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

Optimal. Leaf size=153 \[ \frac {c^2 \left (a^2 x^2+1\right )^2}{60 a^2}+\frac {2 c^2 \left (a^2 x^2+1\right )}{45 a^2}+\frac {4 c^2 \log \left (a^2 x^2+1\right )}{45 a^2}+\frac {c^2 \left (a^2 x^2+1\right )^3 \tan ^{-1}(a x)^2}{6 a^2}-\frac {c^2 x \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)}{15 a}-\frac {4 c^2 x \left (a^2 x^2+1\right ) \tan ^{-1}(a x)}{45 a}-\frac {8 c^2 x \tan ^{-1}(a x)}{45 a} \]

[Out]

2/45*c^2*(a^2*x^2+1)/a^2+1/60*c^2*(a^2*x^2+1)^2/a^2-8/45*c^2*x*arctan(a*x)/a-4/45*c^2*x*(a^2*x^2+1)*arctan(a*x
)/a-1/15*c^2*x*(a^2*x^2+1)^2*arctan(a*x)/a+1/6*c^2*(a^2*x^2+1)^3*arctan(a*x)^2/a^2+4/45*c^2*ln(a^2*x^2+1)/a^2

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Rubi [A]  time = 0.09, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4930, 4878, 4846, 260} \[ \frac {c^2 \left (a^2 x^2+1\right )^2}{60 a^2}+\frac {2 c^2 \left (a^2 x^2+1\right )}{45 a^2}+\frac {4 c^2 \log \left (a^2 x^2+1\right )}{45 a^2}+\frac {c^2 \left (a^2 x^2+1\right )^3 \tan ^{-1}(a x)^2}{6 a^2}-\frac {c^2 x \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)}{15 a}-\frac {4 c^2 x \left (a^2 x^2+1\right ) \tan ^{-1}(a x)}{45 a}-\frac {8 c^2 x \tan ^{-1}(a x)}{45 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c^2*(1 + a^2*x^2))/(45*a^2) + (c^2*(1 + a^2*x^2)^2)/(60*a^2) - (8*c^2*x*ArcTan[a*x])/(45*a) - (4*c^2*x*(1 +
a^2*x^2)*ArcTan[a*x])/(45*a) - (c^2*x*(1 + a^2*x^2)^2*ArcTan[a*x])/(15*a) + (c^2*(1 + a^2*x^2)^3*ArcTan[a*x]^
2)/(6*a^2) + (4*c^2*Log[1 + a^2*x^2])/(45*a^2)

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]

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 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 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]

Rubi steps

\begin {align*} \int x \left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x)^2 \, dx &=\frac {c^2 \left (1+a^2 x^2\right )^3 \tan ^{-1}(a x)^2}{6 a^2}-\frac {\int \left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x) \, dx}{3 a}\\ &=\frac {c^2 \left (1+a^2 x^2\right )^2}{60 a^2}-\frac {c^2 x \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)}{15 a}+\frac {c^2 \left (1+a^2 x^2\right )^3 \tan ^{-1}(a x)^2}{6 a^2}-\frac {(4 c) \int \left (c+a^2 c x^2\right ) \tan ^{-1}(a x) \, dx}{15 a}\\ &=\frac {2 c^2 \left (1+a^2 x^2\right )}{45 a^2}+\frac {c^2 \left (1+a^2 x^2\right )^2}{60 a^2}-\frac {4 c^2 x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}{45 a}-\frac {c^2 x \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)}{15 a}+\frac {c^2 \left (1+a^2 x^2\right )^3 \tan ^{-1}(a x)^2}{6 a^2}-\frac {\left (8 c^2\right ) \int \tan ^{-1}(a x) \, dx}{45 a}\\ &=\frac {2 c^2 \left (1+a^2 x^2\right )}{45 a^2}+\frac {c^2 \left (1+a^2 x^2\right )^2}{60 a^2}-\frac {8 c^2 x \tan ^{-1}(a x)}{45 a}-\frac {4 c^2 x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}{45 a}-\frac {c^2 x \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)}{15 a}+\frac {c^2 \left (1+a^2 x^2\right )^3 \tan ^{-1}(a x)^2}{6 a^2}+\frac {1}{45} \left (8 c^2\right ) \int \frac {x}{1+a^2 x^2} \, dx\\ &=\frac {2 c^2 \left (1+a^2 x^2\right )}{45 a^2}+\frac {c^2 \left (1+a^2 x^2\right )^2}{60 a^2}-\frac {8 c^2 x \tan ^{-1}(a x)}{45 a}-\frac {4 c^2 x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}{45 a}-\frac {c^2 x \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)}{15 a}+\frac {c^2 \left (1+a^2 x^2\right )^3 \tan ^{-1}(a x)^2}{6 a^2}+\frac {4 c^2 \log \left (1+a^2 x^2\right )}{45 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 84, normalized size = 0.55 \[ \frac {c^2 \left (3 a^4 x^4+14 a^2 x^2+16 \log \left (a^2 x^2+1\right )+30 \left (a^2 x^2+1\right )^3 \tan ^{-1}(a x)^2-4 a x \left (3 a^4 x^4+10 a^2 x^2+15\right ) \tan ^{-1}(a x)\right )}{180 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(c + a^2*c*x^2)^2*ArcTan[a*x]^2,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] + 30*(1 + a^2*x^2)^3*ArcTan[a*x
]^2 + 16*Log[1 + a^2*x^2]))/(180*a^2)

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fricas [A]  time = 0.64, size = 123, normalized size = 0.80 \[ \frac {3 \, a^{4} c^{2} x^{4} + 14 \, a^{2} c^{2} x^{2} + 30 \, {\left (a^{6} c^{2} x^{6} + 3 \, a^{4} c^{2} x^{4} + 3 \, a^{2} c^{2} x^{2} + c^{2}\right )} \arctan \left (a x\right )^{2} + 16 \, c^{2} \log \left (a^{2} x^{2} + 1\right ) - 4 \, {\left (3 \, a^{5} c^{2} x^{5} + 10 \, a^{3} c^{2} x^{3} + 15 \, a c^{2} x\right )} \arctan \left (a x\right )}{180 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/180*(3*a^4*c^2*x^4 + 14*a^2*c^2*x^2 + 30*(a^6*c^2*x^6 + 3*a^4*c^2*x^4 + 3*a^2*c^2*x^2 + c^2)*arctan(a*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^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.05, size = 142, normalized size = 0.93 \[ \frac {a^{4} c^{2} \arctan \left (a x \right )^{2} x^{6}}{6}+\frac {a^{2} c^{2} \arctan \left (a x \right )^{2} x^{4}}{2}+\frac {c^{2} \arctan \left (a x \right )^{2} x^{2}}{2}-\frac {a^{3} c^{2} \arctan \left (a x \right ) x^{5}}{15}-\frac {2 a \,c^{2} \arctan \left (a x \right ) x^{3}}{9}-\frac {c^{2} x \arctan \left (a x \right )}{3 a}+\frac {c^{2} \arctan \left (a x \right )^{2}}{6 a^{2}}+\frac {a^{2} c^{2} x^{4}}{60}+\frac {7 c^{2} x^{2}}{90}+\frac {4 c^{2} \ln \left (a^{2} x^{2}+1\right )}{45 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a^4*c^2*arctan(a*x)^2*x^6+1/2*a^2*c^2*arctan(a*x)^2*x^4+1/2*c^2*arctan(a*x)^2*x^2-1/15*a^3*c^2*arctan(a*x)
*x^5-2/9*a*c^2*arctan(a*x)*x^3-1/3*c^2*x*arctan(a*x)/a+1/6/a^2*c^2*arctan(a*x)^2+1/60*a^2*c^2*x^4+7/90*c^2*x^2
+4/45*c^2*ln(a^2*x^2+1)/a^2

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maxima [A]  time = 0.31, size = 111, normalized size = 0.73 \[ \frac {{\left (a^{2} c x^{2} + c\right )}^{3} \arctan \left (a x\right )^{2}}{6 \, a^{2} c} + \frac {{\left (3 \, a^{2} c^{3} x^{4} + 14 \, c^{3} x^{2} + \frac {16 \, c^{3} \log \left (a^{2} x^{2} + 1\right )}{a^{2}}\right )} a - 4 \, {\left (3 \, a^{4} c^{3} x^{5} + 10 \, a^{2} c^{3} x^{3} + 15 \, c^{3} x\right )} \arctan \left (a x\right )}{180 \, a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(a^2*c*x^2 + c)^3*arctan(a*x)^2/(a^2*c) + 1/180*((3*a^2*c^3*x^4 + 14*c^3*x^2 + 16*c^3*log(a^2*x^2 + 1)/a^2
)*a - 4*(3*a^4*c^3*x^5 + 10*a^2*c^3*x^3 + 15*c^3*x)*arctan(a*x))/(a*c)

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mupad [B]  time = 0.31, size = 135, normalized size = 0.88 \[ \frac {\frac {c^2\,\left (30\,{\mathrm {atan}\left (a\,x\right )}^2+16\,\ln \left (a^2\,x^2+1\right )\right )}{180}-\frac {a\,c^2\,x\,\mathrm {atan}\left (a\,x\right )}{3}}{a^2}+\frac {c^2\,\left (90\,x^2\,{\mathrm {atan}\left (a\,x\right )}^2+14\,x^2\right )}{180}+\frac {a^2\,c^2\,\left (90\,x^4\,{\mathrm {atan}\left (a\,x\right )}^2+3\,x^4\right )}{180}-\frac {a^3\,c^2\,x^5\,\mathrm {atan}\left (a\,x\right )}{15}+\frac {a^4\,c^2\,x^6\,{\mathrm {atan}\left (a\,x\right )}^2}{6}-\frac {2\,a\,c^2\,x^3\,\mathrm {atan}\left (a\,x\right )}{9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((c^2*(16*log(a^2*x^2 + 1) + 30*atan(a*x)^2))/180 - (a*c^2*x*atan(a*x))/3)/a^2 + (c^2*(90*x^2*atan(a*x)^2 + 14
*x^2))/180 + (a^2*c^2*(90*x^4*atan(a*x)^2 + 3*x^4))/180 - (a^3*c^2*x^5*atan(a*x))/15 + (a^4*c^2*x^6*atan(a*x)^
2)/6 - (2*a*c^2*x^3*atan(a*x))/9

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sympy [A]  time = 2.02, size = 158, normalized size = 1.03 \[ \begin {cases} \frac {a^{4} c^{2} x^{6} \operatorname {atan}^{2}{\left (a x \right )}}{6} - \frac {a^{3} c^{2} x^{5} \operatorname {atan}{\left (a x \right )}}{15} + \frac {a^{2} c^{2} x^{4} \operatorname {atan}^{2}{\left (a x \right )}}{2} + \frac {a^{2} c^{2} x^{4}}{60} - \frac {2 a c^{2} x^{3} \operatorname {atan}{\left (a x \right )}}{9} + \frac {c^{2} x^{2} \operatorname {atan}^{2}{\left (a x \right )}}{2} + \frac {7 c^{2} x^{2}}{90} - \frac {c^{2} x \operatorname {atan}{\left (a x \right )}}{3 a} + \frac {4 c^{2} \log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{45 a^{2}} + \frac {c^{2} \operatorname {atan}^{2}{\left (a x \right )}}{6 a^{2}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**4*c**2*x**6*atan(a*x)**2/6 - a**3*c**2*x**5*atan(a*x)/15 + a**2*c**2*x**4*atan(a*x)**2/2 + a**2*
c**2*x**4/60 - 2*a*c**2*x**3*atan(a*x)/9 + c**2*x**2*atan(a*x)**2/2 + 7*c**2*x**2/90 - c**2*x*atan(a*x)/(3*a)
+ 4*c**2*log(x**2 + a**(-2))/(45*a**2) + c**2*atan(a*x)**2/(6*a**2), Ne(a, 0)), (0, True))

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