Integrand size = 13, antiderivative size = 85 \[ \int \frac {\cosh ^3(x)}{a+b \cosh (x)} \, dx=\frac {\left (2 a^2+b^2\right ) x}{2 b^3}-\frac {2 a^3 \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^3 \sqrt {a+b}}-\frac {a \sinh (x)}{b^2}+\frac {\cosh (x) \sinh (x)}{2 b} \]
1/2*(2*a^2+b^2)*x/b^3-a*sinh(x)/b^2+1/2*cosh(x)*sinh(x)/b-2*a^3*arctanh((a -b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))/b^3/(a-b)^(1/2)/(a+b)^(1/2)
Time = 0.11 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.92 \[ \int \frac {\cosh ^3(x)}{a+b \cosh (x)} \, dx=\frac {4 a^2 x+2 b^2 x+\frac {8 a^3 \arctan \left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-4 a b \sinh (x)+b^2 \sinh (2 x)}{4 b^3} \]
(4*a^2*x + 2*b^2*x + (8*a^3*ArcTan[((a - b)*Tanh[x/2])/Sqrt[-a^2 + b^2]])/ Sqrt[-a^2 + b^2] - 4*a*b*Sinh[x] + b^2*Sinh[2*x])/(4*b^3)
Time = 0.53 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {3042, 3272, 3042, 3502, 3042, 3214, 3042, 3138, 221}
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.
\(\displaystyle \int \frac {\cosh ^3(x)}{a+b \cosh (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (\frac {\pi }{2}+i x\right )^3}{a+b \sin \left (\frac {\pi }{2}+i x\right )}dx\) |
\(\Big \downarrow \) 3272 |
\(\displaystyle \frac {\int \frac {-2 a \cosh ^2(x)+b \cosh (x)+a}{a+b \cosh (x)}dx}{2 b}+\frac {\sinh (x) \cosh (x)}{2 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sinh (x) \cosh (x)}{2 b}+\frac {\int \frac {-2 a \sin \left (i x+\frac {\pi }{2}\right )^2+b \sin \left (i x+\frac {\pi }{2}\right )+a}{a+b \sin \left (i x+\frac {\pi }{2}\right )}dx}{2 b}\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {\frac {\int \frac {a b+\left (2 a^2+b^2\right ) \cosh (x)}{a+b \cosh (x)}dx}{b}-\frac {2 a \sinh (x)}{b}}{2 b}+\frac {\sinh (x) \cosh (x)}{2 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sinh (x) \cosh (x)}{2 b}+\frac {-\frac {2 a \sinh (x)}{b}+\frac {\int \frac {a b+\left (2 a^2+b^2\right ) \sin \left (i x+\frac {\pi }{2}\right )}{a+b \sin \left (i x+\frac {\pi }{2}\right )}dx}{b}}{2 b}\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle \frac {\frac {\frac {x \left (2 a^2+b^2\right )}{b}-\frac {2 a^3 \int \frac {1}{a+b \cosh (x)}dx}{b}}{b}-\frac {2 a \sinh (x)}{b}}{2 b}+\frac {\sinh (x) \cosh (x)}{2 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sinh (x) \cosh (x)}{2 b}+\frac {-\frac {2 a \sinh (x)}{b}+\frac {\frac {x \left (2 a^2+b^2\right )}{b}-\frac {2 a^3 \int \frac {1}{a+b \sin \left (i x+\frac {\pi }{2}\right )}dx}{b}}{b}}{2 b}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {\frac {\frac {x \left (2 a^2+b^2\right )}{b}-\frac {4 a^3 \int \frac {1}{-\left ((a-b) \tanh ^2\left (\frac {x}{2}\right )\right )+a+b}d\tanh \left (\frac {x}{2}\right )}{b}}{b}-\frac {2 a \sinh (x)}{b}}{2 b}+\frac {\sinh (x) \cosh (x)}{2 b}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {\frac {x \left (2 a^2+b^2\right )}{b}-\frac {4 a^3 \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b \sqrt {a-b} \sqrt {a+b}}}{b}-\frac {2 a \sinh (x)}{b}}{2 b}+\frac {\sinh (x) \cosh (x)}{2 b}\) |
(Cosh[x]*Sinh[x])/(2*b) + ((((2*a^2 + b^2)*x)/b - (4*a^3*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]))/b - (2*a*Sinh[x] )/b)/(2*b)
3.1.55.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*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_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d *(m + n) + b^2*(b*c*(m - 2) + a*d*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Si n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m ] || IntegersQ[2*m, 2*n]) && !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(152\) vs. \(2(71)=142\).
Time = 0.18 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.80
method | result | size |
default | \(-\frac {2 a^{3} \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {-2 a -b}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {\left (-2 a^{2}-b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b^{3}}-\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {-2 a -b}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {\left (2 a^{2}+b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 b^{3}}\) | \(153\) |
risch | \(\frac {x \,a^{2}}{b^{3}}+\frac {x}{2 b}+\frac {{\mathrm e}^{2 x}}{8 b}-\frac {a \,{\mathrm e}^{x}}{2 b^{2}}+\frac {a \,{\mathrm e}^{-x}}{2 b^{2}}-\frac {{\mathrm e}^{-2 x}}{8 b}+\frac {a^{3} \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, b^{3}}-\frac {a^{3} \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, b^{3}}\) | \(171\) |
-2*a^3/b^3/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/ 2))+1/2/b/(tanh(1/2*x)-1)^2-1/2*(-2*a-b)/b^2/(tanh(1/2*x)-1)+1/2/b^3*(-2*a ^2-b^2)*ln(tanh(1/2*x)-1)-1/2/b/(tanh(1/2*x)+1)^2-1/2*(-2*a-b)/b^2/(tanh(1 /2*x)+1)+1/2*(2*a^2+b^2)/b^3*ln(tanh(1/2*x)+1)
Leaf count of result is larger than twice the leaf count of optimal. 417 vs. \(2 (71) = 142\).
Time = 0.28 (sec) , antiderivative size = 903, normalized size of antiderivative = 10.62 \[ \int \frac {\cosh ^3(x)}{a+b \cosh (x)} \, dx=\text {Too large to display} \]
[1/8*((a^2*b^2 - b^4)*cosh(x)^4 + (a^2*b^2 - b^4)*sinh(x)^4 - a^2*b^2 + b^ 4 + 4*(2*a^4 - a^2*b^2 - b^4)*x*cosh(x)^2 - 4*(a^3*b - a*b^3)*cosh(x)^3 - 4*(a^3*b - a*b^3 - (a^2*b^2 - b^4)*cosh(x))*sinh(x)^3 + 2*(3*(a^2*b^2 - b^ 4)*cosh(x)^2 + 2*(2*a^4 - a^2*b^2 - b^4)*x - 6*(a^3*b - a*b^3)*cosh(x))*si nh(x)^2 + 8*(a^3*cosh(x)^2 + 2*a^3*cosh(x)*sinh(x) + a^3*sinh(x)^2)*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)) + 4*(a^3*b - a*b^3)*cosh(x) + 4*(a^3*b - a*b^3 + (a^2*b^2 - b^4)* cosh(x)^3 + 2*(2*a^4 - a^2*b^2 - b^4)*x*cosh(x) - 3*(a^3*b - a*b^3)*cosh(x )^2)*sinh(x))/((a^2*b^3 - b^5)*cosh(x)^2 + 2*(a^2*b^3 - b^5)*cosh(x)*sinh( x) + (a^2*b^3 - b^5)*sinh(x)^2), 1/8*((a^2*b^2 - b^4)*cosh(x)^4 + (a^2*b^2 - b^4)*sinh(x)^4 - a^2*b^2 + b^4 + 4*(2*a^4 - a^2*b^2 - b^4)*x*cosh(x)^2 - 4*(a^3*b - a*b^3)*cosh(x)^3 - 4*(a^3*b - a*b^3 - (a^2*b^2 - b^4)*cosh(x) )*sinh(x)^3 + 2*(3*(a^2*b^2 - b^4)*cosh(x)^2 + 2*(2*a^4 - a^2*b^2 - b^4)*x - 6*(a^3*b - a*b^3)*cosh(x))*sinh(x)^2 + 16*(a^3*cosh(x)^2 + 2*a^3*cosh(x )*sinh(x) + a^3*sinh(x)^2)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*co sh(x) + b*sinh(x) + a)/(a^2 - b^2)) + 4*(a^3*b - a*b^3)*cosh(x) + 4*(a^3*b - a*b^3 + (a^2*b^2 - b^4)*cosh(x)^3 + 2*(2*a^4 - a^2*b^2 - b^4)*x*cosh(x) - 3*(a^3*b - a*b^3)*cosh(x)^2)*sinh(x))/((a^2*b^3 - b^5)*cosh(x)^2 + 2...
Leaf count of result is larger than twice the leaf count of optimal. 4559 vs. \(2 (73) = 146\).
Time = 157.92 (sec) , antiderivative size = 4559, normalized size of antiderivative = 53.64 \[ \int \frac {\cosh ^3(x)}{a+b \cosh (x)} \, dx=\text {Too large to display} \]
Piecewise((zoo*(-x*sinh(x)**2/2 + x*cosh(x)**2/2 + sinh(x)*cosh(x)/2), Eq( a, 0) & Eq(b, 0)), (3*x*tanh(x/2)**4/(2*b*tanh(x/2)**4 - 4*b*tanh(x/2)**2 + 2*b) - 6*x*tanh(x/2)**2/(2*b*tanh(x/2)**4 - 4*b*tanh(x/2)**2 + 2*b) + 3* x/(2*b*tanh(x/2)**4 - 4*b*tanh(x/2)**2 + 2*b) - 2*tanh(x/2)**5/(2*b*tanh(x /2)**4 - 4*b*tanh(x/2)**2 + 2*b) + 10*tanh(x/2)**3/(2*b*tanh(x/2)**4 - 4*b *tanh(x/2)**2 + 2*b) - 4*tanh(x/2)/(2*b*tanh(x/2)**4 - 4*b*tanh(x/2)**2 + 2*b), Eq(a, b)), (3*x*tanh(x/2)**5/(2*b*tanh(x/2)**5 - 4*b*tanh(x/2)**3 + 2*b*tanh(x/2)) - 6*x*tanh(x/2)**3/(2*b*tanh(x/2)**5 - 4*b*tanh(x/2)**3 + 2 *b*tanh(x/2)) + 3*x*tanh(x/2)/(2*b*tanh(x/2)**5 - 4*b*tanh(x/2)**3 + 2*b*t anh(x/2)) - 4*tanh(x/2)**4/(2*b*tanh(x/2)**5 - 4*b*tanh(x/2)**3 + 2*b*tanh (x/2)) + 10*tanh(x/2)**2/(2*b*tanh(x/2)**5 - 4*b*tanh(x/2)**3 + 2*b*tanh(x /2)) - 2/(2*b*tanh(x/2)**5 - 4*b*tanh(x/2)**3 + 2*b*tanh(x/2)), Eq(a, -b)) , ((-2*sinh(x)**3/3 + sinh(x)*cosh(x)**2)/a, Eq(b, 0)), (2*a**3*x*sqrt(a/( a - b) + b/(a - b))*tanh(x/2)**4/(2*a*b**3*sqrt(a/(a - b) + b/(a - b))*tan h(x/2)**4 - 4*a*b**3*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**2 + 2*a*b**3*s qrt(a/(a - b) + b/(a - b)) - 2*b**4*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)* *4 + 4*b**4*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**2 - 2*b**4*sqrt(a/(a - b) + b/(a - b))) - 4*a**3*x*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**2/(2*a* b**3*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**4 - 4*a*b**3*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**2 + 2*a*b**3*sqrt(a/(a - b) + b/(a - b)) - 2*b**4...
Exception generated. \[ \int \frac {\cosh ^3(x)}{a+b \cosh (x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f or more de
Time = 0.26 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.08 \[ \int \frac {\cosh ^3(x)}{a+b \cosh (x)} \, dx=-\frac {2 \, a^{3} \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} b^{3}} + \frac {b e^{\left (2 \, x\right )} - 4 \, a e^{x}}{8 \, b^{2}} + \frac {{\left (2 \, a^{2} + b^{2}\right )} x}{2 \, b^{3}} + \frac {{\left (4 \, a b e^{x} - b^{2}\right )} e^{\left (-2 \, x\right )}}{8 \, b^{3}} \]
-2*a^3*arctan((b*e^x + a)/sqrt(-a^2 + b^2))/(sqrt(-a^2 + b^2)*b^3) + 1/8*( b*e^(2*x) - 4*a*e^x)/b^2 + 1/2*(2*a^2 + b^2)*x/b^3 + 1/8*(4*a*b*e^x - b^2) *e^(-2*x)/b^3
Time = 1.92 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.96 \[ \int \frac {\cosh ^3(x)}{a+b \cosh (x)} \, dx=\frac {{\mathrm {e}}^{2\,x}}{8\,b}-\frac {{\mathrm {e}}^{-2\,x}}{8\,b}-\frac {a\,{\mathrm {e}}^x}{2\,b^2}+\frac {a\,{\mathrm {e}}^{-x}}{2\,b^2}+\frac {x\,\left (2\,a^2+b^2\right )}{2\,b^3}+\frac {a^3\,\ln \left (\frac {2\,a^3\,{\mathrm {e}}^x}{b^4}-\frac {2\,a^3\,\left (b+a\,{\mathrm {e}}^x\right )}{b^4\,\sqrt {a+b}\,\sqrt {a-b}}\right )}{b^3\,\sqrt {a+b}\,\sqrt {a-b}}-\frac {a^3\,\ln \left (\frac {2\,a^3\,{\mathrm {e}}^x}{b^4}+\frac {2\,a^3\,\left (b+a\,{\mathrm {e}}^x\right )}{b^4\,\sqrt {a+b}\,\sqrt {a-b}}\right )}{b^3\,\sqrt {a+b}\,\sqrt {a-b}} \]
exp(2*x)/(8*b) - exp(-2*x)/(8*b) - (a*exp(x))/(2*b^2) + (a*exp(-x))/(2*b^2 ) + (x*(2*a^2 + b^2))/(2*b^3) + (a^3*log((2*a^3*exp(x))/b^4 - (2*a^3*(b + a*exp(x)))/(b^4*(a + b)^(1/2)*(a - b)^(1/2))))/(b^3*(a + b)^(1/2)*(a - b)^ (1/2)) - (a^3*log((2*a^3*exp(x))/b^4 + (2*a^3*(b + a*exp(x)))/(b^4*(a + b) ^(1/2)*(a - b)^(1/2))))/(b^3*(a + b)^(1/2)*(a - b)^(1/2))