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

Optimal. Leaf size=124 \[ -\frac {c \tan ^{-1}(a x)^2}{12 a^4}+\frac {c x \tan ^{-1}(a x)}{6 a^3}+\frac {1}{6} a^2 c x^6 \tan ^{-1}(a x)^2-\frac {c x^2}{180 a^2}-\frac {7 c \log \left (a^2 x^2+1\right )}{90 a^4}-\frac {1}{15} a c x^5 \tan ^{-1}(a x)+\frac {1}{4} c x^4 \tan ^{-1}(a x)^2-\frac {c x^3 \tan ^{-1}(a x)}{18 a}+\frac {c x^4}{60} \]

[Out]

-1/180*c*x^2/a^2+1/60*c*x^4+1/6*c*x*arctan(a*x)/a^3-1/18*c*x^3*arctan(a*x)/a-1/15*a*c*x^5*arctan(a*x)-1/12*c*a
rctan(a*x)^2/a^4+1/4*c*x^4*arctan(a*x)^2+1/6*a^2*c*x^6*arctan(a*x)^2-7/90*c*ln(a^2*x^2+1)/a^4

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Rubi [A]  time = 0.43, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4950, 4852, 4916, 266, 43, 4846, 260, 4884} \[ -\frac {c x^2}{180 a^2}-\frac {7 c \log \left (a^2 x^2+1\right )}{90 a^4}+\frac {1}{6} a^2 c x^6 \tan ^{-1}(a x)^2+\frac {c x \tan ^{-1}(a x)}{6 a^3}-\frac {c \tan ^{-1}(a x)^2}{12 a^4}-\frac {1}{15} a c x^5 \tan ^{-1}(a x)+\frac {1}{4} c x^4 \tan ^{-1}(a x)^2-\frac {c x^3 \tan ^{-1}(a x)}{18 a}+\frac {c x^4}{60} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*x^2)/(180*a^2) + (c*x^4)/60 + (c*x*ArcTan[a*x])/(6*a^3) - (c*x^3*ArcTan[a*x])/(18*a) - (a*c*x^5*ArcTan[a*x
])/15 - (c*ArcTan[a*x]^2)/(12*a^4) + (c*x^4*ArcTan[a*x]^2)/4 + (a^2*c*x^6*ArcTan[a*x]^2)/6 - (7*c*Log[1 + a^2*
x^2])/(90*a^4)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 4852

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
n[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 4884

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

Rule 4916

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

Rule 4950

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

Rubi steps

\begin {align*} \int x^3 \left (c+a^2 c x^2\right ) \tan ^{-1}(a x)^2 \, dx &=c \int x^3 \tan ^{-1}(a x)^2 \, dx+\left (a^2 c\right ) \int x^5 \tan ^{-1}(a x)^2 \, dx\\ &=\frac {1}{4} c x^4 \tan ^{-1}(a x)^2+\frac {1}{6} a^2 c x^6 \tan ^{-1}(a x)^2-\frac {1}{2} (a c) \int \frac {x^4 \tan ^{-1}(a x)}{1+a^2 x^2} \, dx-\frac {1}{3} \left (a^3 c\right ) \int \frac {x^6 \tan ^{-1}(a x)}{1+a^2 x^2} \, dx\\ &=\frac {1}{4} c x^4 \tan ^{-1}(a x)^2+\frac {1}{6} a^2 c x^6 \tan ^{-1}(a x)^2-\frac {c \int x^2 \tan ^{-1}(a x) \, dx}{2 a}+\frac {c \int \frac {x^2 \tan ^{-1}(a x)}{1+a^2 x^2} \, dx}{2 a}-\frac {1}{3} (a c) \int x^4 \tan ^{-1}(a x) \, dx+\frac {1}{3} (a c) \int \frac {x^4 \tan ^{-1}(a x)}{1+a^2 x^2} \, dx\\ &=-\frac {c x^3 \tan ^{-1}(a x)}{6 a}-\frac {1}{15} a c x^5 \tan ^{-1}(a x)+\frac {1}{4} c x^4 \tan ^{-1}(a x)^2+\frac {1}{6} a^2 c x^6 \tan ^{-1}(a x)^2+\frac {1}{6} c \int \frac {x^3}{1+a^2 x^2} \, dx+\frac {c \int \tan ^{-1}(a x) \, dx}{2 a^3}-\frac {c \int \frac {\tan ^{-1}(a x)}{1+a^2 x^2} \, dx}{2 a^3}+\frac {c \int x^2 \tan ^{-1}(a x) \, dx}{3 a}-\frac {c \int \frac {x^2 \tan ^{-1}(a x)}{1+a^2 x^2} \, dx}{3 a}+\frac {1}{15} \left (a^2 c\right ) \int \frac {x^5}{1+a^2 x^2} \, dx\\ &=\frac {c x \tan ^{-1}(a x)}{2 a^3}-\frac {c x^3 \tan ^{-1}(a x)}{18 a}-\frac {1}{15} a c x^5 \tan ^{-1}(a x)-\frac {c \tan ^{-1}(a x)^2}{4 a^4}+\frac {1}{4} c x^4 \tan ^{-1}(a x)^2+\frac {1}{6} a^2 c x^6 \tan ^{-1}(a x)^2+\frac {1}{12} c \operatorname {Subst}\left (\int \frac {x}{1+a^2 x} \, dx,x,x^2\right )-\frac {1}{9} c \int \frac {x^3}{1+a^2 x^2} \, dx-\frac {c \int \tan ^{-1}(a x) \, dx}{3 a^3}+\frac {c \int \frac {\tan ^{-1}(a x)}{1+a^2 x^2} \, dx}{3 a^3}-\frac {c \int \frac {x}{1+a^2 x^2} \, dx}{2 a^2}+\frac {1}{30} \left (a^2 c\right ) \operatorname {Subst}\left (\int \frac {x^2}{1+a^2 x} \, dx,x,x^2\right )\\ &=\frac {c x \tan ^{-1}(a x)}{6 a^3}-\frac {c x^3 \tan ^{-1}(a x)}{18 a}-\frac {1}{15} a c x^5 \tan ^{-1}(a x)-\frac {c \tan ^{-1}(a x)^2}{12 a^4}+\frac {1}{4} c x^4 \tan ^{-1}(a x)^2+\frac {1}{6} a^2 c x^6 \tan ^{-1}(a x)^2-\frac {c \log \left (1+a^2 x^2\right )}{4 a^4}-\frac {1}{18} c \operatorname {Subst}\left (\int \frac {x}{1+a^2 x} \, dx,x,x^2\right )+\frac {1}{12} c \operatorname {Subst}\left (\int \left (\frac {1}{a^2}-\frac {1}{a^2 \left (1+a^2 x\right )}\right ) \, dx,x,x^2\right )+\frac {c \int \frac {x}{1+a^2 x^2} \, dx}{3 a^2}+\frac {1}{30} \left (a^2 c\right ) \operatorname {Subst}\left (\int \left (-\frac {1}{a^4}+\frac {x}{a^2}+\frac {1}{a^4 \left (1+a^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=\frac {c x^2}{20 a^2}+\frac {c x^4}{60}+\frac {c x \tan ^{-1}(a x)}{6 a^3}-\frac {c x^3 \tan ^{-1}(a x)}{18 a}-\frac {1}{15} a c x^5 \tan ^{-1}(a x)-\frac {c \tan ^{-1}(a x)^2}{12 a^4}+\frac {1}{4} c x^4 \tan ^{-1}(a x)^2+\frac {1}{6} a^2 c x^6 \tan ^{-1}(a x)^2-\frac {2 c \log \left (1+a^2 x^2\right )}{15 a^4}-\frac {1}{18} c \operatorname {Subst}\left (\int \left (\frac {1}{a^2}-\frac {1}{a^2 \left (1+a^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=-\frac {c x^2}{180 a^2}+\frac {c x^4}{60}+\frac {c x \tan ^{-1}(a x)}{6 a^3}-\frac {c x^3 \tan ^{-1}(a x)}{18 a}-\frac {1}{15} a c x^5 \tan ^{-1}(a x)-\frac {c \tan ^{-1}(a x)^2}{12 a^4}+\frac {1}{4} c x^4 \tan ^{-1}(a x)^2+\frac {1}{6} a^2 c x^6 \tan ^{-1}(a x)^2-\frac {7 c \log \left (1+a^2 x^2\right )}{90 a^4}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 89, normalized size = 0.72 \[ \frac {c \left (3 a^4 x^4-a^2 x^2-14 \log \left (a^2 x^2+1\right )+15 \left (2 a^6 x^6+3 a^4 x^4-1\right ) \tan ^{-1}(a x)^2-2 a x \left (6 a^4 x^4+5 a^2 x^2-15\right ) \tan ^{-1}(a x)\right )}{180 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(-(a^2*x^2) + 3*a^4*x^4 - 2*a*x*(-15 + 5*a^2*x^2 + 6*a^4*x^4)*ArcTan[a*x] + 15*(-1 + 3*a^4*x^4 + 2*a^6*x^6)
*ArcTan[a*x]^2 - 14*Log[1 + a^2*x^2]))/(180*a^4)

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fricas [A]  time = 1.34, size = 97, normalized size = 0.78 \[ \frac {3 \, a^{4} c x^{4} - a^{2} c x^{2} + 15 \, {\left (2 \, a^{6} c x^{6} + 3 \, a^{4} c x^{4} - c\right )} \arctan \left (a x\right )^{2} - 2 \, {\left (6 \, a^{5} c x^{5} + 5 \, a^{3} c x^{3} - 15 \, a c x\right )} \arctan \left (a x\right ) - 14 \, c \log \left (a^{2} x^{2} + 1\right )}{180 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

sage0*x

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maple [A]  time = 0.04, size = 107, normalized size = 0.86 \[ -\frac {c \,x^{2}}{180 a^{2}}+\frac {c \,x^{4}}{60}+\frac {c x \arctan \left (a x \right )}{6 a^{3}}-\frac {c \,x^{3} \arctan \left (a x \right )}{18 a}-\frac {a c \,x^{5} \arctan \left (a x \right )}{15}-\frac {c \arctan \left (a x \right )^{2}}{12 a^{4}}+\frac {c \,x^{4} \arctan \left (a x \right )^{2}}{4}+\frac {a^{2} c \,x^{6} \arctan \left (a x \right )^{2}}{6}-\frac {7 c \ln \left (a^{2} x^{2}+1\right )}{90 a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/180*c*x^2/a^2+1/60*c*x^4+1/6*c*x*arctan(a*x)/a^3-1/18*c*x^3*arctan(a*x)/a-1/15*a*c*x^5*arctan(a*x)-1/12*c*a
rctan(a*x)^2/a^4+1/4*c*x^4*arctan(a*x)^2+1/6*a^2*c*x^6*arctan(a*x)^2-7/90*c*ln(a^2*x^2+1)/a^4

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maxima [A]  time = 0.44, size = 116, normalized size = 0.94 \[ -\frac {1}{90} \, a {\left (\frac {6 \, a^{4} c x^{5} + 5 \, a^{2} c x^{3} - 15 \, c x}{a^{4}} + \frac {15 \, c \arctan \left (a x\right )}{a^{5}}\right )} \arctan \left (a x\right ) + \frac {1}{12} \, {\left (2 \, a^{2} c x^{6} + 3 \, c x^{4}\right )} \arctan \left (a x\right )^{2} + \frac {3 \, a^{4} c x^{4} - a^{2} c x^{2} + 15 \, c \arctan \left (a x\right )^{2} - 14 \, c \log \left (a^{2} x^{2} + 1\right )}{180 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/90*a*((6*a^4*c*x^5 + 5*a^2*c*x^3 - 15*c*x)/a^4 + 15*c*arctan(a*x)/a^5)*arctan(a*x) + 1/12*(2*a^2*c*x^6 + 3*
c*x^4)*arctan(a*x)^2 + 1/180*(3*a^4*c*x^4 - a^2*c*x^2 + 15*c*arctan(a*x)^2 - 14*c*log(a^2*x^2 + 1))/a^4

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mupad [B]  time = 0.57, size = 102, normalized size = 0.82 \[ -\frac {c\,\left (14\,\ln \left (a^2\,x^2+1\right )+a^2\,x^2-3\,a^4\,x^4+15\,{\mathrm {atan}\left (a\,x\right )}^2+10\,a^3\,x^3\,\mathrm {atan}\left (a\,x\right )+12\,a^5\,x^5\,\mathrm {atan}\left (a\,x\right )-30\,a\,x\,\mathrm {atan}\left (a\,x\right )-45\,a^4\,x^4\,{\mathrm {atan}\left (a\,x\right )}^2-30\,a^6\,x^6\,{\mathrm {atan}\left (a\,x\right )}^2\right )}{180\,a^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c*(14*log(a^2*x^2 + 1) + a^2*x^2 - 3*a^4*x^4 + 15*atan(a*x)^2 + 10*a^3*x^3*atan(a*x) + 12*a^5*x^5*atan(a*x)
- 30*a*x*atan(a*x) - 45*a^4*x^4*atan(a*x)^2 - 30*a^6*x^6*atan(a*x)^2))/(180*a^4)

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sympy [A]  time = 1.95, size = 121, normalized size = 0.98 \[ \begin {cases} \frac {a^{2} c x^{6} \operatorname {atan}^{2}{\left (a x \right )}}{6} - \frac {a c x^{5} \operatorname {atan}{\left (a x \right )}}{15} + \frac {c x^{4} \operatorname {atan}^{2}{\left (a x \right )}}{4} + \frac {c x^{4}}{60} - \frac {c x^{3} \operatorname {atan}{\left (a x \right )}}{18 a} - \frac {c x^{2}}{180 a^{2}} + \frac {c x \operatorname {atan}{\left (a x \right )}}{6 a^{3}} - \frac {7 c \log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{90 a^{4}} - \frac {c \operatorname {atan}^{2}{\left (a x \right )}}{12 a^{4}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a**2*c*x**2+c)*atan(a*x)**2,x)

[Out]

Piecewise((a**2*c*x**6*atan(a*x)**2/6 - a*c*x**5*atan(a*x)/15 + c*x**4*atan(a*x)**2/4 + c*x**4/60 - c*x**3*ata
n(a*x)/(18*a) - c*x**2/(180*a**2) + c*x*atan(a*x)/(6*a**3) - 7*c*log(x**2 + a**(-2))/(90*a**4) - c*atan(a*x)**
2/(12*a**4), Ne(a, 0)), (0, True))

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