3.154 \(\int \frac {\coth (c+d x)}{(a+b \text {sech}^2(c+d x))^2} \, dx\)

Optimal. Leaf size=83 \[ \frac {b^2}{2 a^2 d (a+b) \left (a \cosh ^2(c+d x)+b\right )}+\frac {b (2 a+b) \log \left (a \cosh ^2(c+d x)+b\right )}{2 a^2 d (a+b)^2}+\frac {\log (\sinh (c+d x))}{d (a+b)^2} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 83, 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 = {4138, 446, 88} \[ \frac {b^2}{2 a^2 d (a+b) \left (a \cosh ^2(c+d x)+b\right )}+\frac {b (2 a+b) \log \left (a \cosh ^2(c+d x)+b\right )}{2 a^2 d (a+b)^2}+\frac {\log (\sinh (c+d x))}{d (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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 4138

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> Module[{ff =
FreeFactors[Cos[e + f*x], x]}, -Dist[(f*ff^(m + n*p - 1))^(-1), Subst[Int[((1 - ff^2*x^2)^((m - 1)/2)*(b + a*
(ff*x)^n)^p)/x^(m + n*p), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {\coth (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {x^5}{\left (1-x^2\right ) \left (b+a x^2\right )^2} \, dx,x,\cosh (c+d x)\right )}{d}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {x^2}{(1-x) (b+a x)^2} \, dx,x,\cosh ^2(c+d x)\right )}{2 d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (-\frac {1}{(a+b)^2 (-1+x)}+\frac {b^2}{a (a+b) (b+a x)^2}-\frac {b (2 a+b)}{a (a+b)^2 (b+a x)}\right ) \, dx,x,\cosh ^2(c+d x)\right )}{2 d}\\ &=\frac {b^2}{2 a^2 (a+b) d \left (b+a \cosh ^2(c+d x)\right )}+\frac {b (2 a+b) \log \left (b+a \cosh ^2(c+d x)\right )}{2 a^2 (a+b)^2 d}+\frac {\log (\sinh (c+d x))}{(a+b)^2 d}\\ \end {align*}

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Mathematica [A]  time = 0.30, size = 115, normalized size = 1.39 \[ \frac {a \sinh ^2(c+d x) \left (2 a^2 \log (\sinh (c+d x))+b (2 a+b) \log \left (a \sinh ^2(c+d x)+a+b\right )\right )+(a+b) \left (2 a^2 \log (\sinh (c+d x))+b \left ((2 a+b) \log \left (a \sinh ^2(c+d x)+a+b\right )+b\right )\right )}{a^2 d (a+b)^2 (a \cosh (2 (c+d x))+a+2 b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.55, size = 1031, normalized size = 12.42 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.56, size = 246, normalized size = 2.96 \[ \frac {\frac {{\left (2 \, a b + b^{2}\right )} \log \left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )}{a^{4} + 2 \, a^{3} b + a^{2} b^{2}} + \frac {2 \, e^{\left (2 \, c\right )} \log \left ({\left | -e^{\left (2 \, d x + 2 \, c\right )} + 1 \right |}\right )}{a^{2} e^{\left (2 \, c\right )} + 2 \, a b e^{\left (2 \, c\right )} + b^{2} e^{\left (2 \, c\right )}} - \frac {2 \, d x}{a^{2}} - \frac {2 \, a b e^{\left (4 \, d x + 4 \, c\right )} + b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 4 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 6 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a b + b^{2}}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} {\left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((2*a*b + b^2)*log(a*e^(4*d*x + 4*c) + 2*a*e^(2*d*x + 2*c) + 4*b*e^(2*d*x + 2*c) + a)/(a^4 + 2*a^3*b + a^2
*b^2) + 2*e^(2*c)*log(abs(-e^(2*d*x + 2*c) + 1))/(a^2*e^(2*c) + 2*a*b*e^(2*c) + b^2*e^(2*c)) - 2*d*x/a^2 - (2*
a*b*e^(4*d*x + 4*c) + b^2*e^(4*d*x + 4*c) + 4*a*b*e^(2*d*x + 2*c) + 6*b^2*e^(2*d*x + 2*c) + 2*a*b + b^2)/((a^3
+ 2*a^2*b + a*b^2)*(a*e^(4*d*x + 4*c) + 2*a*e^(2*d*x + 2*c) + 4*b*e^(2*d*x + 2*c) + a)))/d

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maple [B]  time = 0.42, size = 292, normalized size = 3.52 \[ -\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}}-\frac {2 b^{2} \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a \left (a +b \right )^{2} \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +b \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}+\frac {b \ln \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +b \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}{d a \left (a +b \right )^{2}}+\frac {b^{2} \ln \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +b \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}{2 d \,a^{2} \left (a +b \right )^{2}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a +b \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.35, size = 209, normalized size = 2.52 \[ \frac {2 \, b^{2} e^{\left (-2 \, d x - 2 \, c\right )}}{{\left (a^{4} + a^{3} b + 2 \, {\left (a^{4} + 3 \, a^{3} b + 2 \, a^{2} b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{4} + a^{3} b\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} + \frac {{\left (2 \, a b + b^{2}\right )} \log \left (2 \, {\left (a + 2 \, b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a e^{\left (-4 \, d x - 4 \, c\right )} + a\right )}{2 \, {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} d} + \frac {d x + c}{a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^4\,\mathrm {coth}\left (c+d\,x\right )}{{\left (a\,{\mathrm {cosh}\left (c+d\,x\right )}^2+b\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((cosh(c + d*x)^4*coth(c + d*x))/(b + a*cosh(c + d*x)^2)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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