3.17 \(\int \frac {(a+b \tan ^{-1}(c+d x))^3}{c e+d e x} \, dx\)

Optimal. Leaf size=279 \[ -\frac {3 b^2 \text {Li}_3\left (1-\frac {2}{i (c+d x)+1}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{2 d e}+\frac {3 b^2 \text {Li}_3\left (\frac {2}{i (c+d x)+1}-1\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{2 d e}-\frac {3 i b \text {Li}_2\left (1-\frac {2}{i (c+d x)+1}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e}+\frac {3 i b \text {Li}_2\left (\frac {2}{i (c+d x)+1}-1\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e}+\frac {2 \tanh ^{-1}\left (1-\frac {2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^3}{d e}+\frac {3 i b^3 \text {Li}_4\left (1-\frac {2}{i (c+d x)+1}\right )}{4 d e}-\frac {3 i b^3 \text {Li}_4\left (\frac {2}{i (c+d x)+1}-1\right )}{4 d e} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.46, antiderivative size = 279, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {5043, 12, 4850, 4988, 4884, 4994, 4998, 6610} \[ -\frac {3 b^2 \text {PolyLog}\left (3,1-\frac {2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{2 d e}+\frac {3 b^2 \text {PolyLog}\left (3,-1+\frac {2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{2 d e}-\frac {3 i b \text {PolyLog}\left (2,1-\frac {2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e}+\frac {3 i b \text {PolyLog}\left (2,-1+\frac {2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e}+\frac {3 i b^3 \text {PolyLog}\left (4,1-\frac {2}{1+i (c+d x)}\right )}{4 d e}-\frac {3 i b^3 \text {PolyLog}\left (4,-1+\frac {2}{1+i (c+d x)}\right )}{4 d e}+\frac {2 \tanh ^{-1}\left (1-\frac {2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^3}{d e} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c + d*x])^3/(c*e + d*e*x),x]

[Out]

(2*(a + b*ArcTan[c + d*x])^3*ArcTanh[1 - 2/(1 + I*(c + d*x))])/(d*e) - (((3*I)/2)*b*(a + b*ArcTan[c + d*x])^2*
PolyLog[2, 1 - 2/(1 + I*(c + d*x))])/(d*e) + (((3*I)/2)*b*(a + b*ArcTan[c + d*x])^2*PolyLog[2, -1 + 2/(1 + I*(
c + d*x))])/(d*e) - (3*b^2*(a + b*ArcTan[c + d*x])*PolyLog[3, 1 - 2/(1 + I*(c + d*x))])/(2*d*e) + (3*b^2*(a +
b*ArcTan[c + d*x])*PolyLog[3, -1 + 2/(1 + I*(c + d*x))])/(2*d*e) + (((3*I)/4)*b^3*PolyLog[4, 1 - 2/(1 + I*(c +
d*x))])/(d*e) - (((3*I)/4)*b^3*PolyLog[4, -1 + 2/(1 + I*(c + d*x))])/(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 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 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 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 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 4998

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

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 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 )^3}{c e+d e x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^3}{e x} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^3}{x} \, dx,x,c+d x\right )}{d e}\\ &=\frac {2 \left (a+b \tan ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (1-\frac {2}{1+i (c+d x)}\right )}{d e}-\frac {(6 b) \operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2 \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 )^3 \tanh ^{-1}\left (1-\frac {2}{1+i (c+d x)}\right )}{d e}-\frac {(3 b) \operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2 \log \left (2-\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2 \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 )^3 \tanh ^{-1}\left (1-\frac {2}{1+i (c+d x)}\right )}{d e}-\frac {3 i b \left (a+b \tan ^{-1}(c+d x)\right )^2 \text {Li}_2\left (1-\frac {2}{1+i (c+d x)}\right )}{2 d e}+\frac {3 i b \left (a+b \tan ^{-1}(c+d x)\right )^2 \text {Li}_2\left (-1+\frac {2}{1+i (c+d x)}\right )}{2 d e}+\frac {\left (3 i b^2\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right ) \text {Li}_2\left (1-\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e}-\frac {\left (3 i b^2\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right ) \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 )^3 \tanh ^{-1}\left (1-\frac {2}{1+i (c+d x)}\right )}{d e}-\frac {3 i b \left (a+b \tan ^{-1}(c+d x)\right )^2 \text {Li}_2\left (1-\frac {2}{1+i (c+d x)}\right )}{2 d e}+\frac {3 i b \left (a+b \tan ^{-1}(c+d x)\right )^2 \text {Li}_2\left (-1+\frac {2}{1+i (c+d x)}\right )}{2 d e}-\frac {3 b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_3\left (1-\frac {2}{1+i (c+d x)}\right )}{2 d e}+\frac {3 b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_3\left (-1+\frac {2}{1+i (c+d x)}\right )}{2 d e}+\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (1-\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{2 d e}-\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (-1+\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{2 d e}\\ &=\frac {2 \left (a+b \tan ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (1-\frac {2}{1+i (c+d x)}\right )}{d e}-\frac {3 i b \left (a+b \tan ^{-1}(c+d x)\right )^2 \text {Li}_2\left (1-\frac {2}{1+i (c+d x)}\right )}{2 d e}+\frac {3 i b \left (a+b \tan ^{-1}(c+d x)\right )^2 \text {Li}_2\left (-1+\frac {2}{1+i (c+d x)}\right )}{2 d e}-\frac {3 b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_3\left (1-\frac {2}{1+i (c+d x)}\right )}{2 d e}+\frac {3 b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_3\left (-1+\frac {2}{1+i (c+d x)}\right )}{2 d e}+\frac {3 i b^3 \text {Li}_4\left (1-\frac {2}{1+i (c+d x)}\right )}{4 d e}-\frac {3 i b^3 \text {Li}_4\left (-1+\frac {2}{1+i (c+d x)}\right )}{4 d e}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.15, size = 252, normalized size = 0.90 \[ \frac {6 b^2 \text {Li}_3\left (-\frac {c+d x+i}{c+d x-i}\right ) \left (a+b \tan ^{-1}(c+d x)\right )-6 b^2 \text {Li}_3\left (\frac {c+d x+i}{c+d x-i}\right ) \left (a+b \tan ^{-1}(c+d x)\right )+6 i b \text {Li}_2\left (-\frac {c+d x+i}{c+d x-i}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2-6 i b \text {Li}_2\left (\frac {c+d x+i}{c+d x-i}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2+8 \tanh ^{-1}\left (\frac {c+d x+i}{c+d x-i}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^3-3 i b^3 \text {Li}_4\left (-\frac {c+d x+i}{c+d x-i}\right )+3 i b^3 \text {Li}_4\left (\frac {c+d x+i}{c+d x-i}\right )}{4 d e} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c + d*x])^3/(c*e + d*e*x),x]

[Out]

(8*(a + b*ArcTan[c + d*x])^3*ArcTanh[(I + c + d*x)/(-I + c + d*x)] + (6*I)*b*(a + b*ArcTan[c + d*x])^2*PolyLog
[2, -((I + c + d*x)/(-I + c + d*x))] - (6*I)*b*(a + b*ArcTan[c + d*x])^2*PolyLog[2, (I + c + d*x)/(-I + c + d*
x)] + 6*b^2*(a + b*ArcTan[c + d*x])*PolyLog[3, -((I + c + d*x)/(-I + c + d*x))] - 6*b^2*(a + b*ArcTan[c + d*x]
)*PolyLog[3, (I + c + d*x)/(-I + c + d*x)] - (3*I)*b^3*PolyLog[4, -((I + c + d*x)/(-I + c + d*x))] + (3*I)*b^3
*PolyLog[4, (I + c + d*x)/(-I + c + d*x)])/(4*d*e)

________________________________________________________________________________________

fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \arctan \left (d x + c\right )^{3} + 3 \, a b^{2} \arctan \left (d x + c\right )^{2} + 3 \, a^{2} b \arctan \left (d x + c\right ) + a^{3}}{d e x + c e}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(d*x+c))^3/(d*e*x+c*e),x, algorithm="fricas")

[Out]

integral((b^3*arctan(d*x + c)^3 + 3*a*b^2*arctan(d*x + c)^2 + 3*a^2*b*arctan(d*x + c) + a^3)/(d*e*x + c*e), x)

________________________________________________________________________________________

giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(d*x+c))^3/(d*e*x+c*e),x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

maple [C]  time = 0.28, size = 2894, normalized size = 10.37 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(d*x+c))^3/(d*e*x+c*e),x)

[Out]

1/d*a^3/e*ln(d*x+c)+1/2*I/d*b^3/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)^3-3/2*I/d
*a*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+3/2*I/d*a*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)^2-3/2*I/d*a*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)^2-3/2*I/d*a*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)^2-3/2*I/d*a^2*b/e*dilog(1-I*(d*x+c))+3/2*I/d*a^2*b/e*dilog(1+I*(d*x+c))+1/2*I/d*b^3/e*Pi*arctan(d
*x+c)^3+3/d*a*b^2/e*ln(d*x+c)*arctan(d*x+c)^2-3/d*a*b^2/e*arctan(d*x+c)^2*ln((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)+
3/d*a*b^2/e*arctan(d*x+c)^2*ln(1+(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))+3/d*a*b^2/e*arctan(d*x+c)^2*ln(1-(1+I*(d*x
+c))/(1+(d*x+c)^2)^(1/2))-3*I/d*b^3/e*arctan(d*x+c)^2*polylog(2,(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))+3/2*I/d*b^3
/e*arctan(d*x+c)^2*polylog(2,-(1+I*(d*x+c))^2/(1+(d*x+c)^2))-3*I/d*b^3/e*arctan(d*x+c)^2*polylog(2,-(1+I*(d*x+
c))/(1+(d*x+c)^2)^(1/2))+3/d*a^2*b/e*ln(d*x+c)*arctan(d*x+c)-6*I/d*a*b^2/e*arctan(d*x+c)*polylog(2,(1+I*(d*x+c
))/(1+(d*x+c)^2)^(1/2))+3*I/d*a*b^2/e*arctan(d*x+c)*polylog(2,-(1+I*(d*x+c))^2/(1+(d*x+c)^2))-1/2*I/d*b^3/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)^3+1/2*I/d*b^3/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)^3+1/2*I/d*b^3/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*arctan(d*x+c)^3-3/2*I/d*a^2*b/e*l
n(d*x+c)*ln(1-I*(d*x+c))-6*I/d*a*b^2/e*arctan(d*x+c)*polylog(2,-(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))+3/2*I/d*a^2
*b/e*ln(d*x+c)*ln(1+I*(d*x+c))+3/2*I/d*a*b^2/e*Pi*arctan(d*x+c)^2-1/2*I/d*b^3/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)^3-3/2*I/d*a*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)^2-1/2*I/d*b^3/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)^3-1/2*I/d*b^3/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)^3+3/2*I/d*a*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*arcta
n(d*x+c)^2+1/2*I/d*b^3/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)^3+6*I/d*b^3/e*polylog(4,(1+I
*(d*x+c))/(1+(d*x+c)^2)^(1/2))+6*I/d*b^3/e*polylog(4,-(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))-3/4*I/d*b^3/e*polylog
(4,-(1+I*(d*x+c))^2/(1+(d*x+c)^2))-3/2/d*b^3/e*arctan(d*x+c)*polylog(3,-(1+I*(d*x+c))^2/(1+(d*x+c)^2))+1/d*b^3
/e*ln(d*x+c)*arctan(d*x+c)^3-1/d*b^3/e*arctan(d*x+c)^3*ln((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)+1/d*b^3/e*arctan(d*
x+c)^3*ln(1+(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))+3/2*I/d*a*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*arctan(d*x+c)^2+6/d*b^3/e*arctan(d*x+c)*polylog(3,-(1+I*(d*x+c))/(1+(d*x+c)^2)
^(1/2))+1/d*b^3/e*arctan(d*x+c)^3*ln(1-(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))+6/d*b^3/e*arctan(d*x+c)*polylog(3,(1
+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))+6/d*a*b^2/e*polylog(3,-(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))+6/d*a*b^2/e*polylog
(3,(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))-3/2/d*a*b^2/e*polylog(3,-(1+I*(d*x+c))^2/(1+(d*x+c)^2))+3/2*I/d*a*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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {a^{3} \log \left (d e x + c e\right )}{d e} + \int \frac {28 \, b^{3} \arctan \left (d x + c\right )^{3} + 3 \, b^{3} \arctan \left (d x + c\right ) \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )^{2} + 96 \, a b^{2} \arctan \left (d x + c\right )^{2} + 96 \, a^{2} b \arctan \left (d x + c\right )}{32 \, {\left (d e x + c e\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(d*x+c))^3/(d*e*x+c*e),x, algorithm="maxima")

[Out]

a^3*log(d*e*x + c*e)/(d*e) + integrate(1/32*(28*b^3*arctan(d*x + c)^3 + 3*b^3*arctan(d*x + c)*log(d^2*x^2 + 2*
c*d*x + c^2 + 1)^2 + 96*a*b^2*arctan(d*x + c)^2 + 96*a^2*b*arctan(d*x + c))/(d*e*x + c*e), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {atan}\left (c+d\,x\right )\right )}^3}{c\,e+d\,e\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c + d*x))^3/(c*e + d*e*x),x)

[Out]

int((a + b*atan(c + d*x))^3/(c*e + d*e*x), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a^{3}}{c + d x}\, dx + \int \frac {b^{3} \operatorname {atan}^{3}{\left (c + d x \right )}}{c + d x}\, dx + \int \frac {3 a b^{2} \operatorname {atan}^{2}{\left (c + d x \right )}}{c + d x}\, dx + \int \frac {3 a^{2} b \operatorname {atan}{\left (c + d x \right )}}{c + d x}\, dx}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(d*x+c))**3/(d*e*x+c*e),x)

[Out]

(Integral(a**3/(c + d*x), x) + Integral(b**3*atan(c + d*x)**3/(c + d*x), x) + Integral(3*a*b**2*atan(c + d*x)*
*2/(c + d*x), x) + Integral(3*a**2*b*atan(c + d*x)/(c + d*x), x))/e

________________________________________________________________________________________