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

Optimal. Leaf size=169 \[ \frac {1}{2} c \log \left (a^2 x^2+1\right )+\frac {1}{2} a^2 c x^2 \tan ^{-1}(a x)^2-\frac {1}{2} c \text {Li}_3\left (1-\frac {2}{i a x+1}\right )+\frac {1}{2} c \text {Li}_3\left (\frac {2}{i a x+1}-1\right )-i c \text {Li}_2\left (1-\frac {2}{i a x+1}\right ) \tan ^{-1}(a x)+i c \text {Li}_2\left (\frac {2}{i a x+1}-1\right ) \tan ^{-1}(a x)+\frac {1}{2} c \tan ^{-1}(a x)^2-a c x \tan ^{-1}(a x)+2 c \tan ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac {2}{1+i a x}\right ) \]

[Out]

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

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Rubi [A]  time = 0.31, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4950, 4850, 4988, 4884, 4994, 6610, 4852, 4916, 4846, 260} \[ -\frac {1}{2} c \text {PolyLog}\left (3,1-\frac {2}{1+i a x}\right )+\frac {1}{2} c \text {PolyLog}\left (3,-1+\frac {2}{1+i a x}\right )-i c \tan ^{-1}(a x) \text {PolyLog}\left (2,1-\frac {2}{1+i a x}\right )+i c \tan ^{-1}(a x) \text {PolyLog}\left (2,-1+\frac {2}{1+i a x}\right )+\frac {1}{2} c \log \left (a^2 x^2+1\right )+\frac {1}{2} a^2 c x^2 \tan ^{-1}(a x)^2+\frac {1}{2} c \tan ^{-1}(a x)^2-a c x \tan ^{-1}(a x)+2 c \tan ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac {2}{1+i a x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 4850

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

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]))

Rule 4988

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

Rule 4994

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

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin {align*} \int \frac {\left (c+a^2 c x^2\right ) \tan ^{-1}(a x)^2}{x} \, dx &=c \int \frac {\tan ^{-1}(a x)^2}{x} \, dx+\left (a^2 c\right ) \int x \tan ^{-1}(a x)^2 \, dx\\ &=\frac {1}{2} a^2 c x^2 \tan ^{-1}(a x)^2+2 c \tan ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac {2}{1+i a x}\right )-(4 a c) \int \frac {\tan ^{-1}(a x) \tanh ^{-1}\left (1-\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx-\left (a^3 c\right ) \int \frac {x^2 \tan ^{-1}(a x)}{1+a^2 x^2} \, dx\\ &=\frac {1}{2} a^2 c x^2 \tan ^{-1}(a x)^2+2 c \tan ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac {2}{1+i a x}\right )-(a c) \int \tan ^{-1}(a x) \, dx+(a c) \int \frac {\tan ^{-1}(a x)}{1+a^2 x^2} \, dx+(2 a c) \int \frac {\tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx-(2 a c) \int \frac {\tan ^{-1}(a x) \log \left (2-\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx\\ &=-a c x \tan ^{-1}(a x)+\frac {1}{2} c \tan ^{-1}(a x)^2+\frac {1}{2} a^2 c x^2 \tan ^{-1}(a x)^2+2 c \tan ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac {2}{1+i a x}\right )-i c \tan ^{-1}(a x) \text {Li}_2\left (1-\frac {2}{1+i a x}\right )+i c \tan ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1+i a x}\right )+(i a c) \int \frac {\text {Li}_2\left (1-\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx-(i a c) \int \frac {\text {Li}_2\left (-1+\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx+\left (a^2 c\right ) \int \frac {x}{1+a^2 x^2} \, dx\\ &=-a c x \tan ^{-1}(a x)+\frac {1}{2} c \tan ^{-1}(a x)^2+\frac {1}{2} a^2 c x^2 \tan ^{-1}(a x)^2+2 c \tan ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac {2}{1+i a x}\right )+\frac {1}{2} c \log \left (1+a^2 x^2\right )-i c \tan ^{-1}(a x) \text {Li}_2\left (1-\frac {2}{1+i a x}\right )+i c \tan ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1+i a x}\right )-\frac {1}{2} c \text {Li}_3\left (1-\frac {2}{1+i a x}\right )+\frac {1}{2} c \text {Li}_3\left (-1+\frac {2}{1+i a x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 177, normalized size = 1.05 \[ \frac {1}{2} c \log \left (a^2 x^2+1\right )+\frac {1}{2} c \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^2+\frac {1}{2} c \text {Li}_3\left (\frac {-a x-i}{a x-i}\right )-\frac {1}{2} c \text {Li}_3\left (\frac {a x+i}{a x-i}\right )+i c \text {Li}_2\left (\frac {-a x-i}{a x-i}\right ) \tan ^{-1}(a x)-i c \text {Li}_2\left (\frac {a x+i}{a x-i}\right ) \tan ^{-1}(a x)-a c x \tan ^{-1}(a x)+2 c \tan ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac {2 i}{-a x+i}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{2} c x^{2} + c\right )} \arctan \left (a x\right )^{2}}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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maple [C]  time = 3.75, size = 1078, normalized size = 6.38 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{8} \, a^{2} c x^{2} \arctan \left (a x\right )^{2} - \frac {1}{32} \, a^{2} c x^{2} \log \left (a^{2} x^{2} + 1\right )^{2} + 12 \, a^{4} c \int \frac {x^{4} \arctan \left (a x\right )^{2}}{16 \, {\left (a^{2} x^{3} + x\right )}}\,{d x} + a^{4} c \int \frac {x^{4} \log \left (a^{2} x^{2} + 1\right )^{2}}{16 \, {\left (a^{2} x^{3} + x\right )}}\,{d x} + 2 \, a^{4} c \int \frac {x^{4} \log \left (a^{2} x^{2} + 1\right )}{16 \, {\left (a^{2} x^{3} + x\right )}}\,{d x} - 4 \, a^{3} c \int \frac {x^{3} \arctan \left (a x\right )}{16 \, {\left (a^{2} x^{3} + x\right )}}\,{d x} + 24 \, a^{2} c \int \frac {x^{2} \arctan \left (a x\right )^{2}}{16 \, {\left (a^{2} x^{3} + x\right )}}\,{d x} + \frac {1}{48} \, c \log \left (a^{2} x^{2} + 1\right )^{3} + 12 \, c \int \frac {\arctan \left (a x\right )^{2}}{16 \, {\left (a^{2} x^{3} + x\right )}}\,{d x} + c \int \frac {\log \left (a^{2} x^{2} + 1\right )^{2}}{16 \, {\left (a^{2} x^{3} + x\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a^2*c*x^2*arctan(a*x)^2 - 1/32*a^2*c*x^2*log(a^2*x^2 + 1)^2 + 12*a^4*c*integrate(1/16*x^4*arctan(a*x)^2/(a
^2*x^3 + x), x) + a^4*c*integrate(1/16*x^4*log(a^2*x^2 + 1)^2/(a^2*x^3 + x), x) + 2*a^4*c*integrate(1/16*x^4*l
og(a^2*x^2 + 1)/(a^2*x^3 + x), x) - 4*a^3*c*integrate(1/16*x^3*arctan(a*x)/(a^2*x^3 + x), x) + 24*a^2*c*integr
ate(1/16*x^2*arctan(a*x)^2/(a^2*x^3 + x), x) + 1/48*c*log(a^2*x^2 + 1)^3 + 12*c*integrate(1/16*arctan(a*x)^2/(
a^2*x^3 + x), x) + c*integrate(1/16*log(a^2*x^2 + 1)^2/(a^2*x^3 + x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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