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

Optimal. Leaf size=337 \[ \frac{3 i b^2 (d e-c f) \text{PolyLog}\left (2,1-\frac{2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d^2}+\frac{3 b^3 (d e-c f) \text{PolyLog}\left (3,1-\frac{2}{1+i (c+d x)}\right )}{2 d^2}-\frac{3 i b^3 f \text{PolyLog}\left (2,1-\frac{2}{1+i (c+d x)}\right )}{2 d^2}-\frac{3 b^2 f \log \left (\frac{2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d^2}+\frac{i (d e-c f) \left (a+b \tan ^{-1}(c+d x)\right )^3}{d^2}-\frac{(-c f+d e+f) (d e-(c+1) f) \left (a+b \tan ^{-1}(c+d x)\right )^3}{2 d^2 f}+\frac{3 b (d e-c f) \log \left (\frac{2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{d^2}-\frac{3 i b f \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d^2}-\frac{3 b f (c+d x) \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d^2}+\frac{(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^3}{2 f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.631683, antiderivative size = 337, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 11, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.611, Rules used = {5047, 4864, 4846, 4920, 4854, 2402, 2315, 4984, 4884, 4994, 6610} \[ \frac{3 i b^2 (d e-c f) \text{PolyLog}\left (2,1-\frac{2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d^2}+\frac{3 b^3 (d e-c f) \text{PolyLog}\left (3,1-\frac{2}{1+i (c+d x)}\right )}{2 d^2}-\frac{3 i b^3 f \text{PolyLog}\left (2,1-\frac{2}{1+i (c+d x)}\right )}{2 d^2}-\frac{3 b^2 f \log \left (\frac{2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d^2}+\frac{i (d e-c f) \left (a+b \tan ^{-1}(c+d x)\right )^3}{d^2}-\frac{(-c f+d e+f) (d e-(c+1) f) \left (a+b \tan ^{-1}(c+d x)\right )^3}{2 d^2 f}+\frac{3 b (d e-c f) \log \left (\frac{2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{d^2}-\frac{3 i b f \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d^2}-\frac{3 b f (c+d x) \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d^2}+\frac{(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^3}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5047

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

Rule 4864

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(a
+ b*ArcTan[c*x])^p)/(e*(q + 1)), x] - Dist[(b*c*p)/(e*(q + 1)), Int[ExpandIntegrand[(a + b*ArcTan[c*x])^(p -
1), (d + e*x)^(q + 1)/(1 + c^2*x^2), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 1] && IntegerQ[q] && N
eQ[q, -1]

Rule 4846

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x])^p, x] - Dist[b*c*p, Int[
(x*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 4920

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

Rule 4854

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^p*Lo
g[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTan[c*x])^(p - 1)*Log[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 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 4984

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> I
nt[ExpandIntegrand[(a + b*ArcTan[c*x])^p/(d + e*x^2), (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& IGtQ[p, 0] && EqQ[e, c^2*d] && IGtQ[m, 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 (e+f x) \left (a+b \tan ^{-1}(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int \left (\frac{d e-c f}{d}+\frac{f x}{d}\right ) \left (a+b \tan ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac{(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^3}{2 f}-\frac{(3 b) \operatorname{Subst}\left (\int \left (\frac{f^2 \left (a+b \tan ^{-1}(x)\right )^2}{d^2}+\frac{((d e-f-c f) (d e+f-c f)+2 f (d e-c f) x) \left (a+b \tan ^{-1}(x)\right )^2}{d^2 \left (1+x^2\right )}\right ) \, dx,x,c+d x\right )}{2 f}\\ &=\frac{(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^3}{2 f}-\frac{(3 b) \operatorname{Subst}\left (\int \frac{((d e-f-c f) (d e+f-c f)+2 f (d e-c f) x) \left (a+b \tan ^{-1}(x)\right )^2}{1+x^2} \, dx,x,c+d x\right )}{2 d^2 f}-\frac{(3 b f) \operatorname{Subst}\left (\int \left (a+b \tan ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{2 d^2}\\ &=-\frac{3 b f (c+d x) \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d^2}+\frac{(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^3}{2 f}-\frac{(3 b) \operatorname{Subst}\left (\int \left (\frac{(d e+f-c f) (d e-(1+c) f) \left (a+b \tan ^{-1}(x)\right )^2}{1+x^2}-\frac{2 f (-d e+c f) x \left (a+b \tan ^{-1}(x)\right )^2}{1+x^2}\right ) \, dx,x,c+d x\right )}{2 d^2 f}+\frac{\left (3 b^2 f\right ) \operatorname{Subst}\left (\int \frac{x \left (a+b \tan ^{-1}(x)\right )}{1+x^2} \, dx,x,c+d x\right )}{d^2}\\ &=-\frac{3 i b f \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d^2}-\frac{3 b f (c+d x) \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d^2}+\frac{(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^3}{2 f}-\frac{\left (3 b^2 f\right ) \operatorname{Subst}\left (\int \frac{a+b \tan ^{-1}(x)}{i-x} \, dx,x,c+d x\right )}{d^2}-\frac{(3 b (d e-c f)) \operatorname{Subst}\left (\int \frac{x \left (a+b \tan ^{-1}(x)\right )^2}{1+x^2} \, dx,x,c+d x\right )}{d^2}-\frac{(3 b (d e+f-c f) (d e-(1+c) f)) \operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(x)\right )^2}{1+x^2} \, dx,x,c+d x\right )}{2 d^2 f}\\ &=-\frac{3 i b f \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d^2}-\frac{3 b f (c+d x) \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d^2}+\frac{i (d e-c f) \left (a+b \tan ^{-1}(c+d x)\right )^3}{d^2}-\frac{(d e+f-c f) (d e-(1+c) f) \left (a+b \tan ^{-1}(c+d x)\right )^3}{2 d^2 f}+\frac{(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^3}{2 f}-\frac{3 b^2 f \left (a+b \tan ^{-1}(c+d x)\right ) \log \left (\frac{2}{1+i (c+d x)}\right )}{d^2}+\frac{\left (3 b^3 f\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d^2}+\frac{(3 b (d e-c f)) \operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(x)\right )^2}{i-x} \, dx,x,c+d x\right )}{d^2}\\ &=-\frac{3 i b f \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d^2}-\frac{3 b f (c+d x) \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d^2}+\frac{i (d e-c f) \left (a+b \tan ^{-1}(c+d x)\right )^3}{d^2}-\frac{(d e+f-c f) (d e-(1+c) f) \left (a+b \tan ^{-1}(c+d x)\right )^3}{2 d^2 f}+\frac{(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^3}{2 f}-\frac{3 b^2 f \left (a+b \tan ^{-1}(c+d x)\right ) \log \left (\frac{2}{1+i (c+d x)}\right )}{d^2}+\frac{3 b (d e-c f) \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (\frac{2}{1+i (c+d x)}\right )}{d^2}-\frac{\left (3 i b^3 f\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+i (c+d x)}\right )}{d^2}-\frac{\left (6 b^2 (d e-c f)\right ) \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^2}\\ &=-\frac{3 i b f \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d^2}-\frac{3 b f (c+d x) \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d^2}+\frac{i (d e-c f) \left (a+b \tan ^{-1}(c+d x)\right )^3}{d^2}-\frac{(d e+f-c f) (d e-(1+c) f) \left (a+b \tan ^{-1}(c+d x)\right )^3}{2 d^2 f}+\frac{(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^3}{2 f}-\frac{3 b^2 f \left (a+b \tan ^{-1}(c+d x)\right ) \log \left (\frac{2}{1+i (c+d x)}\right )}{d^2}+\frac{3 b (d e-c f) \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (\frac{2}{1+i (c+d x)}\right )}{d^2}-\frac{3 i b^3 f \text{Li}_2\left (1-\frac{2}{1+i (c+d x)}\right )}{2 d^2}+\frac{3 i b^2 (d e-c f) \left (a+b \tan ^{-1}(c+d x)\right ) \text{Li}_2\left (1-\frac{2}{1+i (c+d x)}\right )}{d^2}-\frac{\left (3 i b^3 (d e-c f)\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^2}\\ &=-\frac{3 i b f \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d^2}-\frac{3 b f (c+d x) \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d^2}+\frac{i (d e-c f) \left (a+b \tan ^{-1}(c+d x)\right )^3}{d^2}-\frac{(d e+f-c f) (d e-(1+c) f) \left (a+b \tan ^{-1}(c+d x)\right )^3}{2 d^2 f}+\frac{(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^3}{2 f}-\frac{3 b^2 f \left (a+b \tan ^{-1}(c+d x)\right ) \log \left (\frac{2}{1+i (c+d x)}\right )}{d^2}+\frac{3 b (d e-c f) \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (\frac{2}{1+i (c+d x)}\right )}{d^2}-\frac{3 i b^3 f \text{Li}_2\left (1-\frac{2}{1+i (c+d x)}\right )}{2 d^2}+\frac{3 i b^2 (d e-c f) \left (a+b \tan ^{-1}(c+d x)\right ) \text{Li}_2\left (1-\frac{2}{1+i (c+d x)}\right )}{d^2}+\frac{3 b^3 (d e-c f) \text{Li}_3\left (1-\frac{2}{1+i (c+d x)}\right )}{2 d^2}\\ \end{align*}

Mathematica [A]  time = 0.678918, size = 592, normalized size = 1.76 \[ \frac{6 a b^2 d e \left (\tan ^{-1}(c+d x) \left ((c+d x-i) \tan ^{-1}(c+d x)+2 \log \left (1+e^{2 i \tan ^{-1}(c+d x)}\right )\right )-i \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c+d x)}\right )\right )-6 a b^2 c f \left (\tan ^{-1}(c+d x) \left ((c+d x-i) \tan ^{-1}(c+d x)+2 \log \left (1+e^{2 i \tan ^{-1}(c+d x)}\right )\right )-i \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c+d x)}\right )\right )+2 b^3 d e \left (-3 i \tan ^{-1}(c+d x) \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c+d x)}\right )+\frac{3}{2} \text{PolyLog}\left (3,-e^{2 i \tan ^{-1}(c+d x)}\right )+\tan ^{-1}(c+d x)^2 \left ((c+d x-i) \tan ^{-1}(c+d x)+3 \log \left (1+e^{2 i \tan ^{-1}(c+d x)}\right )\right )\right )+b^3 f \left (3 i \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c+d x)}\right )+\tan ^{-1}(c+d x) \left (\left ((c+d x)^2+1\right ) \tan ^{-1}(c+d x)^2-3 (c+d x) \tan ^{-1}(c+d x)+3 i \tan ^{-1}(c+d x)-6 \log \left (1+e^{2 i \tan ^{-1}(c+d x)}\right )\right )\right )-2 b^3 c f \left (-3 i \tan ^{-1}(c+d x) \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c+d x)}\right )+\frac{3}{2} \text{PolyLog}\left (3,-e^{2 i \tan ^{-1}(c+d x)}\right )+\tan ^{-1}(c+d x)^2 \left ((c+d x-i) \tan ^{-1}(c+d x)+3 \log \left (1+e^{2 i \tan ^{-1}(c+d x)}\right )\right )\right )+a^2 (c+d x) (-2 a c f+2 a d e-3 b f)-3 a^2 b (d e-c f) \log \left ((c+d x)^2+1\right )-3 a^2 b (c+d x) \tan ^{-1}(c+d x) (c f-d (2 e+f x))+3 a^2 b f \tan ^{-1}(c+d x)+a^3 f (c+d x)^2+6 a b^2 f \left (-\log \left (\frac{1}{\sqrt{(c+d x)^2+1}}\right )+\frac{1}{2} \left ((c+d x)^2+1\right ) \tan ^{-1}(c+d x)^2-(c+d x) \tan ^{-1}(c+d x)\right )}{2 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*(2*a*d*e - 3*b*f - 2*a*c*f)*(c + d*x) + a^3*f*(c + d*x)^2 + 3*a^2*b*f*ArcTan[c + d*x] - 3*a^2*b*(c + d*x)
*(c*f - d*(2*e + f*x))*ArcTan[c + d*x] + 6*a*b^2*f*(-((c + d*x)*ArcTan[c + d*x]) + ((1 + (c + d*x)^2)*ArcTan[c
+ d*x]^2)/2 - Log[1/Sqrt[1 + (c + d*x)^2]]) - 3*a^2*b*(d*e - c*f)*Log[1 + (c + d*x)^2] + 6*a*b^2*d*e*(ArcTan[
c + d*x]*((-I + c + d*x)*ArcTan[c + d*x] + 2*Log[1 + E^((2*I)*ArcTan[c + d*x])]) - I*PolyLog[2, -E^((2*I)*ArcT
an[c + d*x])]) - 6*a*b^2*c*f*(ArcTan[c + d*x]*((-I + c + d*x)*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*f*(ArcTan[c + d*x]*((3*I)*ArcTan[c + d*x] - 3*(c + d
*x)*ArcTan[c + d*x] + (1 + (c + d*x)^2)*ArcTan[c + d*x]^2 - 6*Log[1 + E^((2*I)*ArcTan[c + d*x])]) + (3*I)*Poly
Log[2, -E^((2*I)*ArcTan[c + d*x])]) + 2*b^3*d*e*(ArcTan[c + d*x]^2*((-I + c + d*x)*ArcTan[c + d*x] + 3*Log[1 +
E^((2*I)*ArcTan[c + d*x])]) - (3*I)*ArcTan[c + d*x]*PolyLog[2, -E^((2*I)*ArcTan[c + d*x])] + (3*PolyLog[3, -E
^((2*I)*ArcTan[c + d*x])])/2) - 2*b^3*c*f*(ArcTan[c + d*x]^2*((-I + c + d*x)*ArcTan[c + d*x] + 3*Log[1 + E^((2
*I)*ArcTan[c + d*x])]) - (3*I)*ArcTan[c + d*x]*PolyLog[2, -E^((2*I)*ArcTan[c + d*x])] + (3*PolyLog[3, -E^((2*I
)*ArcTan[c + d*x])])/2))/(2*d^2)

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Maple [C]  time = 1.261, size = 16362, normalized size = 48.6 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/8*b^3*c^2*e*arctan(d*x + c)^3*arctan((d^2*x + c*d)/d)/d + 3*a*b^2*c^2*e*arctan(d*x + c)^2*arctan((d^2*x + c*
d)/d)/d - (3*arctan(d*x + c)*arctan((d^2*x + c*d)/d)^2/d - arctan((d^2*x + c*d)/d)^3/d)*a*b^2*c^2*e - 7/32*(6*
arctan(d*x + c)^2*arctan((d^2*x + c*d)/d)^2/d - 4*arctan(d*x + c)*arctan((d^2*x + c*d)/d)^3/d + arctan((d^2*x
+ c*d)/d)^4/d)*b^3*c^2*e + 7/8*b^3*e*arctan(d*x + c)^3*arctan((d^2*x + c*d)/d)/d + 56*b^3*d^2*f*integrate(1/64
*x^3*arctan(d*x + c)^3/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 6*b^3*d^2*f*integrate(1/64*x^3*arctan(d*x + c)*log(
d^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 192*a*b^2*d^2*f*integrate(1/64*x^3*arctan(d
*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 56*b^3*d^2*e*integrate(1/64*x^2*arctan(d*x + c)^3/(d^2*x^2 + 2*c
*d*x + c^2 + 1), x) + 112*b^3*c*d*f*integrate(1/64*x^2*arctan(d*x + c)^3/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 1
2*b^3*d^2*f*integrate(1/64*x^3*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/(d^2*x^2 + 2*c*d*x + c^2 + 1),
x) + 6*b^3*d^2*e*integrate(1/64*x^2*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 + 2*c*d*x + c
^2 + 1), x) + 12*b^3*c*d*f*integrate(1/64*x^2*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 + 2*
c*d*x + c^2 + 1), x) + 192*a*b^2*d^2*e*integrate(1/64*x^2*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x)
+ 384*a*b^2*c*d*f*integrate(1/64*x^2*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 112*b^3*c*d*e*integ
rate(1/64*x*arctan(d*x + c)^3/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 56*b^3*c^2*f*integrate(1/64*x*arctan(d*x + c
)^3/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 24*b^3*d^2*e*integrate(1/64*x^2*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x
+ c^2 + 1)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 12*b^3*c*d*f*integrate(1/64*x^2*arctan(d*x + c)*log(d^2*x^2 + 2
*c*d*x + c^2 + 1)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 12*b^3*c*d*e*integrate(1/64*x*arctan(d*x + c)*log(d^2*x^
2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 6*b^3*c^2*f*integrate(1/64*x*arctan(d*x + c)*log(
d^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 384*a*b^2*c*d*e*integrate(1/64*x*arctan(d*x
+ c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 192*a*b^2*c^2*f*integrate(1/64*x*arctan(d*x + c)^2/(d^2*x^2 + 2*c*
d*x + c^2 + 1), x) + 24*b^3*c*d*e*integrate(1/64*x*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/(d^2*x^2 +
2*c*d*x + c^2 + 1), x) + 6*b^3*c^2*e*integrate(1/64*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x
^2 + 2*c*d*x + c^2 + 1), x) + 1/2*a^3*f*x^2 + 3*a*b^2*e*arctan(d*x + c)^2*arctan((d^2*x + c*d)/d)/d - 12*b^3*d
*f*integrate(1/64*x^2*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 3*b^3*d*f*integrate(1/64*x^2*log(d
^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) - 24*b^3*d*e*integrate(1/64*x*arctan(d*x + c)^
2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 6*b^3*d*e*integrate(1/64*x*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 +
2*c*d*x + c^2 + 1), x) - (3*arctan(d*x + c)*arctan((d^2*x + c*d)/d)^2/d - arctan((d^2*x + c*d)/d)^3/d)*a*b^2*
e - 7/32*(6*arctan(d*x + c)^2*arctan((d^2*x + c*d)/d)^2/d - 4*arctan(d*x + c)*arctan((d^2*x + c*d)/d)^3/d + ar
ctan((d^2*x + c*d)/d)^4/d)*b^3*e + 3/2*(x^2*arctan(d*x + c) - d*(x/d^2 + (c^2 - 1)*arctan((d^2*x + c*d)/d)/d^3
- c*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/d^3))*a^2*b*f + a^3*e*x + 56*b^3*f*integrate(1/64*x*arctan(d*x + c)^3/(d
^2*x^2 + 2*c*d*x + c^2 + 1), x) + 6*b^3*f*integrate(1/64*x*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2/
(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 192*a*b^2*f*integrate(1/64*x*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 +
1), x) + 6*b^3*e*integrate(1/64*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 + 2*c*d*x + c^2 +
1), x) + 3/2*(2*(d*x + c)*arctan(d*x + c) - log((d*x + c)^2 + 1))*a^2*b*e/d + 1/16*(b^3*f*x^2 + 2*b^3*e*x)*arc
tan(d*x + c)^3 - 3/64*(b^3*f*x^2 + 2*b^3*e*x)*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (a^{3} f x + a^{3} e +{\left (b^{3} f x + b^{3} e\right )} \arctan \left (d x + c\right )^{3} + 3 \,{\left (a b^{2} f x + a b^{2} e\right )} \arctan \left (d x + c\right )^{2} + 3 \,{\left (a^{2} b f x + a^{2} b e\right )} \arctan \left (d x + c\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \operatorname{atan}{\left (c + d x \right )}\right )^{3} \left (e + f x\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (f x + e\right )}{\left (b \arctan \left (d x + c\right ) + a\right )}^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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