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

Optimal. Leaf size=133 \[ 3 b^2 c^2 \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-\frac {3}{2} i b c^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {1}{2} c^2 \left (a+b \tan ^{-1}(c x)\right )^3-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{2 x^2}-\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2}{2 x}-\frac {3}{2} i b^3 c^2 \text {Li}_2\left (\frac {2}{1-i c x}-1\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.28, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4852, 4918, 4924, 4868, 2447, 4884} \[ -\frac {3}{2} i b^3 c^2 \text {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )+3 b^2 c^2 \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-\frac {3}{2} i b c^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {1}{2} c^2 \left (a+b \tan ^{-1}(c x)\right )^3-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{2 x^2}-\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x])^3/x^3,x]

[Out]

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

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

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 4868

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[((a + b*ArcTan[c*x]
)^p*Log[2 - 2/(1 + (e*x)/d)])/d, x] - Dist[(b*c*p)/d, Int[((a + b*ArcTan[c*x])^(p - 1)*Log[2 - 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 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 4924

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.34, size = 176, normalized size = 1.32 \[ -\frac {a \left (a (a+3 b c x)-6 b^2 c^2 x^2 \log \left (\frac {c x}{\sqrt {c^2 x^2+1}}\right )\right )+3 b^2 \tan ^{-1}(c x)^2 \left (a c^2 x^2+a+b c x (1+i c x)\right )+3 b \tan ^{-1}(c x) \left (a \left (a c^2 x^2+a+2 b c x\right )-2 b^2 c^2 x^2 \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )\right )+3 i b^3 c^2 x^2 \text {Li}_2\left (e^{2 i \tan ^{-1}(c x)}\right )+b^3 \left (c^2 x^2+1\right ) \tan ^{-1}(c x)^3}{2 x^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \arctan \left (c x\right )^{3} + 3 \, a b^{2} \arctan \left (c x\right )^{2} + 3 \, a^{2} b \arctan \left (c x\right ) + a^{3}}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

maple [B]  time = 0.02, size = 457, normalized size = 3.44 \[ -\frac {3 i c^{2} b^{3} \ln \left (c x +i\right )^{2}}{8}+\frac {3 i c^{2} b^{3} \ln \left (c x -i\right )^{2}}{8}-\frac {3 i c^{2} b^{3} \dilog \left (-i c x +1\right )}{2}+\frac {3 i c^{2} b^{3} \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{4}-\frac {3 i c^{2} b^{3} \dilog \left (\frac {i \left (c x -i\right )}{2}\right )}{4}+\frac {3 i c^{2} b^{3} \dilog \left (i c x +1\right )}{2}+3 c^{2} a \,b^{2} \ln \left (c x \right )-\frac {3 c^{2} a \,b^{2} \arctan \left (c x \right )^{2}}{2}-\frac {3 c \,a^{2} b}{2 x}-\frac {3 c^{2} b^{3} \arctan \left (c x \right ) \ln \left (c^{2} x^{2}+1\right )}{2}+3 c^{2} b^{3} \arctan \left (c x \right ) \ln \left (c x \right )-\frac {3 c^{2} a^{2} b \arctan \left (c x \right )}{2}-\frac {3 c^{2} a \,b^{2} \ln \left (c^{2} x^{2}+1\right )}{2}-\frac {3 a \,b^{2} \arctan \left (c x \right )^{2}}{2 x^{2}}-\frac {3 a^{2} b \arctan \left (c x \right )}{2 x^{2}}-\frac {3 c \,b^{3} \arctan \left (c x \right )^{2}}{2 x}-\frac {a^{3}}{2 x^{2}}+\frac {3 i c^{2} b^{3} \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {3 i c^{2} b^{3} \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{4}-\frac {3 i c^{2} b^{3} \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {3 i c^{2} b^{3} \ln \left (c x -i\right ) \ln \left (c^{2} x^{2}+1\right )}{4}-\frac {3 i c^{2} b^{3} \ln \left (c x +i\right ) \ln \left (\frac {i \left (c x -i\right )}{2}\right )}{4}+\frac {3 i c^{2} b^{3} \ln \left (c x +i\right ) \ln \left (c^{2} x^{2}+1\right )}{4}-\frac {b^{3} \arctan \left (c x \right )^{3}}{2 x^{2}}-\frac {c^{2} b^{3} \arctan \left (c x \right )^{3}}{2}-\frac {3 c a \,b^{2} \arctan \left (c x \right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))^3/x^3,x)

[Out]

int((a + b*atan(c*x))^3/x^3, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________