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

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.58 \[ \int \sqrt {c+a^2 c x^2} \arctan (a x) \, dx=\frac {\sqrt {c \left (1+a^2 x^2\right )} \left (\sqrt {1+a^2 x^2} (-1+a x \arctan (a x))+\arctan (a x) \left (\log \left (1-i e^{i \arctan (a x)}\right )-\log \left (1+i e^{i \arctan (a x)}\right )\right )+i \operatorname {PolyLog}\left (2,-i e^{i \arctan (a x)}\right )-i \operatorname {PolyLog}\left (2,i e^{i \arctan (a x)}\right )\right )}{2 a \sqrt {1+a^2 x^2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.77, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {5413, 5425, 5421}

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.

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

\(\Big \downarrow \) 5413

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

\(\Big \downarrow \) 5425

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

\(\Big \downarrow \) 5421

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 5413

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

rule 5421

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]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && 
GtQ[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]
 

Maple [A] (verified)

Time = 1.31 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.73

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

Input:

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

Output:

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

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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