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

Optimal. Leaf size=95 \[ \frac{e (c+d x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d}+\frac{e \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d}-a b e x+\frac{b^2 e \log \left ((c+d x)^2+1\right )}{2 d}-\frac{b^2 e (c+d x) \tan ^{-1}(c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.119431, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5043, 12, 4852, 4916, 4846, 260, 4884} \[ \frac{e (c+d x)^2 \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d}+\frac{e \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d}-a b e x+\frac{b^2 e \log \left ((c+d x)^2+1\right )}{2 d}-\frac{b^2 e (c+d x) \tan ^{-1}(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5043

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((f*x)/d)^m*(a + b*ArcTan[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f, 0
] && IGtQ[p, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 4916

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2/
e, Int[(f*x)^(m - 2)*(a + b*ArcTan[c*x])^p, x], x] - Dist[(d*f^2)/e, Int[((f*x)^(m - 2)*(a + b*ArcTan[c*x])^p)
/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 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 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]

Rubi steps

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

Mathematica [A]  time = 0.0576755, size = 107, normalized size = 1.13 \[ \frac{e \left (2 b \tan ^{-1}(c+d x) \left (a \left (c^2+2 c d x+d^2 x^2+1\right )-b (c+d x)\right )+a (c+d x) (a c+a d x-2 b)+b^2 \left (c^2+2 c d x+d^2 x^2+1\right ) \tan ^{-1}(c+d x)^2+b^2 \log \left ((c+d x)^2+1\right )\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.046, size = 220, normalized size = 2.3 \begin{align*}{\frac{{a}^{2}{x}^{2}de}{2}}+x{a}^{2}ce+{\frac{{a}^{2}{c}^{2}e}{2\,d}}+{\frac{d \left ( \arctan \left ( dx+c \right ) \right ) ^{2}{x}^{2}{b}^{2}e}{2}}+ \left ( \arctan \left ( dx+c \right ) \right ) ^{2}x{b}^{2}ce+{\frac{ \left ( \arctan \left ( dx+c \right ) \right ) ^{2}{b}^{2}{c}^{2}e}{2\,d}}+{\frac{e{b}^{2} \left ( \arctan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-\arctan \left ( dx+c \right ) x{b}^{2}e-{\frac{\arctan \left ( dx+c \right ){b}^{2}ce}{d}}+{\frac{e{b}^{2}\ln \left ( 1+ \left ( dx+c \right ) ^{2} \right ) }{2\,d}}+d\arctan \left ( dx+c \right ){x}^{2}abe+2\,\arctan \left ( dx+c \right ) xabce+{\frac{\arctan \left ( dx+c \right ) ab{c}^{2}e}{d}}+{\frac{eab\arctan \left ( dx+c \right ) }{d}}-abex-{\frac{abce}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 5.20193, size = 294, normalized size = 3.09 \begin{align*} \frac{1}{2} \, a^{2} d e x^{2} +{\left (x^{2} \arctan \left (d x + c\right ) - d{\left (\frac{x}{d^{2}} + \frac{{\left (c^{2} - 1\right )} \arctan \left (\frac{d^{2} x + c d}{d}\right )}{d^{3}} - \frac{c \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{3}}\right )}\right )} a b d e + a^{2} c e x + \frac{{\left (2 \,{\left (d x + c\right )} \arctan \left (d x + c\right ) - \log \left ({\left (d x + c\right )}^{2} + 1\right )\right )} a b c e}{d} + \frac{b^{2} e \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right ) +{\left (b^{2} d^{2} e x^{2} + 2 \, b^{2} c d e x +{\left (b^{2} c^{2} + b^{2}\right )} e\right )} \arctan \left (d x + c\right )^{2} - 2 \,{\left (b^{2} d e x + b^{2} c e\right )} \arctan \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.72326, size = 335, normalized size = 3.53 \begin{align*} \frac{a^{2} d^{2} e x^{2} + 2 \,{\left (a^{2} c - a b\right )} d e x + b^{2} e \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right ) +{\left (b^{2} d^{2} e x^{2} + 2 \, b^{2} c d e x +{\left (b^{2} c^{2} + b^{2}\right )} e\right )} \arctan \left (d x + c\right )^{2} + 2 \,{\left (a b d^{2} e x^{2} +{\left (2 \, a b c - b^{2}\right )} d e x +{\left (a b c^{2} - b^{2} c + a b\right )} e\right )} \arctan \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.04498, size = 240, normalized size = 2.53 \begin{align*} \begin{cases} a^{2} c e x + \frac{a^{2} d e x^{2}}{2} + \frac{a b c^{2} e \operatorname{atan}{\left (c + d x \right )}}{d} + 2 a b c e x \operatorname{atan}{\left (c + d x \right )} + a b d e x^{2} \operatorname{atan}{\left (c + d x \right )} - a b e x + \frac{a b e \operatorname{atan}{\left (c + d x \right )}}{d} + \frac{b^{2} c^{2} e \operatorname{atan}^{2}{\left (c + d x \right )}}{2 d} + b^{2} c e x \operatorname{atan}^{2}{\left (c + d x \right )} - \frac{b^{2} c e \operatorname{atan}{\left (c + d x \right )}}{d} + \frac{b^{2} d e x^{2} \operatorname{atan}^{2}{\left (c + d x \right )}}{2} - b^{2} e x \operatorname{atan}{\left (c + d x \right )} + \frac{b^{2} e \log{\left (\frac{c^{2}}{d^{2}} + \frac{2 c x}{d} + x^{2} + \frac{1}{d^{2}} \right )}}{2 d} + \frac{b^{2} e \operatorname{atan}^{2}{\left (c + d x \right )}}{2 d} & \text{for}\: d \neq 0 \\c e x \left (a + b \operatorname{atan}{\left (c \right )}\right )^{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.23824, size = 294, normalized size = 3.09 \begin{align*} \frac{b^{2} d^{2} x^{2} \arctan \left (d x + c\right )^{2} e + 2 \, a b d^{2} x^{2} \arctan \left (d x + c\right ) e + 2 \, b^{2} c d x \arctan \left (d x + c\right )^{2} e + a^{2} d^{2} x^{2} e + 4 \, a b c d x \arctan \left (d x + c\right ) e + b^{2} c^{2} \arctan \left (d x + c\right )^{2} e + 2 \, a^{2} c d x e + 2 \, a b c^{2} \arctan \left (d x + c\right ) e - 2 \, b^{2} d x \arctan \left (d x + c\right ) e - 2 \, a b d x e - 2 \, b^{2} c \arctan \left (d x + c\right ) e + b^{2} \arctan \left (d x + c\right )^{2} e + 2 \, a b \arctan \left (d x + c\right ) e + b^{2} e \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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