3.1292 \(\int \frac{(a+b \tan ^{-1}(c x)) (d+e \log (1+c^2 x^2))}{x^2} \, dx\)

Optimal. Leaf size=100 \[ -\frac{1}{2} b c e \text{PolyLog}\left (2,\frac{1}{c^2 x^2+1}\right )-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{x}+\frac{c e \left (a+b \tan ^{-1}(c x)\right )^2}{b}+\frac{1}{2} b c \log \left (1-\frac{1}{c^2 x^2+1}\right ) \left (e \log \left (c^2 x^2+1\right )+d\right ) \]

[Out]

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

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Rubi [A]  time = 0.24991, antiderivative size = 92, normalized size of antiderivative = 0.92, number of steps used = 8, number of rules used = 8, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {5017, 2475, 2411, 2344, 2301, 2316, 2315, 4884} \[ -\frac{1}{2} b c e \text{PolyLog}\left (2,-c^2 x^2\right )-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{x}+\frac{c e \left (a+b \tan ^{-1}(c x)\right )^2}{b}-\frac{b c \left (e \log \left (c^2 x^2+1\right )+d\right )^2}{4 e}+b c d \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*e*(a + b*ArcTan[c*x])^2)/b + b*c*d*Log[x] - ((a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]))/x - (b*c*(d + e*
Log[1 + c^2*x^2])^2)/(4*e) - (b*c*e*PolyLog[2, -(c^2*x^2)])/2

Rule 5017

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*(e_.))*(x_)^(m_.), x_Symbol] :> Simp
[(x^(m + 1)*(d + e*Log[f + g*x^2])*(a + b*ArcTan[c*x]))/(m + 1), x] + (-Dist[(b*c)/(m + 1), Int[(x^(m + 1)*(d
+ e*Log[f + g*x^2]))/(1 + c^2*x^2), x], x] - Dist[(2*e*g)/(m + 1), Int[(x^(m + 2)*(a + b*ArcTan[c*x]))/(f + g*
x^2), x], x]) /; FreeQ[{a, b, c, d, e, f, g}, x] && ILtQ[m/2, 0]

Rule 2475

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q,
x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 2344

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Dist[1/d, Int[(a + b*
Log[c*x^n])^p/x, x], x] - Dist[e/d, Int[(a + b*Log[c*x^n])^p/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e, n}, x]
&& IGtQ[p, 0]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2316

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[((a + b*Log[-((c*d)/e)])*Log[d + e*
x])/e, x] + Dist[b, Int[Log[-((e*x)/d)]/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[-((c*d)/e), 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 4884

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan[c*x])^(p +
1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && NeQ[p, -1]

Rubi steps

\begin{align*} \int \frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^2} \, dx &=-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x}+(b c) \int \frac{d+e \log \left (1+c^2 x^2\right )}{x \left (1+c^2 x^2\right )} \, dx+\left (2 c^2 e\right ) \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx\\ &=\frac{c e \left (a+b \tan ^{-1}(c x)\right )^2}{b}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x}+\frac{1}{2} (b c) \operatorname{Subst}\left (\int \frac{d+e \log \left (1+c^2 x\right )}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )\\ &=\frac{c e \left (a+b \tan ^{-1}(c x)\right )^2}{b}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x}+\frac{b \operatorname{Subst}\left (\int \frac{d+e \log (x)}{x \left (-\frac{1}{c^2}+\frac{x}{c^2}\right )} \, dx,x,1+c^2 x^2\right )}{2 c}\\ &=\frac{c e \left (a+b \tan ^{-1}(c x)\right )^2}{b}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x}+\frac{b \operatorname{Subst}\left (\int \frac{d+e \log (x)}{-\frac{1}{c^2}+\frac{x}{c^2}} \, dx,x,1+c^2 x^2\right )}{2 c}-\frac{1}{2} (b c) \operatorname{Subst}\left (\int \frac{d+e \log (x)}{x} \, dx,x,1+c^2 x^2\right )\\ &=\frac{c e \left (a+b \tan ^{-1}(c x)\right )^2}{b}+b c d \log (x)-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x}-\frac{b c \left (d+e \log \left (1+c^2 x^2\right )\right )^2}{4 e}+\frac{(b e) \operatorname{Subst}\left (\int \frac{\log (x)}{-\frac{1}{c^2}+\frac{x}{c^2}} \, dx,x,1+c^2 x^2\right )}{2 c}\\ &=\frac{c e \left (a+b \tan ^{-1}(c x)\right )^2}{b}+b c d \log (x)-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x}-\frac{b c \left (d+e \log \left (1+c^2 x^2\right )\right )^2}{4 e}-\frac{1}{2} b c e \text{Li}_2\left (-c^2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0897007, size = 113, normalized size = 1.13 \[ b c \left (\frac{1}{2} \left (e \text{PolyLog}\left (2,c^2 x^2+1\right )+\log \left (-c^2 x^2\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )\right )-\frac{\left (e \log \left (c^2 x^2+1\right )+d\right )^2}{4 e}\right )-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{x}+\frac{c e \left (a+b \tan ^{-1}(c x)\right )^2}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*e*(a + b*ArcTan[c*x])^2)/b - ((a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]))/x + b*c*(-(d + e*Log[1 + c^2*x^
2])^2/(4*e) + (Log[-(c^2*x^2)]*(d + e*Log[1 + c^2*x^2]) + e*PolyLog[2, 1 + c^2*x^2])/2)

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Maple [F]  time = 3.487, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\arctan \left ( cx \right ) \right ) \left ( d+e\ln \left ({c}^{2}{x}^{2}+1 \right ) \right ) }{{x}^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \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 +{\left (2 \, c \arctan \left (c x\right ) - \frac{\log \left (c^{2} x^{2} + 1\right )}{x}\right )} a e + b e \int \frac{\arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right )}{x^{2}}\,{d x} - \frac{a d}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b d \arctan \left (c x\right ) + a d +{\left (b e \arctan \left (c x\right ) + a e\right )} \log \left (c^{2} x^{2} + 1\right )}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))*(d+e*log(c^2*x^2+1))/x^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 140.782, size = 160, normalized size = 1.6 \begin{align*} - \frac{a d}{x} + \frac{2 a e \operatorname{atan}{\left (\frac{x}{\sqrt{\frac{1}{c^{2}}}} \right )}}{\sqrt{\frac{1}{c^{2}}}} - \frac{a e \log{\left (c^{2} x^{2} + 1 \right )}}{x} - b c^{3} e \left (\begin{cases} 0 & \text{for}\: c^{2} = 0 \\\frac{\log{\left (c^{2} x^{2} + 1 \right )}^{2}}{4 c^{2}} & \text{otherwise} \end{cases}\right ) + 4 b c^{2} e \left (\begin{cases} 0 & \text{for}\: c = 0 \\\frac{\operatorname{atan}^{2}{\left (c x \right )}}{4 c} & \text{otherwise} \end{cases}\right ) - \frac{b c d \log{\left (c^{2} + \frac{1}{x^{2}} \right )}}{2} - \frac{b c e \operatorname{Li}_{2}\left (c^{2} x^{2} e^{i \pi }\right )}{2} - \frac{b d \operatorname{atan}{\left (c x \right )}}{x} - \frac{b e \log{\left (c^{2} x^{2} + 1 \right )} \operatorname{atan}{\left (c x \right )}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x))*(d+e*ln(c**2*x**2+1))/x**2,x)

[Out]

-a*d/x + 2*a*e*atan(x/sqrt(c**(-2)))/sqrt(c**(-2)) - a*e*log(c**2*x**2 + 1)/x - b*c**3*e*Piecewise((0, Eq(c**2
, 0)), (log(c**2*x**2 + 1)**2/(4*c**2), True)) + 4*b*c**2*e*Piecewise((0, Eq(c, 0)), (atan(c*x)**2/(4*c), True
)) - b*c*d*log(c**2 + x**(-2))/2 - b*c*e*polylog(2, c**2*x**2*exp_polar(I*pi))/2 - b*d*atan(c*x)/x - b*e*log(c
**2*x**2 + 1)*atan(c*x)/x

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (c x\right ) + a\right )}{\left (e \log \left (c^{2} x^{2} + 1\right ) + d\right )}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))*(d+e*log(c^2*x^2+1))/x^2,x, algorithm="giac")

[Out]

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