3.12 \(\int \frac {a+b \tanh ^{-1}(c+d x)}{c e+d e x} \, dx\)

Optimal. Leaf size=54 \[ \frac {a \log (c+d x)}{d e}-\frac {b \text {Li}_2(-c-d x)}{2 d e}+\frac {b \text {Li}_2(c+d x)}{2 d e} \]

[Out]

a*ln(d*x+c)/d/e-1/2*b*polylog(2,-d*x-c)/d/e+1/2*b*polylog(2,d*x+c)/d/e

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6107, 12, 5912} \[ -\frac {b \text {PolyLog}(2,-c-d x)}{2 d e}+\frac {b \text {PolyLog}(2,c+d x)}{2 d e}+\frac {a \log (c+d x)}{d e} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c + d*x])/(c*e + d*e*x),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 5912

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

Rule 6107

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 54, normalized size = 1.00 \[ \frac {a \log (c+d x)}{d e}-\frac {b \text {Li}_2(-c-d x)}{2 d e}+\frac {b \text {Li}_2(c+d x)}{2 d e} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c + d*x])/(c*e + d*e*x),x]

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {artanh}\left (d x + c\right ) + a}{d e x + c e}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arctanh(d*x + c) + a)/(d*e*x + c*e), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {artanh}\left (d x + c\right ) + a}{d e x + c e}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arctanh(d*x + c) + a)/(d*e*x + c*e), x)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 89, normalized size = 1.65 \[ \frac {a \ln \left (d x +c \right )}{d e}+\frac {b \ln \left (d x +c \right ) \arctanh \left (d x +c \right )}{d e}-\frac {b \dilog \left (d x +c \right )}{2 d e}-\frac {b \dilog \left (d x +c +1\right )}{2 d e}-\frac {b \ln \left (d x +c \right ) \ln \left (d x +c +1\right )}{2 d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(d*x+c))/(d*e*x+c*e),x)

[Out]

1/d*a/e*ln(d*x+c)+1/d*b/e*ln(d*x+c)*arctanh(d*x+c)-1/2/d*b/e*dilog(d*x+c)-1/2/d*b/e*dilog(d*x+c+1)-1/2/d*b/e*l
n(d*x+c)*ln(d*x+c+1)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, b \int \frac {\log \left (d x + c + 1\right ) - \log \left (-d x - c + 1\right )}{d e x + c e}\,{d x} + \frac {a \log \left (d e x + c e\right )}{d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {a+b\,\mathrm {atanh}\left (c+d\,x\right )}{c\,e+d\,e\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c + d*x))/(c*e + d*e*x),x)

[Out]

int((a + b*atanh(c + d*x))/(c*e + d*e*x), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a}{c + d x}\, dx + \int \frac {b \operatorname {atanh}{\left (c + d x \right )}}{c + d x}\, dx}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(d*x+c))/(d*e*x+c*e),x)

[Out]

(Integral(a/(c + d*x), x) + Integral(b*atanh(c + d*x)/(c + d*x), x))/e

________________________________________________________________________________________