3.107 \(\int \cosh ^2(a+b x) \coth ^3(a+b x) \, dx\)

Optimal. Leaf size=43 \[ \frac {\sinh ^2(a+b x)}{2 b}-\frac {\text {csch}^2(a+b x)}{2 b}+\frac {2 \log (\sinh (a+b x))}{b} \]

[Out]

-1/2*csch(b*x+a)^2/b+2*ln(sinh(b*x+a))/b+1/2*sinh(b*x+a)^2/b

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Rubi [A]  time = 0.04, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2590, 266, 43} \[ \frac {\sinh ^2(a+b x)}{2 b}-\frac {\text {csch}^2(a+b x)}{2 b}+\frac {2 \log (\sinh (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

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

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 35, normalized size = 0.81 \[ -\frac {-\sinh ^2(a+b x)+\text {csch}^2(a+b x)-4 \log (\sinh (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Csch[a + b*x]^2 - 4*Log[Sinh[a + b*x]] - Sinh[a + b*x]^2)/b

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fricas [B]  time = 0.45, size = 743, normalized size = 17.28 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(cosh(b*x + a)^8 + 8*cosh(b*x + a)*sinh(b*x + a)^7 + sinh(b*x + a)^8 - 2*(8*b*x + 1)*cosh(b*x + a)^6 - 2*(
8*b*x - 14*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^6 + 4*(14*cosh(b*x + a)^3 - 3*(8*b*x + 1)*cosh(b*x + a))*sinh(b*
x + a)^5 + 2*(16*b*x - 7)*cosh(b*x + a)^4 + 2*(35*cosh(b*x + a)^4 - 15*(8*b*x + 1)*cosh(b*x + a)^2 + 16*b*x -
7)*sinh(b*x + a)^4 + 8*(7*cosh(b*x + a)^5 - 5*(8*b*x + 1)*cosh(b*x + a)^3 + (16*b*x - 7)*cosh(b*x + a))*sinh(b
*x + a)^3 - 2*(8*b*x + 1)*cosh(b*x + a)^2 + 2*(14*cosh(b*x + a)^6 - 15*(8*b*x + 1)*cosh(b*x + a)^4 + 6*(16*b*x
- 7)*cosh(b*x + a)^2 - 8*b*x - 1)*sinh(b*x + a)^2 + 16*(cosh(b*x + a)^6 + 6*cosh(b*x + a)*sinh(b*x + a)^5 + s
inh(b*x + a)^6 + (15*cosh(b*x + a)^2 - 2)*sinh(b*x + a)^4 - 2*cosh(b*x + a)^4 + 4*(5*cosh(b*x + a)^3 - 2*cosh(
b*x + a))*sinh(b*x + a)^3 + (15*cosh(b*x + a)^4 - 12*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + cosh(b*x + a)^2 +
2*(3*cosh(b*x + a)^5 - 4*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a))*log(2*sinh(b*x + a)/(cosh(b*x + a) -
sinh(b*x + a))) + 4*(2*cosh(b*x + a)^7 - 3*(8*b*x + 1)*cosh(b*x + a)^5 + 2*(16*b*x - 7)*cosh(b*x + a)^3 - (8*b
*x + 1)*cosh(b*x + a))*sinh(b*x + a) + 1)/(b*cosh(b*x + a)^6 + 6*b*cosh(b*x + a)*sinh(b*x + a)^5 + b*sinh(b*x
+ a)^6 - 2*b*cosh(b*x + a)^4 + (15*b*cosh(b*x + a)^2 - 2*b)*sinh(b*x + a)^4 + 4*(5*b*cosh(b*x + a)^3 - 2*b*cos
h(b*x + a))*sinh(b*x + a)^3 + b*cosh(b*x + a)^2 + (15*b*cosh(b*x + a)^4 - 12*b*cosh(b*x + a)^2 + b)*sinh(b*x +
a)^2 + 2*(3*b*cosh(b*x + a)^5 - 4*b*cosh(b*x + a)^3 + b*cosh(b*x + a))*sinh(b*x + a))

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giac [B]  time = 0.21, size = 99, normalized size = 2.30 \[ -\frac {16 \, b x - {\left (8 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )} + \frac {8 \, {\left (3 \, e^{\left (4 \, b x + 4 \, a\right )} - 4 \, e^{\left (2 \, b x + 2 \, a\right )} + 3\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{2}} - e^{\left (2 \, b x + 2 \, a\right )} - 16 \, \log \left ({\left | e^{\left (2 \, b x + 2 \, a\right )} - 1 \right |}\right )}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(16*b*x - (8*e^(2*b*x + 2*a) + 1)*e^(-2*b*x - 2*a) + 8*(3*e^(4*b*x + 4*a) - 4*e^(2*b*x + 2*a) + 3)/(e^(2*
b*x + 2*a) - 1)^2 - e^(2*b*x + 2*a) - 16*log(abs(e^(2*b*x + 2*a) - 1)))/b

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maple [A]  time = 0.17, size = 48, normalized size = 1.12 \[ \frac {\cosh ^{4}\left (b x +a \right )}{2 b \sinh \left (b x +a \right )^{2}}+\frac {2 \ln \left (\sinh \left (b x +a \right )\right )}{b}-\frac {\coth ^{2}\left (b x +a \right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/b*cosh(b*x+a)^4/sinh(b*x+a)^2+2*ln(sinh(b*x+a))/b-coth(b*x+a)^2/b

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maxima [B]  time = 0.34, size = 120, normalized size = 2.79 \[ \frac {2 \, {\left (b x + a\right )}}{b} + \frac {e^{\left (-2 \, b x - 2 \, a\right )}}{8 \, b} + \frac {2 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{b} + \frac {2 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{b} - \frac {2 \, e^{\left (-2 \, b x - 2 \, a\right )} + 15 \, e^{\left (-4 \, b x - 4 \, a\right )} - 1}{8 \, b {\left (e^{\left (-2 \, b x - 2 \, a\right )} - 2 \, e^{\left (-4 \, b x - 4 \, a\right )} + e^{\left (-6 \, b x - 6 \, a\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b*x + a)/b + 1/8*e^(-2*b*x - 2*a)/b + 2*log(e^(-b*x - a) + 1)/b + 2*log(e^(-b*x - a) - 1)/b - 1/8*(2*e^(-2*
b*x - 2*a) + 15*e^(-4*b*x - 4*a) - 1)/(b*(e^(-2*b*x - 2*a) - 2*e^(-4*b*x - 4*a) + e^(-6*b*x - 6*a)))

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mupad [B]  time = 1.46, size = 97, normalized size = 2.26 \[ \frac {2\,\ln \left ({\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1\right )}{b}-2\,x-\frac {2}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )}-\frac {2}{b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}+\frac {{\mathrm {e}}^{-2\,a-2\,b\,x}}{8\,b}+\frac {{\mathrm {e}}^{2\,a+2\,b\,x}}{8\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*log(exp(2*a)*exp(2*b*x) - 1))/b - 2*x - 2/(b*(exp(2*a + 2*b*x) - 1)) - 2/(b*(exp(4*a + 4*b*x) - 2*exp(2*a +
2*b*x) + 1)) + exp(- 2*a - 2*b*x)/(8*b) + exp(2*a + 2*b*x)/(8*b)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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