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]

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

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Rubi [A]  time = 0.05, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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 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]

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*}

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Mathematica [A]  time = 0.11, 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]

-1/144*(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])/(c*(1 + c*x)^3)

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fricas [A]  time = 0.43, size = 91, normalized size = 1.14 \[ -\frac {6 \, b c^{2} x^{2} + 18 \, b c x - 3 \, {\left (b c^{3} x^{3} + 3 \, b c^{2} x^{2} + 3 \, b c x - 7 \, b\right )} \log \left (-\frac {c x + 1}{c x - 1}\right ) + 48 \, a + 20 \, b}{144 \, {\left (c^{4} x^{3} + 3 \, c^{3} x^{2} + 3 \, c^{2} x + c\right )}} \]

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)

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giac [B]  time = 0.20, size = 161, normalized size = 2.01 \[ \frac {1}{288} \, c {\left (\frac {6 \, {\left (c x - 1\right )}^{3} {\left (\frac {3 \, {\left (c x + 1\right )}^{2} b}{{\left (c x - 1\right )}^{2}} - \frac {3 \, {\left (c x + 1\right )} b}{c x - 1} + b\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{{\left (c x + 1\right )}^{3} c^{2}} + \frac {{\left (c x - 1\right )}^{3} {\left (\frac {36 \, {\left (c x + 1\right )}^{2} a}{{\left (c x - 1\right )}^{2}} - \frac {36 \, {\left (c x + 1\right )} a}{c x - 1} + 12 \, a + \frac {18 \, {\left (c x + 1\right )}^{2} b}{{\left (c x - 1\right )}^{2}} - \frac {9 \, {\left (c x + 1\right )} b}{c x - 1} + 2 \, b\right )}}{{\left (c x + 1\right )}^{3} c^{2}}\right )} \]

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]

1/288*c*(6*(c*x - 1)^3*(3*(c*x + 1)^2*b/(c*x - 1)^2 - 3*(c*x + 1)*b/(c*x - 1) + b)*log(-(c*x + 1)/(c*x - 1))/(
(c*x + 1)^3*c^2) + (c*x - 1)^3*(36*(c*x + 1)^2*a/(c*x - 1)^2 - 36*(c*x + 1)*a/(c*x - 1) + 12*a + 18*(c*x + 1)^
2*b/(c*x - 1)^2 - 9*(c*x + 1)*b/(c*x - 1) + 2*b)/((c*x + 1)^3*c^2))

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maple [A]  time = 0.04, size = 95, normalized size = 1.19 \[ -\frac {a}{3 c \left (c x +1\right )^{3}}-\frac {b \arctanh \left (c x \right )}{3 c \left (c x +1\right )^{3}}-\frac {b \ln \left (c x -1\right )}{48 c}-\frac {b}{18 c \left (c x +1\right )^{3}}-\frac {b}{24 c \left (c x +1\right )^{2}}-\frac {b}{24 c \left (c x +1\right )}+\frac {b \ln \left (c x +1\right )}{48 c} \]

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)

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maxima [A]  time = 0.31, size = 132, normalized size = 1.65 \[ -\frac {1}{144} \, {\left (c {\left (\frac {2 \, {\left (3 \, c^{2} x^{2} + 9 \, c x + 10\right )}}{c^{5} x^{3} + 3 \, c^{4} x^{2} + 3 \, c^{3} x + c^{2}} - \frac {3 \, \log \left (c x + 1\right )}{c^{2}} + \frac {3 \, \log \left (c x - 1\right )}{c^{2}}\right )} + \frac {48 \, \operatorname {artanh}\left (c x\right )}{c^{4} x^{3} + 3 \, c^{3} x^{2} + 3 \, c^{2} x + c}\right )} b - \frac {a}{3 \, {\left (c^{4} x^{3} + 3 \, c^{3} x^{2} + 3 \, c^{2} x + c\right )}} \]

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)

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mupad [B]  time = 1.10, size = 139, normalized size = 1.74 \[ \frac {\frac {b\,c^2\,x^3}{8}-\frac {b\,x}{8}-\frac {b\,\mathrm {atanh}\left (c\,x\right )}{3\,c}-\frac {12\,a+5\,b}{36\,c}+\frac {b\,c^3\,x^4}{24}+\frac {c\,x^2\,\left (24\,a+7\,b\right )}{72}+\frac {b\,c\,x^2\,\mathrm {atanh}\left (c\,x\right )}{3}}{-c^5\,x^5-3\,c^4\,x^4-2\,c^3\,x^3+2\,c^2\,x^2+3\,c\,x+1}-\frac {b\,\ln \left (c^2\,x^2-1\right )}{48\,c}+\frac {b\,\ln \left (c\,x+1\right )}{24\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))/(c*x + 1)^4,x)

[Out]

((b*c^2*x^3)/8 - (b*x)/8 - (b*atanh(c*x))/(3*c) - (12*a + 5*b)/(36*c) + (b*c^3*x^4)/24 + (c*x^2*(24*a + 7*b))/
72 + (b*c*x^2*atanh(c*x))/3)/(3*c*x + 2*c^2*x^2 - 2*c^3*x^3 - 3*c^4*x^4 - c^5*x^5 + 1) - (b*log(c^2*x^2 - 1))/
(48*c) + (b*log(c*x + 1))/(24*c)

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sympy [A]  time = 2.76, size = 294, normalized size = 3.68 \[ \begin {cases} - \frac {24 a}{72 c^{4} x^{3} + 216 c^{3} x^{2} + 216 c^{2} x + 72 c} + \frac {3 b c^{3} x^{3} \operatorname {atanh}{\left (c x \right )}}{72 c^{4} x^{3} + 216 c^{3} x^{2} + 216 c^{2} x + 72 c} + \frac {9 b c^{2} x^{2} \operatorname {atanh}{\left (c x \right )}}{72 c^{4} x^{3} + 216 c^{3} x^{2} + 216 c^{2} x + 72 c} - \frac {3 b c^{2} x^{2}}{72 c^{4} x^{3} + 216 c^{3} x^{2} + 216 c^{2} x + 72 c} + \frac {9 b c x \operatorname {atanh}{\left (c x \right )}}{72 c^{4} x^{3} + 216 c^{3} x^{2} + 216 c^{2} x + 72 c} - \frac {9 b c x}{72 c^{4} x^{3} + 216 c^{3} x^{2} + 216 c^{2} x + 72 c} - \frac {21 b \operatorname {atanh}{\left (c x \right )}}{72 c^{4} x^{3} + 216 c^{3} x^{2} + 216 c^{2} x + 72 c} - \frac {10 b}{72 c^{4} x^{3} + 216 c^{3} x^{2} + 216 c^{2} x + 72 c} & \text {for}\: c \neq 0 \\a x & \text {otherwise} \end {cases} \]

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))

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