3.1.83 \(\int \frac {(a+b \arctan (c x^2))^2}{x^2} \, dx\) [83]

3.1.83.1 Optimal result

Integrand size = 16, antiderivative size = 1164 \[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^2} \, dx =\text {Too large to display} \]

output

-1/4*(2*a+I*b*ln(1-I*c*x^2))^2/x-1/2*b^2*ln(1-I*c*x^2)*ln(1+I*c*x^2)/x-1/2 
*(-1)^(1/4)*b^2*polylog(2,1-2^(1/2)*((-1)^(1/4)+x*c^(1/2))/(1+(-1)^(1/4)*x 
*c^(1/2)))*c^(1/2)-1/2*(-1)^(3/4)*b^2*polylog(2,1+2^(1/2)*((-1)^(3/4)+x*c^ 
(1/2))/(1+(-1)^(3/4)*x*c^(1/2)))*c^(1/2)-1/2*(-1)^(3/4)*b^2*polylog(2,1-(1 
+I)*(1+(-1)^(1/4)*x*c^(1/2))/(1+(-1)^(3/4)*x*c^(1/2)))*c^(1/2)-1/2*(-1)^(1 
/4)*b^2*polylog(2,1+(-1+I)*(1+(-1)^(3/4)*x*c^(1/2))/(1+(-1)^(1/4)*x*c^(1/2 
)))*c^(1/2)+(-1)^(1/4)*b^2*polylog(2,1-2/(1-(-1)^(1/4)*x*c^(1/2)))*c^(1/2) 
+(-1)^(1/4)*b^2*polylog(2,1-2/(1+(-1)^(1/4)*x*c^(1/2)))*c^(1/2)+(-1)^(3/4) 
*b^2*polylog(2,1-2/(1-(-1)^(3/4)*x*c^(1/2)))*c^(1/2)+(-1)^(3/4)*b^2*polylo 
g(2,1-2/(1+(-1)^(3/4)*x*c^(1/2)))*c^(1/2)+(-1)^(1/4)*b^2*arctan((-1)^(3/4) 
*x*c^(1/2))^2*c^(1/2)-(-1)^(3/4)*b^2*arctanh((-1)^(3/4)*x*c^(1/2))^2*c^(1/ 
2)+1/4*b^2*ln(1+I*c*x^2)^2/x-2*(-1)^(3/4)*b^2*arctanh((-1)^(3/4)*x*c^(1/2) 
)*ln(2/(1+(-1)^(3/4)*x*c^(1/2)))*c^(1/2)-2*(-1)^(1/4)*a*b*arctanh((-1)^(3/ 
4)*x*c^(1/2))*c^(1/2)-2*(-1)^(3/4)*b^2*arctan((-1)^(3/4)*x*c^(1/2))*ln(2/( 
1-(-1)^(1/4)*x*c^(1/2)))*c^(1/2)+2*(-1)^(3/4)*b^2*arctan((-1)^(3/4)*x*c^(1 
/2))*ln(2/(1+(-1)^(1/4)*x*c^(1/2)))*c^(1/2)+2*(-1)^(3/4)*b^2*arctanh((-1)^ 
(3/4)*x*c^(1/2))*ln(2/(1-(-1)^(3/4)*x*c^(1/2)))*c^(1/2)+I*a*b*ln(1+I*c*x^2 
)/x-(-1)^(3/4)*b^2*arctanh((-1)^(3/4)*x*c^(1/2))*ln(1-I*c*x^2)*c^(1/2)-(-1 
)^(1/4)*b*arctan((-1)^(3/4)*x*c^(1/2))*(2*a+I*b*ln(1-I*c*x^2))*c^(1/2)+(-1 
)^(3/4)*b^2*arctan((-1)^(3/4)*x*c^(1/2))*ln(1+I*c*x^2)*c^(1/2)+(-1)^(3/...
 

3.1.83.2 Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(4697\) vs. \(2(1164)=2328\).

Time = 35.92 (sec) , antiderivative size = 4697, normalized size of antiderivative = 4.04 \[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^2} \, dx=\text {Result too large to show} \]

input

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

output

-(a^2/x) + (a*b*(c*x^2)^(3/2)*((-2*ArcTan[c*x^2])/Sqrt[c*x^2] + Sqrt[2]*(A 
rcTan[(-1 + c*x^2)/(Sqrt[2]*Sqrt[c*x^2])] + ArcTanh[(Sqrt[2]*Sqrt[c*x^2])/ 
(1 + c*x^2)])))/(c*x^3) + (b^2*(c*x^2)^(3/2)*((-2*ArcTan[c*x^2]^2)/Sqrt[c* 
x^2] + 4*((ArcTan[c*x^2]*(-2*ArcTan[1 - Sqrt[2]*Sqrt[c*x^2]] + 2*ArcTan[1 
+ Sqrt[2]*Sqrt[c*x^2]] - Log[1 + c*x^2 - Sqrt[2]*Sqrt[c*x^2]] + Log[1 + c* 
x^2 + Sqrt[2]*Sqrt[c*x^2]]))/(2*Sqrt[2]) - ((ArcTan[1 - Sqrt[2]*Sqrt[c*x^2 
]] + ArcTan[1 + Sqrt[2]*Sqrt[c*x^2]])*Log[1 + c*x^2 - Sqrt[2]*Sqrt[c*x^2]] 
 - (ArcTan[1 - Sqrt[2]*Sqrt[c*x^2]] + ArcTan[1 + Sqrt[2]*Sqrt[c*x^2]])*Log 
[1 + c*x^2 + Sqrt[2]*Sqrt[c*x^2]] - (Sqrt[c*x^2]*(1 + (1 - Sqrt[2]*Sqrt[c* 
x^2])^2)^(3/2)*(2*(-5*ArcTan[2 + I]*ArcTan[1 - Sqrt[2]*Sqrt[c*x^2]] + 4*Ar 
cTan[1 - Sqrt[2]*Sqrt[c*x^2]]^2 + ((1 + 2*I)*Sqrt[1 + I]*ArcTan[1 - Sqrt[2 
]*Sqrt[c*x^2]]^2)/E^(I*ArcTan[2 + I]) + ((1 - 2*I)*Sqrt[1 - I]*ArcTan[1 - 
Sqrt[2]*Sqrt[c*x^2]]^2)/E^ArcTanh[1 + 2*I] - (5*I)*ArcTan[1 - Sqrt[2]*Sqrt 
[c*x^2]]*ArcTanh[1 + 2*I] + (5*I)*(-ArcTan[2 + I] + ArcTan[1 - Sqrt[2]*Sqr 
t[c*x^2]])*Log[1 - E^((2*I)*(-ArcTan[2 + I] + ArcTan[1 - Sqrt[2]*Sqrt[c*x^ 
2]]))] + 5*((-I)*ArcTan[1 - Sqrt[2]*Sqrt[c*x^2]] + ArcTanh[1 + 2*I])*Log[1 
 - E^((2*I)*ArcTan[1 - Sqrt[2]*Sqrt[c*x^2]] - 2*ArcTanh[1 + 2*I])] + (5*I) 
*ArcTan[2 + I]*Log[-Sin[ArcTan[2 + I] - ArcTan[1 - Sqrt[2]*Sqrt[c*x^2]]]] 
- 5*ArcTanh[1 + 2*I]*Log[Sin[ArcTan[1 - Sqrt[2]*Sqrt[c*x^2]] + I*ArcTanh[1 
 + 2*I]]]) + 5*PolyLog[2, E^((2*I)*(-ArcTan[2 + I] + ArcTan[1 - Sqrt[2]...
 

3.1.83.3 Rubi [A] (verified)

Time = 1.86 (sec) , antiderivative size = 1164, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5365, 2009}

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.

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

\(\Big \downarrow \) 5365

\(\displaystyle \int \left (\frac {\left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{4 x^2}+\frac {b \log \left (1+i c x^2\right ) \left (b \log \left (1-i c x^2\right )-2 i a\right )}{2 x^2}-\frac {b^2 \log ^2\left (1+i c x^2\right )}{4 x^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \sqrt [4]{-1} \sqrt {c} \arctan \left ((-1)^{3/4} \sqrt {c} x\right )^2 b^2-(-1)^{3/4} \sqrt {c} \text {arctanh}\left ((-1)^{3/4} \sqrt {c} x\right )^2 b^2+\frac {\log ^2\left (i c x^2+1\right ) b^2}{4 x}-2 (-1)^{3/4} \sqrt {c} \arctan \left ((-1)^{3/4} \sqrt {c} x\right ) \log \left (\frac {2}{1-\sqrt [4]{-1} \sqrt {c} x}\right ) b^2+2 (-1)^{3/4} \sqrt {c} \arctan \left ((-1)^{3/4} \sqrt {c} x\right ) \log \left (\frac {2}{\sqrt [4]{-1} \sqrt {c} x+1}\right ) b^2-(-1)^{3/4} \sqrt {c} \arctan \left ((-1)^{3/4} \sqrt {c} x\right ) \log \left (\frac {\sqrt {2} \left (\sqrt {c} x+\sqrt [4]{-1}\right )}{\sqrt [4]{-1} \sqrt {c} x+1}\right ) b^2+2 (-1)^{3/4} \sqrt {c} \text {arctanh}\left ((-1)^{3/4} \sqrt {c} x\right ) \log \left (\frac {2}{1-(-1)^{3/4} \sqrt {c} x}\right ) b^2-2 (-1)^{3/4} \sqrt {c} \text {arctanh}\left ((-1)^{3/4} \sqrt {c} x\right ) \log \left (\frac {2}{(-1)^{3/4} \sqrt {c} x+1}\right ) b^2+(-1)^{3/4} \sqrt {c} \text {arctanh}\left ((-1)^{3/4} \sqrt {c} x\right ) \log \left (-\frac {\sqrt {2} \left (\sqrt {c} x+(-1)^{3/4}\right )}{(-1)^{3/4} \sqrt {c} x+1}\right ) b^2+(-1)^{3/4} \sqrt {c} \text {arctanh}\left ((-1)^{3/4} \sqrt {c} x\right ) \log \left (\frac {(1+i) \left (\sqrt [4]{-1} \sqrt {c} x+1\right )}{(-1)^{3/4} \sqrt {c} x+1}\right ) b^2-(-1)^{3/4} \sqrt {c} \arctan \left ((-1)^{3/4} \sqrt {c} x\right ) \log \left (\frac {(1-i) \left ((-1)^{3/4} \sqrt {c} x+1\right )}{\sqrt [4]{-1} \sqrt {c} x+1}\right ) b^2-(-1)^{3/4} \sqrt {c} \text {arctanh}\left ((-1)^{3/4} \sqrt {c} x\right ) \log \left (1-i c x^2\right ) b^2+(-1)^{3/4} \sqrt {c} \arctan \left ((-1)^{3/4} \sqrt {c} x\right ) \log \left (i c x^2+1\right ) b^2+(-1)^{3/4} \sqrt {c} \text {arctanh}\left ((-1)^{3/4} \sqrt {c} x\right ) \log \left (i c x^2+1\right ) b^2-\frac {\log \left (1-i c x^2\right ) \log \left (i c x^2+1\right ) b^2}{2 x}+\sqrt [4]{-1} \sqrt {c} \operatorname {PolyLog}\left (2,1-\frac {2}{1-\sqrt [4]{-1} \sqrt {c} x}\right ) b^2+\sqrt [4]{-1} \sqrt {c} \operatorname {PolyLog}\left (2,1-\frac {2}{\sqrt [4]{-1} \sqrt {c} x+1}\right ) b^2-\frac {1}{2} \sqrt [4]{-1} \sqrt {c} \operatorname {PolyLog}\left (2,1-\frac {\sqrt {2} \left (\sqrt {c} x+\sqrt [4]{-1}\right )}{\sqrt [4]{-1} \sqrt {c} x+1}\right ) b^2+(-1)^{3/4} \sqrt {c} \operatorname {PolyLog}\left (2,1-\frac {2}{1-(-1)^{3/4} \sqrt {c} x}\right ) b^2+(-1)^{3/4} \sqrt {c} \operatorname {PolyLog}\left (2,1-\frac {2}{(-1)^{3/4} \sqrt {c} x+1}\right ) b^2-\frac {1}{2} (-1)^{3/4} \sqrt {c} \operatorname {PolyLog}\left (2,\frac {\sqrt {2} \left (\sqrt {c} x+(-1)^{3/4}\right )}{(-1)^{3/4} \sqrt {c} x+1}+1\right ) b^2-\frac {1}{2} (-1)^{3/4} \sqrt {c} \operatorname {PolyLog}\left (2,1-\frac {(1+i) \left (\sqrt [4]{-1} \sqrt {c} x+1\right )}{(-1)^{3/4} \sqrt {c} x+1}\right ) b^2-\frac {1}{2} \sqrt [4]{-1} \sqrt {c} \operatorname {PolyLog}\left (2,1-\frac {(1-i) \left ((-1)^{3/4} \sqrt {c} x+1\right )}{\sqrt [4]{-1} \sqrt {c} x+1}\right ) b^2-2 \sqrt [4]{-1} a \sqrt {c} \text {arctanh}\left ((-1)^{3/4} \sqrt {c} x\right ) b-\sqrt [4]{-1} \sqrt {c} \arctan \left ((-1)^{3/4} \sqrt {c} x\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right ) b+\frac {i a \log \left (i c x^2+1\right ) b}{x}-\frac {\left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{4 x}\)

input

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

output

(-1)^(1/4)*b^2*Sqrt[c]*ArcTan[(-1)^(3/4)*Sqrt[c]*x]^2 - 2*(-1)^(1/4)*a*b*S 
qrt[c]*ArcTanh[(-1)^(3/4)*Sqrt[c]*x] - (-1)^(3/4)*b^2*Sqrt[c]*ArcTanh[(-1) 
^(3/4)*Sqrt[c]*x]^2 - 2*(-1)^(3/4)*b^2*Sqrt[c]*ArcTan[(-1)^(3/4)*Sqrt[c]*x 
]*Log[2/(1 - (-1)^(1/4)*Sqrt[c]*x)] + 2*(-1)^(3/4)*b^2*Sqrt[c]*ArcTan[(-1) 
^(3/4)*Sqrt[c]*x]*Log[2/(1 + (-1)^(1/4)*Sqrt[c]*x)] - (-1)^(3/4)*b^2*Sqrt[ 
c]*ArcTan[(-1)^(3/4)*Sqrt[c]*x]*Log[(Sqrt[2]*((-1)^(1/4) + Sqrt[c]*x))/(1 
+ (-1)^(1/4)*Sqrt[c]*x)] + 2*(-1)^(3/4)*b^2*Sqrt[c]*ArcTanh[(-1)^(3/4)*Sqr 
t[c]*x]*Log[2/(1 - (-1)^(3/4)*Sqrt[c]*x)] - 2*(-1)^(3/4)*b^2*Sqrt[c]*ArcTa 
nh[(-1)^(3/4)*Sqrt[c]*x]*Log[2/(1 + (-1)^(3/4)*Sqrt[c]*x)] + (-1)^(3/4)*b^ 
2*Sqrt[c]*ArcTanh[(-1)^(3/4)*Sqrt[c]*x]*Log[-((Sqrt[2]*((-1)^(3/4) + Sqrt[ 
c]*x))/(1 + (-1)^(3/4)*Sqrt[c]*x))] + (-1)^(3/4)*b^2*Sqrt[c]*ArcTanh[(-1)^ 
(3/4)*Sqrt[c]*x]*Log[((1 + I)*(1 + (-1)^(1/4)*Sqrt[c]*x))/(1 + (-1)^(3/4)* 
Sqrt[c]*x)] - (-1)^(3/4)*b^2*Sqrt[c]*ArcTan[(-1)^(3/4)*Sqrt[c]*x]*Log[((1 
- I)*(1 + (-1)^(3/4)*Sqrt[c]*x))/(1 + (-1)^(1/4)*Sqrt[c]*x)] - (-1)^(3/4)* 
b^2*Sqrt[c]*ArcTanh[(-1)^(3/4)*Sqrt[c]*x]*Log[1 - I*c*x^2] - (-1)^(1/4)*b* 
Sqrt[c]*ArcTan[(-1)^(3/4)*Sqrt[c]*x]*(2*a + I*b*Log[1 - I*c*x^2]) - (2*a + 
 I*b*Log[1 - I*c*x^2])^2/(4*x) + (I*a*b*Log[1 + I*c*x^2])/x + (-1)^(3/4)*b 
^2*Sqrt[c]*ArcTan[(-1)^(3/4)*Sqrt[c]*x]*Log[1 + I*c*x^2] + (-1)^(3/4)*b^2* 
Sqrt[c]*ArcTanh[(-1)^(3/4)*Sqrt[c]*x]*Log[1 + I*c*x^2] - (b^2*Log[1 - I*c* 
x^2]*Log[1 + I*c*x^2])/(2*x) + (b^2*Log[1 + I*c*x^2]^2)/(4*x) + (-1)^(1...
 

3.1.83.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5365

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> I 
nt[ExpandIntegrand[x^m*(a + (I*b*Log[1 - I*c*x^n])/2 - (I*b*Log[1 + I*c*x^n 
])/2)^p, x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 1] && IGtQ[n, 0] && Integ 
erQ[m]
 

3.1.83.4 Maple [F]

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

input

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

output

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

3.1.83.5 Fricas [F]

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

input

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

output

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

3.1.83.6 Sympy [F]

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

input

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

output

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

3.1.83.7 Maxima [F]

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

input

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

output

1/2*(c*(2*sqrt(2)*arctan(1/2*sqrt(2)*(2*c*x + sqrt(2)*sqrt(c))/sqrt(c))/sq 
rt(c) + 2*sqrt(2)*arctan(1/2*sqrt(2)*(2*c*x - sqrt(2)*sqrt(c))/sqrt(c))/sq 
rt(c) + sqrt(2)*log(c*x^2 + sqrt(2)*sqrt(c)*x + 1)/sqrt(c) - sqrt(2)*log(c 
*x^2 - sqrt(2)*sqrt(c)*x + 1)/sqrt(c)) - 4*arctan(c*x^2)/x)*a*b - 1/16*(4* 
arctan(c*x^2)^2 - 16*x*integrate(-1/16*(8*c^2*x^4*log(c^2*x^4 + 1) - 16*c* 
x^2*arctan(c*x^2) - 12*(c^2*x^4 + 1)*arctan(c*x^2)^2 - (c^2*x^4 + 1)*log(c 
^2*x^4 + 1)^2)/(c^2*x^6 + x^2), x) - log(c^2*x^4 + 1)^2)*b^2/x - a^2/x
 

3.1.83.8 Giac [F]

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

input

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

output

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

3.1.83.9 Mupad [F(-1)]

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

input

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

output

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