\(\int \frac {\sinh (c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx\) [380]

Optimal result

Integrand size = 25, antiderivative size = 74 \[ \int \frac {\sinh (c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {a \arctan (\sinh (c+d x))}{\left (a^2+b^2\right ) d}+\frac {b \log (\cosh (c+d x))}{\left (a^2+b^2\right ) d}+\frac {a^2 \log (a+b \sinh (c+d x))}{b \left (a^2+b^2\right ) d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2916, 12, 1643, 649, 209, 266} \[ \int \frac {\sinh (c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {a \arctan (\sinh (c+d x))}{d \left (a^2+b^2\right )}+\frac {a^2 \log (a+b \sinh (c+d x))}{b d \left (a^2+b^2\right )}+\frac {b \log (\cosh (c+d x))}{d \left (a^2+b^2\right )} \]

[In]

Int[(Sinh[c + d*x]*Tanh[c + d*x])/(a + b*Sinh[c + d*x]),x]

[Out]

-((a*ArcTan[Sinh[c + d*x]])/((a^2 + b^2)*d)) + (b*Log[Cosh[c + d*x]])/((a^2 + b^2)*d) + (a^2*Log[a + b*Sinh[c
+ d*x]])/(b*(a^2 + b^2)*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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 1643

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2916

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/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]

Rubi steps \begin{align*} \text {integral}& = -\frac {b \text {Subst}\left (\int \frac {x^2}{b^2 (a+x) \left (-b^2-x^2\right )} \, dx,x,b \sinh (c+d x)\right )}{d} \\ & = -\frac {\text {Subst}\left (\int \frac {x^2}{(a+x) \left (-b^2-x^2\right )} \, dx,x,b \sinh (c+d x)\right )}{b d} \\ & = -\frac {\text {Subst}\left (\int \left (-\frac {a^2}{\left (a^2+b^2\right ) (a+x)}+\frac {b^2 (a-x)}{\left (a^2+b^2\right ) \left (b^2+x^2\right )}\right ) \, dx,x,b \sinh (c+d x)\right )}{b d} \\ & = \frac {a^2 \log (a+b \sinh (c+d x))}{b \left (a^2+b^2\right ) d}-\frac {b \text {Subst}\left (\int \frac {a-x}{b^2+x^2} \, dx,x,b \sinh (c+d x)\right )}{\left (a^2+b^2\right ) d} \\ & = \frac {a^2 \log (a+b \sinh (c+d x))}{b \left (a^2+b^2\right ) d}+\frac {b \text {Subst}\left (\int \frac {x}{b^2+x^2} \, dx,x,b \sinh (c+d x)\right )}{\left (a^2+b^2\right ) d}-\frac {(a b) \text {Subst}\left (\int \frac {1}{b^2+x^2} \, dx,x,b \sinh (c+d x)\right )}{\left (a^2+b^2\right ) d} \\ & = -\frac {a \arctan (\sinh (c+d x))}{\left (a^2+b^2\right ) d}+\frac {b \log (\cosh (c+d x))}{\left (a^2+b^2\right ) d}+\frac {a^2 \log (a+b \sinh (c+d x))}{b \left (a^2+b^2\right ) d} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.08 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.05 \[ \int \frac {\sinh (c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {b (i a+b) \log (i-\sinh (c+d x))+b (-i a+b) \log (i+\sinh (c+d x))+2 a^2 \log (a+b \sinh (c+d x))}{2 b \left (a^2+b^2\right ) d} \]

[In]

Integrate[(Sinh[c + d*x]*Tanh[c + d*x])/(a + b*Sinh[c + d*x]),x]

[Out]

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

Maple [A] (verified)

Time = 1.10 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.78

method	result	size
derivativedivides	\(\frac {-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b}+\frac {4 b \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-8 a \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{2}+4 b^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b}+\frac {a^{2} \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{b \left (a^{2}+b^{2}\right )}}{d}\)	\(132\)
default	\(\frac {-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b}+\frac {4 b \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-8 a \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{2}+4 b^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b}+\frac {a^{2} \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{b \left (a^{2}+b^{2}\right )}}{d}\)	\(132\)
risch	\(\frac {x}{b}-\frac {2 b \,d^{2} x}{a^{2} d^{2}+b^{2} d^{2}}-\frac {2 b d c}{a^{2} d^{2}+b^{2} d^{2}}-\frac {2 a^{2} x}{b \left (a^{2}+b^{2}\right )}-\frac {2 a^{2} c}{b d \left (a^{2}+b^{2}\right )}+\frac {i \ln \left ({\mathrm e}^{d x +c}-i\right ) a}{\left (a^{2}+b^{2}\right ) d}+\frac {\ln \left ({\mathrm e}^{d x +c}-i\right ) b}{\left (a^{2}+b^{2}\right ) d}-\frac {i \ln \left ({\mathrm e}^{d x +c}+i\right ) a}{\left (a^{2}+b^{2}\right ) d}+\frac {\ln \left ({\mathrm e}^{d x +c}+i\right ) b}{\left (a^{2}+b^{2}\right ) d}+\frac {a^{2} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{b}-1\right )}{b d \left (a^{2}+b^{2}\right )}\)	\(235\)

[In]

int(sinh(d*x+c)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(-1/b*ln(tanh(1/2*d*x+1/2*c)-1)+8/(4*a^2+4*b^2)*(1/2*b*ln(1+tanh(1/2*d*x+1/2*c)^2)-a*arctan(tanh(1/2*d*x+1
/2*c)))-1/b*ln(tanh(1/2*d*x+1/2*c)+1)+a^2/b/(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))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.50 \[ \int \frac {\sinh (c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {{\left (a^{2} + b^{2}\right )} d x + 2 \, a b \arctan \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) - a^{2} \log \left (\frac {2 \, {\left (b \sinh \left (d x + c\right ) + a\right )}}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) - b^{2} \log \left (\frac {2 \, \cosh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right )}{{\left (a^{2} b + b^{3}\right )} d} \]

[In]

integrate(sinh(d*x+c)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\sinh (c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\int \frac {\sinh {\left (c + d x \right )} \tanh {\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \]

[In]

integrate(sinh(d*x+c)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

Integral(sinh(c + d*x)*tanh(c + d*x)/(a + b*sinh(c + d*x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.49 \[ \int \frac {\sinh (c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {a^{2} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{2} b + b^{3}\right )} d} + \frac {2 \, a \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac {b \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac {d x + c}{b d} \]

[In]

integrate(sinh(d*x+c)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.64 \[ \int \frac {\sinh (c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\frac {2 \, a^{2} \log \left ({\left | b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{a^{2} b + b^{3}} - \frac {{\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )} a}{a^{2} + b^{2}} + \frac {b \log \left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}{a^{2} + b^{2}}}{2 \, d} \]

[In]

integrate(sinh(d*x+c)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 2.09 (sec) , antiderivative size = 174, normalized size of antiderivative = 2.35 \[ \int \frac {\sinh (c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\ln \left ({\mathrm {e}}^{c+d\,x}+1{}\mathrm {i}\right )}{b\,d+a\,d\,1{}\mathrm {i}}-\frac {x}{b}+\frac {a^2\,\ln \left (a^2\,b^3-b^5-a^4\,b+2\,a^5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+b^5\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+a^4\,b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}-2\,a^3\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-a^2\,b^3\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+2\,a\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{d\,a^2\,b+d\,b^3}+\frac {\ln \left (1+{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{a\,d+b\,d\,1{}\mathrm {i}} \]

[In]

int((sinh(c + d*x)*tanh(c + d*x))/(a + b*sinh(c + d*x)),x)

[Out]

log(exp(c + d*x) + 1i)/(a*d*1i + b*d) - x/b + (log(exp(c + d*x)*1i + 1)*1i)/(a*d + b*d*1i) + (a^2*log(a^2*b^3
- b^5 - a^4*b + 2*a^5*exp(d*x)*exp(c) + b^5*exp(2*c)*exp(2*d*x) + a^4*b*exp(2*c)*exp(2*d*x) - 2*a^3*b^2*exp(d*
x)*exp(c) - a^2*b^3*exp(2*c)*exp(2*d*x) + 2*a*b^4*exp(d*x)*exp(c)))/(b^3*d + a^2*b*d)