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

Optimal. Leaf size=131 \[ -\frac{3 i b^3 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{2 c^2}-\frac{3 b^2 \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^2}-\frac{3 i b \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}+\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{2 c^2}+\frac{1}{2} x^2 \left (a+b \tan ^{-1}(c x)\right )^3-\frac{3 b x \left (a+b \tan ^{-1}(c x)\right )^2}{2 c} \]

[Out]

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

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Rubi [A]  time = 0.238823, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {4852, 4916, 4846, 4920, 4854, 2402, 2315, 4884} \[ -\frac{3 i b^3 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{2 c^2}-\frac{3 b^2 \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^2}-\frac{3 i b \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}+\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{2 c^2}+\frac{1}{2} x^2 \left (a+b \tan ^{-1}(c x)\right )^3-\frac{3 b x \left (a+b \tan ^{-1}(c x)\right )^2}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcTan[c*x])^3,x]

[Out]

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

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 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 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 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 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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.283898, size = 152, normalized size = 1.16 \[ \frac{3 i b^3 \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+a \left (a c x (a c x-3 b)+3 b^2 \log \left (c^2 x^2+1\right )\right )+3 b^2 \tan ^{-1}(c x)^2 \left (a c^2 x^2+a+b (-c x+i)\right )+3 b \tan ^{-1}(c x) \left (a \left (a c^2 x^2+a-2 b c x\right )-2 b^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )+b^3 \left (c^2 x^2+1\right ) \tan ^{-1}(c x)^3}{2 c^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x*(a + b*ArcTan[c*x])^3,x]

[Out]

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

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Maple [B]  time = 0.089, size = 352, normalized size = 2.7 \begin{align*}{\frac{{x}^{2}{a}^{3}}{2}}+{\frac{{x}^{2}{b}^{3} \left ( \arctan \left ( cx \right ) \right ) ^{3}}{2}}+{\frac{{b}^{3} \left ( \arctan \left ( cx \right ) \right ) ^{3}}{2\,{c}^{2}}}-{\frac{3\,{b}^{3} \left ( \arctan \left ( cx \right ) \right ) ^{2}x}{2\,c}}+{\frac{3\,{b}^{3}\arctan \left ( cx \right ) \ln \left ({c}^{2}{x}^{2}+1 \right ) }{2\,{c}^{2}}}-{\frac{{\frac{3\,i}{4}}{b}^{3}{\it dilog} \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{{c}^{2}}}-{\frac{{\frac{3\,i}{8}}{b}^{3} \left ( \ln \left ( cx-i \right ) \right ) ^{2}}{{c}^{2}}}-{\frac{{\frac{3\,i}{4}}{b}^{3}\ln \left ({c}^{2}{x}^{2}+1 \right ) \ln \left ( cx+i \right ) }{{c}^{2}}}-{\frac{{\frac{3\,i}{4}}{b}^{3}\ln \left ( cx-i \right ) \ln \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{{c}^{2}}}+{\frac{{\frac{3\,i}{8}}{b}^{3} \left ( \ln \left ( cx+i \right ) \right ) ^{2}}{{c}^{2}}}+{\frac{{\frac{3\,i}{4}}{b}^{3}{\it dilog} \left ({\frac{i}{2}} \left ( cx-i \right ) \right ) }{{c}^{2}}}+{\frac{{\frac{3\,i}{4}}{b}^{3}\ln \left ({c}^{2}{x}^{2}+1 \right ) \ln \left ( cx-i \right ) }{{c}^{2}}}+{\frac{{\frac{3\,i}{4}}{b}^{3}\ln \left ( cx+i \right ) \ln \left ({\frac{i}{2}} \left ( cx-i \right ) \right ) }{{c}^{2}}}+{\frac{3\,{x}^{2}a{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}}{2}}+{\frac{3\,a{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}}{2\,{c}^{2}}}-3\,{\frac{a{b}^{2}x\arctan \left ( cx \right ) }{c}}+{\frac{3\,a{b}^{2}\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2\,{c}^{2}}}+{\frac{3\,b{a}^{2}{x}^{2}\arctan \left ( cx \right ) }{2}}-{\frac{3\,{a}^{2}xb}{2\,c}}+{\frac{3\,{a}^{2}b\arctan \left ( cx \right ) }{2\,{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*a^3+1/2*x^2*b^3*arctan(c*x)^3+1/2/c^2*b^3*arctan(c*x)^3-3/2/c*b^3*arctan(c*x)^2*x+3/2/c^2*b^3*arctan(c
*x)*ln(c^2*x^2+1)-3/4*I/c^2*b^3*dilog(-1/2*I*(c*x+I))-3/8*I/c^2*b^3*ln(c*x-I)^2-3/4*I/c^2*b^3*ln(c^2*x^2+1)*ln
(c*x+I)-3/4*I/c^2*b^3*ln(c*x-I)*ln(-1/2*I*(c*x+I))+3/8*I/c^2*b^3*ln(c*x+I)^2+3/4*I/c^2*b^3*dilog(1/2*I*(c*x-I)
)+3/4*I/c^2*b^3*ln(c^2*x^2+1)*ln(c*x-I)+3/4*I/c^2*b^3*ln(c*x+I)*ln(1/2*I*(c*x-I))+3/2*x^2*a*b^2*arctan(c*x)^2+
3/2/c^2*a*b^2*arctan(c*x)^2-3/c*a*b^2*x*arctan(c*x)+3/2/c^2*a*b^2*ln(c^2*x^2+1)+3/2*b*a^2*x^2*arctan(c*x)-3/2/
c*x*a^2*b+3/2/c^2*a^2*b*arctan(c*x)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*(a + b*atan(c*x))**3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arctan(c*x) + a)^3*x, x)