Integrand size = 25, antiderivative size = 50 \[ \int \frac {\coth (c+d x) \text {csch}(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {\text {csch}(c+d x)}{a d}-\frac {b \log (\sinh (c+d x))}{a^2 d}+\frac {b \log (a+b \sinh (c+d x))}{a^2 d} \] Output:
-csch(d*x+c)/a/d-b*ln(sinh(d*x+c))/a^2/d+b*ln(a+b*sinh(d*x+c))/a^2/d
Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00 \[ \int \frac {\coth (c+d x) \text {csch}(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {\text {csch}(c+d x)}{a d}-\frac {b \log (\sinh (c+d x))}{a^2 d}+\frac {b \log (a+b \sinh (c+d x))}{a^2 d} \] Input:
Integrate[(Coth[c + d*x]*Csch[c + d*x])/(a + b*Sinh[c + d*x]),x]
Output:
-(Csch[c + d*x]/(a*d)) - (b*Log[Sinh[c + d*x]])/(a^2*d) + (b*Log[a + b*Sin h[c + d*x]])/(a^2*d)
Time = 0.29 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 25, 3312, 25, 27, 54, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\coth (c+d x) \text {csch}(c+d x)}{a+b \sinh (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\cos (i c+i d x)}{\sin (i c+i d x)^2 (a-i b \sin (i c+i d x))}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cos (i c+i d x)}{\sin (i c+i d x)^2 (a-i b \sin (i c+i d x))}dx\) |
\(\Big \downarrow \) 3312 |
\(\displaystyle -\frac {\int -\frac {\text {csch}^2(c+d x)}{a+b \sinh (c+d x)}d(b \sinh (c+d x))}{b d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {\text {csch}^2(c+d x)}{a+b \sinh (c+d x)}d(b \sinh (c+d x))}{b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b \int \frac {\text {csch}^2(c+d x)}{b^2 (a+b \sinh (c+d x))}d(b \sinh (c+d x))}{d}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {b \int \left (\frac {\text {csch}^2(c+d x)}{a b^2}-\frac {\text {csch}(c+d x)}{a^2 b}+\frac {1}{a^2 (a+b \sinh (c+d x))}\right )d(b \sinh (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b \left (-\frac {\log (b \sinh (c+d x))}{a^2}+\frac {\log (a+b \sinh (c+d x))}{a^2}-\frac {\text {csch}(c+d x)}{a b}\right )}{d}\) |
Input:
Int[(Coth[c + d*x]*Csch[c + d*x])/(a + b*Sinh[c + d*x]),x]
Output:
(b*(-(Csch[c + d*x]/(a*b)) - Log[b*Sinh[c + d*x]]/a^2 + Log[a + b*Sinh[c + d*x]]/a^2))/d
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*(( c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b*f) Su bst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
Time = 0.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.70
method | result | size |
derivativedivides | \(-\frac {\operatorname {csch}\left (d x +c \right )}{a d}+\frac {b \ln \left (a \,\operatorname {csch}\left (d x +c \right )+b \right )}{d \,a^{2}}\) | \(35\) |
default | \(-\frac {\operatorname {csch}\left (d x +c \right )}{a d}+\frac {b \ln \left (a \,\operatorname {csch}\left (d x +c \right )+b \right )}{d \,a^{2}}\) | \(35\) |
risch | \(-\frac {2 \,{\mathrm e}^{d x +c}}{d a \left ({\mathrm e}^{2 d x +2 c}-1\right )}-\frac {b \ln \left ({\mathrm e}^{2 d x +2 c}-1\right )}{a^{2} d}+\frac {b \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{b}-1\right )}{a^{2} d}\) | \(82\) |
Input:
int(coth(d*x+c)*csch(d*x+c)/(a+b*sinh(d*x+c)),x,method=_RETURNVERBOSE)
Output:
-csch(d*x+c)/a/d+1/d*b/a^2*ln(a*csch(d*x+c)+b)
Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (50) = 100\).
Time = 0.09 (sec) , antiderivative size = 211, normalized size of antiderivative = 4.22 \[ \int \frac {\coth (c+d x) \text {csch}(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {2 \, a \cosh \left (d x + c\right ) - {\left (b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} - b\right )} \log \left (\frac {2 \, {\left (b \sinh \left (d x + c\right ) + a\right )}}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + {\left (b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} - b\right )} \log \left (\frac {2 \, \sinh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + 2 \, a \sinh \left (d x + c\right )}{a^{2} d \cosh \left (d x + c\right )^{2} + 2 \, a^{2} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a^{2} d \sinh \left (d x + c\right )^{2} - a^{2} d} \] Input:
integrate(coth(d*x+c)*csch(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="fricas")
Output:
-(2*a*cosh(d*x + c) - (b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 - b)*log(2*(b*sinh(d*x + c) + a)/(cosh(d*x + c) - sin h(d*x + c))) + (b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*si nh(d*x + c)^2 - b)*log(2*sinh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) + 2*a*sinh(d*x + c))/(a^2*d*cosh(d*x + c)^2 + 2*a^2*d*cosh(d*x + c)*sinh(d*x + c) + a^2*d*sinh(d*x + c)^2 - a^2*d)
\[ \int \frac {\coth (c+d x) \text {csch}(c+d x)}{a+b \sinh (c+d x)} \, dx=\int \frac {\coth {\left (c + d x \right )} \operatorname {csch}{\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \] Input:
integrate(coth(d*x+c)*csch(d*x+c)/(a+b*sinh(d*x+c)),x)
Output:
Integral(coth(c + d*x)*csch(c + d*x)/(a + b*sinh(c + d*x)), x)
Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (50) = 100\).
Time = 0.04 (sec) , antiderivative size = 110, normalized size of antiderivative = 2.20 \[ \int \frac {\coth (c+d x) \text {csch}(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {2 \, e^{\left (-d x - c\right )}}{{\left (a e^{\left (-2 \, d x - 2 \, c\right )} - a\right )} d} + \frac {b \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{a^{2} d} - \frac {b \log \left (e^{\left (-d x - c\right )} + 1\right )}{a^{2} d} - \frac {b \log \left (e^{\left (-d x - c\right )} - 1\right )}{a^{2} d} \] Input:
integrate(coth(d*x+c)*csch(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="maxima")
Output:
2*e^(-d*x - c)/((a*e^(-2*d*x - 2*c) - a)*d) + b*log(-2*a*e^(-d*x - c) + b* e^(-2*d*x - 2*c) - b)/(a^2*d) - b*log(e^(-d*x - c) + 1)/(a^2*d) - b*log(e^ (-d*x - c) - 1)/(a^2*d)
Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (50) = 100\).
Time = 0.13 (sec) , antiderivative size = 110, normalized size of antiderivative = 2.20 \[ \int \frac {\coth (c+d x) \text {csch}(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\frac {b \log \left ({\left | b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{a^{2}} - \frac {b \log \left ({\left | e^{\left (d x + c\right )} - e^{\left (-d x - c\right )} \right |}\right )}{a^{2}} + \frac {b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} - 2 \, a}{a^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}}{d} \] Input:
integrate(coth(d*x+c)*csch(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="giac")
Output:
(b*log(abs(b*(e^(d*x + c) - e^(-d*x - c)) + 2*a))/a^2 - b*log(abs(e^(d*x + c) - e^(-d*x - c)))/a^2 + (b*(e^(d*x + c) - e^(-d*x - c)) - 2*a)/(a^2*(e^ (d*x + c) - e^(-d*x - c))))/d
Time = 1.93 (sec) , antiderivative size = 409, normalized size of antiderivative = 8.18 \[ \int \frac {\coth (c+d x) \text {csch}(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\left (2\,\mathrm {atan}\left (\left (4\,a^3\,b\,d\,{\left (b^2\right )}^{5/2}\,\sqrt {-a^4\,d^2}+4\,a^5\,b\,d\,{\left (b^2\right )}^{3/2}\,\sqrt {-a^4\,d^2}\right )\,\left (\frac {1}{8\,a^3\,b^4\,d^2\,{\left (a^2+b^2\right )}^2}-{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (\frac {1}{16\,a^2\,b^5\,d^2\,{\left (a^2+b^2\right )}^2}-\frac {{\left (a^2+2\,b^2\right )}^2}{16\,a^6\,b^5\,d^2\,{\left (a^2+b^2\right )}^2}\right )+\frac {a^2+2\,b^2}{8\,a^5\,b^4\,d^2\,{\left (a^2+b^2\right )}^2}\right )\right )+2\,\mathrm {atan}\left (-\frac {4\,a^3\,b^5\,\sqrt {-a^4\,d^2}+4\,a\,b^7\,\sqrt {-a^4\,d^2}-4\,b^8\,{\mathrm {e}}^{3\,c}\,{\mathrm {e}}^{3\,d\,x}\,\sqrt {-a^4\,d^2}+4\,b^8\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {-a^4\,d^2}-8\,a\,b^7\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\sqrt {-a^4\,d^2}+4\,a^2\,b^6\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {-a^4\,d^2}-8\,a^3\,b^5\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\sqrt {-a^4\,d^2}-4\,a^2\,b^6\,{\mathrm {e}}^{3\,c}\,{\mathrm {e}}^{3\,d\,x}\,\sqrt {-a^4\,d^2}}{b^4\,\left (4\,a^3\,d\,{\left (b^2\right )}^{3/2}+4\,a^5\,d\,\sqrt {b^2}\right )}\right )\right )\,\sqrt {b^2}}{\sqrt {-a^4\,d^2}}-\frac {1}{a\,d\,\mathrm {sinh}\left (c+d\,x\right )} \] Input:
int(coth(c + d*x)/(sinh(c + d*x)*(a + b*sinh(c + d*x))),x)
Output:
((2*atan((4*a^3*b*d*(b^2)^(5/2)*(-a^4*d^2)^(1/2) + 4*a^5*b*d*(b^2)^(3/2)*( -a^4*d^2)^(1/2))*(1/(8*a^3*b^4*d^2*(a^2 + b^2)^2) - exp(d*x)*exp(c)*(1/(16 *a^2*b^5*d^2*(a^2 + b^2)^2) - (a^2 + 2*b^2)^2/(16*a^6*b^5*d^2*(a^2 + b^2)^ 2)) + (a^2 + 2*b^2)/(8*a^5*b^4*d^2*(a^2 + b^2)^2))) + 2*atan(-(4*a^3*b^5*( -a^4*d^2)^(1/2) + 4*a*b^7*(-a^4*d^2)^(1/2) - 4*b^8*exp(3*c)*exp(3*d*x)*(-a ^4*d^2)^(1/2) + 4*b^8*exp(d*x)*exp(c)*(-a^4*d^2)^(1/2) - 8*a*b^7*exp(2*c)* exp(2*d*x)*(-a^4*d^2)^(1/2) + 4*a^2*b^6*exp(d*x)*exp(c)*(-a^4*d^2)^(1/2) - 8*a^3*b^5*exp(2*c)*exp(2*d*x)*(-a^4*d^2)^(1/2) - 4*a^2*b^6*exp(3*c)*exp(3 *d*x)*(-a^4*d^2)^(1/2))/(b^4*(4*a^3*d*(b^2)^(3/2) + 4*a^5*d*(b^2)^(1/2)))) )*(b^2)^(1/2))/(-a^4*d^2)^(1/2) - 1/(a*d*sinh(c + d*x))
Time = 0.22 (sec) , antiderivative size = 171, normalized size of antiderivative = 3.42 \[ \int \frac {\coth (c+d x) \text {csch}(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {-e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}-1\right ) b -e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}+1\right ) b +e^{2 d x +2 c} \mathrm {log}\left (e^{2 d x +2 c} b +2 e^{d x +c} a -b \right ) b -2 e^{d x +c} a +\mathrm {log}\left (e^{d x +c}-1\right ) b +\mathrm {log}\left (e^{d x +c}+1\right ) b -\mathrm {log}\left (e^{2 d x +2 c} b +2 e^{d x +c} a -b \right ) b}{a^{2} d \left (e^{2 d x +2 c}-1\right )} \] Input:
int(coth(d*x+c)*csch(d*x+c)/(a+b*sinh(d*x+c)),x)
Output:
( - e**(2*c + 2*d*x)*log(e**(c + d*x) - 1)*b - e**(2*c + 2*d*x)*log(e**(c + d*x) + 1)*b + e**(2*c + 2*d*x)*log(e**(2*c + 2*d*x)*b + 2*e**(c + d*x)*a - b)*b - 2*e**(c + d*x)*a + log(e**(c + d*x) - 1)*b + log(e**(c + d*x) + 1)*b - log(e**(2*c + 2*d*x)*b + 2*e**(c + d*x)*a - b)*b)/(a**2*d*(e**(2*c + 2*d*x) - 1))