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

Optimal. Leaf size=76 \[ \frac{(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )}{2 e}-\frac{b \left (d^2-\frac{e^2}{c^2}\right ) \tan ^{-1}(c x)}{2 e}-\frac{b d \log \left (c^2 x^2+1\right )}{2 c}-\frac{b e x}{2 c} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.061302, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {4862, 702, 635, 203, 260} \[ \frac{(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )}{2 e}-\frac{b \left (d^2-\frac{e^2}{c^2}\right ) \tan ^{-1}(c x)}{2 e}-\frac{b d \log \left (c^2 x^2+1\right )}{2 c}-\frac{b e x}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*(a + b*ArcTan[c*x]),x]

[Out]

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

Rule 4862

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(a + b*
ArcTan[c*x]))/(e*(q + 1)), x] - Dist[(b*c)/(e*(q + 1)), Int[(d + e*x)^(q + 1)/(1 + c^2*x^2), x], x] /; FreeQ[{
a, b, c, d, e, q}, x] && NeQ[q, -1]

Rule 702

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)^m, a + c*x^2,
x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

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]

Rubi steps

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

Mathematica [A]  time = 0.0054695, size = 77, normalized size = 1.01 \[ a d x+\frac{1}{2} a e x^2-\frac{b d \log \left (c^2 x^2+1\right )}{2 c}+\frac{b e \tan ^{-1}(c x)}{2 c^2}+b d x \tan ^{-1}(c x)+\frac{1}{2} b e x^2 \tan ^{-1}(c x)-\frac{b e x}{2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.024, size = 68, normalized size = 0.9 \begin{align*}{\frac{a{x}^{2}e}{2}}+adx+{\frac{b\arctan \left ( cx \right ){x}^{2}e}{2}}+b\arctan \left ( cx \right ) xd-{\frac{ebx}{2\,c}}-{\frac{bd\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2\,c}}+{\frac{\arctan \left ( cx \right ) be}{2\,{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.48609, size = 96, normalized size = 1.26 \begin{align*} \frac{1}{2} \, a e x^{2} + \frac{1}{2} \,{\left (x^{2} \arctan \left (c x\right ) - c{\left (\frac{x}{c^{2}} - \frac{\arctan \left (c x\right )}{c^{3}}\right )}\right )} b e + a d x + \frac{{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*e*x^2 + 1/2*(x^2*arctan(c*x) - c*(x/c^2 - arctan(c*x)/c^3))*b*e + a*d*x + 1/2*(2*c*x*arctan(c*x) - log(c
^2*x^2 + 1))*b*d/c

________________________________________________________________________________________

Fricas [A]  time = 2.11601, size = 162, normalized size = 2.13 \begin{align*} \frac{a c^{2} e x^{2} - b c d \log \left (c^{2} x^{2} + 1\right ) +{\left (2 \, a c^{2} d - b c e\right )} x +{\left (b c^{2} e x^{2} + 2 \, b c^{2} d x + b e\right )} \arctan \left (c x\right )}{2 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 0.730579, size = 87, normalized size = 1.14 \begin{align*} \begin{cases} a d x + \frac{a e x^{2}}{2} + b d x \operatorname{atan}{\left (c x \right )} + \frac{b e x^{2} \operatorname{atan}{\left (c x \right )}}{2} - \frac{b d \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{2 c} - \frac{b e x}{2 c} + \frac{b e \operatorname{atan}{\left (c x \right )}}{2 c^{2}} & \text{for}\: c \neq 0 \\a \left (d x + \frac{e x^{2}}{2}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.22245, size = 122, normalized size = 1.61 \begin{align*} \frac{b c^{2} x^{2} \arctan \left (c x\right ) e + 2 \, b c^{2} d x \arctan \left (c x\right ) + a c^{2} x^{2} e + 2 \, a c^{2} d x - \pi b e \mathrm{sgn}\left (c\right ) \mathrm{sgn}\left (x\right ) - b c x e - b c d \log \left (c^{2} x^{2} + 1\right ) + b \arctan \left (c x\right ) e}{2 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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