3.122 \(\int \coth ^2(a+b x) \text {csch}^3(a+b x) \, dx\)

Optimal. Leaf size=55 \[ \frac {\tanh ^{-1}(\cosh (a+b x))}{8 b}-\frac {\coth (a+b x) \text {csch}^3(a+b x)}{4 b}-\frac {\coth (a+b x) \text {csch}(a+b x)}{8 b} \]

[Out]

1/8*arctanh(cosh(b*x+a))/b-1/8*coth(b*x+a)*csch(b*x+a)/b-1/4*coth(b*x+a)*csch(b*x+a)^3/b

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2611, 3768, 3770} \[ \frac {\tanh ^{-1}(\cosh (a+b x))}{8 b}-\frac {\coth (a+b x) \text {csch}^3(a+b x)}{4 b}-\frac {\coth (a+b x) \text {csch}(a+b x)}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[Coth[a + b*x]^2*Csch[a + b*x]^3,x]

[Out]

ArcTanh[Cosh[a + b*x]]/(8*b) - (Coth[a + b*x]*Csch[a + b*x])/(8*b) - (Coth[a + b*x]*Csch[a + b*x]^3)/(4*b)

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \coth ^2(a+b x) \text {csch}^3(a+b x) \, dx &=-\frac {\coth (a+b x) \text {csch}^3(a+b x)}{4 b}+\frac {1}{4} \int \text {csch}^3(a+b x) \, dx\\ &=-\frac {\coth (a+b x) \text {csch}(a+b x)}{8 b}-\frac {\coth (a+b x) \text {csch}^3(a+b x)}{4 b}-\frac {1}{8} \int \text {csch}(a+b x) \, dx\\ &=\frac {\tanh ^{-1}(\cosh (a+b x))}{8 b}-\frac {\coth (a+b x) \text {csch}(a+b x)}{8 b}-\frac {\coth (a+b x) \text {csch}^3(a+b x)}{4 b}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 95, normalized size = 1.73 \[ -\frac {\text {csch}^4\left (\frac {1}{2} (a+b x)\right )}{64 b}-\frac {\text {csch}^2\left (\frac {1}{2} (a+b x)\right )}{32 b}+\frac {\text {sech}^4\left (\frac {1}{2} (a+b x)\right )}{64 b}-\frac {\text {sech}^2\left (\frac {1}{2} (a+b x)\right )}{32 b}-\frac {\log \left (\tanh \left (\frac {1}{2} (a+b x)\right )\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[a + b*x]^2*Csch[a + b*x]^3,x]

[Out]

-1/32*Csch[(a + b*x)/2]^2/b - Csch[(a + b*x)/2]^4/(64*b) - Log[Tanh[(a + b*x)/2]]/(8*b) - Sech[(a + b*x)/2]^2/
(32*b) + Sech[(a + b*x)/2]^4/(64*b)

________________________________________________________________________________________

fricas [B]  time = 0.53, size = 1109, normalized size = 20.16 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(b*x+a)^2*csch(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/8*(2*cosh(b*x + a)^7 + 14*cosh(b*x + a)*sinh(b*x + a)^6 + 2*sinh(b*x + a)^7 + 14*(3*cosh(b*x + a)^2 + 1)*si
nh(b*x + a)^5 + 14*cosh(b*x + a)^5 + 70*(cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a)^4 + 14*(5*cosh(b*x + a
)^4 + 10*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^3 + 14*cosh(b*x + a)^3 + 14*(3*cosh(b*x + a)^5 + 10*cosh(b*x + a)^
3 + 3*cosh(b*x + a))*sinh(b*x + a)^2 - (cosh(b*x + a)^8 + 8*cosh(b*x + a)*sinh(b*x + a)^7 + sinh(b*x + a)^8 +
4*(7*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^6 - 4*cosh(b*x + a)^6 + 8*(7*cosh(b*x + a)^3 - 3*cosh(b*x + a))*sinh(b
*x + a)^5 + 2*(35*cosh(b*x + a)^4 - 30*cosh(b*x + a)^2 + 3)*sinh(b*x + a)^4 + 6*cosh(b*x + a)^4 + 8*(7*cosh(b*
x + a)^5 - 10*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^3 + 4*(7*cosh(b*x + a)^6 - 15*cosh(b*x + a)^4 +
9*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 4*cosh(b*x + a)^2 + 8*(cosh(b*x + a)^7 - 3*cosh(b*x + a)^5 + 3*cosh(
b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*log(cosh(b*x + a) + sinh(b*x + a) + 1) + (cosh(b*x + a)^8 + 8*c
osh(b*x + a)*sinh(b*x + a)^7 + sinh(b*x + a)^8 + 4*(7*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^6 - 4*cosh(b*x + a)^6
+ 8*(7*cosh(b*x + a)^3 - 3*cosh(b*x + a))*sinh(b*x + a)^5 + 2*(35*cosh(b*x + a)^4 - 30*cosh(b*x + a)^2 + 3)*s
inh(b*x + a)^4 + 6*cosh(b*x + a)^4 + 8*(7*cosh(b*x + a)^5 - 10*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a
)^3 + 4*(7*cosh(b*x + a)^6 - 15*cosh(b*x + a)^4 + 9*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 4*cosh(b*x + a)^2 +
8*(cosh(b*x + a)^7 - 3*cosh(b*x + a)^5 + 3*cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*log(cosh(b*x +
a) + sinh(b*x + a) - 1) + 2*(7*cosh(b*x + a)^6 + 35*cosh(b*x + a)^4 + 21*cosh(b*x + a)^2 + 1)*sinh(b*x + a) +
2*cosh(b*x + a))/(b*cosh(b*x + a)^8 + 8*b*cosh(b*x + a)*sinh(b*x + a)^7 + b*sinh(b*x + a)^8 - 4*b*cosh(b*x +
a)^6 + 4*(7*b*cosh(b*x + a)^2 - b)*sinh(b*x + a)^6 + 8*(7*b*cosh(b*x + a)^3 - 3*b*cosh(b*x + a))*sinh(b*x + a)
^5 + 6*b*cosh(b*x + a)^4 + 2*(35*b*cosh(b*x + a)^4 - 30*b*cosh(b*x + a)^2 + 3*b)*sinh(b*x + a)^4 + 8*(7*b*cosh
(b*x + a)^5 - 10*b*cosh(b*x + a)^3 + 3*b*cosh(b*x + a))*sinh(b*x + a)^3 - 4*b*cosh(b*x + a)^2 + 4*(7*b*cosh(b*
x + a)^6 - 15*b*cosh(b*x + a)^4 + 9*b*cosh(b*x + a)^2 - b)*sinh(b*x + a)^2 + 8*(b*cosh(b*x + a)^7 - 3*b*cosh(b
*x + a)^5 + 3*b*cosh(b*x + a)^3 - b*cosh(b*x + a))*sinh(b*x + a) + b)

________________________________________________________________________________________

giac [A]  time = 0.16, size = 80, normalized size = 1.45 \[ -\frac {\frac {2 \, {\left (e^{\left (7 \, b x + 7 \, a\right )} + 7 \, e^{\left (5 \, b x + 5 \, a\right )} + 7 \, e^{\left (3 \, b x + 3 \, a\right )} + e^{\left (b x + a\right )}\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{4}} - \log \left (e^{\left (b x + a\right )} + 1\right ) + \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(b*x+a)^2*csch(b*x+a)^3,x, algorithm="giac")

[Out]

-1/8*(2*(e^(7*b*x + 7*a) + 7*e^(5*b*x + 5*a) + 7*e^(3*b*x + 3*a) + e^(b*x + a))/(e^(2*b*x + 2*a) - 1)^4 - log(
e^(b*x + a) + 1) + log(abs(e^(b*x + a) - 1)))/b

________________________________________________________________________________________

maple [A]  time = 0.34, size = 58, normalized size = 1.05 \[ \frac {-\frac {\cosh \left (b x +a \right )}{3 \sinh \left (b x +a \right )^{4}}-\frac {\left (-\frac {\mathrm {csch}\left (b x +a \right )^{3}}{4}+\frac {3 \,\mathrm {csch}\left (b x +a \right )}{8}\right ) \coth \left (b x +a \right )}{3}+\frac {\arctanh \left ({\mathrm e}^{b x +a}\right )}{4}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(b*x+a)^2*csch(b*x+a)^3,x)

[Out]

1/b*(-1/3/sinh(b*x+a)^4*cosh(b*x+a)-1/3*(-1/4*csch(b*x+a)^3+3/8*csch(b*x+a))*coth(b*x+a)+1/4*arctanh(exp(b*x+a
)))

________________________________________________________________________________________

maxima [B]  time = 0.60, size = 129, normalized size = 2.35 \[ \frac {\log \left (e^{\left (-b x - a\right )} + 1\right )}{8 \, b} - \frac {\log \left (e^{\left (-b x - a\right )} - 1\right )}{8 \, b} + \frac {e^{\left (-b x - a\right )} + 7 \, e^{\left (-3 \, b x - 3 \, a\right )} + 7 \, e^{\left (-5 \, b x - 5 \, a\right )} + e^{\left (-7 \, b x - 7 \, a\right )}}{4 \, b {\left (4 \, e^{\left (-2 \, b x - 2 \, a\right )} - 6 \, e^{\left (-4 \, b x - 4 \, a\right )} + 4 \, e^{\left (-6 \, b x - 6 \, a\right )} - e^{\left (-8 \, b x - 8 \, a\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(b*x+a)^2*csch(b*x+a)^3,x, algorithm="maxima")

[Out]

1/8*log(e^(-b*x - a) + 1)/b - 1/8*log(e^(-b*x - a) - 1)/b + 1/4*(e^(-b*x - a) + 7*e^(-3*b*x - 3*a) + 7*e^(-5*b
*x - 5*a) + e^(-7*b*x - 7*a))/(b*(4*e^(-2*b*x - 2*a) - 6*e^(-4*b*x - 4*a) + 4*e^(-6*b*x - 6*a) - e^(-8*b*x - 8
*a) - 1))

________________________________________________________________________________________

mupad [B]  time = 0.11, size = 219, normalized size = 3.98 \[ \frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {-b^2}}{b}\right )}{4\,\sqrt {-b^2}}-\frac {\frac {{\mathrm {e}}^{a+b\,x}}{b}+\frac {2\,{\mathrm {e}}^{3\,a+3\,b\,x}}{b}+\frac {{\mathrm {e}}^{5\,a+5\,b\,x}}{b}}{6\,{\mathrm {e}}^{4\,a+4\,b\,x}-4\,{\mathrm {e}}^{2\,a+2\,b\,x}-4\,{\mathrm {e}}^{6\,a+6\,b\,x}+{\mathrm {e}}^{8\,a+8\,b\,x}+1}-\frac {3\,{\mathrm {e}}^{a+b\,x}}{2\,b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left (3\,{\mathrm {e}}^{2\,a+2\,b\,x}-3\,{\mathrm {e}}^{4\,a+4\,b\,x}+{\mathrm {e}}^{6\,a+6\,b\,x}-1\right )}-\frac {{\mathrm {e}}^{a+b\,x}}{4\,b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(a + b*x)^2/sinh(a + b*x)^3,x)

[Out]

atan((exp(b*x)*exp(a)*(-b^2)^(1/2))/b)/(4*(-b^2)^(1/2)) - (exp(a + b*x)/b + (2*exp(3*a + 3*b*x))/b + exp(5*a +
5*b*x)/b)/(6*exp(4*a + 4*b*x) - 4*exp(2*a + 2*b*x) - 4*exp(6*a + 6*b*x) + exp(8*a + 8*b*x) + 1) - (3*exp(a +
b*x))/(2*b*(exp(4*a + 4*b*x) - 2*exp(2*a + 2*b*x) + 1)) - (2*exp(a + b*x))/(b*(3*exp(2*a + 2*b*x) - 3*exp(4*a
+ 4*b*x) + exp(6*a + 6*b*x) - 1)) - exp(a + b*x)/(4*b*(exp(2*a + 2*b*x) - 1))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \coth ^{2}{\left (a + b x \right )} \operatorname {csch}^{3}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(b*x+a)**2*csch(b*x+a)**3,x)

[Out]

Integral(coth(a + b*x)**2*csch(a + b*x)**3, x)

________________________________________________________________________________________