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

Optimal. Leaf size=101 \[ \frac{i \log (d (a+b x)) \text{PolyLog}(2,-i c (a+b x))}{2 b}-\frac{i \log (d (a+b x)) \text{PolyLog}(2,i c (a+b x))}{2 b}-\frac{i \text{PolyLog}(3,-i c (a+b x))}{2 b}+\frac{i \text{PolyLog}(3,i c (a+b x))}{2 b} \]

[Out]

((I/2)*Log[d*(a + b*x)]*PolyLog[2, (-I)*c*(a + b*x)])/b - ((I/2)*Log[d*(a + b*x)]*PolyLog[2, I*c*(a + b*x)])/b
- ((I/2)*PolyLog[3, (-I)*c*(a + b*x)])/b + ((I/2)*PolyLog[3, I*c*(a + b*x)])/b

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Rubi [A]  time = 0.266421, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {4848, 2391, 5209, 2444, 2433, 2374, 6589} \[ \frac{i \log (d (a+b x)) \text{PolyLog}(2,-i c (a+b x))}{2 b}-\frac{i \log (d (a+b x)) \text{PolyLog}(2,i c (a+b x))}{2 b}-\frac{i \text{PolyLog}(3,-i c (a+b x))}{2 b}+\frac{i \text{PolyLog}(3,i c (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*Log[d*(a + b*x)]*PolyLog[2, (-I)*c*(a + b*x)])/b - ((I/2)*Log[d*(a + b*x)]*PolyLog[2, I*c*(a + b*x)])/b
- ((I/2)*PolyLog[3, (-I)*c*(a + b*x)])/b + ((I/2)*PolyLog[3, I*c*(a + b*x)])/b

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]

Rule 5209

Int[(ArcTan[v_]*Log[w_])/((a_.) + (b_.)*(x_)), x_Symbol] :> Dist[I/2, Int[(Log[1 - I*v]*Log[w])/(a + b*x), x],
x] - Dist[I/2, Int[(Log[1 + I*v]*Log[w])/(a + b*x), x], x] /; FreeQ[{a, b}, x] && LinearQ[v, x] && LinearQ[w,
x] && EqQ[Simplify[D[v/(a + b*x), x]], 0] && EqQ[Simplify[D[w/(a + b*x), x]], 0]

Rule 2444

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Int[u*(a + b*Log[c*ExpandToSum[v, x]^n])^p
, x] /; FreeQ[{a, b, c, n, p}, x] && LinearQ[v, x] &&  !LinearMatchQ[v, x] &&  !(EqQ[n, 1] && MatchQ[c*v, (e_.
)*((f_) + (g_.)*x) /; FreeQ[{e, f, g}, x]])

Rule 2433

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*
(g_.))*((k_.) + (l_.)*(x_))^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((k*x)/d)^r*(a + b*Log[c*x^n])^p*(f + g*Lo
g[h*((e*i - d*j)/e + (j*x)/e)^m]), x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, j, k, l, n, p, r},
x] && EqQ[e*k - d*l, 0]

Rule 2374

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

Rule 6589

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

Rubi steps

\begin{align*} \int \frac{\tan ^{-1}(c (a+b x)) \log (d (a+b x))}{a+b x} \, dx &=\frac{1}{2} i \int \frac{\log (d (a+b x)) \log (1-i c (a+b x))}{a+b x} \, dx-\frac{1}{2} i \int \frac{\log (d (a+b x)) \log (1+i c (a+b x))}{a+b x} \, dx\\ &=\frac{1}{2} i \int \frac{\log (d (a+b x)) \log (1-i a c-i b c x)}{a+b x} \, dx-\frac{1}{2} i \int \frac{\log (d (a+b x)) \log (1+i a c+i b c x)}{a+b x} \, dx\\ &=\frac{i \operatorname{Subst}\left (\int \frac{\log (d x) \log \left (\frac{i a b c+b (1-i a c)}{b}-i c x\right )}{x} \, dx,x,a+b x\right )}{2 b}-\frac{i \operatorname{Subst}\left (\int \frac{\log (d x) \log \left (\frac{-i a b c+b (1+i a c)}{b}+i c x\right )}{x} \, dx,x,a+b x\right )}{2 b}\\ &=\frac{i \log (d (a+b x)) \text{Li}_2(-i c (a+b x))}{2 b}-\frac{i \log (d (a+b x)) \text{Li}_2(i c (a+b x))}{2 b}-\frac{i \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i c x)}{x} \, dx,x,a+b x\right )}{2 b}+\frac{i \operatorname{Subst}\left (\int \frac{\text{Li}_2(i c x)}{x} \, dx,x,a+b x\right )}{2 b}\\ &=\frac{i \log (d (a+b x)) \text{Li}_2(-i c (a+b x))}{2 b}-\frac{i \log (d (a+b x)) \text{Li}_2(i c (a+b x))}{2 b}-\frac{i \text{Li}_3(-i c (a+b x))}{2 b}+\frac{i \text{Li}_3(i c (a+b x))}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.137435, size = 79, normalized size = 0.78 \[ \frac{i (\log (d (a+b x)) \text{PolyLog}(2,-i c (a+b x))-\log (d (a+b x)) \text{PolyLog}(2,i c (a+b x))-\text{PolyLog}(3,-i c (a+b x))+\text{PolyLog}(3,i c (a+b x)))}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*(Log[d*(a + b*x)]*PolyLog[2, (-I)*c*(a + b*x)] - Log[d*(a + b*x)]*PolyLog[2, I*c*(a + b*x)] - PolyLog[3
, (-I)*c*(a + b*x)] + PolyLog[3, I*c*(a + b*x)]))/b

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Maple [F]  time = 1.936, size = 0, normalized size = 0. \begin{align*} \int{\frac{\arctan \left ( c \left ( bx+a \right ) \right ) \ln \left ( d \left ( bx+a \right ) \right ) }{bx+a}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(c*(b*x+a))*ln(d*(b*x+a))/(b*x+a),x)

[Out]

int(arctan(c*(b*x+a))*ln(d*(b*x+a))/(b*x+a),x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arctan(b*c*x + a*c)*log(b*d*x + a*d)/(b*x + a), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arctan((b*x + a)*c)*log((b*x + a)*d)/(b*x + a), x)