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

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

[Out]

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

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Rubi [A]  time = 0.0250794, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4852, 325, 203} \[ -\frac{a+b \tan ^{-1}(c x)}{4 x^4}+\frac{b c^3}{4 x}+\frac{1}{4} b c^4 \tan ^{-1}(c x)-\frac{b c}{12 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x])/x^5,x]

[Out]

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

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 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])

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*}

Mathematica [C]  time = 0.002897, size = 46, normalized size = 0.96 \[ -\frac{b c \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},-c^2 x^2\right )}{12 x^3}-\frac{a}{4 x^4}-\frac{b \tan ^{-1}(c x)}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c*x])/x^5,x]

[Out]

-a/(4*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 = 0.009, size = 44, normalized size = 0.9 \begin{align*} -{\frac{a}{4\,{x}^{4}}}-{\frac{b\arctan \left ( cx \right ) }{4\,{x}^{4}}}+{\frac{b{c}^{4}\arctan \left ( cx \right ) }{4}}-{\frac{bc}{12\,{x}^{3}}}+{\frac{b{c}^{3}}{4\,x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.45929, size = 62, normalized size = 1.29 \begin{align*} \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{align*}

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 = 2.37558, size = 95, normalized size = 1.98 \begin{align*} \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{align*}

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 = 1.20462, size = 46, normalized size = 0.96 \begin{align*} - \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{align*}

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 [A]  time = 1.7354, size = 88, normalized size = 1.83 \begin{align*} \frac{3 \, b c^{4} i x^{4} \log \left (c i x - 1\right ) - 3 \, b c^{4} i x^{4} \log \left (-c i x - 1\right ) + 6 \, b c^{3} x^{3} - 2 \, b c x - 6 \, b \arctan \left (c x\right ) - 6 \, a}{24 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*b*c^4*i*x^4*log(c*i*x - 1) - 3*b*c^4*i*x^4*log(-c*i*x - 1) + 6*b*c^3*x^3 - 2*b*c*x - 6*b*arctan(c*x) -
6*a)/x^4