Integrand size = 11, antiderivative size = 60 \[ \int \frac {\sinh (x)}{(a+b \sinh (x))^2} \, dx=-\frac {2 b \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}+\frac {a \cosh (x)}{\left (a^2+b^2\right ) (a+b \sinh (x))} \] Output:
-2*b*arctanh((b-a*tanh(1/2*x))/(a^2+b^2)^(1/2))/(a^2+b^2)^(3/2)+a*cosh(x)/ (a^2+b^2)/(a+b*sinh(x))
Time = 0.12 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.13 \[ \int \frac {\sinh (x)}{(a+b \sinh (x))^2} \, dx=-\frac {2 b \arctan \left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\left (-a^2-b^2\right )^{3/2}}+\frac {a \cosh (x)}{\left (a^2+b^2\right ) (a+b \sinh (x))} \] Input:
Integrate[Sinh[x]/(a + b*Sinh[x])^2,x]
Output:
(-2*b*ArcTan[(b - a*Tanh[x/2])/Sqrt[-a^2 - b^2]])/(-a^2 - b^2)^(3/2) + (a* Cosh[x])/((a^2 + b^2)*(a + b*Sinh[x]))
Result contains complex when optimal does not.
Time = 0.33 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.23, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.818, Rules used = {3042, 26, 3233, 26, 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.
\(\displaystyle \int \frac {\sinh (x)}{(a+b \sinh (x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {i \sin (i x)}{(a-i b \sin (i x))^2}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {\sin (i x)}{(a-i b \sin (i x))^2}dx\) |
\(\Big \downarrow \) 3233 |
\(\displaystyle -i \left (\frac {i a \cosh (x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}-\frac {\int -\frac {i b}{a+b \sinh (x)}dx}{a^2+b^2}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \left (\frac {i \int \frac {b}{a+b \sinh (x)}dx}{a^2+b^2}+\frac {i a \cosh (x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -i \left (\frac {i b \int \frac {1}{a+b \sinh (x)}dx}{a^2+b^2}+\frac {i a \cosh (x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -i \left (\frac {i b \int \frac {1}{a-i b \sin (i x)}dx}{a^2+b^2}+\frac {i a \cosh (x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}\right )\) |
\(\Big \downarrow \) 3139 |
\(\displaystyle -i \left (\frac {2 i b \int \frac {1}{-a \tanh ^2\left (\frac {x}{2}\right )+2 b \tanh \left (\frac {x}{2}\right )+a}d\tanh \left (\frac {x}{2}\right )}{a^2+b^2}+\frac {i a \cosh (x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle -i \left (\frac {i a \cosh (x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}-\frac {4 i b \int \frac {1}{4 \left (a^2+b^2\right )-\left (2 b-2 a \tanh \left (\frac {x}{2}\right )\right )^2}d\left (2 b-2 a \tanh \left (\frac {x}{2}\right )\right )}{a^2+b^2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -i \left (\frac {i a \cosh (x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}-\frac {2 i b \text {arctanh}\left (\frac {2 b-2 a \tanh \left (\frac {x}{2}\right )}{2 \sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}\right )\) |
Input:
Int[Sinh[x]/(a + b*Sinh[x])^2,x]
Output:
(-I)*(((-2*I)*b*ArcTanh[(2*b - 2*a*Tanh[x/2])/(2*Sqrt[a^2 + b^2])])/(a^2 + b^2)^(3/2) + (I*a*Cosh[x])/((a^2 + b^2)*(a + b*Sinh[x])))
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_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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_)])^(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]
Time = 0.15 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.62
method | result | size |
default | \(\frac {8 b \tanh \left (\frac {x}{2}\right )+8 a}{\left (-4 a^{2}-4 b^{2}\right ) \left (\tanh \left (\frac {x}{2}\right )^{2} a -2 b \tanh \left (\frac {x}{2}\right )-a \right )}-\frac {8 b \,\operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (-4 a^{2}-4 b^{2}\right ) \sqrt {a^{2}+b^{2}}}\) | \(97\) |
risch | \(-\frac {2 a \left ({\mathrm e}^{x} a -b \right )}{b \left (a^{2}+b^{2}\right ) \left (b \,{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} a -b \right )}+\frac {b \ln \left ({\mathrm e}^{x}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}-a^{4}-2 a^{2} b^{2}-b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}-\frac {b \ln \left ({\mathrm e}^{x}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}+a^{4}+2 a^{2} b^{2}+b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\) | \(155\) |
Input:
int(sinh(x)/(a+b*sinh(x))^2,x,method=_RETURNVERBOSE)
Output:
4*(2*b*tanh(1/2*x)+2*a)/(-4*a^2-4*b^2)/(tanh(1/2*x)^2*a-2*b*tanh(1/2*x)-a) -8*b/(-4*a^2-4*b^2)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tanh(1/2*x)-2*b)/(a^2 +b^2)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 341 vs. \(2 (56) = 112\).
Time = 0.08 (sec) , antiderivative size = 341, normalized size of antiderivative = 5.68 \[ \int \frac {\sinh (x)}{(a+b \sinh (x))^2} \, dx=-\frac {2 \, a^{3} b + 2 \, a b^{3} + {\left (b^{3} \cosh \left (x\right )^{2} + b^{3} \sinh \left (x\right )^{2} + 2 \, a b^{2} \cosh \left (x\right ) - b^{3} + 2 \, {\left (b^{3} \cosh \left (x\right ) + a b^{2}\right )} \sinh \left (x\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \, {\left (b \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) - b}\right ) - 2 \, {\left (a^{4} + a^{2} b^{2}\right )} \cosh \left (x\right ) - 2 \, {\left (a^{4} + a^{2} b^{2}\right )} \sinh \left (x\right )}{a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6} - {\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} \cosh \left (x\right )^{2} - {\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} \sinh \left (x\right )^{2} - 2 \, {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \cosh \left (x\right ) - 2 \, {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5} + {\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )} \] Input:
integrate(sinh(x)/(a+b*sinh(x))^2,x, algorithm="fricas")
Output:
-(2*a^3*b + 2*a*b^3 + (b^3*cosh(x)^2 + b^3*sinh(x)^2 + 2*a*b^2*cosh(x) - b ^3 + 2*(b^3*cosh(x) + a*b^2)*sinh(x))*sqrt(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*sinh(x) + a))/(b*cosh(x)^2 + b*sinh( x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)*sinh(x) - b)) - 2*(a^4 + a^2*b^2)*c osh(x) - 2*(a^4 + a^2*b^2)*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))
Timed out. \[ \int \frac {\sinh (x)}{(a+b \sinh (x))^2} \, dx=\text {Timed out} \] Input:
integrate(sinh(x)/(a+b*sinh(x))**2,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 117 vs. \(2 (56) = 112\).
Time = 0.11 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.95 \[ \int \frac {\sinh (x)}{(a+b \sinh (x))^2} \, dx=\frac {b \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (a^{2} e^{\left (-x\right )} + a b\right )}}{a^{2} b^{2} + b^{4} + 2 \, {\left (a^{3} b + a b^{3}\right )} e^{\left (-x\right )} - {\left (a^{2} b^{2} + b^{4}\right )} e^{\left (-2 \, x\right )}} \] Input:
integrate(sinh(x)/(a+b*sinh(x))^2,x, algorithm="maxima")
Output:
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))
Time = 0.13 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.65 \[ \int \frac {\sinh (x)}{(a+b \sinh (x))^2} \, dx=\frac {b \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (a^{2} e^{x} - a b\right )}}{{\left (a^{2} b + b^{3}\right )} {\left (b e^{\left (2 \, x\right )} + 2 \, a e^{x} - b\right )}} \] Input:
integrate(sinh(x)/(a+b*sinh(x))^2,x, algorithm="giac")
Output:
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*(a^2*e^x - a*b)/((a^2*b + b^3)*(b*e^(2*x) + 2*a*e^x - b))
Time = 1.92 (sec) , antiderivative size = 142, normalized size of antiderivative = 2.37 \[ \int \frac {\sinh (x)}{(a+b \sinh (x))^2} \, dx=\frac {\frac {2\,a\,b}{a^2\,b+b^3}-\frac {2\,a^2\,{\mathrm {e}}^x}{a^2\,b+b^3}}{2\,a\,{\mathrm {e}}^x-b+b\,{\mathrm {e}}^{2\,x}}-\frac {b\,\ln \left (-\frac {2\,{\mathrm {e}}^x}{a^2+b^2}-\frac {2\,\left (b-a\,{\mathrm {e}}^x\right )}{{\left (a^2+b^2\right )}^{3/2}}\right )}{{\left (a^2+b^2\right )}^{3/2}}+\frac {b\,\ln \left (\frac {2\,\left (b-a\,{\mathrm {e}}^x\right )}{{\left (a^2+b^2\right )}^{3/2}}-\frac {2\,{\mathrm {e}}^x}{a^2+b^2}\right )}{{\left (a^2+b^2\right )}^{3/2}} \] Input:
int(sinh(x)/(a + b*sinh(x))^2,x)
Output:
((2*a*b)/(a^2*b + b^3) - (2*a^2*exp(x))/(a^2*b + b^3))/(2*a*exp(x) - b + b *exp(2*x)) - (b*log(- (2*exp(x))/(a^2 + b^2) - (2*(b - a*exp(x)))/(a^2 + b ^2)^(3/2)))/(a^2 + b^2)^(3/2) + (b*log((2*(b - a*exp(x)))/(a^2 + b^2)^(3/2 ) - (2*exp(x))/(a^2 + b^2)))/(a^2 + b^2)^(3/2)
Time = 0.17 (sec) , antiderivative size = 226, normalized size of antiderivative = 3.77 \[ \int \frac {\sinh (x)}{(a+b \sinh (x))^2} \, dx=\frac {2 e^{2 x} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) b^{2} i +4 e^{x} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a b i -2 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) b^{2} i +e^{2 x} a^{3}+e^{2 x} a \,b^{2}+a^{3}+a \,b^{2}}{e^{2 x} a^{4} b +2 e^{2 x} a^{2} b^{3}+e^{2 x} b^{5}+2 e^{x} a^{5}+4 e^{x} a^{3} b^{2}+2 e^{x} a \,b^{4}-a^{4} b -2 a^{2} b^{3}-b^{5}} \] Input:
int(sinh(x)/(a+b*sinh(x))^2,x)
Output:
(2*e**(2*x)*sqrt(a**2 + b**2)*atan((e**x*b*i + a*i)/sqrt(a**2 + b**2))*b** 2*i + 4*e**x*sqrt(a**2 + b**2)*atan((e**x*b*i + a*i)/sqrt(a**2 + b**2))*a* b*i - 2*sqrt(a**2 + b**2)*atan((e**x*b*i + a*i)/sqrt(a**2 + b**2))*b**2*i + e**(2*x)*a**3 + e**(2*x)*a*b**2 + a**3 + a*b**2)/(e**(2*x)*a**4*b + 2*e* *(2*x)*a**2*b**3 + e**(2*x)*b**5 + 2*e**x*a**5 + 4*e**x*a**3*b**2 + 2*e**x *a*b**4 - a**4*b - 2*a**2*b**3 - b**5)