\(\int \frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{x^3} \, dx\) [206]

Optimal result

Integrand size = 22, antiderivative size = 240 \[ \int \frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{x^3} \, dx=-\frac {a \sqrt {c+a^2 c x^2}}{2 x}-\frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{2 x^2}-\frac {a^2 c \sqrt {1+a^2 x^2} \arctan (a x) \text {arctanh}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}+\frac {i a^2 c \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{2 \sqrt {c+a^2 c x^2}}-\frac {i a^2 c \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{2 \sqrt {c+a^2 c x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.88 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.69 \[ \int \frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{x^3} \, dx=\frac {a^2 \sqrt {c \left (1+a^2 x^2\right )} \left (-2 \cot \left (\frac {1}{2} \arctan (a x)\right )-\arctan (a x) \csc ^2\left (\frac {1}{2} \arctan (a x)\right )+4 \arctan (a x) \log \left (1-e^{i \arctan (a x)}\right )-4 \arctan (a x) \log \left (1+e^{i \arctan (a x)}\right )+4 i \operatorname {PolyLog}\left (2,-e^{i \arctan (a x)}\right )-4 i \operatorname {PolyLog}\left (2,e^{i \arctan (a x)}\right )+\arctan (a x) \sec ^2\left (\frac {1}{2} \arctan (a x)\right )-2 \tan \left (\frac {1}{2} \arctan (a x)\right )\right )}{8 \sqrt {1+a^2 x^2}} \] Input:

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

Output:

(a^2*Sqrt[c*(1 + a^2*x^2)]*(-2*Cot[ArcTan[a*x]/2] - ArcTan[a*x]*Csc[ArcTan 
[a*x]/2]^2 + 4*ArcTan[a*x]*Log[1 - E^(I*ArcTan[a*x])] - 4*ArcTan[a*x]*Log[ 
1 + E^(I*ArcTan[a*x])] + (4*I)*PolyLog[2, -E^(I*ArcTan[a*x])] - (4*I)*Poly 
Log[2, E^(I*ArcTan[a*x])] + ArcTan[a*x]*Sec[ArcTan[a*x]/2]^2 - 2*Tan[ArcTa 
n[a*x]/2]))/(8*Sqrt[1 + a^2*x^2])
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {5481, 242, 5497, 242, 5493, 5489}

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.

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

\(\Big \downarrow \) 5481

\(\displaystyle -c \int \frac {\arctan (a x)}{x^3 \sqrt {a^2 c x^2+c}}dx+a c \int \frac {1}{x^2 \sqrt {a^2 c x^2+c}}dx-\frac {\arctan (a x) \sqrt {a^2 c x^2+c}}{x^2}\)

\(\Big \downarrow \) 242

\(\displaystyle -c \int \frac {\arctan (a x)}{x^3 \sqrt {a^2 c x^2+c}}dx-\frac {\arctan (a x) \sqrt {a^2 c x^2+c}}{x^2}-\frac {a \sqrt {a^2 c x^2+c}}{x}\)

\(\Big \downarrow \) 5497

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

\(\Big \downarrow \) 242

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

\(\Big \downarrow \) 5493

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

\(\Big \downarrow \) 5489

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 242

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ 
(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, p}, x 
] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
 

rule 5481

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*Sqrt[(d_) + (e_.)* 
(x_)^2], x_Symbol] :> Simp[(f*x)^(m + 1)*Sqrt[d + e*x^2]*((a + b*ArcTan[c*x 
])/(f*(m + 2))), x] + (Simp[d/(m + 2)   Int[(f*x)^m*((a + b*ArcTan[c*x])/Sq 
rt[d + e*x^2]), x], x] - Simp[b*c*(d/(f*(m + 2)))   Int[(f*x)^(m + 1)/Sqrt[ 
d + e*x^2], x], x]) /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && 
NeQ[m, -2]
 

rule 5489

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((x_)*Sqrt[(d_) + (e_.)*(x_)^2]), x_ 
Symbol] :> Simp[(-2/Sqrt[d])*(a + b*ArcTan[c*x])*ArcTanh[Sqrt[1 + I*c*x]/Sq 
rt[1 - I*c*x]], x] + (Simp[I*(b/Sqrt[d])*PolyLog[2, -Sqrt[1 + I*c*x]/Sqrt[1 
 - I*c*x]], x] - Simp[I*(b/Sqrt[d])*PolyLog[2, Sqrt[1 + I*c*x]/Sqrt[1 - I*c 
*x]], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[d, 0]
 

rule 5493

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*Sqrt[(d_) + (e_.)*(x_)^2 
]), x_Symbol] :> Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]   Int[(a + b*ArcTan 
[c*x])^p/(x*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 5497

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) 
+ (e_.)*(x_)^2], x_Symbol] :> Simp[(f*x)^(m + 1)*Sqrt[d + e*x^2]*((a + b*Ar 
cTan[c*x])^p/(d*f*(m + 1))), x] + (-Simp[b*c*(p/(f*(m + 1)))   Int[(f*x)^(m 
 + 1)*((a + b*ArcTan[c*x])^(p - 1)/Sqrt[d + e*x^2]), x], x] - Simp[c^2*((m 
+ 2)/(f^2*(m + 1)))   Int[(f*x)^(m + 2)*((a + b*ArcTan[c*x])^p/Sqrt[d + e*x 
^2]), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d] && GtQ[p, 0] 
 && LtQ[m, -1] && NeQ[m, -2]
 

Maple [A] (verified)

Time = 1.34 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.70

method	result	size
default	\(-\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (a x +\arctan \left (a x \right )\right )}{2 x^{2}}+\frac {i a^{2} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (i \arctan \left (a x \right ) \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-i \arctan \left (a x \right ) \ln \left (1-\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-\operatorname {polylog}\left (2, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+\operatorname {polylog}\left (2, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{2 \sqrt {a^{2} x^{2}+1}}\)	\(169\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{x^3} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a^2*c*x^2+c)^(1/2)*arctan(a*x)/x^3,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. \[ \int \frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{x^3} \, dx=\int \frac {\mathrm {atan}\left (a\,x\right )\,\sqrt {c\,a^2\,x^2+c}}{x^3} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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