3.1119 \(\int \frac{(d+e x^2) (a+b \tan ^{-1}(c x))}{x^2} \, dx\)

Optimal. Leaf size=57 \[ -\frac{d \left (a+b \tan ^{-1}(c x)\right )}{x}+e x \left (a+b \tan ^{-1}(c x)\right )-\frac{b \left (c^2 d+e\right ) \log \left (c^2 x^2+1\right )}{2 c}+b c d \log (x) \]

[Out]

-((d*(a + b*ArcTan[c*x]))/x) + e*x*(a + b*ArcTan[c*x]) + b*c*d*Log[x] - (b*(c^2*d + e)*Log[1 + c^2*x^2])/(2*c)

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Rubi [A]  time = 0.0765387, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {14, 4976, 446, 72} \[ -\frac{d \left (a+b \tan ^{-1}(c x)\right )}{x}+e x \left (a+b \tan ^{-1}(c x)\right )-\frac{b \left (c^2 d+e\right ) \log \left (c^2 x^2+1\right )}{2 c}+b c d \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d*(a + b*ArcTan[c*x]))/x) + e*x*(a + b*ArcTan[c*x]) + b*c*d*Log[x] - (b*(c^2*d + e)*Log[1 + c^2*x^2])/(2*c)

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 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 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

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

Mathematica [A]  time = 0.0049527, size = 73, normalized size = 1.28 \[ -\frac{a d}{x}+a e x-\frac{1}{2} b c d \log \left (c^2 x^2+1\right )-\frac{b e \log \left (c^2 x^2+1\right )}{2 c}+b c d \log (x)-\frac{b d \tan ^{-1}(c x)}{x}+b e x \tan ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*d)/x) + a*e*x - (b*d*ArcTan[c*x])/x + b*e*x*ArcTan[c*x] + b*c*d*Log[x] - (b*c*d*Log[1 + c^2*x^2])/2 - (b*
e*Log[1 + c^2*x^2])/(2*c)

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Maple [A]  time = 0.04, size = 72, normalized size = 1.3 \begin{align*} aex-{\frac{ad}{x}}+bex\arctan \left ( cx \right ) -{\frac{\arctan \left ( cx \right ) bd}{x}}-{\frac{bcd\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2}}-{\frac{be\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2\,c}}+cbd\ln \left ( cx \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*e*x-a*d/x+b*e*x*arctan(c*x)-b*arctan(c*x)*d/x-1/2*b*c*d*ln(c^2*x^2+1)-1/2*b*e*ln(c^2*x^2+1)/c+c*b*d*ln(c*x)

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Maxima [A]  time = 0.944748, size = 99, normalized size = 1.74 \begin{align*} -\frac{1}{2} \,{\left (c{\left (\log \left (c^{2} x^{2} + 1\right ) - \log \left (x^{2}\right )\right )} + \frac{2 \, \arctan \left (c x\right )}{x}\right )} b d + a e x + \frac{{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b e}{2 \, c} - \frac{a d}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(c*(log(c^2*x^2 + 1) - log(x^2)) + 2*arctan(c*x)/x)*b*d + a*e*x + 1/2*(2*c*x*arctan(c*x) - log(c^2*x^2 +
1))*b*e/c - a*d/x

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Fricas [A]  time = 1.66704, size = 174, normalized size = 3.05 \begin{align*} \frac{2 \, b c^{2} d x \log \left (x\right ) + 2 \, a c e x^{2} - 2 \, a c d -{\left (b c^{2} d + b e\right )} x \log \left (c^{2} x^{2} + 1\right ) + 2 \,{\left (b c e x^{2} - b c d\right )} \arctan \left (c x\right )}{2 \, c x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b*c^2*d*x*log(x) + 2*a*c*e*x^2 - 2*a*c*d - (b*c^2*d + b*e)*x*log(c^2*x^2 + 1) + 2*(b*c*e*x^2 - b*c*d)*a
rctan(c*x))/(c*x)

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Sympy [A]  time = 1.43454, size = 80, normalized size = 1.4 \begin{align*} \begin{cases} - \frac{a d}{x} + a e x + b c d \log{\left (x \right )} - \frac{b c d \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{2} - \frac{b d \operatorname{atan}{\left (c x \right )}}{x} + b e x \operatorname{atan}{\left (c x \right )} - \frac{b e \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{2 c} & \text{for}\: c \neq 0 \\a \left (- \frac{d}{x} + e x\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**2,x)

[Out]

Piecewise((-a*d/x + a*e*x + b*c*d*log(x) - b*c*d*log(x**2 + c**(-2))/2 - b*d*atan(c*x)/x + b*e*x*atan(c*x) - b
*e*log(x**2 + c**(-2))/(2*c), Ne(c, 0)), (a*(-d/x + e*x), True))

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Giac [A]  time = 1.09957, size = 120, normalized size = 2.11 \begin{align*} \frac{2 \, b c x^{2} \arctan \left (c x\right ) e - b c^{2} d x \log \left (c^{2} x^{2} + 1\right ) + 2 \, b c^{2} d x \log \left (x\right ) + 2 \, a c x^{2} e - 2 \, b c d \arctan \left (c x\right ) - b x e \log \left (c^{2} x^{2} + 1\right ) - 2 \, a c d}{2 \, c x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b*c*x^2*arctan(c*x)*e - b*c^2*d*x*log(c^2*x^2 + 1) + 2*b*c^2*d*x*log(x) + 2*a*c*x^2*e - 2*b*c*d*arctan(
c*x) - b*x*e*log(c^2*x^2 + 1) - 2*a*c*d)/(c*x)