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

Optimal. Leaf size=48 \[ \frac{e (c+d x)^2 \left (a+b \tan ^{-1}(c+d x)\right )}{2 d}+\frac{b e \tan ^{-1}(c+d x)}{2 d}-\frac{b e x}{2} \]

[Out]

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

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Rubi [A]  time = 0.0302586, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {5043, 12, 4852, 321, 203} \[ \frac{e (c+d x)^2 \left (a+b \tan ^{-1}(c+d x)\right )}{2 d}+\frac{b e \tan ^{-1}(c+d x)}{2 d}-\frac{b e x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.0120132, size = 40, normalized size = 0.83 \[ \frac{e \left ((c+d x)^2 \left (a+b \tan ^{-1}(c+d x)\right )+b \left (\tan ^{-1}(c+d x)-d x\right )\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.036, size = 92, normalized size = 1.9 \begin{align*}{\frac{d{x}^{2}ae}{2}}+xace+{\frac{a{c}^{2}e}{2\,d}}+{\frac{d\arctan \left ( dx+c \right ){x}^{2}be}{2}}+\arctan \left ( dx+c \right ) xbce+{\frac{\arctan \left ( dx+c \right ) b{c}^{2}e}{2\,d}}-{\frac{bex}{2}}-{\frac{bce}{2\,d}}+{\frac{be\arctan \left ( dx+c \right ) }{2\,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)),x)

[Out]

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

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Maxima [B]  time = 1.49881, size = 162, normalized size = 3.38 \begin{align*} \frac{1}{2} \, a d e x^{2} + \frac{1}{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 )} b d e + a 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 )} b c e}{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)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.54816, size = 139, normalized size = 2.9 \begin{align*} \frac{a d^{2} e x^{2} +{\left (2 \, a c - b\right )} d e x +{\left (b d^{2} e x^{2} + 2 \, b c d e x +{\left (b c^{2} + 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)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.56704, size = 95, normalized size = 1.98 \begin{align*} \begin{cases} a c e x + \frac{a d e x^{2}}{2} + \frac{b c^{2} e \operatorname{atan}{\left (c + d x \right )}}{2 d} + b c e x \operatorname{atan}{\left (c + d x \right )} + \frac{b d e x^{2} \operatorname{atan}{\left (c + d x \right )}}{2} - \frac{b e x}{2} + \frac{b e \operatorname{atan}{\left (c + d x \right )}}{2 d} & \text{for}\: d \neq 0 \\c e x \left (a + b \operatorname{atan}{\left (c \right )}\right ) & \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)),x)

[Out]

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

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Giac [B]  time = 1.13934, size = 178, normalized size = 3.71 \begin{align*} \frac{2 \, b d^{2} x^{2} \arctan \left (d x + c\right ) e + 2 \, a d^{2} x^{2} e + 4 \, b c d x \arctan \left (d x + c\right ) e + \pi b c^{2} e \mathrm{sgn}\left (d x + c\right ) - \pi b c^{2} e + 4 \, a c d x e - 2 \, b c^{2} \arctan \left (\frac{1}{d x + c}\right ) e - 2 \, b d x e + \pi b e \mathrm{sgn}\left (d x + c\right ) - \pi b e - 2 \, b \arctan \left (\frac{1}{d x + c}\right ) e}{4 \, 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)),x, algorithm="giac")

[Out]

1/4*(2*b*d^2*x^2*arctan(d*x + c)*e + 2*a*d^2*x^2*e + 4*b*c*d*x*arctan(d*x + c)*e + pi*b*c^2*e*sgn(d*x + c) - p
i*b*c^2*e + 4*a*c*d*x*e - 2*b*c^2*arctan(1/(d*x + c))*e - 2*b*d*x*e + pi*b*e*sgn(d*x + c) - pi*b*e - 2*b*arcta
n(1/(d*x + c))*e)/d