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

Optimal. Leaf size=117 \[ -\frac{c^2 \left (a^2 x^2+1\right )^2}{20 a}-\frac{2 c^2 \left (a^2 x^2+1\right )}{15 a}-\frac{4 c^2 \log \left (a^2 x^2+1\right )}{15 a}+\frac{1}{5} c^2 x \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)+\frac{4}{15} c^2 x \left (a^2 x^2+1\right ) \tan ^{-1}(a x)+\frac{8}{15} c^2 x \tan ^{-1}(a x) \]

[Out]

(-2*c^2*(1 + a^2*x^2))/(15*a) - (c^2*(1 + a^2*x^2)^2)/(20*a) + (8*c^2*x*ArcTan[a*x])/15 + (4*c^2*x*(1 + a^2*x^
2)*ArcTan[a*x])/15 + (c^2*x*(1 + a^2*x^2)^2*ArcTan[a*x])/5 - (4*c^2*Log[1 + a^2*x^2])/(15*a)

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Rubi [A]  time = 0.0454285, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {4878, 4846, 260} \[ -\frac{c^2 \left (a^2 x^2+1\right )^2}{20 a}-\frac{2 c^2 \left (a^2 x^2+1\right )}{15 a}-\frac{4 c^2 \log \left (a^2 x^2+1\right )}{15 a}+\frac{1}{5} c^2 x \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)+\frac{4}{15} c^2 x \left (a^2 x^2+1\right ) \tan ^{-1}(a x)+\frac{8}{15} c^2 x \tan ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[(c + a^2*c*x^2)^2*ArcTan[a*x],x]

[Out]

(-2*c^2*(1 + a^2*x^2))/(15*a) - (c^2*(1 + a^2*x^2)^2)/(20*a) + (8*c^2*x*ArcTan[a*x])/15 + (4*c^2*x*(1 + a^2*x^
2)*ArcTan[a*x])/15 + (c^2*x*(1 + a^2*x^2)^2*ArcTan[a*x])/5 - (4*c^2*Log[1 + a^2*x^2])/(15*a)

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 \left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x) \, dx &=-\frac{c^2 \left (1+a^2 x^2\right )^2}{20 a}+\frac{1}{5} c^2 x \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)+\frac{1}{5} (4 c) \int \left (c+a^2 c x^2\right ) \tan ^{-1}(a x) \, dx\\ &=-\frac{2 c^2 \left (1+a^2 x^2\right )}{15 a}-\frac{c^2 \left (1+a^2 x^2\right )^2}{20 a}+\frac{4}{15} c^2 x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)+\frac{1}{5} c^2 x \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)+\frac{1}{15} \left (8 c^2\right ) \int \tan ^{-1}(a x) \, dx\\ &=-\frac{2 c^2 \left (1+a^2 x^2\right )}{15 a}-\frac{c^2 \left (1+a^2 x^2\right )^2}{20 a}+\frac{8}{15} c^2 x \tan ^{-1}(a x)+\frac{4}{15} c^2 x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)+\frac{1}{5} c^2 x \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)-\frac{1}{15} \left (8 a c^2\right ) \int \frac{x}{1+a^2 x^2} \, dx\\ &=-\frac{2 c^2 \left (1+a^2 x^2\right )}{15 a}-\frac{c^2 \left (1+a^2 x^2\right )^2}{20 a}+\frac{8}{15} c^2 x \tan ^{-1}(a x)+\frac{4}{15} c^2 x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)+\frac{1}{5} c^2 x \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)-\frac{4 c^2 \log \left (1+a^2 x^2\right )}{15 a}\\ \end{align*}

Mathematica [A]  time = 0.0558637, size = 65, normalized size = 0.56 \[ \frac{c^2 \left (-3 a^4 x^4-14 a^2 x^2-16 \log \left (a^2 x^2+1\right )+4 a x \left (3 a^4 x^4+10 a^2 x^2+15\right ) \tan ^{-1}(a x)\right )}{60 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(-14*a^2*x^2 - 3*a^4*x^4 + 4*a*x*(15 + 10*a^2*x^2 + 3*a^4*x^4)*ArcTan[a*x] - 16*Log[1 + a^2*x^2]))/(60*a)

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Maple [A]  time = 0.025, size = 79, normalized size = 0.7 \begin{align*}{\frac{{a}^{4}{c}^{2}\arctan \left ( ax \right ){x}^{5}}{5}}+{\frac{2\,{a}^{2}{c}^{2}\arctan \left ( ax \right ){x}^{3}}{3}}+{c}^{2}x\arctan \left ( ax \right ) -{\frac{{a}^{3}{c}^{2}{x}^{4}}{20}}-{\frac{7\,{c}^{2}{x}^{2}a}{30}}-{\frac{4\,{c}^{2}\ln \left ({a}^{2}{x}^{2}+1 \right ) }{15\,a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a^4*c^2*arctan(a*x)*x^5+2/3*a^2*c^2*arctan(a*x)*x^3+c^2*x*arctan(a*x)-1/20*a^3*c^2*x^4-7/30*c^2*x^2*a-4/15
*c^2*ln(a^2*x^2+1)/a

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Maxima [A]  time = 0.980269, size = 104, normalized size = 0.89 \begin{align*} -\frac{1}{60} \,{\left (3 \, a^{2} c^{2} x^{4} + 14 \, c^{2} x^{2} + \frac{16 \, c^{2} \log \left (a^{2} x^{2} + 1\right )}{a^{2}}\right )} a + \frac{1}{15} \,{\left (3 \, a^{4} c^{2} x^{5} + 10 \, a^{2} c^{2} x^{3} + 15 \, c^{2} x\right )} \arctan \left (a x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(3*a^2*c^2*x^4 + 14*c^2*x^2 + 16*c^2*log(a^2*x^2 + 1)/a^2)*a + 1/15*(3*a^4*c^2*x^5 + 10*a^2*c^2*x^3 + 15
*c^2*x)*arctan(a*x)

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Fricas [A]  time = 1.57001, size = 176, normalized size = 1.5 \begin{align*} -\frac{3 \, a^{4} c^{2} x^{4} + 14 \, a^{2} c^{2} x^{2} + 16 \, c^{2} \log \left (a^{2} x^{2} + 1\right ) - 4 \,{\left (3 \, a^{5} c^{2} x^{5} + 10 \, a^{3} c^{2} x^{3} + 15 \, a c^{2} x\right )} \arctan \left (a x\right )}{60 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(3*a^4*c^2*x^4 + 14*a^2*c^2*x^2 + 16*c^2*log(a^2*x^2 + 1) - 4*(3*a^5*c^2*x^5 + 10*a^3*c^2*x^3 + 15*a*c^2
*x)*arctan(a*x))/a

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Sympy [A]  time = 1.46037, size = 88, normalized size = 0.75 \begin{align*} \begin{cases} \frac{a^{4} c^{2} x^{5} \operatorname{atan}{\left (a x \right )}}{5} - \frac{a^{3} c^{2} x^{4}}{20} + \frac{2 a^{2} c^{2} x^{3} \operatorname{atan}{\left (a x \right )}}{3} - \frac{7 a c^{2} x^{2}}{30} + c^{2} x \operatorname{atan}{\left (a x \right )} - \frac{4 c^{2} \log{\left (x^{2} + \frac{1}{a^{2}} \right )}}{15 a} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**4*c**2*x**5*atan(a*x)/5 - a**3*c**2*x**4/20 + 2*a**2*c**2*x**3*atan(a*x)/3 - 7*a*c**2*x**2/30 +
c**2*x*atan(a*x) - 4*c**2*log(x**2 + a**(-2))/(15*a), Ne(a, 0)), (0, True))

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Giac [A]  time = 1.18454, size = 111, normalized size = 0.95 \begin{align*} \frac{1}{15} \,{\left (3 \, a^{4} c^{2} x^{5} + 10 \, a^{2} c^{2} x^{3} + 15 \, c^{2} x\right )} \arctan \left (a x\right ) - \frac{4 \, c^{2} \log \left (a^{2} x^{2} + 1\right )}{15 \, a} - \frac{3 \, a^{7} c^{2} x^{4} + 14 \, a^{5} c^{2} x^{2}}{60 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*a^4*c^2*x^5 + 10*a^2*c^2*x^3 + 15*c^2*x)*arctan(a*x) - 4/15*c^2*log(a^2*x^2 + 1)/a - 1/60*(3*a^7*c^2*x
^4 + 14*a^5*c^2*x^2)/a^4