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

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

[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)]

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Rubi [A]  time = 0.600576, antiderivative size = 198, normalized size of antiderivative = 1., 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 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 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 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 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 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]

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 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 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]

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*}

Mathematica [A]  time = 0.655673, size = 265, normalized size = 1.34 \[ -\frac{-4 i b^3 c^4 x^4 \text{PolyLog}\left (2,e^{2 i \tan ^{-1}(c x)}\right )+b \tan ^{-1}(c x) \left (a^2 \left (3-3 c^4 x^4\right )+a b \left (2 c x-6 c^3 x^3\right )+b^2 c^2 x^2 \left (c^2 x^2+1\right )+8 b^2 c^4 x^4 \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )\right )-3 a^2 b c^3 x^3+a^2 b c x+a^3+a b^2 c^4 x^4+a b^2 c^2 x^2+8 a b^2 c^4 x^4 \log \left (\frac{c x}{\sqrt{c^2 x^2+1}}\right )+b^2 \tan ^{-1}(c x)^2 \left (a \left (3-3 c^4 x^4\right )+b c x \left (-4 i c^3 x^3-3 c^2 x^2+1\right )\right )+b^3 c^3 x^3-b^3 \left (c^4 x^4-1\right ) \tan ^{-1}(c x)^3}{4 x^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(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)*ArcTan[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])])/(4
*x^4)

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Maple [B]  time = 0.1, size = 550, normalized size = 2.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{atan}{\left (c x \right )}\right )^{3}}{x^{5}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (c x\right ) + a\right )}^{3}}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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