\(\int \frac {\arctan (a x)^3}{\sqrt {c+a^2 c x^2}} \, dx\) [439]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 21, antiderivative size = 368 \[ \int \frac {\arctan (a x)^3}{\sqrt {c+a^2 c x^2}} \, dx=-\frac {2 i \sqrt {1+a^2 x^2} \arctan \left (e^{i \arctan (a x)}\right ) \arctan (a x)^3}{a \sqrt {c+a^2 c x^2}}+\frac {3 i \sqrt {1+a^2 x^2} \arctan (a x)^2 \operatorname {PolyLog}\left (2,-i e^{i \arctan (a x)}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {3 i \sqrt {1+a^2 x^2} \arctan (a x)^2 \operatorname {PolyLog}\left (2,i e^{i \arctan (a x)}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {6 \sqrt {1+a^2 x^2} \arctan (a x) \operatorname {PolyLog}\left (3,-i e^{i \arctan (a x)}\right )}{a \sqrt {c+a^2 c x^2}}+\frac {6 \sqrt {1+a^2 x^2} \arctan (a x) \operatorname {PolyLog}\left (3,i e^{i \arctan (a x)}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {6 i \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (4,-i e^{i \arctan (a x)}\right )}{a \sqrt {c+a^2 c x^2}}+\frac {6 i \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (4,i e^{i \arctan (a x)}\right )}{a \sqrt {c+a^2 c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 368, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5010, 5008, 4266, 2611, 6744, 2320, 6724} \[ \int \frac {\arctan (a x)^3}{\sqrt {c+a^2 c x^2}} \, dx=\frac {3 i \sqrt {a^2 x^2+1} \arctan (a x)^2 \operatorname {PolyLog}\left (2,-i e^{i \arctan (a x)}\right )}{a \sqrt {a^2 c x^2+c}}-\frac {3 i \sqrt {a^2 x^2+1} \arctan (a x)^2 \operatorname {PolyLog}\left (2,i e^{i \arctan (a x)}\right )}{a \sqrt {a^2 c x^2+c}}-\frac {6 \sqrt {a^2 x^2+1} \arctan (a x) \operatorname {PolyLog}\left (3,-i e^{i \arctan (a x)}\right )}{a \sqrt {a^2 c x^2+c}}+\frac {6 \sqrt {a^2 x^2+1} \arctan (a x) \operatorname {PolyLog}\left (3,i e^{i \arctan (a x)}\right )}{a \sqrt {a^2 c x^2+c}}-\frac {6 i \sqrt {a^2 x^2+1} \operatorname {PolyLog}\left (4,-i e^{i \arctan (a x)}\right )}{a \sqrt {a^2 c x^2+c}}+\frac {6 i \sqrt {a^2 x^2+1} \operatorname {PolyLog}\left (4,i e^{i \arctan (a x)}\right )}{a \sqrt {a^2 c x^2+c}}-\frac {2 i \sqrt {a^2 x^2+1} \arctan \left (e^{i \arctan (a x)}\right ) \arctan (a x)^3}{a \sqrt {a^2 c x^2+c}} \]

[In]

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

[Out]

((-2*I)*Sqrt[1 + a^2*x^2]*ArcTan[E^(I*ArcTan[a*x])]*ArcTan[a*x]^3)/(a*Sqrt[c + a^2*c*x^2]) + ((3*I)*Sqrt[1 + a
^2*x^2]*ArcTan[a*x]^2*PolyLog[2, (-I)*E^(I*ArcTan[a*x])])/(a*Sqrt[c + a^2*c*x^2]) - ((3*I)*Sqrt[1 + a^2*x^2]*A
rcTan[a*x]^2*PolyLog[2, I*E^(I*ArcTan[a*x])])/(a*Sqrt[c + a^2*c*x^2]) - (6*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyL
og[3, (-I)*E^(I*ArcTan[a*x])])/(a*Sqrt[c + a^2*c*x^2]) + (6*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[3, I*E^(I*Ar
cTan[a*x])])/(a*Sqrt[c + a^2*c*x^2]) - ((6*I)*Sqrt[1 + a^2*x^2]*PolyLog[4, (-I)*E^(I*ArcTan[a*x])])/(a*Sqrt[c
+ a^2*c*x^2]) + ((6*I)*Sqrt[1 + a^2*x^2]*PolyLog[4, I*E^(I*ArcTan[a*x])])/(a*Sqrt[c + a^2*c*x^2])

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 4266

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))]/f), x] + (-Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
 x], x] + Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 5008

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[1/(c*Sqrt[d]), Subst
[Int[(a + b*x)^p*Sec[x], x], x, ArcTan[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0] &
& GtQ[d, 0]

Rule 5010

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[1 + c^2*x^2]/Sq
rt[d + e*x^2], Int[(a + b*ArcTan[c*x])^p/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*
d] && IGtQ[p, 0] &&  !GtQ[d, 0]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 6744

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a + b*x)))^p]/(b*c*p*Log[F])), x] - Dist[f*(m/(b*c*p*Log[F])), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.52 \[ \int \frac {\arctan (a x)^3}{\sqrt {c+a^2 c x^2}} \, dx=-\frac {i \sqrt {c \left (1+a^2 x^2\right )} \left (2 \arctan \left (e^{i \arctan (a x)}\right ) \arctan (a x)^3-3 \arctan (a x)^2 \operatorname {PolyLog}\left (2,-i e^{i \arctan (a x)}\right )+3 \arctan (a x)^2 \operatorname {PolyLog}\left (2,i e^{i \arctan (a x)}\right )-6 i \arctan (a x) \operatorname {PolyLog}\left (3,-i e^{i \arctan (a x)}\right )+6 i \arctan (a x) \operatorname {PolyLog}\left (3,i e^{i \arctan (a x)}\right )+6 \operatorname {PolyLog}\left (4,-i e^{i \arctan (a x)}\right )-6 \operatorname {PolyLog}\left (4,i e^{i \arctan (a x)}\right )\right )}{a c \sqrt {1+a^2 x^2}} \]

[In]

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

[Out]

((-I)*Sqrt[c*(1 + a^2*x^2)]*(2*ArcTan[E^(I*ArcTan[a*x])]*ArcTan[a*x]^3 - 3*ArcTan[a*x]^2*PolyLog[2, (-I)*E^(I*
ArcTan[a*x])] + 3*ArcTan[a*x]^2*PolyLog[2, I*E^(I*ArcTan[a*x])] - (6*I)*ArcTan[a*x]*PolyLog[3, (-I)*E^(I*ArcTa
n[a*x])] + (6*I)*ArcTan[a*x]*PolyLog[3, I*E^(I*ArcTan[a*x])] + 6*PolyLog[4, (-I)*E^(I*ArcTan[a*x])] - 6*PolyLo
g[4, I*E^(I*ArcTan[a*x])]))/(a*c*Sqrt[1 + a^2*x^2])

Maple [F]

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

[In]

int(arctan(a*x)^3/(a^2*c*x^2+c)^(1/2),x)

[Out]

int(arctan(a*x)^3/(a^2*c*x^2+c)^(1/2),x)

Fricas [F]

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

[In]

integrate(arctan(a*x)^3/(a^2*c*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\arctan (a x)^3}{\sqrt {c+a^2 c x^2}} \, dx=\int \frac {\operatorname {atan}^{3}{\left (a x \right )}}{\sqrt {c \left (a^{2} x^{2} + 1\right )}}\, dx \]

[In]

integrate(atan(a*x)**3/(a**2*c*x**2+c)**(1/2),x)

[Out]

Integral(atan(a*x)**3/sqrt(c*(a**2*x**2 + 1)), x)

Maxima [F]

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

[In]

integrate(arctan(a*x)^3/(a^2*c*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(arctan(a*x)^3/(a^2*c*x^2+c)^(1/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\arctan (a x)^3}{\sqrt {c+a^2 c x^2}} \, dx=\int \frac {{\mathrm {atan}\left (a\,x\right )}^3}{\sqrt {c\,a^2\,x^2+c}} \,d x \]

[In]

int(atan(a*x)^3/(c + a^2*c*x^2)^(1/2),x)

[Out]

int(atan(a*x)^3/(c + a^2*c*x^2)^(1/2), x)