\(\int \frac {A+B \cosh (x)}{a+b \sinh (x)} \, dx\) [246]

Optimal result

Integrand size = 15, antiderivative size = 51 \[ \int \frac {A+B \cosh (x)}{a+b \sinh (x)} \, dx=-\frac {2 A \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}+\frac {B \log (a+b \sinh (x))}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 51, 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} \[ \int \frac {A+B \cosh (x)}{a+b \sinh (x)} \, dx=\frac {B \log (a+b \sinh (x))}{b}-\frac {2 A \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}} \]

[In]

Int[(A + B*Cosh[x])/(a + b*Sinh[x]),x]

[Out]

(-2*A*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/Sqrt[a^2 + b^2] + (B*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 {A}{a+b \sinh (x)}+\frac {B \cosh (x)}{a+b \sinh (x)}\right ) \, dx \\ & = A \int \frac {1}{a+b \sinh (x)} \, dx+B \int \frac {\cosh (x)}{a+b \sinh (x)} \, dx \\ & = (2 A) \text {Subst}\left (\int \frac {1}{a+2 b x-a x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )+\frac {B \text {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \sinh (x)\right )}{b} \\ & = \frac {B \log (a+b \sinh (x))}{b}-(4 A) \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 A \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}+\frac {B \log (a+b \sinh (x))}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.16 \[ \int \frac {A+B \cosh (x)}{a+b \sinh (x)} \, dx=\frac {2 A \arctan \left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}+\frac {B \log (a+b \sinh (x))}{b} \]

[In]

Integrate[(A + B*Cosh[x])/(a + b*Sinh[x]),x]

[Out]

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

Maple [A] (verified)

Time = 1.65 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.96

method	result	size
parts	\(\frac {2 A \,\operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}+\frac {B \ln \left (a +b \sinh \left (x \right )\right )}{b}\)	\(49\)
default	\(\frac {B \ln \left (\tanh \left (\frac {x}{2}\right )^{2} a -2 b \tanh \left (\frac {x}{2}\right )-a \right )+\frac {2 A b \,\operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}}{b}-\frac {B \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{b}-\frac {B \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{b}\)	\(92\)
risch	\(\frac {B x}{b}-\frac {2 x B \,a^{2} b}{a^{2} b^{2}+b^{4}}-\frac {2 x B \,b^{3}}{a^{2} b^{2}+b^{4}}+\frac {\ln \left ({\mathrm e}^{x}+\frac {A a b -\sqrt {A^{2} a^{2} b^{2}+A^{2} b^{4}}}{A \,b^{2}}\right ) B \,a^{2}}{\left (a^{2}+b^{2}\right ) b}+\frac {b \ln \left ({\mathrm e}^{x}+\frac {A a b -\sqrt {A^{2} a^{2} b^{2}+A^{2} b^{4}}}{A \,b^{2}}\right ) B}{a^{2}+b^{2}}+\frac {\ln \left ({\mathrm e}^{x}+\frac {A a b -\sqrt {A^{2} a^{2} b^{2}+A^{2} b^{4}}}{A \,b^{2}}\right ) \sqrt {A^{2} a^{2} b^{2}+A^{2} b^{4}}}{\left (a^{2}+b^{2}\right ) b}+\frac {\ln \left ({\mathrm e}^{x}+\frac {A a b +\sqrt {A^{2} a^{2} b^{2}+A^{2} b^{4}}}{A \,b^{2}}\right ) B \,a^{2}}{\left (a^{2}+b^{2}\right ) b}+\frac {b \ln \left ({\mathrm e}^{x}+\frac {A a b +\sqrt {A^{2} a^{2} b^{2}+A^{2} b^{4}}}{A \,b^{2}}\right ) B}{a^{2}+b^{2}}-\frac {\ln \left ({\mathrm e}^{x}+\frac {A a b +\sqrt {A^{2} a^{2} b^{2}+A^{2} b^{4}}}{A \,b^{2}}\right ) \sqrt {A^{2} a^{2} b^{2}+A^{2} b^{4}}}{\left (a^{2}+b^{2}\right ) b}\)	\(396\)

[In]

int((A+B*cosh(x))/(a+b*sinh(x)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (47) = 94\).

Time = 0.30 (sec) , antiderivative size = 170, normalized size of antiderivative = 3.33 \[ \int \frac {A+B \cosh (x)}{a+b \sinh (x)} \, dx=\frac {\sqrt {a^{2} + b^{2}} A b \log \left (\frac {b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \, {\left (b \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) - b}\right ) - {\left (B a^{2} + B b^{2}\right )} x + {\left (B a^{2} + B b^{2}\right )} \log \left (\frac {2 \, {\left (b \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{2} b + b^{3}} \]

[In]

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

[Out]

(sqrt(a^2 + b^2)*A*b*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)) - (B*a^2 + B*b^2)*x + (B*a^2 + B*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 = 15.13 (sec) , antiderivative size = 517, normalized size of antiderivative = 10.14 \[ \int \frac {A+B \cosh (x)}{a+b \sinh (x)} \, dx=\begin {cases} \tilde {\infty } \left (A \log {\left (\tanh {\left (\frac {x}{2} \right )} \right )} + B x - 2 B \log {\left (\tanh {\left (\frac {x}{2} \right )} + 1 \right )} + B \log {\left (\tanh {\left (\frac {x}{2} \right )} \right )}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {A \log {\left (\tanh {\left (\frac {x}{2} \right )} \right )} + B x - 2 B \log {\left (\tanh {\left (\frac {x}{2} \right )} + 1 \right )} + B \log {\left (\tanh {\left (\frac {x}{2} \right )} \right )}}{b} & \text {for}\: a = 0 \\\frac {A x + B \sinh {\left (x \right )}}{a} & \text {for}\: b = 0 \\\frac {2 i A}{b \tanh {\left (\frac {x}{2} \right )} - i b} + \frac {B x \tanh {\left (\frac {x}{2} \right )}}{b \tanh {\left (\frac {x}{2} \right )} - i b} - \frac {i B x}{b \tanh {\left (\frac {x}{2} \right )} - i b} - \frac {2 B \log {\left (\tanh {\left (\frac {x}{2} \right )} + 1 \right )} \tanh {\left (\frac {x}{2} \right )}}{b \tanh {\left (\frac {x}{2} \right )} - i b} + \frac {2 i B \log {\left (\tanh {\left (\frac {x}{2} \right )} + 1 \right )}}{b \tanh {\left (\frac {x}{2} \right )} - i b} + \frac {2 B \log {\left (\tanh {\left (\frac {x}{2} \right )} - i \right )} \tanh {\left (\frac {x}{2} \right )}}{b \tanh {\left (\frac {x}{2} \right )} - i b} - \frac {2 i B \log {\left (\tanh {\left (\frac {x}{2} \right )} - i \right )}}{b \tanh {\left (\frac {x}{2} \right )} - i b} & \text {for}\: a = - i b \\- \frac {2 i A}{b \tanh {\left (\frac {x}{2} \right )} + i b} + \frac {B x \tanh {\left (\frac {x}{2} \right )}}{b \tanh {\left (\frac {x}{2} \right )} + i b} + \frac {i B x}{b \tanh {\left (\frac {x}{2} \right )} + i b} - \frac {2 B \log {\left (\tanh {\left (\frac {x}{2} \right )} + 1 \right )} \tanh {\left (\frac {x}{2} \right )}}{b \tanh {\left (\frac {x}{2} \right )} + i b} - \frac {2 i B \log {\left (\tanh {\left (\frac {x}{2} \right )} + 1 \right )}}{b \tanh {\left (\frac {x}{2} \right )} + i b} + \frac {2 B \log {\left (\tanh {\left (\frac {x}{2} \right )} + i \right )} \tanh {\left (\frac {x}{2} \right )}}{b \tanh {\left (\frac {x}{2} \right )} + i b} + \frac {2 i B \log {\left (\tanh {\left (\frac {x}{2} \right )} + i \right )}}{b \tanh {\left (\frac {x}{2} \right )} + i b} & \text {for}\: a = i b \\- \frac {A \log {\left (\tanh {\left (\frac {x}{2} \right )} - \frac {b}{a} - \frac {\sqrt {a^{2} + b^{2}}}{a} \right )}}{\sqrt {a^{2} + b^{2}}} + \frac {A \log {\left (\tanh {\left (\frac {x}{2} \right )} - \frac {b}{a} + \frac {\sqrt {a^{2} + b^{2}}}{a} \right )}}{\sqrt {a^{2} + b^{2}}} + \frac {B x}{b} - \frac {2 B \log {\left (\tanh {\left (\frac {x}{2} \right )} + 1 \right )}}{b} + \frac {B \log {\left (\tanh {\left (\frac {x}{2} \right )} - \frac {b}{a} - \frac {\sqrt {a^{2} + b^{2}}}{a} \right )}}{b} + \frac {B \log {\left (\tanh {\left (\frac {x}{2} \right )} - \frac {b}{a} + \frac {\sqrt {a^{2} + b^{2}}}{a} \right )}}{b} & \text {otherwise} \end {cases} \]

[In]

integrate((A+B*cosh(x))/(a+b*sinh(x)),x)

[Out]

Piecewise((zoo*(A*log(tanh(x/2)) + B*x - 2*B*log(tanh(x/2) + 1) + B*log(tanh(x/2))), Eq(a, 0) & Eq(b, 0)), ((A
*log(tanh(x/2)) + B*x - 2*B*log(tanh(x/2) + 1) + B*log(tanh(x/2)))/b, Eq(a, 0)), ((A*x + B*sinh(x))/a, Eq(b, 0
)), (2*I*A/(b*tanh(x/2) - I*b) + B*x*tanh(x/2)/(b*tanh(x/2) - I*b) - I*B*x/(b*tanh(x/2) - I*b) - 2*B*log(tanh(
x/2) + 1)*tanh(x/2)/(b*tanh(x/2) - I*b) + 2*I*B*log(tanh(x/2) + 1)/(b*tanh(x/2) - I*b) + 2*B*log(tanh(x/2) - I
)*tanh(x/2)/(b*tanh(x/2) - I*b) - 2*I*B*log(tanh(x/2) - I)/(b*tanh(x/2) - I*b), Eq(a, -I*b)), (-2*I*A/(b*tanh(
x/2) + I*b) + B*x*tanh(x/2)/(b*tanh(x/2) + I*b) + I*B*x/(b*tanh(x/2) + I*b) - 2*B*log(tanh(x/2) + 1)*tanh(x/2)
/(b*tanh(x/2) + I*b) - 2*I*B*log(tanh(x/2) + 1)/(b*tanh(x/2) + I*b) + 2*B*log(tanh(x/2) + I)*tanh(x/2)/(b*tanh
(x/2) + I*b) + 2*I*B*log(tanh(x/2) + I)/(b*tanh(x/2) + I*b), Eq(a, I*b)), (-A*log(tanh(x/2) - b/a - sqrt(a**2
+ b**2)/a)/sqrt(a**2 + b**2) + A*log(tanh(x/2) - b/a + sqrt(a**2 + b**2)/a)/sqrt(a**2 + b**2) + B*x/b - 2*B*lo
g(tanh(x/2) + 1)/b + B*log(tanh(x/2) - b/a - sqrt(a**2 + b**2)/a)/b + B*log(tanh(x/2) - b/a + sqrt(a**2 + b**2
)/a)/b, True))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.33 \[ \int \frac {A+B \cosh (x)}{a+b \sinh (x)} \, dx=\frac {A \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}}} + \frac {B \log \left (b \sinh \left (x\right ) + a\right )}{b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.71 \[ \int \frac {A+B \cosh (x)}{a+b \sinh (x)} \, dx=\frac {A \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}}} - \frac {B x}{b} + \frac {B \log \left ({\left | b e^{\left (2 \, x\right )} + 2 \, a e^{x} - b \right |}\right )}{b} \]

[In]

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

[Out]

A*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) - B*x/b +
B*log(abs(b*e^(2*x) + 2*a*e^x - b))/b

Mupad [B] (verification not implemented)

Time = 3.52 (sec) , antiderivative size = 198, normalized size of antiderivative = 3.88 \[ \int \frac {A+B \cosh (x)}{a+b \sinh (x)} \, dx=\frac {B\,b^3\,\ln \left (8\,A^2\,a\,{\mathrm {e}}^x-4\,A^2\,b+4\,A^2\,b\,{\mathrm {e}}^{2\,x}\right )}{a^2\,b^2+b^4}-\frac {B\,x}{b}-\frac {2\,\mathrm {atan}\left (\frac {A^2\,b^2\,{\mathrm {e}}^x\,\sqrt {-a^2-b^2}}{\left (A\,a^2\,b+A\,b^3\right )\,\sqrt {A^2}}+\frac {A^2\,a\,b\,\sqrt {-a^2-b^2}}{\left (A\,a^2\,b+A\,b^3\right )\,\sqrt {A^2}}\right )\,\sqrt {A^2}}{\sqrt {-a^2-b^2}}+\frac {B\,a^2\,b\,\ln \left (8\,A^2\,a\,{\mathrm {e}}^x-4\,A^2\,b+4\,A^2\,b\,{\mathrm {e}}^{2\,x}\right )}{a^2\,b^2+b^4} \]

[In]

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

[Out]

(B*b^3*log(8*A^2*a*exp(x) - 4*A^2*b + 4*A^2*b*exp(2*x)))/(b^4 + a^2*b^2) - (B*x)/b - (2*atan((A^2*b^2*exp(x)*(
- a^2 - b^2)^(1/2))/((A*b^3 + A*a^2*b)*(A^2)^(1/2)) + (A^2*a*b*(- a^2 - b^2)^(1/2))/((A*b^3 + A*a^2*b)*(A^2)^(
1/2)))*(A^2)^(1/2))/(- a^2 - b^2)^(1/2) + (B*a^2*b*log(8*A^2*a*exp(x) - 4*A^2*b + 4*A^2*b*exp(2*x)))/(b^4 + a^
2*b^2)