3.106 \(\int \frac {a+b \coth ^{-1}(c+d x)}{e+f x} \, dx\)

Optimal. Leaf size=130 \[ \frac {\left (a+b \coth ^{-1}(c+d x)\right ) \log \left (\frac {2 d (e+f x)}{(c+d x+1) (-c f+d e+f)}\right )}{f}-\frac {\log \left (\frac {2}{c+d x+1}\right ) \left (a+b \coth ^{-1}(c+d x)\right )}{f}-\frac {b \text {Li}_2\left (1-\frac {2 d (e+f x)}{(d e-c f+f) (c+d x+1)}\right )}{2 f}+\frac {b \text {Li}_2\left (1-\frac {2}{c+d x+1}\right )}{2 f} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {6112, 5921, 2402, 2315, 2447} \[ -\frac {b \text {PolyLog}\left (2,1-\frac {2 d (e+f x)}{(c+d x+1) (-c f+d e+f)}\right )}{2 f}+\frac {b \text {PolyLog}\left (2,1-\frac {2}{c+d x+1}\right )}{2 f}+\frac {\left (a+b \coth ^{-1}(c+d x)\right ) \log \left (\frac {2 d (e+f x)}{(c+d x+1) (-c f+d e+f)}\right )}{f}-\frac {\log \left (\frac {2}{c+d x+1}\right ) \left (a+b \coth ^{-1}(c+d x)\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 5921

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcCoth[c*x])*Log[2/(1
+ c*x)])/e, x] + (Dist[(b*c)/e, Int[Log[2/(1 + c*x)]/(1 - c^2*x^2), x], x] - Dist[(b*c)/e, Int[Log[(2*c*(d +
e*x))/((c*d + e)*(1 + c*x))]/(1 - c^2*x^2), x], x] + Simp[((a + b*ArcCoth[c*x])*Log[(2*c*(d + e*x))/((c*d + e)
*(1 + c*x))])/e, x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 - e^2, 0]

Rule 6112

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

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 206, normalized size = 1.58 \[ \frac {a \log (e+f x)}{f}-\frac {b \text {Li}_2\left (\frac {d (e+f x)}{d e-c f-f}\right )}{2 f}+\frac {b \text {Li}_2\left (\frac {d (e+f x)}{d e-c f+f}\right )}{2 f}+\frac {b \log (e+f x) \log \left (\frac {f (-c-d x+1)}{-c f+d e+f}\right )}{2 f}-\frac {b \log \left (-\frac {-c-d x+1}{c+d x}\right ) \log (e+f x)}{2 f}-\frac {b \log (e+f x) \log \left (-\frac {f (c+d x+1)}{-c f+d e-f}\right )}{2 f}+\frac {b \log \left (\frac {c+d x+1}{c+d x}\right ) \log (e+f x)}{2 f} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {arcoth}\left (d x + c\right ) + a}{f x + e}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 202, normalized size = 1.55 \[ \frac {a \ln \left (\left (d x +c \right ) f -c f +d e \right )}{f}+\frac {b \ln \left (\left (d x +c \right ) f -c f +d e \right ) \mathrm {arccoth}\left (d x +c \right )}{f}-\frac {b \ln \left (\left (d x +c \right ) f -c f +d e \right ) \ln \left (\frac {\left (d x +c \right ) f +f}{c f -d e +f}\right )}{2 f}-\frac {b \dilog \left (\frac {\left (d x +c \right ) f +f}{c f -d e +f}\right )}{2 f}+\frac {b \ln \left (\left (d x +c \right ) f -c f +d e \right ) \ln \left (\frac {\left (d x +c \right ) f -f}{c f -d e -f}\right )}{2 f}+\frac {b \dilog \left (\frac {\left (d x +c \right ) f -f}{c f -d e -f}\right )}{2 f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*ln((d*x+c)*f-c*f+d*e)/f+b*ln((d*x+c)*f-c*f+d*e)/f*arccoth(d*x+c)-1/2*b/f*ln((d*x+c)*f-c*f+d*e)*ln(((d*x+c)*f
+f)/(c*f-d*e+f))-1/2*b/f*dilog(((d*x+c)*f+f)/(c*f-d*e+f))+1/2*b/f*ln((d*x+c)*f-c*f+d*e)*ln(((d*x+c)*f-f)/(c*f-
d*e-f))+1/2*b/f*dilog(((d*x+c)*f-f)/(c*f-d*e-f))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {acoth}\left (c+d\,x\right )}{e+f\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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