3.1295 \(\int \frac{(a+b \tan ^{-1}(c x)) (d+e \log (1+c^2 x^2))}{x^5} \, dx\)

Optimal. Leaf size=225 \[ -\frac{1}{4} i b c^4 e \text{PolyLog}(2,-i c x)+\frac{1}{4} i b c^4 e \text{PolyLog}(2,i c x)-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{4 x^4}-\frac{a c^2 e}{4 x^2}+\frac{1}{4} a c^4 e \log \left (c^2 x^2+1\right )-\frac{1}{2} a c^4 e \log (x)+\frac{b c^3 \left (e \log \left (c^2 x^2+1\right )+d\right )}{4 x}-\frac{b c \left (e \log \left (c^2 x^2+1\right )+d\right )}{12 x^3}+\frac{1}{4} b c^4 \tan ^{-1}(c x) \left (e \log \left (c^2 x^2+1\right )+d\right )-\frac{b c^2 e \tan ^{-1}(c x)}{4 x^2}-\frac{5 b c^3 e}{12 x}-\frac{11}{12} b c^4 e \tan ^{-1}(c x) \]

[Out]

-(a*c^2*e)/(4*x^2) - (5*b*c^3*e)/(12*x) - (11*b*c^4*e*ArcTan[c*x])/12 - (b*c^2*e*ArcTan[c*x])/(4*x^2) - (a*c^4
*e*Log[x])/2 + (a*c^4*e*Log[1 + c^2*x^2])/4 - (b*c*(d + e*Log[1 + c^2*x^2]))/(12*x^3) + (b*c^3*(d + e*Log[1 +
c^2*x^2]))/(4*x) + (b*c^4*ArcTan[c*x]*(d + e*Log[1 + c^2*x^2]))/4 - ((a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^
2]))/(4*x^4) - (I/4)*b*c^4*e*PolyLog[2, (-I)*c*x] + (I/4)*b*c^4*e*PolyLog[2, I*c*x]

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Rubi [A]  time = 0.255709, antiderivative size = 225, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 10, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {4852, 325, 203, 5021, 1802, 635, 260, 4980, 4848, 2391} \[ -\frac{1}{4} i b c^4 e \text{PolyLog}(2,-i c x)+\frac{1}{4} i b c^4 e \text{PolyLog}(2,i c x)-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{4 x^4}-\frac{a c^2 e}{4 x^2}+\frac{1}{4} a c^4 e \log \left (c^2 x^2+1\right )-\frac{1}{2} a c^4 e \log (x)+\frac{b c^3 \left (e \log \left (c^2 x^2+1\right )+d\right )}{4 x}-\frac{b c \left (e \log \left (c^2 x^2+1\right )+d\right )}{12 x^3}+\frac{1}{4} b c^4 \tan ^{-1}(c x) \left (e \log \left (c^2 x^2+1\right )+d\right )-\frac{b c^2 e \tan ^{-1}(c x)}{4 x^2}-\frac{5 b c^3 e}{12 x}-\frac{11}{12} b c^4 e \tan ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[((a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]))/x^5,x]

[Out]

-(a*c^2*e)/(4*x^2) - (5*b*c^3*e)/(12*x) - (11*b*c^4*e*ArcTan[c*x])/12 - (b*c^2*e*ArcTan[c*x])/(4*x^2) - (a*c^4
*e*Log[x])/2 + (a*c^4*e*Log[1 + c^2*x^2])/4 - (b*c*(d + e*Log[1 + c^2*x^2]))/(12*x^3) + (b*c^3*(d + e*Log[1 +
c^2*x^2]))/(4*x) + (b*c^4*ArcTan[c*x]*(d + e*Log[1 + c^2*x^2]))/4 - ((a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^
2]))/(4*x^4) - (I/4)*b*c^4*e*PolyLog[2, (-I)*c*x] + (I/4)*b*c^4*e*PolyLog[2, I*c*x]

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 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])

Rule 5021

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

Rule 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 260

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

Rule 4980

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With
[{u = ExpandIntegrand[(a + b*ArcTan[c*x])^p, (f*x)^m*(d + e*x^2)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b,
c, d, e, f, m}, x] && IntegerQ[q] && IGtQ[p, 0] && ((EqQ[p, 1] && GtQ[q, 0]) || IntegerQ[m])

Rule 4848

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[(I*b)/2, Int[Log[1 - I*c*x
]/x, x], x] - Dist[(I*b)/2, Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 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^5} \, dx &=-\frac{b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{12 x^3}+\frac{b c^3 \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x}+\frac{1}{4} b c^4 \tan ^{-1}(c x) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x^4}-\left (2 c^2 e\right ) \int \left (\frac{-3 a-b c x+3 b c^3 x^3}{12 x^3 \left (1+c^2 x^2\right )}+\frac{b \left (-1+c^2 x^2\right ) \tan ^{-1}(c x)}{4 x^3}\right ) \, dx\\ &=-\frac{b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{12 x^3}+\frac{b c^3 \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x}+\frac{1}{4} b c^4 \tan ^{-1}(c x) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x^4}-\frac{1}{6} \left (c^2 e\right ) \int \frac{-3 a-b c x+3 b c^3 x^3}{x^3 \left (1+c^2 x^2\right )} \, dx-\frac{1}{2} \left (b c^2 e\right ) \int \frac{\left (-1+c^2 x^2\right ) \tan ^{-1}(c x)}{x^3} \, dx\\ &=-\frac{b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{12 x^3}+\frac{b c^3 \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x}+\frac{1}{4} b c^4 \tan ^{-1}(c x) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x^4}-\frac{1}{6} \left (c^2 e\right ) \int \left (-\frac{3 a}{x^3}-\frac{b c}{x^2}+\frac{3 a c^2}{x}-\frac{c^3 (-4 b+3 a c x)}{1+c^2 x^2}\right ) \, dx-\frac{1}{2} \left (b c^2 e\right ) \int \left (-\frac{\tan ^{-1}(c x)}{x^3}+\frac{c^2 \tan ^{-1}(c x)}{x}\right ) \, dx\\ &=-\frac{a c^2 e}{4 x^2}-\frac{b c^3 e}{6 x}-\frac{1}{2} a c^4 e \log (x)-\frac{b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{12 x^3}+\frac{b c^3 \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x}+\frac{1}{4} b c^4 \tan ^{-1}(c x) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x^4}+\frac{1}{2} \left (b c^2 e\right ) \int \frac{\tan ^{-1}(c x)}{x^3} \, dx-\frac{1}{2} \left (b c^4 e\right ) \int \frac{\tan ^{-1}(c x)}{x} \, dx+\frac{1}{6} \left (c^5 e\right ) \int \frac{-4 b+3 a c x}{1+c^2 x^2} \, dx\\ &=-\frac{a c^2 e}{4 x^2}-\frac{b c^3 e}{6 x}-\frac{b c^2 e \tan ^{-1}(c x)}{4 x^2}-\frac{1}{2} a c^4 e \log (x)-\frac{b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{12 x^3}+\frac{b c^3 \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x}+\frac{1}{4} b c^4 \tan ^{-1}(c x) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x^4}+\frac{1}{4} \left (b c^3 e\right ) \int \frac{1}{x^2 \left (1+c^2 x^2\right )} \, dx-\frac{1}{4} \left (i b c^4 e\right ) \int \frac{\log (1-i c x)}{x} \, dx+\frac{1}{4} \left (i b c^4 e\right ) \int \frac{\log (1+i c x)}{x} \, dx-\frac{1}{3} \left (2 b c^5 e\right ) \int \frac{1}{1+c^2 x^2} \, dx+\frac{1}{2} \left (a c^6 e\right ) \int \frac{x}{1+c^2 x^2} \, dx\\ &=-\frac{a c^2 e}{4 x^2}-\frac{5 b c^3 e}{12 x}-\frac{2}{3} b c^4 e \tan ^{-1}(c x)-\frac{b c^2 e \tan ^{-1}(c x)}{4 x^2}-\frac{1}{2} a c^4 e \log (x)+\frac{1}{4} a c^4 e \log \left (1+c^2 x^2\right )-\frac{b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{12 x^3}+\frac{b c^3 \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x}+\frac{1}{4} b c^4 \tan ^{-1}(c x) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x^4}-\frac{1}{4} i b c^4 e \text{Li}_2(-i c x)+\frac{1}{4} i b c^4 e \text{Li}_2(i c x)-\frac{1}{4} \left (b c^5 e\right ) \int \frac{1}{1+c^2 x^2} \, dx\\ &=-\frac{a c^2 e}{4 x^2}-\frac{5 b c^3 e}{12 x}-\frac{11}{12} b c^4 e \tan ^{-1}(c x)-\frac{b c^2 e \tan ^{-1}(c x)}{4 x^2}-\frac{1}{2} a c^4 e \log (x)+\frac{1}{4} a c^4 e \log \left (1+c^2 x^2\right )-\frac{b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{12 x^3}+\frac{b c^3 \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x}+\frac{1}{4} b c^4 \tan ^{-1}(c x) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{4 x^4}-\frac{1}{4} i b c^4 e \text{Li}_2(-i c x)+\frac{1}{4} i b c^4 e \text{Li}_2(i c x)\\ \end{align*}

Mathematica [A]  time = 0.17545, size = 260, normalized size = 1.16 \[ -\frac{3 i b c^4 e x^4 \text{PolyLog}(2,-i c x)-3 i b c^4 e x^4 \text{PolyLog}(2,i c x)+3 a c^2 e x^2+6 a c^4 e x^4 \log (x)-3 a c^4 e x^4 \log \left (c^2 x^2+1\right )+3 a e \log \left (c^2 x^2+1\right )+3 a d-3 b c^3 d x^3-3 b c^4 d x^4 \tan ^{-1}(c x)+5 b c^3 e x^3-3 b c^3 e x^3 \log \left (c^2 x^2+1\right )+b c e x \log \left (c^2 x^2+1\right )+11 b c^4 e x^4 \tan ^{-1}(c x)+3 b c^2 e x^2 \tan ^{-1}(c x)-3 b c^4 e x^4 \log \left (c^2 x^2+1\right ) \tan ^{-1}(c x)+3 b e \log \left (c^2 x^2+1\right ) \tan ^{-1}(c x)+b c d x+3 b d \tan ^{-1}(c x)}{12 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]))/x^5,x]

[Out]

-(3*a*d + b*c*d*x + 3*a*c^2*e*x^2 - 3*b*c^3*d*x^3 + 5*b*c^3*e*x^3 + 3*b*d*ArcTan[c*x] + 3*b*c^2*e*x^2*ArcTan[c
*x] - 3*b*c^4*d*x^4*ArcTan[c*x] + 11*b*c^4*e*x^4*ArcTan[c*x] + 6*a*c^4*e*x^4*Log[x] + 3*a*e*Log[1 + c^2*x^2] +
b*c*e*x*Log[1 + c^2*x^2] - 3*b*c^3*e*x^3*Log[1 + c^2*x^2] - 3*a*c^4*e*x^4*Log[1 + c^2*x^2] + 3*b*e*ArcTan[c*x
]*Log[1 + c^2*x^2] - 3*b*c^4*e*x^4*ArcTan[c*x]*Log[1 + c^2*x^2] + (3*I)*b*c^4*e*x^4*PolyLog[2, (-I)*c*x] - (3*
I)*b*c^4*e*x^4*PolyLog[2, I*c*x])/(12*x^4)

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Maple [F]  time = 9.67, 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}^{5}}}\, 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^5,x)

[Out]

int((a+b*arctan(c*x))*(d+e*ln(c^2*x^2+1))/x^5,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \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}{4} \,{\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^{2} - \frac{\log \left (c^{2} x^{2} + 1\right )}{x^{4}}\right )} a e - \frac{{\left (6 \, c^{6} x^{4} \int \frac{x \arctan \left (c x\right )}{c^{2} x^{2} + 1}\,{d x} + 8 \, c^{4} x^{4} \arctan \left (c x\right ) - 6 \, c^{2} x^{4} \int \frac{\arctan \left (c x\right )}{c^{2} x^{5} + x^{3}}\,{d x} + 2 \, c^{3} x^{3} -{\left (3 \, c^{3} x^{3} - c x + 3 \,{\left (c^{4} x^{4} - 1\right )} \arctan \left (c x\right )\right )} \log \left (c^{2} x^{2} + 1\right )\right )} b e}{12 \, x^{4}} - \frac{a d}{4 \, x^{4}} \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^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 + 1/4*((c^2*log(c^2*x^2 + 1) - c^2*
log(x^2) - 1/x^2)*c^2 - log(c^2*x^2 + 1)/x^4)*a*e - 1/12*(72*c^6*x^4*integrate(1/12*x*arctan(c*x)/(c^2*x^2 + 1
), x) + 8*c^4*x^4*arctan(c*x) - 72*c^2*x^4*integrate(1/12*arctan(c*x)/(c^2*x^5 + x^3), x) + 2*c^3*x^3 - (3*c^3
*x^3 - c*x + 3*(c^4*x^4 - 1)*arctan(c*x))*log(c^2*x^2 + 1))*b*e/x^4 - 1/4*a*d/x^4

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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^{5}}, 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^5,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^5, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{atan}{\left (c x \right )}\right ) \left (d + e \log{\left (c^{2} x^{2} + 1 \right )}\right )}{x^{5}}\, dx \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**5,x)

[Out]

Integral((a + b*atan(c*x))*(d + e*log(c**2*x**2 + 1))/x**5, 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^{5}}\,{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^5,x, algorithm="giac")

[Out]

integrate((b*arctan(c*x) + a)*(e*log(c^2*x^2 + 1) + d)/x^5, x)