3.20 \(\int \frac{(a+b \tan ^{-1}(c+d x))^3}{(c e+d e x)^4} \, dx\)

Optimal. Leaf size=287 \[ \frac{i b^2 \text{PolyLog}\left (2,-1+\frac{2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d e^4}-\frac{b^3 \text{PolyLog}\left (3,-1+\frac{2}{1-i (c+d x)}\right )}{2 d e^4}-\frac{b^2 \left (a+b \tan ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac{b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac{b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac{i \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac{b \log \left (2-\frac{2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{d e^4}+\frac{b^3 \log (c+d x)}{d e^4}-\frac{b^3 \log \left ((c+d x)^2+1\right )}{2 d e^4} \]

[Out]

-((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)

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Rubi [A]  time = 0.499924, antiderivative size = 287, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 13, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.565, Rules used = {5043, 12, 4852, 4918, 266, 36, 29, 31, 4884, 4924, 4868, 4992, 6610} \[ \frac{i b^2 \text{PolyLog}\left (2,-1+\frac{2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d e^4}-\frac{b^3 \text{PolyLog}\left (3,-1+\frac{2}{1-i (c+d x)}\right )}{2 d e^4}-\frac{b^2 \left (a+b \tan ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac{b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac{b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac{i \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac{b \log \left (2-\frac{2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{d e^4}+\frac{b^3 \log (c+d x)}{d e^4}-\frac{b^3 \log \left ((c+d x)^2+1\right )}{2 d e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((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)

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 4852

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
n[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 4918

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]

Rule 266

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]]

Rule 36

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]

Rule 29

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

Rule 31

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

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 4924

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

Rule 4868

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]

Rule 4992

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(I*(a + b*ArcT
an[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 )^3}{(c e+d e x)^4} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(x)\right )^3}{e^4 x^4} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(x)\right )^3}{x^4} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac{b \operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(x)\right )^2}{x^3 \left (1+x^2\right )} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac{b \operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(x)\right )^2}{x^3} \, dx,x,c+d x\right )}{d e^4}-\frac{b \operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(x)\right )^2}{x \left (1+x^2\right )} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac{b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac{i \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac{(i b) \operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(x)\right )^2}{x (i+x)} \, dx,x,c+d x\right )}{d e^4}+\frac{b^2 \operatorname{Subst}\left (\int \frac{a+b \tan ^{-1}(x)}{x^2 \left (1+x^2\right )} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac{b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac{i \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac{b \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (2-\frac{2}{1-i (c+d x)}\right )}{d e^4}+\frac{b^2 \operatorname{Subst}\left (\int \frac{a+b \tan ^{-1}(x)}{x^2} \, dx,x,c+d x\right )}{d e^4}-\frac{b^2 \operatorname{Subst}\left (\int \frac{a+b \tan ^{-1}(x)}{1+x^2} \, dx,x,c+d x\right )}{d e^4}+\frac{\left (2 b^2\right ) \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^4}\\ &=-\frac{b^2 \left (a+b \tan ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac{b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4}-\frac{b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac{i \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac{b \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (2-\frac{2}{1-i (c+d x)}\right )}{d e^4}+\frac{i b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text{Li}_2\left (-1+\frac{2}{1-i (c+d x)}\right )}{d e^4}-\frac{\left (i b^3\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^4}+\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{x \left (1+x^2\right )} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac{b^2 \left (a+b \tan ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac{b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4}-\frac{b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac{i \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac{b \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (2-\frac{2}{1-i (c+d x)}\right )}{d e^4}+\frac{i b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text{Li}_2\left (-1+\frac{2}{1-i (c+d x)}\right )}{d e^4}-\frac{b^3 \text{Li}_3\left (-1+\frac{2}{1-i (c+d x)}\right )}{2 d e^4}+\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{x (1+x)} \, dx,x,(c+d x)^2\right )}{2 d e^4}\\ &=-\frac{b^2 \left (a+b \tan ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac{b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4}-\frac{b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac{i \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac{b \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (2-\frac{2}{1-i (c+d x)}\right )}{d e^4}+\frac{i b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text{Li}_2\left (-1+\frac{2}{1-i (c+d x)}\right )}{d e^4}-\frac{b^3 \text{Li}_3\left (-1+\frac{2}{1-i (c+d x)}\right )}{2 d e^4}+\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,(c+d x)^2\right )}{2 d e^4}-\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,(c+d x)^2\right )}{2 d e^4}\\ &=-\frac{b^2 \left (a+b \tan ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac{b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4}-\frac{b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac{i \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^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 \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (2-\frac{2}{1-i (c+d x)}\right )}{d e^4}+\frac{i b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text{Li}_2\left (-1+\frac{2}{1-i (c+d x)}\right )}{d e^4}-\frac{b^3 \text{Li}_3\left (-1+\frac{2}{1-i (c+d x)}\right )}{2 d e^4}\\ \end{align*}

Mathematica [A]  time = 0.823884, size = 360, normalized size = 1.25 \[ \frac{24 a b^2 \left (i \text{PolyLog}\left (2,e^{2 i \tan ^{-1}(c+d x)}\right )-\frac{(c+d x)^2+\tan ^{-1}(c+d x)^2}{(c+d x)^3}+\tan ^{-1}(c+d x) \left (-\frac{1}{(c+d x)^2}+i \tan ^{-1}(c+d x)-2 \log \left (1-e^{2 i \tan ^{-1}(c+d x)}\right )-1\right )\right )+b^3 \left (-24 i \tan ^{-1}(c+d x) \text{PolyLog}\left (2,e^{-2 i \tan ^{-1}(c+d x)}\right )-12 \text{PolyLog}\left (3,e^{-2 i \tan ^{-1}(c+d x)}\right )+24 \log \left (\frac{c+d x}{\sqrt{(c+d x)^2+1}}\right )-\frac{8 \tan ^{-1}(c+d x)^3}{(c+d x)^3}-8 i \tan ^{-1}(c+d x)^3-\frac{12 \tan ^{-1}(c+d x)^2}{(c+d x)^2}-12 \tan ^{-1}(c+d x)^2-\frac{24 \tan ^{-1}(c+d x)}{c+d x}-24 \tan ^{-1}(c+d x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c+d x)}\right )+i \pi ^3\right )+12 a^2 b \log \left (c^2+2 c d x+d^2 x^2+1\right )-\frac{12 a^2 b}{(c+d x)^2}-24 a^2 b \log (c+d x)-\frac{24 a^2 b \tan ^{-1}(c+d x)}{(c+d x)^3}-\frac{8 a^3}{(c+d x)^3}}{24 d e^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-8*a^3)/(c + d*x)^3 - (12*a^2*b)/(c + d*x)^2 - (24*a^2*b*ArcTan[c + d*x])/(c + d*x)^3 - 24*a^2*b*Log[c + d*x
] + 12*a^2*b*Log[1 + c^2 + 2*c*d*x + d^2*x^2] + 24*a*b^2*(-(((c + d*x)^2 + ArcTan[c + d*x]^2)/(c + d*x)^3) + A
rcTan[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])]) + b^3*(I*Pi^3 - (24*ArcTan[c + d*x])/(c + d*x) - 12*ArcTan[c + d*x]^2 - (12*ArcTa
n[c + d*x]^2)/(c + d*x)^2 - (8*I)*ArcTan[c + d*x]^3 - (8*ArcTan[c + d*x]^3)/(c + d*x)^3 - 24*ArcTan[c + d*x]^2
*Log[1 - E^((-2*I)*ArcTan[c + d*x])] + 24*Log[(c + d*x)/Sqrt[1 + (c + d*x)^2]] - (24*I)*ArcTan[c + d*x]*PolyLo
g[2, E^((-2*I)*ArcTan[c + d*x])] - 12*PolyLog[3, E^((-2*I)*ArcTan[c + d*x])]))/(24*d*e^4)

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Maple [C]  time = 0.864, size = 7083, normalized size = 24.7 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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^{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\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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

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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 )}^{3}}{{\left (d e x + c e\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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