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

Optimal. Leaf size=264 \[ \frac{3 i b^2 d \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}-\frac{3 i b^3 e \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{2 c^2}+\frac{3 b^3 d \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )}{2 c}-\frac{3 b^2 e \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^2}-\frac{\left (d^2-\frac{e^2}{c^2}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}-\frac{3 i b e \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}+\frac{(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}+\frac{i d \left (a+b \tan ^{-1}(c x)\right )^3}{c}+\frac{3 b d \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c}-\frac{3 b e x \left (a+b \tan ^{-1}(c x)\right )^2}{2 c} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.580084, antiderivative size = 264, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {4864, 4846, 4920, 4854, 2402, 2315, 4984, 4884, 4994, 6610} \[ \frac{3 i b^2 d \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}-\frac{3 i b^3 e \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{2 c^2}+\frac{3 b^3 d \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )}{2 c}-\frac{3 b^2 e \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^2}-\frac{\left (d^2-\frac{e^2}{c^2}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}-\frac{3 i b e \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}+\frac{(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}+\frac{i d \left (a+b \tan ^{-1}(c x)\right )^3}{c}+\frac{3 b d \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c}-\frac{3 b e x \left (a+b \tan ^{-1}(c x)\right )^2}{2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.594318, size = 342, normalized size = 1.3 \[ \frac{6 a b^2 c d \left (\tan ^{-1}(c x) \left ((c x-i) \tan ^{-1}(c x)+2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )-i \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )\right )+b^3 e \left (3 i \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+\tan ^{-1}(c x) \left (\left (c^2 x^2+1\right ) \tan ^{-1}(c x)^2+(-3 c x+3 i) \tan ^{-1}(c x)-6 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )\right )+b^3 c d \left (-6 i \tan ^{-1}(c x) \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+3 \text{PolyLog}\left (3,-e^{2 i \tan ^{-1}(c x)}\right )+2 \tan ^{-1}(c x)^2 \left ((c x-i) \tan ^{-1}(c x)+3 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )\right )+3 a^2 b c^2 x \tan ^{-1}(c x) (2 d+e x)-3 a^2 b c d \log \left (c^2 x^2+1\right )+a^2 c x (2 a c d-3 b e)+3 a^2 b e \tan ^{-1}(c x)+a^3 c^2 e x^2+3 a b^2 e \left (\log \left (c^2 x^2+1\right )+\left (c^2 x^2+1\right ) \tan ^{-1}(c x)^2-2 c x \tan ^{-1}(c x)\right )}{2 c^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.813, size = 7462, normalized size = 28.3 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

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

[Out]

7/32*b^3*d*arctan(c*x)^4/c + 56*b^3*c^2*e*integrate(1/64*x^3*arctan(c*x)^3/(c^2*x^2 + 1), x) + 6*b^3*c^2*e*int
egrate(1/64*x^3*arctan(c*x)*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 192*a*b^2*c^2*e*integrate(1/64*x^3*arctan(c
*x)^2/(c^2*x^2 + 1), x) + 56*b^3*c^2*d*integrate(1/64*x^2*arctan(c*x)^3/(c^2*x^2 + 1), x) + 12*b^3*c^2*e*integ
rate(1/64*x^3*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) + 6*b^3*c^2*d*integrate(1/64*x^2*arctan(c*x)*log(
c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 192*a*b^2*c^2*d*integrate(1/64*x^2*arctan(c*x)^2/(c^2*x^2 + 1), x) + 24*b^3
*c^2*d*integrate(1/64*x^2*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) + 1/2*a^3*e*x^2 + a*b^2*d*arctan(c*x)
^3/c - 12*b^3*c*e*integrate(1/64*x^2*arctan(c*x)^2/(c^2*x^2 + 1), x) + 3*b^3*c*e*integrate(1/64*x^2*log(c^2*x^
2 + 1)^2/(c^2*x^2 + 1), x) - 24*b^3*c*d*integrate(1/64*x*arctan(c*x)^2/(c^2*x^2 + 1), x) + 6*b^3*c*d*integrate
(1/64*x*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 3/2*(x^2*arctan(c*x) - c*(x/c^2 - arctan(c*x)/c^3))*a^2*b*e + a
^3*d*x + 56*b^3*e*integrate(1/64*x*arctan(c*x)^3/(c^2*x^2 + 1), x) + 6*b^3*e*integrate(1/64*x*arctan(c*x)*log(
c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 192*a*b^2*e*integrate(1/64*x*arctan(c*x)^2/(c^2*x^2 + 1), x) + 6*b^3*d*inte
grate(1/64*arctan(c*x)*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 3/2*(2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*a^2*b
*d/c + 1/16*(b^3*e*x^2 + 2*b^3*d*x)*arctan(c*x)^3 - 3/64*(b^3*e*x^2 + 2*b^3*d*x)*arctan(c*x)*log(c^2*x^2 + 1)^
2

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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