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

Optimal. Leaf size=41 \[ -\frac{a+b \tan ^{-1}\left (c x^2\right )}{4 x^4}-\frac{1}{4} b c^2 \tan ^{-1}\left (c x^2\right )-\frac{b c}{4 x^2} \]

[Out]

-(b*c)/(4*x^2) - (b*c^2*ArcTan[c*x^2])/4 - (a + b*ArcTan[c*x^2])/(4*x^4)

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Rubi [A]  time = 0.0247925, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5033, 275, 325, 203} \[ -\frac{a+b \tan ^{-1}\left (c x^2\right )}{4 x^4}-\frac{1}{4} b c^2 \tan ^{-1}\left (c x^2\right )-\frac{b c}{4 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x^2])/x^5,x]

[Out]

-(b*c)/(4*x^2) - (b*c^2*ArcTan[c*x^2])/4 - (a + b*ArcTan[c*x^2])/(4*x^4)

Rule 5033

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

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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

Mathematica [C]  time = 0.0067644, size = 48, normalized size = 1.17 \[ -\frac{b c \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},-c^2 x^4\right )}{4 x^2}-\frac{a}{4 x^4}-\frac{b \tan ^{-1}\left (c x^2\right )}{4 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.024, size = 39, normalized size = 1. \begin{align*} -{\frac{a}{4\,{x}^{4}}}-{\frac{b\arctan \left ( c{x}^{2} \right ) }{4\,{x}^{4}}}-{\frac{bc}{4\,{x}^{2}}}-{\frac{b{c}^{2}\arctan \left ( c{x}^{2} \right ) }{4}} \end{align*}

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/x^4-1/4*b/x^4*arctan(c*x^2)-1/4*b*c/x^2-1/4*b*c^2*arctan(c*x^2)

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Maxima [A]  time = 1.50057, size = 47, normalized size = 1.15 \begin{align*} -\frac{1}{4} \,{\left ({\left (c \arctan \left (c x^{2}\right ) + \frac{1}{x^{2}}\right )} c + \frac{\arctan \left (c x^{2}\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^2))/x^5,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.68126, size = 76, normalized size = 1.85 \begin{align*} -\frac{b c x^{2} +{\left (b c^{2} x^{4} + b\right )} \arctan \left (c x^{2}\right ) + a}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 36.2676, size = 42, normalized size = 1.02 \begin{align*} - \frac{a}{4 x^{4}} - \frac{b c^{2} \operatorname{atan}{\left (c x^{2} \right )}}{4} - \frac{b c}{4 x^{2}} - \frac{b \operatorname{atan}{\left (c x^{2} \right )}}{4 x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/(4*x**4) - b*c**2*atan(c*x**2)/4 - b*c/(4*x**2) - b*atan(c*x**2)/(4*x**4)

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Giac [B]  time = 1.20103, size = 100, normalized size = 2.44 \begin{align*} \frac{b c^{5} i x^{4} \log \left (c i x^{2} + 1\right ) - b c^{5} i x^{4} \log \left (-c i x^{2} + 1\right ) - 2 \, b c^{4} x^{2} - 2 \, b c^{3} \arctan \left (c x^{2}\right ) - 2 \, a c^{3}}{8 \, c^{3} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b*c^5*i*x^4*log(c*i*x^2 + 1) - b*c^5*i*x^4*log(-c*i*x^2 + 1) - 2*b*c^4*x^2 - 2*b*c^3*arctan(c*x^2) - 2*a*
c^3)/(c^3*x^4)