3.1289 \(\int x (a+b \tan ^{-1}(c x)) (d+e \log (1+c^2 x^2)) \, dx\)

Optimal. Leaf size=137 \[ \frac{e \left (c^2 x^2+1\right ) \log \left (c^2 x^2+1\right ) \left (a+b \tan ^{-1}(c x)\right )}{2 c^2}+\frac{1}{2} d x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{2} e x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{b (d-e) \tan ^{-1}(c x)}{2 c^2}-\frac{b e x \log \left (c^2 x^2+1\right )}{2 c}-\frac{b e \tan ^{-1}(c x)}{c^2}-\frac{b x (d-e)}{2 c}+\frac{b e x}{c} \]

[Out]

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

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Rubi [A]  time = 0.110792, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {2454, 2389, 2295, 5019, 321, 203, 2448} \[ \frac{e \left (c^2 x^2+1\right ) \log \left (c^2 x^2+1\right ) \left (a+b \tan ^{-1}(c x)\right )}{2 c^2}+\frac{1}{2} d x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{2} e x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{b (d-e) \tan ^{-1}(c x)}{2 c^2}-\frac{b e x \log \left (c^2 x^2+1\right )}{2 c}-\frac{b e \tan ^{-1}(c x)}{c^2}-\frac{b x (d-e)}{2 c}+\frac{b e x}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 5019

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*(e_.))*(x_)^(m_.), x_Symbol] :> With
[{u = IntHide[x^m*(d + e*Log[f + g*x^2]), x]}, Dist[a + b*ArcTan[c*x], u, x] - Dist[b*c, Int[ExpandIntegrand[u
/(1 + c^2*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ[(m + 1)/2, 0]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 2448

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)], x_Symbol] :> Simp[x*Log[c*(d + e*x^n)^p], x] - Dist[e*n*p, Int[
x^n/(d + e*x^n), x], x] /; FreeQ[{c, d, e, n, p}, x]

Rubi steps

\begin{align*} \int x \left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx &=\frac{1}{2} d x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{2} e x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{e \left (1+c^2 x^2\right ) \left (a+b \tan ^{-1}(c x)\right ) \log \left (1+c^2 x^2\right )}{2 c^2}-(b c) \int \left (\frac{(d-e) x^2}{2 \left (1+c^2 x^2\right )}+\frac{e \log \left (1+c^2 x^2\right )}{2 c^2}\right ) \, dx\\ &=\frac{1}{2} d x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{2} e x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{e \left (1+c^2 x^2\right ) \left (a+b \tan ^{-1}(c x)\right ) \log \left (1+c^2 x^2\right )}{2 c^2}-\frac{1}{2} (b c (d-e)) \int \frac{x^2}{1+c^2 x^2} \, dx-\frac{(b e) \int \log \left (1+c^2 x^2\right ) \, dx}{2 c}\\ &=-\frac{b (d-e) x}{2 c}+\frac{1}{2} d x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{2} e x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac{b e x \log \left (1+c^2 x^2\right )}{2 c}+\frac{e \left (1+c^2 x^2\right ) \left (a+b \tan ^{-1}(c x)\right ) \log \left (1+c^2 x^2\right )}{2 c^2}+\frac{(b (d-e)) \int \frac{1}{1+c^2 x^2} \, dx}{2 c}+(b c e) \int \frac{x^2}{1+c^2 x^2} \, dx\\ &=-\frac{b (d-e) x}{2 c}+\frac{b e x}{c}+\frac{b (d-e) \tan ^{-1}(c x)}{2 c^2}+\frac{1}{2} d x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{2} e x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac{b e x \log \left (1+c^2 x^2\right )}{2 c}+\frac{e \left (1+c^2 x^2\right ) \left (a+b \tan ^{-1}(c x)\right ) \log \left (1+c^2 x^2\right )}{2 c^2}-\frac{(b e) \int \frac{1}{1+c^2 x^2} \, dx}{c}\\ &=-\frac{b (d-e) x}{2 c}+\frac{b e x}{c}+\frac{b (d-e) \tan ^{-1}(c x)}{2 c^2}-\frac{b e \tan ^{-1}(c x)}{c^2}+\frac{1}{2} d x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{2} e x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac{b e x \log \left (1+c^2 x^2\right )}{2 c}+\frac{e \left (1+c^2 x^2\right ) \left (a+b \tan ^{-1}(c x)\right ) \log \left (1+c^2 x^2\right )}{2 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0892493, size = 105, normalized size = 0.77 \[ \frac{e \log \left (c^2 x^2+1\right ) \left (a c^2 x^2+a-b c x\right )+c x (a c x (d-e)-b (d-3 e))+b \tan ^{-1}(c x) \left (c^2 d x^2-e \left (c^2 x^2+3\right )+\left (c^2 e x^2+e\right ) \log \left (c^2 x^2+1\right )+d\right )}{2 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]),x]

[Out]

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

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Maple [C]  time = 0.785, size = 3074, normalized size = 22.4 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arctan(c*x))*(d+e*ln(c^2*x^2+1)),x)

[Out]

-1/4*I/c^2*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^3*Pi*e*arctan(c*x)-1/2*b*d*x/c+1/2*b*d*arctan(c*x)/c^2+1/2*b*arct
an(c*x)*x^2*d-1/2*b*arctan(c*x)*x^2*e-1/2*I/c^2*b*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2*csgn(I*((1+I*c*x)^2/
(c^2*x^2+1)+1))*Pi*e*arctan(c*x)+1/2*I/c^2*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^2*csgn(I*(1+I*c*x)/(c^2*x^2+1)^(1
/2))*Pi*e*arctan(c*x)+1/4*I/c^2*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2*csgn(I*(1+I*
c*x)^2/(c^2*x^2+1))*Pi*e*arctan(c*x)-1/4*I/c^2*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(1+I*c*x)/(c^2*x^2+1)^
(1/2))^2*Pi*e*arctan(c*x)+1/4*I/c^2*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2*csgn(I/(
(1+I*c*x)^2/(c^2*x^2+1)+1)^2)*Pi*e*arctan(c*x)+1/4*I*b*arctan(c*x)*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1))^2*cs
gn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*x^2*e-1/2*I*b*arctan(c*x)*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*((
1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2*x^2*e+1/2*I*b*arctan(c*x)*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^2*csgn(I*(1+I*c*x)
/(c^2*x^2+1)^(1/2))*x^2*e+1/4*I*b*arctan(c*x)*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1
)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2*x^2*e-1/4*I*b*arctan(c*x)*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(1+I*c*
x)/(c^2*x^2+1)^(1/2))^2*x^2*e+1/4*I*b*arctan(c*x)*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1
)^2)^2*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*x^2*e-1/2*a*x^2*e-1/4*I/c^2*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+
I*c*x)^2/(c^2*x^2+1)+1)^2)^3*Pi*e*arctan(c*x)-1/4/c^2*b*e*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*csgn(I*(1+I
*c*x)^2/(c^2*x^2+1))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)-5/2/c^2*b*e*arctan(c*x)+3/2
*b*e*x/c+1/2*x^2*a*d+1/4/c^2*b*e*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^3-1/4/c^2*b*e*Pi*csgn(I*(1+I*c*x)^2/
(c^2*x^2+1))^3-1/4/c^2*b*e*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^3+1/c*b*ln((1+I*c*
x)^2/(c^2*x^2+1)+1)*x*e-1/c*b*ln(2)*x*e-1/c^2*b*ln((1+I*c*x)^2/(c^2*x^2+1)+1)*e*arctan(c*x)+1/c^2*b*ln(2)*e*ar
ctan(c*x)-b*arctan(c*x)*ln((1+I*c*x)^2/(c^2*x^2+1)+1)*x^2*e+b*arctan(c*x)*ln(2)*x^2*e-1/2*a*e/c^2+1/2*x^2*a*e*
ln(c^2*x^2+1)+1/2*a*e/c^2*ln(c^2*x^2+1)+1/4*I*b*arctan(c*x)*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^3*x^2*e-1
/4*I*b*arctan(c*x)*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^3*x^2*e-1/4*I*b*arctan(c*x)*Pi*csgn(I*(1+I*c*x)^2/(c^2*x
^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^3*x^2*e-1/4*I/c*b*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^3*x*e+1/4*I/c*
b*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^3*x*e+1/4*I/c*b*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1
)+1)^2)^3*x*e+1/4*I/c^2*b*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^3*Pi*e*arctan(c*x)-1/4*I*b*arctan(c*x)*Pi*csgn
(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*csgn(I/((1+I*c*x)^2/
(c^2*x^2+1)+1)^2)*x^2*e+1/4*I/c*b*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)
^2/(c^2*x^2+1)+1)^2)*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*x*e-1/4*I/c^2*b*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+
I*c*x)^2/(c^2*x^2+1)+1)^2)*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*Pi*e*arctan(c
*x)-1/4*I/c*b*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1))^2*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*x*e+1/2*I/c*b*Pi*
csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2*x*e-1/2*I/c*b*Pi*csgn(I*(1+I*c*x)^
2/(c^2*x^2+1))^2*csgn(I*(1+I*c*x)/(c^2*x^2+1)^(1/2))*x*e-1/4*I/c*b*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(
1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2*x*e+1/4*I/c*b*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(
I*(1+I*c*x)/(c^2*x^2+1)^(1/2))^2*x*e-1/4*I/c*b*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2
)^2*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*x*e+1/4*I/c^2*b*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*csgn(I*((1+I*c
*x)^2/(c^2*x^2+1)+1))^2*Pi*e*arctan(c*x)-I/c^2*b*e*ln(2)-1/2*I/c^2*b*d+3/2*I/c^2*b*e+1/4/c^2*b*e*Pi*csgn(I*((1
+I*c*x)^2/(c^2*x^2+1)+1))^2*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)-1/2/c^2*b*e*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+
1)+1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2+1/2/c^2*b*e*Pi*csgn(I*(1+I*c*x)/(c^2*x^2+1)^(1/2))*csgn(I*(1+I*
c*x)^2/(c^2*x^2+1))^2+1/4/c^2*b*e*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)
^2/(c^2*x^2+1)+1)^2)^2-1/4/c^2*b*e*Pi*csgn(I*(1+I*c*x)/(c^2*x^2+1)^(1/2))^2*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))+1/
4/c^2*b*e*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^
2)^2+1/c^2*b*e*(arctan(c*x)*x*c-I*arctan(c*x)-1)*(c*x+I)*ln((1+I*c*x)/(c^2*x^2+1)^(1/2))

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Maxima [A]  time = 1.4501, size = 201, normalized size = 1.47 \begin{align*} \frac{1}{2} \, a d x^{2} + \frac{1}{2} \,{\left (x^{2} \arctan \left (c x\right ) - c{\left (\frac{x}{c^{2}} - \frac{\arctan \left (c x\right )}{c^{3}}\right )}\right )} b d - \frac{{\left (x \log \left (c^{2} x^{2} + 1\right ) - 3 \, x + \frac{2 \, \arctan \left (c x\right )}{c}\right )} b e}{2 \, c} - \frac{{\left (c^{2} x^{2} -{\left (c^{2} x^{2} + 1\right )} \log \left (c^{2} x^{2} + 1\right ) + 1\right )} b e \arctan \left (c x\right )}{2 \, c^{2}} - \frac{{\left (c^{2} x^{2} -{\left (c^{2} x^{2} + 1\right )} \log \left (c^{2} x^{2} + 1\right ) + 1\right )} a e}{2 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctan(c*x))*(d+e*log(c^2*x^2+1)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.44882, size = 262, normalized size = 1.91 \begin{align*} \frac{{\left (a c^{2} d - a c^{2} e\right )} x^{2} -{\left (b c d - 3 \, b c e\right )} x +{\left ({\left (b c^{2} d - b c^{2} e\right )} x^{2} + b d - 3 \, b e\right )} \arctan \left (c x\right ) +{\left (a c^{2} e x^{2} - b c e x + a e +{\left (b c^{2} e x^{2} + b e\right )} \arctan \left (c x\right )\right )} \log \left (c^{2} x^{2} + 1\right )}{2 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctan(c*x))*(d+e*log(c^2*x^2+1)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 6.0819, size = 202, normalized size = 1.47 \begin{align*} \begin{cases} \frac{a d x^{2}}{2} + \frac{a e x^{2} \log{\left (c^{2} x^{2} + 1 \right )}}{2} - \frac{a e x^{2}}{2} + \frac{a e \log{\left (c^{2} x^{2} + 1 \right )}}{2 c^{2}} + \frac{b d x^{2} \operatorname{atan}{\left (c x \right )}}{2} + \frac{b e x^{2} \log{\left (c^{2} x^{2} + 1 \right )} \operatorname{atan}{\left (c x \right )}}{2} - \frac{b e x^{2} \operatorname{atan}{\left (c x \right )}}{2} - \frac{b d x}{2 c} - \frac{b e x \log{\left (c^{2} x^{2} + 1 \right )}}{2 c} + \frac{3 b e x}{2 c} + \frac{b d \operatorname{atan}{\left (c x \right )}}{2 c^{2}} + \frac{b e \log{\left (c^{2} x^{2} + 1 \right )} \operatorname{atan}{\left (c x \right )}}{2 c^{2}} - \frac{3 b e \operatorname{atan}{\left (c x \right )}}{2 c^{2}} & \text{for}\: c \neq 0 \\\frac{a d x^{2}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*atan(c*x))*(d+e*ln(c**2*x**2+1)),x)

[Out]

Piecewise((a*d*x**2/2 + a*e*x**2*log(c**2*x**2 + 1)/2 - a*e*x**2/2 + a*e*log(c**2*x**2 + 1)/(2*c**2) + b*d*x**
2*atan(c*x)/2 + b*e*x**2*log(c**2*x**2 + 1)*atan(c*x)/2 - b*e*x**2*atan(c*x)/2 - b*d*x/(2*c) - b*e*x*log(c**2*
x**2 + 1)/(2*c) + 3*b*e*x/(2*c) + b*d*atan(c*x)/(2*c**2) + b*e*log(c**2*x**2 + 1)*atan(c*x)/(2*c**2) - 3*b*e*a
tan(c*x)/(2*c**2), Ne(c, 0)), (a*d*x**2/2, True))

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Giac [B]  time = 1.31502, size = 398, normalized size = 2.91 \begin{align*} -\frac{3 \, b \arctan \left (c x\right ) e}{2 \, c^{2}} + \frac{{\left (\pi b e + 2 \, a e\right )} \log \left (c^{2} x^{2} + 1\right )}{4 \, c^{2}} + \frac{\pi b c^{2} x^{2} e \log \left (c^{2} x^{2} + 1\right ) \mathrm{sgn}\left (c\right ) \mathrm{sgn}\left (x\right ) - \pi b c^{2} x^{2} e \mathrm{sgn}\left (c\right ) \mathrm{sgn}\left (x\right ) - 2 \, b c^{2} x^{2} \arctan \left (\frac{1}{c x}\right ) e \log \left (c^{2} x^{2} + 1\right ) + 2 \, b c^{2} d x^{2} \arctan \left (c x\right ) + 2 \, b c^{2} x^{2} \arctan \left (\frac{1}{c x}\right ) e + 2 \, a c^{2} x^{2} e \log \left (c^{2} x^{2} + 1\right ) + 2 \, a c^{2} d x^{2} - 2 \, a c^{2} x^{2} e + \pi b e \log \left (c^{2} x^{2} + 1\right ) \mathrm{sgn}\left (c\right ) \mathrm{sgn}\left (x\right ) - 2 \, b c x e \log \left (c^{2} x^{2} + 1\right ) - 2 \, \pi b d \mathrm{sgn}\left (c\right ) \mathrm{sgn}\left (x\right ) - 2 \, b c d x + 6 \, b c x e - \pi b e \log \left (c^{2} x^{2} + 1\right ) - 2 \, b \arctan \left (\frac{1}{c x}\right ) e \log \left (c^{2} x^{2} + 1\right ) + 2 \, b d \arctan \left (c x\right )}{4 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctan(c*x))*(d+e*log(c^2*x^2+1)),x, algorithm="giac")

[Out]

-3/2*b*arctan(c*x)*e/c^2 + 1/4*(pi*b*e + 2*a*e)*log(c^2*x^2 + 1)/c^2 + 1/4*(pi*b*c^2*x^2*e*log(c^2*x^2 + 1)*sg
n(c)*sgn(x) - pi*b*c^2*x^2*e*sgn(c)*sgn(x) - 2*b*c^2*x^2*arctan(1/(c*x))*e*log(c^2*x^2 + 1) + 2*b*c^2*d*x^2*ar
ctan(c*x) + 2*b*c^2*x^2*arctan(1/(c*x))*e + 2*a*c^2*x^2*e*log(c^2*x^2 + 1) + 2*a*c^2*d*x^2 - 2*a*c^2*x^2*e + p
i*b*e*log(c^2*x^2 + 1)*sgn(c)*sgn(x) - 2*b*c*x*e*log(c^2*x^2 + 1) - 2*pi*b*d*sgn(c)*sgn(x) - 2*b*c*d*x + 6*b*c
*x*e - pi*b*e*log(c^2*x^2 + 1) - 2*b*arctan(1/(c*x))*e*log(c^2*x^2 + 1) + 2*b*d*arctan(c*x))/c^2