3.1123 \(\int \frac{(d+e x^2) (a+b \tan ^{-1}(c x))}{x^6} \, dx\)

Optimal. Leaf size=110 \[ -\frac{d \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{e \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}+\frac{b c \left (3 c^2 d-5 e\right )}{30 x^2}-\frac{1}{30} b c^3 \left (3 c^2 d-5 e\right ) \log \left (c^2 x^2+1\right )+\frac{1}{15} b c^3 \log (x) \left (3 c^2 d-5 e\right )-\frac{b c d}{20 x^4} \]

[Out]

-(b*c*d)/(20*x^4) + (b*c*(3*c^2*d - 5*e))/(30*x^2) - (d*(a + b*ArcTan[c*x]))/(5*x^5) - (e*(a + b*ArcTan[c*x]))
/(3*x^3) + (b*c^3*(3*c^2*d - 5*e)*Log[x])/15 - (b*c^3*(3*c^2*d - 5*e)*Log[1 + c^2*x^2])/30

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Rubi [A]  time = 0.128825, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {14, 4976, 12, 446, 77} \[ -\frac{d \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{e \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}+\frac{b c \left (3 c^2 d-5 e\right )}{30 x^2}-\frac{1}{30} b c^3 \left (3 c^2 d-5 e\right ) \log \left (c^2 x^2+1\right )+\frac{1}{15} b c^3 \log (x) \left (3 c^2 d-5 e\right )-\frac{b c d}{20 x^4} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)*(a + b*ArcTan[c*x]))/x^6,x]

[Out]

-(b*c*d)/(20*x^4) + (b*c*(3*c^2*d - 5*e))/(30*x^2) - (d*(a + b*ArcTan[c*x]))/(5*x^5) - (e*(a + b*ArcTan[c*x]))
/(3*x^3) + (b*c^3*(3*c^2*d - 5*e)*Log[x])/15 - (b*c^3*(3*c^2*d - 5*e)*Log[1 + c^2*x^2])/30

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 4976

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^q, x]}, Dist[a + b*ArcTan[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/(1 + c^
2*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && ((IGtQ[q, 0] &&  !(ILtQ[(m - 1)/2, 0] && GtQ[m +
2*q + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[q, 0] && GtQ[m + 2*q + 3, 0])) || (ILtQ[(m + 2*q + 1)/2, 0] &&
!ILtQ[(m - 1)/2, 0]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \frac{\left (d+e x^2\right ) \left (a+b \tan ^{-1}(c x)\right )}{x^6} \, dx &=-\frac{d \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{e \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-(b c) \int \frac{-3 d-5 e x^2}{15 x^5 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{d \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{e \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{1}{15} (b c) \int \frac{-3 d-5 e x^2}{x^5 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{d \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{e \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{1}{30} (b c) \operatorname{Subst}\left (\int \frac{-3 d-5 e x}{x^3 \left (1+c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac{d \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{e \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{1}{30} (b c) \operatorname{Subst}\left (\int \left (-\frac{3 d}{x^3}+\frac{3 c^2 d-5 e}{x^2}+\frac{-3 c^4 d+5 c^2 e}{x}+\frac{3 c^6 d-5 c^4 e}{1+c^2 x}\right ) \, dx,x,x^2\right )\\ &=-\frac{b c d}{20 x^4}+\frac{b c \left (3 c^2 d-5 e\right )}{30 x^2}-\frac{d \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{e \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}+\frac{1}{15} b c^3 \left (3 c^2 d-5 e\right ) \log (x)-\frac{1}{30} b c^3 \left (3 c^2 d-5 e\right ) \log \left (1+c^2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0422192, size = 123, normalized size = 1.12 \[ -\frac{a d}{5 x^5}-\frac{a e}{3 x^3}+\frac{1}{10} b c d \left (\frac{c^2}{x^2}-c^4 \log \left (c^2 x^2+1\right )+2 c^4 \log (x)-\frac{1}{2 x^4}\right )+\frac{1}{6} b c e \left (c^2 \log \left (c^2 x^2+1\right )-2 c^2 \log (x)-\frac{1}{x^2}\right )-\frac{b d \tan ^{-1}(c x)}{5 x^5}-\frac{b e \tan ^{-1}(c x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)*(a + b*ArcTan[c*x]))/x^6,x]

[Out]

-(a*d)/(5*x^5) - (a*e)/(3*x^3) - (b*d*ArcTan[c*x])/(5*x^5) - (b*e*ArcTan[c*x])/(3*x^3) + (b*c*e*(-x^(-2) - 2*c
^2*Log[x] + c^2*Log[1 + c^2*x^2]))/6 + (b*c*d*(-1/(2*x^4) + c^2/x^2 + 2*c^4*Log[x] - c^4*Log[1 + c^2*x^2]))/10

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Maple [A]  time = 0.044, size = 120, normalized size = 1.1 \begin{align*} -{\frac{ad}{5\,{x}^{5}}}-{\frac{ae}{3\,{x}^{3}}}-{\frac{\arctan \left ( cx \right ) bd}{5\,{x}^{5}}}-{\frac{b\arctan \left ( cx \right ) e}{3\,{x}^{3}}}-{\frac{{c}^{5}b\ln \left ({c}^{2}{x}^{2}+1 \right ) d}{10}}+{\frac{b{c}^{3}e\ln \left ({c}^{2}{x}^{2}+1 \right ) }{6}}+{\frac{{c}^{5}bd\ln \left ( cx \right ) }{5}}-{\frac{{c}^{3}b\ln \left ( cx \right ) e}{3}}+{\frac{b{c}^{3}d}{10\,{x}^{2}}}-{\frac{bce}{6\,{x}^{2}}}-{\frac{bcd}{20\,{x}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)*(a+b*arctan(c*x))/x^6,x)

[Out]

-1/5*a*d/x^5-1/3*a*e/x^3-1/5*b*arctan(c*x)*d/x^5-1/3*b*arctan(c*x)*e/x^3-1/10*c^5*b*ln(c^2*x^2+1)*d+1/6*b*c^3*
e*ln(c^2*x^2+1)+1/5*c^5*b*d*ln(c*x)-1/3*c^3*b*ln(c*x)*e+1/10*c^3*b*d/x^2-1/6*c*b*e/x^2-1/20*b*c*d/x^4

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Maxima [A]  time = 0.955741, size = 157, normalized size = 1.43 \begin{align*} -\frac{1}{20} \,{\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) - \frac{2 \, c^{2} x^{2} - 1}{x^{4}}\right )} c + \frac{4 \, \arctan \left (c x\right )}{x^{5}}\right )} b d + \frac{1}{6} \,{\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac{1}{x^{2}}\right )} c - \frac{2 \, \arctan \left (c x\right )}{x^{3}}\right )} b e - \frac{a e}{3 \, x^{3}} - \frac{a d}{5 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arctan(c*x))/x^6,x, algorithm="maxima")

[Out]

-1/20*((2*c^4*log(c^2*x^2 + 1) - 2*c^4*log(x^2) - (2*c^2*x^2 - 1)/x^4)*c + 4*arctan(c*x)/x^5)*b*d + 1/6*((c^2*
log(c^2*x^2 + 1) - c^2*log(x^2) - 1/x^2)*c - 2*arctan(c*x)/x^3)*b*e - 1/3*a*e/x^3 - 1/5*a*d/x^5

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Fricas [A]  time = 1.85409, size = 269, normalized size = 2.45 \begin{align*} -\frac{2 \,{\left (3 \, b c^{5} d - 5 \, b c^{3} e\right )} x^{5} \log \left (c^{2} x^{2} + 1\right ) - 4 \,{\left (3 \, b c^{5} d - 5 \, b c^{3} e\right )} x^{5} \log \left (x\right ) + 3 \, b c d x + 20 \, a e x^{2} - 2 \,{\left (3 \, b c^{3} d - 5 \, b c e\right )} x^{3} + 12 \, a d + 4 \,{\left (5 \, b e x^{2} + 3 \, b d\right )} \arctan \left (c x\right )}{60 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arctan(c*x))/x^6,x, algorithm="fricas")

[Out]

-1/60*(2*(3*b*c^5*d - 5*b*c^3*e)*x^5*log(c^2*x^2 + 1) - 4*(3*b*c^5*d - 5*b*c^3*e)*x^5*log(x) + 3*b*c*d*x + 20*
a*e*x^2 - 2*(3*b*c^3*d - 5*b*c*e)*x^3 + 12*a*d + 4*(5*b*e*x^2 + 3*b*d)*arctan(c*x))/x^5

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Sympy [A]  time = 3.17923, size = 153, normalized size = 1.39 \begin{align*} \begin{cases} - \frac{a d}{5 x^{5}} - \frac{a e}{3 x^{3}} + \frac{b c^{5} d \log{\left (x \right )}}{5} - \frac{b c^{5} d \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{10} + \frac{b c^{3} d}{10 x^{2}} - \frac{b c^{3} e \log{\left (x \right )}}{3} + \frac{b c^{3} e \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{6} - \frac{b c d}{20 x^{4}} - \frac{b c e}{6 x^{2}} - \frac{b d \operatorname{atan}{\left (c x \right )}}{5 x^{5}} - \frac{b e \operatorname{atan}{\left (c x \right )}}{3 x^{3}} & \text{for}\: c \neq 0 \\a \left (- \frac{d}{5 x^{5}} - \frac{e}{3 x^{3}}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)*(a+b*atan(c*x))/x**6,x)

[Out]

Piecewise((-a*d/(5*x**5) - a*e/(3*x**3) + b*c**5*d*log(x)/5 - b*c**5*d*log(x**2 + c**(-2))/10 + b*c**3*d/(10*x
**2) - b*c**3*e*log(x)/3 + b*c**3*e*log(x**2 + c**(-2))/6 - b*c*d/(20*x**4) - b*c*e/(6*x**2) - b*d*atan(c*x)/(
5*x**5) - b*e*atan(c*x)/(3*x**3), Ne(c, 0)), (a*(-d/(5*x**5) - e/(3*x**3)), True))

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Giac [A]  time = 1.10962, size = 174, normalized size = 1.58 \begin{align*} -\frac{6 \, b c^{5} d x^{5} \log \left (c^{2} x^{2} + 1\right ) - 12 \, b c^{5} d x^{5} \log \left (x\right ) - 10 \, b c^{3} x^{5} e \log \left (c^{2} x^{2} + 1\right ) + 20 \, b c^{3} x^{5} e \log \left (x\right ) - 6 \, b c^{3} d x^{3} + 10 \, b c x^{3} e + 20 \, b x^{2} \arctan \left (c x\right ) e + 3 \, b c d x + 20 \, a x^{2} e + 12 \, b d \arctan \left (c x\right ) + 12 \, a d}{60 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arctan(c*x))/x^6,x, algorithm="giac")

[Out]

-1/60*(6*b*c^5*d*x^5*log(c^2*x^2 + 1) - 12*b*c^5*d*x^5*log(x) - 10*b*c^3*x^5*e*log(c^2*x^2 + 1) + 20*b*c^3*x^5
*e*log(x) - 6*b*c^3*d*x^3 + 10*b*c*x^3*e + 20*b*x^2*arctan(c*x)*e + 3*b*c*d*x + 20*a*x^2*e + 12*b*d*arctan(c*x
) + 12*a*d)/x^5