\(\int \frac {(a+b \arctan (c+d x))^2}{c e+d e x} \, dx\) [10]

Optimal result

Integrand size = 23, antiderivative size = 183 \[ \int \frac {(a+b \arctan (c+d x))^2}{c e+d e x} \, dx=\frac {2 (a+b \arctan (c+d x))^2 \text {arctanh}\left (1-\frac {2}{1+i (c+d x)}\right )}{d e}-\frac {i b (a+b \arctan (c+d x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i (c+d x)}\right )}{d e}+\frac {i b (a+b \arctan (c+d x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i (c+d x)}\right )}{d e}-\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i (c+d x)}\right )}{2 d e}+\frac {b^2 \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i (c+d x)}\right )}{2 d e} \]

[Out]

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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {5151, 12, 4942, 5108, 5004, 5114, 6745} \[ \int \frac {(a+b \arctan (c+d x))^2}{c e+d e x} \, dx=\frac {2 \text {arctanh}\left (1-\frac {2}{1+i (c+d x)}\right ) (a+b \arctan (c+d x))^2}{d e}-\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2}{i (c+d x)+1}\right ) (a+b \arctan (c+d x))}{d e}+\frac {i b \operatorname {PolyLog}\left (2,\frac {2}{i (c+d x)+1}-1\right ) (a+b \arctan (c+d x))}{d e}-\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{i (c+d x)+1}\right )}{2 d e}+\frac {b^2 \operatorname {PolyLog}\left (3,\frac {2}{i (c+d x)+1}-1\right )}{2 d e} \]

[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 12

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

Rule 4942

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 5004

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 5108

Int[(ArcTanh[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/2, Int[L
og[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 5114

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*(a + b*Ar
cTan[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 5151

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 6745

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*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(a+b \arctan (x))^2}{e x} \, dx,x,c+d x\right )}{d} \\ & = \frac {\text {Subst}\left (\int \frac {(a+b \arctan (x))^2}{x} \, dx,x,c+d x\right )}{d e} \\ & = \frac {2 (a+b \arctan (c+d x))^2 \text {arctanh}\left (1-\frac {2}{1+i (c+d x)}\right )}{d e}-\frac {(4 b) \text {Subst}\left (\int \frac {(a+b \arctan (x)) \text {arctanh}\left (1-\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e} \\ & = \frac {2 (a+b \arctan (c+d x))^2 \text {arctanh}\left (1-\frac {2}{1+i (c+d x)}\right )}{d e}-\frac {(2 b) \text {Subst}\left (\int \frac {(a+b \arctan (x)) \log \left (2-\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e}+\frac {(2 b) \text {Subst}\left (\int \frac {(a+b \arctan (x)) \log \left (\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e} \\ & = \frac {2 (a+b \arctan (c+d x))^2 \text {arctanh}\left (1-\frac {2}{1+i (c+d x)}\right )}{d e}-\frac {i b (a+b \arctan (c+d x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i (c+d x)}\right )}{d e}+\frac {i b (a+b \arctan (c+d x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i (c+d x)}\right )}{d e}+\frac {\left (i b^2\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (2,1-\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e}-\frac {\left (i b^2\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (2,-1+\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e} \\ & = \frac {2 (a+b \arctan (c+d x))^2 \text {arctanh}\left (1-\frac {2}{1+i (c+d x)}\right )}{d e}-\frac {i b (a+b \arctan (c+d x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i (c+d x)}\right )}{d e}+\frac {i b (a+b \arctan (c+d x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i (c+d x)}\right )}{d e}-\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i (c+d x)}\right )}{2 d e}+\frac {b^2 \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i (c+d x)}\right )}{2 d e} \\ \end{align*}

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(381\) vs. \(2(183)=366\).

Time = 0.30 (sec) , antiderivative size = 381, normalized size of antiderivative = 2.08 \[ \int \frac {(a+b \arctan (c+d x))^2}{c e+d e x} \, dx=\frac {-6 i a b \pi ^2-i b^2 \pi ^3+24 i a b \pi \arctan (c+d x)-48 i a b \arctan (c+d x)^2+16 i b^2 \arctan (c+d x)^3-a b \pi \log (16777216)+24 b^2 \arctan (c+d x)^2 \log \left (1-e^{-2 i \arctan (c+d x)}\right )+24 a b \pi \log \left (1+e^{-2 i \arctan (c+d x)}\right )-48 a b \arctan (c+d x) \log \left (1+e^{-2 i \arctan (c+d x)}\right )+48 a b \arctan (c+d x) \log \left (1-e^{2 i \arctan (c+d x)}\right )-24 b^2 \arctan (c+d x)^2 \log \left (1+e^{2 i \arctan (c+d x)}\right )+24 a^2 \log (c+d x)+12 a b \pi \log \left (1+c^2+2 c d x+d^2 x^2\right )-24 i a b \operatorname {PolyLog}\left (2,-e^{-2 i \arctan (c+d x)}\right )+24 i b^2 \arctan (c+d x) \operatorname {PolyLog}\left (2,e^{-2 i \arctan (c+d x)}\right )+24 i b^2 \arctan (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c+d x)}\right )-24 i a b \operatorname {PolyLog}\left (2,e^{2 i \arctan (c+d x)}\right )+12 b^2 \operatorname {PolyLog}\left (3,e^{-2 i \arctan (c+d x)}\right )-12 b^2 \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c+d x)}\right )}{24 d e} \]

[In]

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

[Out]

((-6*I)*a*b*Pi^2 - I*b^2*Pi^3 + (24*I)*a*b*Pi*ArcTan[c + d*x] - (48*I)*a*b*ArcTan[c + d*x]^2 + (16*I)*b^2*ArcT
an[c + d*x]^3 - a*b*Pi*Log[16777216] + 24*b^2*ArcTan[c + d*x]^2*Log[1 - E^((-2*I)*ArcTan[c + d*x])] + 24*a*b*P
i*Log[1 + E^((-2*I)*ArcTan[c + d*x])] - 48*a*b*ArcTan[c + d*x]*Log[1 + E^((-2*I)*ArcTan[c + d*x])] + 48*a*b*Ar
cTan[c + d*x]*Log[1 - E^((2*I)*ArcTan[c + d*x])] - 24*b^2*ArcTan[c + d*x]^2*Log[1 + E^((2*I)*ArcTan[c + d*x])]
+ 24*a^2*Log[c + d*x] + 12*a*b*Pi*Log[1 + c^2 + 2*c*d*x + d^2*x^2] - (24*I)*a*b*PolyLog[2, -E^((-2*I)*ArcTan[
c + d*x])] + (24*I)*b^2*ArcTan[c + d*x]*PolyLog[2, E^((-2*I)*ArcTan[c + d*x])] + (24*I)*b^2*ArcTan[c + d*x]*Po
lyLog[2, -E^((2*I)*ArcTan[c + d*x])] - (24*I)*a*b*PolyLog[2, E^((2*I)*ArcTan[c + d*x])] + 12*b^2*PolyLog[3, E^
((-2*I)*ArcTan[c + d*x])] - 12*b^2*PolyLog[3, -E^((2*I)*ArcTan[c + d*x])])/(24*d*e)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 2.56 (sec) , antiderivative size = 1154, normalized size of antiderivative = 6.31

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1154\)
default	\(\text {Expression too large to display}\)	\(1154\)
parts	\(\text {Expression too large to display}\)	\(1159\)

[In]

int((a+b*arctan(d*x+c))^2/(d*e*x+c*e),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F]

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

[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)

Sympy [F]

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

[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

Maxima [F]

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

[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)

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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