \log (x)-\frac {a^2 c \arctan (a x)}{x}-\frac {1}{3} a^3 c \log \left (a^2 x^2+1\right )-\frac {c \arctan (a x)}{3 x^3}-\frac {a c}{6 x^2} \]

[In]

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

[Out]

-1/6*(a*c)/x^2 - (c*ArcTan[a*x])/(3*x^3) - (a^2*c*ArcTan[a*x])/x + (2*a^3*c*Log[x])/3 - (a^3*c*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 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 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 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 5070

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.92 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^4} \, dx=\frac {c \left (-2 \left (1+3 a^2 x^2\right ) \arctan (a x)+a x \left (-1+4 a^2 x^2 \log (x)-2 a^2 x^2 \log \left (1+a^2 x^2\right )\right )\right )}{6 x^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.89

method	result	size
parts	\(-\frac {a^{2} c \arctan \left (a x \right )}{x}-\frac {c \arctan \left (a x \right )}{3 x^{3}}-\frac {c a \left (a^{2} \ln \left (a^{2} x^{2}+1\right )+\frac {1}{2 x^{2}}-2 a^{2} \ln \left (x \right )\right )}{3}\)	\(56\)
derivativedivides	\(a^{3} \left (-\frac {c \arctan \left (a x \right )}{a x}-\frac {c \arctan \left (a x \right )}{3 a^{3} x^{3}}-\frac {c \left (\ln \left (a^{2} x^{2}+1\right )+\frac {1}{2 a^{2} x^{2}}-2 \ln \left (a x \right )\right )}{3}\right )\)	\(60\)
default	\(a^{3} \left (-\frac {c \arctan \left (a x \right )}{a x}-\frac {c \arctan \left (a x \right )}{3 a^{3} x^{3}}-\frac {c \left (\ln \left (a^{2} x^{2}+1\right )+\frac {1}{2 a^{2} x^{2}}-2 \ln \left (a x \right )\right )}{3}\right )\)	\(60\)
parallelrisch	\(\frac {4 c \,a^{3} \ln \left (x \right ) x^{3}-2 c \,a^{3} \ln \left (a^{2} x^{2}+1\right ) x^{3}+a^{3} c \,x^{3}-6 a^{2} c \,x^{2} \arctan \left (a x \right )-a c x -2 c \arctan \left (a x \right )}{6 x^{3}}\)	\(70\)
risch	\(\frac {i c \left (3 a^{2} x^{2}+1\right ) \ln \left (i a x +1\right )}{6 x^{3}}+\frac {c \left (4 \ln \left (x \right ) a^{3} x^{3}-2 \ln \left (-3 a^{2} x^{2}-3\right ) a^{3} x^{3}-3 i a^{2} x^{2} \ln \left (-i a x +1\right )-a x -i \ln \left (-i a x +1\right )\right )}{6 x^{3}}\)	\(95\)
meijerg	\(\frac {a^{3} c \left (4 \ln \left (x \right )+4 \ln \left (a \right )-\frac {4 \arctan \left (\sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}-2 \ln \left (a^{2} x^{2}+1\right )\right )}{4}+\frac {a^{3} c \left (-\frac {2}{a^{2} x^{2}}+\frac {4}{9}-\frac {4 \ln \left (x \right )}{3}-\frac {4 \ln \left (a \right )}{3}+\frac {-\frac {4 a^{2} x^{2}}{9}+\frac {4}{3}}{a^{2} x^{2}}-\frac {4 \arctan \left (\sqrt {a^{2} x^{2}}\right )}{3 a^{2} x^{2} \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (a^{2} x^{2}+1\right )}{3}\right )}{4}\)	\(131\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.90 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^4} \, dx=-\frac {2 \, a^{3} c x^{3} \log \left (a^{2} x^{2} + 1\right ) - 4 \, a^{3} c x^{3} \log \left (x\right ) + a c x + 2 \, {\left (3 \, a^{2} c x^{2} + c\right )} \arctan \left (a x\right )}{6 \, x^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.97 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^4} \, dx=\begin {cases} \frac {2 a^{3} c \log {\left (x \right )}}{3} - \frac {a^{3} c \log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{3} - \frac {a^{2} c \operatorname {atan}{\left (a x \right )}}{x} - \frac {a c}{6 x^{2}} - \frac {c \operatorname {atan}{\left (a x \right )}}{3 x^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.89 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^4} \, dx=-\frac {1}{6} \, {\left (2 \, a^{2} c \log \left (a^{2} x^{2} + 1\right ) - 2 \, a^{2} c \log \left (x^{2}\right ) + \frac {c}{x^{2}}\right )} a - \frac {{\left (3 \, a^{2} c x^{2} + c\right )} \arctan \left (a x\right )}{3 \, x^{3}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.90 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^4} \, dx=\frac {c\,\left (4\,a^3\,\ln \left (x\right )-2\,a^3\,\ln \left (a^2\,x^2+1\right )\right )}{6}-\frac {\frac {c\,\mathrm {atan}\left (a\,x\right )}{3}+\frac {a\,c\,x}{6}+a^2\,c\,x^2\,\mathrm {atan}\left (a\,x\right )}{x^3} \]

[In]

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

[Out]

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