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

Optimal. Leaf size=196 \[ -\frac {1}{2} a^2 c \text {Li}_3\left (1-\frac {2}{i a x+1}\right )+\frac {1}{2} a^2 c \text {Li}_3\left (\frac {2}{i a x+1}-1\right )-i a^2 c \text {Li}_2\left (1-\frac {2}{i a x+1}\right ) \tan ^{-1}(a x)+i a^2 c \text {Li}_2\left (\frac {2}{i a x+1}-1\right ) \tan ^{-1}(a x)-\frac {1}{2} a^2 c \log \left (a^2 x^2+1\right )+a^2 c \log (x)-\frac {1}{2} a^2 c \tan ^{-1}(a x)^2+2 a^2 c \tan ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac {2}{1+i a x}\right )-\frac {c \tan ^{-1}(a x)^2}{2 x^2}-\frac {a c \tan ^{-1}(a x)}{x} \]

[Out]

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

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Rubi [A]  time = 0.33, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 12, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4950, 4852, 4918, 266, 36, 29, 31, 4884, 4850, 4988, 4994, 6610} \[ -\frac {1}{2} a^2 c \text {PolyLog}\left (3,1-\frac {2}{1+i a x}\right )+\frac {1}{2} a^2 c \text {PolyLog}\left (3,-1+\frac {2}{1+i a x}\right )-i a^2 c \tan ^{-1}(a x) \text {PolyLog}\left (2,1-\frac {2}{1+i a x}\right )+i a^2 c \tan ^{-1}(a x) \text {PolyLog}\left (2,-1+\frac {2}{1+i a x}\right )-\frac {1}{2} a^2 c \log \left (a^2 x^2+1\right )+a^2 c \log (x)-\frac {1}{2} a^2 c \tan ^{-1}(a x)^2+2 a^2 c \tan ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac {2}{1+i a x}\right )-\frac {c \tan ^{-1}(a x)^2}{2 x^2}-\frac {a c \tan ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 4918

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d,
Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Dist[e/(d*f^2), 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] && LtQ[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^3} \, dx &=c \int \frac {\tan ^{-1}(a x)^2}{x^3} \, dx+\left (a^2 c\right ) \int \frac {\tan ^{-1}(a x)^2}{x} \, dx\\ &=-\frac {c \tan ^{-1}(a x)^2}{2 x^2}+2 a^2 c \tan ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac {2}{1+i a x}\right )+(a c) \int \frac {\tan ^{-1}(a x)}{x^2 \left (1+a^2 x^2\right )} \, dx-\left (4 a^3 c\right ) \int \frac {\tan ^{-1}(a x) \tanh ^{-1}\left (1-\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx\\ &=-\frac {c \tan ^{-1}(a x)^2}{2 x^2}+2 a^2 c \tan ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac {2}{1+i a x}\right )+(a c) \int \frac {\tan ^{-1}(a x)}{x^2} \, dx-\left (a^3 c\right ) \int \frac {\tan ^{-1}(a x)}{1+a^2 x^2} \, dx+\left (2 a^3 c\right ) \int \frac {\tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx-\left (2 a^3 c\right ) \int \frac {\tan ^{-1}(a x) \log \left (2-\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx\\ &=-\frac {a c \tan ^{-1}(a x)}{x}-\frac {1}{2} a^2 c \tan ^{-1}(a x)^2-\frac {c \tan ^{-1}(a x)^2}{2 x^2}+2 a^2 c \tan ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac {2}{1+i a x}\right )-i a^2 c \tan ^{-1}(a x) \text {Li}_2\left (1-\frac {2}{1+i a x}\right )+i a^2 c \tan ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1+i a x}\right )+\left (a^2 c\right ) \int \frac {1}{x \left (1+a^2 x^2\right )} \, dx+\left (i a^3 c\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx-\left (i a^3 c\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx\\ &=-\frac {a c \tan ^{-1}(a x)}{x}-\frac {1}{2} a^2 c \tan ^{-1}(a x)^2-\frac {c \tan ^{-1}(a x)^2}{2 x^2}+2 a^2 c \tan ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac {2}{1+i a x}\right )-i a^2 c \tan ^{-1}(a x) \text {Li}_2\left (1-\frac {2}{1+i a x}\right )+i a^2 c \tan ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1+i a x}\right )-\frac {1}{2} a^2 c \text {Li}_3\left (1-\frac {2}{1+i a x}\right )+\frac {1}{2} a^2 c \text {Li}_3\left (-1+\frac {2}{1+i a x}\right )+\frac {1}{2} \left (a^2 c\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1+a^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {a c \tan ^{-1}(a x)}{x}-\frac {1}{2} a^2 c \tan ^{-1}(a x)^2-\frac {c \tan ^{-1}(a x)^2}{2 x^2}+2 a^2 c \tan ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac {2}{1+i a x}\right )-i a^2 c \tan ^{-1}(a x) \text {Li}_2\left (1-\frac {2}{1+i a x}\right )+i a^2 c \tan ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1+i a x}\right )-\frac {1}{2} a^2 c \text {Li}_3\left (1-\frac {2}{1+i a x}\right )+\frac {1}{2} a^2 c \text {Li}_3\left (-1+\frac {2}{1+i a x}\right )+\frac {1}{2} \left (a^2 c\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{2} \left (a^4 c\right ) \operatorname {Subst}\left (\int \frac {1}{1+a^2 x} \, dx,x,x^2\right )\\ &=-\frac {a c \tan ^{-1}(a x)}{x}-\frac {1}{2} a^2 c \tan ^{-1}(a x)^2-\frac {c \tan ^{-1}(a x)^2}{2 x^2}+2 a^2 c \tan ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac {2}{1+i a x}\right )+a^2 c \log (x)-\frac {1}{2} a^2 c \log \left (1+a^2 x^2\right )-i a^2 c \tan ^{-1}(a x) \text {Li}_2\left (1-\frac {2}{1+i a x}\right )+i a^2 c \tan ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1+i a x}\right )-\frac {1}{2} a^2 c \text {Li}_3\left (1-\frac {2}{1+i a x}\right )+\frac {1}{2} a^2 c \text {Li}_3\left (-1+\frac {2}{1+i a x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 208, normalized size = 1.06 \[ \frac {1}{2} a^2 c \text {Li}_3\left (\frac {-a x-i}{a x-i}\right )-\frac {1}{2} a^2 c \text {Li}_3\left (\frac {a x+i}{a x-i}\right )+i a^2 c \text {Li}_2\left (\frac {-a x-i}{a x-i}\right ) \tan ^{-1}(a x)-i a^2 c \text {Li}_2\left (\frac {a x+i}{a x-i}\right ) \tan ^{-1}(a x)-\frac {1}{2} a^2 c \log \left (a^2 x^2+1\right )+\frac {c \left (-a^2 x^2-1\right ) \tan ^{-1}(a x)^2}{2 x^2}+a^2 c \log (x)+2 a^2 c \tan ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac {2 i}{-a x+i}\right )-\frac {a c \tan ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{2} c x^{2} + c\right )} \arctan \left (a x\right )^{2}}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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maple [C]  time = 6.32, size = 1167, normalized size = 5.95 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (72 \, a^{4} c \int \frac {x^{4} \arctan \left (a x\right )^{2}}{a^{2} x^{5} + x^{3}}\,{d x} + a^{2} c \log \left (a^{2} x^{2} + 1\right )^{3} - 3 \, {\left (a^{2} {\left (\frac {\log \left (a^{2} x^{2} + 1\right )^{2}}{a^{2}} - \frac {2 \, {\left (2 \, \log \left (a^{2} x^{2} + 1\right ) \log \relax (x) + {\rm Li}_2\left (-a^{2} x^{2}\right )\right )}}{a^{2}}\right )} - 2 \, {\left (\log \left (a^{2} x^{2} + 1\right ) - 2 \, \log \relax (x)\right )} \log \left (a^{2} x^{2} + 1\right )\right )} a^{2} c - 2 \, {\left (\log \left (a^{2} x^{2} + 1\right )^{3} - 3 \, \log \left (a^{2} x^{2} + 1\right )^{2} \log \left (-a^{2} x^{2}\right ) - 6 \, {\rm Li}_2\left (a^{2} x^{2} + 1\right ) \log \left (a^{2} x^{2} + 1\right ) + 6 \, {\rm Li}_{3}(a^{2} x^{2} + 1)\right )} a^{2} c + 144 \, a^{2} c \int \frac {x^{2} \arctan \left (a x\right )^{2}}{a^{2} x^{5} + x^{3}}\,{d x} + 12 \, {\left ({\left (\arctan \left (a x\right )^{2} - \log \left (a^{2} x^{2} + 1\right ) + 2 \, \log \relax (x)\right )} a - 2 \, {\left (a \arctan \left (a x\right ) + \frac {1}{x}\right )} \arctan \left (a x\right )\right )} a c - {\left (3 \, {\left (\log \left (a^{2} x^{2} + 1\right )^{2} \log \left (-a^{2} x^{2}\right ) + 2 \, {\rm Li}_2\left (a^{2} x^{2} + 1\right ) \log \left (a^{2} x^{2} + 1\right ) - 2 \, {\rm Li}_{3}(a^{2} x^{2} + 1)\right )} a^{2} - 6 \, {\left (\log \left (a^{2} x^{2} + 1\right ) \log \left (-a^{2} x^{2}\right ) + {\rm Li}_2\left (a^{2} x^{2} + 1\right )\right )} a^{2} - \frac {a^{2} x^{2} \log \left (a^{2} x^{2} + 1\right )^{3} - 3 \, {\left (a^{2} x^{2} + 1\right )} \log \left (a^{2} x^{2} + 1\right )^{2}}{x^{2}}\right )} c + 72 \, c \int \frac {\arctan \left (a x\right )^{2}}{a^{2} x^{5} + x^{3}}\,{d x}\right )} x^{2} - 12 \, c \arctan \left (a x\right )^{2} + 3 \, c \log \left (a^{2} x^{2} + 1\right )^{2}}{96 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*((1152*a^4*c*integrate(1/16*x^4*arctan(a*x)^2/(a^2*x^5 + x^3), x) + a^2*c*log(a^2*x^2 + 1)^3 + 2304*a^2*c
*integrate(1/16*x^2*arctan(a*x)^2/(a^2*x^5 + x^3), x) + 192*a^2*c*integrate(1/16*x^2*log(a^2*x^2 + 1)^2/(a^2*x
^5 + x^3), x) - 192*a^2*c*integrate(1/16*x^2*log(a^2*x^2 + 1)/(a^2*x^5 + x^3), x) + 384*a*c*integrate(1/16*x*a
rctan(a*x)/(a^2*x^5 + x^3), x) + 1152*c*integrate(1/16*arctan(a*x)^2/(a^2*x^5 + x^3), x) + 96*c*integrate(1/16
*log(a^2*x^2 + 1)^2/(a^2*x^5 + x^3), x))*x^2 - 12*c*arctan(a*x)^2 + 3*c*log(a^2*x^2 + 1)^2)/x^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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