3.1294 \(\int \frac{(a+b \tan ^{-1}(c x)) (d+e \log (1+c^2 x^2))}{x^4} \, dx\)

Optimal. Leaf size=189 \[ \frac{1}{6} b c^3 e \text{PolyLog}\left (2,\frac{1}{c^2 x^2+1}\right )-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{3 x^3}-\frac{c^3 e \left (a+b \tan ^{-1}(c x)\right )^2}{3 b}-\frac{2 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{3 x}-\frac{1}{6} b c^3 \log \left (1-\frac{1}{c^2 x^2+1}\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )-\frac{b c \left (c^2 x^2+1\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{6 x^2}-\frac{1}{3} b c^3 e \log \left (c^2 x^2+1\right )+b c^3 e \log (x) \]

[Out]

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

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Rubi [A]  time = 0.428442, antiderivative size = 186, normalized size of antiderivative = 0.98, number of steps used = 17, number of rules used = 16, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.615, Rules used = {5017, 2475, 2411, 2347, 2344, 2301, 2316, 2315, 2314, 31, 4918, 4852, 266, 36, 29, 4884} \[ \frac{1}{6} b c^3 e \text{PolyLog}\left (2,-c^2 x^2\right )-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{3 x^3}-\frac{c^3 e \left (a+b \tan ^{-1}(c x)\right )^2}{3 b}-\frac{2 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{3 x}+\frac{b c^3 \left (e \log \left (c^2 x^2+1\right )+d\right )^2}{12 e}-\frac{b c \left (c^2 x^2+1\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{6 x^2}-\frac{1}{3} b c^3 d \log (x)-\frac{1}{3} b c^3 e \log \left (c^2 x^2+1\right )+b c^3 e \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c^2*e*(a + b*ArcTan[c*x]))/(3*x) - (c^3*e*(a + b*ArcTan[c*x])^2)/(3*b) - (b*c^3*d*Log[x])/3 + b*c^3*e*Log[
x] - (b*c^3*e*Log[1 + c^2*x^2])/3 - (b*c*(1 + c^2*x^2)*(d + e*Log[1 + c^2*x^2]))/(6*x^2) - ((a + b*ArcTan[c*x]
)*(d + e*Log[1 + c^2*x^2]))/(3*x^3) + (b*c^3*(d + e*Log[1 + c^2*x^2])^2)/(12*e) + (b*c^3*e*PolyLog[2, -(c^2*x^
2)])/6

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 2347

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/(x_), x_Symbol] :> Dist[1/d, Int[((
d + e*x)^(q + 1)*(a + b*Log[c*x^n])^p)/x, x], x] - Dist[e/d, Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; F
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]

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 2314

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 4918

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

Rule 4852

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
n[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 266

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

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^4} \, dx &=-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{3 x^3}+\frac{1}{3} (b c) \int \frac{d+e \log \left (1+c^2 x^2\right )}{x^3 \left (1+c^2 x^2\right )} \, dx+\frac{1}{3} \left (2 c^2 e\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^2 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{3 x^3}+\frac{1}{6} (b c) \operatorname{Subst}\left (\int \frac{d+e \log \left (1+c^2 x\right )}{x^2 \left (1+c^2 x\right )} \, dx,x,x^2\right )+\frac{1}{3} \left (2 c^2 e\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^2} \, dx-\frac{1}{3} \left (2 c^4 e\right ) \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx\\ &=-\frac{2 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{3 x}-\frac{c^3 e \left (a+b \tan ^{-1}(c x)\right )^2}{3 b}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{3 x^3}+\frac{b \operatorname{Subst}\left (\int \frac{d+e \log (x)}{x \left (-\frac{1}{c^2}+\frac{x}{c^2}\right )^2} \, dx,x,1+c^2 x^2\right )}{6 c}+\frac{1}{3} \left (2 b c^3 e\right ) \int \frac{1}{x \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{2 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{3 x}-\frac{c^3 e \left (a+b \tan ^{-1}(c x)\right )^2}{3 b}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{3 x^3}+\frac{b \operatorname{Subst}\left (\int \frac{d+e \log (x)}{\left (-\frac{1}{c^2}+\frac{x}{c^2}\right )^2} \, dx,x,1+c^2 x^2\right )}{6 c}-\frac{1}{6} (b c) \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 )+\frac{1}{3} \left (b c^3 e\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac{2 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{3 x}-\frac{c^3 e \left (a+b \tan ^{-1}(c x)\right )^2}{3 b}-\frac{b c \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{6 x^2}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{3 x^3}-\frac{1}{6} (b c) \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 )+\frac{1}{6} \left (b c^3\right ) \operatorname{Subst}\left (\int \frac{d+e \log (x)}{x} \, dx,x,1+c^2 x^2\right )+\frac{1}{6} (b c e) \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{c^2}+\frac{x}{c^2}} \, dx,x,1+c^2 x^2\right )+\frac{1}{3} \left (b c^3 e\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )-\frac{1}{3} \left (b c^5 e\right ) \operatorname{Subst}\left (\int \frac{1}{1+c^2 x} \, dx,x,x^2\right )\\ &=-\frac{2 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{3 x}-\frac{c^3 e \left (a+b \tan ^{-1}(c x)\right )^2}{3 b}-\frac{1}{3} b c^3 d \log (x)+b c^3 e \log (x)-\frac{1}{3} b c^3 e \log \left (1+c^2 x^2\right )-\frac{b c \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{6 x^2}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{3 x^3}+\frac{b c^3 \left (d+e \log \left (1+c^2 x^2\right )\right )^2}{12 e}-\frac{1}{6} (b c e) \operatorname{Subst}\left (\int \frac{\log (x)}{-\frac{1}{c^2}+\frac{x}{c^2}} \, dx,x,1+c^2 x^2\right )\\ &=-\frac{2 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{3 x}-\frac{c^3 e \left (a+b \tan ^{-1}(c x)\right )^2}{3 b}-\frac{1}{3} b c^3 d \log (x)+b c^3 e \log (x)-\frac{1}{3} b c^3 e \log \left (1+c^2 x^2\right )-\frac{b c \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{6 x^2}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{3 x^3}+\frac{b c^3 \left (d+e \log \left (1+c^2 x^2\right )\right )^2}{12 e}+\frac{1}{6} b c^3 e \text{Li}_2\left (-c^2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.140352, size = 181, normalized size = 0.96 \[ \frac{1}{12} \left (-2 b c^3 \left (e \text{PolyLog}\left (2,c^2 x^2+1\right )+\log \left (-c^2 x^2\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )\right )-\frac{4 \left (a+b \tan ^{-1}(c x)\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{x^3}-\frac{4 c^3 e \left (a+b \tan ^{-1}(c x)\right )^2}{b}-\frac{8 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{x}+\frac{b c^3 \left (e \log \left (c^2 x^2+1\right )+d\right )^2}{e}-\frac{2 b c \left (e \log \left (c^2 x^2+1\right )+d\right )}{x^2}+6 b c^3 e \left (2 \log (x)-\log \left (c^2 x^2+1\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-8*c^2*e*(a + b*ArcTan[c*x]))/x - (4*c^3*e*(a + b*ArcTan[c*x])^2)/b + 6*b*c^3*e*(2*Log[x] - Log[1 + c^2*x^2]
) - (2*b*c*(d + e*Log[1 + c^2*x^2]))/x^2 - (4*(a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]))/x^3 + (b*c^3*(d +
e*Log[1 + c^2*x^2])^2)/e - 2*b*c^3*(Log[-(c^2*x^2)]*(d + e*Log[1 + c^2*x^2]) + e*PolyLog[2, 1 + c^2*x^2]))/12

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Maple [F]  time = 6.51, 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}^{4}}}\, 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^4,x)

[Out]

int((a+b*arctan(c*x))*(d+e*ln(c^2*x^2+1))/x^4,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 d - \frac{1}{3} \,{\left (2 \,{\left (c \arctan \left (c x\right ) + \frac{1}{x}\right )} c^{2} + \frac{\log \left (c^{2} x^{2} + 1\right )}{x^{3}}\right )} a e + b e \int \frac{\arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right )}{x^{4}}\,{d x} - \frac{a d}{3 \, x^{3}} \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^4,x, algorithm="maxima")

[Out]

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*d - 1/3*(2*(c*arctan(c*x) + 1/x)*c
^2 + log(c^2*x^2 + 1)/x^3)*a*e + b*e*integrate(arctan(c*x)*log(c^2*x^2 + 1)/x^4, x) - 1/3*a*d/x^3

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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^{4}}, 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^4,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^4, x)

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Sympy [A]  time = 53.4573, size = 428, normalized size = 2.26 \begin{align*} - \frac{2 a c^{2} e \operatorname{atan}{\left (\frac{x}{\sqrt{\frac{1}{c^{2}}}} \right )}}{3 \sqrt{\frac{1}{c^{2}}}} - \frac{2 a c^{2} e}{3 x} - \frac{a d}{3 x^{3}} - \frac{a e \log{\left (c^{2} x^{2} + 1 \right )}}{3 x^{3}} - 2 b c^{7} e \left (\begin{cases} \frac{x^{2}}{12 c^{2}} - \frac{\log{\left (c^{2} x^{2} + 1 \right )}}{12 c^{4}} & \text{for}\: c = 0 \\\frac{\log{\left (c^{2} x^{2} + 1 \right )}^{2}}{24 c^{4}} & \text{otherwise} \end{cases}\right ) + \frac{b c^{5} d \left (\begin{cases} x^{2} & \text{for}\: c^{2} = 0 \\\frac{\log{\left (c^{2} x^{2} + 1 \right )}}{c^{2}} & \text{otherwise} \end{cases}\right )}{6} + \frac{b c^{5} e \left (\begin{cases} x^{2} & \text{for}\: c^{2} = 0 \\\frac{\log{\left (c^{2} x^{2} + 1 \right )}}{c^{2}} & \text{otherwise} \end{cases}\right ) \log{\left (c^{2} x^{2} + 1 \right )}}{6} - \frac{b c^{3} d \log{\left (x^{2} \right )}}{6} + \frac{b c^{3} e \log{\left (x \right )}}{3} - \frac{b c^{3} e \log{\left (c^{2} x^{2} + 1 \right )}}{6} - \frac{b c^{3} e \log{\left (6 c^{2} \sqrt{\frac{1}{c^{2}}} + \frac{6 \sqrt{\frac{1}{c^{2}}}}{x^{2}} \right )}}{3} + \frac{b c^{3} e \operatorname{atan}^{2}{\left (\frac{x}{\sqrt{\frac{1}{c^{2}}}} \right )}}{3} + \frac{b c^{3} e \operatorname{Li}_{2}\left (c^{2} x^{2} e^{i \pi }\right )}{6} - \frac{2 b c^{2} e \operatorname{atan}{\left (c x \right )} \operatorname{atan}{\left (\frac{x}{\sqrt{\frac{1}{c^{2}}}} \right )}}{3 \sqrt{\frac{1}{c^{2}}}} - \frac{2 b c^{2} e \operatorname{atan}{\left (c x \right )}}{3 x} - \frac{b c d}{6 x^{2}} - \frac{b c e \log{\left (c^{2} x^{2} + 1 \right )}}{6 x^{2}} - \frac{b d \operatorname{atan}{\left (c x \right )}}{3 x^{3}} - \frac{b e \log{\left (c^{2} x^{2} + 1 \right )} \operatorname{atan}{\left (c x \right )}}{3 x^{3}} \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**4,x)

[Out]

-2*a*c**2*e*atan(x/sqrt(c**(-2)))/(3*sqrt(c**(-2))) - 2*a*c**2*e/(3*x) - a*d/(3*x**3) - a*e*log(c**2*x**2 + 1)
/(3*x**3) - 2*b*c**7*e*Piecewise((x**2/(12*c**2) - log(c**2*x**2 + 1)/(12*c**4), Eq(c, 0)), (log(c**2*x**2 + 1
)**2/(24*c**4), True)) + b*c**5*d*Piecewise((x**2, Eq(c**2, 0)), (log(c**2*x**2 + 1)/c**2, True))/6 + b*c**5*e
*Piecewise((x**2, Eq(c**2, 0)), (log(c**2*x**2 + 1)/c**2, True))*log(c**2*x**2 + 1)/6 - b*c**3*d*log(x**2)/6 +
b*c**3*e*log(x)/3 - b*c**3*e*log(c**2*x**2 + 1)/6 - b*c**3*e*log(6*c**2*sqrt(c**(-2)) + 6*sqrt(c**(-2))/x**2)
/3 + b*c**3*e*atan(x/sqrt(c**(-2)))**2/3 + b*c**3*e*polylog(2, c**2*x**2*exp_polar(I*pi))/6 - 2*b*c**2*e*atan(
c*x)*atan(x/sqrt(c**(-2)))/(3*sqrt(c**(-2))) - 2*b*c**2*e*atan(c*x)/(3*x) - b*c*d/(6*x**2) - b*c*e*log(c**2*x*
*2 + 1)/(6*x**2) - b*d*atan(c*x)/(3*x**3) - b*e*log(c**2*x**2 + 1)*atan(c*x)/(3*x**3)

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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^{4}}\,{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^4,x, algorithm="giac")

[Out]

integrate((b*arctan(c*x) + a)*(e*log(c^2*x^2 + 1) + d)/x^4, x)