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

Optimal. Leaf size=160 \[ -\frac {i c \text {Li}_2\left (1-\frac {2}{i a x+1}\right )}{2 a^2}+\frac {c \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^3}{4 a^2}-\frac {c x \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^2}{4 a}+\frac {c \left (a^2 x^2+1\right ) \tan ^{-1}(a x)}{4 a^2}-\frac {i c \tan ^{-1}(a x)^2}{2 a^2}-\frac {c \log \left (\frac {2}{1+i a x}\right ) \tan ^{-1}(a x)}{a^2}-\frac {c x}{4 a}-\frac {c x \tan ^{-1}(a x)^2}{2 a} \]

[Out]

-1/4*c*x/a+1/4*c*(a^2*x^2+1)*arctan(a*x)/a^2-1/2*I*c*arctan(a*x)^2/a^2-1/2*c*x*arctan(a*x)^2/a-1/4*c*x*(a^2*x^
2+1)*arctan(a*x)^2/a+1/4*c*(a^2*x^2+1)^2*arctan(a*x)^3/a^2-c*arctan(a*x)*ln(2/(1+I*a*x))/a^2-1/2*I*c*polylog(2
,1-2/(1+I*a*x))/a^2

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Rubi [A]  time = 0.13, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {4930, 4880, 4846, 4920, 4854, 2402, 2315, 8} \[ -\frac {i c \text {PolyLog}\left (2,1-\frac {2}{1+i a x}\right )}{2 a^2}+\frac {c \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^3}{4 a^2}-\frac {c x \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^2}{4 a}+\frac {c \left (a^2 x^2+1\right ) \tan ^{-1}(a x)}{4 a^2}-\frac {i c \tan ^{-1}(a x)^2}{2 a^2}-\frac {c \log \left (\frac {2}{1+i a x}\right ) \tan ^{-1}(a x)}{a^2}-\frac {c x}{4 a}-\frac {c x \tan ^{-1}(a x)^2}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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 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 4880

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> -Simp[(b*p*(d + e*x^2)^q
*(a + b*ArcTan[c*x])^(p - 1))/(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])^p, x], x] + Dist[(b^2*d*p*(p - 1))/(2*q*(2*q + 1)), Int[(d + e*x^2)^(q - 1)*(a + b*ArcTan[c*x])^(
p - 2), x], x] + Simp[(x*(d + e*x^2)^q*(a + b*ArcTan[c*x])^p)/(2*q + 1), x]) /; FreeQ[{a, b, c, d, e}, x] && E
qQ[e, c^2*d] && GtQ[q, 0] && GtQ[p, 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 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 ) \tan ^{-1}(a x)^3 \, dx &=\frac {c \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^3}{4 a^2}-\frac {3 \int \left (c+a^2 c x^2\right ) \tan ^{-1}(a x)^2 \, dx}{4 a}\\ &=\frac {c \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}{4 a^2}-\frac {c x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^2}{4 a}+\frac {c \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^3}{4 a^2}-\frac {c \int 1 \, dx}{4 a}-\frac {c \int \tan ^{-1}(a x)^2 \, dx}{2 a}\\ &=-\frac {c x}{4 a}+\frac {c \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}{4 a^2}-\frac {c x \tan ^{-1}(a x)^2}{2 a}-\frac {c x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^2}{4 a}+\frac {c \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^3}{4 a^2}+c \int \frac {x \tan ^{-1}(a x)}{1+a^2 x^2} \, dx\\ &=-\frac {c x}{4 a}+\frac {c \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}{4 a^2}-\frac {i c \tan ^{-1}(a x)^2}{2 a^2}-\frac {c x \tan ^{-1}(a x)^2}{2 a}-\frac {c x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^2}{4 a}+\frac {c \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^3}{4 a^2}-\frac {c \int \frac {\tan ^{-1}(a x)}{i-a x} \, dx}{a}\\ &=-\frac {c x}{4 a}+\frac {c \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}{4 a^2}-\frac {i c \tan ^{-1}(a x)^2}{2 a^2}-\frac {c x \tan ^{-1}(a x)^2}{2 a}-\frac {c x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^2}{4 a}+\frac {c \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^3}{4 a^2}-\frac {c \tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )}{a^2}+\frac {c \int \frac {\log \left (\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx}{a}\\ &=-\frac {c x}{4 a}+\frac {c \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}{4 a^2}-\frac {i c \tan ^{-1}(a x)^2}{2 a^2}-\frac {c x \tan ^{-1}(a x)^2}{2 a}-\frac {c x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^2}{4 a}+\frac {c \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^3}{4 a^2}-\frac {c \tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )}{a^2}-\frac {(i c) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i a x}\right )}{a^2}\\ &=-\frac {c x}{4 a}+\frac {c \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}{4 a^2}-\frac {i c \tan ^{-1}(a x)^2}{2 a^2}-\frac {c x \tan ^{-1}(a x)^2}{2 a}-\frac {c x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^2}{4 a}+\frac {c \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^3}{4 a^2}-\frac {c \tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )}{a^2}-\frac {i c \text {Li}_2\left (1-\frac {2}{1+i a x}\right )}{2 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 101, normalized size = 0.63 \[ \frac {c \left (-\left (a^3 x^3+3 a x-2 i\right ) \tan ^{-1}(a x)^2+\left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^3+\tan ^{-1}(a x) \left (a^2 x^2-4 \log \left (1+e^{2 i \tan ^{-1}(a x)}\right )+1\right )+2 i \text {Li}_2\left (-e^{2 i \tan ^{-1}(a x)}\right )-a x\right )}{4 a^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{2} c x^{3} + c x\right )} \arctan \left (a x\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

sage0*x

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maple [A]  time = 0.12, size = 276, normalized size = 1.72 \[ \frac {a^{2} c \arctan \left (a x \right )^{3} x^{4}}{4}+\frac {c \arctan \left (a x \right )^{3} x^{2}}{2}-\frac {a c \arctan \left (a x \right )^{2} x^{3}}{4}-\frac {3 c x \arctan \left (a x \right )^{2}}{4 a}+\frac {c \arctan \left (a x \right )^{3}}{4 a^{2}}+\frac {c \arctan \left (a x \right ) x^{2}}{4}+\frac {c \arctan \left (a x \right ) \ln \left (a^{2} x^{2}+1\right )}{2 a^{2}}-\frac {c x}{4 a}+\frac {c \arctan \left (a x \right )}{4 a^{2}}-\frac {i c \ln \left (a x +i\right ) \ln \left (a^{2} x^{2}+1\right )}{4 a^{2}}+\frac {i c \dilog \left (\frac {i \left (a x -i\right )}{2}\right )}{4 a^{2}}-\frac {i c \ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )}{4 a^{2}}+\frac {i c \ln \left (a x -i\right ) \ln \left (a^{2} x^{2}+1\right )}{4 a^{2}}+\frac {i c \ln \left (a x +i\right )^{2}}{8 a^{2}}-\frac {i c \dilog \left (-\frac {i \left (a x +i\right )}{2}\right )}{4 a^{2}}+\frac {i c \ln \left (a x +i\right ) \ln \left (\frac {i \left (a x -i\right )}{2}\right )}{4 a^{2}}-\frac {i c \ln \left (a x -i\right )^{2}}{8 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,{\mathrm {atan}\left (a\,x\right )}^3\,\left (c\,a^2\,x^2+c\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*atan(a*x)^3*(c + a^2*c*x^2),x)

[Out]

int(x*atan(a*x)^3*(c + a^2*c*x^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ c \left (\int x \operatorname {atan}^{3}{\left (a x \right )}\, dx + \int a^{2} x^{3} \operatorname {atan}^{3}{\left (a x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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