3.260 \(\int x (c+a^2 c x^2) \tan ^{-1}(a x)^2 \, dx\)

Optimal. Leaf size=96 \[ \frac{c \left (a^2 x^2+1\right )}{12 a^2}+\frac{c \log \left (a^2 x^2+1\right )}{6 a^2}+\frac{c \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^2}{4 a^2}-\frac{c x \left (a^2 x^2+1\right ) \tan ^{-1}(a x)}{6 a}-\frac{c x \tan ^{-1}(a x)}{3 a} \]

[Out]

(c*(1 + a^2*x^2))/(12*a^2) - (c*x*ArcTan[a*x])/(3*a) - (c*x*(1 + a^2*x^2)*ArcTan[a*x])/(6*a) + (c*(1 + a^2*x^2
)^2*ArcTan[a*x]^2)/(4*a^2) + (c*Log[1 + a^2*x^2])/(6*a^2)

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Rubi [A]  time = 0.0525261, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {4930, 4878, 4846, 260} \[ \frac{c \left (a^2 x^2+1\right )}{12 a^2}+\frac{c \log \left (a^2 x^2+1\right )}{6 a^2}+\frac{c \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^2}{4 a^2}-\frac{c x \left (a^2 x^2+1\right ) \tan ^{-1}(a x)}{6 a}-\frac{c x \tan ^{-1}(a x)}{3 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(1 + a^2*x^2))/(12*a^2) - (c*x*ArcTan[a*x])/(3*a) - (c*x*(1 + a^2*x^2)*ArcTan[a*x])/(6*a) + (c*(1 + a^2*x^2
)^2*ArcTan[a*x]^2)/(4*a^2) + (c*Log[1 + a^2*x^2])/(6*a^2)

Rule 4930

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[((d + e*x^2)^
(q + 1)*(a + b*ArcTan[c*x])^p)/(2*e*(q + 1)), x] - Dist[(b*p)/(2*c*(q + 1)), Int[(d + e*x^2)^q*(a + b*ArcTan[c
*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && EqQ[e, c^2*d] && GtQ[p, 0] && NeQ[q, -1]

Rule 4878

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> -Simp[(b*(d + e*x^2)^q)/(2*c*
q*(2*q + 1)), x] + (Dist[(2*d*q)/(2*q + 1), Int[(d + e*x^2)^(q - 1)*(a + b*ArcTan[c*x]), x], x] + Simp[(x*(d +
e*x^2)^q*(a + b*ArcTan[c*x]))/(2*q + 1), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[q, 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 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 x \left (c+a^2 c x^2\right ) \tan ^{-1}(a x)^2 \, dx &=\frac{c \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^2}{4 a^2}-\frac{\int \left (c+a^2 c x^2\right ) \tan ^{-1}(a x) \, dx}{2 a}\\ &=\frac{c \left (1+a^2 x^2\right )}{12 a^2}-\frac{c x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}{6 a}+\frac{c \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^2}{4 a^2}-\frac{c \int \tan ^{-1}(a x) \, dx}{3 a}\\ &=\frac{c \left (1+a^2 x^2\right )}{12 a^2}-\frac{c x \tan ^{-1}(a x)}{3 a}-\frac{c x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}{6 a}+\frac{c \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^2}{4 a^2}+\frac{1}{3} c \int \frac{x}{1+a^2 x^2} \, dx\\ &=\frac{c \left (1+a^2 x^2\right )}{12 a^2}-\frac{c x \tan ^{-1}(a x)}{3 a}-\frac{c x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}{6 a}+\frac{c \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^2}{4 a^2}+\frac{c \log \left (1+a^2 x^2\right )}{6 a^2}\\ \end{align*}

Mathematica [A]  time = 0.0296227, size = 64, normalized size = 0.67 \[ \frac{c \left (a^2 x^2+2 \log \left (a^2 x^2+1\right )-2 a x \left (a^2 x^2+3\right ) \tan ^{-1}(a x)+3 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^2\right )}{12 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(a^2*x^2 - 2*a*x*(3 + a^2*x^2)*ArcTan[a*x] + 3*(1 + a^2*x^2)^2*ArcTan[a*x]^2 + 2*Log[1 + a^2*x^2]))/(12*a^2
)

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Maple [A]  time = 0.033, size = 85, normalized size = 0.9 \begin{align*}{\frac{{a}^{2}c \left ( \arctan \left ( ax \right ) \right ) ^{2}{x}^{4}}{4}}+{\frac{c \left ( \arctan \left ( ax \right ) \right ) ^{2}{x}^{2}}{2}}-{\frac{ac\arctan \left ( ax \right ){x}^{3}}{6}}-{\frac{cx\arctan \left ( ax \right ) }{2\,a}}+{\frac{c \left ( \arctan \left ( ax \right ) \right ) ^{2}}{4\,{a}^{2}}}+{\frac{c{x}^{2}}{12}}+{\frac{c\ln \left ({a}^{2}{x}^{2}+1 \right ) }{6\,{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^2*c*arctan(a*x)^2*x^4+1/2*c*arctan(a*x)^2*x^2-1/6*a*c*arctan(a*x)*x^3-1/2*c*x*arctan(a*x)/a+1/4/a^2*c*ar
ctan(a*x)^2+1/12*c*x^2+1/6*c*ln(a^2*x^2+1)/a^2

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Maxima [A]  time = 1.00575, size = 117, normalized size = 1.22 \begin{align*} \frac{{\left (a^{2} c x^{2} + c\right )}^{2} \arctan \left (a x\right )^{2}}{4 \, a^{2} c} + \frac{{\left (c^{2} x^{2} + \frac{2 \, c^{2} \log \left (a^{2} x^{2} + 1\right )}{a^{2}}\right )} a - 2 \,{\left (a^{2} c^{2} x^{3} + 3 \, c^{2} x\right )} \arctan \left (a x\right )}{12 \, a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a^2*c*x^2 + c)^2*arctan(a*x)^2/(a^2*c) + 1/12*((c^2*x^2 + 2*c^2*log(a^2*x^2 + 1)/a^2)*a - 2*(a^2*c^2*x^3
+ 3*c^2*x)*arctan(a*x))/(a*c)

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Fricas [A]  time = 2.20166, size = 177, normalized size = 1.84 \begin{align*} \frac{a^{2} c x^{2} + 3 \,{\left (a^{4} c x^{4} + 2 \, a^{2} c x^{2} + c\right )} \arctan \left (a x\right )^{2} - 2 \,{\left (a^{3} c x^{3} + 3 \, a c x\right )} \arctan \left (a x\right ) + 2 \, c \log \left (a^{2} x^{2} + 1\right )}{12 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(a^2*c*x^2 + 3*(a^4*c*x^4 + 2*a^2*c*x^2 + c)*arctan(a*x)^2 - 2*(a^3*c*x^3 + 3*a*c*x)*arctan(a*x) + 2*c*lo
g(a^2*x^2 + 1))/a^2

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Sympy [A]  time = 1.49577, size = 94, normalized size = 0.98 \begin{align*} \begin{cases} \frac{a^{2} c x^{4} \operatorname{atan}^{2}{\left (a x \right )}}{4} - \frac{a c x^{3} \operatorname{atan}{\left (a x \right )}}{6} + \frac{c x^{2} \operatorname{atan}^{2}{\left (a x \right )}}{2} + \frac{c x^{2}}{12} - \frac{c x \operatorname{atan}{\left (a x \right )}}{2 a} + \frac{c \log{\left (x^{2} + \frac{1}{a^{2}} \right )}}{6 a^{2}} + \frac{c \operatorname{atan}^{2}{\left (a x \right )}}{4 a^{2}} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a**2*c*x**2+c)*atan(a*x)**2,x)

[Out]

Piecewise((a**2*c*x**4*atan(a*x)**2/4 - a*c*x**3*atan(a*x)/6 + c*x**2*atan(a*x)**2/2 + c*x**2/12 - c*x*atan(a*
x)/(2*a) + c*log(x**2 + a**(-2))/(6*a**2) + c*atan(a*x)**2/(4*a**2), Ne(a, 0)), (0, True))

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Giac [A]  time = 1.14065, size = 112, normalized size = 1.17 \begin{align*} \frac{1}{4} \,{\left (a^{2} c x^{4} + 2 \, c x^{2}\right )} \arctan \left (a x\right )^{2} - \frac{2 \, a^{3} c x^{3} \arctan \left (a x\right ) - a^{2} c x^{2} + 6 \, a c x \arctan \left (a x\right ) - 3 \, c \arctan \left (a x\right )^{2} - 2 \, c \log \left (a^{2} x^{2} + 1\right )}{12 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a^2*c*x^4 + 2*c*x^2)*arctan(a*x)^2 - 1/12*(2*a^3*c*x^3*arctan(a*x) - a^2*c*x^2 + 6*a*c*x*arctan(a*x) - 3*
c*arctan(a*x)^2 - 2*c*log(a^2*x^2 + 1))/a^2