\(\int \sqrt {c+a^2 c x^2} \arctan (a x) \, dx\) [203]

Optimal result

Integrand size = 19, antiderivative size = 244 \[ \int \sqrt {c+a^2 c x^2} \arctan (a x) \, dx=-\frac {\sqrt {c+a^2 c x^2}}{2 a}+\frac {1}{2} x \sqrt {c+a^2 c x^2} \arctan (a x)-\frac {i c \sqrt {1+a^2 x^2} \arctan (a x) \arctan \left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a \sqrt {c+a^2 c x^2}}+\frac {i c \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{2 a \sqrt {c+a^2 c x^2}}-\frac {i c \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{2 a \sqrt {c+a^2 c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4998, 5010, 5006} \[ \int \sqrt {c+a^2 c x^2} \arctan (a x) \, dx=-\frac {i c \sqrt {a^2 x^2+1} \arctan \left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right ) \arctan (a x)}{a \sqrt {a^2 c x^2+c}}+\frac {1}{2} x \arctan (a x) \sqrt {a^2 c x^2+c}+\frac {i c \sqrt {a^2 x^2+1} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{2 a \sqrt {a^2 c x^2+c}}-\frac {i c \sqrt {a^2 x^2+1} \operatorname {PolyLog}\left (2,\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{2 a \sqrt {a^2 c x^2+c}}-\frac {\sqrt {a^2 c x^2+c}}{2 a} \]

[In]

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

[Out]

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

Rule 4998

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

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]

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

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.58 \[ \int \sqrt {c+a^2 c x^2} \arctan (a x) \, dx=\frac {\sqrt {c \left (1+a^2 x^2\right )} \left (\sqrt {1+a^2 x^2} (-1+a x \arctan (a x))+\arctan (a x) \left (\log \left (1-i e^{i \arctan (a x)}\right )-\log \left (1+i e^{i \arctan (a x)}\right )\right )+i \operatorname {PolyLog}\left (2,-i e^{i \arctan (a x)}\right )-i \operatorname {PolyLog}\left (2,i e^{i \arctan (a x)}\right )\right )}{2 a \sqrt {1+a^2 x^2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.73

method	result	size
default	\(\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (x \arctan \left (a x \right ) a -1\right )}{2 a}-\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (\arctan \left (a x \right ) \ln \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-\arctan \left (a x \right ) \ln \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-i \operatorname {dilog}\left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+i \operatorname {dilog}\left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{2 a \sqrt {a^{2} x^{2}+1}}\)	\(178\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \sqrt {c+a^2 c x^2} \arctan (a x) \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [F(-1)]

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

[In]

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

[Out]

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