3.1.11 \(\int \frac {a+b \text {ArcTan}(c x)}{x^5} \, dx\) [11]

Optimal. Leaf size=48 \[ -\frac {b c}{12 x^3}+\frac {b c^3}{4 x}+\frac {1}{4} b c^4 \text {ArcTan}(c x)-\frac {a+b \text {ArcTan}(c x)}{4 x^4} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4946, 331, 209} \begin {gather*} -\frac {a+b \text {ArcTan}(c x)}{4 x^4}+\frac {1}{4} b c^4 \text {ArcTan}(c x)+\frac {b c^3}{4 x}-\frac {b c}{12 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x])/x^5,x]

[Out]

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

Rule 209

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

Rule 331

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 4946

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^
n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.00, size = 46, normalized size = 0.96 \begin {gather*} -\frac {a}{4 x^4}-\frac {b \text {ArcTan}(c x)}{4 x^4}-\frac {b c \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-c^2 x^2\right )}{12 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c*x])/x^5,x]

[Out]

-1/4*a/x^4 - (b*ArcTan[c*x])/(4*x^4) - (b*c*Hypergeometric2F1[-3/2, 1, -1/2, -(c^2*x^2)])/(12*x^3)

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Maple [A]
time = 0.13, size = 53, normalized size = 1.10

method	result	size
derivativedivides	\(c^{4} \left (-\frac {a}{4 c^{4} x^{4}}-\frac {b \arctan \left (c x \right )}{4 c^{4} x^{4}}+\frac {b \arctan \left (c x \right )}{4}-\frac {b}{12 c^{3} x^{3}}+\frac {b}{4 c x}\right )\)	\(53\)
default	\(c^{4} \left (-\frac {a}{4 c^{4} x^{4}}-\frac {b \arctan \left (c x \right )}{4 c^{4} x^{4}}+\frac {b \arctan \left (c x \right )}{4}-\frac {b}{12 c^{3} x^{3}}+\frac {b}{4 c x}\right )\)	\(53\)
risch	\(\frac {i b \ln \left (i c x +1\right )}{8 x^{4}}-\frac {-3 i c^{4} b \ln \left (-c x -i\right ) x^{4}+3 i c^{4} b \ln \left (-c x +i\right ) x^{4}-6 b \,c^{3} x^{3}+3 i b \ln \left (-i c x +1\right )+2 x b c +6 a}{24 x^{4}}\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x))/x^5,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.47, size = 46, normalized size = 0.96 \begin {gather*} \frac {1}{12} \, {\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 )} b - \frac {a}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.63, size = 41, normalized size = 0.85 \begin {gather*} \frac {3 \, b c^{3} x^{3} - b c x + 3 \, {\left (b c^{4} x^{4} - b\right )} \arctan \left (c x\right ) - 3 \, a}{12 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.30, size = 46, normalized size = 0.96 \begin {gather*} - \frac {a}{4 x^{4}} + \frac {b c^{4} \operatorname {atan}{\left (c x \right )}}{4} + \frac {b c^{3}}{4 x} - \frac {b c}{12 x^{3}} - \frac {b \operatorname {atan}{\left (c x \right )}}{4 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/(4*x**4) + b*c**4*atan(c*x)/4 + b*c**3/(4*x) - b*c/(12*x**3) - b*atan(c*x)/(4*x**4)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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Mupad [B]
time = 0.36, size = 42, normalized size = 0.88 \begin {gather*} \frac {b\,c^4\,\mathrm {atan}\left (c\,x\right )}{4}-\frac {-b\,c^3\,x^3+\frac {b\,c\,x}{3}+a}{4\,x^4}-\frac {b\,\mathrm {atan}\left (c\,x\right )}{4\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))/x^5,x)

[Out]

(b*c^4*atan(c*x))/4 - (a - b*c^3*x^3 + (b*c*x)/3)/(4*x^4) - (b*atan(c*x))/(4*x^4)

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