3.34 \(\int \frac {(a+b \tan ^{-1}(c x))^3}{x^5} \, dx\)

Optimal. Leaf size=198 \[ -2 b^2 c^4 \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-\frac {b^2 c^2 \left (a+b \tan ^{-1}(c x)\right )}{4 x^2}+\frac {1}{4} c^4 \left (a+b \tan ^{-1}(c x)\right )^3+i b c^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {3 b c^3 \left (a+b \tan ^{-1}(c x)\right )^2}{4 x}-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{4 x^4}-\frac {b c \left (a+b \tan ^{-1}(c x)\right )^2}{4 x^3}+i b^3 c^4 \text {Li}_2\left (\frac {2}{1-i c x}-1\right )-\frac {1}{4} b^3 c^4 \tan ^{-1}(c x)-\frac {b^3 c^3}{4 x} \]

[Out]

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

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Rubi [A]  time = 0.60, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {4852, 4918, 325, 203, 4924, 4868, 2447, 4884} \[ i b^3 c^4 \text {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )-\frac {b^2 c^2 \left (a+b \tan ^{-1}(c x)\right )}{4 x^2}-2 b^2 c^4 \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{4} c^4 \left (a+b \tan ^{-1}(c x)\right )^3+i b c^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {3 b c^3 \left (a+b \tan ^{-1}(c x)\right )^2}{4 x}-\frac {b c \left (a+b \tan ^{-1}(c x)\right )^2}{4 x^3}-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{4 x^4}-\frac {b^3 c^3}{4 x}-\frac {1}{4} b^3 c^4 \tan ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x])^3/x^5,x]

[Out]

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, 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^5} \, dx &=-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{4 x^4}+\frac {1}{4} (3 b c) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x^4 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{4 x^4}+\frac {1}{4} (3 b c) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x^4} \, dx-\frac {1}{4} \left (3 b c^3\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x^2 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {b c \left (a+b \tan ^{-1}(c x)\right )^2}{4 x^3}-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{4 x^4}+\frac {1}{2} \left (b^2 c^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{x^3 \left (1+c^2 x^2\right )} \, dx-\frac {1}{4} \left (3 b c^3\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x^2} \, dx+\frac {1}{4} \left (3 b c^5\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx\\ &=-\frac {b c \left (a+b \tan ^{-1}(c x)\right )^2}{4 x^3}+\frac {3 b c^3 \left (a+b \tan ^{-1}(c x)\right )^2}{4 x}+\frac {1}{4} c^4 \left (a+b \tan ^{-1}(c x)\right )^3-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{4 x^4}+\frac {1}{2} \left (b^2 c^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{x^3} \, dx-\frac {1}{2} \left (b^2 c^4\right ) \int \frac {a+b \tan ^{-1}(c x)}{x \left (1+c^2 x^2\right )} \, dx-\frac {1}{2} \left (3 b^2 c^4\right ) \int \frac {a+b \tan ^{-1}(c x)}{x \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {b^2 c^2 \left (a+b \tan ^{-1}(c x)\right )}{4 x^2}+i b c^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {b c \left (a+b \tan ^{-1}(c x)\right )^2}{4 x^3}+\frac {3 b c^3 \left (a+b \tan ^{-1}(c x)\right )^2}{4 x}+\frac {1}{4} c^4 \left (a+b \tan ^{-1}(c x)\right )^3-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{4 x^4}+\frac {1}{4} \left (b^3 c^3\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx-\frac {1}{2} \left (i b^2 c^4\right ) \int \frac {a+b \tan ^{-1}(c x)}{x (i+c x)} \, dx-\frac {1}{2} \left (3 i b^2 c^4\right ) \int \frac {a+b \tan ^{-1}(c x)}{x (i+c x)} \, dx\\ &=-\frac {b^3 c^3}{4 x}-\frac {b^2 c^2 \left (a+b \tan ^{-1}(c x)\right )}{4 x^2}+i b c^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {b c \left (a+b \tan ^{-1}(c x)\right )^2}{4 x^3}+\frac {3 b c^3 \left (a+b \tan ^{-1}(c x)\right )^2}{4 x}+\frac {1}{4} c^4 \left (a+b \tan ^{-1}(c x)\right )^3-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{4 x^4}-2 b^2 c^4 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )-\frac {1}{4} \left (b^3 c^5\right ) \int \frac {1}{1+c^2 x^2} \, dx+\frac {1}{2} \left (b^3 c^5\right ) \int \frac {\log \left (2-\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx+\frac {1}{2} \left (3 b^3 c^5\right ) \int \frac {\log \left (2-\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx\\ &=-\frac {b^3 c^3}{4 x}-\frac {1}{4} b^3 c^4 \tan ^{-1}(c x)-\frac {b^2 c^2 \left (a+b \tan ^{-1}(c x)\right )}{4 x^2}+i b c^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {b c \left (a+b \tan ^{-1}(c x)\right )^2}{4 x^3}+\frac {3 b c^3 \left (a+b \tan ^{-1}(c x)\right )^2}{4 x}+\frac {1}{4} c^4 \left (a+b \tan ^{-1}(c x)\right )^3-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{4 x^4}-2 b^2 c^4 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )+i b^3 c^4 \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.72, size = 265, normalized size = 1.34 \[ -\frac {a^3+b \tan ^{-1}(c x) \left (a^2 \left (3-3 c^4 x^4\right )+a b \left (2 c x-6 c^3 x^3\right )+8 b^2 c^4 x^4 \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )+b^2 c^2 x^2 \left (c^2 x^2+1\right )\right )-3 a^2 b c^3 x^3+a^2 b c x+a b^2 c^4 x^4+a b^2 c^2 x^2+8 a b^2 c^4 x^4 \log \left (\frac {c x}{\sqrt {c^2 x^2+1}}\right )+b^2 \tan ^{-1}(c x)^2 \left (a \left (3-3 c^4 x^4\right )+b c x \left (-4 i c^3 x^3-3 c^2 x^2+1\right )\right )-4 i b^3 c^4 x^4 \text {Li}_2\left (e^{2 i \tan ^{-1}(c x)}\right )-b^3 \left (c^4 x^4-1\right ) \tan ^{-1}(c x)^3+b^3 c^3 x^3}{4 x^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTan[c*x])^3/x^5,x]

[Out]

-1/4*(a^3 + a^2*b*c*x + a*b^2*c^2*x^2 - 3*a^2*b*c^3*x^3 + b^3*c^3*x^3 + a*b^2*c^4*x^4 + b^2*(b*c*x*(1 - 3*c^2*
x^2 - (4*I)*c^3*x^3) + a*(3 - 3*c^4*x^4))*ArcTan[c*x]^2 - b^3*(-1 + c^4*x^4)*ArcTan[c*x]^3 + b*ArcTan[c*x]*(b^
2*c^2*x^2*(1 + c^2*x^2) + a*b*(2*c*x - 6*c^3*x^3) + a^2*(3 - 3*c^4*x^4) + 8*b^2*c^4*x^4*Log[1 - E^((2*I)*ArcTa
n[c*x])]) + 8*a*b^2*c^4*x^4*Log[(c*x)/Sqrt[1 + c^2*x^2]] - (4*I)*b^3*c^4*x^4*PolyLog[2, E^((2*I)*ArcTan[c*x])]
)/x^4

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \arctan \left (c x\right )^{3} + 3 \, a b^{2} \arctan \left (c x\right )^{2} + 3 \, a^{2} b \arctan \left (c x\right ) + a^{3}}{x^{5}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))^3/x^5,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^5, 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+b*arctan(c*x))^3/x^5,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.02, size = 550, normalized size = 2.78 \[ i c^{4} b^{3} \dilog \left (-i c x +1\right )-\frac {i c^{4} b^{3} \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{2}-i c^{4} b^{3} \dilog \left (i c x +1\right )+\frac {i c^{4} b^{3} \dilog \left (\frac {i \left (c x -i\right )}{2}\right )}{2}+\frac {i c^{4} b^{3} \ln \left (c x +i\right )^{2}}{4}-\frac {i c^{4} b^{3} \ln \left (c x -i\right )^{2}}{4}-\frac {c \,a^{2} b}{4 x^{3}}-\frac {c^{2} a \,b^{2}}{4 x^{2}}+\frac {3 c^{3} a^{2} b}{4 x}+c^{4} a \,b^{2} \ln \left (c^{2} x^{2}+1\right )+\frac {3 c^{4} a^{2} b \arctan \left (c x \right )}{4}+\frac {3 c^{4} a \,b^{2} \arctan \left (c x \right )^{2}}{4}-\frac {3 a^{2} b \arctan \left (c x \right )}{4 x^{4}}-\frac {3 a \,b^{2} \arctan \left (c x \right )^{2}}{4 x^{4}}-\frac {c^{2} b^{3} \arctan \left (c x \right )}{4 x^{2}}-\frac {c \,b^{3} \arctan \left (c x \right )^{2}}{4 x^{3}}+\frac {3 c^{3} b^{3} \arctan \left (c x \right )^{2}}{4 x}+c^{4} b^{3} \arctan \left (c x \right ) \ln \left (c^{2} x^{2}+1\right )-2 c^{4} b^{3} \arctan \left (c x \right ) \ln \left (c x \right )-2 c^{4} a \,b^{2} \ln \left (c x \right )-\frac {b^{3} c^{3}}{4 x}-\frac {b^{3} c^{4} \arctan \left (c x \right )}{4}-\frac {c a \,b^{2} \arctan \left (c x \right )}{2 x^{3}}+\frac {3 c^{3} a \,b^{2} \arctan \left (c x \right )}{2 x}-\frac {a^{3}}{4 x^{4}}+\frac {i c^{4} b^{3} \ln \left (c x -i\right ) \ln \left (c^{2} x^{2}+1\right )}{2}-\frac {i c^{4} b^{3} \ln \left (c x +i\right ) \ln \left (c^{2} x^{2}+1\right )}{2}-\frac {i c^{4} b^{3} \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{2}-i c^{4} b^{3} \ln \left (c x \right ) \ln \left (i c x +1\right )+\frac {i c^{4} b^{3} \ln \left (c x +i\right ) \ln \left (\frac {i \left (c x -i\right )}{2}\right )}{2}+i c^{4} b^{3} \ln \left (c x \right ) \ln \left (-i c x +1\right )-\frac {b^{3} \arctan \left (c x \right )^{3}}{4 x^{4}}+\frac {c^{4} b^{3} \arctan \left (c x \right )^{3}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x))^3/x^5,x)

[Out]

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

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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((a+b*arctan(c*x))^3/x^5,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^3}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))^3/x^5,x)

[Out]

int((a + b*atan(c*x))^3/x^5, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right )^{3}}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x))**3/x**5,x)

[Out]

Integral((a + b*atan(c*x))**3/x**5, x)

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