3.1122 \(\int \frac{(d+e x^2) (a+b \tan ^{-1}(c x))}{x^5} \, dx\)

Optimal. Leaf size=82 \[ -\frac{d \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{e \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+\frac{b c \left (c^2 d-2 e\right )}{4 x}+\frac{1}{4} b c^2 \left (c^2 d-2 e\right ) \tan ^{-1}(c x)-\frac{b c d}{12 x^3} \]

[Out]

-(b*c*d)/(12*x^3) + (b*c*(c^2*d - 2*e))/(4*x) + (b*c^2*(c^2*d - 2*e)*ArcTan[c*x])/4 - (d*(a + b*ArcTan[c*x]))/
(4*x^4) - (e*(a + b*ArcTan[c*x]))/(2*x^2)

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Rubi [A]  time = 0.0904979, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {14, 4976, 12, 453, 325, 203} \[ -\frac{d \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{e \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+\frac{b c \left (c^2 d-2 e\right )}{4 x}+\frac{1}{4} b c^2 \left (c^2 d-2 e\right ) \tan ^{-1}(c x)-\frac{b c d}{12 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c*d)/(12*x^3) + (b*c*(c^2*d - 2*e))/(4*x) + (b*c^2*(c^2*d - 2*e)*ArcTan[c*x])/4 - (d*(a + b*ArcTan[c*x]))/
(4*x^4) - (e*(a + b*ArcTan[c*x]))/(2*x^2)

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 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -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{\left (d+e x^2\right ) \left (a+b \tan ^{-1}(c x)\right )}{x^5} \, dx &=-\frac{d \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{e \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-(b c) \int \frac{-d-2 e x^2}{4 x^4 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{d \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{e \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac{1}{4} (b c) \int \frac{-d-2 e x^2}{x^4 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{b c d}{12 x^3}-\frac{d \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{e \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac{1}{4} \left (b c \left (c^2 d-2 e\right )\right ) \int \frac{1}{x^2 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{b c d}{12 x^3}+\frac{b c \left (c^2 d-2 e\right )}{4 x}-\frac{d \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{e \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+\frac{1}{4} \left (b c^3 \left (c^2 d-2 e\right )\right ) \int \frac{1}{1+c^2 x^2} \, dx\\ &=-\frac{b c d}{12 x^3}+\frac{b c \left (c^2 d-2 e\right )}{4 x}+\frac{1}{4} b c^2 \left (c^2 d-2 e\right ) \tan ^{-1}(c x)-\frac{d \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{e \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}\\ \end{align*}

Mathematica [C]  time = 0.0052514, size = 97, normalized size = 1.18 \[ -\frac{b c d \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},-c^2 x^2\right )}{12 x^3}-\frac{b c e \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},-c^2 x^2\right )}{2 x}-\frac{a d}{4 x^4}-\frac{a e}{2 x^2}-\frac{b d \tan ^{-1}(c x)}{4 x^4}-\frac{b e \tan ^{-1}(c x)}{2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*d)/(4*x^4) - (a*e)/(2*x^2) - (b*d*ArcTan[c*x])/(4*x^4) - (b*e*ArcTan[c*x])/(2*x^2) - (b*c*d*Hypergeometric
2F1[-3/2, 1, -1/2, -(c^2*x^2)])/(12*x^3) - (b*c*e*Hypergeometric2F1[-1/2, 1, 1/2, -(c^2*x^2)])/(2*x)

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Maple [A]  time = 0.044, size = 86, normalized size = 1.1 \begin{align*} -{\frac{ae}{2\,{x}^{2}}}-{\frac{ad}{4\,{x}^{4}}}-{\frac{b\arctan \left ( cx \right ) e}{2\,{x}^{2}}}-{\frac{\arctan \left ( cx \right ) bd}{4\,{x}^{4}}}+{\frac{{c}^{4}b\arctan \left ( cx \right ) d}{4}}-{\frac{b{c}^{2}e\arctan \left ( cx \right ) }{2}}+{\frac{b{c}^{3}d}{4\,x}}-{\frac{bce}{2\,x}}-{\frac{bcd}{12\,{x}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.43652, size = 108, normalized size = 1.32 \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 d - \frac{1}{2} \,{\left ({\left (c \arctan \left (c x\right ) + \frac{1}{x}\right )} c + \frac{\arctan \left (c x\right )}{x^{2}}\right )} b e - \frac{a e}{2 \, x^{2}} - \frac{a d}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(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*d - 1/2*((c*arctan(c*x) + 1/x)*c + ar
ctan(c*x)/x^2)*b*e - 1/2*a*e/x^2 - 1/4*a*d/x^4

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Fricas [A]  time = 1.68077, size = 177, normalized size = 2.16 \begin{align*} -\frac{b c d x + 6 \, a e x^{2} - 3 \,{\left (b c^{3} d - 2 \, b c e\right )} x^{3} + 3 \, a d - 3 \,{\left ({\left (b c^{4} d - 2 \, b c^{2} e\right )} x^{4} - 2 \, b e x^{2} - b d\right )} \arctan \left (c x\right )}{12 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.49037, size = 99, normalized size = 1.21 \begin{align*} - \frac{a d}{4 x^{4}} - \frac{a e}{2 x^{2}} + \frac{b c^{4} d \operatorname{atan}{\left (c x \right )}}{4} + \frac{b c^{3} d}{4 x} - \frac{b c^{2} e \operatorname{atan}{\left (c x \right )}}{2} - \frac{b c d}{12 x^{3}} - \frac{b c e}{2 x} - \frac{b d \operatorname{atan}{\left (c x \right )}}{4 x^{4}} - \frac{b e \operatorname{atan}{\left (c x \right )}}{2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*d/(4*x**4) - a*e/(2*x**2) + b*c**4*d*atan(c*x)/4 + b*c**3*d/(4*x) - b*c**2*e*atan(c*x)/2 - b*c*d/(12*x**3)
- b*c*e/(2*x) - b*d*atan(c*x)/(4*x**4) - b*e*atan(c*x)/(2*x**2)

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Giac [A]  time = 1.16072, size = 143, normalized size = 1.74 \begin{align*} -\frac{3 \, \pi b c^{4} d x^{4} \mathrm{sgn}\left (c\right ) \mathrm{sgn}\left (x\right ) - 3 \, b c^{4} d x^{4} \arctan \left (c x\right ) + 6 \, b c^{2} x^{4} \arctan \left (c x\right ) e - 3 \, b c^{3} d x^{3} + 6 \, b c x^{3} e + 6 \, b x^{2} \arctan \left (c x\right ) e + b c d x + 6 \, a x^{2} e + 3 \, b d \arctan \left (c x\right ) + 3 \, a d}{12 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*pi*b*c^4*d*x^4*sgn(c)*sgn(x) - 3*b*c^4*d*x^4*arctan(c*x) + 6*b*c^2*x^4*arctan(c*x)*e - 3*b*c^3*d*x^3
+ 6*b*c*x^3*e + 6*b*x^2*arctan(c*x)*e + b*c*d*x + 6*a*x^2*e + 3*b*d*arctan(c*x) + 3*a*d)/x^4