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

Optimal. Leaf size=171 \[ \frac{i b^2 d \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{c}-\frac{\left (d^2-\frac{e^2}{c^2}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{2 e}+\frac{(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 e}+\frac{i d \left (a+b \tan ^{-1}(c x)\right )^2}{c}+\frac{2 b d \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}-\frac{a b e x}{c}+\frac{b^2 e \log \left (c^2 x^2+1\right )}{2 c^2}-\frac{b^2 e x \tan ^{-1}(c x)}{c} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.297722, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.562, Rules used = {4864, 4846, 260, 4984, 4884, 4920, 4854, 2402, 2315} \[ \frac{i b^2 d \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{c}-\frac{\left (d^2-\frac{e^2}{c^2}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{2 e}+\frac{(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 e}+\frac{i d \left (a+b \tan ^{-1}(c x)\right )^2}{c}+\frac{2 b d \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}-\frac{a b e x}{c}+\frac{b^2 e \log \left (c^2 x^2+1\right )}{2 c^2}-\frac{b^2 e x \tan ^{-1}(c x)}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*b*e*x)/c) - (b^2*e*x*ArcTan[c*x])/c + (I*d*(a + b*ArcTan[c*x])^2)/c - ((d^2 - e^2/c^2)*(a + b*ArcTan[c*x]
)^2)/(2*e) + ((d + e*x)^2*(a + b*ArcTan[c*x])^2)/(2*e) + (2*b*d*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/c + (b
^2*e*Log[1 + c^2*x^2])/(2*c^2) + (I*b^2*d*PolyLog[2, 1 - 2/(1 + I*c*x)])/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 260

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

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 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]

Rubi steps

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

Mathematica [A]  time = 0.269229, size = 172, normalized size = 1.01 \[ \frac{-2 i b^2 c d \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+2 a^2 c^2 d x+a^2 c^2 e x^2+2 b \tan ^{-1}(c x) \left (a \left (2 c^2 d x+c^2 e x^2+e\right )+2 b c d \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-b c e x\right )-2 a b c d \log \left (c^2 x^2+1\right )-2 a b c e x+b^2 e \log \left (c^2 x^2+1\right )+b^2 (c x-i) \tan ^{-1}(c x)^2 (2 c d+c e x+i e)}{2 c^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.091, size = 360, normalized size = 2.1 \begin{align*}{\frac{{a}^{2}{x}^{2}e}{2}}+{a}^{2}dx+{\frac{{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}{x}^{2}e}{2}}+{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}xd-{\frac{{b}^{2}\arctan \left ( cx \right ) \ln \left ({c}^{2}{x}^{2}+1 \right ) d}{c}}+{\frac{{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}e}{2\,{c}^{2}}}-{\frac{{b}^{2}ex\arctan \left ( cx \right ) }{c}}+{\frac{{b}^{2}e\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2\,{c}^{2}}}+{\frac{{\frac{i}{2}}{b}^{2}d\ln \left ( cx-i \right ) \ln \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{c}}+{\frac{{\frac{i}{2}}{b}^{2}d\ln \left ({c}^{2}{x}^{2}+1 \right ) \ln \left ( cx+i \right ) }{c}}-{\frac{{\frac{i}{2}}{b}^{2}d{\it dilog} \left ({\frac{i}{2}} \left ( cx-i \right ) \right ) }{c}}-{\frac{{\frac{i}{4}}{b}^{2}d \left ( \ln \left ( cx+i \right ) \right ) ^{2}}{c}}-{\frac{{\frac{i}{2}}{b}^{2}d\ln \left ({c}^{2}{x}^{2}+1 \right ) \ln \left ( cx-i \right ) }{c}}+{\frac{{\frac{i}{2}}{b}^{2}d{\it dilog} \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{c}}-{\frac{{\frac{i}{2}}{b}^{2}d\ln \left ( cx+i \right ) \ln \left ({\frac{i}{2}} \left ( cx-i \right ) \right ) }{c}}+{\frac{{\frac{i}{4}}{b}^{2}d \left ( \ln \left ( cx-i \right ) \right ) ^{2}}{c}}+ab\arctan \left ( cx \right ){x}^{2}e+2\,ab\arctan \left ( cx \right ) xd-{\frac{abex}{c}}-{\frac{abd\ln \left ({c}^{2}{x}^{2}+1 \right ) }{c}}+{\frac{abe\arctan \left ( cx \right ) }{{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 12 \, b^{2} c^{2} e \int \frac{x^{3} \arctan \left (c x\right )^{2}}{16 \,{\left (c^{2} x^{2} + 1\right )}}\,{d x} + b^{2} c^{2} e \int \frac{x^{3} \log \left (c^{2} x^{2} + 1\right )^{2}}{16 \,{\left (c^{2} x^{2} + 1\right )}}\,{d x} + 12 \, b^{2} c^{2} d \int \frac{x^{2} \arctan \left (c x\right )^{2}}{16 \,{\left (c^{2} x^{2} + 1\right )}}\,{d x} + 2 \, b^{2} c^{2} e \int \frac{x^{3} \log \left (c^{2} x^{2} + 1\right )}{16 \,{\left (c^{2} x^{2} + 1\right )}}\,{d x} + b^{2} c^{2} d \int \frac{x^{2} \log \left (c^{2} x^{2} + 1\right )^{2}}{16 \,{\left (c^{2} x^{2} + 1\right )}}\,{d x} + 4 \, b^{2} c^{2} d \int \frac{x^{2} \log \left (c^{2} x^{2} + 1\right )}{16 \,{\left (c^{2} x^{2} + 1\right )}}\,{d x} + \frac{1}{2} \, a^{2} e x^{2} + \frac{b^{2} d \arctan \left (c x\right )^{3}}{4 \, c} - 4 \, b^{2} c e \int \frac{x^{2} \arctan \left (c x\right )}{16 \,{\left (c^{2} x^{2} + 1\right )}}\,{d x} - 8 \, b^{2} c d \int \frac{x \arctan \left (c x\right )}{16 \,{\left (c^{2} x^{2} + 1\right )}}\,{d x} +{\left (x^{2} \arctan \left (c x\right ) - c{\left (\frac{x}{c^{2}} - \frac{\arctan \left (c x\right )}{c^{3}}\right )}\right )} a b e + a^{2} d x + 12 \, b^{2} e \int \frac{x \arctan \left (c x\right )^{2}}{16 \,{\left (c^{2} x^{2} + 1\right )}}\,{d x} + b^{2} e \int \frac{x \log \left (c^{2} x^{2} + 1\right )^{2}}{16 \,{\left (c^{2} x^{2} + 1\right )}}\,{d x} + b^{2} d \int \frac{\log \left (c^{2} x^{2} + 1\right )^{2}}{16 \,{\left (c^{2} x^{2} + 1\right )}}\,{d x} + \frac{{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} a b d}{c} + \frac{1}{8} \,{\left (b^{2} e x^{2} + 2 \, b^{2} d x\right )} \arctan \left (c x\right )^{2} - \frac{1}{32} \,{\left (b^{2} e x^{2} + 2 \, b^{2} d x\right )} \log \left (c^{2} x^{2} + 1\right )^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

12*b^2*c^2*e*integrate(1/16*x^3*arctan(c*x)^2/(c^2*x^2 + 1), x) + b^2*c^2*e*integrate(1/16*x^3*log(c^2*x^2 + 1
)^2/(c^2*x^2 + 1), x) + 12*b^2*c^2*d*integrate(1/16*x^2*arctan(c*x)^2/(c^2*x^2 + 1), x) + 2*b^2*c^2*e*integrat
e(1/16*x^3*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) + b^2*c^2*d*integrate(1/16*x^2*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1),
x) + 4*b^2*c^2*d*integrate(1/16*x^2*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) + 1/2*a^2*e*x^2 + 1/4*b^2*d*arctan(c*x
)^3/c - 4*b^2*c*e*integrate(1/16*x^2*arctan(c*x)/(c^2*x^2 + 1), x) - 8*b^2*c*d*integrate(1/16*x*arctan(c*x)/(c
^2*x^2 + 1), x) + (x^2*arctan(c*x) - c*(x/c^2 - arctan(c*x)/c^3))*a*b*e + a^2*d*x + 12*b^2*e*integrate(1/16*x*
arctan(c*x)^2/(c^2*x^2 + 1), x) + b^2*e*integrate(1/16*x*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + b^2*d*integrat
e(1/16*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + (2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*a*b*d/c + 1/8*(b^2*e*x^2
+ 2*b^2*d*x)*arctan(c*x)^2 - 1/32*(b^2*e*x^2 + 2*b^2*d*x)*log(c^2*x^2 + 1)^2

________________________________________________________________________________________

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

[Out]

integral(a^2*e*x + a^2*d + (b^2*e*x + b^2*d)*arctan(c*x)^2 + 2*(a*b*e*x + a*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 )^{2} \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))**2,x)

[Out]

Integral((a + b*atan(c*x))**2*(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 )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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