3.1.25 \(\int (d+e x) (a+b \arctan (c x^2))^2 \, dx\) [25]

3.1.25.1 Optimal result

Integrand size = 18, antiderivative size = 1325 \[ \int (d+e x) \left (a+b \arctan \left (c x^2\right )\right )^2 \, dx =\text {Too large to display} \]

output

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

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

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(4824\) vs. \(2(1325)=2650\).

Time = 36.63 (sec) , antiderivative size = 4824, normalized size of antiderivative = 3.64 \[ \int (d+e x) \left (a+b \arctan \left (c x^2\right )\right )^2 \, dx=\text {Result too large to show} \]

input

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

output

a^2*d*x + (a^2*e*x^2)/2 + (a*b*d*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]*S 
qrt[c*x^2])/(1 + c*x^2)])))/(c*x) + (a*b*e*(c*x^2*ArcTan[c*x^2] + Log[1/Sq 
rt[1 + c^2*x^4]]))/c + (b^2*e*((-I)*ArcTan[c*x^2]^2 + c*x^2*ArcTan[c*x^2]^ 
2 + 2*ArcTan[c*x^2]*Log[1 + E^((2*I)*ArcTan[c*x^2])] - I*PolyLog[2, -E^((2 
*I)*ArcTan[c*x^2])]))/(2*c) + (b^2*d*Sqrt[c*x^2]*(2*Sqrt[c*x^2]*ArcTan[c*x 
^2]^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]*Sqr 
t[c*x^2])^2)^(3/2)*(2*(-5*ArcTan[2 + I]*ArcTan[1 - Sqrt[2]*Sqrt[c*x^2]] + 
4*ArcTan[1 - Sqrt[2]*Sqrt[c*x^2]]^2 + ((1 + 2*I)*Sqrt[1 + I]*ArcTan[1 - Sq 
rt[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] 
*Sqrt[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])*L 
og[1 - E^((2*I)*ArcTan[1 - Sqrt[2]*Sqrt[c*x^2]] - 2*ArcTanh[1 + 2*I])] ...
 

3.1.25.3 Rubi [A] (verified)

Time = 2.31 (sec) , antiderivative size = 1325, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {5397, 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.

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

\(\Big \downarrow \) 5397

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

\(\Big \downarrow \) 2009

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

input

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

output

a^2*d*x - (2*(-1)^(3/4)*a*b*d*ArcTan[(-1)^(3/4)*Sqrt[c]*x])/Sqrt[c] + ((-1 
)^(3/4)*b^2*d*ArcTan[(-1)^(3/4)*Sqrt[c]*x]^2)/Sqrt[c] + ((I/2)*e*(a + b*Ar 
cTan[c*x^2])^2)/c + (e*x^2*(a + b*ArcTan[c*x^2])^2)/2 + (2*(-1)^(3/4)*a*b* 
d*ArcTanh[(-1)^(3/4)*Sqrt[c]*x])/Sqrt[c] - ((-1)^(1/4)*b^2*d*ArcTanh[(-1)^ 
(3/4)*Sqrt[c]*x]^2)/Sqrt[c] + (2*(-1)^(1/4)*b^2*d*ArcTan[(-1)^(3/4)*Sqrt[c 
]*x]*Log[2/(1 - (-1)^(1/4)*Sqrt[c]*x)])/Sqrt[c] - (2*(-1)^(1/4)*b^2*d*ArcT 
an[(-1)^(3/4)*Sqrt[c]*x]*Log[2/(1 + (-1)^(1/4)*Sqrt[c]*x)])/Sqrt[c] + ((-1 
)^(1/4)*b^2*d*ArcTan[(-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*d*ArcTanh 
[(-1)^(3/4)*Sqrt[c]*x]*Log[2/(1 - (-1)^(3/4)*Sqrt[c]*x)])/Sqrt[c] - (2*(-1 
)^(1/4)*b^2*d*ArcTanh[(-1)^(3/4)*Sqrt[c]*x]*Log[2/(1 + (-1)^(3/4)*Sqrt[c]* 
x)])/Sqrt[c] + ((-1)^(1/4)*b^2*d*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*d*ArcTanh[(-1)^(3/4)*Sqrt[c]*x]*Log[((1 + I)*(1 + (-1)^(1/4)*S 
qrt[c]*x))/(1 + (-1)^(3/4)*Sqrt[c]*x)])/Sqrt[c] + ((-1)^(1/4)*b^2*d*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*d*x*Log[1 - I*c*x^2] + ((-1)^(1/4)*b^2* 
d*ArcTan[(-1)^(3/4)*Sqrt[c]*x]*Log[1 - I*c*x^2])/Sqrt[c] - ((-1)^(1/4)*b^2 
*d*ArcTanh[(-1)^(3/4)*Sqrt[c]*x]*Log[1 - I*c*x^2])/Sqrt[c] - (b^2*d*x*Log[ 
1 - I*c*x^2]^2)/4 + (b*e*(a + b*ArcTan[c*x^2])*Log[2/(1 + I*c*x^2)])/c ...
 

3.1.25.3.1 Defintions of rubi rules used

rule 2009

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

rule 5397

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(m_.), 
 x_Symbol] :> Int[ExpandIntegrand[(a + b*ArcTan[c*x^n])^p, (d + e*x)^m, x], 
 x] /; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 1] && IGtQ[m, 0]
 

3.1.25.4 Maple [F]

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

input

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

output

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

3.1.25.5 Fricas [F]

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

input

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

output

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

3.1.25.6 Sympy [F]

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

input

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

output

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

3.1.25.7 Maxima [F]

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

input

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

output

12*b^2*c^2*e*integrate(1/16*x^5*arctan(c*x^2)^2/(c^2*x^4 + 1), x) + b^2*c^ 
2*e*integrate(1/16*x^5*log(c^2*x^4 + 1)^2/(c^2*x^4 + 1), x) + 12*b^2*c^2*d 
*integrate(1/16*x^4*arctan(c*x^2)^2/(c^2*x^4 + 1), x) + 4*b^2*c^2*e*integr 
ate(1/16*x^5*log(c^2*x^4 + 1)/(c^2*x^4 + 1), x) + b^2*c^2*d*integrate(1/16 
*x^4*log(c^2*x^4 + 1)^2/(c^2*x^4 + 1), x) + 8*b^2*c^2*d*integrate(1/16*x^4 
*log(c^2*x^4 + 1)/(c^2*x^4 + 1), x) + 1/2*a^2*e*x^2 + 1/8*b^2*e*arctan(c*x 
^2)^3/c - 8*b^2*c*e*integrate(1/16*x^3*arctan(c*x^2)/(c^2*x^4 + 1), x) - 1 
6*b^2*c*d*integrate(1/16*x^2*arctan(c*x^2)/(c^2*x^4 + 1), x) - 1/2*(c*(2*s 
qrt(2)*arctan(1/2*sqrt(2)*(2*c*x + sqrt(2)*sqrt(c))/sqrt(c))/c^(3/2) + 2*s 
qrt(2)*arctan(1/2*sqrt(2)*(2*c*x - sqrt(2)*sqrt(c))/sqrt(c))/c^(3/2) - sqr 
t(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*d + a^2*d*x + b^2*e*i 
ntegrate(1/16*x*log(c^2*x^4 + 1)^2/(c^2*x^4 + 1), x) + 12*b^2*d*integrate( 
1/16*arctan(c*x^2)^2/(c^2*x^4 + 1), x) + b^2*d*integrate(1/16*log(c^2*x^4 
+ 1)^2/(c^2*x^4 + 1), x) + 1/2*(2*c*x^2*arctan(c*x^2) - log(c^2*x^4 + 1))* 
a*b*e/c + 1/8*(b^2*e*x^2 + 2*b^2*d*x)*arctan(c*x^2)^2 - 1/32*(b^2*e*x^2 + 
2*b^2*d*x)*log(c^2*x^4 + 1)^2
 

3.1.25.8 Giac [F]

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

input

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

output

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

3.1.25.9 Mupad [F(-1)]

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

input

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

output

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