3.4.34 \(\int \frac {\arctan (a x)^2}{\sqrt {c+a^2 c x^2}} \, dx\) [334]

3.4.34.1 Optimal result

Integrand size = 21, antiderivative size = 256 \[ \int \frac {\arctan (a x)^2}{\sqrt {c+a^2 c x^2}} \, dx=-\frac {2 i \sqrt {1+a^2 x^2} \arctan \left (e^{i \arctan (a x)}\right ) \arctan (a x)^2}{a \sqrt {c+a^2 c x^2}}+\frac {2 i \sqrt {1+a^2 x^2} \arctan (a x) \operatorname {PolyLog}\left (2,-i e^{i \arctan (a x)}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {2 i \sqrt {1+a^2 x^2} \arctan (a x) \operatorname {PolyLog}\left (2,i e^{i \arctan (a x)}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {2 \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (3,-i e^{i \arctan (a x)}\right )}{a \sqrt {c+a^2 c x^2}}+\frac {2 \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (3,i e^{i \arctan (a x)}\right )}{a \sqrt {c+a^2 c x^2}} \]

output

-2*I*arctan((1+I*a*x)/(a^2*x^2+1)^(1/2))*arctan(a*x)^2*(a^2*x^2+1)^(1/2)/a 
/(a^2*c*x^2+c)^(1/2)+2*I*arctan(a*x)*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)-2*I*arctan(a*x)*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)-2*poly 
log(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)+2*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)
 

3.4.34.2 Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.55 \[ \int \frac {\arctan (a x)^2}{\sqrt {c+a^2 c x^2}} \, dx=\frac {2 \sqrt {c \left (1+a^2 x^2\right )} \left (-i \arctan \left (e^{i \arctan (a x)}\right ) \arctan (a x)^2+i \arctan (a x) \operatorname {PolyLog}\left (2,-i e^{i \arctan (a x)}\right )-i \arctan (a x) \operatorname {PolyLog}\left (2,i e^{i \arctan (a x)}\right )-\operatorname {PolyLog}\left (3,-i e^{i \arctan (a x)}\right )+\operatorname {PolyLog}\left (3,i e^{i \arctan (a x)}\right )\right )}{a c \sqrt {1+a^2 x^2}} \]

input

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

output

(2*Sqrt[c*(1 + a^2*x^2)]*((-I)*ArcTan[E^(I*ArcTan[a*x])]*ArcTan[a*x]^2 + I 
*ArcTan[a*x]*PolyLog[2, (-I)*E^(I*ArcTan[a*x])] - I*ArcTan[a*x]*PolyLog[2, 
 I*E^(I*ArcTan[a*x])] - PolyLog[3, (-I)*E^(I*ArcTan[a*x])] + PolyLog[3, I* 
E^(I*ArcTan[a*x])]))/(a*c*Sqrt[1 + a^2*x^2])
 

3.4.34.3 Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.56, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5425, 5423, 3042, 4669, 3011, 2720, 7143}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\arctan (a x)^2}{\sqrt {a^2 c x^2+c}} \, dx\)

\(\Big \downarrow \) 5425

\(\displaystyle \frac {\sqrt {a^2 x^2+1} \int \frac {\arctan (a x)^2}{\sqrt {a^2 x^2+1}}dx}{\sqrt {a^2 c x^2+c}}\)

\(\Big \downarrow \) 5423

\(\displaystyle \frac {\sqrt {a^2 x^2+1} \int \sqrt {a^2 x^2+1} \arctan (a x)^2d\arctan (a x)}{a \sqrt {a^2 c x^2+c}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {a^2 x^2+1} \int \arctan (a x)^2 \csc \left (\arctan (a x)+\frac {\pi }{2}\right )d\arctan (a x)}{a \sqrt {a^2 c x^2+c}}\)

\(\Big \downarrow \) 4669

\(\displaystyle \frac {\sqrt {a^2 x^2+1} \left (-2 \int \arctan (a x) \log \left (1-i e^{i \arctan (a x)}\right )d\arctan (a x)+2 \int \arctan (a x) \log \left (1+i e^{i \arctan (a x)}\right )d\arctan (a x)-2 i \arctan \left (e^{i \arctan (a x)}\right ) \arctan (a x)^2\right )}{a \sqrt {a^2 c x^2+c}}\)

\(\Big \downarrow \) 3011

\(\displaystyle \frac {\sqrt {a^2 x^2+1} \left (2 \left (i \arctan (a x) \operatorname {PolyLog}\left (2,-i e^{i \arctan (a x)}\right )-i \int \operatorname {PolyLog}\left (2,-i e^{i \arctan (a x)}\right )d\arctan (a x)\right )-2 \left (i \arctan (a x) \operatorname {PolyLog}\left (2,i e^{i \arctan (a x)}\right )-i \int \operatorname {PolyLog}\left (2,i e^{i \arctan (a x)}\right )d\arctan (a x)\right )-2 i \arctan \left (e^{i \arctan (a x)}\right ) \arctan (a x)^2\right )}{a \sqrt {a^2 c x^2+c}}\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\sqrt {a^2 x^2+1} \left (2 \left (i \arctan (a x) \operatorname {PolyLog}\left (2,-i e^{i \arctan (a x)}\right )-\int e^{-i \arctan (a x)} \operatorname {PolyLog}\left (2,-i e^{i \arctan (a x)}\right )de^{i \arctan (a x)}\right )-2 \left (i \arctan (a x) \operatorname {PolyLog}\left (2,i e^{i \arctan (a x)}\right )-\int e^{-i \arctan (a x)} \operatorname {PolyLog}\left (2,i e^{i \arctan (a x)}\right )de^{i \arctan (a x)}\right )-2 i \arctan \left (e^{i \arctan (a x)}\right ) \arctan (a x)^2\right )}{a \sqrt {a^2 c x^2+c}}\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {\sqrt {a^2 x^2+1} \left (2 \left (i \arctan (a x) \operatorname {PolyLog}\left (2,-i e^{i \arctan (a x)}\right )-\operatorname {PolyLog}\left (3,-i e^{i \arctan (a x)}\right )\right )-2 \left (i \arctan (a x) \operatorname {PolyLog}\left (2,i e^{i \arctan (a x)}\right )-\operatorname {PolyLog}\left (3,i e^{i \arctan (a x)}\right )\right )-2 i \arctan \left (e^{i \arctan (a x)}\right ) \arctan (a x)^2\right )}{a \sqrt {a^2 c x^2+c}}\)

input

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

output

(Sqrt[1 + a^2*x^2]*((-2*I)*ArcTan[E^(I*ArcTan[a*x])]*ArcTan[a*x]^2 + 2*(I* 
ArcTan[a*x]*PolyLog[2, (-I)*E^(I*ArcTan[a*x])] - PolyLog[3, (-I)*E^(I*ArcT 
an[a*x])]) - 2*(I*ArcTan[a*x]*PolyLog[2, I*E^(I*ArcTan[a*x])] - PolyLog[3, 
 I*E^(I*ArcTan[a*x])])))/(a*Sqrt[c + a^2*c*x^2])
 

3.4.34.3.1 Defintions of rubi rules used

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{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 3011

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] + Simp[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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4669

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] + (-Si 
mp[d*(m/f)   Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))], x], 
 x] + Simp[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 5423

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_S 
ymbol] :> Simp[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] && Gt 
Q[d, 0]
 

rule 5425

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_S 
ymbol] :> Simp[Sqrt[1 + c^2*x^2]/Sqrt[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 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> 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]
 

3.4.34.4 Maple [F]

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

input

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

output

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

3.4.34.5 Fricas [F]

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

input

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

output

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

3.4.34.6 Sympy [F]

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

input

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

output

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

3.4.34.7 Maxima [F]

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

input

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

output

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

3.4.34.8 Giac [F]

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

input

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

output

sage0*x
 

3.4.34.9 Mupad [F(-1)]

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

input

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

output

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