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

Optimal. Leaf size=156 \[ -\frac {2 i c \text {Li}_2\left (1-\frac {2}{i a x+1}\right )}{15 a^3}-\frac {2 i c \tan ^{-1}(a x)^2}{15 a^3}-\frac {c \tan ^{-1}(a x)}{30 a^3}-\frac {4 c \log \left (\frac {2}{1+i a x}\right ) \tan ^{-1}(a x)}{15 a^3}+\frac {1}{5} a^2 c x^5 \tan ^{-1}(a x)^2+\frac {c x}{30 a^2}-\frac {1}{10} a c x^4 \tan ^{-1}(a x)+\frac {1}{3} c x^3 \tan ^{-1}(a x)^2-\frac {2 c x^2 \tan ^{-1}(a x)}{15 a}+\frac {c x^3}{30} \]

[Out]

1/30*c*x/a^2+1/30*c*x^3-1/30*c*arctan(a*x)/a^3-2/15*c*x^2*arctan(a*x)/a-1/10*a*c*x^4*arctan(a*x)-2/15*I*c*arct
an(a*x)^2/a^3+1/3*c*x^3*arctan(a*x)^2+1/5*a^2*c*x^5*arctan(a*x)^2-4/15*c*arctan(a*x)*ln(2/(1+I*a*x))/a^3-2/15*
I*c*polylog(2,1-2/(1+I*a*x))/a^3

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Rubi [A]  time = 0.41, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4950, 4852, 4916, 321, 203, 4920, 4854, 2402, 2315, 302} \[ -\frac {2 i c \text {PolyLog}\left (2,1-\frac {2}{1+i a x}\right )}{15 a^3}+\frac {1}{5} a^2 c x^5 \tan ^{-1}(a x)^2+\frac {c x}{30 a^2}-\frac {2 i c \tan ^{-1}(a x)^2}{15 a^3}-\frac {c \tan ^{-1}(a x)}{30 a^3}-\frac {4 c \log \left (\frac {2}{1+i a x}\right ) \tan ^{-1}(a x)}{15 a^3}-\frac {1}{10} a c x^4 \tan ^{-1}(a x)+\frac {1}{3} c x^3 \tan ^{-1}(a x)^2-\frac {2 c x^2 \tan ^{-1}(a x)}{15 a}+\frac {c x^3}{30} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x)/(30*a^2) + (c*x^3)/30 - (c*ArcTan[a*x])/(30*a^3) - (2*c*x^2*ArcTan[a*x])/(15*a) - (a*c*x^4*ArcTan[a*x])/
10 - (((2*I)/15)*c*ArcTan[a*x]^2)/a^3 + (c*x^3*ArcTan[a*x]^2)/3 + (a^2*c*x^5*ArcTan[a*x]^2)/5 - (4*c*ArcTan[a*
x]*Log[2/(1 + I*a*x)])/(15*a^3) - (((2*I)/15)*c*PolyLog[2, 1 - 2/(1 + I*a*x)])/a^3

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 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 4854

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^p*Lo
g[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTan[c*x])^(p - 1)*Log[2/(1 + (e*x)/d)])/(1 + c^2*x^
2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0]

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 4920

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

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^2 \left (c+a^2 c x^2\right ) \tan ^{-1}(a x)^2 \, dx &=c \int x^2 \tan ^{-1}(a x)^2 \, dx+\left (a^2 c\right ) \int x^4 \tan ^{-1}(a x)^2 \, dx\\ &=\frac {1}{3} c x^3 \tan ^{-1}(a x)^2+\frac {1}{5} a^2 c x^5 \tan ^{-1}(a x)^2-\frac {1}{3} (2 a c) \int \frac {x^3 \tan ^{-1}(a x)}{1+a^2 x^2} \, dx-\frac {1}{5} \left (2 a^3 c\right ) \int \frac {x^5 \tan ^{-1}(a x)}{1+a^2 x^2} \, dx\\ &=\frac {1}{3} c x^3 \tan ^{-1}(a x)^2+\frac {1}{5} a^2 c x^5 \tan ^{-1}(a x)^2-\frac {(2 c) \int x \tan ^{-1}(a x) \, dx}{3 a}+\frac {(2 c) \int \frac {x \tan ^{-1}(a x)}{1+a^2 x^2} \, dx}{3 a}-\frac {1}{5} (2 a c) \int x^3 \tan ^{-1}(a x) \, dx+\frac {1}{5} (2 a c) \int \frac {x^3 \tan ^{-1}(a x)}{1+a^2 x^2} \, dx\\ &=-\frac {c x^2 \tan ^{-1}(a x)}{3 a}-\frac {1}{10} a c x^4 \tan ^{-1}(a x)-\frac {i c \tan ^{-1}(a x)^2}{3 a^3}+\frac {1}{3} c x^3 \tan ^{-1}(a x)^2+\frac {1}{5} a^2 c x^5 \tan ^{-1}(a x)^2+\frac {1}{3} c \int \frac {x^2}{1+a^2 x^2} \, dx-\frac {(2 c) \int \frac {\tan ^{-1}(a x)}{i-a x} \, dx}{3 a^2}+\frac {(2 c) \int x \tan ^{-1}(a x) \, dx}{5 a}-\frac {(2 c) \int \frac {x \tan ^{-1}(a x)}{1+a^2 x^2} \, dx}{5 a}+\frac {1}{10} \left (a^2 c\right ) \int \frac {x^4}{1+a^2 x^2} \, dx\\ &=\frac {c x}{3 a^2}-\frac {2 c x^2 \tan ^{-1}(a x)}{15 a}-\frac {1}{10} a c x^4 \tan ^{-1}(a x)-\frac {2 i c \tan ^{-1}(a x)^2}{15 a^3}+\frac {1}{3} c x^3 \tan ^{-1}(a x)^2+\frac {1}{5} a^2 c x^5 \tan ^{-1}(a x)^2-\frac {2 c \tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )}{3 a^3}-\frac {1}{5} c \int \frac {x^2}{1+a^2 x^2} \, dx-\frac {c \int \frac {1}{1+a^2 x^2} \, dx}{3 a^2}+\frac {(2 c) \int \frac {\tan ^{-1}(a x)}{i-a x} \, dx}{5 a^2}+\frac {(2 c) \int \frac {\log \left (\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx}{3 a^2}+\frac {1}{10} \left (a^2 c\right ) \int \left (-\frac {1}{a^4}+\frac {x^2}{a^2}+\frac {1}{a^4 \left (1+a^2 x^2\right )}\right ) \, dx\\ &=\frac {c x}{30 a^2}+\frac {c x^3}{30}-\frac {c \tan ^{-1}(a x)}{3 a^3}-\frac {2 c x^2 \tan ^{-1}(a x)}{15 a}-\frac {1}{10} a c x^4 \tan ^{-1}(a x)-\frac {2 i c \tan ^{-1}(a x)^2}{15 a^3}+\frac {1}{3} c x^3 \tan ^{-1}(a x)^2+\frac {1}{5} a^2 c x^5 \tan ^{-1}(a x)^2-\frac {4 c \tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )}{15 a^3}-\frac {(2 i c) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i a x}\right )}{3 a^3}+\frac {c \int \frac {1}{1+a^2 x^2} \, dx}{10 a^2}+\frac {c \int \frac {1}{1+a^2 x^2} \, dx}{5 a^2}-\frac {(2 c) \int \frac {\log \left (\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx}{5 a^2}\\ &=\frac {c x}{30 a^2}+\frac {c x^3}{30}-\frac {c \tan ^{-1}(a x)}{30 a^3}-\frac {2 c x^2 \tan ^{-1}(a x)}{15 a}-\frac {1}{10} a c x^4 \tan ^{-1}(a x)-\frac {2 i c \tan ^{-1}(a x)^2}{15 a^3}+\frac {1}{3} c x^3 \tan ^{-1}(a x)^2+\frac {1}{5} a^2 c x^5 \tan ^{-1}(a x)^2-\frac {4 c \tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )}{15 a^3}-\frac {i c \text {Li}_2\left (1-\frac {2}{1+i a x}\right )}{3 a^3}+\frac {(2 i c) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i a x}\right )}{5 a^3}\\ &=\frac {c x}{30 a^2}+\frac {c x^3}{30}-\frac {c \tan ^{-1}(a x)}{30 a^3}-\frac {2 c x^2 \tan ^{-1}(a x)}{15 a}-\frac {1}{10} a c x^4 \tan ^{-1}(a x)-\frac {2 i c \tan ^{-1}(a x)^2}{15 a^3}+\frac {1}{3} c x^3 \tan ^{-1}(a x)^2+\frac {1}{5} a^2 c x^5 \tan ^{-1}(a x)^2-\frac {4 c \tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )}{15 a^3}-\frac {2 i c \text {Li}_2\left (1-\frac {2}{1+i a x}\right )}{15 a^3}\\ \end {align*}

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Mathematica [A]  time = 0.64, size = 104, normalized size = 0.67 \[ \frac {c \left (a^3 x^3+2 \left (3 a^5 x^5+5 a^3 x^3+2 i\right ) \tan ^{-1}(a x)^2-\tan ^{-1}(a x) \left (3 a^4 x^4+4 a^2 x^2+8 \log \left (1+e^{2 i \tan ^{-1}(a x)}\right )+1\right )+4 i \text {Li}_2\left (-e^{2 i \tan ^{-1}(a x)}\right )+a x\right )}{30 a^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c*(a*x + a^3*x^3 + 2*(2*I + 5*a^3*x^3 + 3*a^5*x^5)*ArcTan[a*x]^2 - ArcTan[a*x]*(1 + 4*a^2*x^2 + 3*a^4*x^4 + 8
*Log[1 + E^((2*I)*ArcTan[a*x])]) + (4*I)*PolyLog[2, -E^((2*I)*ArcTan[a*x])]))/(30*a^3)

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{2} c x^{4} + c x^{2}\right )} \arctan \left (a x\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^2*c*x^4 + c*x^2)*arctan(a*x)^2, x)

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

[Out]

sage0*x

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maple [A]  time = 0.12, size = 258, normalized size = 1.65 \[ \frac {a^{2} c \,x^{5} \arctan \left (a x \right )^{2}}{5}+\frac {c \,x^{3} \arctan \left (a x \right )^{2}}{3}-\frac {a c \,x^{4} \arctan \left (a x \right )}{10}-\frac {2 c \,x^{2} \arctan \left (a x \right )}{15 a}+\frac {2 c \arctan \left (a x \right ) \ln \left (a^{2} x^{2}+1\right )}{15 a^{3}}+\frac {c \,x^{3}}{30}+\frac {c x}{30 a^{2}}-\frac {c \arctan \left (a x \right )}{30 a^{3}}+\frac {i c \ln \left (a x -i\right ) \ln \left (a^{2} x^{2}+1\right )}{15 a^{3}}-\frac {i c \ln \left (a x -i\right )^{2}}{30 a^{3}}+\frac {i c \ln \left (a x +i\right )^{2}}{30 a^{3}}-\frac {i c \ln \left (a x +i\right ) \ln \left (a^{2} x^{2}+1\right )}{15 a^{3}}-\frac {i c \dilog \left (-\frac {i \left (a x +i\right )}{2}\right )}{15 a^{3}}+\frac {i c \ln \left (a x +i\right ) \ln \left (\frac {i \left (a x -i\right )}{2}\right )}{15 a^{3}}+\frac {i c \dilog \left (\frac {i \left (a x -i\right )}{2}\right )}{15 a^{3}}-\frac {i c \ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )}{15 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a^2*c*x^5*arctan(a*x)^2+1/3*c*x^3*arctan(a*x)^2-1/10*a*c*x^4*arctan(a*x)-2/15*c*x^2*arctan(a*x)/a+2/15/a^3
*c*arctan(a*x)*ln(a^2*x^2+1)+1/30*c*x^3+1/30*c*x/a^2-1/30*c*arctan(a*x)/a^3+1/15*I/a^3*c*ln(a*x-I)*ln(a^2*x^2+
1)-1/15*I/a^3*c*dilog(-1/2*I*(I+a*x))+1/15*I/a^3*c*dilog(1/2*I*(a*x-I))-1/30*I/a^3*c*ln(a*x-I)^2+1/30*I/a^3*c*
ln(I+a*x)^2-1/15*I/a^3*c*ln(I+a*x)*ln(a^2*x^2+1)+1/15*I/a^3*c*ln(I+a*x)*ln(1/2*I*(a*x-I))-1/15*I/a^3*c*ln(a*x-
I)*ln(-1/2*I*(I+a*x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{60} \, {\left (3 \, a^{2} c x^{5} + 5 \, c x^{3}\right )} \arctan \left (a x\right )^{2} - \frac {1}{240} \, {\left (3 \, a^{2} c x^{5} + 5 \, c x^{3}\right )} \log \left (a^{2} x^{2} + 1\right )^{2} + \int \frac {180 \, {\left (a^{4} c x^{6} + 2 \, a^{2} c x^{4} + c x^{2}\right )} \arctan \left (a x\right )^{2} + 15 \, {\left (a^{4} c x^{6} + 2 \, a^{2} c x^{4} + c x^{2}\right )} \log \left (a^{2} x^{2} + 1\right )^{2} - 8 \, {\left (3 \, a^{3} c x^{5} + 5 \, a c x^{3}\right )} \arctan \left (a x\right ) + 4 \, {\left (3 \, a^{4} c x^{6} + 5 \, a^{2} c x^{4}\right )} \log \left (a^{2} x^{2} + 1\right )}{240 \, {\left (a^{2} x^{2} + 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(3*a^2*c*x^5 + 5*c*x^3)*arctan(a*x)^2 - 1/240*(3*a^2*c*x^5 + 5*c*x^3)*log(a^2*x^2 + 1)^2 + integrate(1/24
0*(180*(a^4*c*x^6 + 2*a^2*c*x^4 + c*x^2)*arctan(a*x)^2 + 15*(a^4*c*x^6 + 2*a^2*c*x^4 + c*x^2)*log(a^2*x^2 + 1)
^2 - 8*(3*a^3*c*x^5 + 5*a*c*x^3)*arctan(a*x) + 4*(3*a^4*c*x^6 + 5*a^2*c*x^4)*log(a^2*x^2 + 1))/(a^2*x^2 + 1),
x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,{\mathrm {atan}\left (a\,x\right )}^2\,\left (c\,a^2\,x^2+c\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ c \left (\int x^{2} \operatorname {atan}^{2}{\left (a x \right )}\, dx + \int a^{2} x^{4} \operatorname {atan}^{2}{\left (a x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*(Integral(x**2*atan(a*x)**2, x) + Integral(a**2*x**4*atan(a*x)**2, x))

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