\(\int \frac {(c+a^2 c x^2) \arctan (a x)}{x^3} \, dx\) [155]

Optimal result

Integrand size = 18, antiderivative size = 70 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^3} \, dx=-\frac {a c}{2 x}-\frac {1}{2} a^2 c \arctan (a x)-\frac {c \arctan (a x)}{2 x^2}+\frac {1}{2} i a^2 c \operatorname {PolyLog}(2,-i a x)-\frac {1}{2} i a^2 c \operatorname {PolyLog}(2,i a x) \]

[Out]

-1/2*a*c/x-1/2*a^2*c*arctan(a*x)-1/2*c*arctan(a*x)/x^2+1/2*I*a^2*c*polylog(2,-I*a*x)-1/2*I*a^2*c*polylog(2,I*a
*x)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5070, 4946, 331, 209, 4940, 2438} \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^3} \, dx=-\frac {1}{2} a^2 c \arctan (a x)+\frac {1}{2} i a^2 c \operatorname {PolyLog}(2,-i a x)-\frac {1}{2} i a^2 c \operatorname {PolyLog}(2,i a x)-\frac {c \arctan (a x)}{2 x^2}-\frac {a c}{2 x} \]

[In]

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

[Out]

-1/2*(a*c)/x - (a^2*c*ArcTan[a*x])/2 - (c*ArcTan[a*x])/(2*x^2) + (I/2)*a^2*c*PolyLog[2, (-I)*a*x] - (I/2)*a^2*
c*PolyLog[2, I*a*x]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 4940

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[I*(b/2), Int[Log[1 - I*c*x
]/x, x], x] - Dist[I*(b/2), Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

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^3} \, dx+\left (a^2 c\right ) \int \frac {\arctan (a x)}{x} \, dx \\ & = -\frac {c \arctan (a x)}{2 x^2}+\frac {1}{2} (a c) \int \frac {1}{x^2 \left (1+a^2 x^2\right )} \, dx+\frac {1}{2} \left (i a^2 c\right ) \int \frac {\log (1-i a x)}{x} \, dx-\frac {1}{2} \left (i a^2 c\right ) \int \frac {\log (1+i a x)}{x} \, dx \\ & = -\frac {a c}{2 x}-\frac {c \arctan (a x)}{2 x^2}+\frac {1}{2} i a^2 c \operatorname {PolyLog}(2,-i a x)-\frac {1}{2} i a^2 c \operatorname {PolyLog}(2,i a x)-\frac {1}{2} \left (a^3 c\right ) \int \frac {1}{1+a^2 x^2} \, dx \\ & = -\frac {a c}{2 x}-\frac {1}{2} a^2 c \arctan (a x)-\frac {c \arctan (a x)}{2 x^2}+\frac {1}{2} i a^2 c \operatorname {PolyLog}(2,-i a x)-\frac {1}{2} i a^2 c \operatorname {PolyLog}(2,i a x) \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.01 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.06 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^3} \, dx=-\frac {c \arctan (a x)}{2 x^2}-\frac {a c \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-a^2 x^2\right )}{2 x}+\frac {1}{2} i a^2 c \operatorname {PolyLog}(2,-i a x)-\frac {1}{2} i a^2 c \operatorname {PolyLog}(2,i a x) \]

[In]

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

[Out]

-1/2*(c*ArcTan[a*x])/x^2 - (a*c*Hypergeometric2F1[-1/2, 1, 1/2, -(a^2*x^2)])/(2*x) + (I/2)*a^2*c*PolyLog[2, (-
I)*a*x] - (I/2)*a^2*c*PolyLog[2, I*a*x]

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.36

method	result	size
parts	\(c \arctan \left (a x \right ) a^{2} \ln \left (x \right )-\frac {c \arctan \left (a x \right )}{2 x^{2}}-\frac {c a \left (\frac {1}{x}+a \arctan \left (a x \right )+2 a^{2} \left (-\frac {i \ln \left (x \right ) \left (\ln \left (i a x +1\right )-\ln \left (-i a x +1\right )\right )}{2 a}-\frac {i \left (\operatorname {dilog}\left (i a x +1\right )-\operatorname {dilog}\left (-i a x +1\right )\right )}{2 a}\right )\right )}{2}\)	\(95\)
derivativedivides	\(a^{2} \left (-\frac {c \arctan \left (a x \right )}{2 a^{2} x^{2}}+c \arctan \left (a x \right ) \ln \left (a x \right )-\frac {c \left (\frac {1}{a x}+\arctan \left (a x \right )-i \ln \left (a x \right ) \ln \left (i a x +1\right )+i \ln \left (a x \right ) \ln \left (-i a x +1\right )-i \operatorname {dilog}\left (i a x +1\right )+i \operatorname {dilog}\left (-i a x +1\right )\right )}{2}\right )\)	\(96\)
default	\(a^{2} \left (-\frac {c \arctan \left (a x \right )}{2 a^{2} x^{2}}+c \arctan \left (a x \right ) \ln \left (a x \right )-\frac {c \left (\frac {1}{a x}+\arctan \left (a x \right )-i \ln \left (a x \right ) \ln \left (i a x +1\right )+i \ln \left (a x \right ) \ln \left (-i a x +1\right )-i \operatorname {dilog}\left (i a x +1\right )+i \operatorname {dilog}\left (-i a x +1\right )\right )}{2}\right )\)	\(96\)
meijerg	\(\frac {c \,a^{2} \left (-\frac {2 i a x \operatorname {polylog}\left (2, i \sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}+\frac {2 i a x \operatorname {polylog}\left (2, -i \sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}\right )}{4}+\frac {c \,a^{2} \left (-\frac {2}{a x}-\frac {2 \left (a^{2} x^{2}+1\right ) \arctan \left (a x \right )}{a^{2} x^{2}}\right )}{4}\)	\(101\)
risch	\(-\frac {i c \,a^{2} \operatorname {dilog}\left (-i a x +1\right )}{2}+\frac {i c \,a^{2} \ln \left (-i a x \right )}{4}-\frac {a c}{2 x}-\frac {i c \,a^{2} \ln \left (-i a x +1\right )}{4}-\frac {i c \ln \left (-i a x +1\right )}{4 x^{2}}+\frac {i c \,a^{2} \operatorname {dilog}\left (i a x +1\right )}{2}-\frac {i c \,a^{2} \ln \left (i a x \right )}{4}+\frac {i c \,a^{2} \ln \left (i a x +1\right )}{4}+\frac {i c \ln \left (i a x +1\right )}{4 x^{2}}\)	\(125\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((a^2*c*x^2 + c)*arctan(a*x)/x^3, x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.36 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^3} \, dx=-\frac {\pi a^{2} c x^{2} \log \left (a^{2} x^{2} + 1\right ) - 4 \, a^{2} c x^{2} \arctan \left (a x\right ) \log \left (a x\right ) + 2 i \, a^{2} c x^{2} {\rm Li}_2\left (i \, a x + 1\right ) - 2 i \, a^{2} c x^{2} {\rm Li}_2\left (-i \, a x + 1\right ) + 2 \, a c x + 2 \, {\left (a^{2} c x^{2} + c\right )} \arctan \left (a x\right )}{4 \, x^{2}} \]

[In]

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

[Out]

-1/4*(pi*a^2*c*x^2*log(a^2*x^2 + 1) - 4*a^2*c*x^2*arctan(a*x)*log(a*x) + 2*I*a^2*c*x^2*dilog(I*a*x + 1) - 2*I*
a^2*c*x^2*dilog(-I*a*x + 1) + 2*a*c*x + 2*(a^2*c*x^2 + c)*arctan(a*x))/x^2

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.66 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.01 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^3} \, dx=\left \{\begin {array}{cl} 0 & \text {\ if\ \ }a=0\\ -\frac {c\,\mathrm {atan}\left (a\,x\right )}{2\,x^2}-\frac {c\,\left (a^3\,\mathrm {atan}\left (a\,x\right )+\frac {a^2}{x}\right )}{2\,a}-\frac {a^2\,c\,{\mathrm {Li}}_{\mathrm {2}}\left (1-a\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {a^2\,c\,{\mathrm {Li}}_{\mathrm {2}}\left (1+a\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} & \text {\ if\ \ }a\neq 0 \end {array}\right . \]

[In]

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

[Out]

piecewise(a == 0, 0, a ~= 0, - (c*atan(a*x))/(2*x^2) - (a^2*c*dilog(- a*x*1i + 1)*1i)/2 + (a^2*c*dilog(a*x*1i
+ 1)*1i)/2 - (c*(a^3*atan(a*x) + a^2/x))/(2*a))