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

Optimal. Leaf size=222 \[ \frac {i (d e-c f) \left (a+b \tan ^{-1}(c+d x)\right )^2}{d^2}-\frac {(-c f+d e+f) (d e-(c+1) f) \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d^2 f}+\frac {2 b (d e-c f) \log \left (\frac {2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d^2}+\frac {(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 f}-\frac {a b f x}{d}+\frac {i b^2 (d e-c f) \text {Li}_2\left (1-\frac {2}{i (c+d x)+1}\right )}{d^2}+\frac {b^2 f \log \left ((c+d x)^2+1\right )}{2 d^2}-\frac {b^2 f (c+d x) \tan ^{-1}(c+d x)}{d^2} \]

[Out]

-a*b*f*x/d-b^2*f*(d*x+c)*arctan(d*x+c)/d^2+I*(-c*f+d*e)*(a+b*arctan(d*x+c))^2/d^2-1/2*(-c*f+d*e+f)*(d*e-(1+c)*
f)*(a+b*arctan(d*x+c))^2/d^2/f+1/2*(f*x+e)^2*(a+b*arctan(d*x+c))^2/f+2*b*(-c*f+d*e)*(a+b*arctan(d*x+c))*ln(2/(
1+I*(d*x+c)))/d^2+1/2*b^2*f*ln(1+(d*x+c)^2)/d^2+I*b^2*(-c*f+d*e)*polylog(2,1-2/(1+I*(d*x+c)))/d^2

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Rubi [A]  time = 0.37, antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {5047, 4864, 4846, 260, 4984, 4884, 4920, 4854, 2402, 2315} \[ \frac {i b^2 (d e-c f) \text {PolyLog}\left (2,1-\frac {2}{1+i (c+d x)}\right )}{d^2}+\frac {i (d e-c f) \left (a+b \tan ^{-1}(c+d x)\right )^2}{d^2}-\frac {(-c f+d e+f) (d e-(c+1) f) \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d^2 f}+\frac {2 b (d e-c f) \log \left (\frac {2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d^2}+\frac {(e+f x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 f}-\frac {a b f x}{d}+\frac {b^2 f \log \left ((c+d x)^2+1\right )}{2 d^2}-\frac {b^2 f (c+d x) \tan ^{-1}(c+d x)}{d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 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 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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.45, size = 264, normalized size = 1.19 \[ \frac {-a^2 c^2 f+2 a^2 c d e+2 a^2 d^2 e x+a^2 d^2 f x^2-2 b \tan ^{-1}(c+d x) \left (a \left (c^2 f-2 c d e-2 d^2 e x-f \left (d^2 x^2+1\right )\right )-2 b (d e-c f) \log \left (1+e^{2 i \tan ^{-1}(c+d x)}\right )+b f (c+d x)\right )+4 a b d e \log \left (\frac {1}{\sqrt {(c+d x)^2+1}}\right )-4 a b c f \log \left (\frac {1}{\sqrt {(c+d x)^2+1}}\right )-2 a b c f-2 a b d f x-2 i b^2 (d e-c f) \text {Li}_2\left (-e^{2 i \tan ^{-1}(c+d x)}\right )+b^2 (c+d x-i) \tan ^{-1}(c+d x)^2 (-c f+2 d e+d f x+i f)-2 b^2 f \log \left (\frac {1}{\sqrt {(c+d x)^2+1}}\right )}{2 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*a^2*c*d*e - 2*a*b*c*f - a^2*c^2*f + 2*a^2*d^2*e*x - 2*a*b*d*f*x + a^2*d^2*f*x^2 + b^2*(-I + c + d*x)*(2*d*e
+ I*f - c*f + d*f*x)*ArcTan[c + d*x]^2 - 2*b*ArcTan[c + d*x]*(b*f*(c + d*x) + a*(-2*c*d*e + c^2*f - 2*d^2*e*x
- f*(1 + d^2*x^2)) - 2*b*(d*e - c*f)*Log[1 + E^((2*I)*ArcTan[c + d*x])]) + 4*a*b*d*e*Log[1/Sqrt[1 + (c + d*x)
^2]] - 2*b^2*f*Log[1/Sqrt[1 + (c + d*x)^2]] - 4*a*b*c*f*Log[1/Sqrt[1 + (c + d*x)^2]] - (2*I)*b^2*(d*e - c*f)*P
olyLog[2, -E^((2*I)*ArcTan[c + d*x])])/(2*d^2)

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (a^{2} f x + a^{2} e + {\left (b^{2} f x + b^{2} e\right )} \arctan \left (d x + c\right )^{2} + 2 \, {\left (a b f x + a b e\right )} \arctan \left (d x + c\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [B]  time = 0.14, size = 748, normalized size = 3.37 \[ -\frac {i b^{2} \ln \left (d x +c -i\right )^{2} c f}{4 d^{2}}+\frac {i b^{2} \ln \left (d x +c +i\right )^{2} c f}{4 d^{2}}-\frac {i b^{2} \ln \left (1+\left (d x +c \right )^{2}\right ) \ln \left (d x +c -i\right ) e}{2 d}-\frac {i b^{2} \ln \left (d x +c +i\right ) \ln \left (\frac {i \left (d x +c -i\right )}{2}\right ) e}{2 d}+\frac {i b^{2} \ln \left (1+\left (d x +c \right )^{2}\right ) \ln \left (d x +c +i\right ) e}{2 d}+\frac {i b^{2} \ln \left (-\frac {i \left (d x +c +i\right )}{2}\right ) \ln \left (d x +c -i\right ) e}{2 d}-\frac {i b^{2} \dilog \left (-\frac {i \left (d x +c +i\right )}{2}\right ) c f}{2 d^{2}}+\frac {i b^{2} \dilog \left (\frac {i \left (d x +c -i\right )}{2}\right ) c f}{2 d^{2}}-\frac {i b^{2} \ln \left (1+\left (d x +c \right )^{2}\right ) \ln \left (d x +c +i\right ) c f}{2 d^{2}}+\frac {i b^{2} \ln \left (d x +c +i\right ) \ln \left (\frac {i \left (d x +c -i\right )}{2}\right ) c f}{2 d^{2}}+\frac {i b^{2} \ln \left (1+\left (d x +c \right )^{2}\right ) \ln \left (d x +c -i\right ) c f}{2 d^{2}}-\frac {i b^{2} \ln \left (-\frac {i \left (d x +c +i\right )}{2}\right ) \ln \left (d x +c -i\right ) c f}{2 d^{2}}+\frac {i b^{2} \ln \left (d x +c -i\right )^{2} e}{4 d}-\frac {i b^{2} \dilog \left (\frac {i \left (d x +c -i\right )}{2}\right ) e}{2 d}+\frac {i b^{2} \dilog \left (-\frac {i \left (d x +c +i\right )}{2}\right ) e}{2 d}-\frac {i b^{2} \ln \left (d x +c +i\right )^{2} e}{4 d}-\frac {a b f x}{d}+\frac {b^{2} f \ln \left (1+\left (d x +c \right )^{2}\right )}{2 d^{2}}+\frac {b^{2} \arctan \left (d x +c \right )^{2} f}{2 d^{2}}+\frac {b^{2} \arctan \left (d x +c \right )^{2} f \,x^{2}}{2}+\arctan \left (d x +c \right )^{2} x \,b^{2} e +\frac {a^{2} c e}{d}-\frac {a^{2} f \,c^{2}}{2 d^{2}}+\frac {a b \ln \left (1+\left (d x +c \right )^{2}\right ) c f}{d^{2}}+\frac {2 \arctan \left (d x +c \right ) a b c e}{d}+\frac {b^{2} \ln \left (1+\left (d x +c \right )^{2}\right ) \arctan \left (d x +c \right ) c f}{d^{2}}-\frac {a b f \arctan \left (d x +c \right ) c^{2}}{d^{2}}+\frac {a b f \arctan \left (d x +c \right )}{d^{2}}+\frac {\arctan \left (d x +c \right )^{2} b^{2} c e}{d}-\frac {b^{2} \arctan \left (d x +c \right )^{2} f \,c^{2}}{2 d^{2}}-\frac {b^{2} \arctan \left (d x +c \right ) f x}{d}-\frac {b^{2} \arctan \left (d x +c \right ) f c}{d^{2}}+2 \arctan \left (d x +c \right ) x a b e +a b f \arctan \left (d x +c \right ) x^{2}-\frac {b^{2} \ln \left (1+\left (d x +c \right )^{2}\right ) \arctan \left (d x +c \right ) e}{d}-\frac {a b \ln \left (1+\left (d x +c \right )^{2}\right ) e}{d}-\frac {a b c f}{d^{2}}+\frac {a^{2} x^{2} f}{2}+a^{2} x e \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*b^2*c^2*e*arctan(d*x + c)^2*arctan((d^2*x + c*d)/d)/d - 1/4*(3*arctan(d*x + c)*arctan((d^2*x + c*d)/d)^2/d
- arctan((d^2*x + c*d)/d)^3/d)*b^2*c^2*e + 12*b^2*d^2*f*integrate(1/16*x^3*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d
*x + c^2 + 1), x) + b^2*d^2*f*integrate(1/16*x^3*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^2*d^2*e*integrate(1/16*x^2*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 24*b^2*c*d*f*i
ntegrate(1/16*x^2*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 2*b^2*d^2*f*integrate(1/16*x^3*log(d^2
*x^2 + 2*c*d*x + c^2 + 1)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + b^2*d^2*e*integrate(1/16*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) + 2*b^2*c*d*f*integrate(1/16*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^2*c*d*e*integrate(1/16*x*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x
+ c^2 + 1), x) + 12*b^2*c^2*f*integrate(1/16*x*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 4*b^2*d^
2*e*integrate(1/16*x^2*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 2*b^2*c*d*f*integr
ate(1/16*x^2*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 2*b^2*c*d*e*integrate(1/16*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) + b^2*c^2*f*integrate(1/16*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) + 4*b^2*c*d*e*integrate(1/16*x*log(d^2*x^2 + 2*c*d*x
+ c^2 + 1)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + b^2*c^2*e*integrate(1/16*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^2*f*x^2 + 3/4*b^2*e*arctan(d*x + c)^2*arctan((d^2*x + c*d)/d)/d - 4*b^
2*d*f*integrate(1/16*x^2*arctan(d*x + c)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) - 8*b^2*d*e*integrate(1/16*x*arctan
(d*x + c)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) - 1/4*(3*arctan(d*x + c)*arctan((d^2*x + c*d)/d)^2/d - arctan((d^2
*x + c*d)/d)^3/d)*b^2*e + (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*b*f + a^2*e*x + 12*b^2*f*integrate(1/16*x*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*
x + c^2 + 1), x) + b^2*f*integrate(1/16*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)
+ b^2*e*integrate(1/16*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) + (2*(d*x + c)*ar
ctan(d*x + c) - log((d*x + c)^2 + 1))*a*b*e/d + 1/8*(b^2*f*x^2 + 2*b^2*e*x)*arctan(d*x + c)^2 - 1/32*(b^2*f*x^
2 + 2*b^2*e*x)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (e+f\,x\right )\,{\left (a+b\,\mathrm {atan}\left (c+d\,x\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \operatorname {atan}{\left (c + d x \right )}\right )^{2} \left (e + f x\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*atan(c + d*x))**2*(e + f*x), x)

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