\(\int \frac {A+B \sinh (x)}{(a+b \sinh (x))^2} \, dx\) [130]

Optimal result

Integrand size = 15, antiderivative size = 74 \[ \int \frac {A+B \sinh (x)}{(a+b \sinh (x))^2} \, dx=-\frac {2 (a A+b B) \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}-\frac {(A b-a B) \cosh (x)}{\left (a^2+b^2\right ) (a+b \sinh (x))} \] Output:

-2*(A*a+B*b)*arctanh((b-a*tanh(1/2*x))/(a^2+b^2)^(1/2))/(a^2+b^2)^(3/2)-(A 
*b-B*a)*cosh(x)/(a^2+b^2)/(a+b*sinh(x))
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.11 \[ \int \frac {A+B \sinh (x)}{(a+b \sinh (x))^2} \, dx=\frac {\frac {2 (a A+b B) \arctan \left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}+\frac {(-A b+a B) \cosh (x)}{a+b \sinh (x)}}{a^2+b^2} \] Input:

Integrate[(A + B*Sinh[x])/(a + b*Sinh[x])^2,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {3042, 3233, 25, 27, 3042, 3139, 1083, 219}

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.

\(\displaystyle \int \frac {A+B \sinh (x)}{(a+b \sinh (x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A-i B \sin (i x)}{(a-i b \sin (i x))^2}dx\)

\(\Big \downarrow \) 3233

\(\displaystyle -\frac {\int -\frac {a A+b B}{a+b \sinh (x)}dx}{a^2+b^2}-\frac {\cosh (x) (A b-a B)}{\left (a^2+b^2\right ) (a+b \sinh (x))}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {a A+b B}{a+b \sinh (x)}dx}{a^2+b^2}-\frac {\cosh (x) (A b-a B)}{\left (a^2+b^2\right ) (a+b \sinh (x))}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(a A+b B) \int \frac {1}{a+b \sinh (x)}dx}{a^2+b^2}-\frac {\cosh (x) (A b-a B)}{\left (a^2+b^2\right ) (a+b \sinh (x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\cosh (x) (A b-a B)}{\left (a^2+b^2\right ) (a+b \sinh (x))}+\frac {(a A+b B) \int \frac {1}{a-i b \sin (i x)}dx}{a^2+b^2}\)

\(\Big \downarrow \) 3139

\(\displaystyle \frac {2 (a A+b B) \int \frac {1}{-a \tanh ^2\left (\frac {x}{2}\right )+2 b \tanh \left (\frac {x}{2}\right )+a}d\tanh \left (\frac {x}{2}\right )}{a^2+b^2}-\frac {\cosh (x) (A b-a B)}{\left (a^2+b^2\right ) (a+b \sinh (x))}\)

\(\Big \downarrow \) 1083

\(\displaystyle -\frac {4 (a A+b B) \int \frac {1}{4 \left (a^2+b^2\right )-\left (2 b-2 a \tanh \left (\frac {x}{2}\right )\right )^2}d\left (2 b-2 a \tanh \left (\frac {x}{2}\right )\right )}{a^2+b^2}-\frac {\cosh (x) (A b-a B)}{\left (a^2+b^2\right ) (a+b \sinh (x))}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {2 (a A+b B) \text {arctanh}\left (\frac {2 b-2 a \tanh \left (\frac {x}{2}\right )}{2 \sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}-\frac {\cosh (x) (A b-a B)}{\left (a^2+b^2\right ) (a+b \sinh (x))}\)

Input:

Int[(A + B*Sinh[x])/(a + b*Sinh[x])^2,x]
 

Output:

(-2*(a*A + b*B)*ArcTanh[(2*b - 2*a*Tanh[x/2])/(2*Sqrt[a^2 + b^2])])/(a^2 + 
 b^2)^(3/2) - ((A*b - a*B)*Cosh[x])/((a^2 + b^2)*(a + b*Sinh[x]))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1083

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3139

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]
 

rule 3233

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + 
 f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) 
  Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*( 
m + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c 
- a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
 

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.53

method	result	size
default	\(-\frac {2 \left (-\frac {b \left (A b -a B \right ) \tanh \left (\frac {x}{2}\right )}{a \left (a^{2}+b^{2}\right )}-\frac {A b -a B}{a^{2}+b^{2}}\right )}{\tanh \left (\frac {x}{2}\right )^{2} a -2 b \tanh \left (\frac {x}{2}\right )-a}+\frac {2 \left (A a +b B \right ) \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\)	\(113\)
risch	\(\frac {2 \left (A b -a B \right ) \left ({\mathrm e}^{x} a -b \right )}{b \left (a^{2}+b^{2}\right ) \left (b \,{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} a -b \right )}+\frac {\ln \left ({\mathrm e}^{x}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}-a^{4}-2 a^{2} b^{2}-b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right ) A a}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}+\frac {\ln \left ({\mathrm e}^{x}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}-a^{4}-2 a^{2} b^{2}-b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right ) b B}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}-\frac {\ln \left ({\mathrm e}^{x}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}+a^{4}+2 a^{2} b^{2}+b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right ) A a}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}-\frac {\ln \left ({\mathrm e}^{x}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}+a^{4}+2 a^{2} b^{2}+b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right ) b B}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\)	\(279\)

Input:

int((A+B*sinh(x))/(a+b*sinh(x))^2,x,method=_RETURNVERBOSE)
 

Output:

-2*(-b*(A*b-B*a)/a/(a^2+b^2)*tanh(1/2*x)-(A*b-B*a)/(a^2+b^2))/(tanh(1/2*x) 
^2*a-2*b*tanh(1/2*x)-a)+2*(A*a+B*b)/(a^2+b^2)^(3/2)*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. 444 vs. \(2 (69) = 138\).

Time = 0.14 (sec) , antiderivative size = 444, normalized size of antiderivative = 6.00 \[ \int \frac {A+B \sinh (x)}{(a+b \sinh (x))^2} \, dx=-\frac {2 \, B a^{3} b - 2 \, A a^{2} b^{2} + 2 \, B a b^{3} - 2 \, A b^{4} - {\left (A a b^{2} + B b^{3} - {\left (A a b^{2} + B b^{3}\right )} \cosh \left (x\right )^{2} - {\left (A a b^{2} + B b^{3}\right )} \sinh \left (x\right )^{2} - 2 \, {\left (A a^{2} b + B a b^{2}\right )} \cosh \left (x\right ) - 2 \, {\left (A a^{2} b + B a b^{2} + {\left (A a b^{2} + B b^{3}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \, {\left (b \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) - b}\right ) - 2 \, {\left (B a^{4} - A a^{3} b + B a^{2} b^{2} - A a b^{3}\right )} \cosh \left (x\right ) - 2 \, {\left (B a^{4} - A a^{3} b + B a^{2} b^{2} - A a b^{3}\right )} \sinh \left (x\right )}{a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6} - {\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} \cosh \left (x\right )^{2} - {\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} \sinh \left (x\right )^{2} - 2 \, {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \cosh \left (x\right ) - 2 \, {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5} + {\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )} \] Input:

integrate((A+B*sinh(x))/(a+b*sinh(x))^2,x, algorithm="fricas")
 

Output:

-(2*B*a^3*b - 2*A*a^2*b^2 + 2*B*a*b^3 - 2*A*b^4 - (A*a*b^2 + B*b^3 - (A*a* 
b^2 + B*b^3)*cosh(x)^2 - (A*a*b^2 + B*b^3)*sinh(x)^2 - 2*(A*a^2*b + B*a*b^ 
2)*cosh(x) - 2*(A*a^2*b + B*a*b^2 + (A*a*b^2 + B*b^3)*cosh(x))*sinh(x))*sq 
rt(a^2 + b^2)*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*si 
nh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)*s 
inh(x) - b)) - 2*(B*a^4 - A*a^3*b + B*a^2*b^2 - A*a*b^3)*cosh(x) - 2*(B*a^ 
4 - A*a^3*b + B*a^2*b^2 - A*a*b^3)*sinh(x))/(a^4*b^2 + 2*a^2*b^4 + b^6 - ( 
a^4*b^2 + 2*a^2*b^4 + b^6)*cosh(x)^2 - (a^4*b^2 + 2*a^2*b^4 + b^6)*sinh(x) 
^2 - 2*(a^5*b + 2*a^3*b^3 + a*b^5)*cosh(x) - 2*(a^5*b + 2*a^3*b^3 + a*b^5 
+ (a^4*b^2 + 2*a^2*b^4 + b^6)*cosh(x))*sinh(x))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \sinh (x)}{(a+b \sinh (x))^2} \, dx=\text {Timed out} \] Input:

integrate((A+B*sinh(x))/(a+b*sinh(x))**2,x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 229 vs. \(2 (69) = 138\).

Time = 0.12 (sec) , antiderivative size = 229, normalized size of antiderivative = 3.09 \[ \int \frac {A+B \sinh (x)}{(a+b \sinh (x))^2} \, dx=A {\left (\frac {a \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (a e^{\left (-x\right )} + b\right )}}{a^{2} b + b^{3} + 2 \, {\left (a^{3} + a b^{2}\right )} e^{\left (-x\right )} - {\left (a^{2} b + b^{3}\right )} e^{\left (-2 \, x\right )}}\right )} + B {\left (\frac {b \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (a^{2} e^{\left (-x\right )} + a b\right )}}{a^{2} b^{2} + b^{4} + 2 \, {\left (a^{3} b + a b^{3}\right )} e^{\left (-x\right )} - {\left (a^{2} b^{2} + b^{4}\right )} e^{\left (-2 \, x\right )}}\right )} \] Input:

integrate((A+B*sinh(x))/(a+b*sinh(x))^2,x, algorithm="maxima")
 

Output:

A*(a*log((b*e^(-x) - a - sqrt(a^2 + b^2))/(b*e^(-x) - a + sqrt(a^2 + b^2)) 
)/(a^2 + b^2)^(3/2) - 2*(a*e^(-x) + b)/(a^2*b + b^3 + 2*(a^3 + a*b^2)*e^(- 
x) - (a^2*b + b^3)*e^(-2*x))) + B*(b*log((b*e^(-x) - a - sqrt(a^2 + b^2))/ 
(b*e^(-x) - a + sqrt(a^2 + b^2)))/(a^2 + b^2)^(3/2) + 2*(a^2*e^(-x) + a*b) 
/(a^2*b^2 + b^4 + 2*(a^3*b + a*b^3)*e^(-x) - (a^2*b^2 + b^4)*e^(-2*x)))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.61 \[ \int \frac {A+B \sinh (x)}{(a+b \sinh (x))^2} \, dx=\frac {{\left (A a + B b\right )} \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (B a^{2} e^{x} - A a b e^{x} - B a b + A b^{2}\right )}}{{\left (a^{2} b + b^{3}\right )} {\left (b e^{\left (2 \, x\right )} + 2 \, a e^{x} - b\right )}} \] Input:

integrate((A+B*sinh(x))/(a+b*sinh(x))^2,x, algorithm="giac")
 

Output:

(A*a + B*b)*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)))/(a^2 + b^2)^(3/2) - 2*(B*a^2*e^x - A*a*b*e^x - B*a*b 
+ A*b^2)/((a^2*b + b^3)*(b*e^(2*x) + 2*a*e^x - b))
 

Mupad [B] (verification not implemented)

Time = 1.79 (sec) , antiderivative size = 223, normalized size of antiderivative = 3.01 \[ \int \frac {A+B \sinh (x)}{(a+b \sinh (x))^2} \, dx=\frac {\ln \left (\frac {2\,\left (b-a\,{\mathrm {e}}^x\right )\,\left (A\,a+B\,b\right )}{b\,{\left (a^2+b^2\right )}^{3/2}}-\frac {2\,{\mathrm {e}}^x\,\left (A\,a+B\,b\right )}{a^2\,b+b^3}\right )\,\left (A\,a+B\,b\right )}{{\left (a^2+b^2\right )}^{3/2}}-\frac {\ln \left (-\frac {2\,{\mathrm {e}}^x\,\left (A\,a+B\,b\right )}{a^2\,b+b^3}-\frac {2\,\left (b-a\,{\mathrm {e}}^x\right )\,\left (A\,a+B\,b\right )}{b\,{\left (a^2+b^2\right )}^{3/2}}\right )\,\left (A\,a+B\,b\right )}{{\left (a^2+b^2\right )}^{3/2}}-\frac {\frac {2\,\left (A\,b^3-B\,a\,b^2\right )}{b\,\left (a^2\,b+b^3\right )}+\frac {2\,{\mathrm {e}}^x\,\left (B\,a^2\,b^2-A\,a\,b^3\right )}{b^2\,\left (a^2\,b+b^3\right )}}{2\,a\,{\mathrm {e}}^x-b+b\,{\mathrm {e}}^{2\,x}} \] Input:

int((A + B*sinh(x))/(a + b*sinh(x))^2,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.57 \[ \int \frac {A+B \sinh (x)}{(a+b \sinh (x))^2} \, dx=\frac {2 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) i}{a^{2}+b^{2}} \] Input:

int((A+B*sinh(x))/(a+b*sinh(x))^2,x)
 

Output:

(2*sqrt(a**2 + b**2)*atan((e**x*b*i + a*i)/sqrt(a**2 + b**2))*i)/(a**2 + b 
**2)