3.479 \(\int \frac {\coth (c+d x) \text {csch}^2(c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=72 \[ \frac {b^2 \log (\sinh (c+d x))}{a^3 d}-\frac {b^2 \log (a+b \sinh (c+d x))}{a^3 d}+\frac {b \text {csch}(c+d x)}{a^2 d}-\frac {\text {csch}^2(c+d x)}{2 a d} \]

[Out]

b*csch(d*x+c)/a^2/d-1/2*csch(d*x+c)^2/a/d+b^2*ln(sinh(d*x+c))/a^3/d-b^2*ln(a+b*sinh(d*x+c))/a^3/d

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Rubi [A]  time = 0.11, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2833, 12, 44} \[ \frac {b^2 \log (\sinh (c+d x))}{a^3 d}-\frac {b^2 \log (a+b \sinh (c+d x))}{a^3 d}+\frac {b \text {csch}(c+d x)}{a^2 d}-\frac {\text {csch}^2(c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Csch[c + d*x])/(a^2*d) - Csch[c + d*x]^2/(2*a*d) + (b^2*Log[Sinh[c + d*x]])/(a^3*d) - (b^2*Log[a + b*Sinh[c
+ d*x]])/(a^3*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 2833

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 60, normalized size = 0.83 \[ \frac {-a^2 \text {csch}^2(c+d x)+2 b^2 (\log (\sinh (c+d x))-\log (a+b \sinh (c+d x)))+2 a b \text {csch}(c+d x)}{2 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*b*Csch[c + d*x] - a^2*Csch[c + d*x]^2 + 2*b^2*(Log[Sinh[c + d*x]] - Log[a + b*Sinh[c + d*x]]))/(2*a^3*d)

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fricas [B]  time = 1.54, size = 545, normalized size = 7.57 \[ \frac {2 \, a b \cosh \left (d x + c\right )^{3} + 2 \, a b \sinh \left (d x + c\right )^{3} - 2 \, a^{2} \cosh \left (d x + c\right )^{2} - 2 \, a b \cosh \left (d x + c\right ) + 2 \, {\left (3 \, a b \cosh \left (d x + c\right ) - a^{2}\right )} \sinh \left (d x + c\right )^{2} - {\left (b^{2} \cosh \left (d x + c\right )^{4} + 4 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + b^{2} \sinh \left (d x + c\right )^{4} - 2 \, b^{2} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, b^{2} \cosh \left (d x + c\right )^{2} - b^{2}\right )} \sinh \left (d x + c\right )^{2} + b^{2} + 4 \, {\left (b^{2} \cosh \left (d x + c\right )^{3} - b^{2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )} \log \left (\frac {2 \, {\left (b \sinh \left (d x + c\right ) + a\right )}}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + {\left (b^{2} \cosh \left (d x + c\right )^{4} + 4 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + b^{2} \sinh \left (d x + c\right )^{4} - 2 \, b^{2} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, b^{2} \cosh \left (d x + c\right )^{2} - b^{2}\right )} \sinh \left (d x + c\right )^{2} + b^{2} + 4 \, {\left (b^{2} \cosh \left (d x + c\right )^{3} - b^{2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )} \log \left (\frac {2 \, \sinh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + 2 \, {\left (3 \, a b \cosh \left (d x + c\right )^{2} - 2 \, a^{2} \cosh \left (d x + c\right ) - a b\right )} \sinh \left (d x + c\right )}{a^{3} d \cosh \left (d x + c\right )^{4} + 4 \, a^{3} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a^{3} d \sinh \left (d x + c\right )^{4} - 2 \, a^{3} d \cosh \left (d x + c\right )^{2} + a^{3} d + 2 \, {\left (3 \, a^{3} d \cosh \left (d x + c\right )^{2} - a^{3} d\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left (a^{3} d \cosh \left (d x + c\right )^{3} - a^{3} d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.21, size = 134, normalized size = 1.86 \[ \frac {\frac {b^{2} \log \left (e^{\left (d x + c\right )} + 1\right )}{a^{3}} - \frac {b^{2} \log \left ({\left | b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} - b \right |}\right )}{a^{3}} + \frac {b^{2} \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{a^{3}} + \frac {2 \, {\left (a b e^{\left (3 \, d x + 3 \, c\right )} - a^{2} e^{\left (2 \, d x + 2 \, c\right )} - a b e^{\left (d x + c\right )}\right )}}{a^{3} {\left (e^{\left (d x + c\right )} + 1\right )}^{2} {\left (e^{\left (d x + c\right )} - 1\right )}^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 73, normalized size = 1.01 \[ -\frac {b^{2} \ln \left (a +b \sinh \left (d x +c \right )\right )}{a^{3} d}-\frac {1}{2 d a \sinh \left (d x +c \right )^{2}}+\frac {b^{2} \ln \left (\sinh \left (d x +c \right )\right )}{a^{3} d}+\frac {b}{d \,a^{2} \sinh \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^2*ln(a+b*sinh(d*x+c))/a^3/d-1/2/d/a/sinh(d*x+c)^2+b^2*ln(sinh(d*x+c))/a^3/d+1/d/a^2*b/sinh(d*x+c)

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maxima [B]  time = 0.32, size = 161, normalized size = 2.24 \[ -\frac {2 \, {\left (b e^{\left (-d x - c\right )} - a e^{\left (-2 \, d x - 2 \, c\right )} - b e^{\left (-3 \, d x - 3 \, c\right )}\right )}}{{\left (2 \, a^{2} e^{\left (-2 \, d x - 2 \, c\right )} - a^{2} e^{\left (-4 \, d x - 4 \, c\right )} - a^{2}\right )} d} - \frac {b^{2} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{a^{3} d} + \frac {b^{2} \log \left (e^{\left (-d x - c\right )} + 1\right )}{a^{3} d} + \frac {b^{2} \log \left (e^{\left (-d x - c\right )} - 1\right )}{a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.00, size = 470, normalized size = 6.53 \[ -\frac {\frac {2}{a\,d}-\frac {2\,b\,{\mathrm {e}}^{c+d\,x}}{a^2\,d}}{{\mathrm {e}}^{2\,c+2\,d\,x}-1}-\frac {\left (2\,\mathrm {atan}\left (-\frac {4\,a^3\,b^5\,\sqrt {-a^6\,d^2}+4\,a\,b^7\,\sqrt {-a^6\,d^2}-4\,b^8\,{\mathrm {e}}^{3\,c}\,{\mathrm {e}}^{3\,d\,x}\,\sqrt {-a^6\,d^2}+4\,b^8\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {-a^6\,d^2}-8\,a\,b^7\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\sqrt {-a^6\,d^2}+4\,a^2\,b^6\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {-a^6\,d^2}-8\,a^3\,b^5\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\sqrt {-a^6\,d^2}-4\,a^2\,b^6\,{\mathrm {e}}^{3\,c}\,{\mathrm {e}}^{3\,d\,x}\,\sqrt {-a^6\,d^2}}{4\,a^4\,b\,d\,{\left (b^4\right )}^{3/2}+4\,a^6\,b^3\,d\,\sqrt {b^4}}\right )+2\,\mathrm {atan}\left (\left (4\,a^4\,b^5\,d\,\sqrt {b^4}\,\sqrt {-a^6\,d^2}+4\,a^6\,b^3\,d\,\sqrt {b^4}\,\sqrt {-a^6\,d^2}\right )\,\left (\frac {1}{8\,a^5\,b^5\,d^2\,{\left (a^2+b^2\right )}^2}-{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (\frac {1}{16\,a^4\,b^6\,d^2\,{\left (a^2+b^2\right )}^2}-\frac {{\left (a^2+2\,b^2\right )}^2}{16\,a^8\,b^6\,d^2\,{\left (a^2+b^2\right )}^2}\right )+\frac {a^2+2\,b^2}{8\,a^7\,b^5\,d^2\,{\left (a^2+b^2\right )}^2}\right )\right )\right )\,\sqrt {b^4}}{\sqrt {-a^6\,d^2}}-\frac {2}{a\,d\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (2/(a*d) - (2*b*exp(c + d*x))/(a^2*d))/(exp(2*c + 2*d*x) - 1) - ((2*atan(-(4*a^3*b^5*(-a^6*d^2)^(1/2) + 4*a*
b^7*(-a^6*d^2)^(1/2) - 4*b^8*exp(3*c)*exp(3*d*x)*(-a^6*d^2)^(1/2) + 4*b^8*exp(d*x)*exp(c)*(-a^6*d^2)^(1/2) - 8
*a*b^7*exp(2*c)*exp(2*d*x)*(-a^6*d^2)^(1/2) + 4*a^2*b^6*exp(d*x)*exp(c)*(-a^6*d^2)^(1/2) - 8*a^3*b^5*exp(2*c)*
exp(2*d*x)*(-a^6*d^2)^(1/2) - 4*a^2*b^6*exp(3*c)*exp(3*d*x)*(-a^6*d^2)^(1/2))/(4*a^4*b*d*(b^4)^(3/2) + 4*a^6*b
^3*d*(b^4)^(1/2))) + 2*atan((4*a^4*b^5*d*(b^4)^(1/2)*(-a^6*d^2)^(1/2) + 4*a^6*b^3*d*(b^4)^(1/2)*(-a^6*d^2)^(1/
2))*(1/(8*a^5*b^5*d^2*(a^2 + b^2)^2) - exp(d*x)*exp(c)*(1/(16*a^4*b^6*d^2*(a^2 + b^2)^2) - (a^2 + 2*b^2)^2/(16
*a^8*b^6*d^2*(a^2 + b^2)^2)) + (a^2 + 2*b^2)/(8*a^7*b^5*d^2*(a^2 + b^2)^2))))*(b^4)^(1/2))/(-a^6*d^2)^(1/2) -
2/(a*d*(exp(4*c + 4*d*x) - 2*exp(2*c + 2*d*x) + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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