3.186 \(\int \frac {\coth (c+d x)}{(a+b \tanh ^2(c+d x))^2} \, dx\)

Optimal. Leaf size=95 \[ -\frac {b (2 a+b) \log \left (a+b \tanh ^2(c+d x)\right )}{2 a^2 d (a+b)^2}+\frac {\log (\tanh (c+d x))}{a^2 d}+\frac {b}{2 a d (a+b) \left (a+b \tanh ^2(c+d x)\right )}+\frac {\log (\cosh (c+d x))}{d (a+b)^2} \]

[Out]

ln(cosh(d*x+c))/(a+b)^2/d+ln(tanh(d*x+c))/a^2/d-1/2*b*(2*a+b)*ln(a+b*tanh(d*x+c)^2)/a^2/(a+b)^2/d+1/2*b/a/(a+b
)/d/(a+b*tanh(d*x+c)^2)

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Rubi [A]  time = 0.15, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3670, 446, 72} \[ -\frac {b (2 a+b) \log \left (a+b \tanh ^2(c+d x)\right )}{2 a^2 d (a+b)^2}+\frac {\log (\tanh (c+d x))}{a^2 d}+\frac {b}{2 a d (a+b) \left (a+b \tanh ^2(c+d x)\right )}+\frac {\log (\cosh (c+d x))}{d (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[Cosh[c + d*x]]/((a + b)^2*d) + Log[Tanh[c + d*x]]/(a^2*d) - (b*(2*a + b)*Log[a + b*Tanh[c + d*x]^2])/(2*a^
2*(a + b)^2*d) + b/(2*a*(a + b)*d*(a + b*Tanh[c + d*x]^2))

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rubi steps

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

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Mathematica [A]  time = 2.16, size = 83, normalized size = 0.87 \[ \frac {\frac {\frac {b \left (\frac {a (a+b)}{a+b \tanh ^2(c+d x)}-(2 a+b) \log \left (a+b \tanh ^2(c+d x)\right )\right )}{(a+b)^2}+2 \log (\tanh (c+d x))}{a^2}+\frac {2 \log (\cosh (c+d x))}{(a+b)^2}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*Log[Cosh[c + d*x]])/(a + b)^2 + (2*Log[Tanh[c + d*x]] + (b*(-((2*a + b)*Log[a + b*Tanh[c + d*x]^2]) + (a*(
a + b))/(a + b*Tanh[c + d*x]^2)))/(a + b)^2)/a^2)/(2*d)

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fricas [B]  time = 0.56, size = 1148, normalized size = 12.08 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(a^3 + a^2*b)*d*x*cosh(d*x + c)^4 + 8*(a^3 + a^2*b)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 2*(a^3 + a^2*b
)*d*x*sinh(d*x + c)^4 + 2*(a^3 + a^2*b)*d*x - 4*(a*b^2 - (a^3 - a^2*b)*d*x)*cosh(d*x + c)^2 + 4*(3*(a^3 + a^2*
b)*d*x*cosh(d*x + c)^2 - a*b^2 + (a^3 - a^2*b)*d*x)*sinh(d*x + c)^2 + ((2*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)
^4 + 4*(2*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)*sinh(d*x + c)^3 + (2*a^2*b + 3*a*b^2 + b^3)*sinh(d*x + c)^4 + 2
*a^2*b + 3*a*b^2 + b^3 + 2*(2*a^2*b - a*b^2 - b^3)*cosh(d*x + c)^2 + 2*(2*a^2*b - a*b^2 - b^3 + 3*(2*a^2*b + 3
*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*((2*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^3 + (2*a^2*b - a*b
^2 - b^3)*cosh(d*x + c))*sinh(d*x + c))*log(2*((a + b)*cosh(d*x + c)^2 + (a + b)*sinh(d*x + c)^2 + a - b)/(cos
h(d*x + c)^2 - 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2)) - 2*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x
+ c)^4 + 4*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*si
nh(d*x + c)^4 + a^3 + 3*a^2*b + 3*a*b^2 + b^3 + 2*(a^3 + a^2*b - a*b^2 - b^3)*cosh(d*x + c)^2 + 2*(a^3 + a^2*b
- a*b^2 - b^3 + 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*((a^3 + 3*a^2*b + 3*a*
b^2 + b^3)*cosh(d*x + c)^3 + (a^3 + a^2*b - a*b^2 - b^3)*cosh(d*x + c))*sinh(d*x + c))*log(2*sinh(d*x + c)/(co
sh(d*x + c) - sinh(d*x + c))) + 8*((a^3 + a^2*b)*d*x*cosh(d*x + c)^3 - (a*b^2 - (a^3 - a^2*b)*d*x)*cosh(d*x +
c))*sinh(d*x + c))/((a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*d*cosh(d*x + c)^4 + 4*(a^5 + 3*a^4*b + 3*a^3*b^2 + a
^2*b^3)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*d*sinh(d*x + c)^4 + 2*(a^5 + a
^4*b - a^3*b^2 - a^2*b^3)*d*cosh(d*x + c)^2 + 2*(3*(a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*d*cosh(d*x + c)^2 + (
a^5 + a^4*b - a^3*b^2 - a^2*b^3)*d)*sinh(d*x + c)^2 + (a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*d + 4*((a^5 + 3*a^
4*b + 3*a^3*b^2 + a^2*b^3)*d*cosh(d*x + c)^3 + (a^5 + a^4*b - a^3*b^2 - a^2*b^3)*d*cosh(d*x + c))*sinh(d*x + c
))

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giac [B]  time = 0.31, size = 195, normalized size = 2.05 \[ -\frac {\frac {{\left (2 \, a b + b^{2}\right )} \log \left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}{a^{4} + 2 \, a^{3} b + a^{2} b^{2}} + \frac {2 \, {\left (d x + c\right )}}{a^{2} + 2 \, a b + b^{2}} - \frac {4 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )}}{{\left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )} {\left (a + b\right )}^{2} a} - \frac {2 \, \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right )}{a^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.46, size = 325, normalized size = 3.42 \[ -\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \left (a +b \right )^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \left (a +b \right )^{2}}-\frac {2 b^{2} \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a +b \right )^{2} a \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +4 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a \right )}-\frac {2 b^{3} \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a +b \right )^{2} a^{2} \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +4 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a \right )}-\frac {b \ln \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +4 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a \right )}{d \left (a +b \right )^{2} a}-\frac {b^{2} \ln \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +4 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a \right )}{2 d \left (a +b \right )^{2} a^{2}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d/(a+b)^2*ln(tanh(1/2*d*x+1/2*c)-1)-1/d/(a+b)^2*ln(tanh(1/2*d*x+1/2*c)+1)-2/d*b^2/(a+b)^2/a*tanh(1/2*d*x+1/
2*c)^2/(tanh(1/2*d*x+1/2*c)^4*a+2*tanh(1/2*d*x+1/2*c)^2*a+4*tanh(1/2*d*x+1/2*c)^2*b+a)-2/d*b^3/(a+b)^2/a^2*tan
h(1/2*d*x+1/2*c)^2/(tanh(1/2*d*x+1/2*c)^4*a+2*tanh(1/2*d*x+1/2*c)^2*a+4*tanh(1/2*d*x+1/2*c)^2*b+a)-1/d*b/(a+b)
^2/a*ln(tanh(1/2*d*x+1/2*c)^4*a+2*tanh(1/2*d*x+1/2*c)^2*a+4*tanh(1/2*d*x+1/2*c)^2*b+a)-1/2/d*b^2/(a+b)^2/a^2*l
n(tanh(1/2*d*x+1/2*c)^4*a+2*tanh(1/2*d*x+1/2*c)^2*a+4*tanh(1/2*d*x+1/2*c)^2*b+a)+1/d/a^2*ln(tanh(1/2*d*x+1/2*c
))

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maxima [B]  time = 0.33, size = 235, normalized size = 2.47 \[ \frac {2 \, b^{2} e^{\left (-2 \, d x - 2 \, c\right )}}{{\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3} + 2 \, {\left (a^{4} + a^{3} b - a^{2} b^{2} - a b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} - \frac {{\left (2 \, a b + b^{2}\right )} \log \left (2 \, {\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + a + b\right )}{2 \, {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} d} + \frac {d x + c}{{\left (a^{2} + 2 \, a b + b^{2}\right )} d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{a^{2} d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {coth}\left (c+d\,x\right )}{{\left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\coth {\left (c + d x \right )}}{\left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(coth(c + d*x)/(a + b*tanh(c + d*x)**2)**2, x)

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