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]

1/35*c^3*(a^2*x^2+1)/a^2+3/280*c^3*(a^2*x^2+1)^2/a^2+1/168*c^3*(a^2*x^2+1)^3/a^2-4/35*c^3*x*arctan(a*x)/a-2/35
*c^3*x*(a^2*x^2+1)*arctan(a*x)/a-3/70*c^3*x*(a^2*x^2+1)^2*arctan(a*x)/a-1/28*c^3*x*(a^2*x^2+1)^3*arctan(a*x)/a
+1/8*c^3*(a^2*x^2+1)^4*arctan(a*x)^2/a^2+2/35*c^3*ln(a^2*x^2+1)/a^2

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Rubi [A]  time = 0.12, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4930, 4878, 4846, 260} \[ \frac {c^3 \left (a^2 x^2+1\right )^3}{168 a^2}+\frac {3 c^3 \left (a^2 x^2+1\right )^2}{280 a^2}+\frac {c^3 \left (a^2 x^2+1\right )}{35 a^2}+\frac {2 c^3 \log \left (a^2 x^2+1\right )}{35 a^2}+\frac {c^3 \left (a^2 x^2+1\right )^4 \tan ^{-1}(a x)^2}{8 a^2}-\frac {c^3 x \left (a^2 x^2+1\right )^3 \tan ^{-1}(a x)}{28 a}-\frac {3 c^3 x \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)}{70 a}-\frac {2 c^3 x \left (a^2 x^2+1\right ) \tan ^{-1}(a x)}{35 a}-\frac {4 c^3 x \tan ^{-1}(a x)}{35 a} \]

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 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 )^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*}

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Mathematica [A]  time = 0.09, size = 100, normalized size = 0.50 \[ \frac {c^3 \left (5 a^6 x^6+24 a^4 x^4+57 a^2 x^2+48 \log \left (a^2 x^2+1\right )+105 \left (a^2 x^2+1\right )^4 \tan ^{-1}(a x)^2-6 a x \left (5 a^6 x^6+21 a^4 x^4+35 a^2 x^2+35\right ) \tan ^{-1}(a x)\right )}{840 a^2} \]

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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fricas [A]  time = 0.66, size = 156, normalized size = 0.78 \[ \frac {5 \, a^{6} c^{3} x^{6} + 24 \, a^{4} c^{3} x^{4} + 57 \, a^{2} c^{3} x^{2} + 48 \, c^{3} \log \left (a^{2} x^{2} + 1\right ) + 105 \, {\left (a^{8} c^{3} x^{8} + 4 \, a^{6} c^{3} x^{6} + 6 \, a^{4} c^{3} x^{4} + 4 \, a^{2} c^{3} x^{2} + c^{3}\right )} \arctan \left (a x\right )^{2} - 6 \, {\left (5 \, a^{7} c^{3} x^{7} + 21 \, a^{5} c^{3} x^{5} + 35 \, a^{3} c^{3} x^{3} + 35 \, a c^{3} x\right )} \arctan \left (a x\right )}{840 \, a^{2}} \]

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

[Out]

sage0*x

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maple [A]  time = 0.04, size = 185, normalized size = 0.92 \[ \frac {a^{6} c^{3} \arctan \left (a x \right )^{2} x^{8}}{8}+\frac {a^{4} c^{3} \arctan \left (a x \right )^{2} x^{6}}{2}+\frac {3 a^{2} c^{3} \arctan \left (a x \right )^{2} x^{4}}{4}+\frac {c^{3} \arctan \left (a x \right )^{2} x^{2}}{2}-\frac {a^{5} c^{3} \arctan \left (a x \right ) x^{7}}{28}-\frac {3 a^{3} c^{3} \arctan \left (a x \right ) x^{5}}{20}-\frac {a \,c^{3} \arctan \left (a x \right ) x^{3}}{4}-\frac {c^{3} x \arctan \left (a x \right )}{4 a}+\frac {c^{3} \arctan \left (a x \right )^{2}}{8 a^{2}}+\frac {a^{4} c^{3} x^{6}}{168}+\frac {a^{2} x^{4} c^{3}}{35}+\frac {19 x^{2} c^{3}}{280}+\frac {2 c^{3} \ln \left (a^{2} x^{2}+1\right )}{35 a^{2}} \]

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 = 0.32, size = 133, normalized size = 0.66 \[ \frac {{\left (a^{2} c x^{2} + c\right )}^{4} \arctan \left (a x\right )^{2}}{8 \, a^{2} c} + \frac {{\left (5 \, a^{4} c^{4} x^{6} + 24 \, a^{2} c^{4} x^{4} + 57 \, c^{4} x^{2} + \frac {48 \, c^{4} \log \left (a^{2} x^{2} + 1\right )}{a^{2}}\right )} a - 6 \, {\left (5 \, a^{6} c^{4} x^{7} + 21 \, a^{4} c^{4} x^{5} + 35 \, a^{2} c^{4} x^{3} + 35 \, c^{4} x\right )} \arctan \left (a x\right )}{840 \, a c} \]

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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mupad [B]  time = 0.46, size = 156, normalized size = 0.78 \[ {\mathrm {atan}\left (a\,x\right )}^2\,\left (\frac {c^3}{8\,a^2}+\frac {c^3\,x^2}{2}+\frac {3\,a^2\,c^3\,x^4}{4}+\frac {a^4\,c^3\,x^6}{2}+\frac {a^6\,c^3\,x^8}{8}\right )+\frac {19\,c^3\,x^2}{280}-a^2\,\mathrm {atan}\left (a\,x\right )\,\left (\frac {c^3\,x}{4\,a^3}+\frac {3\,a\,c^3\,x^5}{20}+\frac {c^3\,x^3}{4\,a}+\frac {a^3\,c^3\,x^7}{28}\right )+\frac {2\,c^3\,\ln \left (a^2\,x^2+1\right )}{35\,a^2}+\frac {a^2\,c^3\,x^4}{35}+\frac {a^4\,c^3\,x^6}{168} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 3.29, size = 207, normalized size = 1.04 \[ \begin {cases} \frac {a^{6} c^{3} x^{8} \operatorname {atan}^{2}{\left (a x \right )}}{8} - \frac {a^{5} c^{3} x^{7} \operatorname {atan}{\left (a x \right )}}{28} + \frac {a^{4} c^{3} x^{6} \operatorname {atan}^{2}{\left (a x \right )}}{2} + \frac {a^{4} c^{3} x^{6}}{168} - \frac {3 a^{3} c^{3} x^{5} \operatorname {atan}{\left (a x \right )}}{20} + \frac {3 a^{2} c^{3} x^{4} \operatorname {atan}^{2}{\left (a x \right )}}{4} + \frac {a^{2} c^{3} x^{4}}{35} - \frac {a c^{3} x^{3} \operatorname {atan}{\left (a x \right )}}{4} + \frac {c^{3} x^{2} \operatorname {atan}^{2}{\left (a x \right )}}{2} + \frac {19 c^{3} x^{2}}{280} - \frac {c^{3} x \operatorname {atan}{\left (a x \right )}}{4 a} + \frac {2 c^{3} \log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{35 a^{2}} + \frac {c^{3} \operatorname {atan}^{2}{\left (a x \right )}}{8 a^{2}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

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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