3.62 \(\int \frac{a+b \tanh ^{-1}(c x)}{(d+c d x)^3} \, dx\)

Optimal. Leaf size=77 \[ -\frac{a+b \tanh ^{-1}(c x)}{2 c d^3 (c x+1)^2}-\frac{b}{8 c d^3 (c x+1)}-\frac{b}{8 c d^3 (c x+1)^2}+\frac{b \tanh ^{-1}(c x)}{8 c d^3} \]

[Out]

-b/(8*c*d^3*(1 + c*x)^2) - b/(8*c*d^3*(1 + c*x)) + (b*ArcTanh[c*x])/(8*c*d^3) - (a + b*ArcTanh[c*x])/(2*c*d^3*
(1 + c*x)^2)

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Rubi [A]  time = 0.0550454, antiderivative size = 77, 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} \[ -\frac{a+b \tanh ^{-1}(c x)}{2 c d^3 (c x+1)^2}-\frac{b}{8 c d^3 (c x+1)}-\frac{b}{8 c d^3 (c x+1)^2}+\frac{b \tanh ^{-1}(c x)}{8 c d^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])/(d + c*d*x)^3,x]

[Out]

-b/(8*c*d^3*(1 + c*x)^2) - b/(8*c*d^3*(1 + c*x)) + (b*ArcTanh[c*x])/(8*c*d^3) - (a + b*ArcTanh[c*x])/(2*c*d^3*
(1 + c*x)^2)

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)^3} \, dx &=-\frac{a+b \tanh ^{-1}(c x)}{2 c d^3 (1+c x)^2}+\frac{b \int \frac{1}{(d+c d x)^2 \left (1-c^2 x^2\right )} \, dx}{2 d}\\ &=-\frac{a+b \tanh ^{-1}(c x)}{2 c d^3 (1+c x)^2}+\frac{b \int \frac{1}{\left (\frac{1}{d}-\frac{c x}{d}\right ) (d+c d x)^3} \, dx}{2 d}\\ &=-\frac{a+b \tanh ^{-1}(c x)}{2 c d^3 (1+c x)^2}+\frac{b \int \left (\frac{1}{2 d^2 (1+c x)^3}+\frac{1}{4 d^2 (1+c x)^2}-\frac{1}{4 d^2 \left (-1+c^2 x^2\right )}\right ) \, dx}{2 d}\\ &=-\frac{b}{8 c d^3 (1+c x)^2}-\frac{b}{8 c d^3 (1+c x)}-\frac{a+b \tanh ^{-1}(c x)}{2 c d^3 (1+c x)^2}-\frac{b \int \frac{1}{-1+c^2 x^2} \, dx}{8 d^3}\\ &=-\frac{b}{8 c d^3 (1+c x)^2}-\frac{b}{8 c d^3 (1+c x)}+\frac{b \tanh ^{-1}(c x)}{8 c d^3}-\frac{a+b \tanh ^{-1}(c x)}{2 c d^3 (1+c x)^2}\\ \end{align*}

Mathematica [A]  time = 0.0730421, size = 86, normalized size = 1.12 \[ \frac{-8 a+b c^2 x^2 \log (c x+1)-2 b c x+2 b c x \log (c x+1)-b (c x+1)^2 \log (1-c x)+b \log (c x+1)-8 b \tanh ^{-1}(c x)-4 b}{16 c d^3 (c x+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x])/(d + c*d*x)^3,x]

[Out]

(-8*a - 4*b - 2*b*c*x - 8*b*ArcTanh[c*x] - b*(1 + c*x)^2*Log[1 - c*x] + b*Log[1 + c*x] + 2*b*c*x*Log[1 + c*x]
+ b*c^2*x^2*Log[1 + c*x])/(16*c*d^3*(1 + c*x)^2)

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Maple [A]  time = 0.033, size = 100, normalized size = 1.3 \begin{align*} -{\frac{a}{2\,c{d}^{3} \left ( cx+1 \right ) ^{2}}}-{\frac{b{\it Artanh} \left ( cx \right ) }{2\,c{d}^{3} \left ( cx+1 \right ) ^{2}}}-{\frac{b\ln \left ( cx-1 \right ) }{16\,c{d}^{3}}}-{\frac{b}{8\,c{d}^{3} \left ( cx+1 \right ) ^{2}}}-{\frac{b}{8\,c{d}^{3} \left ( cx+1 \right ) }}+{\frac{b\ln \left ( cx+1 \right ) }{16\,c{d}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/c*a/d^3/(c*x+1)^2-1/2/c*b/d^3*arctanh(c*x)/(c*x+1)^2-1/16/c*b/d^3*ln(c*x-1)-1/8*b/c/d^3/(c*x+1)^2-1/8*b/c
/d^3/(c*x+1)+1/16/c*b/d^3*ln(c*x+1)

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Maxima [A]  time = 0.975595, size = 181, normalized size = 2.35 \begin{align*} -\frac{1}{16} \,{\left (c{\left (\frac{2 \,{\left (c x + 2\right )}}{c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}} - \frac{\log \left (c x + 1\right )}{c^{2} d^{3}} + \frac{\log \left (c x - 1\right )}{c^{2} d^{3}}\right )} + \frac{8 \, \operatorname{artanh}\left (c x\right )}{c^{3} d^{3} x^{2} + 2 \, c^{2} d^{3} x + c d^{3}}\right )} b - \frac{a}{2 \,{\left (c^{3} d^{3} x^{2} + 2 \, c^{2} d^{3} x + c d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/(c*d*x+d)^3,x, algorithm="maxima")

[Out]

-1/16*(c*(2*(c*x + 2)/(c^4*d^3*x^2 + 2*c^3*d^3*x + c^2*d^3) - log(c*x + 1)/(c^2*d^3) + log(c*x - 1)/(c^2*d^3))
+ 8*arctanh(c*x)/(c^3*d^3*x^2 + 2*c^2*d^3*x + c*d^3))*b - 1/2*a/(c^3*d^3*x^2 + 2*c^2*d^3*x + c*d^3)

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Fricas [A]  time = 2.11686, size = 163, normalized size = 2.12 \begin{align*} -\frac{2 \, b c x -{\left (b c^{2} x^{2} + 2 \, b c x - 3 \, b\right )} \log \left (-\frac{c x + 1}{c x - 1}\right ) + 8 \, a + 4 \, b}{16 \,{\left (c^{3} d^{3} x^{2} + 2 \, c^{2} d^{3} x + c d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/(c*d*x+d)^3,x, algorithm="fricas")

[Out]

-1/16*(2*b*c*x - (b*c^2*x^2 + 2*b*c*x - 3*b)*log(-(c*x + 1)/(c*x - 1)) + 8*a + 4*b)/(c^3*d^3*x^2 + 2*c^2*d^3*x
+ c*d^3)

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Sympy [A]  time = 4.16484, size = 289, normalized size = 3.75 \begin{align*} \begin{cases} - \frac{12 a}{24 c^{3} d^{3} x^{2} + 48 c^{2} d^{3} x + 24 c d^{3}} + \frac{3 b c^{2} x^{2} \operatorname{atanh}{\left (c x \right )}}{24 c^{3} d^{3} x^{2} + 48 c^{2} d^{3} x + 24 c d^{3}} + \frac{b c^{2} x^{2}}{24 c^{3} d^{3} x^{2} + 48 c^{2} d^{3} x + 24 c d^{3}} + \frac{6 b c x \operatorname{atanh}{\left (c x \right )}}{24 c^{3} d^{3} x^{2} + 48 c^{2} d^{3} x + 24 c d^{3}} - \frac{b c x}{24 c^{3} d^{3} x^{2} + 48 c^{2} d^{3} x + 24 c d^{3}} - \frac{9 b \operatorname{atanh}{\left (c x \right )}}{24 c^{3} d^{3} x^{2} + 48 c^{2} d^{3} x + 24 c d^{3}} - \frac{5 b}{24 c^{3} d^{3} x^{2} + 48 c^{2} d^{3} x + 24 c d^{3}} & \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)**3,x)

[Out]

Piecewise((-12*a/(24*c**3*d**3*x**2 + 48*c**2*d**3*x + 24*c*d**3) + 3*b*c**2*x**2*atanh(c*x)/(24*c**3*d**3*x**
2 + 48*c**2*d**3*x + 24*c*d**3) + b*c**2*x**2/(24*c**3*d**3*x**2 + 48*c**2*d**3*x + 24*c*d**3) + 6*b*c*x*atanh
(c*x)/(24*c**3*d**3*x**2 + 48*c**2*d**3*x + 24*c*d**3) - b*c*x/(24*c**3*d**3*x**2 + 48*c**2*d**3*x + 24*c*d**3
) - 9*b*atanh(c*x)/(24*c**3*d**3*x**2 + 48*c**2*d**3*x + 24*c*d**3) - 5*b/(24*c**3*d**3*x**2 + 48*c**2*d**3*x
+ 24*c*d**3), 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 = 1.24967, size = 157, normalized size = 2.04 \begin{align*} -\frac{b \log \left (-\frac{c x + 1}{c x - 1}\right )}{4 \,{\left (c^{3} d^{3} x^{2} + 2 \, c^{2} d^{3} x + c d^{3}\right )}} - \frac{b c x + 4 \, a + 2 \, b}{8 \,{\left (c^{3} d^{3} x^{2} + 2 \, c^{2} d^{3} x + c d^{3}\right )}} + \frac{b \log \left (c x + 1\right )}{16 \, c d^{3}} - \frac{b \log \left (c x - 1\right )}{16 \, c d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/(c*d*x+d)^3,x, algorithm="giac")

[Out]

-1/4*b*log(-(c*x + 1)/(c*x - 1))/(c^3*d^3*x^2 + 2*c^2*d^3*x + c*d^3) - 1/8*(b*c*x + 4*a + 2*b)/(c^3*d^3*x^2 +
2*c^2*d^3*x + c*d^3) + 1/16*b*log(c*x + 1)/(c*d^3) - 1/16*b*log(c*x - 1)/(c*d^3)