3.433 \(\int \frac {\cosh ^2(c+d x) \coth (c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=57 \[ -\frac {\left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{a b^2 d}+\frac {\log (\sinh (c+d x))}{a d}+\frac {\sinh (c+d x)}{b d} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 57, 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 = {2837, 12, 894} \[ -\frac {\left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{a b^2 d}+\frac {\log (\sinh (c+d x))}{a d}+\frac {\sinh (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[Sinh[c + d*x]]/(a*d) - ((a^2 + b^2)*Log[a + b*Sinh[c + d*x]])/(a*b^2*d) + Sinh[c + d*x]/(b*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 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 48, normalized size = 0.84 \[ \frac {-\left (\frac {a}{b^2}+\frac {1}{a}\right ) \log (a+b \sinh (c+d x))+\frac {\log (\sinh (c+d x))}{a}+\frac {\sinh (c+d x)}{b}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.74, size = 203, normalized size = 3.56 \[ \frac {2 \, a^{2} d x \cosh \left (d x + c\right ) + a b \cosh \left (d x + c\right )^{2} + a b \sinh \left (d x + c\right )^{2} - a b - 2 \, {\left ({\left (a^{2} + b^{2}\right )} \cosh \left (d x + c\right ) + {\left (a^{2} + b^{2}\right )} \sinh \left (d x + c\right )\right )} \log \left (\frac {2 \, {\left (b \sinh \left (d x + c\right ) + a\right )}}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + 2 \, {\left (b^{2} \cosh \left (d x + c\right ) + b^{2} \sinh \left (d x + c\right )\right )} \log \left (\frac {2 \, \sinh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + 2 \, {\left (a^{2} d x + a b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{2 \, {\left (a b^{2} d \cosh \left (d x + c\right ) + a b^{2} d \sinh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a^2*d*x*cosh(d*x + c) + a*b*cosh(d*x + c)^2 + a*b*sinh(d*x + c)^2 - a*b - 2*((a^2 + b^2)*cosh(d*x + c)
+ (a^2 + b^2)*sinh(d*x + c))*log(2*(b*sinh(d*x + c) + a)/(cosh(d*x + c) - sinh(d*x + c))) + 2*(b^2*cosh(d*x +
c) + b^2*sinh(d*x + c))*log(2*sinh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) + 2*(a^2*d*x + a*b*cosh(d*x + c))
*sinh(d*x + c))/(a*b^2*d*cosh(d*x + c) + a*b^2*d*sinh(d*x + c))

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giac [A]  time = 0.19, size = 108, normalized size = 1.89 \[ \frac {\frac {2 \, a d x}{b^{2}} + \frac {e^{\left (d x + c\right )}}{b} - \frac {e^{\left (-d x - c\right )}}{b} + \frac {2 \, \log \left (e^{\left (d x + c\right )} + 1\right )}{a} + \frac {2 \, \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{a} - \frac {2 \, {\left (a^{2} + b^{2}\right )} \log \left ({\left | b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} - b \right |}\right )}{a b^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.14, size = 178, normalized size = 3.12 \[ -\frac {1}{d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,b^{2}}-\frac {1}{d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {a \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,b^{2}}-\frac {a \ln \left (\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right )}{d \,b^{2}}-\frac {\ln \left (\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right )}{d a}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.31, size = 130, normalized size = 2.28 \[ -\frac {{\left (d x + c\right )} a}{b^{2} d} + \frac {e^{\left (d x + c\right )}}{2 \, b d} - \frac {e^{\left (-d x - c\right )}}{2 \, b d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{a d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{a d} - \frac {{\left (a^{2} + b^{2}\right )} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{a b^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.49, size = 360, normalized size = 6.32 \[ \frac {{\mathrm {e}}^{c+d\,x}}{2\,b\,d}-\frac {{\mathrm {e}}^{-c-d\,x}}{2\,b\,d}-\frac {\ln \left (8\,a^5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-16\,b^5-16\,a^2\,b^3-4\,a^4\,b+16\,b^5\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+4\,a^4\,b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+32\,a^3\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+16\,a^2\,b^3\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+32\,a\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{a\,d}+\frac {\ln \left (4\,a^6+16\,b^6+32\,a^2\,b^4+20\,a^4\,b^2-4\,a^6\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}-16\,b^6\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}-32\,a^2\,b^4\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}-20\,a^4\,b^2\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\right )}{a\,d}+\frac {a\,x}{b^2}-\frac {a\,\ln \left (8\,a^5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-16\,b^5-16\,a^2\,b^3-4\,a^4\,b+16\,b^5\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+4\,a^4\,b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+32\,a^3\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+16\,a^2\,b^3\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+32\,a\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{b^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(c + d*x)/(2*b*d) - exp(- c - d*x)/(2*b*d) - log(8*a^5*exp(d*x)*exp(c) - 16*b^5 - 16*a^2*b^3 - 4*a^4*b + 16
*b^5*exp(2*c)*exp(2*d*x) + 4*a^4*b*exp(2*c)*exp(2*d*x) + 32*a^3*b^2*exp(d*x)*exp(c) + 16*a^2*b^3*exp(2*c)*exp(
2*d*x) + 32*a*b^4*exp(d*x)*exp(c))/(a*d) + log(4*a^6 + 16*b^6 + 32*a^2*b^4 + 20*a^4*b^2 - 4*a^6*exp(2*c)*exp(2
*d*x) - 16*b^6*exp(2*c)*exp(2*d*x) - 32*a^2*b^4*exp(2*c)*exp(2*d*x) - 20*a^4*b^2*exp(2*c)*exp(2*d*x))/(a*d) +
(a*x)/b^2 - (a*log(8*a^5*exp(d*x)*exp(c) - 16*b^5 - 16*a^2*b^3 - 4*a^4*b + 16*b^5*exp(2*c)*exp(2*d*x) + 4*a^4*
b*exp(2*c)*exp(2*d*x) + 32*a^3*b^2*exp(d*x)*exp(c) + 16*a^2*b^3*exp(2*c)*exp(2*d*x) + 32*a*b^4*exp(d*x)*exp(c)
))/(b^2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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