\(\int \frac {b+c+\cosh (x)}{a+b \sinh (x)} \, dx\) [566]

Optimal result

Integrand size = 14, antiderivative size = 52 \[ \int \frac {b+c+\cosh (x)}{a+b \sinh (x)} \, dx=-\frac {2 (b+c) \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}+\frac {\log (a+b \sinh (x))}{b} \]

[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 = 52, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4486, 2739, 632, 212, 2747, 31} \[ \int \frac {b+c+\cosh (x)}{a+b \sinh (x)} \, dx=\frac {\log (a+b \sinh (x))}{b}-\frac {2 (b+c) \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}} \]

[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 = 0.16 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.15 \[ \int \frac {b+c+\cosh (x)}{a+b \sinh (x)} \, dx=\frac {2 (b+c) \arctan \left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}+\frac {\log (a+b \sinh (x))}{b} \]

[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 = 0.69 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.96

method	result	size
parts	\(\frac {\ln \left (a +b \sinh \left (x \right )\right )}{b}+\frac {2 \left (b +c \right ) \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}\)	\(50\)
default	\(-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{b}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{b}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )^{2} a -2 b \tanh \left (\frac {x}{2}\right )-a \right )-\frac {2 \left (-b^{2}-c b \right ) \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}}{b}\)	\(98\)
risch	\(\frac {x}{b}-\frac {2 x \,a^{2} b}{a^{2} b^{2}+b^{4}}-\frac {2 x \,b^{3}}{a^{2} b^{2}+b^{4}}+\frac {\ln \left ({\mathrm e}^{x}-\frac {-a \,b^{2}-c a b +\sqrt {a^{2} b^{4}+2 a^{2} b^{3} c +a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{b^{2} \left (b +c \right )}\right ) a^{2}}{\left (a^{2}+b^{2}\right ) b}+\frac {b \ln \left ({\mathrm e}^{x}-\frac {-a \,b^{2}-c a b +\sqrt {a^{2} b^{4}+2 a^{2} b^{3} c +a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{b^{2} \left (b +c \right )}\right )}{a^{2}+b^{2}}+\frac {\ln \left ({\mathrm e}^{x}-\frac {-a \,b^{2}-c a b +\sqrt {a^{2} b^{4}+2 a^{2} b^{3} c +a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{b^{2} \left (b +c \right )}\right ) \sqrt {a^{2} b^{4}+2 a^{2} b^{3} c +a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{\left (a^{2}+b^{2}\right ) b}+\frac {\ln \left ({\mathrm e}^{x}+\frac {a \,b^{2}+c a b +\sqrt {a^{2} b^{4}+2 a^{2} b^{3} c +a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{b^{2} \left (b +c \right )}\right ) a^{2}}{\left (a^{2}+b^{2}\right ) b}+\frac {b \ln \left ({\mathrm e}^{x}+\frac {a \,b^{2}+c a b +\sqrt {a^{2} b^{4}+2 a^{2} b^{3} c +a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{b^{2} \left (b +c \right )}\right )}{a^{2}+b^{2}}-\frac {\ln \left ({\mathrm e}^{x}+\frac {a \,b^{2}+c a b +\sqrt {a^{2} b^{4}+2 a^{2} b^{3} c +a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{b^{2} \left (b +c \right )}\right ) \sqrt {a^{2} b^{4}+2 a^{2} b^{3} c +a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{\left (a^{2}+b^{2}\right ) b}\)	\(634\)

[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. 167 vs. \(2 (48) = 96\).

Time = 0.27 (sec) , antiderivative size = 167, normalized size of antiderivative = 3.21 \[ \int \frac {b+c+\cosh (x)}{a+b \sinh (x)} \, dx=\frac {\sqrt {a^{2} + b^{2}} {\left (b^{2} + b c\right )} \log \left (\frac {b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \, {\left (b \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) - b}\right ) - {\left (a^{2} + b^{2}\right )} x + {\left (a^{2} + b^{2}\right )} \log \left (\frac {2 \, {\left (b \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{2} b + b^{3}} \]

[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 = 30.48 (sec) , antiderivative size = 585, normalized size of antiderivative = 11.25 \[ \int \frac {b+c+\cosh (x)}{a+b \sinh (x)} \, dx=\text {Too large to display} \]

[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(t
anh(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, Eq
(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*tan
h(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. 122 vs. \(2 (48) = 96\).

Time = 0.28 (sec) , antiderivative size = 122, normalized size of antiderivative = 2.35 \[ \int \frac {b+c+\cosh (x)}{a+b \sinh (x)} \, dx=\frac {b \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}}} + \frac {c \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}}} + \frac {\log \left (b \sinh \left (x\right ) + a\right )}{b} \]

[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 = 87, normalized size of antiderivative = 1.67 \[ \int \frac {b+c+\cosh (x)}{a+b \sinh (x)} \, dx=\frac {{\left (b + c\right )} \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}}} - \frac {x}{b} + \frac {\log \left ({\left | b e^{\left (2 \, x\right )} + 2 \, a e^{x} - b \right |}\right )}{b} \]

[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 = 2.56 (sec) , antiderivative size = 178, normalized size of antiderivative = 3.42 \[ \int \frac {b+c+\cosh (x)}{a+b \sinh (x)} \, dx=\frac {\ln \left (a^2\,{\mathrm {e}}^x-b\,\sqrt {a^2+b^2}+b^2\,{\mathrm {e}}^x+a\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}\right )\,\left (b^2\,\sqrt {a^2+b^2}+a^2+b^2+b\,c\,\sqrt {a^2+b^2}\right )}{a^2\,b+b^3}-\frac {\ln \left (b\,\sqrt {a^2+b^2}+a^2\,{\mathrm {e}}^x+b^2\,{\mathrm {e}}^x-a\,{\mathrm {e}}^x\,\sqrt {a^2+b^2}\right )\,\left (b^2\,\sqrt {a^2+b^2}-a^2-b^2+b\,c\,\sqrt {a^2+b^2}\right )}{a^2\,b+b^3}-\frac {x}{b} \]

[In]

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

[Out]

(log(a^2*exp(x) - b*(a^2 + b^2)^(1/2) + 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*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) - x/b