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]

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

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Rubi [A]  time = 0.06, antiderivative size = 77, normalized size of antiderivative = 1.00, 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 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)}{(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*}

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Mathematica [A]  time = 0.07, 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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fricas [A]  time = 0.41, size = 75, normalized size = 0.97 \[ -\frac {2 \, b c x - {\left (b c^{2} x^{2} + 2 \, b c x - 3 \, b\right )} \log \left (-\frac {c x + 1}{c x - 1}\right ) + 8 \, a + 4 \, b}{16 \, {\left (c^{3} d^{3} x^{2} + 2 \, c^{2} d^{3} x + c d^{3}\right )}} \]

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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giac [A]  time = 0.22, size = 118, normalized size = 1.53 \[ \frac {1}{32} \, c {\left (\frac {2 \, {\left (c x - 1\right )}^{2} {\left (\frac {2 \, {\left (c x + 1\right )} b}{c x - 1} - b\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{{\left (c x + 1\right )}^{2} c^{2} d^{3}} + \frac {{\left (c x - 1\right )}^{2} {\left (\frac {8 \, {\left (c x + 1\right )} a}{c x - 1} - 4 \, a + \frac {4 \, {\left (c x + 1\right )} b}{c x - 1} - b\right )}}{{\left (c x + 1\right )}^{2} c^{2} d^{3}}\right )} \]

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]

1/32*c*(2*(c*x - 1)^2*(2*(c*x + 1)*b/(c*x - 1) - b)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^2*c^2*d^3) + (c*x - 1
)^2*(8*(c*x + 1)*a/(c*x - 1) - 4*a + 4*(c*x + 1)*b/(c*x - 1) - b)/((c*x + 1)^2*c^2*d^3))

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maple [A]  time = 0.04, size = 100, normalized size = 1.30 \[ -\frac {a}{2 c \,d^{3} \left (c x +1\right )^{2}}-\frac {b \arctanh \left (c x \right )}{2 c \,d^{3} \left (c x +1\right )^{2}}-\frac {b \ln \left (c x -1\right )}{16 c \,d^{3}}-\frac {b}{8 c \,d^{3} \left (c x +1\right )^{2}}-\frac {b}{8 c \,d^{3} \left (c x +1\right )}+\frac {b \ln \left (c x +1\right )}{16 c \,d^{3}} \]

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 = 0.32, size = 134, normalized size = 1.74 \[ -\frac {1}{16} \, {\left (c {\left (\frac {2 \, {\left (c x + 2\right )}}{c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}} - \frac {\log \left (c x + 1\right )}{c^{2} d^{3}} + \frac {\log \left (c x - 1\right )}{c^{2} d^{3}}\right )} + \frac {8 \, \operatorname {artanh}\left (c x\right )}{c^{3} d^{3} x^{2} + 2 \, c^{2} d^{3} x + c d^{3}}\right )} b - \frac {a}{2 \, {\left (c^{3} d^{3} x^{2} + 2 \, c^{2} d^{3} x + c d^{3}\right )}} \]

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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mupad [B]  time = 1.09, size = 123, normalized size = 1.60 \[ \frac {c^2\,\left (\frac {a\,x^2}{2}+\frac {b\,x^2}{4}-\frac {b\,x^2\,\ln \left (c^2\,x^2-1\right )}{16}+\frac {b\,x^2\,\ln \left (c\,x+1\right )}{8}\right )-\frac {b\,\ln \left (c^2\,x^2-1\right )}{16}-\frac {b\,\mathrm {atanh}\left (c\,x\right )}{2}+\frac {b\,\ln \left (c\,x+1\right )}{8}+c\,\left (a\,x+\frac {3\,b\,x}{8}+\frac {b\,x\,\ln \left (c\,x+1\right )}{4}-\frac {b\,x\,\ln \left (c^2\,x^2-1\right )}{8}\right )}{c\,d^3\,{\left (c\,x+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.15, size = 250, normalized size = 3.25 \[ \begin {cases} - \frac {4 a}{8 c^{3} d^{3} x^{2} + 16 c^{2} d^{3} x + 8 c d^{3}} + \frac {b c^{2} x^{2} \operatorname {atanh}{\left (c x \right )}}{8 c^{3} d^{3} x^{2} + 16 c^{2} d^{3} x + 8 c d^{3}} + \frac {2 b c x \operatorname {atanh}{\left (c x \right )}}{8 c^{3} d^{3} x^{2} + 16 c^{2} d^{3} x + 8 c d^{3}} - \frac {b c x}{8 c^{3} d^{3} x^{2} + 16 c^{2} d^{3} x + 8 c d^{3}} - \frac {3 b \operatorname {atanh}{\left (c x \right )}}{8 c^{3} d^{3} x^{2} + 16 c^{2} d^{3} x + 8 c d^{3}} - \frac {2 b}{8 c^{3} d^{3} x^{2} + 16 c^{2} d^{3} x + 8 c d^{3}} & \text {for}\: d \neq 0 \\\tilde {\infty } \left (a x + b x \operatorname {atanh}{\left (c x \right )} + \frac {b \log {\left (x - \frac {1}{c} \right )}}{c} + \frac {b \operatorname {atanh}{\left (c x \right )}}{c}\right ) & \text {otherwise} \end {cases} \]

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((-4*a/(8*c**3*d**3*x**2 + 16*c**2*d**3*x + 8*c*d**3) + b*c**2*x**2*atanh(c*x)/(8*c**3*d**3*x**2 + 16
*c**2*d**3*x + 8*c*d**3) + 2*b*c*x*atanh(c*x)/(8*c**3*d**3*x**2 + 16*c**2*d**3*x + 8*c*d**3) - b*c*x/(8*c**3*d
**3*x**2 + 16*c**2*d**3*x + 8*c*d**3) - 3*b*atanh(c*x)/(8*c**3*d**3*x**2 + 16*c**2*d**3*x + 8*c*d**3) - 2*b/(8
*c**3*d**3*x**2 + 16*c**2*d**3*x + 8*c*d**3), Ne(d, 0)), (zoo*(a*x + b*x*atanh(c*x) + b*log(x - 1/c)/c + b*ata
nh(c*x)/c), True))

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