3.1.0 ∫ (a+b arctan(c+dx))3 ______________________________

Integrand size = 23, antiderivative size = 287 \[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^4} \, dx=-\frac {b^2 (a+b \arctan (c+d x))}{d e^4 (c+d x)}-\frac {b (a+b \arctan (c+d x))^2}{2 d e^4}-\frac {b (a+b \arctan (c+d x))^2}{2 d e^4 (c+d x)^2}+\frac {i (a+b \arctan (c+d x))^3}{3 d e^4}-\frac {(a+b \arctan (c+d x))^3}{3 d e^4 (c+d x)^3}+\frac {b^3 \log (c+d x)}{d e^4}-\frac {b^3 \log \left (1+(c+d x)^2\right )}{2 d e^4}-\frac {b (a+b \arctan (c+d x))^2 \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^4}+\frac {i b^2 (a+b \arctan (c+d x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i (c+d x)}\right )}{d e^4}-\frac {b^3 \operatorname {PolyLog}\left (3,-1+\frac {2}{1-i (c+d x)}\right )}{2 d e^4} \]

-b^2*(a+b*arctan(d*x+c))/d/e^4/(d*x+c)-1/2*b*(a+b*arctan(d*x+c))^2/d/e^4-1/2*b*(a+b*arctan(d*x+c))^2/d/e^4/(d*

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

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

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

Rubi [A] (veriﬁed)

Time = 0.35 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {5151, 12, 4946, 5038, 272, 36, 29, 31, 5004, 5044, 4988, 5112, 6745} \[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^4} \, dx=\frac {i b^2 \operatorname {PolyLog}\left (2,\frac {2}{1-i (c+d x)}-1\right ) (a+b \arctan (c+d x))}{d e^4}-\frac {b^2 (a+b \arctan (c+d x))}{d e^4 (c+d x)}-\frac {b (a+b \arctan (c+d x))^2}{2 d e^4 (c+d x)^2}-\frac {b (a+b \arctan (c+d x))^2}{2 d e^4}-\frac {(a+b \arctan (c+d x))^3}{3 d e^4 (c+d x)^3}+\frac {i (a+b \arctan (c+d x))^3}{3 d e^4}-\frac {b \log \left (2-\frac {2}{1-i (c+d x)}\right ) (a+b \arctan (c+d x))^2}{d e^4}-\frac {b^3 \operatorname {PolyLog}\left (3,\frac {2}{1-i (c+d x)}-1\right )}{2 d e^4}+\frac {b^3 \log (c+d x)}{d e^4}-\frac {b^3 \log \left ((c+d x)^2+1\right )}{2 d e^4} \]

Int[(a + b*ArcTan[c + d*x])^3/(c*e + d*e*x)^4,x]

-((b^2*(a + b*ArcTan[c + d*x]))/(d*e^4*(c + d*x))) - (b*(a + b*ArcTan[c + d*x])^2)/(2*d*e^4) - (b*(a + b*ArcTa

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

3*d*e^4*(c + d*x)^3) + (b^3*Log[c + d*x])/(d*e^4) - (b^3*Log[1 + (c + d*x)^2])/(2*d*e^4) - (b*(a + b*ArcTan[c

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

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

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

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -

Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a

+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^

n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),

x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[(a + b*ArcTan[c*x])

^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Dist[b*c*(p/d), Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[2 - 2/(1 + e*(x/d))

]/(1 + c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0]

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]

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d,

Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Dist[e/(d*f^2), Int[(f*x)^(m + 2)*((a + b*ArcTan[c*x])^p/(d + e*x

^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]

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

an[c*x])^(p + 1)/(b*d*(p + 1))), x] + Dist[I/d, Int[(a + b*ArcTan[c*x])^p/(x*(I + c*x)), x], x] /; FreeQ[{a, b

, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[p, 0]

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

n[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

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

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

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

Mathematica [A] (veriﬁed)

Time = 1.22 (sec) , antiderivative size = 371, normalized size of antiderivative = 1.29 \[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^4} \, dx=\frac {-\frac {2 a^3}{(c+d x)^3}-\frac {3 a^2 b}{(c+d x)^2}-\frac {6 a^2 b \arctan (c+d x)}{(c+d x)^3}-6 a^2 b \log (c+d x)+3 a^2 b \log \left (1+c^2+2 c d x+d^2 x^2\right )+6 a b^2 \left (-\frac {(c+d x)^2+\arctan (c+d x)^2}{(c+d x)^3}+\arctan (c+d x) \left (-1-\frac {1}{(c+d x)^2}+i \arctan (c+d x)-2 \log \left (1-e^{2 i \arctan (c+d x)}\right )\right )+i \operatorname {PolyLog}\left (2,e^{2 i \arctan (c+d x)}\right )\right )+6 b^3 \left (\frac {i \pi ^3}{24}-\frac {\arctan (c+d x)}{c+d x}-\frac {1}{2} \arctan (c+d x)^2-\frac {\arctan (c+d x)^2}{2 (c+d x)^2}-\frac {1}{3} i \arctan (c+d x)^3-\frac {\arctan (c+d x)^3}{3 (c+d x)^3}-\arctan (c+d x)^2 \log \left (1-e^{-2 i \arctan (c+d x)}\right )+\log (c+d x)+\log \left (\frac {1}{\sqrt {1+(c+d x)^2}}\right )-i \arctan (c+d x) \operatorname {PolyLog}\left (2,e^{-2 i \arctan (c+d x)}\right )-\frac {1}{2} \operatorname {PolyLog}\left (3,e^{-2 i \arctan (c+d x)}\right )\right )}{6 d e^4} \]

Integrate[(a + b*ArcTan[c + d*x])^3/(c*e + d*e*x)^4,x]

((-2*a^3)/(c + d*x)^3 - (3*a^2*b)/(c + d*x)^2 - (6*a^2*b*ArcTan[c + d*x])/(c + d*x)^3 - 6*a^2*b*Log[c + d*x] +

3*a^2*b*Log[1 + c^2 + 2*c*d*x + d^2*x^2] + 6*a*b^2*(-(((c + d*x)^2 + ArcTan[c + d*x]^2)/(c + d*x)^3) + ArcTan

[c + d*x]*(-1 - (c + d*x)^(-2) + I*ArcTan[c + d*x] - 2*Log[1 - E^((2*I)*ArcTan[c + d*x])]) + I*PolyLog[2, E^((

2*I)*ArcTan[c + d*x])]) + 6*b^3*((I/24)*Pi^3 - ArcTan[c + d*x]/(c + d*x) - ArcTan[c + d*x]^2/2 - ArcTan[c + d*

x]^2/(2*(c + d*x)^2) - (I/3)*ArcTan[c + d*x]^3 - ArcTan[c + d*x]^3/(3*(c + d*x)^3) - ArcTan[c + d*x]^2*Log[1 -

E^((-2*I)*ArcTan[c + d*x])] + Log[c + d*x] + Log[1/Sqrt[1 + (c + d*x)^2]] - I*ArcTan[c + d*x]*PolyLog[2, E^((

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

Maple [C] (warning: unable to verify)

Time = 8.42 (sec) , antiderivative size = 2465, normalized size of antiderivative = 8.59


method	result	size

derivativedivides	\(\text {Expression too large to display}\)	\(2465\)
default	\(\text {Expression too large to display}\)	\(2465\)
parts	\(\text {Expression too large to display}\)	\(2473\)

int((a+b*arctan(d*x+c))^3/(d*e*x+c*e)^4,x,method=_RETURNVERBOSE)

1/d*(-1/3*a^3/e^4/(d*x+c)^3+b^3/e^4*(-1/3/(d*x+c)^3*arctan(d*x+c)^3-1/2/(d*x+c)^2*arctan(d*x+c)^2-ln(d*x+c)*ar

ctan(d*x+c)^2+1/2*arctan(d*x+c)^2*ln(1+(d*x+c)^2)-arctan(d*x+c)^2*ln((1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))+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*arct

an(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*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))+1/12*arctan(d*x+c)*(6*I*Pi*arctan(d*x+c)*csgn(I*(1+(1+I*(d*x

+c))^2/(1+(d*x+c)^2))^2)^2*csgn(I*(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))*(d*x+c)+4*I*arctan(d*x+c)^2*(d*x+c)-3*I*P

i*arctan(d*x+c)*csgn(I*(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+I*(d*x+c))^2/

(1+(d*x+c)^2))^2)^2*(d*x+c)-3*I*Pi*arctan(d*x+c)*csgn(I*(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)))^2*(d*x+c)-6*I*Pi*arctan(d*x+c)*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)))*(d*x+c)-6*I*Pi

*arctan(d*x+c)*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2))^2*csgn(I*(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))*(d*x+c)-6*I*P

i*arctan(d*x+c)*(d*x+c)+3*I*Pi*arctan(d*x+c)*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2))*csgn(I*(1+I*(d*x+c))/(1+(d*

x+c)^2)^(1/2))^2*(d*x+c)+3*I*Pi*arctan(d*x+c)*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2))^3*(d*x+c)+6*I*Pi*arctan(d*

x+c)*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*(d*x+c)+6*I*Pi*arctan(d*x+c)*

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)))*(d*x+c)+3*I*Pi*arctan(d*x+c)*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))^

2)^3*(d*x+c)-6*I*Pi*arctan(d*x+c)*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*

(d*x+c)-12*I*(d*x+c)-6*I*Pi*arctan(d*x+c)*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*(d*x+c)-3*I*Pi*arctan(d*x+c)*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))^

2)^2*csgn(I/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2))^2)*(d*x+c)+6*I*Pi*arctan(d*x+c)*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))*(d*x+c)+3*I*Pi*arctan(

d*x+c)*csgn(I*(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+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))^2)*(d*x+c)+6*I*Pi*arctan(d*x+c)*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*(d*x+c)-6*I*Pi*arctan(d*x+c)*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+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))*(d*x+c)-12*ln(2)*

arctan(d*x+c)*(d*x+c)-12-3*I*Pi*arctan(d*x+c)*csgn(I*(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2))^2)^3*(d*x+c)-6*(d*x+c)*

arctan(d*x+c))/(d*x+c)+ln((1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2)-1)+ln(1+(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2)))+3*a*b^

2/e^4*(-1/3/(d*x+c)^3*arctan(d*x+c)^2-1/3/(d*x+c)^2*arctan(d*x+c)-2/3*ln(d*x+c)*arctan(d*x+c)+1/3*arctan(d*x+c

)*ln(1+(d*x+c)^2)+1/6*I*(ln(d*x+c-I)*ln(1+(d*x+c)^2)-1/2*ln(d*x+c-I)^2-dilog(-1/2*I*(d*x+c+I))-ln(d*x+c-I)*ln(

-1/2*I*(d*x+c+I)))-1/6*I*(ln(d*x+c+I)*ln(1+(d*x+c)^2)-1/2*ln(d*x+c+I)^2-dilog(1/2*I*(d*x+c-I))-ln(d*x+c+I)*ln(

1/2*I*(d*x+c-I)))-1/3/(d*x+c)-1/3*arctan(d*x+c)-1/3*I*ln(d*x+c)*ln(1+I*(d*x+c))+1/3*I*ln(d*x+c)*ln(1-I*(d*x+c)

)-1/3*I*dilog(1+I*(d*x+c))+1/3*I*dilog(1-I*(d*x+c)))+3*a^2*b/e^4*(-1/3/(d*x+c)^3*arctan(d*x+c)-1/6/(d*x+c)^2-1

Fricas [F]

\[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^4} \, dx=\int { \frac {{\left (b \arctan \left (d x + c\right ) + a\right )}^{3}}{{\left (d e x + c e\right )}^{4}} \,d x } \]

integrate((a+b*arctan(d*x+c))^3/(d*e*x+c*e)^4,x, algorithm="fricas")

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^4*e^4*x^4 + 4*

c*d^3*e^4*x^3 + 6*c^2*d^2*e^4*x^2 + 4*c^3*d*e^4*x + c^4*e^4), x)

Sympy [F]

\[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^4} \, dx=\frac {\int \frac {a^{3}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {b^{3} \operatorname {atan}^{3}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {3 a b^{2} \operatorname {atan}^{2}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {3 a^{2} b \operatorname {atan}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx}{e^{4}} \]

integrate((a+b*atan(d*x+c))**3/(d*e*x+c*e)**4,x)

(Integral(a**3/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(b**3*atan(c +

d*x)**3/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(3*a*b**2*atan(c + d

*x)**2/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(3*a**2*b*atan(c + d*x

)/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x))/e**4

Maxima [F]

\[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^4} \, dx=\int { \frac {{\left (b \arctan \left (d x + c\right ) + a\right )}^{3}}{{\left (d e x + c e\right )}^{4}} \,d x } \]

integrate((a+b*arctan(d*x+c))^3/(d*e*x+c*e)^4,x, algorithm="maxima")

-1/2*(d*(1/(d^4*e^4*x^2 + 2*c*d^3*e^4*x + c^2*d^2*e^4) - log(d^2*x^2 + 2*c*d*x + c^2 + 1)/(d^2*e^4) + 2*log(d*

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

/3*a^3/(d^4*e^4*x^3 + 3*c*d^3*e^4*x^2 + 3*c^2*d^2*e^4*x + c^3*d*e^4) - 1/96*(4*b^3*arctan(d*x + c)^3 - 3*b^3*a

rctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2 - 96*(d^4*e^4*x^3 + 3*c*d^3*e^4*x^2 + 3*c^2*d^2*e^4*x + c^3*

d*e^4)*integrate(1/32*(28*(b^3*d^2*x^2 + 2*b^3*c*d*x + b^3*c^2 + b^3)*arctan(d*x + c)^3 + 4*(24*a*b^2*d^2*x^2

+ 24*a*b^2*c^2 + b^3*c + 24*a*b^2 + (48*a*b^2*c + b^3)*d*x)*arctan(d*x + c)^2 - 4*(b^3*d^2*x^2 + 2*b^3*c*d*x +

b^3*c^2)*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1) - (b^3*d*x + b^3*c - 3*(b^3*d^2*x^2 + 2*b^3*c*d*x +

b^3*c^2 + b^3)*arctan(d*x + c))*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2)/(d^6*e^4*x^6 + 6*c*d^5*e^4*x^5 + (15*c^2

+ 1)*d^4*e^4*x^4 + 4*(5*c^3 + c)*d^3*e^4*x^3 + 3*(5*c^4 + 2*c^2)*d^2*e^4*x^2 + 2*(3*c^5 + 2*c^3)*d*e^4*x + (c^

6 + c^4)*e^4), x))/(d^4*e^4*x^3 + 3*c*d^3*e^4*x^2 + 3*c^2*d^2*e^4*x + c^3*d*e^4)

Giac [F(-1)]

Timed out. \[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^4} \, dx=\text {Timed out} \]

integrate((a+b*arctan(d*x+c))^3/(d*e*x+c*e)^4,x, algorithm="giac")

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^4} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c+d\,x\right )\right )}^3}{{\left (c\,e+d\,e\,x\right )}^4} \,d x \]

int((a + b*atan(c + d*x))^3/(c*e + d*e*x)^4,x)

int((a + b*atan(c + d*x))^3/(c*e + d*e*x)^4, x)

\(\int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^4} \, dx\) [20]

Optimal result