3.23 \(\int \frac {(a+b \tan ^{-1}(c x))^2}{x^5} \, dx\)

Optimal. Leaf size=116 \[ \frac {1}{4} c^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {b c^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{4 x^4}-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{6 x^3}-\frac {2}{3} b^2 c^4 \log (x)-\frac {b^2 c^2}{12 x^2}+\frac {1}{3} b^2 c^4 \log \left (c^2 x^2+1\right ) \]

[Out]

-1/12*b^2*c^2/x^2-1/6*b*c*(a+b*arctan(c*x))/x^3+1/2*b*c^3*(a+b*arctan(c*x))/x+1/4*c^4*(a+b*arctan(c*x))^2-1/4*
(a+b*arctan(c*x))^2/x^4-2/3*b^2*c^4*ln(x)+1/3*b^2*c^4*ln(c^2*x^2+1)

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Rubi [A]  time = 0.22, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {4852, 4918, 266, 44, 36, 29, 31, 4884} \[ \frac {1}{4} c^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {b c^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x}-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{6 x^3}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{4 x^4}-\frac {b^2 c^2}{12 x^2}+\frac {1}{3} b^2 c^4 \log \left (c^2 x^2+1\right )-\frac {2}{3} b^2 c^4 \log (x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x])^2/x^5,x]

[Out]

-(b^2*c^2)/(12*x^2) - (b*c*(a + b*ArcTan[c*x]))/(6*x^3) + (b*c^3*(a + b*ArcTan[c*x]))/(2*x) + (c^4*(a + b*ArcT
an[c*x])^2)/4 - (a + b*ArcTan[c*x])^2/(4*x^4) - (2*b^2*c^4*Log[x])/3 + (b^2*c^4*Log[1 + c^2*x^2])/3

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 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 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 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]

Rubi steps

\begin {align*} \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x^5} \, dx &=-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{4 x^4}+\frac {1}{2} (b c) \int \frac {a+b \tan ^{-1}(c x)}{x^4 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{4 x^4}+\frac {1}{2} (b c) \int \frac {a+b \tan ^{-1}(c x)}{x^4} \, dx-\frac {1}{2} \left (b c^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x^2 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{6 x^3}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{4 x^4}+\frac {1}{6} \left (b^2 c^2\right ) \int \frac {1}{x^3 \left (1+c^2 x^2\right )} \, dx-\frac {1}{2} \left (b c^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x^2} \, dx+\frac {1}{2} \left (b c^5\right ) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx\\ &=-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{6 x^3}+\frac {b c^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x}+\frac {1}{4} c^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{4 x^4}+\frac {1}{12} \left (b^2 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \left (1+c^2 x\right )} \, dx,x,x^2\right )-\frac {1}{2} \left (b^2 c^4\right ) \int \frac {1}{x \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{6 x^3}+\frac {b c^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x}+\frac {1}{4} c^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{4 x^4}+\frac {1}{12} \left (b^2 c^2\right ) \operatorname {Subst}\left (\int \left (\frac {1}{x^2}-\frac {c^2}{x}+\frac {c^4}{1+c^2 x}\right ) \, dx,x,x^2\right )-\frac {1}{4} \left (b^2 c^4\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {b^2 c^2}{12 x^2}-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{6 x^3}+\frac {b c^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x}+\frac {1}{4} c^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{4 x^4}-\frac {1}{6} b^2 c^4 \log (x)+\frac {1}{12} b^2 c^4 \log \left (1+c^2 x^2\right )-\frac {1}{4} \left (b^2 c^4\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{4} \left (b^2 c^6\right ) \operatorname {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right )\\ &=-\frac {b^2 c^2}{12 x^2}-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{6 x^3}+\frac {b c^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x}+\frac {1}{4} c^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{4 x^4}-\frac {2}{3} b^2 c^4 \log (x)+\frac {1}{3} b^2 c^4 \log \left (1+c^2 x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 128, normalized size = 1.10 \[ \frac {-3 a^2+6 a b c^3 x^3+2 b \tan ^{-1}(c x) \left (3 a \left (c^4 x^4-1\right )+b c x \left (3 c^2 x^2-1\right )\right )-2 a b c x-8 b^2 c^4 x^4 \log (x)+3 b^2 \left (c^4 x^4-1\right ) \tan ^{-1}(c x)^2-b^2 c^2 x^2+4 b^2 c^4 x^4 \log \left (c^2 x^2+1\right )}{12 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c*x])^2/x^5,x]

[Out]

(-3*a^2 - 2*a*b*c*x - b^2*c^2*x^2 + 6*a*b*c^3*x^3 + 2*b*(b*c*x*(-1 + 3*c^2*x^2) + 3*a*(-1 + c^4*x^4))*ArcTan[c
*x] + 3*b^2*(-1 + c^4*x^4)*ArcTan[c*x]^2 - 8*b^2*c^4*x^4*Log[x] + 4*b^2*c^4*x^4*Log[1 + c^2*x^2])/(12*x^4)

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fricas [A]  time = 0.45, size = 135, normalized size = 1.16 \[ \frac {4 \, b^{2} c^{4} x^{4} \log \left (c^{2} x^{2} + 1\right ) - 8 \, b^{2} c^{4} x^{4} \log \relax (x) + 6 \, a b c^{3} x^{3} - b^{2} c^{2} x^{2} - 2 \, a b c x + 3 \, {\left (b^{2} c^{4} x^{4} - b^{2}\right )} \arctan \left (c x\right )^{2} - 3 \, a^{2} + 2 \, {\left (3 \, a b c^{4} x^{4} + 3 \, b^{2} c^{3} x^{3} - b^{2} c x - 3 \, a b\right )} \arctan \left (c x\right )}{12 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))^2/x^5,x, algorithm="fricas")

[Out]

1/12*(4*b^2*c^4*x^4*log(c^2*x^2 + 1) - 8*b^2*c^4*x^4*log(x) + 6*a*b*c^3*x^3 - b^2*c^2*x^2 - 2*a*b*c*x + 3*(b^2
*c^4*x^4 - b^2)*arctan(c*x)^2 - 3*a^2 + 2*(3*a*b*c^4*x^4 + 3*b^2*c^3*x^3 - b^2*c*x - 3*a*b)*arctan(c*x))/x^4

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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))^2/x^5,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.02, size = 147, normalized size = 1.27 \[ -\frac {a^{2}}{4 x^{4}}-\frac {b^{2} \arctan \left (c x \right )^{2}}{4 x^{4}}-\frac {c \,b^{2} \arctan \left (c x \right )}{6 x^{3}}+\frac {c^{3} b^{2} \arctan \left (c x \right )}{2 x}+\frac {c^{4} b^{2} \arctan \left (c x \right )^{2}}{4}-\frac {b^{2} c^{2}}{12 x^{2}}-\frac {2 c^{4} b^{2} \ln \left (c x \right )}{3}+\frac {b^{2} c^{4} \ln \left (c^{2} x^{2}+1\right )}{3}-\frac {a b \arctan \left (c x \right )}{2 x^{4}}-\frac {a b c}{6 x^{3}}+\frac {c^{3} a b}{2 x}+\frac {c^{4} a b \arctan \left (c x \right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x))^2/x^5,x)

[Out]

-1/4*a^2/x^4-1/4*b^2/x^4*arctan(c*x)^2-1/6*c*b^2*arctan(c*x)/x^3+1/2*c^3*b^2*arctan(c*x)/x+1/4*c^4*b^2*arctan(
c*x)^2-1/12*b^2*c^2/x^2-2/3*c^4*b^2*ln(c*x)+1/3*b^2*c^4*ln(c^2*x^2+1)-1/2*a*b/x^4*arctan(c*x)-1/6*a*b*c/x^3+1/
2*c^3*a*b/x+1/2*c^4*a*b*arctan(c*x)

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maxima [A]  time = 0.42, size = 152, normalized size = 1.31 \[ \frac {1}{6} \, {\left ({\left (3 \, c^{3} \arctan \left (c x\right ) + \frac {3 \, c^{2} x^{2} - 1}{x^{3}}\right )} c - \frac {3 \, \arctan \left (c x\right )}{x^{4}}\right )} a b + \frac {1}{12} \, {\left (2 \, {\left (3 \, c^{3} \arctan \left (c x\right ) + \frac {3 \, c^{2} x^{2} - 1}{x^{3}}\right )} c \arctan \left (c x\right ) - \frac {{\left (3 \, c^{2} x^{2} \arctan \left (c x\right )^{2} - 4 \, c^{2} x^{2} \log \left (c^{2} x^{2} + 1\right ) + 8 \, c^{2} x^{2} \log \relax (x) + 1\right )} c^{2}}{x^{2}}\right )} b^{2} - \frac {b^{2} \arctan \left (c x\right )^{2}}{4 \, x^{4}} - \frac {a^{2}}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))^2/x^5,x, algorithm="maxima")

[Out]

1/6*((3*c^3*arctan(c*x) + (3*c^2*x^2 - 1)/x^3)*c - 3*arctan(c*x)/x^4)*a*b + 1/12*(2*(3*c^3*arctan(c*x) + (3*c^
2*x^2 - 1)/x^3)*c*arctan(c*x) - (3*c^2*x^2*arctan(c*x)^2 - 4*c^2*x^2*log(c^2*x^2 + 1) + 8*c^2*x^2*log(x) + 1)*
c^2/x^2)*b^2 - 1/4*b^2*arctan(c*x)^2/x^4 - 1/4*a^2/x^4

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mupad [B]  time = 2.26, size = 171, normalized size = 1.47 \[ \frac {b^2\,c^4\,{\mathrm {atan}\left (c\,x\right )}^2}{4}-\frac {2\,b^2\,c^4\,\ln \relax (x)}{3}-\frac {\frac {b^2\,{\mathrm {atan}\left (c\,x\right )}^2}{4}+\frac {a^2}{4}+x\,\left (\frac {c\,\mathrm {atan}\left (c\,x\right )\,b^2}{6}+\frac {a\,c\,b}{6}\right )-x^3\,\left (\frac {b^2\,c^3\,\mathrm {atan}\left (c\,x\right )}{2}+\frac {a\,b\,c^3}{2}\right )+\frac {b^2\,c^2\,x^2}{12}+\frac {a\,b\,\mathrm {atan}\left (c\,x\right )}{2}}{x^4}+\frac {b^2\,c^4\,\ln \left (c\,x+1{}\mathrm {i}\right )}{3}+\frac {b^2\,c^4\,\ln \left (1+c\,x\,1{}\mathrm {i}\right )}{3}+\frac {a\,b\,c^4\,\ln \left (c\,x+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4}-\frac {a\,b\,c^4\,\ln \left (1+c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))^2/x^5,x)

[Out]

(b^2*c^4*atan(c*x)^2)/4 - (2*b^2*c^4*log(x))/3 - ((b^2*atan(c*x)^2)/4 + a^2/4 + x*((b^2*c*atan(c*x))/6 + (a*b*
c)/6) - x^3*((b^2*c^3*atan(c*x))/2 + (a*b*c^3)/2) + (b^2*c^2*x^2)/12 + (a*b*atan(c*x))/2)/x^4 + (b^2*c^4*log(c
*x + 1i))/3 + (b^2*c^4*log(c*x*1i + 1))/3 + (a*b*c^4*log(c*x + 1i)*1i)/4 - (a*b*c^4*log(c*x*1i + 1)*1i)/4

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sympy [A]  time = 1.78, size = 170, normalized size = 1.47 \[ \begin {cases} - \frac {a^{2}}{4 x^{4}} + \frac {a b c^{4} \operatorname {atan}{\left (c x \right )}}{2} + \frac {a b c^{3}}{2 x} - \frac {a b c}{6 x^{3}} - \frac {a b \operatorname {atan}{\left (c x \right )}}{2 x^{4}} - \frac {2 b^{2} c^{4} \log {\relax (x )}}{3} + \frac {b^{2} c^{4} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{3} + \frac {b^{2} c^{4} \operatorname {atan}^{2}{\left (c x \right )}}{4} + \frac {b^{2} c^{3} \operatorname {atan}{\left (c x \right )}}{2 x} - \frac {b^{2} c^{2}}{12 x^{2}} - \frac {b^{2} c \operatorname {atan}{\left (c x \right )}}{6 x^{3}} - \frac {b^{2} \operatorname {atan}^{2}{\left (c x \right )}}{4 x^{4}} & \text {for}\: c \neq 0 \\- \frac {a^{2}}{4 x^{4}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x))**2/x**5,x)

[Out]

Piecewise((-a**2/(4*x**4) + a*b*c**4*atan(c*x)/2 + a*b*c**3/(2*x) - a*b*c/(6*x**3) - a*b*atan(c*x)/(2*x**4) -
2*b**2*c**4*log(x)/3 + b**2*c**4*log(x**2 + c**(-2))/3 + b**2*c**4*atan(c*x)**2/4 + b**2*c**3*atan(c*x)/(2*x)
- b**2*c**2/(12*x**2) - b**2*c*atan(c*x)/(6*x**3) - b**2*atan(c*x)**2/(4*x**4), Ne(c, 0)), (-a**2/(4*x**4), Tr
ue))

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