Integrand size = 27, antiderivative size = 130 \[ \int \frac {\text {csch}^3(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {b \arctan (\sinh (c+d x))}{\left (a^2+b^2\right ) d}+\frac {b \text {csch}(c+d x)}{a^2 d}-\frac {\text {csch}^2(c+d x)}{2 a d}+\frac {a \log (\cosh (c+d x))}{\left (a^2+b^2\right ) d}-\frac {\left (a^2-b^2\right ) \log (\sinh (c+d x))}{a^3 d}-\frac {b^4 \log (a+b \sinh (c+d x))}{a^3 \left (a^2+b^2\right ) d} \]
b*arctan(sinh(d*x+c))/(a^2+b^2)/d+b*csch(d*x+c)/a^2/d-1/2*csch(d*x+c)^2/a/ d+a*ln(cosh(d*x+c))/(a^2+b^2)/d-(a^2-b^2)*ln(sinh(d*x+c))/a^3/d-b^4*ln(a+b *sinh(d*x+c))/a^3/(a^2+b^2)/d
Time = 0.31 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.26 \[ \int \frac {\text {csch}^3(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\frac {2 b \text {csch}(c+d x)}{a^2}-\frac {\text {csch}^2(c+d x)}{a}-\frac {2 (a-b) (a+b) \log (\sinh (c+d x))}{a^3}+\frac {\left (a-\sqrt {-b^2}\right ) \log \left (\sqrt {-b^2}-b \sinh (c+d x)\right )}{a^2+b^2}-\frac {2 b^4 \log (a+b \sinh (c+d x))}{a^3 \left (a^2+b^2\right )}+\frac {\left (a+\sqrt {-b^2}\right ) \log \left (\sqrt {-b^2}+b \sinh (c+d x)\right )}{a^2+b^2}}{2 d} \]
((2*b*Csch[c + d*x])/a^2 - Csch[c + d*x]^2/a - (2*(a - b)*(a + b)*Log[Sinh [c + d*x]])/a^3 + ((a - Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Sinh[c + d*x]])/(a^ 2 + b^2) - (2*b^4*Log[a + b*Sinh[c + d*x]])/(a^3*(a^2 + b^2)) + ((a + Sqrt [-b^2])*Log[Sqrt[-b^2] + b*Sinh[c + d*x]])/(a^2 + b^2))/(2*d)
Time = 0.41 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3042, 26, 3316, 26, 27, 615, 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.
| \(\displaystyle \int \frac {\text {csch}^3(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {i}{\sin (i c+i d x)^3 \cos (i c+i d x) (a-i b \sin (i c+i d x))}dx\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle -i \int \frac {1}{\cos (i c+i d x) \sin (i c+i d x)^3 (a-i b \sin (i c+i d x))}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {i b \int -\frac {i \text {csch}^3(c+d x)}{(a+b \sinh (c+d x)) \left (\sinh ^2(c+d x) b^2+b^2\right )}d(b \sinh (c+d x))}{d}\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {b \int \frac {\text {csch}^3(c+d x)}{(a+b \sinh (c+d x)) \left (\sinh ^2(c+d x) b^2+b^2\right )}d(b \sinh (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^4 \int \frac {\text {csch}^3(c+d x)}{b^3 (a+b \sinh (c+d x)) \left (\sinh ^2(c+d x) b^2+b^2\right )}d(b \sinh (c+d x))}{d}\) |
| \(\Big \downarrow \) 615 |
| \(\displaystyle \frac {b^4 \int \left (\frac {\text {csch}^3(c+d x)}{a b^5}-\frac {\text {csch}^2(c+d x)}{a^2 b^4}+\frac {\left (b^2-a^2\right ) \text {csch}(c+d x)}{a^3 b^5}-\frac {1}{a^3 \left (a^2+b^2\right ) (a+b \sinh (c+d x))}+\frac {b^2+a \sinh (c+d x) b}{b^4 \left (a^2+b^2\right ) \left (\sinh ^2(c+d x) b^2+b^2\right )}\right )d(b \sinh (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^4 \left (\frac {\arctan (\sinh (c+d x))}{b^3 \left (a^2+b^2\right )}+\frac {\text {csch}(c+d x)}{a^2 b^3}+\frac {a \log \left (b^2 \sinh ^2(c+d x)+b^2\right )}{2 b^4 \left (a^2+b^2\right )}-\frac {\log (a+b \sinh (c+d x))}{a^3 \left (a^2+b^2\right )}-\frac {\left (a^2-b^2\right ) \log (b \sinh (c+d x))}{a^3 b^4}-\frac {\text {csch}^2(c+d x)}{2 a b^4}\right )}{d}\) |
(b^4*(ArcTan[Sinh[c + d*x]]/(b^3*(a^2 + b^2)) + Csch[c + d*x]/(a^2*b^3) - Csch[c + d*x]^2/(2*a*b^4) - ((a^2 - b^2)*Log[b*Sinh[c + d*x]])/(a^3*b^4) - Log[a + b*Sinh[c + d*x]]/(a^3*(a^2 + b^2)) + (a*Log[b^2 + b^2*Sinh[c + d* x]^2])/(2*b^4*(a^2 + b^2))))/d
3.5.94.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^m*(c + (d/b)*x)^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]
Time = 7.07 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.43
| method | result | size |
| derivativedivides | \(\frac {-\frac {\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}{2}+2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{2}}-\frac {b^{4} \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{a^{3} \left (a^{2}+b^{2}\right )}+\frac {4 a \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+8 b \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{2}+4 b^{2}}-\frac {1}{8 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (-4 a^{2}+4 b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{3}}+\frac {b}{2 a^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\) | \(186\) |
| default | \(\frac {-\frac {\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}{2}+2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{2}}-\frac {b^{4} \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{a^{3} \left (a^{2}+b^{2}\right )}+\frac {4 a \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+8 b \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{2}+4 b^{2}}-\frac {1}{8 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (-4 a^{2}+4 b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{3}}+\frac {b}{2 a^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\) | \(186\) |
| risch | \(-\frac {2 a \,d^{2} x}{a^{2} d^{2}+b^{2} d^{2}}-\frac {2 a d c}{a^{2} d^{2}+b^{2} d^{2}}+\frac {2 x}{a}+\frac {2 c}{d a}-\frac {2 b^{2} x}{a^{3}}-\frac {2 b^{2} c}{d \,a^{3}}+\frac {2 b^{4} x}{a^{3} \left (a^{2}+b^{2}\right )}+\frac {2 b^{4} c}{d \,a^{3} \left (a^{2}+b^{2}\right )}-\frac {2 \,{\mathrm e}^{d x +c} \left (-b \,{\mathrm e}^{2 d x +2 c}+a \,{\mathrm e}^{d x +c}+b \right )}{a^{2} d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}-\frac {i \ln \left ({\mathrm e}^{d x +c}-i\right ) b}{\left (a^{2}+b^{2}\right ) d}+\frac {\ln \left ({\mathrm e}^{d x +c}+i\right ) a}{\left (a^{2}+b^{2}\right ) d}+\frac {i \ln \left ({\mathrm e}^{d x +c}+i\right ) b}{\left (a^{2}+b^{2}\right ) d}+\frac {\ln \left ({\mathrm e}^{d x +c}-i\right ) a}{\left (a^{2}+b^{2}\right ) d}-\frac {\ln \left ({\mathrm e}^{2 d x +2 c}-1\right )}{d a}+\frac {b^{2} \ln \left ({\mathrm e}^{2 d x +2 c}-1\right )}{d \,a^{3}}-\frac {b^{4} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{b}-1\right )}{d \,a^{3} \left (a^{2}+b^{2}\right )}\) | \(358\) |
1/d*(-1/4/a^2*(1/2*tanh(1/2*d*x+1/2*c)^2*a+2*b*tanh(1/2*d*x+1/2*c))-b^4/a^ 3/(a^2+b^2)*ln(tanh(1/2*d*x+1/2*c)^2*a-2*b*tanh(1/2*d*x+1/2*c)-a)+1/4/(a^2 +b^2)*(4*a*ln(1+tanh(1/2*d*x+1/2*c)^2)+8*b*arctan(tanh(1/2*d*x+1/2*c)))-1/ 8/a/tanh(1/2*d*x+1/2*c)^2+1/4/a^3*(-4*a^2+4*b^2)*ln(tanh(1/2*d*x+1/2*c))+1 /2*b/a^2/tanh(1/2*d*x+1/2*c))
Leaf count of result is larger than twice the leaf count of optimal. 1035 vs. \(2 (128) = 256\).
Time = 0.35 (sec) , antiderivative size = 1035, normalized size of antiderivative = 7.96 \[ \int \frac {\text {csch}^3(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=\text {Too large to display} \]
(2*(a^3*b + a*b^3)*cosh(d*x + c)^3 + 2*(a^3*b + a*b^3)*sinh(d*x + c)^3 - 2 *(a^4 + a^2*b^2)*cosh(d*x + c)^2 - 2*(a^4 + a^2*b^2 - 3*(a^3*b + a*b^3)*co sh(d*x + c))*sinh(d*x + c)^2 + 2*(a^3*b*cosh(d*x + c)^4 + 4*a^3*b*cosh(d*x + c)*sinh(d*x + c)^3 + a^3*b*sinh(d*x + c)^4 - 2*a^3*b*cosh(d*x + c)^2 + a^3*b + 2*(3*a^3*b*cosh(d*x + c)^2 - a^3*b)*sinh(d*x + c)^2 + 4*(a^3*b*cos h(d*x + c)^3 - a^3*b*cosh(d*x + c))*sinh(d*x + c))*arctan(cosh(d*x + c) + sinh(d*x + c)) - 2*(a^3*b + a*b^3)*cosh(d*x + c) - (b^4*cosh(d*x + c)^4 + 4*b^4*cosh(d*x + c)*sinh(d*x + c)^3 + b^4*sinh(d*x + c)^4 - 2*b^4*cosh(d*x + c)^2 + b^4 + 2*(3*b^4*cosh(d*x + c)^2 - b^4)*sinh(d*x + c)^2 + 4*(b^4*c osh(d*x + c)^3 - b^4*cosh(d*x + c))*sinh(d*x + c))*log(2*(b*sinh(d*x + c) + a)/(cosh(d*x + c) - sinh(d*x + c))) + (a^4*cosh(d*x + c)^4 + 4*a^4*cosh( d*x + c)*sinh(d*x + c)^3 + a^4*sinh(d*x + c)^4 - 2*a^4*cosh(d*x + c)^2 + a ^4 + 2*(3*a^4*cosh(d*x + c)^2 - a^4)*sinh(d*x + c)^2 + 4*(a^4*cosh(d*x + c )^3 - a^4*cosh(d*x + c))*sinh(d*x + c))*log(2*cosh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) - ((a^4 - b^4)*cosh(d*x + c)^4 + 4*(a^4 - b^4)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^4 - b^4)*sinh(d*x + c)^4 + a^4 - b^4 - 2*(a^4 - b^4)*cosh(d*x + c)^2 - 2*(a^4 - b^4 - 3*(a^4 - b^4)*cosh(d*x + c)^2)*sinh (d*x + c)^2 + 4*((a^4 - b^4)*cosh(d*x + c)^3 - (a^4 - b^4)*cosh(d*x + c))* sinh(d*x + c))*log(2*sinh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) - 2*(a ^3*b + a*b^3 - 3*(a^3*b + a*b^3)*cosh(d*x + c)^2 + 2*(a^4 + a^2*b^2)*co...
Timed out. \[ \int \frac {\text {csch}^3(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.82 \[ \int \frac {\text {csch}^3(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {b^{4} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{5} + a^{3} b^{2}\right )} d} - \frac {2 \, b \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac {a \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{2} + b^{2}\right )} d} - \frac {2 \, {\left (b e^{\left (-d x - c\right )} - a e^{\left (-2 \, d x - 2 \, c\right )} - b e^{\left (-3 \, d x - 3 \, c\right )}\right )}}{{\left (2 \, a^{2} e^{\left (-2 \, d x - 2 \, c\right )} - a^{2} e^{\left (-4 \, d x - 4 \, c\right )} - a^{2}\right )} d} - \frac {{\left (a^{2} - b^{2}\right )} \log \left (e^{\left (-d x - c\right )} + 1\right )}{a^{3} d} - \frac {{\left (a^{2} - b^{2}\right )} \log \left (e^{\left (-d x - c\right )} - 1\right )}{a^{3} d} \]
-b^4*log(-2*a*e^(-d*x - c) + b*e^(-2*d*x - 2*c) - b)/((a^5 + a^3*b^2)*d) - 2*b*arctan(e^(-d*x - c))/((a^2 + b^2)*d) + a*log(e^(-2*d*x - 2*c) + 1)/(( a^2 + b^2)*d) - 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) - (a^2 - b^2) *log(e^(-d*x - c) + 1)/(a^3*d) - (a^2 - b^2)*log(e^(-d*x - c) - 1)/(a^3*d)
Leaf count of result is larger than twice the leaf count of optimal. 263 vs. \(2 (128) = 256\).
Time = 0.29 (sec) , antiderivative size = 263, normalized size of antiderivative = 2.02 \[ \int \frac {\text {csch}^3(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {\frac {2 \, b^{5} \log \left ({\left | b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{a^{5} b + a^{3} b^{3}} - \frac {{\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )} b}{a^{2} + b^{2}} - \frac {a \log \left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}{a^{2} + b^{2}} + \frac {2 \, {\left (a^{2} - b^{2}\right )} \log \left ({\left | e^{\left (d x + c\right )} - e^{\left (-d x - c\right )} \right |}\right )}{a^{3}} - \frac {3 \, a^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} - 3 \, b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4 \, a b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} - 4 \, a^{2}}{a^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2}}}{2 \, d} \]
-1/2*(2*b^5*log(abs(b*(e^(d*x + c) - e^(-d*x - c)) + 2*a))/(a^5*b + a^3*b^ 3) - (pi + 2*arctan(1/2*(e^(2*d*x + 2*c) - 1)*e^(-d*x - c)))*b/(a^2 + b^2) - a*log((e^(d*x + c) - e^(-d*x - c))^2 + 4)/(a^2 + b^2) + 2*(a^2 - b^2)*l og(abs(e^(d*x + c) - e^(-d*x - c)))/a^3 - (3*a^2*(e^(d*x + c) - e^(-d*x - c))^2 - 3*b^2*(e^(d*x + c) - e^(-d*x - c))^2 + 4*a*b*(e^(d*x + c) - e^(-d* x - c)) - 4*a^2)/(a^3*(e^(d*x + c) - e^(-d*x - c))^2))/d
Time = 4.63 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.51 \[ \int \frac {\text {csch}^3(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\ln \left ({\mathrm {e}}^{c+d\,x}+1{}\mathrm {i}\right )}{a\,d-b\,d\,1{}\mathrm {i}}-\frac {\frac {2}{a\,d}-\frac {2\,b\,{\mathrm {e}}^{c+d\,x}}{a^2\,d}}{{\mathrm {e}}^{2\,c+2\,d\,x}-1}-\frac {2}{a\,d\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {b^4\,\ln \left (2\,a\,{\mathrm {e}}^{c+d\,x}-b+b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{d\,a^5+d\,a^3\,b^2}-\frac {\ln \left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )\,\left (a^2-b^2\right )}{a^3\,d}+\frac {\ln \left (1+{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{-b\,d+a\,d\,1{}\mathrm {i}} \]
log(exp(c + d*x) + 1i)/(a*d - b*d*1i) - (2/(a*d) - (2*b*exp(c + d*x))/(a^2 *d))/(exp(2*c + 2*d*x) - 1) + (log(exp(c + d*x)*1i + 1)*1i)/(a*d*1i - b*d) - 2/(a*d*(exp(4*c + 4*d*x) - 2*exp(2*c + 2*d*x) + 1)) - (b^4*log(2*a*exp( c + d*x) - b + b*exp(2*c + 2*d*x)))/(a^5*d + a^3*b^2*d) - (log(exp(2*c + 2 *d*x) - 1)*(a^2 - b^2))/(a^3*d)