3.10 \(\int \frac{(a+b \tan ^{-1}(c+d x))^2}{c e+d e x} \, dx\)

Optimal. Leaf size=183 \[ -\frac{i b \text{PolyLog}\left (2,1-\frac{2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d e}+\frac{i b \text{PolyLog}\left (2,-1+\frac{2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d e}-\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{1+i (c+d x)}\right )}{2 d e}+\frac{b^2 \text{PolyLog}\left (3,-1+\frac{2}{1+i (c+d x)}\right )}{2 d e}+\frac{2 \tanh ^{-1}\left (1-\frac{2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{d e} \]

[Out]

(2*(a + b*ArcTan[c + d*x])^2*ArcTanh[1 - 2/(1 + I*(c + d*x))])/(d*e) - (I*b*(a + b*ArcTan[c + d*x])*PolyLog[2,
1 - 2/(1 + I*(c + d*x))])/(d*e) + (I*b*(a + b*ArcTan[c + d*x])*PolyLog[2, -1 + 2/(1 + I*(c + d*x))])/(d*e) -
(b^2*PolyLog[3, 1 - 2/(1 + I*(c + d*x))])/(2*d*e) + (b^2*PolyLog[3, -1 + 2/(1 + I*(c + d*x))])/(2*d*e)

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Rubi [A]  time = 0.338879, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {5043, 12, 4850, 4988, 4884, 4994, 6610} \[ -\frac{i b \text{PolyLog}\left (2,1-\frac{2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d e}+\frac{i b \text{PolyLog}\left (2,-1+\frac{2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d e}-\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{1+i (c+d x)}\right )}{2 d e}+\frac{b^2 \text{PolyLog}\left (3,-1+\frac{2}{1+i (c+d x)}\right )}{2 d e}+\frac{2 \tanh ^{-1}\left (1-\frac{2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*ArcTan[c + d*x])^2*ArcTanh[1 - 2/(1 + I*(c + d*x))])/(d*e) - (I*b*(a + b*ArcTan[c + d*x])*PolyLog[2,
1 - 2/(1 + I*(c + d*x))])/(d*e) + (I*b*(a + b*ArcTan[c + d*x])*PolyLog[2, -1 + 2/(1 + I*(c + d*x))])/(d*e) -
(b^2*PolyLog[3, 1 - 2/(1 + I*(c + d*x))])/(2*d*e) + (b^2*PolyLog[3, -1 + 2/(1 + I*(c + d*x))])/(2*d*e)

Rule 5043

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((f*x)/d)^m*(a + b*ArcTan[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f, 0
] && IGtQ[p, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 4850

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)/(x_), x_Symbol] :> Simp[2*(a + b*ArcTan[c*x])^p*ArcTanh[1 - 2/(1 +
I*c*x)], x] - Dist[2*b*c*p, Int[((a + b*ArcTan[c*x])^(p - 1)*ArcTanh[1 - 2/(1 + I*c*x)])/(1 + c^2*x^2), x], x
] /; FreeQ[{a, b, c}, x] && IGtQ[p, 1]

Rule 4988

Int[(ArcTanh[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/2, Int[(
Log[1 + u]*(a + b*ArcTan[c*x])^p)/(d + e*x^2), x], x] - Dist[1/2, Int[(Log[1 - u]*(a + b*ArcTan[c*x])^p)/(d +
e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[u^2 - (1 - (2*I)/(I - c*x))^
2, 0]

Rule 4884

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

Rule 4994

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a + b*Arc
Tan[c*x])^p*PolyLog[2, 1 - u])/(2*c*d), x] + Dist[(b*p*I)/2, Int[((a + b*ArcTan[c*x])^(p - 1)*PolyLog[2, 1 - u
])/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[(1 - u)^2 - (1 - (2*
I)/(I - c*x))^2, 0]

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin{align*} \int \frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{c e+d e x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(x)\right )^2}{e x} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(x)\right )^2}{x} \, dx,x,c+d x\right )}{d e}\\ &=\frac{2 \left (a+b \tan ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i (c+d x)}\right )}{d e}-\frac{(4 b) \operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(x)\right ) \tanh ^{-1}\left (1-\frac{2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e}\\ &=\frac{2 \left (a+b \tan ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i (c+d x)}\right )}{d e}-\frac{(2 b) \operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(x)\right ) \log \left (2-\frac{2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e}+\frac{(2 b) \operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(x)\right ) \log \left (\frac{2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e}\\ &=\frac{2 \left (a+b \tan ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i (c+d x)}\right )}{d e}-\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 )}{d e}+\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 )}{d e}+\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (1-\frac{2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e}-\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-1+\frac{2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e}\\ &=\frac{2 \left (a+b \tan ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i (c+d x)}\right )}{d e}-\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 )}{d e}+\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 )}{d e}-\frac{b^2 \text{Li}_3\left (1-\frac{2}{1+i (c+d x)}\right )}{2 d e}+\frac{b^2 \text{Li}_3\left (-1+\frac{2}{1+i (c+d x)}\right )}{2 d e}\\ \end{align*}

Mathematica [A]  time = 0.0626482, size = 170, normalized size = 0.93 \[ \frac{2 i b \text{PolyLog}\left (2,-\frac{c+d x+i}{c+d x-i}\right ) \left (a+b \tan ^{-1}(c+d x)\right )-2 i b \text{PolyLog}\left (2,\frac{c+d x+i}{c+d x-i}\right ) \left (a+b \tan ^{-1}(c+d x)\right )+b^2 \text{PolyLog}\left (3,-\frac{c+d x+i}{c+d x-i}\right )-b^2 \text{PolyLog}\left (3,\frac{c+d x+i}{c+d x-i}\right )+4 \tanh ^{-1}\left (\frac{c+d x+i}{c+d x-i}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(a + b*ArcTan[c + d*x])^2*ArcTanh[(I + c + d*x)/(-I + c + d*x)] + (2*I)*b*(a + b*ArcTan[c + d*x])*PolyLog[2
, -((I + c + d*x)/(-I + c + d*x))] - (2*I)*b*(a + b*ArcTan[c + d*x])*PolyLog[2, (I + c + d*x)/(-I + c + d*x)]
+ b^2*PolyLog[3, -((I + c + d*x)/(-I + c + d*x))] - b^2*PolyLog[3, (I + c + d*x)/(-I + c + d*x)])/(2*d*e)

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Maple [C]  time = 0.588, size = 1433, normalized size = 7.8 \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/(d*e*x+c*e),x)

[Out]

-1/2*I/d*b^2/e*Pi*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1))*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/((1+I*(d*
x+c))^2/(1+(d*x+c)^2)+1))^2*arctan(d*x+c)^2-1/2*I/d*b^2/e*Pi*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/((1+I*(d
*x+c))^2/(1+(d*x+c)^2)+1))*csgn(((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))^2*arctan(
d*x+c)^2+1/2*I/d*b^2/e*Pi*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))*csgn(((1
+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))*arctan(d*x+c)^2-1/2*I/d*b^2/e*Pi*csgn(I/((1+
I*(d*x+c))^2/(1+(d*x+c)^2)+1))*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))^2*a
rctan(d*x+c)^2+1/2*I/d*b^2/e*Pi*csgn(((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))^3*ar
ctan(d*x+c)^2+2/d*b^2/e*polylog(3,-(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))+2/d*b^2/e*polylog(3,(1+I*(d*x+c))/(1+(d*
x+c)^2)^(1/2))-1/2/d*b^2/e*polylog(3,-(1+I*(d*x+c))^2/(1+(d*x+c)^2))+I/d*b^2/e*arctan(d*x+c)*polylog(2,-(1+I*(
d*x+c))^2/(1+(d*x+c)^2))+2/d*a*b/e*ln(d*x+c)*arctan(d*x+c)+1/2*I/d*b^2/e*Pi*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)
^2)-1))*csgn(I/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/((1+I*(d*x+c))^2/(1
+(d*x+c)^2)+1))*arctan(d*x+c)^2-I/d*a*b/e*dilog(1-I*(d*x+c))+1/2*I/d*b^2/e*Pi*arctan(d*x+c)^2-2*I/d*b^2/e*arct
an(d*x+c)*polylog(2,(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))-2*I/d*b^2/e*arctan(d*x+c)*polylog(2,-(1+I*(d*x+c))/(1+(
d*x+c)^2)^(1/2))+1/d*a^2/e*ln(d*x+c)+I/d*a*b/e*dilog(1+I*(d*x+c))+1/d*b^2/e*ln(d*x+c)*arctan(d*x+c)^2-1/d*b^2/
e*arctan(d*x+c)^2*ln((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)+1/d*b^2/e*arctan(d*x+c)^2*ln(1-(1+I*(d*x+c))/(1+(d*x+c)^
2)^(1/2))+1/d*b^2/e*arctan(d*x+c)^2*ln(1+(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))+I/d*a*b/e*ln(d*x+c)*ln(1+I*(d*x+c)
)-1/2*I/d*b^2/e*Pi*csgn(((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))^2*arctan(d*x+c)^2
-I/d*a*b/e*ln(d*x+c)*ln(1-I*(d*x+c))+1/2*I/d*b^2/e*Pi*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/((1+I*(d*x+c))^
2/(1+(d*x+c)^2)+1))^3*arctan(d*x+c)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{a^{2} \log \left (d e x + c e\right )}{d e} + \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 (d e x + c 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/(d*e*x+c*e),x, algorithm="maxima")

[Out]

a^2*log(d*e*x + c*e)/(d*e) + 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))/(d*e*x + c*e), x)

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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}}{d e x + c e}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{2}}{c + d x}\, dx + \int \frac{b^{2} \operatorname{atan}^{2}{\left (c + d x \right )}}{c + d x}\, dx + \int \frac{2 a b \operatorname{atan}{\left (c + d x \right )}}{c + d x}\, dx}{e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (d x + c\right ) + a\right )}^{2}}{d e x + c e}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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