Integrand size = 19, antiderivative size = 54 \[ \int \frac {\sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {x}{b}+\frac {2 a \text {arctanh}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d} \]
Time = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.19 \[ \int \frac {\sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\frac {c}{d}+x-\frac {2 a \arctan \left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2} d}}{b} \]
Result contains complex when optimal does not.
Time = 0.31 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3042, 26, 3214, 3042, 3139, 1083, 217}
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.
\(\displaystyle \int \frac {\sinh (c+d x)}{a+b \sinh (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {i \sin (i c+i d x)}{a-i b \sin (i c+i d x)}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {\sin (i c+i d x)}{a-i b \sin (i c+i d x)}dx\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle -i \left (\frac {i x}{b}-\frac {i a \int \frac {1}{a+b \sinh (c+d x)}dx}{b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -i \left (\frac {i x}{b}-\frac {i a \int \frac {1}{a-i b \sin (i c+i d x)}dx}{b}\right )\) |
\(\Big \downarrow \) 3139 |
\(\displaystyle -i \left (\frac {i x}{b}-\frac {2 a \int \frac {1}{-a \tanh ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tanh \left (\frac {1}{2} (c+d x)\right )+a}d\left (i \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{b d}\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle -i \left (\frac {4 a \int \frac {1}{\tanh ^2\left (\frac {1}{2} (c+d x)\right )-4 \left (a^2+b^2\right )}d\left (2 i a \tanh \left (\frac {1}{2} (c+d x)\right )-2 i b\right )}{b d}+\frac {i x}{b}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle -i \left (\frac {i x}{b}-\frac {2 i a \text {arctanh}\left (\frac {\tanh \left (\frac {1}{2} (c+d x)\right )}{2 \sqrt {a^2+b^2}}\right )}{b d \sqrt {a^2+b^2}}\right )\) |
(-I)*((I*x)/b - ((2*I)*a*ArcTanh[Tanh[(c + d*x)/2]/(2*Sqrt[a^2 + b^2])])/( b*Sqrt[a^2 + b^2]*d))
3.3.26.3.1 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_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ [a^2 - b^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Time = 0.81 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.52
method | result | size |
derivativedivides | \(\frac {-\frac {2 a \,\operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{b \sqrt {a^{2}+b^{2}}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b}}{d}\) | \(82\) |
default | \(\frac {-\frac {2 a \,\operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{b \sqrt {a^{2}+b^{2}}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b}}{d}\) | \(82\) |
risch | \(\frac {x}{b}+\frac {a \ln \left ({\mathrm e}^{d x +c}+\frac {a \sqrt {a^{2}+b^{2}}+a^{2}+b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, d b}-\frac {a \ln \left ({\mathrm e}^{d x +c}+\frac {a \sqrt {a^{2}+b^{2}}-a^{2}-b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, d b}\) | \(124\) |
1/d*(-2*a/b/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tanh(1/2*d*x+1/2*c)-2*b)/(a^2 +b^2)^(1/2))-1/b*ln(tanh(1/2*d*x+1/2*c)-1)+1/b*ln(tanh(1/2*d*x+1/2*c)+1))
Leaf count of result is larger than twice the leaf count of optimal. 186 vs. \(2 (51) = 102\).
Time = 0.25 (sec) , antiderivative size = 186, normalized size of antiderivative = 3.44 \[ \int \frac {\sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {{\left (a^{2} + b^{2}\right )} d x + \sqrt {a^{2} + b^{2}} a \log \left (\frac {b^{2} \cosh \left (d x + c\right )^{2} + b^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{b \cosh \left (d x + c\right )^{2} + b \sinh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) + 2 \, {\left (b \cosh \left (d x + c\right ) + a\right )} \sinh \left (d x + c\right ) - b}\right )}{{\left (a^{2} b + b^{3}\right )} d} \]
((a^2 + b^2)*d*x + sqrt(a^2 + b^2)*a*log((b^2*cosh(d*x + c)^2 + b^2*sinh(d *x + c)^2 + 2*a*b*cosh(d*x + c) + 2*a^2 + b^2 + 2*(b^2*cosh(d*x + c) + a*b )*sinh(d*x + c) + 2*sqrt(a^2 + b^2)*(b*cosh(d*x + c) + b*sinh(d*x + c) + a ))/(b*cosh(d*x + c)^2 + b*sinh(d*x + c)^2 + 2*a*cosh(d*x + c) + 2*(b*cosh( d*x + c) + a)*sinh(d*x + c) - b)))/((a^2*b + b^3)*d)
Result contains complex when optimal does not.
Time = 31.85 (sec) , antiderivative size = 269, normalized size of antiderivative = 4.98 \[ \int \frac {\sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=\begin {cases} \tilde {\infty } x & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac {x}{b} & \text {for}\: a = 0 \\\frac {\cosh {\left (c + d x \right )}}{a d} & \text {for}\: b = 0 \\\frac {x \sinh {\left (c \right )}}{a + b \sinh {\left (c \right )}} & \text {for}\: d = 0 \\\frac {d x \tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{b d \tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} - i b d} - \frac {i d x}{b d \tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} - i b d} - \frac {2}{b d \tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} - i b d} & \text {for}\: a = - i b \\\frac {d x \tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{b d \tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} + i b d} + \frac {i d x}{b d \tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} + i b d} - \frac {2}{b d \tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} + i b d} & \text {for}\: a = i b \\\frac {a \log {\left (\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} - \frac {b}{a} - \frac {\sqrt {a^{2} + b^{2}}}{a} \right )}}{b d \sqrt {a^{2} + b^{2}}} - \frac {a \log {\left (\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} - \frac {b}{a} + \frac {\sqrt {a^{2} + b^{2}}}{a} \right )}}{b d \sqrt {a^{2} + b^{2}}} + \frac {x}{b} & \text {otherwise} \end {cases} \]
Piecewise((zoo*x, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (x/b, Eq(a, 0)), (cosh( c + d*x)/(a*d), Eq(b, 0)), (x*sinh(c)/(a + b*sinh(c)), Eq(d, 0)), (d*x*tan h(c/2 + d*x/2)/(b*d*tanh(c/2 + d*x/2) - I*b*d) - I*d*x/(b*d*tanh(c/2 + d*x /2) - I*b*d) - 2/(b*d*tanh(c/2 + d*x/2) - I*b*d), Eq(a, -I*b)), (d*x*tanh( c/2 + d*x/2)/(b*d*tanh(c/2 + d*x/2) + I*b*d) + I*d*x/(b*d*tanh(c/2 + d*x/2 ) + I*b*d) - 2/(b*d*tanh(c/2 + d*x/2) + I*b*d), Eq(a, I*b)), (a*log(tanh(c /2 + d*x/2) - b/a - sqrt(a**2 + b**2)/a)/(b*d*sqrt(a**2 + b**2)) - a*log(t anh(c/2 + d*x/2) - b/a + sqrt(a**2 + b**2)/a)/(b*d*sqrt(a**2 + b**2)) + x/ b, True))
Time = 0.30 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.57 \[ \int \frac {\sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {a \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} b d} + \frac {d x + c}{b d} \]
-a*log((b*e^(-d*x - c) - a - sqrt(a^2 + b^2))/(b*e^(-d*x - c) - a + sqrt(a ^2 + b^2)))/(sqrt(a^2 + b^2)*b*d) + (d*x + c)/(b*d)
Time = 0.30 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.56 \[ \int \frac {\sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {\frac {a \log \left (\frac {{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b} - \frac {d x + c}{b}}{d} \]
-(a*log(abs(2*b*e^(d*x + c) + 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*e^(d*x + c) + 2*a + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*b) - (d*x + c)/b)/d
Time = 1.60 (sec) , antiderivative size = 121, normalized size of antiderivative = 2.24 \[ \int \frac {\sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {x}{b}-\frac {a\,\ln \left (\frac {2\,a\,{\mathrm {e}}^{c+d\,x}}{b^2}-\frac {2\,a\,\left (b-a\,{\mathrm {e}}^{c+d\,x}\right )}{b^2\,\sqrt {a^2+b^2}}\right )}{b\,d\,\sqrt {a^2+b^2}}+\frac {a\,\ln \left (\frac {2\,a\,{\mathrm {e}}^{c+d\,x}}{b^2}+\frac {2\,a\,\left (b-a\,{\mathrm {e}}^{c+d\,x}\right )}{b^2\,\sqrt {a^2+b^2}}\right )}{b\,d\,\sqrt {a^2+b^2}} \]