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

Optimal. Leaf size=382 \[ \frac {i \left (-\left (1-3 c^2\right ) f^2-6 c d e f+3 d^2 e^2\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d^3}-\frac {(d e-c f) \left (-\left (3-c^2\right ) f^2-2 c d e f+d^2 e^2\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d^3 f}+\frac {2 b \left (-\left (1-3 c^2\right ) f^2-6 c d e f+3 d^2 e^2\right ) \log \left (\frac {2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{3 d^3}-\frac {b f^2 (c+d x)^2 \left (a+b \tan ^{-1}(c+d x)\right )}{3 d^3}-\frac {2 a b f x (d e-c f)}{d^2}+\frac {(e+f x)^3 \left (a+b \tan ^{-1}(c+d x)\right )^2}{3 f}+\frac {i b^2 \left (-\left (1-3 c^2\right ) f^2-6 c d e f+3 d^2 e^2\right ) \text {Li}_2\left (1-\frac {2}{i (c+d x)+1}\right )}{3 d^3}+\frac {b^2 f (d e-c f) \log \left ((c+d x)^2+1\right )}{d^3}-\frac {2 b^2 f (c+d x) (d e-c f) \tan ^{-1}(c+d x)}{d^3}-\frac {b^2 f^2 \tan ^{-1}(c+d x)}{3 d^3}+\frac {b^2 f^2 x}{3 d^2} \]

[Out]

1/3*b^2*f^2*x/d^2-2*a*b*f*(-c*f+d*e)*x/d^2-1/3*b^2*f^2*arctan(d*x+c)/d^3-2*b^2*f*(-c*f+d*e)*(d*x+c)*arctan(d*x
+c)/d^3-1/3*b*f^2*(d*x+c)^2*(a+b*arctan(d*x+c))/d^3+1/3*I*(3*d^2*e^2-6*c*d*e*f-(-3*c^2+1)*f^2)*(a+b*arctan(d*x
+c))^2/d^3-1/3*(-c*f+d*e)*(d^2*e^2-2*c*d*e*f-(-c^2+3)*f^2)*(a+b*arctan(d*x+c))^2/d^3/f+1/3*(f*x+e)^3*(a+b*arct
an(d*x+c))^2/f+2/3*b*(3*d^2*e^2-6*c*d*e*f-(-3*c^2+1)*f^2)*(a+b*arctan(d*x+c))*ln(2/(1+I*(d*x+c)))/d^3+b^2*f*(-
c*f+d*e)*ln(1+(d*x+c)^2)/d^3+1/3*I*b^2*(3*d^2*e^2-6*c*d*e*f-(-3*c^2+1)*f^2)*polylog(2,1-2/(1+I*(d*x+c)))/d^3

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Rubi [A]  time = 0.57, antiderivative size = 382, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.650, Rules used = {5047, 4864, 4846, 260, 4852, 321, 203, 4984, 4884, 4920, 4854, 2402, 2315} \[ \frac {i b^2 \left (-\left (1-3 c^2\right ) f^2-6 c d e f+3 d^2 e^2\right ) \text {PolyLog}\left (2,1-\frac {2}{1+i (c+d x)}\right )}{3 d^3}+\frac {i \left (-\left (1-3 c^2\right ) f^2-6 c d e f+3 d^2 e^2\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d^3}-\frac {(d e-c f) \left (-\left (3-c^2\right ) f^2-2 c d e f+d^2 e^2\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d^3 f}+\frac {2 b \left (-\left (1-3 c^2\right ) f^2-6 c d e f+3 d^2 e^2\right ) \log \left (\frac {2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{3 d^3}-\frac {2 a b f x (d e-c f)}{d^2}-\frac {b f^2 (c+d x)^2 \left (a+b \tan ^{-1}(c+d x)\right )}{3 d^3}+\frac {(e+f x)^3 \left (a+b \tan ^{-1}(c+d x)\right )^2}{3 f}+\frac {b^2 f (d e-c f) \log \left ((c+d x)^2+1\right )}{d^3}-\frac {2 b^2 f (c+d x) (d e-c f) \tan ^{-1}(c+d x)}{d^3}-\frac {b^2 f^2 \tan ^{-1}(c+d x)}{3 d^3}+\frac {b^2 f^2 x}{3 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 4852

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

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

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Mathematica [B]  time = 4.23, size = 801, normalized size = 2.10 \[ \frac {1}{3} a^2 f^2 x^3+a^2 e f x^2+a^2 e^2 x+\frac {a b \left (-d f x (6 d e-4 c f+d f x)+2 \left (f^2 c^3-3 d e f c^2+3 \left (d^2 e^2-f^2\right ) c+3 d e f+d^3 x \left (3 e^2+3 f x e+f^2 x^2\right )\right ) \tan ^{-1}(c+d x)+\left (-3 d^2 e^2+6 c d f e+\left (1-3 c^2\right ) f^2\right ) \log \left ((c+d x)^2+1\right )\right )}{3 d^3}+\frac {b^2 e^2 \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 {Li}_2\left (-e^{2 i \tan ^{-1}(c+d x)}\right )\right )}{d}+\frac {b^2 e f \left (\left (-c^2+2 i c+d^2 x^2+1\right ) \tan ^{-1}(c+d x)^2-2 \left (2 \log \left (1+e^{2 i \tan ^{-1}(c+d x)}\right ) c+c+d x\right ) \tan ^{-1}(c+d x)+\log \left ((c+d x)^2+1\right )+2 i c \text {Li}_2\left (-e^{2 i \tan ^{-1}(c+d x)}\right )\right )}{d^2}+\frac {b^2 f^2 \left ((c+d x)^2+1\right )^{3/2} \left (\frac {3 (c+d x) \tan ^{-1}(c+d x)^2 c^2}{\sqrt {(c+d x)^2+1}}-3 i \tan ^{-1}(c+d x)^2 \cos \left (3 \tan ^{-1}(c+d x)\right ) c^2+6 \tan ^{-1}(c+d x) \cos \left (3 \tan ^{-1}(c+d x)\right ) \log \left (1+e^{2 i \tan ^{-1}(c+d x)}\right ) c^2+3 \tan ^{-1}(c+d x)^2 \sin \left (3 \tan ^{-1}(c+d x)\right ) c^2+\frac {6 (c+d x) \tan ^{-1}(c+d x) c}{\sqrt {(c+d x)^2+1}}+6 \cos \left (3 \tan ^{-1}(c+d x)\right ) \log \left (\frac {1}{\sqrt {(c+d x)^2+1}}\right ) c+6 \tan ^{-1}(c+d x) \sin \left (3 \tan ^{-1}(c+d x)\right ) c+\frac {3 (c+d x) \tan ^{-1}(c+d x)^2}{\sqrt {(c+d x)^2+1}}+i \tan ^{-1}(c+d x)^2 \cos \left (3 \tan ^{-1}(c+d x)\right )-2 \tan ^{-1}(c+d x) \cos \left (3 \tan ^{-1}(c+d x)\right ) \log \left (1+e^{2 i \tan ^{-1}(c+d x)}\right )+\frac {\left (-9 i c^2-12 c+3 i\right ) \tan ^{-1}(c+d x)^2+2 \left (\left (9 c^2-3\right ) \log \left (1+e^{2 i \tan ^{-1}(c+d x)}\right )-2\right ) \tan ^{-1}(c+d x)+18 c \log \left (\frac {1}{\sqrt {(c+d x)^2+1}}\right )}{\sqrt {(c+d x)^2+1}}-\frac {4 i \left (3 c^2-1\right ) \text {Li}_2\left (-e^{2 i \tan ^{-1}(c+d x)}\right )}{\left ((c+d x)^2+1\right )^{3/2}}-\tan ^{-1}(c+d x)^2 \sin \left (3 \tan ^{-1}(c+d x)\right )+\sin \left (3 \tan ^{-1}(c+d x)\right )+\frac {c+d x}{\sqrt {(c+d x)^2+1}}\right )}{12 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

a^2*e^2*x + a^2*e*f*x^2 + (a^2*f^2*x^3)/3 + (a*b*(-(d*f*x*(6*d*e - 4*c*f + d*f*x)) + 2*(3*d*e*f - 3*c^2*d*e*f
+ c^3*f^2 + 3*c*(d^2*e^2 - f^2) + d^3*x*(3*e^2 + 3*e*f*x + f^2*x^2))*ArcTan[c + d*x] + (-3*d^2*e^2 + 6*c*d*e*f
+ (1 - 3*c^2)*f^2)*Log[1 + (c + d*x)^2]))/(3*d^3) + (b^2*e^2*(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])]))/d + (b^2*e*f*((1 + (2*I)
*c - c^2 + d^2*x^2)*ArcTan[c + d*x]^2 - 2*ArcTan[c + d*x]*(c + d*x + 2*c*Log[1 + E^((2*I)*ArcTan[c + d*x])]) +
Log[1 + (c + d*x)^2] + (2*I)*c*PolyLog[2, -E^((2*I)*ArcTan[c + d*x])]))/d^2 + (b^2*f^2*(1 + (c + d*x)^2)^(3/2
)*((c + d*x)/Sqrt[1 + (c + d*x)^2] + (6*c*(c + d*x)*ArcTan[c + d*x])/Sqrt[1 + (c + d*x)^2] + (3*(c + d*x)*ArcT
an[c + d*x]^2)/Sqrt[1 + (c + d*x)^2] + (3*c^2*(c + d*x)*ArcTan[c + d*x]^2)/Sqrt[1 + (c + d*x)^2] + I*ArcTan[c
+ d*x]^2*Cos[3*ArcTan[c + d*x]] - (3*I)*c^2*ArcTan[c + d*x]^2*Cos[3*ArcTan[c + d*x]] - 2*ArcTan[c + d*x]*Cos[3
*ArcTan[c + d*x]]*Log[1 + E^((2*I)*ArcTan[c + d*x])] + 6*c^2*ArcTan[c + d*x]*Cos[3*ArcTan[c + d*x]]*Log[1 + E^
((2*I)*ArcTan[c + d*x])] + 6*c*Cos[3*ArcTan[c + d*x]]*Log[1/Sqrt[1 + (c + d*x)^2]] + ((3*I - 12*c - (9*I)*c^2)
*ArcTan[c + d*x]^2 + 2*ArcTan[c + d*x]*(-2 + (-3 + 9*c^2)*Log[1 + E^((2*I)*ArcTan[c + d*x])]) + 18*c*Log[1/Sqr
t[1 + (c + d*x)^2]])/Sqrt[1 + (c + d*x)^2] - ((4*I)*(-1 + 3*c^2)*PolyLog[2, -E^((2*I)*ArcTan[c + d*x])])/(1 +
(c + d*x)^2)^(3/2) + Sin[3*ArcTan[c + d*x]] + 6*c*ArcTan[c + d*x]*Sin[3*ArcTan[c + d*x]] - ArcTan[c + d*x]^2*S
in[3*ArcTan[c + d*x]] + 3*c^2*ArcTan[c + d*x]^2*Sin[3*ArcTan[c + d*x]]))/(12*d^3)

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (a^{2} f^{2} x^{2} + 2 \, a^{2} e f x + a^{2} e^{2} + {\left (b^{2} f^{2} x^{2} + 2 \, b^{2} e f x + b^{2} e^{2}\right )} \arctan \left (d x + c\right )^{2} + 2 \, {\left (a b f^{2} x^{2} + 2 \, a b e f x + a b e^{2}\right )} \arctan \left (d x + c\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

sage0*x

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maple [B]  time = 0.15, size = 1622, normalized size = 4.25 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*x*e^2+1/3/d^3*a*b*f^2*ln(1+(d*x+c)^2)+5/3/d^3*a*b*f^2*c^2+1/3*a^2/f*e^3+1/3/d^3*b^2*f^2*c+a^2*f*x^2*e+1/3*
b^2*f^2*arctan(d*x+c)^2*x^3+arctan(d*x+c)^2*x*b^2*e^2-1/2*I/d^3*b^2*dilog(1/2*I*(d*x+c-I))*c^2*f^2+1/6*I/d^3*b
^2*ln(1+(d*x+c)^2)*ln(d*x+c-I)*f^2-1/6*I/d^3*b^2*ln(1+(d*x+c)^2)*ln(I+d*x+c)*f^2+2/d*arctan(d*x+c)*a*b*c*e^2-1
/d^3*a*b*f^2*ln(1+(d*x+c)^2)*c^2-2/d^3*a*b*f^2*arctan(d*x+c)*c-2/d*b^2*f*arctan(d*x+c)*e*x+2/3/d^3*a*b*f^2*arc
tan(d*x+c)*c^3-1/d^2*b^2*f*arctan(d*x+c)^2*c^2*e+1/3*a^2*f^2*x^3+4/3*a*b/d^2*f^2*c*x-2*a*b/d*f*e*x+1/6*I/d^3*b
^2*ln(I+d*x+c)*ln(1/2*I*(d*x+c-I))*f^2-1/4*I/d^3*b^2*ln(I+d*x+c)^2*c^2*f^2+1/4*I/d^3*b^2*ln(d*x+c-I)^2*c^2*f^2
-2/d^2*b^2*f*arctan(d*x+c)*e*c+4/3/d^2*b^2*f^2*arctan(d*x+c)*c*x+2*a*b*f*arctan(d*x+c)*e*x^2-1/d^3*b^2*f^2*arc
tan(d*x+c)*ln(1+(d*x+c)^2)*c^2+2/d^2*a*b*f*arctan(d*x+c)*e-1/6*I/d^3*b^2*ln(d*x+c-I)*ln(-1/2*I*(I+d*x+c))*f^2+
1/2*I/d^3*b^2*dilog(-1/2*I*(I+d*x+c))*c^2*f^2+1/2*I/d*b^2*ln(1+(d*x+c)^2)*ln(I+d*x+c)*e^2-1/2*I/d*b^2*ln(I+d*x
+c)*ln(1/2*I*(d*x+c-I))*e^2-1/2*I/d*b^2*ln(1+(d*x+c)^2)*ln(d*x+c-I)*e^2+1/2*I/d*b^2*ln(d*x+c-I)*ln(-1/2*I*(I+d
*x+c))*e^2+1/3*b^2*f^2*x/d^2-1/3*b^2*f^2*arctan(d*x+c)/d^3-I/d^2*b^2*ln(d*x+c-I)*ln(-1/2*I*(I+d*x+c))*c*e*f-I/
d^2*b^2*ln(1+(d*x+c)^2)*ln(I+d*x+c)*c*e*f+I/d^2*b^2*ln(I+d*x+c)*ln(1/2*I*(d*x+c-I))*c*e*f+I/d^2*b^2*ln(1+(d*x+
c)^2)*ln(d*x+c-I)*c*e*f-1/d^3*b^2*f^2*arctan(d*x+c)^2*c+1/d*arctan(d*x+c)^2*b^2*c*e^2+1/3/d^3*b^2*f^2*arctan(d
*x+c)*ln(1+(d*x+c)^2)+1/d^2*b^2*f*ln(1+(d*x+c)^2)*e+1/d^2*b^2*f*arctan(d*x+c)^2*e-1/2*I/d^3*b^2*ln(1+(d*x+c)^2
)*ln(d*x+c-I)*c^2*f^2+1/2*I/d^3*b^2*ln(d*x+c-I)*ln(-1/2*I*(I+d*x+c))*c^2*f^2-1/2*I/d^2*b^2*ln(d*x+c-I)^2*c*e*f
+1/2*I/d^3*b^2*ln(1+(d*x+c)^2)*ln(I+d*x+c)*c^2*f^2-I/d^2*b^2*dilog(-1/2*I*(I+d*x+c))*c*e*f+I/d^2*b^2*dilog(1/2
*I*(d*x+c-I))*c*e*f+1/2*I/d^2*b^2*ln(I+d*x+c)^2*c*e*f-1/2*I/d^3*b^2*ln(I+d*x+c)*ln(1/2*I*(d*x+c-I))*c^2*f^2+1/
3/d^3*b^2*f^2*arctan(d*x+c)^2*c^3+5/3/d^3*b^2*f^2*arctan(d*x+c)*c^2-1/3/d*b^2*f^2*arctan(d*x+c)*x^2-1/d^3*b^2*
f^2*ln(1+(d*x+c)^2)*c+b^2*f*arctan(d*x+c)^2*e*x^2-1/3/d*a*b*f^2*x^2+2/3*a*b*f^2*arctan(d*x+c)*x^3+2*arctan(d*x
+c)*x*a*b*e^2+1/6*I/d^3*b^2*dilog(1/2*I*(d*x+c-I))*f^2-1/6*I/d^3*b^2*dilog(-1/2*I*(I+d*x+c))*f^2+1/12*I/d^3*b^
2*ln(I+d*x+c)^2*f^2+1/2*I/d*b^2*dilog(-1/2*I*(I+d*x+c))*e^2-1/2*I/d*b^2*dilog(1/2*I*(d*x+c-I))*e^2+1/4*I/d*b^2
*ln(d*x+c-I)^2*e^2-1/4*I/d*b^2*ln(I+d*x+c)^2*e^2-1/12*I/d^3*b^2*ln(d*x+c-I)^2*f^2-1/d*e^2*a*b*ln(1+(d*x+c)^2)-
1/d*e^2*b^2*arctan(d*x+c)*ln(1+(d*x+c)^2)-2/d^2*a*b*f*c*e-2/d^2*a*b*f*arctan(d*x+c)*e*c^2+2/d^2*a*b*f*ln(1+(d*
x+c)^2)*c*e+2/d^2*b^2*f*arctan(d*x+c)*ln(1+(d*x+c)^2)*c*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)^2*(a+b*arctan(d*x+c))^2,x, algorithm="maxima")

[Out]

3/4*b^2*c^2*e^2*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^2 + 1/3*a^2*f^2*x^3 + 36*b^2*d^2*f^2*integrate(1/48*x^4*arctan(d*x
+ c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 3*b^2*d^2*f^2*integrate(1/48*x^4*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) + 72*b^2*d^2*e*f*integrate(1/48*x^3*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x +
c^2 + 1), x) + 72*b^2*c*d*f^2*integrate(1/48*x^3*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 4*b^2*
d^2*f^2*integrate(1/48*x^4*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^2*d^2*e*f*
integrate(1/48*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) + 6*b^2*c*d*f^2*integr
ate(1/48*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) + 36*b^2*d^2*e^2*integrate(1
/48*x^2*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 144*b^2*c*d*e*f*integrate(1/48*x^2*arctan(d*x +
c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 36*b^2*c^2*f^2*integrate(1/48*x^2*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*
x + c^2 + 1), x) + 12*b^2*d^2*e*f*integrate(1/48*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) + 4*b^2*c*d*f^2*integrate(1/48*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
) + 3*b^2*d^2*e^2*integrate(1/48*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) + 12
*b^2*c*d*e*f*integrate(1/48*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) + 3*b^2*c
^2*f^2*integrate(1/48*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) + 72*b^2*c*d*e^
2*integrate(1/48*x*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 72*b^2*c^2*e*f*integrate(1/48*x*arcta
n(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 12*b^2*d^2*e^2*integrate(1/48*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) + 12*b^2*c*d*e*f*integrate(1/48*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) + 6*b^2*c*d*e^2*integrate(1/48*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) + 6*b^2*c^2*e*f*integrate(1/48*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) + 12*b^2*c*d*e^2*integrate(1/48*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) + 3*b^2*c^2*e^2*integrate(1/48*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) +
a^2*e*f*x^2 + 3/4*b^2*e^2*arctan(d*x + c)^2*arctan((d^2*x + c*d)/d)/d - 8*b^2*d*f^2*integrate(1/48*x^3*arctan(
d*x + c)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) - 24*b^2*d*e*f*integrate(1/48*x^2*arctan(d*x + c)/(d^2*x^2 + 2*c*d*
x + c^2 + 1), x) - 24*b^2*d*e^2*integrate(1/48*x*arctan(d*x + c)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) - 1/4*(3*ar
ctan(d*x + c)*arctan((d^2*x + c*d)/d)^2/d - arctan((d^2*x + c*d)/d)^3/d)*b^2*e^2 + 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*b*e*f + 1/3*(2*x^3
*arctan(d*x + c) - d*((d*x^2 - 4*c*x)/d^3 - 2*(c^3 - 3*c)*arctan((d^2*x + c*d)/d)/d^4 + (3*c^2 - 1)*log(d^2*x^
2 + 2*c*d*x + c^2 + 1)/d^4))*a*b*f^2 + a^2*e^2*x + 36*b^2*f^2*integrate(1/48*x^2*arctan(d*x + c)^2/(d^2*x^2 +
2*c*d*x + c^2 + 1), x) + 3*b^2*f^2*integrate(1/48*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) + 72*b^2*e*f*integrate(1/48*x*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 6*b^2*e*f*int
egrate(1/48*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*b^2*e^2*integrate(1/48*
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)*arctan(d*x + c) - log((d*x
+ c)^2 + 1))*a*b*e^2/d + 1/12*(b^2*f^2*x^3 + 3*b^2*e*f*x^2 + 3*b^2*e^2*x)*arctan(d*x + c)^2 - 1/48*(b^2*f^2*x
^3 + 3*b^2*e*f*x^2 + 3*b^2*e^2*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 \[ \int {\left (e+f\,x\right )}^2\,{\left (a+b\,\mathrm {atan}\left (c+d\,x\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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