\(\int \frac {b+c+\cosh (x)}{a-b \sinh (x)} \, dx\) [567]

Optimal result

Integrand size = 15, antiderivative size = 53 \[ \int \frac {b+c+\cosh (x)}{a-b \sinh (x)} \, dx=\frac {2 (b+c) \text {arctanh}\left (\frac {b+a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}-\frac {\log (a-b \sinh (x))}{b} \]

[Out]

-ln(a-b*sinh(x))/b+2*(b+c)*arctanh((b+a*tanh(1/2*x))/(a^2+b^2)^(1/2))/(a^2+b^2)^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4486, 2739, 632, 212, 2747, 31} \[ \int \frac {b+c+\cosh (x)}{a-b \sinh (x)} \, dx=\frac {2 (b+c) \text {arctanh}\left (\frac {a \tanh \left (\frac {x}{2}\right )+b}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}-\frac {\log (a-b \sinh (x))}{b} \]

[In]

Int[(b + c + Cosh[x])/(a - b*Sinh[x]),x]

[Out]

(2*(b + c)*ArcTanh[(b + a*Tanh[x/2])/Sqrt[a^2 + b^2]])/Sqrt[a^2 + b^2] - Log[a - b*Sinh[x]]/b

Rule 31

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

Rule 212

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

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[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]

Rule 2747

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 4486

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\left (1+\frac {b}{c}\right ) c}{a-b \sinh (x)}+\frac {\cosh (x)}{a-b \sinh (x)}\right ) \, dx \\ & = (b+c) \int \frac {1}{a-b \sinh (x)} \, dx+\int \frac {\cosh (x)}{a-b \sinh (x)} \, dx \\ & = -\frac {\text {Subst}\left (\int \frac {1}{a+x} \, dx,x,-b \sinh (x)\right )}{b}+(2 (b+c)) \text {Subst}\left (\int \frac {1}{a-2 b x-a x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right ) \\ & = -\frac {\log (a-b \sinh (x))}{b}-(4 (b+c)) \text {Subst}\left (\int \frac {1}{4 \left (a^2+b^2\right )-x^2} \, dx,x,-2 b-2 a \tanh \left (\frac {x}{2}\right )\right ) \\ & = \frac {2 (b+c) \text {arctanh}\left (\frac {b+a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}-\frac {\log (a-b \sinh (x))}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.17 \[ \int \frac {b+c+\cosh (x)}{a-b \sinh (x)} \, dx=-\frac {2 (b+c) \arctan \left (\frac {b+a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}-\frac {\log (-a+b \sinh (x))}{b} \]

[In]

Integrate[(b + c + Cosh[x])/(a - b*Sinh[x]),x]

[Out]

(-2*(b + c)*ArcTan[(b + a*Tanh[x/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2] - Log[-a + b*Sinh[x]]/b

Maple [A] (verified)

Time = 0.71 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.98

method	result	size
parts	\(-\frac {\ln \left (a -b \sinh \left (x \right )\right )}{b}+\frac {2 \left (b +c \right ) \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}\)	\(52\)
default	\(\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{b}+\frac {-\ln \left (\tanh \left (\frac {x}{2}\right )^{2} a +2 b \tanh \left (\frac {x}{2}\right )-a \right )-\frac {2 \left (-b^{2}-c b \right ) \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}}{b}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{b}\)	\(96\)
risch	\(-\frac {x}{b}+\frac {2 x \,a^{2} b}{a^{2} b^{2}+b^{4}}+\frac {2 x \,b^{3}}{a^{2} b^{2}+b^{4}}-\frac {\ln \left ({\mathrm e}^{x}+\frac {-a \,b^{2}-c a b +\sqrt {a^{2} b^{4}+2 a^{2} b^{3} c +a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{b^{2} \left (b +c \right )}\right ) a^{2}}{\left (a^{2}+b^{2}\right ) b}-\frac {b \ln \left ({\mathrm e}^{x}+\frac {-a \,b^{2}-c a b +\sqrt {a^{2} b^{4}+2 a^{2} b^{3} c +a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{b^{2} \left (b +c \right )}\right )}{a^{2}+b^{2}}+\frac {\ln \left ({\mathrm e}^{x}+\frac {-a \,b^{2}-c a b +\sqrt {a^{2} b^{4}+2 a^{2} b^{3} c +a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{b^{2} \left (b +c \right )}\right ) \sqrt {a^{2} b^{4}+2 a^{2} b^{3} c +a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{\left (a^{2}+b^{2}\right ) b}-\frac {\ln \left ({\mathrm e}^{x}-\frac {a \,b^{2}+c a b +\sqrt {a^{2} b^{4}+2 a^{2} b^{3} c +a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{b^{2} \left (b +c \right )}\right ) a^{2}}{\left (a^{2}+b^{2}\right ) b}-\frac {b \ln \left ({\mathrm e}^{x}-\frac {a \,b^{2}+c a b +\sqrt {a^{2} b^{4}+2 a^{2} b^{3} c +a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{b^{2} \left (b +c \right )}\right )}{a^{2}+b^{2}}-\frac {\ln \left ({\mathrm e}^{x}-\frac {a \,b^{2}+c a b +\sqrt {a^{2} b^{4}+2 a^{2} b^{3} c +a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{b^{2} \left (b +c \right )}\right ) \sqrt {a^{2} b^{4}+2 a^{2} b^{3} c +a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{\left (a^{2}+b^{2}\right ) b}\)	\(639\)

[In]

int((b+c+cosh(x))/(a-b*sinh(x)),x,method=_RETURNVERBOSE)

[Out]

-ln(a-b*sinh(x))/b+2*(b+c)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tanh(1/2*x)+2*b)/(a^2+b^2)^(1/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 174 vs. \(2 (47) = 94\).

Time = 0.26 (sec) , antiderivative size = 174, normalized size of antiderivative = 3.28 \[ \int \frac {b+c+\cosh (x)}{a-b \sinh (x)} \, dx=\frac {\sqrt {a^{2} + b^{2}} {\left (b^{2} + b c\right )} \log \left (\frac {b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} - 2 \, a b \cosh \left (x\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (x\right ) - a b\right )} \sinh \left (x\right ) + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) - a\right )}}{b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} - 2 \, a \cosh \left (x\right ) + 2 \, {\left (b \cosh \left (x\right ) - a\right )} \sinh \left (x\right ) - b}\right ) + {\left (a^{2} + b^{2}\right )} x - {\left (a^{2} + b^{2}\right )} \log \left (\frac {2 \, {\left (b \sinh \left (x\right ) - a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{2} b + b^{3}} \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 30.30 (sec) , antiderivative size = 586, normalized size of antiderivative = 11.06 \[ \int \frac {b+c+\cosh (x)}{a-b \sinh (x)} \, dx=\text {Too large to display} \]

[In]

integrate((b+c+cosh(x))/(a-b*sinh(x)),x)

[Out]

Piecewise((zoo*(c*log(tanh(x/2)) + x - 2*log(tanh(x/2) + 1) + log(tanh(x/2))), Eq(a, 0) & Eq(b, 0)), (-(b*log(
tanh(x/2)) + c*log(tanh(x/2)) + x - 2*log(tanh(x/2) + 1) + log(tanh(x/2)))/b, Eq(a, 0)), ((c*x + sinh(x))/a, E
q(b, 0)), (2*I*b/(b*tanh(x/2) + I*b) + 2*I*c/(b*tanh(x/2) + I*b) - x*tanh(x/2)/(b*tanh(x/2) + I*b) - I*x/(b*ta
nh(x/2) + I*b) + 2*log(tanh(x/2) + 1)*tanh(x/2)/(b*tanh(x/2) + I*b) + 2*I*log(tanh(x/2) + 1)/(b*tanh(x/2) + I*
b) - 2*log(tanh(x/2) + I)*tanh(x/2)/(b*tanh(x/2) + I*b) - 2*I*log(tanh(x/2) + I)/(b*tanh(x/2) + I*b), Eq(a, -I
*b)), (-2*I*b/(b*tanh(x/2) - I*b) - 2*I*c/(b*tanh(x/2) - I*b) - x*tanh(x/2)/(b*tanh(x/2) - I*b) + I*x/(b*tanh(
x/2) - I*b) + 2*log(tanh(x/2) + 1)*tanh(x/2)/(b*tanh(x/2) - I*b) - 2*I*log(tanh(x/2) + 1)/(b*tanh(x/2) - I*b)
- 2*log(tanh(x/2) - I)*tanh(x/2)/(b*tanh(x/2) - I*b) + 2*I*log(tanh(x/2) - I)/(b*tanh(x/2) - I*b), Eq(a, I*b))
, (-b*log(tanh(x/2) + b/a - sqrt(a**2 + b**2)/a)/sqrt(a**2 + b**2) + b*log(tanh(x/2) + b/a + sqrt(a**2 + b**2)
/a)/sqrt(a**2 + b**2) - c*log(tanh(x/2) + b/a - sqrt(a**2 + b**2)/a)/sqrt(a**2 + b**2) + c*log(tanh(x/2) + b/a
+ sqrt(a**2 + b**2)/a)/sqrt(a**2 + b**2) - x/b + 2*log(tanh(x/2) + 1)/b - log(tanh(x/2) + b/a - sqrt(a**2 + b
**2)/a)/b - log(tanh(x/2) + b/a + sqrt(a**2 + b**2)/a)/b, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (47) = 94\).

Time = 0.30 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.25 \[ \int \frac {b+c+\cosh (x)}{a-b \sinh (x)} \, dx=-\frac {b \log \left (\frac {b e^{\left (-x\right )} + a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} + a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}}} - \frac {c \log \left (\frac {b e^{\left (-x\right )} + a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} + a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}}} - \frac {\log \left (b \sinh \left (x\right ) - a\right )}{b} \]

[In]

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

[Out]

-b*log((b*e^(-x) + a - sqrt(a^2 + b^2))/(b*e^(-x) + a + sqrt(a^2 + b^2)))/sqrt(a^2 + b^2) - c*log((b*e^(-x) +
a - sqrt(a^2 + b^2))/(b*e^(-x) + a + sqrt(a^2 + b^2)))/sqrt(a^2 + b^2) - log(b*sinh(x) - a)/b

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.66 \[ \int \frac {b+c+\cosh (x)}{a-b \sinh (x)} \, dx=-\frac {{\left (b + c\right )} \log \left (\frac {{\left | 2 \, b e^{x} - 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} - 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}}} + \frac {x}{b} - \frac {\log \left ({\left | b e^{\left (2 \, x\right )} - 2 \, a e^{x} - b \right |}\right )}{b} \]

[In]

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

[Out]

-(b + c)*log(abs(2*b*e^x - 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*e^x - 2*a + 2*sqrt(a^2 + b^2)))/sqrt(a^2 + b^2) +
x/b - log(abs(b*e^(2*x) - 2*a*e^x - b))/b

Mupad [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 177, normalized size of antiderivative = 3.34 \[ \int \frac {b+c+\cosh (x)}{a-b \sinh (x)} \, dx=\frac {x}{b}+\frac {\ln \left (b\,\sqrt {a^2+b^2}+a^2\,{\mathrm {e}}^x+b^2\,{\mathrm {e}}^x+a\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}\right )\,\left (b^2\,\sqrt {a^2+b^2}-a^2-b^2+b\,c\,\sqrt {a^2+b^2}\right )}{a^2\,b+b^3}-\frac {\ln \left (b\,\sqrt {a^2+b^2}-a^2\,{\mathrm {e}}^x-b^2\,{\mathrm {e}}^x+a\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}\right )\,\left (b^2\,\sqrt {a^2+b^2}+a^2+b^2+b\,c\,\sqrt {a^2+b^2}\right )}{a^2\,b+b^3} \]

[In]

int((b + c + cosh(x))/(a - b*sinh(x)),x)

[Out]

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