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

Optimal. Leaf size=104 \[ \frac{i b^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x^3}\right )}{3 c}+\frac{1}{3} x^3 \left (a+b \tan ^{-1}\left (c x^3\right )\right )^2+\frac{i \left (a+b \tan ^{-1}\left (c x^3\right )\right )^2}{3 c}+\frac{2 b \log \left (\frac{2}{1+i c x^3}\right ) \left (a+b \tan ^{-1}\left (c x^3\right )\right )}{3 c} \]

[Out]

((I/3)*(a + b*ArcTan[c*x^3])^2)/c + (x^3*(a + b*ArcTan[c*x^3])^2)/3 + (2*b*(a + b*ArcTan[c*x^3])*Log[2/(1 + I*
c*x^3)])/(3*c) + ((I/3)*b^2*PolyLog[2, 1 - 2/(1 + I*c*x^3)])/c

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Rubi [B]  time = 0.644147, antiderivative size = 255, normalized size of antiderivative = 2.45, number of steps used = 28, number of rules used = 12, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {5035, 2454, 2389, 2296, 2295, 6715, 2430, 43, 2416, 2394, 2393, 2391} \[ -\frac{i b^2 \text{PolyLog}\left (2,\frac{1}{2} \left (1-i c x^3\right )\right )}{6 c}+\frac{i b^2 \text{PolyLog}\left (2,\frac{1}{2} \left (1+i c x^3\right )\right )}{6 c}-\frac{1}{6} b x^3 \log \left (1+i c x^3\right ) \left (2 i a-b \log \left (1-i c x^3\right )\right )+\frac{i \left (1-i c x^3\right ) \left (2 a+i b \log \left (1-i c x^3\right )\right )^2}{12 c}+\frac{i b \log \left (\frac{1}{2} \left (1+i c x^3\right )\right ) \left (2 i a-b \log \left (1-i c x^3\right )\right )}{6 c}+\frac{i b^2 \left (1+i c x^3\right ) \log ^2\left (1+i c x^3\right )}{12 c}+\frac{i b^2 \log \left (\frac{1}{2} \left (1-i c x^3\right )\right ) \log \left (1+i c x^3\right )}{6 c} \]

Warning: Unable to verify antiderivative.

[In]

Int[x^2*(a + b*ArcTan[c*x^3])^2,x]

[Out]

((I/12)*(1 - I*c*x^3)*(2*a + I*b*Log[1 - I*c*x^3])^2)/c + ((I/6)*b*((2*I)*a - b*Log[1 - I*c*x^3])*Log[(1 + I*c
*x^3)/2])/c + ((I/6)*b^2*Log[(1 - I*c*x^3)/2]*Log[1 + I*c*x^3])/c - (b*x^3*((2*I)*a - b*Log[1 - I*c*x^3])*Log[
1 + I*c*x^3])/6 + ((I/12)*b^2*(1 + I*c*x^3)*Log[1 + I*c*x^3]^2)/c - ((I/6)*b^2*PolyLog[2, (1 - I*c*x^3)/2])/c
+ ((I/6)*b^2*PolyLog[2, (1 + I*c*x^3)/2])/c

Rule 5035

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

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 2430

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*
(g_.)), x_Symbol] :> Simp[x*(a + b*Log[c*(d + e*x)^n])^p*(f + g*Log[h*(i + j*x)^m]), x] + (-Dist[g*j*m, Int[(x
*(a + b*Log[c*(d + e*x)^n])^p)/(i + j*x), x], x] - Dist[b*e*n*p, Int[(x*(a + b*Log[c*(d + e*x)^n])^(p - 1)*(f
+ g*Log[h*(i + j*x)^m]))/(d + e*x), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, i, j, m, n}, x] && IGtQ[p, 0]

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 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rule 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2391

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

Rubi steps

\begin{align*} \int x^2 \left (a+b \tan ^{-1}\left (c x^3\right )\right )^2 \, dx &=\int \left (\frac{1}{4} x^2 \left (2 a+i b \log \left (1-i c x^3\right )\right )^2+\frac{1}{2} b x^2 \left (-2 i a+b \log \left (1-i c x^3\right )\right ) \log \left (1+i c x^3\right )-\frac{1}{4} b^2 x^2 \log ^2\left (1+i c x^3\right )\right ) \, dx\\ &=\frac{1}{4} \int x^2 \left (2 a+i b \log \left (1-i c x^3\right )\right )^2 \, dx+\frac{1}{2} b \int x^2 \left (-2 i a+b \log \left (1-i c x^3\right )\right ) \log \left (1+i c x^3\right ) \, dx-\frac{1}{4} b^2 \int x^2 \log ^2\left (1+i c x^3\right ) \, dx\\ &=\frac{1}{12} \operatorname{Subst}\left (\int (2 a+i b \log (1-i c x))^2 \, dx,x,x^3\right )+\frac{1}{6} b \operatorname{Subst}\left (\int (-2 i a+b \log (1-i c x)) \log (1+i c x) \, dx,x,x^3\right )-\frac{1}{12} b^2 \operatorname{Subst}\left (\int \log ^2(1+i c x) \, dx,x,x^3\right )\\ &=-\frac{1}{6} b x^3 \left (2 i a-b \log \left (1-i c x^3\right )\right ) \log \left (1+i c x^3\right )+\frac{i \operatorname{Subst}\left (\int (2 a+i b \log (x))^2 \, dx,x,1-i c x^3\right )}{12 c}+\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \log ^2(x) \, dx,x,1+i c x^3\right )}{12 c}-\frac{1}{6} (i b c) \operatorname{Subst}\left (\int \frac{x (-2 i a+b \log (1-i c x))}{1+i c x} \, dx,x,x^3\right )+\frac{1}{6} \left (i b^2 c\right ) \operatorname{Subst}\left (\int \frac{x \log (1+i c x)}{1-i c x} \, dx,x,x^3\right )\\ &=\frac{i \left (1-i c x^3\right ) \left (2 a+i b \log \left (1-i c x^3\right )\right )^2}{12 c}-\frac{1}{6} b x^3 \left (2 i a-b \log \left (1-i c x^3\right )\right ) \log \left (1+i c x^3\right )+\frac{i b^2 \left (1+i c x^3\right ) \log ^2\left (1+i c x^3\right )}{12 c}+\frac{b \operatorname{Subst}\left (\int (2 a+i b \log (x)) \, dx,x,1-i c x^3\right )}{6 c}-\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \log (x) \, dx,x,1+i c x^3\right )}{6 c}-\frac{1}{6} (i b c) \operatorname{Subst}\left (\int \left (-\frac{i (-2 i a+b \log (1-i c x))}{c}+\frac{-2 i a+b \log (1-i c x)}{c (-i+c x)}\right ) \, dx,x,x^3\right )+\frac{1}{6} \left (i b^2 c\right ) \operatorname{Subst}\left (\int \left (\frac{i \log (1+i c x)}{c}+\frac{\log (1+i c x)}{c (i+c x)}\right ) \, dx,x,x^3\right )\\ &=-\frac{1}{3} i a b x^3-\frac{b^2 x^3}{6}+\frac{i \left (1-i c x^3\right ) \left (2 a+i b \log \left (1-i c x^3\right )\right )^2}{12 c}-\frac{i b^2 \left (1+i c x^3\right ) \log \left (1+i c x^3\right )}{6 c}-\frac{1}{6} b x^3 \left (2 i a-b \log \left (1-i c x^3\right )\right ) \log \left (1+i c x^3\right )+\frac{i b^2 \left (1+i c x^3\right ) \log ^2\left (1+i c x^3\right )}{12 c}-\frac{1}{6} (i b) \operatorname{Subst}\left (\int \frac{-2 i a+b \log (1-i c x)}{-i+c x} \, dx,x,x^3\right )-\frac{1}{6} b \operatorname{Subst}\left (\int (-2 i a+b \log (1-i c x)) \, dx,x,x^3\right )+\frac{1}{6} \left (i b^2\right ) \operatorname{Subst}\left (\int \frac{\log (1+i c x)}{i+c x} \, dx,x,x^3\right )-\frac{1}{6} b^2 \operatorname{Subst}\left (\int \log (1+i c x) \, dx,x,x^3\right )+\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \log (x) \, dx,x,1-i c x^3\right )}{6 c}\\ &=-\frac{1}{3} b^2 x^3+\frac{i b^2 \left (1-i c x^3\right ) \log \left (1-i c x^3\right )}{6 c}+\frac{i \left (1-i c x^3\right ) \left (2 a+i b \log \left (1-i c x^3\right )\right )^2}{12 c}+\frac{i b \left (2 i a-b \log \left (1-i c x^3\right )\right ) \log \left (\frac{1}{2} \left (1+i c x^3\right )\right )}{6 c}-\frac{i b^2 \left (1+i c x^3\right ) \log \left (1+i c x^3\right )}{6 c}+\frac{i b^2 \log \left (\frac{1}{2} \left (1-i c x^3\right )\right ) \log \left (1+i c x^3\right )}{6 c}-\frac{1}{6} b x^3 \left (2 i a-b \log \left (1-i c x^3\right )\right ) \log \left (1+i c x^3\right )+\frac{i b^2 \left (1+i c x^3\right ) \log ^2\left (1+i c x^3\right )}{12 c}-\frac{1}{6} b^2 \operatorname{Subst}\left (\int \log (1-i c x) \, dx,x,x^3\right )+\frac{1}{6} b^2 \operatorname{Subst}\left (\int \frac{\log \left (\frac{1}{2} i (-i+c x)\right )}{1-i c x} \, dx,x,x^3\right )+\frac{1}{6} b^2 \operatorname{Subst}\left (\int \frac{\log \left (-\frac{1}{2} i (i+c x)\right )}{1+i c x} \, dx,x,x^3\right )+\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \log (x) \, dx,x,1+i c x^3\right )}{6 c}\\ &=-\frac{1}{6} b^2 x^3+\frac{i b^2 \left (1-i c x^3\right ) \log \left (1-i c x^3\right )}{6 c}+\frac{i \left (1-i c x^3\right ) \left (2 a+i b \log \left (1-i c x^3\right )\right )^2}{12 c}+\frac{i b \left (2 i a-b \log \left (1-i c x^3\right )\right ) \log \left (\frac{1}{2} \left (1+i c x^3\right )\right )}{6 c}+\frac{i b^2 \log \left (\frac{1}{2} \left (1-i c x^3\right )\right ) \log \left (1+i c x^3\right )}{6 c}-\frac{1}{6} b x^3 \left (2 i a-b \log \left (1-i c x^3\right )\right ) \log \left (1+i c x^3\right )+\frac{i b^2 \left (1+i c x^3\right ) \log ^2\left (1+i c x^3\right )}{12 c}+\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{2}\right )}{x} \, dx,x,1-i c x^3\right )}{6 c}-\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{2}\right )}{x} \, dx,x,1+i c x^3\right )}{6 c}-\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \log (x) \, dx,x,1-i c x^3\right )}{6 c}\\ &=\frac{i \left (1-i c x^3\right ) \left (2 a+i b \log \left (1-i c x^3\right )\right )^2}{12 c}+\frac{i b \left (2 i a-b \log \left (1-i c x^3\right )\right ) \log \left (\frac{1}{2} \left (1+i c x^3\right )\right )}{6 c}+\frac{i b^2 \log \left (\frac{1}{2} \left (1-i c x^3\right )\right ) \log \left (1+i c x^3\right )}{6 c}-\frac{1}{6} b x^3 \left (2 i a-b \log \left (1-i c x^3\right )\right ) \log \left (1+i c x^3\right )+\frac{i b^2 \left (1+i c x^3\right ) \log ^2\left (1+i c x^3\right )}{12 c}-\frac{i b^2 \text{Li}_2\left (\frac{1}{2} \left (1-i c x^3\right )\right )}{6 c}+\frac{i b^2 \text{Li}_2\left (\frac{1}{2} \left (1+i c x^3\right )\right )}{6 c}\\ \end{align*}

Mathematica [A]  time = 0.0706193, size = 107, normalized size = 1.03 \[ \frac{-i b^2 \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}\left (c x^3\right )}\right )+a \left (a c x^3-b \log \left (c^2 x^6+1\right )\right )+2 b \tan ^{-1}\left (c x^3\right ) \left (a c x^3+b \log \left (1+e^{2 i \tan ^{-1}\left (c x^3\right )}\right )\right )+b^2 \left (c x^3-i\right ) \tan ^{-1}\left (c x^3\right )^2}{3 c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(b^2*(-I + c*x^3)*ArcTan[c*x^3]^2 + 2*b*ArcTan[c*x^3]*(a*c*x^3 + b*Log[1 + E^((2*I)*ArcTan[c*x^3])]) + a*(a*c*
x^3 - b*Log[1 + c^2*x^6]) - I*b^2*PolyLog[2, -E^((2*I)*ArcTan[c*x^3])])/(3*c)

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Maple [A]  time = 0.09, size = 148, normalized size = 1.4 \begin{align*}{\frac{{x}^{3}{b}^{2} \left ( \arctan \left ( c{x}^{3} \right ) \right ) ^{2}}{3}}+{\frac{2\,{x}^{3}ab\arctan \left ( c{x}^{3} \right ) }{3}}+{\frac{{x}^{3}{a}^{2}}{3}}-{\frac{{\frac{i}{3}} \left ( \arctan \left ( c{x}^{3} \right ) \right ) ^{2}{b}^{2}}{c}}+{\frac{2\,\arctan \left ( c{x}^{3} \right ){b}^{2}}{3\,c}\ln \left ({\frac{ \left ( 1+ic{x}^{3} \right ) ^{2}}{{c}^{2}{x}^{6}+1}}+1 \right ) }-{\frac{{\frac{i}{3}}{b}^{2}}{c}{\it polylog} \left ( 2,-{\frac{ \left ( 1+ic{x}^{3} \right ) ^{2}}{{c}^{2}{x}^{6}+1}} \right ) }-{\frac{ab\ln \left ({c}^{2}{x}^{6}+1 \right ) }{3\,c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*b^2*arctan(c*x^3)^2+2/3*x^3*a*b*arctan(c*x^3)+1/3*x^3*a^2-1/3*I/c*arctan(c*x^3)^2*b^2+2/3/c*arctan(c*x
^3)*ln((1+I*c*x^3)^2/(c^2*x^6+1)+1)*b^2-1/3*I/c*polylog(2,-(1+I*c*x^3)^2/(c^2*x^6+1))*b^2-1/3/c*a*b*ln(c^2*x^6
+1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{3} \, a^{2} x^{3} + \frac{1}{48} \,{\left (4 \, x^{3} \arctan \left (c x^{3}\right )^{2} - x^{3} \log \left (c^{2} x^{6} + 1\right )^{2} + 576 \, c^{2} \int \frac{x^{8} \arctan \left (c x^{3}\right )^{2}}{16 \,{\left (c^{2} x^{6} + 1\right )}}\,{d x} + 48 \, c^{2} \int \frac{x^{8} \log \left (c^{2} x^{6} + 1\right )^{2}}{16 \,{\left (c^{2} x^{6} + 1\right )}}\,{d x} + 192 \, c^{2} \int \frac{x^{8} \log \left (c^{2} x^{6} + 1\right )}{16 \,{\left (c^{2} x^{6} + 1\right )}}\,{d x} + \frac{4 \, \arctan \left (c x^{3}\right )^{3}}{c} - 384 \, c \int \frac{x^{5} \arctan \left (c x^{3}\right )}{16 \,{\left (c^{2} x^{6} + 1\right )}}\,{d x} + 48 \, \int \frac{x^{2} \log \left (c^{2} x^{6} + 1\right )^{2}}{16 \,{\left (c^{2} x^{6} + 1\right )}}\,{d x}\right )} b^{2} + \frac{{\left (2 \, c x^{3} \arctan \left (c x^{3}\right ) - \log \left (c^{2} x^{6} + 1\right )\right )} a b}{3 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a^2*x^3 + 1/48*(4*x^3*arctan(c*x^3)^2 - x^3*log(c^2*x^6 + 1)^2 + 576*c^2*integrate(1/16*x^8*arctan(c*x^3)^
2/(c^2*x^6 + 1), x) + 48*c^2*integrate(1/16*x^8*log(c^2*x^6 + 1)^2/(c^2*x^6 + 1), x) + 192*c^2*integrate(1/16*
x^8*log(c^2*x^6 + 1)/(c^2*x^6 + 1), x) + 4*arctan(c*x^3)^3/c - 384*c*integrate(1/16*x^5*arctan(c*x^3)/(c^2*x^6
+ 1), x) + 48*integrate(1/16*x^2*log(c^2*x^6 + 1)^2/(c^2*x^6 + 1), x))*b^2 + 1/3*(2*c*x^3*arctan(c*x^3) - log
(c^2*x^6 + 1))*a*b/c

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (b^{2} x^{2} \arctan \left (c x^{3}\right )^{2} + 2 \, a b x^{2} \arctan \left (c x^{3}\right ) + a^{2} x^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \left (a + b \operatorname{atan}{\left (c x^{3} \right )}\right )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*(a + b*atan(c*x**3))**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \arctan \left (c x^{3}\right ) + a\right )}^{2} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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