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

Optimal. Leaf size=57 \[ -\frac{a+b \tanh ^{-1}(c x)}{c d^2 (c x+1)}-\frac{b}{2 c d^2 (c x+1)}+\frac{b \tanh ^{-1}(c x)}{2 c d^2} \]

[Out]

-b/(2*c*d^2*(1 + c*x)) + (b*ArcTanh[c*x])/(2*c*d^2) - (a + b*ArcTanh[c*x])/(c*d^2*(1 + c*x))

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Rubi [A]  time = 0.0453211, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {5926, 627, 44, 207} \[ -\frac{a+b \tanh ^{-1}(c x)}{c d^2 (c x+1)}-\frac{b}{2 c d^2 (c x+1)}+\frac{b \tanh ^{-1}(c x)}{2 c d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-b/(2*c*d^2*(1 + c*x)) + (b*ArcTanh[c*x])/(2*c*d^2) - (a + b*ArcTanh[c*x])/(c*d^2*(1 + c*x))

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 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 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])

Rubi steps

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

Mathematica [A]  time = 0.0680515, size = 64, normalized size = 1.12 \[ \frac{-4 a-(b c x+b) \log (1-c x)+b \log (c x+1)+b c x \log (c x+1)-4 b \tanh ^{-1}(c x)-2 b}{4 c d^2 (c x+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a - 2*b - 4*b*ArcTanh[c*x] - (b + b*c*x)*Log[1 - c*x] + b*Log[1 + c*x] + b*c*x*Log[1 + c*x])/(4*c*d^2*(1 +
c*x))

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Maple [A]  time = 0.035, size = 84, normalized size = 1.5 \begin{align*} -{\frac{a}{c{d}^{2} \left ( cx+1 \right ) }}-{\frac{b{\it Artanh} \left ( cx \right ) }{c{d}^{2} \left ( cx+1 \right ) }}-{\frac{b\ln \left ( cx-1 \right ) }{4\,c{d}^{2}}}-{\frac{b}{2\,c{d}^{2} \left ( cx+1 \right ) }}+{\frac{b\ln \left ( cx+1 \right ) }{4\,c{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/c*a/d^2/(c*x+1)-1/c*b/d^2*arctanh(c*x)/(c*x+1)-1/4/c*b/d^2*ln(c*x-1)-1/2*b/c/d^2/(c*x+1)+1/4/c*b/d^2*ln(c*x
+1)

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Maxima [A]  time = 0.952915, size = 130, normalized size = 2.28 \begin{align*} -\frac{1}{4} \,{\left (c{\left (\frac{2}{c^{3} d^{2} x + c^{2} d^{2}} - \frac{\log \left (c x + 1\right )}{c^{2} d^{2}} + \frac{\log \left (c x - 1\right )}{c^{2} d^{2}}\right )} + \frac{4 \, \operatorname{artanh}\left (c x\right )}{c^{2} d^{2} x + c d^{2}}\right )} b - \frac{a}{c^{2} d^{2} x + c d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(c*(2/(c^3*d^2*x + c^2*d^2) - log(c*x + 1)/(c^2*d^2) + log(c*x - 1)/(c^2*d^2)) + 4*arctanh(c*x)/(c^2*d^2*
x + c*d^2))*b - a/(c^2*d^2*x + c*d^2)

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Fricas [A]  time = 2.07068, size = 104, normalized size = 1.82 \begin{align*} \frac{{\left (b c x - b\right )} \log \left (-\frac{c x + 1}{c x - 1}\right ) - 4 \, a - 2 \, b}{4 \,{\left (c^{2} d^{2} x + c d^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((b*c*x - b)*log(-(c*x + 1)/(c*x - 1)) - 4*a - 2*b)/(c^2*d^2*x + c*d^2)

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Sympy [A]  time = 4.21529, size = 121, normalized size = 2.12 \begin{align*} \begin{cases} - \frac{2 a}{2 c^{2} d^{2} x + 2 c d^{2}} + \frac{b c x \operatorname{atanh}{\left (c x \right )}}{2 c^{2} d^{2} x + 2 c d^{2}} - \frac{b \operatorname{atanh}{\left (c x \right )}}{2 c^{2} d^{2} x + 2 c d^{2}} - \frac{b}{2 c^{2} d^{2} x + 2 c d^{2}} & \text{for}\: d \neq 0 \\\tilde{\infty } \left (a x + b x \operatorname{atanh}{\left (c x \right )} + \frac{b \log{\left (x - \frac{1}{c} \right )}}{c} + \frac{b \operatorname{atanh}{\left (c x \right )}}{c}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*a/(2*c**2*d**2*x + 2*c*d**2) + b*c*x*atanh(c*x)/(2*c**2*d**2*x + 2*c*d**2) - b*atanh(c*x)/(2*c**
2*d**2*x + 2*c*d**2) - b/(2*c**2*d**2*x + 2*c*d**2), Ne(d, 0)), (zoo*(a*x + b*x*atanh(c*x) + b*log(x - 1/c)/c
+ b*atanh(c*x)/c), True))

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Giac [A]  time = 1.13747, size = 131, normalized size = 2.3 \begin{align*} -\frac{1}{4} \,{\left (c d^{2}{\left (\frac{\log \left ({\left | -\frac{2 \, d}{c d x + d} + 1 \right |}\right )}{c^{2} d^{4}} + \frac{2}{{\left (c d x + d\right )} c^{2} d^{3}}\right )} + \frac{2 \, \log \left (-\frac{c x + 1}{c x - 1}\right )}{{\left (c d x + d\right )} c d}\right )} b - \frac{a}{{\left (c d x + d\right )} c d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(c*d^2*(log(abs(-2*d/(c*d*x + d) + 1))/(c^2*d^4) + 2/((c*d*x + d)*c^2*d^3)) + 2*log(-(c*x + 1)/(c*x - 1))
/((c*d*x + d)*c*d))*b - a/((c*d*x + d)*c*d)