3.55 \(\int \frac {a+b \tanh ^{-1}(c x)}{x (d+c d x)^2} \, dx\)

Optimal. Leaf size=124 \[ \frac {a+b \tanh ^{-1}(c x)}{d^2 (c x+1)}+\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}+\frac {a \log (x)}{d^2}-\frac {b \text {Li}_2(-c x)}{2 d^2}+\frac {b \text {Li}_2(c x)}{2 d^2}-\frac {b \text {Li}_2\left (1-\frac {2}{c x+1}\right )}{2 d^2}+\frac {b}{2 d^2 (c x+1)}-\frac {b \tanh ^{-1}(c x)}{2 d^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.17, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {5940, 5912, 5926, 627, 44, 207, 5918, 2402, 2315} \[ -\frac {b \text {PolyLog}(2,-c x)}{2 d^2}+\frac {b \text {PolyLog}(2,c x)}{2 d^2}-\frac {b \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{2 d^2}+\frac {a+b \tanh ^{-1}(c x)}{d^2 (c x+1)}+\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}+\frac {a \log (x)}{d^2}+\frac {b}{2 d^2 (c x+1)}-\frac {b \tanh ^{-1}(c x)}{2 d^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])/(x*(d + c*d*x)^2),x]

[Out]

b/(2*d^2*(1 + c*x)) - (b*ArcTanh[c*x])/(2*d^2) + (a + b*ArcTanh[c*x])/(d^2*(1 + c*x)) + (a*Log[x])/d^2 + ((a +
b*ArcTanh[c*x])*Log[2/(1 + c*x)])/d^2 - (b*PolyLog[2, -(c*x)])/(2*d^2) + (b*PolyLog[2, c*x])/(2*d^2) - (b*Pol
yLog[2, 1 - 2/(1 + c*x)])/(2*d^2)

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

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 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 5918

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

Rule 5926

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

Rule 5940

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*ArcTanh[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[
p, 0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.42, size = 101, normalized size = 0.81 \[ \frac {\frac {4 a}{c x+1}-4 a \log (c x+1)+4 a \log (x)+b \left (-2 \text {Li}_2\left (e^{-2 \tanh ^{-1}(c x)}\right )-\sinh \left (2 \tanh ^{-1}(c x)\right )+\cosh \left (2 \tanh ^{-1}(c x)\right )+2 \tanh ^{-1}(c x) \left (2 \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )-\sinh \left (2 \tanh ^{-1}(c x)\right )+\cosh \left (2 \tanh ^{-1}(c x)\right )\right )\right )}{4 d^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTanh[c*x])/(x*(d + c*d*x)^2),x]

[Out]

((4*a)/(1 + c*x) + 4*a*Log[x] - 4*a*Log[1 + c*x] + b*(Cosh[2*ArcTanh[c*x]] - 2*PolyLog[2, E^(-2*ArcTanh[c*x])]
+ 2*ArcTanh[c*x]*(Cosh[2*ArcTanh[c*x]] + 2*Log[1 - E^(-2*ArcTanh[c*x])] - Sinh[2*ArcTanh[c*x]]) - Sinh[2*ArcT
anh[c*x]]))/(4*d^2)

________________________________________________________________________________________

fricas [F]  time = 0.40, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {artanh}\left (c x\right ) + a}{c^{2} d^{2} x^{3} + 2 \, c d^{2} x^{2} + d^{2} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x/(c*d*x+d)^2,x, algorithm="fricas")

[Out]

integral((b*arctanh(c*x) + a)/(c^2*d^2*x^3 + 2*c*d^2*x^2 + d^2*x), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {artanh}\left (c x\right ) + a}{{\left (c d x + d\right )}^{2} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x/(c*d*x+d)^2,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)/((c*d*x + d)^2*x), x)

________________________________________________________________________________________

maple [A]  time = 0.06, size = 221, normalized size = 1.78 \[ \frac {a \ln \left (c x \right )}{d^{2}}+\frac {a}{d^{2} \left (c x +1\right )}-\frac {a \ln \left (c x +1\right )}{d^{2}}+\frac {b \arctanh \left (c x \right ) \ln \left (c x \right )}{d^{2}}+\frac {b \arctanh \left (c x \right )}{d^{2} \left (c x +1\right )}-\frac {b \arctanh \left (c x \right ) \ln \left (c x +1\right )}{d^{2}}+\frac {b \ln \left (c x +1\right )^{2}}{4 d^{2}}-\frac {b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{2 d^{2}}+\frac {b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{2 d^{2}}+\frac {b \dilog \left (\frac {1}{2}+\frac {c x}{2}\right )}{2 d^{2}}+\frac {b \ln \left (c x -1\right )}{4 d^{2}}+\frac {b}{2 d^{2} \left (c x +1\right )}-\frac {b \ln \left (c x +1\right )}{4 d^{2}}-\frac {b \dilog \left (c x \right )}{2 d^{2}}-\frac {b \dilog \left (c x +1\right )}{2 d^{2}}-\frac {b \ln \left (c x \right ) \ln \left (c x +1\right )}{2 d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))/x/(c*d*x+d)^2,x)

[Out]

a/d^2*ln(c*x)+a/d^2/(c*x+1)-a/d^2*ln(c*x+1)+b/d^2*arctanh(c*x)*ln(c*x)+b/d^2*arctanh(c*x)/(c*x+1)-b/d^2*arctan
h(c*x)*ln(c*x+1)+1/4*b/d^2*ln(c*x+1)^2-1/2*b/d^2*ln(-1/2*c*x+1/2)*ln(c*x+1)+1/2*b/d^2*ln(-1/2*c*x+1/2)*ln(1/2+
1/2*c*x)+1/2*b/d^2*dilog(1/2+1/2*c*x)+1/4*b/d^2*ln(c*x-1)+1/2*b/d^2/(c*x+1)-1/4*b/d^2*ln(c*x+1)-1/2*b/d^2*dilo
g(c*x)-1/2*b/d^2*dilog(c*x+1)-1/2*b/d^2*ln(c*x)*ln(c*x+1)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a {\left (\frac {1}{c d^{2} x + d^{2}} - \frac {\log \left (c x + 1\right )}{d^{2}} + \frac {\log \relax (x)}{d^{2}}\right )} + \frac {1}{2} \, b \int \frac {\log \left (c x + 1\right ) - \log \left (-c x + 1\right )}{c^{2} d^{2} x^{3} + 2 \, c d^{2} x^{2} + d^{2} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x/(c*d*x+d)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {atanh}\left (c\,x\right )}{x\,{\left (d+c\,d\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))/(x*(d + c*d*x)^2),x)

[Out]

int((a + b*atanh(c*x))/(x*(d + c*d*x)^2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a}{c^{2} x^{3} + 2 c x^{2} + x}\, dx + \int \frac {b \operatorname {atanh}{\left (c x \right )}}{c^{2} x^{3} + 2 c x^{2} + x}\, dx}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))/x/(c*d*x+d)**2,x)

[Out]

(Integral(a/(c**2*x**3 + 2*c*x**2 + x), x) + Integral(b*atanh(c*x)/(c**2*x**3 + 2*c*x**2 + x), x))/d**2

________________________________________________________________________________________