\(\int \frac {x \arctan (a x)^2}{\sqrt {c+a^2 c x^2}} \, dx\) [333]

Optimal result

Integrand size = 22, antiderivative size = 220 \[ \int \frac {x \arctan (a x)^2}{\sqrt {c+a^2 c x^2}} \, dx=\frac {\sqrt {c+a^2 c x^2} \arctan (a x)^2}{a^2 c}+\frac {4 i \sqrt {1+a^2 x^2} \arctan (a x) \arctan \left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a^2 \sqrt {c+a^2 c x^2}}-\frac {2 i \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a^2 \sqrt {c+a^2 c x^2}}+\frac {2 i \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a^2 \sqrt {c+a^2 c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {5050, 5010, 5006} \[ \int \frac {x \arctan (a x)^2}{\sqrt {c+a^2 c x^2}} \, dx=\frac {\arctan (a x)^2 \sqrt {a^2 c x^2+c}}{a^2 c}+\frac {4 i \sqrt {a^2 x^2+1} \arctan \left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right ) \arctan (a x)}{a^2 \sqrt {a^2 c x^2+c}}-\frac {2 i \sqrt {a^2 x^2+1} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{a^2 \sqrt {a^2 c x^2+c}}+\frac {2 i \sqrt {a^2 x^2+1} \operatorname {PolyLog}\left (2,\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{a^2 \sqrt {a^2 c x^2+c}} \]

[In]

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

[Out]

(Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2)/(a^2*c) + ((4*I)*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*ArcTan[Sqrt[1 + I*a*x]/Sqrt
[1 - I*a*x]])/(a^2*Sqrt[c + a^2*c*x^2]) - ((2*I)*Sqrt[1 + a^2*x^2]*PolyLog[2, ((-I)*Sqrt[1 + I*a*x])/Sqrt[1 -
I*a*x]])/(a^2*Sqrt[c + a^2*c*x^2]) + ((2*I)*Sqrt[1 + a^2*x^2]*PolyLog[2, (I*Sqrt[1 + I*a*x])/Sqrt[1 - I*a*x]])
/(a^2*Sqrt[c + a^2*c*x^2])

Rule 5006

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[-2*I*(a + b*ArcTan[c*x])*(
ArcTan[Sqrt[1 + I*c*x]/Sqrt[1 - I*c*x]]/(c*Sqrt[d])), x] + (Simp[I*b*(PolyLog[2, (-I)*(Sqrt[1 + I*c*x]/Sqrt[1
- I*c*x])]/(c*Sqrt[d])), x] - Simp[I*b*(PolyLog[2, I*(Sqrt[1 + I*c*x]/Sqrt[1 - I*c*x])]/(c*Sqrt[d])), x]) /; F
reeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && 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 5050

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

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

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.57 \[ \int \frac {x \arctan (a x)^2}{\sqrt {c+a^2 c x^2}} \, dx=\frac {\sqrt {c \left (1+a^2 x^2\right )} \left (\arctan (a x)^2-\frac {2 \left (\arctan (a x) \left (\log \left (1-i e^{i \arctan (a x)}\right )-\log \left (1+i e^{i \arctan (a x)}\right )\right )+i \left (\operatorname {PolyLog}\left (2,-i e^{i \arctan (a x)}\right )-\operatorname {PolyLog}\left (2,i e^{i \arctan (a x)}\right )\right )\right )}{\sqrt {1+a^2 x^2}}\right )}{a^2 c} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.23 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.82

method	result	size
default	\(\frac {\arctan \left (a x \right )^{2} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{a^{2} c}+\frac {2 \left (\arctan \left (a x \right ) \ln \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-\arctan \left (a x \right ) \ln \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-i \operatorname {dilog}\left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+i \operatorname {dilog}\left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{\sqrt {a^{2} x^{2}+1}\, a^{2} c}\)	\(180\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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