Integrand size = 13, antiderivative size = 57 \[ \int \frac {\sinh ^2(x)}{a+b \sinh (x)} \, dx=-\frac {a x}{b^2}-\frac {2 a^2 \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^2 \sqrt {a^2+b^2}}+\frac {\cosh (x)}{b} \] Output:
-a*x/b^2-2*a^2*arctanh((b-a*tanh(1/2*x))/(a^2+b^2)^(1/2))/b^2/(a^2+b^2)^(1 /2)+cosh(x)/b
Time = 0.23 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.07 \[ \int \frac {\sinh ^2(x)}{a+b \sinh (x)} \, dx=\frac {a \left (-x+\frac {2 a \arctan \left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}\right )+b \cosh (x)}{b^2} \] Input:
Integrate[Sinh[x]^2/(a + b*Sinh[x]),x]
Output:
(a*(-x + (2*a*ArcTan[(b - a*Tanh[x/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2] ) + b*Cosh[x])/b^2
Result contains complex when optimal does not.
Time = 0.46 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.26, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.923, Rules used = {3042, 25, 3225, 26, 27, 3042, 26, 3214, 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.
\(\displaystyle \int \frac {\sinh ^2(x)}{a+b \sinh (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\sin (i x)^2}{a-i b \sin (i x)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\sin (i x)^2}{a-i b \sin (i x)}dx\) |
\(\Big \downarrow \) 3225 |
\(\displaystyle \frac {\cosh (x)}{b}-\frac {i \int -\frac {i a \sinh (x)}{a+b \sinh (x)}dx}{b}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\cosh (x)}{b}-\frac {\int \frac {a \sinh (x)}{a+b \sinh (x)}dx}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\cosh (x)}{b}-\frac {a \int \frac {\sinh (x)}{a+b \sinh (x)}dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\cosh (x)}{b}-\frac {a \int -\frac {i \sin (i x)}{a-i b \sin (i x)}dx}{b}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\cosh (x)}{b}+\frac {i a \int \frac {\sin (i x)}{a-i b \sin (i x)}dx}{b}\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle \frac {\cosh (x)}{b}+\frac {i a \left (\frac {i x}{b}-\frac {i a \int \frac {1}{a+b \sinh (x)}dx}{b}\right )}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\cosh (x)}{b}+\frac {i a \left (\frac {i x}{b}-\frac {i a \int \frac {1}{a-i b \sin (i x)}dx}{b}\right )}{b}\) |
\(\Big \downarrow \) 3139 |
\(\displaystyle \frac {\cosh (x)}{b}+\frac {i a \left (\frac {i x}{b}-\frac {2 i a \int \frac {1}{-a \tanh ^2\left (\frac {x}{2}\right )+2 b \tanh \left (\frac {x}{2}\right )+a}d\tanh \left (\frac {x}{2}\right )}{b}\right )}{b}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {\cosh (x)}{b}+\frac {i a \left (\frac {4 i a \int \frac {1}{4 \left (a^2+b^2\right )-\left (2 b-2 a \tanh \left (\frac {x}{2}\right )\right )^2}d\left (2 b-2 a \tanh \left (\frac {x}{2}\right )\right )}{b}+\frac {i x}{b}\right )}{b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\cosh (x)}{b}+\frac {i a \left (\frac {2 i a \text {arctanh}\left (\frac {2 b-2 a \tanh \left (\frac {x}{2}\right )}{2 \sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2}}+\frac {i x}{b}\right )}{b}\) |
Input:
Int[Sinh[x]^2/(a + b*Sinh[x]),x]
Output:
(I*a*((I*x)/b + ((2*I)*a*ArcTanh[(2*b - 2*a*Tanh[x/2])/(2*Sqrt[a^2 + b^2]) ])/(b*Sqrt[a^2 + b^2])))/b + Cosh[x]/b
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_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2/((c_.) + (d_.)*sin[(e_.) + (f _.)*(x_)]), x_Symbol] :> Simp[(-b^2)*(Cos[e + f*x]/(d*f)), x] + Simp[1/d Int[Simp[a^2*d - b*(b*c - 2*a*d)*Sin[e + f*x], x]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Time = 0.16 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.61
method | result | size |
default | \(-\frac {1}{b \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {a \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{b^{2}}+\frac {1}{b \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {a \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{b^{2}}+\frac {2 a^{2} \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{b^{2} \sqrt {a^{2}+b^{2}}}\) | \(92\) |
risch | \(-\frac {a x}{b^{2}}+\frac {{\mathrm e}^{x}}{2 b}+\frac {{\mathrm e}^{-x}}{2 b}+\frac {a^{2} \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}+b^{2}}-a^{2}-b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, b^{2}}-\frac {a^{2} \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}+b^{2}}+a^{2}+b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, b^{2}}\) | \(132\) |
Input:
int(sinh(x)^2/(a+b*sinh(x)),x,method=_RETURNVERBOSE)
Output:
-1/b/(tanh(1/2*x)-1)+a/b^2*ln(tanh(1/2*x)-1)+1/b/(tanh(1/2*x)+1)-a/b^2*ln( tanh(1/2*x)+1)+2*a^2/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. 238 vs. \(2 (53) = 106\).
Time = 0.10 (sec) , antiderivative size = 238, normalized size of antiderivative = 4.18 \[ \int \frac {\sinh ^2(x)}{a+b \sinh (x)} \, dx=\frac {a^{2} b + b^{3} - 2 \, {\left (a^{3} + a b^{2}\right )} x \cosh \left (x\right ) + {\left (a^{2} b + b^{3}\right )} \cosh \left (x\right )^{2} + {\left (a^{2} b + b^{3}\right )} \sinh \left (x\right )^{2} + 2 \, {\left (a^{2} \cosh \left (x\right ) + a^{2} \sinh \left (x\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \, {\left (b \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) - b}\right ) - 2 \, {\left ({\left (a^{3} + a b^{2}\right )} x - {\left (a^{2} b + b^{3}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )}{2 \, {\left ({\left (a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right ) + {\left (a^{2} b^{2} + b^{4}\right )} \sinh \left (x\right )\right )}} \] Input:
integrate(sinh(x)^2/(a+b*sinh(x)),x, algorithm="fricas")
Output:
1/2*(a^2*b + b^3 - 2*(a^3 + a*b^2)*x*cosh(x) + (a^2*b + b^3)*cosh(x)^2 + ( a^2*b + b^3)*sinh(x)^2 + 2*(a^2*cosh(x) + a^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*cos h(x) + a*b)*sinh(x) - 2*sqrt(a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*co sh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)*sinh(x) - b)) - 2* ((a^3 + a*b^2)*x - (a^2*b + b^3)*cosh(x))*sinh(x))/((a^2*b^2 + b^4)*cosh(x ) + (a^2*b^2 + b^4)*sinh(x))
Timed out. \[ \int \frac {\sinh ^2(x)}{a+b \sinh (x)} \, dx=\text {Timed out} \] Input:
integrate(sinh(x)**2/(a+b*sinh(x)),x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.47 \[ \int \frac {\sinh ^2(x)}{a+b \sinh (x)} \, dx=\frac {a^{2} \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} b^{2}} - \frac {a x}{b^{2}} + \frac {e^{\left (-x\right )}}{2 \, b} + \frac {e^{x}}{2 \, b} \] Input:
integrate(sinh(x)^2/(a+b*sinh(x)),x, algorithm="maxima")
Output:
a^2*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^2) - a*x/b^2 + 1/2*e^(-x)/b + 1/2*e^x/b
Time = 0.13 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.51 \[ \int \frac {\sinh ^2(x)}{a+b \sinh (x)} \, dx=\frac {a^{2} \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b^{2}} - \frac {a x}{b^{2}} + \frac {e^{\left (-x\right )}}{2 \, b} + \frac {e^{x}}{2 \, b} \] Input:
integrate(sinh(x)^2/(a+b*sinh(x)),x, algorithm="giac")
Output:
a^2*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^2) - a*x/b^2 + 1/2*e^(-x)/b + 1/2*e^x/b
Time = 1.83 (sec) , antiderivative size = 129, normalized size of antiderivative = 2.26 \[ \int \frac {\sinh ^2(x)}{a+b \sinh (x)} \, dx=\frac {{\mathrm {e}}^{-x}}{2\,b}+\frac {{\mathrm {e}}^x}{2\,b}-\frac {a\,x}{b^2}-\frac {a^2\,\ln \left (-\frac {2\,a^2\,{\mathrm {e}}^x}{b^3}-\frac {2\,a^2\,\left (b-a\,{\mathrm {e}}^x\right )}{b^3\,\sqrt {a^2+b^2}}\right )}{b^2\,\sqrt {a^2+b^2}}+\frac {a^2\,\ln \left (\frac {2\,a^2\,\left (b-a\,{\mathrm {e}}^x\right )}{b^3\,\sqrt {a^2+b^2}}-\frac {2\,a^2\,{\mathrm {e}}^x}{b^3}\right )}{b^2\,\sqrt {a^2+b^2}} \] Input:
int(sinh(x)^2/(a + b*sinh(x)),x)
Output:
exp(-x)/(2*b) + exp(x)/(2*b) - (a*x)/b^2 - (a^2*log(- (2*a^2*exp(x))/b^3 - (2*a^2*(b - a*exp(x)))/(b^3*(a^2 + b^2)^(1/2))))/(b^2*(a^2 + b^2)^(1/2)) + (a^2*log((2*a^2*(b - a*exp(x)))/(b^3*(a^2 + b^2)^(1/2)) - (2*a^2*exp(x)) /b^3))/(b^2*(a^2 + b^2)^(1/2))
Time = 0.16 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.33 \[ \int \frac {\sinh ^2(x)}{a+b \sinh (x)} \, dx=\frac {2 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a^{2} i +\cosh \left (x \right ) a^{2} b +\cosh \left (x \right ) b^{3}-a^{3} x -a \,b^{2} x}{b^{2} \left (a^{2}+b^{2}\right )} \] Input:
int(sinh(x)^2/(a+b*sinh(x)),x)
Output:
(2*sqrt(a**2 + b**2)*atan((e**x*b*i + a*i)/sqrt(a**2 + b**2))*a**2*i + cos h(x)*a**2*b + cosh(x)*b**3 - a**3*x - a*b**2*x)/(b**2*(a**2 + b**2))