3.147 \(\int \coth ^2(c+b x) \sinh (a+b x) \, dx\)

Optimal. Leaf size=46 \[ -\frac {\cosh (a-c) \tanh ^{-1}(\cosh (b x+c))}{b}-\frac {\sinh (a-c) \text {csch}(b x+c)}{b}+\frac {\cosh (a+b x)}{b} \]

[Out]

-arctanh(cosh(b*x+c))*cosh(a-c)/b+cosh(b*x+a)/b-csch(b*x+c)*sinh(a-c)/b

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Rubi [A]  time = 0.05, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5622, 5621, 2638, 3770, 2606, 8} \[ -\frac {\cosh (a-c) \tanh ^{-1}(\cosh (b x+c))}{b}-\frac {\sinh (a-c) \text {csch}(b x+c)}{b}+\frac {\cosh (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Coth[c + b*x]^2*Sinh[a + b*x],x]

[Out]

-((ArcTanh[Cosh[c + b*x]]*Cosh[a - c])/b) + Cosh[a + b*x]/b - (Csch[c + b*x]*Sinh[a - c])/b

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2638

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

Rule 3770

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

Rule 5621

Int[Cosh[v_]*Coth[w_]^(n_.), x_Symbol] :> Int[Sinh[v]*Coth[w]^(n - 1), x] + Dist[Cosh[v - w], Int[Csch[w]*Coth
[w]^(n - 1), x], x] /; GtQ[n, 0] && NeQ[w, v] && FreeQ[v - w, x]

Rule 5622

Int[Coth[w_]^(n_.)*Sinh[v_], x_Symbol] :> Int[Cosh[v]*Coth[w]^(n - 1), x] + Dist[Sinh[v - w], Int[Csch[w]*Coth
[w]^(n - 1), x], x] /; GtQ[n, 0] && NeQ[w, v] && FreeQ[v - w, x]

Rubi steps

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

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Mathematica [C]  time = 0.10, size = 110, normalized size = 2.39 \[ -\frac {\sinh (a-c) \text {csch}(b x+c)}{b}-\frac {2 i \cosh (a-c) \tan ^{-1}\left (\frac {(\cosh (c)-\sinh (c)) \left (\sinh (c) \sinh \left (\frac {b x}{2}\right )+\cosh (c) \cosh \left (\frac {b x}{2}\right )\right )}{i \cosh (c) \cosh \left (\frac {b x}{2}\right )-i \sinh (c) \cosh \left (\frac {b x}{2}\right )}\right )}{b}+\frac {\sinh (a) \sinh (b x)}{b}+\frac {\cosh (a) \cosh (b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[c + b*x]^2*Sinh[a + b*x],x]

[Out]

((-2*I)*ArcTan[((Cosh[c] - Sinh[c])*(Cosh[c]*Cosh[(b*x)/2] + Sinh[c]*Sinh[(b*x)/2]))/(I*Cosh[c]*Cosh[(b*x)/2]
- I*Cosh[(b*x)/2]*Sinh[c])]*Cosh[a - c])/b + (Cosh[a]*Cosh[b*x])/b - (Csch[c + b*x]*Sinh[a - c])/b + (Sinh[a]*
Sinh[b*x])/b

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fricas [B]  time = 0.45, size = 1237, normalized size = 26.89 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.14, size = 121, normalized size = 2.63 \[ -\frac {{\left (e^{\left (2 \, a + c\right )} + e^{\left (3 \, c\right )}\right )} e^{\left (-a - 2 \, c\right )} \log \left (e^{\left (b x + c\right )} + 1\right ) - {\left (e^{\left (2 \, a + c\right )} + e^{\left (3 \, c\right )}\right )} e^{\left (-a - 2 \, c\right )} \log \left ({\left | e^{\left (b x + c\right )} - 1 \right |}\right ) + \frac {{\left (2 \, e^{\left (2 \, b x + 2 \, a\right )} - 3 \, e^{\left (2 \, b x + 2 \, c\right )} + 1\right )} e^{\left (-a\right )}}{e^{\left (3 \, b x + 2 \, c\right )} - e^{\left (b x\right )}} - e^{\left (b x + a\right )}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.23, size = 197, normalized size = 4.28 \[ \frac {{\mathrm e}^{b x +a}}{2 b}+\frac {{\mathrm e}^{-b x -a}}{2 b}+\frac {{\mathrm e}^{b x +a} \left ({\mathrm e}^{2 a}-{\mathrm e}^{2 c}\right )}{b \left (-{\mathrm e}^{2 b x +2 a +2 c}+{\mathrm e}^{2 a}\right )}+\frac {\ln \left ({\mathrm e}^{b x +a}-{\mathrm e}^{a -c}\right ) {\mathrm e}^{-a -c} {\mathrm e}^{2 a}}{2 b}+\frac {\ln \left ({\mathrm e}^{b x +a}-{\mathrm e}^{a -c}\right ) {\mathrm e}^{-a -c} {\mathrm e}^{2 c}}{2 b}-\frac {\ln \left ({\mathrm e}^{b x +a}+{\mathrm e}^{a -c}\right ) {\mathrm e}^{-a -c} {\mathrm e}^{2 a}}{2 b}-\frac {\ln \left ({\mathrm e}^{b x +a}+{\mathrm e}^{a -c}\right ) {\mathrm e}^{-a -c} {\mathrm e}^{2 c}}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*exp(b*x+a)/b+1/2*exp(-b*x-a)/b+1/b*exp(b*x+a)*(exp(2*a)-exp(2*c))/(-exp(2*b*x+2*a+2*c)+exp(2*a))+1/2*ln(ex
p(b*x+a)-exp(a-c))/b*exp(-a-c)*exp(2*a)+1/2*ln(exp(b*x+a)-exp(a-c))/b*exp(-a-c)*exp(2*c)-1/2*ln(exp(b*x+a)+exp
(a-c))/b*exp(-a-c)*exp(2*a)-1/2*ln(exp(b*x+a)+exp(a-c))/b*exp(-a-c)*exp(2*c)

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maxima [B]  time = 0.33, size = 140, normalized size = 3.04 \[ -\frac {{\left (e^{\left (2 \, a\right )} + e^{\left (2 \, c\right )}\right )} e^{\left (-a - c\right )} \log \left (e^{\left (-b x\right )} + e^{c}\right )}{2 \, b} + \frac {{\left (e^{\left (2 \, a\right )} + e^{\left (2 \, c\right )}\right )} e^{\left (-a - c\right )} \log \left (e^{\left (-b x\right )} - e^{c}\right )}{2 \, b} + \frac {e^{\left (-b x - a\right )}}{2 \, b} - \frac {{\left (3 \, e^{\left (2 \, a\right )} - 2 \, e^{\left (2 \, c\right )}\right )} e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (2 \, c\right )}}{2 \, b {\left (e^{\left (-b x - a + 2 \, c\right )} - e^{\left (-3 \, b x - a\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.59, size = 181, normalized size = 3.93 \[ \frac {{\mathrm {e}}^{a+b\,x}}{2\,b}+\frac {{\mathrm {e}}^{-a-b\,x}}{2\,b}-\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{-a}\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{b\,x}\,\left (\sqrt {-b^2}+{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-2\,c}\,\sqrt {-b^2}\right )}{b\,\sqrt {{\mathrm {e}}^{-2\,a}\,{\mathrm {e}}^{2\,c}\,\left (2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-2\,c}+{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-4\,c}+1\right )}}\right )\,\sqrt {{\mathrm {e}}^{2\,c-2\,a}\,\left (2\,{\mathrm {e}}^{2\,a-2\,c}+{\mathrm {e}}^{4\,a-4\,c}+1\right )}}{\sqrt {-b^2}}+\frac {{\mathrm {e}}^{a+b\,x}\,\left ({\mathrm {e}}^{2\,a-2\,c}-1\right )}{b\,\left ({\mathrm {e}}^{2\,a-2\,c}-{\mathrm {e}}^{2\,a+2\,b\,x}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(c + b*x)^2*sinh(a + b*x),x)

[Out]

exp(a + b*x)/(2*b) + exp(- a - b*x)/(2*b) - (atan((exp(-a)*exp(2*c)*exp(b*x)*((-b^2)^(1/2) + exp(2*a)*exp(-2*c
)*(-b^2)^(1/2)))/(b*(exp(-2*a)*exp(2*c)*(2*exp(2*a)*exp(-2*c) + exp(4*a)*exp(-4*c) + 1))^(1/2)))*(exp(2*c - 2*
a)*(2*exp(2*a - 2*c) + exp(4*a - 4*c) + 1))^(1/2))/(-b^2)^(1/2) + (exp(a + b*x)*(exp(2*a - 2*c) - 1))/(b*(exp(
2*a - 2*c) - exp(2*a + 2*b*x)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(b*x+c)**2*sinh(b*x+a),x)

[Out]

Integral(sinh(a + b*x)*coth(b*x + c)**2, x)

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