3.2.76 \(\int \frac {\coth (c+d x)}{a+b \tanh ^2(c+d x)} \, dx\) [176]

Optimal. Leaf size=60 \[ \frac {\log (\cosh (c+d x))}{(a+b) d}+\frac {\log (\tanh (c+d x))}{a d}-\frac {b \log \left (a+b \tanh ^2(c+d x)\right )}{2 a (a+b) d} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 60, 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 = {3751, 457, 84} \begin {gather*} -\frac {b \log \left (a+b \tanh ^2(c+d x)\right )}{2 a d (a+b)}+\frac {\log (\cosh (c+d x))}{d (a+b)}+\frac {\log (\tanh (c+d x))}{a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 84

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 457

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 3751

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

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Mathematica [A]
time = 0.05, size = 54, normalized size = 0.90 \begin {gather*} \frac {2 a \log (\cosh (c+d x))+2 (a+b) \log (\tanh (c+d x))-b \log \left (a+b \tanh ^2(c+d x)\right )}{2 a (a+b) d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 2.90, size = 113, normalized size = 1.88

method	result	size
derivativedivides	\(\frac {-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a +b}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a +b}-\frac {b \ln \left (a \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 a \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 b \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a \right )}{2 a \left (a +b \right )}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}}{d}\)	\(113\)
default	\(\frac {-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a +b}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a +b}-\frac {b \ln \left (a \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 a \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 b \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a \right )}{2 a \left (a +b \right )}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}}{d}\)	\(113\)
risch	\(\frac {x}{a +b}-\frac {2 x}{a}-\frac {2 c}{a d}+\frac {2 b x}{a \left (a +b \right )}+\frac {2 b c}{a d \left (a +b \right )}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}-1\right )}{a d}-\frac {b \ln \left ({\mathrm e}^{4 d x +4 c}+\frac {2 \left (a -b \right ) {\mathrm e}^{2 d x +2 c}}{a +b}+1\right )}{2 a d \left (a +b \right )}\)	\(117\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-1/(a+b)*ln(tanh(1/2*d*x+1/2*c)+1)-1/(a+b)*ln(tanh(1/2*d*x+1/2*c)-1)-1/2*b/a/(a+b)*ln(a*tanh(1/2*d*x+1/2*
c)^4+2*a*tanh(1/2*d*x+1/2*c)^2+4*b*tanh(1/2*d*x+1/2*c)^2+a)+1/a*ln(tanh(1/2*d*x+1/2*c)))

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Maxima [A]
time = 0.27, size = 101, normalized size = 1.68 \begin {gather*} -\frac {b \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^{2} + a b\right )} d} + \frac {d x + c}{{\left (a + b\right )} d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{a d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 118 vs. \(2 (58) = 116\).
time = 0.42, size = 118, normalized size = 1.97 \begin {gather*} -\frac {2 \, a d x + b \log \left (\frac {2 \, {\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{2} + {\left (a + b\right )} \sinh \left (d x + c\right )^{2} + a - b\right )}}{\cosh \left (d x + c\right )^{2} - 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2}}\right ) - 2 \, {\left (a + b\right )} \log \left (\frac {2 \, \sinh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right )}{2 \, {\left (a^{2} + a b\right )} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*a*d*x + b*log(2*((a + b)*cosh(d*x + c)^2 + (a + b)*sinh(d*x + c)^2 + a - b)/(cosh(d*x + c)^2 - 2*cosh(
d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2)) - 2*(a + b)*log(2*sinh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))))/(
(a^2 + a*b)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.44, size = 97, normalized size = 1.62 \begin {gather*} -\frac {\frac {b \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^{2} + a b} + \frac {2 \, {\left (d x + c\right )}}{a + b} - \frac {2 \, \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right )}{a}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.47, size = 194, normalized size = 3.23 \begin {gather*} \frac {\ln \left (12\,a\,b^2+4\,a^2\,b+9\,b^3-9\,b^3\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}-12\,a\,b^2\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}-4\,a^2\,b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\right )}{a\,d}-\frac {b\,\ln \left (5\,a\,b+2\,a^2+3\,b^2+4\,a^2\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+2\,a^2\,{\mathrm {e}}^{4\,c}\,{\mathrm {e}}^{4\,d\,x}-6\,b^2\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+3\,b^2\,{\mathrm {e}}^{4\,c}\,{\mathrm {e}}^{4\,d\,x}+2\,a\,b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+5\,a\,b\,{\mathrm {e}}^{4\,c}\,{\mathrm {e}}^{4\,d\,x}\right )}{2\,d\,a^2+2\,b\,d\,a}-\frac {x}{a+b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(12*a*b^2 + 4*a^2*b + 9*b^3 - 9*b^3*exp(2*c)*exp(2*d*x) - 12*a*b^2*exp(2*c)*exp(2*d*x) - 4*a^2*b*exp(2*c)*e
xp(2*d*x))/(a*d) - (b*log(5*a*b + 2*a^2 + 3*b^2 + 4*a^2*exp(2*c)*exp(2*d*x) + 2*a^2*exp(4*c)*exp(4*d*x) - 6*b^
2*exp(2*c)*exp(2*d*x) + 3*b^2*exp(4*c)*exp(4*d*x) + 2*a*b*exp(2*c)*exp(2*d*x) + 5*a*b*exp(4*c)*exp(4*d*x)))/(2
*a^2*d + 2*a*b*d) - x/(a + b)

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