Integrand size = 27, antiderivative size = 89 \[ \int \frac {\sinh ^2(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {b \arctan (\sinh (c+d x))}{\left (a^2+b^2\right ) d}-\frac {a \log (\cosh (c+d x))}{\left (a^2+b^2\right ) d}-\frac {a^3 \log (a+b \sinh (c+d x))}{b^2 \left (a^2+b^2\right ) d}+\frac {\sinh (c+d x)}{b d} \] Output:
-b*arctan(sinh(d*x+c))/(a^2+b^2)/d-a*ln(cosh(d*x+c))/(a^2+b^2)/d-a^3*ln(a+ b*sinh(d*x+c))/b^2/(a^2+b^2)/d+sinh(d*x+c)/b/d
Result contains complex when optimal does not.
Time = 0.13 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.02 \[ \int \frac {\sinh ^2(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {\frac {\log (i-\sinh (c+d x))}{a+i b}+\frac {\log (i+\sinh (c+d x))}{a-i b}+\frac {2 a^3 \log (a+b \sinh (c+d x))}{b^2 \left (a^2+b^2\right )}-\frac {2 \sinh (c+d x)}{b}}{2 d} \] Input:
Integrate[(Sinh[c + d*x]^2*Tanh[c + d*x])/(a + b*Sinh[c + d*x]),x]
Output:
-1/2*(Log[I - Sinh[c + d*x]]/(a + I*b) + Log[I + Sinh[c + d*x]]/(a - I*b) + (2*a^3*Log[a + b*Sinh[c + d*x]])/(b^2*(a^2 + b^2)) - (2*Sinh[c + d*x])/b )/d
Time = 0.43 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.09, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 26, 3316, 26, 27, 604, 25, 2160, 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.
| \(\displaystyle \int \frac {\sinh ^2(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx\) |
| \(\Big \downarrow \) 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 |
| \(\displaystyle i \int \frac {\sin (i c+i d x)^3}{\cos (i c+i d x) (a-i b \sin (i c+i d x))}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle -\frac {i b \int \frac {i \sinh ^3(c+d x)}{(a+b \sinh (c+d x)) \left (\sinh ^2(c+d x) b^2+b^2\right )}d(b \sinh (c+d x))}{d}\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {b \int \frac {\sinh ^3(c+d x)}{(a+b \sinh (c+d x)) \left (\sinh ^2(c+d x) b^2+b^2\right )}d(b \sinh (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {b^3 \sinh ^3(c+d x)}{(a+b \sinh (c+d x)) \left (\sinh ^2(c+d x) b^2+b^2\right )}d(b \sinh (c+d x))}{b^2 d}\) |
| \(\Big \downarrow \) 604 |
| \(\displaystyle \frac {\int -\frac {\sinh (c+d x) b^3+a \sinh ^2(c+d x) b^2+a b^2}{(a+b \sinh (c+d x)) \left (\sinh ^2(c+d x) b^2+b^2\right )}d(b \sinh (c+d x))+a+b \sinh (c+d x)}{b^2 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\int \frac {\sinh (c+d x) b^3+a \sinh ^2(c+d x) b^2+a b^2}{(a+b \sinh (c+d x)) \left (\sinh ^2(c+d x) b^2+b^2\right )}d(b \sinh (c+d x))+a+b \sinh (c+d x)}{b^2 d}\) |
| \(\Big \downarrow \) 2160 |
| \(\displaystyle \frac {-\int \left (\frac {a^3}{\left (a^2+b^2\right ) (a+b \sinh (c+d x))}+\frac {b^4+a \sinh (c+d x) b^3}{\left (a^2+b^2\right ) \left (\sinh ^2(c+d x) b^2+b^2\right )}\right )d(b \sinh (c+d x))+a+b \sinh (c+d x)}{b^2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {b^3 \arctan (\sinh (c+d x))}{a^2+b^2}-\frac {a b^2 \log \left (b^2 \sinh ^2(c+d x)+b^2\right )}{2 \left (a^2+b^2\right )}-\frac {a^3 \log (a+b \sinh (c+d x))}{a^2+b^2}+a+b \sinh (c+d x)}{b^2 d}\) |
Input:
Int[(Sinh[c + d*x]^2*Tanh[c + d*x])/(a + b*Sinh[c + d*x]),x]
Output:
(a - (b^3*ArcTan[Sinh[c + d*x]])/(a^2 + b^2) - (a^3*Log[a + b*Sinh[c + d*x ]])/(a^2 + b^2) - (a*b^2*Log[b^2 + b^2*Sinh[c + d*x]^2])/(2*(a^2 + b^2)) + b*Sinh[c + d*x])/(b^2*d)
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[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[(c + d*x)^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*d^(m - 1)*(m + n + 2*p + 1))), x] + Simp[1/(b*d^m*(m + n + 2*p + 1)) Int[(c + d*x)^n*(a + b* x^2)^p*ExpandToSum[b*d^m*(m + n + 2*p + 1)*x^m - b*(m + n + 2*p + 1)*(c + d *x)^m - (c + d*x)^(m - 2)*(a*d^2*(m + n - 1) - b*c^2*(m + n + 2*p + 1) - 2* b*c*d*(m + n + p)*x), x], x], x] /; FreeQ[{a, b, c, d, n, p}, x] && IGtQ[m, 1] && NeQ[m + n + 2*p + 1, 0] && IntegerQ[2*p]
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
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 = 1.09 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.90
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{2}}-\frac {a^{3} \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{b^{2} \left (a^{2}+b^{2}\right )}-\frac {1}{b \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {a \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}+\frac {-8 a \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-16 b \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a^{2}+8 b^{2}}}{d}\) | \(169\) |
| default | \(\frac {-\frac {1}{b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{2}}-\frac {a^{3} \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{b^{2} \left (a^{2}+b^{2}\right )}-\frac {1}{b \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {a \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}+\frac {-8 a \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-16 b \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a^{2}+8 b^{2}}}{d}\) | \(169\) |
| risch | \(-\frac {a x}{b^{2}}+\frac {{\mathrm e}^{d x +c}}{2 b d}-\frac {{\mathrm e}^{-d x -c}}{2 b d}+\frac {2 d^{2} a x}{a^{2} d^{2}+b^{2} d^{2}}+\frac {2 d a c}{a^{2} d^{2}+b^{2} d^{2}}+\frac {2 a^{3} x}{b^{2} \left (a^{2}+b^{2}\right )}+\frac {2 a^{3} c}{b^{2} d \left (a^{2}+b^{2}\right )}+\frac {i \ln \left ({\mathrm e}^{d x +c}-i\right ) b}{\left (a^{2}+b^{2}\right ) d}-\frac {\ln \left ({\mathrm e}^{d x +c}-i\right ) a}{\left (a^{2}+b^{2}\right ) d}-\frac {i \ln \left ({\mathrm e}^{d x +c}+i\right ) b}{\left (a^{2}+b^{2}\right ) d}-\frac {\ln \left ({\mathrm e}^{d x +c}+i\right ) a}{\left (a^{2}+b^{2}\right ) d}-\frac {a^{3} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{b}-1\right )}{b^{2} d \left (a^{2}+b^{2}\right )}\) | \(271\) |
Input:
int(sinh(d*x+c)^2*tanh(d*x+c)/(a+b*sinh(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(-1/b/(tanh(1/2*d*x+1/2*c)-1)+a/b^2*ln(tanh(1/2*d*x+1/2*c)-1)-a^3/b^2/ (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/b/(1+tan h(1/2*d*x+1/2*c))+a/b^2*ln(1+tanh(1/2*d*x+1/2*c))+16/(8*a^2+8*b^2)*(-1/2*a *ln(1+tanh(1/2*d*x+1/2*c)^2)-b*arctan(tanh(1/2*d*x+1/2*c))))
Leaf count of result is larger than twice the leaf count of optimal. 288 vs. \(2 (89) = 178\).
Time = 0.10 (sec) , antiderivative size = 288, normalized size of antiderivative = 3.24 \[ \int \frac {\sinh ^2(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {2 \, {\left (a^{3} + a b^{2}\right )} d x \cosh \left (d x + c\right ) - a^{2} b - b^{3} + {\left (a^{2} b + b^{3}\right )} \cosh \left (d x + c\right )^{2} + {\left (a^{2} b + b^{3}\right )} \sinh \left (d x + c\right )^{2} - 4 \, {\left (b^{3} \cosh \left (d x + c\right ) + b^{3} \sinh \left (d x + c\right )\right )} \arctan \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) - 2 \, {\left (a^{3} \cosh \left (d x + c\right ) + a^{3} \sinh \left (d x + c\right )\right )} \log \left (\frac {2 \, {\left (b \sinh \left (d x + c\right ) + a\right )}}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) - 2 \, {\left (a b^{2} \cosh \left (d x + c\right ) + a b^{2} \sinh \left (d x + c\right )\right )} \log \left (\frac {2 \, \cosh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + 2 \, {\left ({\left (a^{3} + a b^{2}\right )} d x + {\left (a^{2} b + b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{2 \, {\left ({\left (a^{2} b^{2} + b^{4}\right )} d \cosh \left (d x + c\right ) + {\left (a^{2} b^{2} + b^{4}\right )} d \sinh \left (d x + c\right )\right )}} \] Input:
integrate(sinh(d*x+c)^2*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="fricas ")
Output:
1/2*(2*(a^3 + a*b^2)*d*x*cosh(d*x + c) - a^2*b - b^3 + (a^2*b + b^3)*cosh( d*x + c)^2 + (a^2*b + b^3)*sinh(d*x + c)^2 - 4*(b^3*cosh(d*x + c) + b^3*si nh(d*x + c))*arctan(cosh(d*x + c) + sinh(d*x + c)) - 2*(a^3*cosh(d*x + c) + a^3*sinh(d*x + c))*log(2*(b*sinh(d*x + c) + a)/(cosh(d*x + c) - sinh(d*x + c))) - 2*(a*b^2*cosh(d*x + c) + a*b^2*sinh(d*x + c))*log(2*cosh(d*x + c )/(cosh(d*x + c) - sinh(d*x + c))) + 2*((a^3 + a*b^2)*d*x + (a^2*b + b^3)* cosh(d*x + c))*sinh(d*x + c))/((a^2*b^2 + b^4)*d*cosh(d*x + c) + (a^2*b^2 + b^4)*d*sinh(d*x + c))
\[ \int \frac {\sinh ^2(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\int \frac {\sinh ^{2}{\left (c + d x \right )} \tanh {\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \] Input:
integrate(sinh(d*x+c)**2*tanh(d*x+c)/(a+b*sinh(d*x+c)),x)
Output:
Integral(sinh(c + d*x)**2*tanh(c + d*x)/(a + b*sinh(c + d*x)), x)
Time = 0.13 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.65 \[ \int \frac {\sinh ^2(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {a^{3} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{2} b^{2} + b^{4}\right )} d} + \frac {2 \, b \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{2} + b^{2}\right )} d} - \frac {a \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{2} + b^{2}\right )} d} - \frac {{\left (d x + c\right )} a}{b^{2} d} + \frac {e^{\left (d x + c\right )}}{2 \, b d} - \frac {e^{\left (-d x - c\right )}}{2 \, b d} \] Input:
integrate(sinh(d*x+c)^2*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="maxima ")
Output:
-a^3*log(-2*a*e^(-d*x - c) + b*e^(-2*d*x - 2*c) - b)/((a^2*b^2 + b^4)*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) - (d*x + c)*a/(b^2*d) + 1/2*e^(d*x + c)/(b*d) - 1/2*e^(-d*x - c)/(b*d)
Time = 0.15 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.63 \[ \int \frac {\sinh ^2(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {\frac {2 \, a^{3} \log \left ({\left | b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{a^{2} b^{2} + b^{4}} + \frac {{\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )} b}{a^{2} + b^{2}} + \frac {a \log \left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}{a^{2} + b^{2}} - \frac {e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}}{b}}{2 \, d} \] Input:
integrate(sinh(d*x+c)^2*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="giac")
Output:
-1/2*(2*a^3*log(abs(b*(e^(d*x + c) - e^(-d*x - c)) + 2*a))/(a^2*b^2 + b^4) + (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) - (e^(d*x + c) - e^ (-d*x - c))/b)/d
Time = 2.70 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.80 \[ \int \frac {\sinh ^2(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {{\mathrm {e}}^{c+d\,x}}{2\,b\,d}-\frac {\ln \left ({\mathrm {e}}^{c+d\,x}+1{}\mathrm {i}\right )}{a\,d-b\,d\,1{}\mathrm {i}}-\frac {a^3\,\ln \left (2\,a^4\,b^3-b^7-a^2\,b^5-a^6\,b+2\,a^7\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+b^7\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+a^6\,b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+2\,a^3\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-4\,a^5\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+a^2\,b^5\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}-2\,a^4\,b^3\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+2\,a\,b^6\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{d\,a^2\,b^2+d\,b^4}-\frac {{\mathrm {e}}^{-c-d\,x}}{2\,b\,d}+\frac {a\,x}{b^2}-\frac {\ln \left (1+{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{-b\,d+a\,d\,1{}\mathrm {i}} \] Input:
int((sinh(c + d*x)^2*tanh(c + d*x))/(a + b*sinh(c + d*x)),x)
Output:
exp(c + d*x)/(2*b*d) - (log(exp(c + d*x)*1i + 1)*1i)/(a*d*1i - b*d) - log( exp(c + d*x) + 1i)/(a*d - b*d*1i) - (a^3*log(2*a^4*b^3 - b^7 - a^2*b^5 - a ^6*b + 2*a^7*exp(d*x)*exp(c) + b^7*exp(2*c)*exp(2*d*x) + a^6*b*exp(2*c)*ex p(2*d*x) + 2*a^3*b^4*exp(d*x)*exp(c) - 4*a^5*b^2*exp(d*x)*exp(c) + a^2*b^5 *exp(2*c)*exp(2*d*x) - 2*a^4*b^3*exp(2*c)*exp(2*d*x) + 2*a*b^6*exp(d*x)*ex p(c)))/(b^4*d + a^2*b^2*d) - exp(- c - d*x)/(2*b*d) + (a*x)/b^2
Time = 0.19 (sec) , antiderivative size = 181, normalized size of antiderivative = 2.03 \[ \int \frac {\sinh ^2(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {-4 e^{d x +c} \mathit {atan} \left (e^{d x +c}\right ) b^{3}+e^{2 d x +2 c} a^{2} b +e^{2 d x +2 c} b^{3}-2 e^{d x +c} \mathrm {log}\left (e^{2 d x +2 c}+1\right ) a \,b^{2}-2 e^{d x +c} \mathrm {log}\left (e^{2 d x +2 c} b +2 e^{d x +c} a -b \right ) a^{3}+2 e^{d x +c} a^{3} d x +2 e^{d x +c} a \,b^{2} d x -a^{2} b -b^{3}}{2 e^{d x +c} b^{2} d \left (a^{2}+b^{2}\right )} \] Input:
int(sinh(d*x+c)^2*tanh(d*x+c)/(a+b*sinh(d*x+c)),x)
Output:
( - 4*e**(c + d*x)*atan(e**(c + d*x))*b**3 + e**(2*c + 2*d*x)*a**2*b + e** (2*c + 2*d*x)*b**3 - 2*e**(c + d*x)*log(e**(2*c + 2*d*x) + 1)*a*b**2 - 2*e **(c + d*x)*log(e**(2*c + 2*d*x)*b + 2*e**(c + d*x)*a - b)*a**3 + 2*e**(c + d*x)*a**3*d*x + 2*e**(c + d*x)*a*b**2*d*x - a**2*b - b**3)/(2*e**(c + d* x)*b**2*d*(a**2 + b**2))