3.276 \(\int x (c+a^2 c x^2)^3 \tan ^{-1}(a x)^2 \, dx\)

Optimal. Leaf size=200 \[ \frac{c^3 \left (a^2 x^2+1\right )^3}{168 a^2}+\frac{3 c^3 \left (a^2 x^2+1\right )^2}{280 a^2}+\frac{c^3 \left (a^2 x^2+1\right )}{35 a^2}+\frac{2 c^3 \log \left (a^2 x^2+1\right )}{35 a^2}+\frac{c^3 \left (a^2 x^2+1\right )^4 \tan ^{-1}(a x)^2}{8 a^2}-\frac{c^3 x \left (a^2 x^2+1\right )^3 \tan ^{-1}(a x)}{28 a}-\frac{3 c^3 x \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)}{70 a}-\frac{2 c^3 x \left (a^2 x^2+1\right ) \tan ^{-1}(a x)}{35 a}-\frac{4 c^3 x \tan ^{-1}(a x)}{35 a} \]

[Out]

(c^3*(1 + a^2*x^2))/(35*a^2) + (3*c^3*(1 + a^2*x^2)^2)/(280*a^2) + (c^3*(1 + a^2*x^2)^3)/(168*a^2) - (4*c^3*x*
ArcTan[a*x])/(35*a) - (2*c^3*x*(1 + a^2*x^2)*ArcTan[a*x])/(35*a) - (3*c^3*x*(1 + a^2*x^2)^2*ArcTan[a*x])/(70*a
) - (c^3*x*(1 + a^2*x^2)^3*ArcTan[a*x])/(28*a) + (c^3*(1 + a^2*x^2)^4*ArcTan[a*x]^2)/(8*a^2) + (2*c^3*Log[1 +
a^2*x^2])/(35*a^2)

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Rubi [A]  time = 0.121294, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4930, 4878, 4846, 260} \[ \frac{c^3 \left (a^2 x^2+1\right )^3}{168 a^2}+\frac{3 c^3 \left (a^2 x^2+1\right )^2}{280 a^2}+\frac{c^3 \left (a^2 x^2+1\right )}{35 a^2}+\frac{2 c^3 \log \left (a^2 x^2+1\right )}{35 a^2}+\frac{c^3 \left (a^2 x^2+1\right )^4 \tan ^{-1}(a x)^2}{8 a^2}-\frac{c^3 x \left (a^2 x^2+1\right )^3 \tan ^{-1}(a x)}{28 a}-\frac{3 c^3 x \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)}{70 a}-\frac{2 c^3 x \left (a^2 x^2+1\right ) \tan ^{-1}(a x)}{35 a}-\frac{4 c^3 x \tan ^{-1}(a x)}{35 a} \]

Antiderivative was successfully verified.

[In]

Int[x*(c + a^2*c*x^2)^3*ArcTan[a*x]^2,x]

[Out]

(c^3*(1 + a^2*x^2))/(35*a^2) + (3*c^3*(1 + a^2*x^2)^2)/(280*a^2) + (c^3*(1 + a^2*x^2)^3)/(168*a^2) - (4*c^3*x*
ArcTan[a*x])/(35*a) - (2*c^3*x*(1 + a^2*x^2)*ArcTan[a*x])/(35*a) - (3*c^3*x*(1 + a^2*x^2)^2*ArcTan[a*x])/(70*a
) - (c^3*x*(1 + a^2*x^2)^3*ArcTan[a*x])/(28*a) + (c^3*(1 + a^2*x^2)^4*ArcTan[a*x]^2)/(8*a^2) + (2*c^3*Log[1 +
a^2*x^2])/(35*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 )^3 \tan ^{-1}(a x)^2 \, dx &=\frac{c^3 \left (1+a^2 x^2\right )^4 \tan ^{-1}(a x)^2}{8 a^2}-\frac{\int \left (c+a^2 c x^2\right )^3 \tan ^{-1}(a x) \, dx}{4 a}\\ &=\frac{c^3 \left (1+a^2 x^2\right )^3}{168 a^2}-\frac{c^3 x \left (1+a^2 x^2\right )^3 \tan ^{-1}(a x)}{28 a}+\frac{c^3 \left (1+a^2 x^2\right )^4 \tan ^{-1}(a x)^2}{8 a^2}-\frac{(3 c) \int \left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x) \, dx}{14 a}\\ &=\frac{3 c^3 \left (1+a^2 x^2\right )^2}{280 a^2}+\frac{c^3 \left (1+a^2 x^2\right )^3}{168 a^2}-\frac{3 c^3 x \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)}{70 a}-\frac{c^3 x \left (1+a^2 x^2\right )^3 \tan ^{-1}(a x)}{28 a}+\frac{c^3 \left (1+a^2 x^2\right )^4 \tan ^{-1}(a x)^2}{8 a^2}-\frac{\left (6 c^2\right ) \int \left (c+a^2 c x^2\right ) \tan ^{-1}(a x) \, dx}{35 a}\\ &=\frac{c^3 \left (1+a^2 x^2\right )}{35 a^2}+\frac{3 c^3 \left (1+a^2 x^2\right )^2}{280 a^2}+\frac{c^3 \left (1+a^2 x^2\right )^3}{168 a^2}-\frac{2 c^3 x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}{35 a}-\frac{3 c^3 x \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)}{70 a}-\frac{c^3 x \left (1+a^2 x^2\right )^3 \tan ^{-1}(a x)}{28 a}+\frac{c^3 \left (1+a^2 x^2\right )^4 \tan ^{-1}(a x)^2}{8 a^2}-\frac{\left (4 c^3\right ) \int \tan ^{-1}(a x) \, dx}{35 a}\\ &=\frac{c^3 \left (1+a^2 x^2\right )}{35 a^2}+\frac{3 c^3 \left (1+a^2 x^2\right )^2}{280 a^2}+\frac{c^3 \left (1+a^2 x^2\right )^3}{168 a^2}-\frac{4 c^3 x \tan ^{-1}(a x)}{35 a}-\frac{2 c^3 x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}{35 a}-\frac{3 c^3 x \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)}{70 a}-\frac{c^3 x \left (1+a^2 x^2\right )^3 \tan ^{-1}(a x)}{28 a}+\frac{c^3 \left (1+a^2 x^2\right )^4 \tan ^{-1}(a x)^2}{8 a^2}+\frac{1}{35} \left (4 c^3\right ) \int \frac{x}{1+a^2 x^2} \, dx\\ &=\frac{c^3 \left (1+a^2 x^2\right )}{35 a^2}+\frac{3 c^3 \left (1+a^2 x^2\right )^2}{280 a^2}+\frac{c^3 \left (1+a^2 x^2\right )^3}{168 a^2}-\frac{4 c^3 x \tan ^{-1}(a x)}{35 a}-\frac{2 c^3 x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}{35 a}-\frac{3 c^3 x \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)}{70 a}-\frac{c^3 x \left (1+a^2 x^2\right )^3 \tan ^{-1}(a x)}{28 a}+\frac{c^3 \left (1+a^2 x^2\right )^4 \tan ^{-1}(a x)^2}{8 a^2}+\frac{2 c^3 \log \left (1+a^2 x^2\right )}{35 a^2}\\ \end{align*}

Mathematica [A]  time = 0.0801937, size = 100, normalized size = 0.5 \[ \frac{c^3 \left (5 a^6 x^6+24 a^4 x^4+57 a^2 x^2+48 \log \left (a^2 x^2+1\right )-6 a x \left (5 a^6 x^6+21 a^4 x^4+35 a^2 x^2+35\right ) \tan ^{-1}(a x)+105 \left (a^2 x^2+1\right )^4 \tan ^{-1}(a x)^2\right )}{840 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(c + a^2*c*x^2)^3*ArcTan[a*x]^2,x]

[Out]

(c^3*(57*a^2*x^2 + 24*a^4*x^4 + 5*a^6*x^6 - 6*a*x*(35 + 35*a^2*x^2 + 21*a^4*x^4 + 5*a^6*x^6)*ArcTan[a*x] + 105
*(1 + a^2*x^2)^4*ArcTan[a*x]^2 + 48*Log[1 + a^2*x^2]))/(840*a^2)

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Maple [A]  time = 0.034, size = 185, normalized size = 0.9 \begin{align*}{\frac{{a}^{6}{c}^{3} \left ( \arctan \left ( ax \right ) \right ) ^{2}{x}^{8}}{8}}+{\frac{{a}^{4}{c}^{3} \left ( \arctan \left ( ax \right ) \right ) ^{2}{x}^{6}}{2}}+{\frac{3\,{a}^{2}{c}^{3} \left ( \arctan \left ( ax \right ) \right ) ^{2}{x}^{4}}{4}}+{\frac{{c}^{3} \left ( \arctan \left ( ax \right ) \right ) ^{2}{x}^{2}}{2}}-{\frac{{a}^{5}{c}^{3}\arctan \left ( ax \right ){x}^{7}}{28}}-{\frac{3\,{a}^{3}{c}^{3}\arctan \left ( ax \right ){x}^{5}}{20}}-{\frac{a{c}^{3}\arctan \left ( ax \right ){x}^{3}}{4}}-{\frac{{c}^{3}x\arctan \left ( ax \right ) }{4\,a}}+{\frac{{c}^{3} \left ( \arctan \left ( ax \right ) \right ) ^{2}}{8\,{a}^{2}}}+{\frac{{a}^{4}{c}^{3}{x}^{6}}{168}}+{\frac{{a}^{2}{x}^{4}{c}^{3}}{35}}+{\frac{19\,{x}^{2}{c}^{3}}{280}}+{\frac{2\,{c}^{3}\ln \left ({a}^{2}{x}^{2}+1 \right ) }{35\,{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a^2*c*x^2+c)^3*arctan(a*x)^2,x)

[Out]

1/8*a^6*c^3*arctan(a*x)^2*x^8+1/2*a^4*c^3*arctan(a*x)^2*x^6+3/4*a^2*c^3*arctan(a*x)^2*x^4+1/2*c^3*arctan(a*x)^
2*x^2-1/28*a^5*c^3*arctan(a*x)*x^7-3/20*a^3*c^3*arctan(a*x)*x^5-1/4*a*c^3*arctan(a*x)*x^3-1/4*c^3*x*arctan(a*x
)/a+1/8/a^2*c^3*arctan(a*x)^2+1/168*a^4*c^3*x^6+1/35*a^2*x^4*c^3+19/280*x^2*c^3+2/35*c^3*ln(a^2*x^2+1)/a^2

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Maxima [A]  time = 0.995014, size = 180, normalized size = 0.9 \begin{align*} \frac{{\left (a^{2} c x^{2} + c\right )}^{4} \arctan \left (a x\right )^{2}}{8 \, a^{2} c} + \frac{{\left (5 \, a^{4} c^{4} x^{6} + 24 \, a^{2} c^{4} x^{4} + 57 \, c^{4} x^{2} + \frac{48 \, c^{4} \log \left (a^{2} x^{2} + 1\right )}{a^{2}}\right )} a - 6 \,{\left (5 \, a^{6} c^{4} x^{7} + 21 \, a^{4} c^{4} x^{5} + 35 \, a^{2} c^{4} x^{3} + 35 \, c^{4} x\right )} \arctan \left (a x\right )}{840 \, a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a^2*c*x^2+c)^3*arctan(a*x)^2,x, algorithm="maxima")

[Out]

1/8*(a^2*c*x^2 + c)^4*arctan(a*x)^2/(a^2*c) + 1/840*((5*a^4*c^4*x^6 + 24*a^2*c^4*x^4 + 57*c^4*x^2 + 48*c^4*log
(a^2*x^2 + 1)/a^2)*a - 6*(5*a^6*c^4*x^7 + 21*a^4*c^4*x^5 + 35*a^2*c^4*x^3 + 35*c^4*x)*arctan(a*x))/(a*c)

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Fricas [A]  time = 2.23156, size = 343, normalized size = 1.72 \begin{align*} \frac{5 \, a^{6} c^{3} x^{6} + 24 \, a^{4} c^{3} x^{4} + 57 \, a^{2} c^{3} x^{2} + 48 \, c^{3} \log \left (a^{2} x^{2} + 1\right ) + 105 \,{\left (a^{8} c^{3} x^{8} + 4 \, a^{6} c^{3} x^{6} + 6 \, a^{4} c^{3} x^{4} + 4 \, a^{2} c^{3} x^{2} + c^{3}\right )} \arctan \left (a x\right )^{2} - 6 \,{\left (5 \, a^{7} c^{3} x^{7} + 21 \, a^{5} c^{3} x^{5} + 35 \, a^{3} c^{3} x^{3} + 35 \, a c^{3} x\right )} \arctan \left (a x\right )}{840 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a^2*c*x^2+c)^3*arctan(a*x)^2,x, algorithm="fricas")

[Out]

1/840*(5*a^6*c^3*x^6 + 24*a^4*c^3*x^4 + 57*a^2*c^3*x^2 + 48*c^3*log(a^2*x^2 + 1) + 105*(a^8*c^3*x^8 + 4*a^6*c^
3*x^6 + 6*a^4*c^3*x^4 + 4*a^2*c^3*x^2 + c^3)*arctan(a*x)^2 - 6*(5*a^7*c^3*x^7 + 21*a^5*c^3*x^5 + 35*a^3*c^3*x^
3 + 35*a*c^3*x)*arctan(a*x))/a^2

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Sympy [A]  time = 4.81666, size = 207, normalized size = 1.03 \begin{align*} \begin{cases} \frac{a^{6} c^{3} x^{8} \operatorname{atan}^{2}{\left (a x \right )}}{8} - \frac{a^{5} c^{3} x^{7} \operatorname{atan}{\left (a x \right )}}{28} + \frac{a^{4} c^{3} x^{6} \operatorname{atan}^{2}{\left (a x \right )}}{2} + \frac{a^{4} c^{3} x^{6}}{168} - \frac{3 a^{3} c^{3} x^{5} \operatorname{atan}{\left (a x \right )}}{20} + \frac{3 a^{2} c^{3} x^{4} \operatorname{atan}^{2}{\left (a x \right )}}{4} + \frac{a^{2} c^{3} x^{4}}{35} - \frac{a c^{3} x^{3} \operatorname{atan}{\left (a x \right )}}{4} + \frac{c^{3} x^{2} \operatorname{atan}^{2}{\left (a x \right )}}{2} + \frac{19 c^{3} x^{2}}{280} - \frac{c^{3} x \operatorname{atan}{\left (a x \right )}}{4 a} + \frac{2 c^{3} \log{\left (x^{2} + \frac{1}{a^{2}} \right )}}{35 a^{2}} + \frac{c^{3} \operatorname{atan}^{2}{\left (a x \right )}}{8 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)**3*atan(a*x)**2,x)

[Out]

Piecewise((a**6*c**3*x**8*atan(a*x)**2/8 - a**5*c**3*x**7*atan(a*x)/28 + a**4*c**3*x**6*atan(a*x)**2/2 + a**4*
c**3*x**6/168 - 3*a**3*c**3*x**5*atan(a*x)/20 + 3*a**2*c**3*x**4*atan(a*x)**2/4 + a**2*c**3*x**4/35 - a*c**3*x
**3*atan(a*x)/4 + c**3*x**2*atan(a*x)**2/2 + 19*c**3*x**2/280 - c**3*x*atan(a*x)/(4*a) + 2*c**3*log(x**2 + a**
(-2))/(35*a**2) + c**3*atan(a*x)**2/(8*a**2), Ne(a, 0)), (0, True))

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Giac [A]  time = 1.16338, size = 301, normalized size = 1.5 \begin{align*} \frac{{\left (a^{2} c x^{2} + c\right )}^{4} \arctan \left (a x\right )^{2}}{8 \, a^{2} c} - \frac{5 \,{\left (12 \, x^{7} \arctan \left (a x\right ) - a{\left (\frac{2 \, a^{4} x^{6} - 3 \, a^{2} x^{4} + 6 \, x^{2}}{a^{6}} - \frac{6 \, \log \left (a^{2} x^{2} + 1\right )}{a^{8}}\right )}\right )} a^{6} c^{3} + 63 \,{\left (4 \, x^{5} \arctan \left (a x\right ) - a{\left (\frac{a^{2} x^{4} - 2 \, x^{2}}{a^{4}} + \frac{2 \, \log \left (a^{2} x^{2} + 1\right )}{a^{6}}\right )}\right )} a^{4} c^{3} + 210 \,{\left (2 \, x^{3} \arctan \left (a x\right ) - a{\left (\frac{x^{2}}{a^{2}} - \frac{\log \left (a^{2} x^{2} + 1\right )}{a^{4}}\right )}\right )} a^{2} c^{3} + \frac{210 \,{\left (2 \, a x \arctan \left (a x\right ) - \log \left (a^{2} x^{2} + 1\right )\right )} c^{3}}{a}}{1680 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a^2*c*x^2+c)^3*arctan(a*x)^2,x, algorithm="giac")

[Out]

1/8*(a^2*c*x^2 + c)^4*arctan(a*x)^2/(a^2*c) - 1/1680*(5*(12*x^7*arctan(a*x) - a*((2*a^4*x^6 - 3*a^2*x^4 + 6*x^
2)/a^6 - 6*log(a^2*x^2 + 1)/a^8))*a^6*c^3 + 63*(4*x^5*arctan(a*x) - a*((a^2*x^4 - 2*x^2)/a^4 + 2*log(a^2*x^2 +
1)/a^6))*a^4*c^3 + 210*(2*x^3*arctan(a*x) - a*(x^2/a^2 - log(a^2*x^2 + 1)/a^4))*a^2*c^3 + 210*(2*a*x*arctan(a
*x) - log(a^2*x^2 + 1))*c^3/a)/a