3.1.82 \(\int (a+b \arctan (c x^2))^2 \, dx\) [82]

3.1.82.1 Optimal result

Integrand size = 12, antiderivative size = 1191 \[ \int \left (a+b \arctan \left (c x^2\right )\right )^2 \, dx =\text {Too large to display} \]

output

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

3.1.82.2 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. \(2(1191)=2382\).

Time = 38.10 (sec) , antiderivative size = 4697, normalized size of antiderivative = 3.94 \[ \int \left (a+b \arctan \left (c x^2\right )\right )^2 \, dx=\text {Result too large to show} \]

input

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

output

a^2*x + (a*b*Sqrt[c*x^2]*(2*Sqrt[c*x^2]*ArcTan[c*x^2] - Sqrt[2]*(ArcTan[(- 
1 + c*x^2)/(Sqrt[2]*Sqrt[c*x^2])] - ArcTanh[(Sqrt[2]*Sqrt[c*x^2])/(1 + c*x 
^2)])))/(c*x) + (b^2*Sqrt[c*x^2]*(2*Sqrt[c*x^2]*ArcTan[c*x^2]^2 - 4*((ArcT 
an[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]*S 
qrt[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*ArcTan[1 - Sqr 
t[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]]*Arc 
Tanh[1 + 2*I] + (5*I)*(-ArcTan[2 + I] + ArcTan[1 - Sqrt[2]*Sqrt[c*x^2]])*L 
og[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]*Sqrt[c*x^2]...
 

3.1.82.3 Rubi [A] (verified)

Time = 1.92 (sec) , antiderivative size = 1191, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5347, 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.

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

\(\Big \downarrow \) 5347

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

\(\Big \downarrow \) 2009

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

input

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

output

a^2*x - (2*(-1)^(3/4)*a*b*ArcTan[(-1)^(3/4)*Sqrt[c]*x])/Sqrt[c] + ((-1)^(3 
/4)*b^2*ArcTan[(-1)^(3/4)*Sqrt[c]*x]^2)/Sqrt[c] + (2*(-1)^(3/4)*a*b*ArcTan 
h[(-1)^(3/4)*Sqrt[c]*x])/Sqrt[c] - ((-1)^(1/4)*b^2*ArcTanh[(-1)^(3/4)*Sqrt 
[c]*x]^2)/Sqrt[c] + (2*(-1)^(1/4)*b^2*ArcTan[(-1)^(3/4)*Sqrt[c]*x]*Log[2/( 
1 - (-1)^(1/4)*Sqrt[c]*x)])/Sqrt[c] - (2*(-1)^(1/4)*b^2*ArcTan[(-1)^(3/4)* 
Sqrt[c]*x]*Log[2/(1 + (-1)^(1/4)*Sqrt[c]*x)])/Sqrt[c] + ((-1)^(1/4)*b^2*Ar 
cTan[(-1)^(3/4)*Sqrt[c]*x]*Log[(Sqrt[2]*((-1)^(1/4) + Sqrt[c]*x))/(1 + (-1 
)^(1/4)*Sqrt[c]*x)])/Sqrt[c] + (2*(-1)^(1/4)*b^2*ArcTanh[(-1)^(3/4)*Sqrt[c 
]*x]*Log[2/(1 - (-1)^(3/4)*Sqrt[c]*x)])/Sqrt[c] - (2*(-1)^(1/4)*b^2*ArcTan 
h[(-1)^(3/4)*Sqrt[c]*x]*Log[2/(1 + (-1)^(3/4)*Sqrt[c]*x)])/Sqrt[c] + ((-1) 
^(1/4)*b^2*ArcTanh[(-1)^(3/4)*Sqrt[c]*x]*Log[-((Sqrt[2]*((-1)^(3/4) + Sqrt 
[c]*x))/(1 + (-1)^(3/4)*Sqrt[c]*x))])/Sqrt[c] + ((-1)^(1/4)*b^2*ArcTanh[(- 
1)^(3/4)*Sqrt[c]*x]*Log[((1 + I)*(1 + (-1)^(1/4)*Sqrt[c]*x))/(1 + (-1)^(3/ 
4)*Sqrt[c]*x)])/Sqrt[c] + ((-1)^(1/4)*b^2*ArcTan[(-1)^(3/4)*Sqrt[c]*x]*Log 
[((1 - I)*(1 + (-1)^(3/4)*Sqrt[c]*x))/(1 + (-1)^(1/4)*Sqrt[c]*x)])/Sqrt[c] 
 + I*a*b*x*Log[1 - I*c*x^2] + ((-1)^(1/4)*b^2*ArcTan[(-1)^(3/4)*Sqrt[c]*x] 
*Log[1 - I*c*x^2])/Sqrt[c] - ((-1)^(1/4)*b^2*ArcTanh[(-1)^(3/4)*Sqrt[c]*x] 
*Log[1 - I*c*x^2])/Sqrt[c] - (b^2*x*Log[1 - I*c*x^2]^2)/4 - I*a*b*x*Log[1 
+ I*c*x^2] - ((-1)^(1/4)*b^2*ArcTan[(-1)^(3/4)*Sqrt[c]*x]*Log[1 + I*c*x^2] 
)/Sqrt[c] + ((-1)^(1/4)*b^2*ArcTanh[(-1)^(3/4)*Sqrt[c]*x]*Log[1 + I*c*x...
 

3.1.82.3.1 Defintions of rubi rules used

rule 2009

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

rule 5347

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_)]*(b_.))^(p_), x_Symbol] :> Int[ExpandIn 
tegrand[(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]
 

3.1.82.4 Maple [F]

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

input

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

output

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

3.1.82.5 Fricas [F]

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

input

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

output

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

3.1.82.6 Sympy [F]

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

input

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

output

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

3.1.82.7 Maxima [F]

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

input

integrate((a+b*arctan(c*x^2))^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))/c 
^(3/2) + 2*sqrt(2)*arctan(1/2*sqrt(2)*(2*c*x - sqrt(2)*sqrt(c))/sqrt(c))/c 
^(3/2) - sqrt(2)*log(c*x^2 + sqrt(2)*sqrt(c)*x + 1)/c^(3/2) + sqrt(2)*log( 
c*x^2 - sqrt(2)*sqrt(c)*x + 1)/c^(3/2)) - 4*x*arctan(c*x^2))*a*b + 1/16*(4 
*x*arctan(c*x^2)^2 - x*log(c^2*x^4 + 1)^2 + 16*integrate(1/16*(8*c^2*x^4*l 
og(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^4 + 1), x))*b^2 + a^2*x
 

3.1.82.8 Giac [F]

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

input

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

output

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

3.1.82.9 Mupad [F(-1)]

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

input

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

output

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