\(\int \frac {(c+a^2 c x^2)^3 \arctan (a x)}{x^3} \, dx\) [171]

Optimal result

Integrand size = 20, antiderivative size = 138 \[ \int \frac {\left (c+a^2 c x^2\right )^3 \arctan (a x)}{x^3} \, dx=-\frac {a c^3}{2 x}-\frac {5}{4} a^3 c^3 x-\frac {1}{12} a^5 c^3 x^3+\frac {3}{4} a^2 c^3 \arctan (a x)-\frac {c^3 \arctan (a x)}{2 x^2}+\frac {3}{2} a^4 c^3 x^2 \arctan (a x)+\frac {1}{4} a^6 c^3 x^4 \arctan (a x)+\frac {3}{2} i a^2 c^3 \operatorname {PolyLog}(2,-i a x)-\frac {3}{2} i a^2 c^3 \operatorname {PolyLog}(2,i a x) \]

[Out]

-1/2*a*c^3/x-5/4*a^3*c^3*x-1/12*a^5*c^3*x^3+3/4*a^2*c^3*arctan(a*x)-1/2*c^3*arctan(a*x)/x^2+3/2*a^4*c^3*x^2*ar
ctan(a*x)+1/4*a^6*c^3*x^4*arctan(a*x)+3/2*I*a^2*c^3*polylog(2,-I*a*x)-3/2*I*a^2*c^3*polylog(2,I*a*x)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5068, 4946, 331, 209, 4940, 2438, 327, 308} \[ \int \frac {\left (c+a^2 c x^2\right )^3 \arctan (a x)}{x^3} \, dx=\frac {1}{4} a^6 c^3 x^4 \arctan (a x)-\frac {1}{12} a^5 c^3 x^3+\frac {3}{2} a^4 c^3 x^2 \arctan (a x)-\frac {5}{4} a^3 c^3 x+\frac {3}{4} a^2 c^3 \arctan (a x)+\frac {3}{2} i a^2 c^3 \operatorname {PolyLog}(2,-i a x)-\frac {3}{2} i a^2 c^3 \operatorname {PolyLog}(2,i a x)-\frac {c^3 \arctan (a x)}{2 x^2}-\frac {a c^3}{2 x} \]

[In]

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

[Out]

-1/2*(a*c^3)/x - (5*a^3*c^3*x)/4 - (a^5*c^3*x^3)/12 + (3*a^2*c^3*ArcTan[a*x])/4 - (c^3*ArcTan[a*x])/(2*x^2) +
(3*a^4*c^3*x^2*ArcTan[a*x])/2 + (a^6*c^3*x^4*ArcTan[a*x])/4 + ((3*I)/2)*a^2*c^3*PolyLog[2, (-I)*a*x] - ((3*I)/
2)*a^2*c^3*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 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 327

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

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 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 (\frac {c^3 \arctan (a x)}{x^3}+\frac {3 a^2 c^3 \arctan (a x)}{x}+3 a^4 c^3 x \arctan (a x)+a^6 c^3 x^3 \arctan (a x)\right ) \, dx \\ & = c^3 \int \frac {\arctan (a x)}{x^3} \, dx+\left (3 a^2 c^3\right ) \int \frac {\arctan (a x)}{x} \, dx+\left (3 a^4 c^3\right ) \int x \arctan (a x) \, dx+\left (a^6 c^3\right ) \int x^3 \arctan (a x) \, dx \\ & = -\frac {c^3 \arctan (a x)}{2 x^2}+\frac {3}{2} a^4 c^3 x^2 \arctan (a x)+\frac {1}{4} a^6 c^3 x^4 \arctan (a x)+\frac {1}{2} \left (a c^3\right ) \int \frac {1}{x^2 \left (1+a^2 x^2\right )} \, dx+\frac {1}{2} \left (3 i a^2 c^3\right ) \int \frac {\log (1-i a x)}{x} \, dx-\frac {1}{2} \left (3 i a^2 c^3\right ) \int \frac {\log (1+i a x)}{x} \, dx-\frac {1}{2} \left (3 a^5 c^3\right ) \int \frac {x^2}{1+a^2 x^2} \, dx-\frac {1}{4} \left (a^7 c^3\right ) \int \frac {x^4}{1+a^2 x^2} \, dx \\ & = -\frac {a c^3}{2 x}-\frac {3}{2} a^3 c^3 x-\frac {c^3 \arctan (a x)}{2 x^2}+\frac {3}{2} a^4 c^3 x^2 \arctan (a x)+\frac {1}{4} a^6 c^3 x^4 \arctan (a x)+\frac {3}{2} i a^2 c^3 \operatorname {PolyLog}(2,-i a x)-\frac {3}{2} i a^2 c^3 \operatorname {PolyLog}(2,i a x)-\frac {1}{2} \left (a^3 c^3\right ) \int \frac {1}{1+a^2 x^2} \, dx+\frac {1}{2} \left (3 a^3 c^3\right ) \int \frac {1}{1+a^2 x^2} \, dx-\frac {1}{4} \left (a^7 c^3\right ) \int \left (-\frac {1}{a^4}+\frac {x^2}{a^2}+\frac {1}{a^4 \left (1+a^2 x^2\right )}\right ) \, dx \\ & = -\frac {a c^3}{2 x}-\frac {5}{4} a^3 c^3 x-\frac {1}{12} a^5 c^3 x^3+a^2 c^3 \arctan (a x)-\frac {c^3 \arctan (a x)}{2 x^2}+\frac {3}{2} a^4 c^3 x^2 \arctan (a x)+\frac {1}{4} a^6 c^3 x^4 \arctan (a x)+\frac {3}{2} i a^2 c^3 \operatorname {PolyLog}(2,-i a x)-\frac {3}{2} i a^2 c^3 \operatorname {PolyLog}(2,i a x)-\frac {1}{4} \left (a^3 c^3\right ) \int \frac {1}{1+a^2 x^2} \, dx \\ & = -\frac {a c^3}{2 x}-\frac {5}{4} a^3 c^3 x-\frac {1}{12} a^5 c^3 x^3+\frac {3}{4} a^2 c^3 \arctan (a x)-\frac {c^3 \arctan (a x)}{2 x^2}+\frac {3}{2} a^4 c^3 x^2 \arctan (a x)+\frac {1}{4} a^6 c^3 x^4 \arctan (a x)+\frac {3}{2} i a^2 c^3 \operatorname {PolyLog}(2,-i a x)-\frac {3}{2} i a^2 c^3 \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 = 0.03 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.12 \[ \int \frac {\left (c+a^2 c x^2\right )^3 \arctan (a x)}{x^3} \, dx=-\frac {5}{4} a^3 c^3 x-\frac {1}{12} a^5 c^3 x^3+\frac {5}{4} a^2 c^3 \arctan (a x)-\frac {c^3 \arctan (a x)}{2 x^2}+\frac {3}{2} a^4 c^3 x^2 \arctan (a x)+\frac {1}{4} a^6 c^3 x^4 \arctan (a x)-\frac {a c^3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-a^2 x^2\right )}{2 x}+\frac {3}{2} i a^2 c^3 \operatorname {PolyLog}(2,-i a x)-\frac {3}{2} i a^2 c^3 \operatorname {PolyLog}(2,i a x) \]

[In]

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

[Out]

(-5*a^3*c^3*x)/4 - (a^5*c^3*x^3)/12 + (5*a^2*c^3*ArcTan[a*x])/4 - (c^3*ArcTan[a*x])/(2*x^2) + (3*a^4*c^3*x^2*A
rcTan[a*x])/2 + (a^6*c^3*x^4*ArcTan[a*x])/4 - (a*c^3*Hypergeometric2F1[-1/2, 1, 1/2, -(a^2*x^2)])/(2*x) + ((3*
I)/2)*a^2*c^3*PolyLog[2, (-I)*a*x] - ((3*I)/2)*a^2*c^3*PolyLog[2, I*a*x]

Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.07

method	result	size
derivativedivides	\(a^{2} \left (\frac {a^{4} c^{3} x^{4} \arctan \left (a x \right )}{4}+\frac {3 a^{2} c^{3} x^{2} \arctan \left (a x \right )}{2}-\frac {c^{3} \arctan \left (a x \right )}{2 a^{2} x^{2}}+3 c^{3} \arctan \left (a x \right ) \ln \left (a x \right )-\frac {c^{3} \left (\frac {a^{3} x^{3}}{3}+5 a x -3 \arctan \left (a x \right )+\frac {2}{a x}-6 i \ln \left (a x \right ) \ln \left (i a x +1\right )+6 i \ln \left (a x \right ) \ln \left (-i a x +1\right )-6 i \operatorname {dilog}\left (i a x +1\right )+6 i \operatorname {dilog}\left (-i a x +1\right )\right )}{4}\right )\)	\(148\)
default	\(a^{2} \left (\frac {a^{4} c^{3} x^{4} \arctan \left (a x \right )}{4}+\frac {3 a^{2} c^{3} x^{2} \arctan \left (a x \right )}{2}-\frac {c^{3} \arctan \left (a x \right )}{2 a^{2} x^{2}}+3 c^{3} \arctan \left (a x \right ) \ln \left (a x \right )-\frac {c^{3} \left (\frac {a^{3} x^{3}}{3}+5 a x -3 \arctan \left (a x \right )+\frac {2}{a x}-6 i \ln \left (a x \right ) \ln \left (i a x +1\right )+6 i \ln \left (a x \right ) \ln \left (-i a x +1\right )-6 i \operatorname {dilog}\left (i a x +1\right )+6 i \operatorname {dilog}\left (-i a x +1\right )\right )}{4}\right )\)	\(148\)
parts	\(\frac {a^{6} c^{3} x^{4} \arctan \left (a x \right )}{4}+\frac {3 a^{4} c^{3} x^{2} \arctan \left (a x \right )}{2}+3 c^{3} \arctan \left (a x \right ) a^{2} \ln \left (x \right )-\frac {c^{3} \arctan \left (a x \right )}{2 x^{2}}-\frac {c^{3} a \left (\frac {a^{4} x^{3}}{3}+5 a^{2} x +\frac {2}{x}-3 a \arctan \left (a x \right )+12 a^{2} \left (-\frac {i \ln \left (x \right ) \left (\ln \left (i a x +1\right )-\ln \left (-i a x +1\right )\right )}{2 a}-\frac {i \left (\operatorname {dilog}\left (i a x +1\right )-\operatorname {dilog}\left (-i a x +1\right )\right )}{2 a}\right )\right )}{4}\)	\(149\)
meijerg	\(\frac {a^{2} c^{3} \left (\frac {a x \left (-5 a^{2} x^{2}+15\right )}{15}-\frac {a x \left (-5 a^{4} x^{4}+5\right ) \arctan \left (\sqrt {a^{2} x^{2}}\right )}{5 \sqrt {a^{2} x^{2}}}\right )}{4}+\frac {3 a^{2} c^{3} \left (-2 a x +\frac {2 \left (3 a^{2} x^{2}+3\right ) \arctan \left (a x \right )}{3}\right )}{4}+\frac {3 a^{2} c^{3} \left (-\frac {2 i a x \operatorname {polylog}\left (2, i \sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}+\frac {2 i a x \operatorname {polylog}\left (2, -i \sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}\right )}{4}+\frac {a^{2} c^{3} \left (-\frac {2}{a x}-\frac {2 \left (a^{2} x^{2}+1\right ) \arctan \left (a x \right )}{a^{2} x^{2}}\right )}{4}\)	\(190\)
risch	\(-\frac {i c^{3} a^{2} \ln \left (i a x \right )}{4}+\frac {i c^{3} \ln \left (i a x +1\right )}{4 x^{2}}+\frac {3 i c^{3} a^{4} \ln \left (-i a x +1\right ) x^{2}}{4}+\frac {3 a^{2} c^{3} \arctan \left (a x \right )}{4}-\frac {a^{5} c^{3} x^{3}}{12}-\frac {5 a^{3} c^{3} x}{4}+\frac {i c^{3} a^{6} \ln \left (-i a x +1\right ) x^{4}}{8}+\frac {i c^{3} a^{2} \ln \left (-i a x \right )}{4}-\frac {a \,c^{3}}{2 x}-\frac {3 i c^{3} a^{4} \ln \left (i a x +1\right ) x^{2}}{4}-\frac {i c^{3} a^{6} \ln \left (i a x +1\right ) x^{4}}{8}-\frac {3 i c^{3} a^{2} \operatorname {dilog}\left (-i a x +1\right )}{2}-\frac {i c^{3} \ln \left (-i a x +1\right )}{4 x^{2}}+\frac {3 i c^{3} a^{2} \operatorname {dilog}\left (i a x +1\right )}{2}\)	\(221\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((a^6*c^3*x^6 + 3*a^4*c^3*x^4 + 3*a^2*c^3*x^2 + c^3)*arctan(a*x)/x^3, x)

Sympy [F]

\[ \int \frac {\left (c+a^2 c x^2\right )^3 \arctan (a x)}{x^3} \, dx=c^{3} \left (\int \frac {\operatorname {atan}{\left (a x \right )}}{x^{3}}\, dx + \int \frac {3 a^{2} \operatorname {atan}{\left (a x \right )}}{x}\, dx + \int 3 a^{4} x \operatorname {atan}{\left (a x \right )}\, dx + \int a^{6} x^{3} \operatorname {atan}{\left (a x \right )}\, dx\right ) \]

[In]

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

[Out]

c**3*(Integral(atan(a*x)/x**3, x) + Integral(3*a**2*atan(a*x)/x, x) + Integral(3*a**4*x*atan(a*x), x) + Integr
al(a**6*x**3*atan(a*x), x))

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.12 \[ \int \frac {\left (c+a^2 c x^2\right )^3 \arctan (a x)}{x^3} \, dx=-\frac {a^{5} c^{3} x^{5} + 15 \, a^{3} c^{3} x^{3} + 9 \, \pi a^{2} c^{3} x^{2} \log \left (a^{2} x^{2} + 1\right ) - 36 \, a^{2} c^{3} x^{2} \arctan \left (a x\right ) \log \left (a x\right ) + 18 i \, a^{2} c^{3} x^{2} {\rm Li}_2\left (i \, a x + 1\right ) - 18 i \, a^{2} c^{3} x^{2} {\rm Li}_2\left (-i \, a x + 1\right ) + 6 \, a c^{3} x - 3 \, {\left (a^{6} c^{3} x^{6} + 6 \, a^{4} c^{3} x^{4} + 3 \, a^{2} c^{3} x^{2} - 2 \, c^{3}\right )} \arctan \left (a x\right )}{12 \, x^{2}} \]

[In]

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

[Out]

-1/12*(a^5*c^3*x^5 + 15*a^3*c^3*x^3 + 9*pi*a^2*c^3*x^2*log(a^2*x^2 + 1) - 36*a^2*c^3*x^2*arctan(a*x)*log(a*x)
+ 18*I*a^2*c^3*x^2*dilog(I*a*x + 1) - 18*I*a^2*c^3*x^2*dilog(-I*a*x + 1) + 6*a*c^3*x - 3*(a^6*c^3*x^6 + 6*a^4*
c^3*x^4 + 3*a^2*c^3*x^2 - 2*c^3)*arctan(a*x))/x^2

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.65 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.10 \[ \int \frac {\left (c+a^2 c x^2\right )^3 \arctan (a x)}{x^3} \, dx=\left \{\begin {array}{cl} 0 & \text {\ if\ \ }a=0\\ 3\,a^4\,c^3\,\mathrm {atan}\left (a\,x\right )\,\left (\frac {1}{2\,a^2}+\frac {x^2}{2}\right )-\frac {a^2\,c^3\,\left (3\,\mathrm {atan}\left (a\,x\right )-3\,a\,x+a^3\,x^3\right )}{12}-\frac {c^3\,\mathrm {atan}\left (a\,x\right )}{2\,x^2}-\frac {c^3\,\left (a^3\,\mathrm {atan}\left (a\,x\right )+\frac {a^2}{x}\right )}{2\,a}-\frac {3\,a^3\,c^3\,x}{2}+\frac {a^6\,c^3\,x^4\,\mathrm {atan}\left (a\,x\right )}{4}-\frac {a^2\,c^3\,{\mathrm {Li}}_{\mathrm {2}}\left (1-a\,x\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2}+\frac {a^2\,c^3\,{\mathrm {Li}}_{\mathrm {2}}\left (1+a\,x\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2} & \text {\ if\ \ }a\neq 0 \end {array}\right . \]

[In]

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

[Out]

piecewise(a == 0, 0, a ~= 0, - (3*a^3*c^3*x)/2 - (a^2*c^3*(3*atan(a*x) - 3*a*x + a^3*x^3))/12 - (c^3*atan(a*x)
)/(2*x^2) - (a^2*c^3*dilog(- a*x*1i + 1)*3i)/2 + (a^2*c^3*dilog(a*x*1i + 1)*3i)/2 - (c^3*(a^3*atan(a*x) + a^2/
x))/(2*a) + 3*a^4*c^3*atan(a*x)*(1/(2*a^2) + x^2/2) + (a^6*c^3*x^4*atan(a*x))/4)