3.34 \(\int \frac{(a+b \tan ^{-1}(c+d x))^2}{e+f x} \, dx\)
Optimal. Leaf size=261 \[ -\frac{i b \left (a+b \tan ^{-1}(c+d x)\right ) \text{PolyLog}\left (2,1-\frac{2 d (e+f x)}{(1-i (c+d x)) (-c f+d e+i f)}\right )}{f}+\frac{i b \text{PolyLog}\left (2,1-\frac{2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{f}+\frac{b^2 \text{PolyLog}\left (3,1-\frac{2 d (e+f x)}{(1-i (c+d x)) (-c f+d e+i f)}\right )}{2 f}-\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{1-i (c+d x)}\right )}{2 f}+\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (\frac{2 d (e+f x)}{(1-i (c+d x)) (-c f+d e+i f)}\right )}{f}-\frac{\log \left (\frac{2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{f} \]
[Out]
-(((a + b*ArcTan[c + d*x])^2*Log[2/(1 - I*(c + d*x))])/f) + ((a + b*ArcTan[c + d*x])^2*Log[(2*d*(e + f*x))/((d
*e + I*f - c*f)*(1 - I*(c + d*x)))])/f + (I*b*(a + b*ArcTan[c + d*x])*PolyLog[2, 1 - 2/(1 - I*(c + d*x))])/f -
(I*b*(a + b*ArcTan[c + d*x])*PolyLog[2, 1 - (2*d*(e + f*x))/((d*e + I*f - c*f)*(1 - I*(c + d*x)))])/f - (b^2*
PolyLog[3, 1 - 2/(1 - I*(c + d*x))])/(2*f) + (b^2*PolyLog[3, 1 - (2*d*(e + f*x))/((d*e + I*f - c*f)*(1 - I*(c
+ d*x)))])/(2*f)
________________________________________________________________________________________
Rubi [A] time = 0.164644, antiderivative size = 261, normalized size of antiderivative = 1.,
number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used =
{5047, 4858} \[ -\frac{i b \left (a+b \tan ^{-1}(c+d x)\right ) \text{PolyLog}\left (2,1-\frac{2 d (e+f x)}{(1-i (c+d x)) (-c f+d e+i f)}\right )}{f}+\frac{i b \text{PolyLog}\left (2,1-\frac{2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{f}+\frac{b^2 \text{PolyLog}\left (3,1-\frac{2 d (e+f x)}{(1-i (c+d x)) (-c f+d e+i f)}\right )}{2 f}-\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{1-i (c+d x)}\right )}{2 f}+\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (\frac{2 d (e+f x)}{(1-i (c+d x)) (-c f+d e+i f)}\right )}{f}-\frac{\log \left (\frac{2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{f} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*ArcTan[c + d*x])^2/(e + f*x),x]
[Out]
-(((a + b*ArcTan[c + d*x])^2*Log[2/(1 - I*(c + d*x))])/f) + ((a + b*ArcTan[c + d*x])^2*Log[(2*d*(e + f*x))/((d
*e + I*f - c*f)*(1 - I*(c + d*x)))])/f + (I*b*(a + b*ArcTan[c + d*x])*PolyLog[2, 1 - 2/(1 - I*(c + d*x))])/f -
(I*b*(a + b*ArcTan[c + d*x])*PolyLog[2, 1 - (2*d*(e + f*x))/((d*e + I*f - c*f)*(1 - I*(c + d*x)))])/f - (b^2*
PolyLog[3, 1 - 2/(1 - I*(c + d*x))])/(2*f) + (b^2*PolyLog[3, 1 - (2*d*(e + f*x))/((d*e + I*f - c*f)*(1 - I*(c
+ d*x)))])/(2*f)
Rule 5047
Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcTan[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x]
&& IGtQ[p, 0]
Rule 4858
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^2/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^2*Log[2/
(1 - I*c*x)])/e, x] + (Simp[((a + b*ArcTan[c*x])^2*Log[(2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/e, x] + Sim
p[(I*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 - I*c*x)])/e, x] - Simp[(I*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 -
(2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/e, x] - Simp[(b^2*PolyLog[3, 1 - 2/(1 - I*c*x)])/(2*e), x] + Simp
[(b^2*PolyLog[3, 1 - (2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/(2*e), x]) /; FreeQ[{a, b, c, d, e}, x] && Ne
Q[c^2*d^2 + e^2, 0]
Rubi steps
\begin{align*} \int \frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{e+f x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(x)\right )^2}{\frac{d e-c f}{d}+\frac{f x}{d}} \, dx,x,c+d x\right )}{d}\\ &=-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (\frac{2}{1-i (c+d x)}\right )}{f}+\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{f}+\frac{i b \left (a+b \tan ^{-1}(c+d x)\right ) \text{Li}_2\left (1-\frac{2}{1-i (c+d x)}\right )}{f}-\frac{i b \left (a+b \tan ^{-1}(c+d x)\right ) \text{Li}_2\left (1-\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{f}-\frac{b^2 \text{Li}_3\left (1-\frac{2}{1-i (c+d x)}\right )}{2 f}+\frac{b^2 \text{Li}_3\left (1-\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{2 f}\\ \end{align*}
Mathematica [F] time = 5.44485, size = 0, normalized size = 0. \[ \int \frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{e+f x} \, dx \]
Verification is Not applicable to the result.
[In]
Integrate[(a + b*ArcTan[c + d*x])^2/(e + f*x),x]
[Out]
Integrate[(a + b*ArcTan[c + d*x])^2/(e + f*x), x]
________________________________________________________________________________________
Maple [C] time = 1.339, size = 2149, normalized size = 8.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*arctan(d*x+c))^2/(f*x+e),x)
[Out]
-1/2*d*b^2/f*e/(I*f+c*f-d*e)*polylog(3,(I*f+c*f-d*e)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))+I*b^2/f*arct
an(d*x+c)^2*csgn(I*(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+c*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-d*e*(1+I*(d*x+c))^2/(1
+(d*x+c)^2)-I*f+c*f-d*e)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))^2*Pi+I*a*b*ln(f*(d*x+c)-c*f+d*e)/f*ln((I*f-f*(d*x+
c))/(d*e+I*f-c*f))-1/2*I*b^2/f*arctan(d*x+c)^2*csgn(I*(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+c*f*(1+I*(d*x+c))^2/(
1+(d*x+c)^2)-d*e*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-I*f+c*f-d*e)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))^3*Pi-I*b^2*c/(I
*f+c*f-d*e)*arctan(d*x+c)*polylog(2,(I*f+c*f-d*e)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))-I*a*b*ln(f*(d*x
+c)-c*f+d*e)/f*ln((I*f+f*(d*x+c))/(I*f+c*f-d*e))-1/2*b^2/f*polylog(3,-(1+I*(d*x+c))^2/(1+(d*x+c)^2))+a^2*ln(f*
(d*x+c)-c*f+d*e)/f-b^2/f*arctan(d*x+c)^2*ln(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+c*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2
)-d*e*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-I*f+c*f-d*e)+b^2/(I*f+c*f-d*e)*arctan(d*x+c)*polylog(2,(I*f+c*f-d*e)*(1+I*
(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))+b^2*ln(f*(d*x+c)-c*f+d*e)/f*arctan(d*x+c)^2+1/2*b^2*c/(I*f+c*f-d*e)*po
lylog(3,(I*f+c*f-d*e)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))+1/2*I*b^2/(I*f+c*f-d*e)*polylog(3,(I*f+c*f-
d*e)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))+2*I*d*b^2/f*e*arctan(d*x+c)*polylog(2,(I*f+c*f-d*e)*(1+I*(d*
x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))/(2*I*f+2*c*f-2*d*e)+1/2*I*b^2/f*arctan(d*x+c)^2*csgn(I/((1+I*(d*x+c))^2/(
1+(d*x+c)^2)+1))*csgn(I*(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+c*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-d*e*(1+I*(d*x+c))
^2/(1+(d*x+c)^2)-I*f+c*f-d*e))*csgn(I*(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+c*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-d*e
*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-I*f+c*f-d*e)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))*Pi-d*b^2/f*e/(I*f+c*f-d*e)*arct
an(d*x+c)^2*ln(1-(I*f+c*f-d*e)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))-1/2*I*b^2/f*arctan(d*x+c)^2*csgn(I
*(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+c*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-d*e*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-I*f+c*
f-d*e))*csgn(I*(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+c*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-d*e*(1+I*(d*x+c))^2/(1+(d*
x+c)^2)-I*f+c*f-d*e)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))^2*Pi-1/2*I*b^2/f*arctan(d*x+c)^2*csgn(I/((1+I*(d*x+c))
^2/(1+(d*x+c)^2)+1))*csgn(I*(I*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)+c*f*(1+I*(d*x+c))^2/(1+(d*x+c)^2)-d*e*(1+I*(d*x
+c))^2/(1+(d*x+c)^2)-I*f+c*f-d*e)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))^2*Pi+b^2*c/(I*f+c*f-d*e)*arctan(d*x+c)^2*
ln(1-(I*f+c*f-d*e)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))+I*a*b/f*dilog((I*f-f*(d*x+c))/(d*e+I*f-c*f))-I
*b^2/f*Pi*arctan(d*x+c)^2-I*a*b/f*dilog((I*f+f*(d*x+c))/(I*f+c*f-d*e))+2*a*b*ln(f*(d*x+c)-c*f+d*e)/f*arctan(d*
x+c)+I*b^2/f*arctan(d*x+c)*polylog(2,-(1+I*(d*x+c))^2/(1+(d*x+c)^2))+I*b^2/(I*f+c*f-d*e)*arctan(d*x+c)^2*ln(1-
(I*f+c*f-d*e)*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(d*e+I*f-c*f))
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{a^{2} \log \left (f x + e\right )}{f} + \int \frac{12 \, b^{2} \arctan \left (d x + c\right )^{2} + b^{2} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )^{2} + 32 \, a b \arctan \left (d x + c\right )}{16 \,{\left (f x + e\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctan(d*x+c))^2/(f*x+e),x, algorithm="maxima")
[Out]
a^2*log(f*x + e)/f + integrate(1/16*(12*b^2*arctan(d*x + c)^2 + b^2*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2 + 32*a*
b*arctan(d*x + c))/(f*x + e), x)
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \arctan \left (d x + c\right )^{2} + 2 \, a b \arctan \left (d x + c\right ) + a^{2}}{f x + e}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctan(d*x+c))^2/(f*x+e),x, algorithm="fricas")
[Out]
integral((b^2*arctan(d*x + c)^2 + 2*a*b*arctan(d*x + c) + a^2)/(f*x + e), x)
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{atan}{\left (c + d x \right )}\right )^{2}}{e + f x}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*atan(d*x+c))**2/(f*x+e),x)
[Out]
Integral((a + b*atan(c + d*x))**2/(e + f*x), x)
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (d x + c\right ) + a\right )}^{2}}{f x + e}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctan(d*x+c))^2/(f*x+e),x, algorithm="giac")
[Out]
integrate((b*arctan(d*x + c) + a)^2/(f*x + e), x)