3.66 \(\int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^4} \, dx\)

Optimal. Leaf size=80 \[ -\frac{a+b \tanh ^{-1}(c x)}{3 c (c x+1)^3}-\frac{b}{24 c (c x+1)}-\frac{b}{24 c (c x+1)^2}-\frac{b}{18 c (c x+1)^3}+\frac{b \tanh ^{-1}(c x)}{24 c} \]

[Out]

-b/(18*c*(1 + c*x)^3) - b/(24*c*(1 + c*x)^2) - b/(24*c*(1 + c*x)) + (b*ArcTanh[c*x])/(24*c) - (a + b*ArcTanh[c
*x])/(3*c*(1 + c*x)^3)

________________________________________________________________________________________

Rubi [A]  time = 0.053593, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5926, 627, 44, 207} \[ -\frac{a+b \tanh ^{-1}(c x)}{3 c (c x+1)^3}-\frac{b}{24 c (c x+1)}-\frac{b}{24 c (c x+1)^2}-\frac{b}{18 c (c x+1)^3}+\frac{b \tanh ^{-1}(c x)}{24 c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])/(1 + c*x)^4,x]

[Out]

-b/(18*c*(1 + c*x)^3) - b/(24*c*(1 + c*x)^2) - b/(24*c*(1 + c*x)) + (b*ArcTanh[c*x])/(24*c) - (a + b*ArcTanh[c
*x])/(3*c*(1 + c*x)^3)

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)}{(1+c x)^4} \, dx &=-\frac{a+b \tanh ^{-1}(c x)}{3 c (1+c x)^3}+\frac{1}{3} b \int \frac{1}{(1+c x)^3 \left (1-c^2 x^2\right )} \, dx\\ &=-\frac{a+b \tanh ^{-1}(c x)}{3 c (1+c x)^3}+\frac{1}{3} b \int \frac{1}{(1-c x) (1+c x)^4} \, dx\\ &=-\frac{a+b \tanh ^{-1}(c x)}{3 c (1+c x)^3}+\frac{1}{3} b \int \left (\frac{1}{2 (1+c x)^4}+\frac{1}{4 (1+c x)^3}+\frac{1}{8 (1+c x)^2}-\frac{1}{8 \left (-1+c^2 x^2\right )}\right ) \, dx\\ &=-\frac{b}{18 c (1+c x)^3}-\frac{b}{24 c (1+c x)^2}-\frac{b}{24 c (1+c x)}-\frac{a+b \tanh ^{-1}(c x)}{3 c (1+c x)^3}-\frac{1}{24} b \int \frac{1}{-1+c^2 x^2} \, dx\\ &=-\frac{b}{18 c (1+c x)^3}-\frac{b}{24 c (1+c x)^2}-\frac{b}{24 c (1+c x)}+\frac{b \tanh ^{-1}(c x)}{24 c}-\frac{a+b \tanh ^{-1}(c x)}{3 c (1+c x)^3}\\ \end{align*}

Mathematica [A]  time = 0.105182, size = 75, normalized size = 0.94 \[ -\frac{48 a+2 b \left (3 c^2 x^2+9 c x+10\right )+3 b (c x+1)^3 \log (1-c x)-3 b (c x+1)^3 \log (c x+1)+48 b \tanh ^{-1}(c x)}{144 c (c x+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x])/(1 + c*x)^4,x]

[Out]

-(48*a + 2*b*(10 + 9*c*x + 3*c^2*x^2) + 48*b*ArcTanh[c*x] + 3*b*(1 + c*x)^3*Log[1 - c*x] - 3*b*(1 + c*x)^3*Log
[1 + c*x])/(144*c*(1 + c*x)^3)

________________________________________________________________________________________

Maple [A]  time = 0.034, size = 95, normalized size = 1.2 \begin{align*} -{\frac{a}{3\,c \left ( cx+1 \right ) ^{3}}}-{\frac{b{\it Artanh} \left ( cx \right ) }{3\,c \left ( cx+1 \right ) ^{3}}}-{\frac{b\ln \left ( cx-1 \right ) }{48\,c}}-{\frac{b}{18\,c \left ( cx+1 \right ) ^{3}}}-{\frac{b}{24\,c \left ( cx+1 \right ) ^{2}}}-{\frac{b}{24\,c \left ( cx+1 \right ) }}+{\frac{b\ln \left ( cx+1 \right ) }{48\,c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))/(c*x+1)^4,x)

[Out]

-1/3/c*a/(c*x+1)^3-1/3/c*b/(c*x+1)^3*arctanh(c*x)-1/48/c*b*ln(c*x-1)-1/18*b/c/(c*x+1)^3-1/24*b/c/(c*x+1)^2-1/2
4*b/c/(c*x+1)+1/48/c*b*ln(c*x+1)

________________________________________________________________________________________

Maxima [A]  time = 0.973127, size = 178, normalized size = 2.22 \begin{align*} -\frac{1}{144} \,{\left (c{\left (\frac{2 \,{\left (3 \, c^{2} x^{2} + 9 \, c x + 10\right )}}{c^{5} x^{3} + 3 \, c^{4} x^{2} + 3 \, c^{3} x + c^{2}} - \frac{3 \, \log \left (c x + 1\right )}{c^{2}} + \frac{3 \, \log \left (c x - 1\right )}{c^{2}}\right )} + \frac{48 \, \operatorname{artanh}\left (c x\right )}{c^{4} x^{3} + 3 \, c^{3} x^{2} + 3 \, c^{2} x + c}\right )} b - \frac{a}{3 \,{\left (c^{4} x^{3} + 3 \, c^{3} x^{2} + 3 \, c^{2} x + c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/(c*x+1)^4,x, algorithm="maxima")

[Out]

-1/144*(c*(2*(3*c^2*x^2 + 9*c*x + 10)/(c^5*x^3 + 3*c^4*x^2 + 3*c^3*x + c^2) - 3*log(c*x + 1)/c^2 + 3*log(c*x -
1)/c^2) + 48*arctanh(c*x)/(c^4*x^3 + 3*c^3*x^2 + 3*c^2*x + c))*b - 1/3*a/(c^4*x^3 + 3*c^3*x^2 + 3*c^2*x + c)

________________________________________________________________________________________

Fricas [A]  time = 2.37688, size = 209, normalized size = 2.61 \begin{align*} -\frac{6 \, b c^{2} x^{2} + 18 \, b c x - 3 \,{\left (b c^{3} x^{3} + 3 \, b c^{2} x^{2} + 3 \, b c x - 7 \, b\right )} \log \left (-\frac{c x + 1}{c x - 1}\right ) + 48 \, a + 20 \, b}{144 \,{\left (c^{4} x^{3} + 3 \, c^{3} x^{2} + 3 \, c^{2} x + c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/(c*x+1)^4,x, algorithm="fricas")

[Out]

-1/144*(6*b*c^2*x^2 + 18*b*c*x - 3*(b*c^3*x^3 + 3*b*c^2*x^2 + 3*b*c*x - 7*b)*log(-(c*x + 1)/(c*x - 1)) + 48*a
+ 20*b)/(c^4*x^3 + 3*c^3*x^2 + 3*c^2*x + c)

________________________________________________________________________________________

Sympy [A]  time = 8.14961, size = 294, normalized size = 3.68 \begin{align*} \begin{cases} - \frac{24 a}{72 c^{4} x^{3} + 216 c^{3} x^{2} + 216 c^{2} x + 72 c} + \frac{3 b c^{3} x^{3} \operatorname{atanh}{\left (c x \right )}}{72 c^{4} x^{3} + 216 c^{3} x^{2} + 216 c^{2} x + 72 c} + \frac{9 b c^{2} x^{2} \operatorname{atanh}{\left (c x \right )}}{72 c^{4} x^{3} + 216 c^{3} x^{2} + 216 c^{2} x + 72 c} - \frac{3 b c^{2} x^{2}}{72 c^{4} x^{3} + 216 c^{3} x^{2} + 216 c^{2} x + 72 c} + \frac{9 b c x \operatorname{atanh}{\left (c x \right )}}{72 c^{4} x^{3} + 216 c^{3} x^{2} + 216 c^{2} x + 72 c} - \frac{9 b c x}{72 c^{4} x^{3} + 216 c^{3} x^{2} + 216 c^{2} x + 72 c} - \frac{21 b \operatorname{atanh}{\left (c x \right )}}{72 c^{4} x^{3} + 216 c^{3} x^{2} + 216 c^{2} x + 72 c} - \frac{10 b}{72 c^{4} x^{3} + 216 c^{3} x^{2} + 216 c^{2} x + 72 c} & \text{for}\: c \neq 0 \\a x & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))/(c*x+1)**4,x)

[Out]

Piecewise((-24*a/(72*c**4*x**3 + 216*c**3*x**2 + 216*c**2*x + 72*c) + 3*b*c**3*x**3*atanh(c*x)/(72*c**4*x**3 +
216*c**3*x**2 + 216*c**2*x + 72*c) + 9*b*c**2*x**2*atanh(c*x)/(72*c**4*x**3 + 216*c**3*x**2 + 216*c**2*x + 72
*c) - 3*b*c**2*x**2/(72*c**4*x**3 + 216*c**3*x**2 + 216*c**2*x + 72*c) + 9*b*c*x*atanh(c*x)/(72*c**4*x**3 + 21
6*c**3*x**2 + 216*c**2*x + 72*c) - 9*b*c*x/(72*c**4*x**3 + 216*c**3*x**2 + 216*c**2*x + 72*c) - 21*b*atanh(c*x
)/(72*c**4*x**3 + 216*c**3*x**2 + 216*c**2*x + 72*c) - 10*b/(72*c**4*x**3 + 216*c**3*x**2 + 216*c**2*x + 72*c)
, Ne(c, 0)), (a*x, True))

________________________________________________________________________________________

Giac [A]  time = 1.15987, size = 157, normalized size = 1.96 \begin{align*} \frac{b \log \left (c x + 1\right )}{48 \, c} - \frac{b \log \left (c x - 1\right )}{48 \, c} - \frac{b \log \left (-\frac{c x + 1}{c x - 1}\right )}{6 \,{\left (c^{4} x^{3} + 3 \, c^{3} x^{2} + 3 \, c^{2} x + c\right )}} - \frac{3 \, b c^{2} x^{2} + 9 \, b c x + 24 \, a + 10 \, b}{72 \,{\left (c^{4} x^{3} + 3 \, c^{3} x^{2} + 3 \, c^{2} x + c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/(c*x+1)^4,x, algorithm="giac")

[Out]

1/48*b*log(c*x + 1)/c - 1/48*b*log(c*x - 1)/c - 1/6*b*log(-(c*x + 1)/(c*x - 1))/(c^4*x^3 + 3*c^3*x^2 + 3*c^2*x
+ c) - 1/72*(3*b*c^2*x^2 + 9*b*c*x + 24*a + 10*b)/(c^4*x^3 + 3*c^3*x^2 + 3*c^2*x + c)