3.146 \(\int \frac{\coth ^3(c+d x)}{a+b \text{sech}^2(c+d x)} \, dx\)

Optimal. Leaf size=73 \[ \frac{b^2 \log \left (a \cosh ^2(c+d x)+b\right )}{2 a d (a+b)^2}-\frac{\text{csch}^2(c+d x)}{2 d (a+b)}+\frac{(a+2 b) \log (\sinh (c+d x))}{d (a+b)^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.121826, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {4138, 446, 88} \[ \frac{b^2 \log \left (a \cosh ^2(c+d x)+b\right )}{2 a d (a+b)^2}-\frac{\text{csch}^2(c+d x)}{2 d (a+b)}+\frac{(a+2 b) \log (\sinh (c+d x))}{d (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Coth[c + d*x]^3/(a + b*Sech[c + d*x]^2),x]

[Out]

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

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]

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 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]))

Rubi steps

\begin{align*} \int \frac{\coth ^3(c+d x)}{a+b \text{sech}^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^5}{\left (1-x^2\right )^2 \left (b+a x^2\right )} \, dx,x,\cosh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{(1-x)^2 (b+a x)} \, dx,x,\cosh ^2(c+d x)\right )}{2 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{(a+b) (-1+x)^2}+\frac{a+2 b}{(a+b)^2 (-1+x)}+\frac{b^2}{(a+b)^2 (b+a x)}\right ) \, dx,x,\cosh ^2(c+d x)\right )}{2 d}\\ &=-\frac{\text{csch}^2(c+d x)}{2 (a+b) d}+\frac{b^2 \log \left (b+a \cosh ^2(c+d x)\right )}{2 a (a+b)^2 d}+\frac{(a+2 b) \log (\sinh (c+d x))}{(a+b)^2 d}\\ \end{align*}

Mathematica [A]  time = 0.221937, size = 100, normalized size = 1.37 \[ -\frac{\text{sech}^2(c+d x) (a \cosh (2 (c+d x))+a+2 b) \left (b^2 \left (-\log \left (a \sinh ^2(c+d x)+a+b\right )\right )+a (a+b) \text{csch}^2(c+d x)-2 a (a+2 b) \log (\sinh (c+d x))\right )}{4 a d (a+b)^2 \left (a+b \text{sech}^2(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[c + d*x]^3/(a + b*Sech[c + d*x]^2),x]

[Out]

-((a + 2*b + a*Cosh[2*(c + d*x)])*(a*(a + b)*Csch[c + d*x]^2 - 2*a*(a + 2*b)*Log[Sinh[c + d*x]] - b^2*Log[a +
b + a*Sinh[c + d*x]^2])*Sech[c + d*x]^2)/(4*a*(a + b)^2*d*(a + b*Sech[c + d*x]^2))

________________________________________________________________________________________

Maple [B]  time = 0.07, size = 199, normalized size = 2.7 \begin{align*} -{\frac{1}{8\,d \left ( a+b \right ) } \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}-{\frac{1}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{{b}^{2}}{2\,da \left ( a+b \right ) ^{2}}\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}a+b \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}+2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) }-{\frac{1}{8\,d \left ( a+b \right ) } \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}+{\frac{a}{d \left ( a+b \right ) ^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+2\,{\frac{\ln \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) b}{d \left ( a+b \right ) ^{2}}}-{\frac{1}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(d*x+c)^3/(a+b*sech(d*x+c)^2),x)

[Out]

-1/8/d*tanh(1/2*d*x+1/2*c)^2/(a+b)-1/d/a*ln(tanh(1/2*d*x+1/2*c)+1)+1/2/d*b^2/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)-1/8/d/(a+b)/tanh(1/2*d*x+
1/2*c)^2+1/d/(a+b)^2*ln(tanh(1/2*d*x+1/2*c))*a+2/d/(a+b)^2*ln(tanh(1/2*d*x+1/2*c))*b-1/d/a*ln(tanh(1/2*d*x+1/2
*c)-1)

________________________________________________________________________________________

Maxima [B]  time = 1.20983, size = 252, normalized size = 3.45 \begin{align*} \frac{b^{2} \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^{3} + 2 \, a^{2} b + a b^{2}\right )} d} + \frac{{\left (a + 2 \, b\right )} \log \left (e^{\left (-d x - c\right )} + 1\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} d} + \frac{{\left (a + 2 \, b\right )} \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 d} + \frac{2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{{\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 )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^3/(a+b*sech(d*x+c)^2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 3.02111, size = 2144, normalized size = 29.37 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^3/(a+b*sech(d*x+c)^2),x, algorithm="fricas")

[Out]

-1/2*(2*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^4 + 8*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 2*
(a^2 + 2*a*b + b^2)*d*x*sinh(d*x + c)^4 + 2*(a^2 + 2*a*b + b^2)*d*x - 4*((a^2 + 2*a*b + b^2)*d*x - a^2 - a*b)*
cosh(d*x + c)^2 + 4*(3*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^2 - (a^2 + 2*a*b + b^2)*d*x + a^2 + a*b)*sinh(d*x
+ c)^2 - (b^2*cosh(d*x + c)^4 + 4*b^2*cosh(d*x + c)*sinh(d*x + c)^3 + b^2*sinh(d*x + c)^4 - 2*b^2*cosh(d*x +
c)^2 + 2*(3*b^2*cosh(d*x + c)^2 - b^2)*sinh(d*x + c)^2 + b^2 + 4*(b^2*cosh(d*x + c)^3 - b^2*cosh(d*x + c))*sin
h(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^2 + 2*a*b)*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b)*cosh(d*x + c)*sinh(d*x + c)^3
+ (a^2 + 2*a*b)*sinh(d*x + c)^4 - 2*(a^2 + 2*a*b)*cosh(d*x + c)^2 + 2*(3*(a^2 + 2*a*b)*cosh(d*x + c)^2 - a^2 -
2*a*b)*sinh(d*x + c)^2 + a^2 + 2*a*b + 4*((a^2 + 2*a*b)*cosh(d*x + c)^3 - (a^2 + 2*a*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^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^3 - (
(a^2 + 2*a*b + b^2)*d*x - a^2 - a*b)*cosh(d*x + c))*sinh(d*x + c))/((a^3 + 2*a^2*b + a*b^2)*d*cosh(d*x + c)^4
+ 4*(a^3 + 2*a^2*b + a*b^2)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^3 + 2*a^2*b + a*b^2)*d*sinh(d*x + c)^4 - 2*(a
^3 + 2*a^2*b + a*b^2)*d*cosh(d*x + c)^2 + 2*(3*(a^3 + 2*a^2*b + a*b^2)*d*cosh(d*x + c)^2 - (a^3 + 2*a^2*b + a*
b^2)*d)*sinh(d*x + c)^2 + (a^3 + 2*a^2*b + a*b^2)*d + 4*((a^3 + 2*a^2*b + a*b^2)*d*cosh(d*x + c)^3 - (a^3 + 2*
a^2*b + a*b^2)*d*cosh(d*x + c))*sinh(d*x + c))

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth ^{3}{\left (c + d x \right )}}{a + b \operatorname{sech}^{2}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)**3/(a+b*sech(d*x+c)**2),x)

[Out]

Integral(coth(c + d*x)**3/(a + b*sech(c + d*x)**2), x)

________________________________________________________________________________________

Giac [B]  time = 2.07635, size = 288, normalized size = 3.95 \begin{align*} \frac{\frac{b^{2} \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^{3} + 2 \, a^{2} b + a b^{2}} - \frac{2 \, d x}{a} + \frac{2 \,{\left (a e^{\left (2 \, c\right )} + 2 \, b e^{\left (2 \, c\right )}\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{3 \, a e^{\left (4 \, d x + 4 \, c\right )} + 6 \, b e^{\left (4 \, d x + 4 \, c\right )} - 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 8 \, b e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a + 6 \, b}{{\left (a^{2} + 2 \, a b + b^{2}\right )}{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^3/(a+b*sech(d*x+c)^2),x, algorithm="giac")

[Out]

1/2*(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^3 + 2*a^2*b + a*b^2) - 2*d*
x/a + 2*(a*e^(2*c) + 2*b*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)) -
(3*a*e^(4*d*x + 4*c) + 6*b*e^(4*d*x + 4*c) - 2*a*e^(2*d*x + 2*c) - 8*b*e^(2*d*x + 2*c) + 3*a + 6*b)/((a^2 + 2
*a*b + b^2)*(e^(2*d*x + 2*c) - 1)^2))/d