Integrand size = 13, antiderivative size = 103 \[ \int (a \cosh (x)+b \sinh (x))^{5/2} \, dx=\frac {2}{5} (b \cosh (x)+a \sinh (x)) (a \cosh (x)+b \sinh (x))^{3/2}-\frac {6 i \left (a^2-b^2\right ) E\left (\left .\frac {1}{2} \left (i x-\tan ^{-1}(a,-i b)\right )\right |2\right ) \sqrt {a \cosh (x)+b \sinh (x)}}{5 \sqrt {\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}}} \]
2/5*(b*cosh(x)+a*sinh(x))*(a*cosh(x)+b*sinh(x))^(3/2)-6/5*I*(a^2-b^2)*(cos (1/2*I*x-1/2*arctan(a,-I*b))^2)^(1/2)/cos(1/2*I*x-1/2*arctan(a,-I*b))*Elli pticE(sin(1/2*I*x-1/2*arctan(a,-I*b)),2^(1/2))*(a*cosh(x)+b*sinh(x))^(1/2) /((a*cosh(x)+b*sinh(x))/(a^2-b^2)^(1/2))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.59 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.87 \[ \int (a \cosh (x)+b \sinh (x))^{5/2} \, dx=\frac {(a \cosh (x)+b \sinh (x)) \left (6 a \left (a^2-b^2\right )+2 a b^2 \cosh (2 x)+b \left (a^2+b^2\right ) \sinh (2 x)\right )-\frac {3 (a-b)^2 (a+b)^2 \left (b \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cosh ^2\left (x+\text {arctanh}\left (\frac {b}{a}\right )\right )\right ) \sinh \left (x+\text {arctanh}\left (\frac {b}{a}\right )\right )+\sqrt {-\sinh ^2\left (x+\text {arctanh}\left (\frac {b}{a}\right )\right )} \left (2 a \cosh \left (x+\text {arctanh}\left (\frac {b}{a}\right )\right )-b \sinh \left (x+\text {arctanh}\left (\frac {b}{a}\right )\right )\right )\right )}{a \sqrt {1-\frac {b^2}{a^2}} \sqrt {-\sinh ^2\left (x+\text {arctanh}\left (\frac {b}{a}\right )\right )}}}{5 b \sqrt {a \cosh (x)+b \sinh (x)}} \]
((a*Cosh[x] + b*Sinh[x])*(6*a*(a^2 - b^2) + 2*a*b^2*Cosh[2*x] + b*(a^2 + b ^2)*Sinh[2*x]) - (3*(a - b)^2*(a + b)^2*(b*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cosh[x + ArcTanh[b/a]]^2]*Sinh[x + ArcTanh[b/a]] + Sqrt[-Sinh[x + ArcTanh[b/a]]^2]*(2*a*Cosh[x + ArcTanh[b/a]] - b*Sinh[x + ArcTanh[b/a]]))) /(a*Sqrt[1 - b^2/a^2]*Sqrt[-Sinh[x + ArcTanh[b/a]]^2]))/(5*b*Sqrt[a*Cosh[x ] + b*Sinh[x]])
Time = 0.39 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3042, 3552, 3042, 3557, 3042, 3119}
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.
\(\displaystyle \int (a \cosh (x)+b \sinh (x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \cos (i x)-i b \sin (i x))^{5/2}dx\) |
\(\Big \downarrow \) 3552 |
\(\displaystyle \frac {3}{5} \left (a^2-b^2\right ) \int \sqrt {a \cosh (x)+b \sinh (x)}dx+\frac {2}{5} (a \sinh (x)+b \cosh (x)) (a \cosh (x)+b \sinh (x))^{3/2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2}{5} (a \sinh (x)+b \cosh (x)) (a \cosh (x)+b \sinh (x))^{3/2}+\frac {3}{5} \left (a^2-b^2\right ) \int \sqrt {a \cos (i x)-i b \sin (i x)}dx\) |
\(\Big \downarrow \) 3557 |
\(\displaystyle \frac {2}{5} (a \sinh (x)+b \cosh (x)) (a \cosh (x)+b \sinh (x))^{3/2}+\frac {3 \left (a^2-b^2\right ) \sqrt {a \cosh (x)+b \sinh (x)} \int \sqrt {\cosh \left (x+i \tan ^{-1}(a,-i b)\right )}dx}{5 \sqrt {\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2}{5} (a \sinh (x)+b \cosh (x)) (a \cosh (x)+b \sinh (x))^{3/2}+\frac {3 \left (a^2-b^2\right ) \sqrt {a \cosh (x)+b \sinh (x)} \int \sqrt {\sin \left (i x-\tan ^{-1}(a,-i b)+\frac {\pi }{2}\right )}dx}{5 \sqrt {\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {2}{5} (a \sinh (x)+b \cosh (x)) (a \cosh (x)+b \sinh (x))^{3/2}-\frac {6 i \left (a^2-b^2\right ) \sqrt {a \cosh (x)+b \sinh (x)} E\left (\left .\frac {1}{2} \left (i x-\tan ^{-1}(a,-i b)\right )\right |2\right )}{5 \sqrt {\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}}}\) |
(2*(b*Cosh[x] + a*Sinh[x])*(a*Cosh[x] + b*Sinh[x])^(3/2))/5 - (((6*I)/5)*( a^2 - b^2)*EllipticE[(I*x - ArcTan[a, (-I)*b])/2, 2]*Sqrt[a*Cosh[x] + b*Si nh[x]])/Sqrt[(a*Cosh[x] + b*Sinh[x])/Sqrt[a^2 - b^2]]
3.6.92.3.1 Defintions of rubi rules used
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x _Symbol] :> Simp[(-(b*Cos[c + d*x] - a*Sin[c + d*x]))*((a*Cos[c + d*x] + b* Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[(n - 1)*((a^2 + b^2)/n) Int[(a*Co s[c + d*x] + b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{a, b, c, d}, x] && N eQ[a^2 + b^2, 0] && !IntegerQ[(n - 1)/2] && GtQ[n, 1]
Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x _Symbol] :> Simp[(a*Cos[c + d*x] + b*Sin[c + d*x])^n/((a*Cos[c + d*x] + b*S in[c + d*x])/Sqrt[a^2 + b^2])^n Int[Cos[c + d*x - ArcTan[a, b]]^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && !(GeQ[n, 1] || LeQ[n, -1]) && !(GtQ[a^2 + b^2, 0] || EqQ[a^2 + b^2, 0])
Time = 0.91 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.41
method | result | size |
default | \(-\frac {\left (\left (a +b \right ) \left (a -b \right )\right )^{\frac {3}{2}} \left (\frac {\cosh \left (x \right )^{3}}{3}-\cosh \left (x \right )\right )}{\sqrt {-\sinh \left (x \right ) \sqrt {a^{2}-b^{2}}}}\) | \(42\) |
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 274, normalized size of antiderivative = 2.66 \[ \int (a \cosh (x)+b \sinh (x))^{5/2} \, dx=-\frac {12 \, {\left (\sqrt {2} {\left (a^{2} - b^{2}\right )} \cosh \left (x\right )^{2} + 2 \, \sqrt {2} {\left (a^{2} - b^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right ) + \sqrt {2} {\left (a^{2} - b^{2}\right )} \sinh \left (x\right )^{2}\right )} \sqrt {a + b} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (a - b\right )}}{a + b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (a - b\right )}}{a + b}, 0, \cosh \left (x\right ) + \sinh \left (x\right )\right )\right ) - {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (x\right )^{4} + 4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right )^{3} + {\left (a^{2} + 2 \, a b + b^{2}\right )} \sinh \left (x\right )^{4} - 12 \, {\left (a^{2} - b^{2}\right )} \cosh \left (x\right )^{2} + 6 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (x\right )^{2} - 2 \, a^{2} + 2 \, b^{2}\right )} \sinh \left (x\right )^{2} - a^{2} + 2 \, a b - b^{2} + 4 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (x\right )^{3} - 6 \, {\left (a^{2} - b^{2}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )\right )} \sqrt {a \cosh \left (x\right ) + b \sinh \left (x\right )}}{10 \, {\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2}\right )}} \]
-1/10*(12*(sqrt(2)*(a^2 - b^2)*cosh(x)^2 + 2*sqrt(2)*(a^2 - b^2)*cosh(x)*s inh(x) + sqrt(2)*(a^2 - b^2)*sinh(x)^2)*sqrt(a + b)*weierstrassZeta(-4*(a - b)/(a + b), 0, weierstrassPInverse(-4*(a - b)/(a + b), 0, cosh(x) + sinh (x))) - ((a^2 + 2*a*b + b^2)*cosh(x)^4 + 4*(a^2 + 2*a*b + b^2)*cosh(x)*sin h(x)^3 + (a^2 + 2*a*b + b^2)*sinh(x)^4 - 12*(a^2 - b^2)*cosh(x)^2 + 6*((a^ 2 + 2*a*b + b^2)*cosh(x)^2 - 2*a^2 + 2*b^2)*sinh(x)^2 - a^2 + 2*a*b - b^2 + 4*((a^2 + 2*a*b + b^2)*cosh(x)^3 - 6*(a^2 - b^2)*cosh(x))*sinh(x))*sqrt( a*cosh(x) + b*sinh(x)))/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)
Timed out. \[ \int (a \cosh (x)+b \sinh (x))^{5/2} \, dx=\text {Timed out} \]
\[ \int (a \cosh (x)+b \sinh (x))^{5/2} \, dx=\int { {\left (a \cosh \left (x\right ) + b \sinh \left (x\right )\right )}^{\frac {5}{2}} \,d x } \]
\[ \int (a \cosh (x)+b \sinh (x))^{5/2} \, dx=\int { {\left (a \cosh \left (x\right ) + b \sinh \left (x\right )\right )}^{\frac {5}{2}} \,d x } \]
Timed out. \[ \int (a \cosh (x)+b \sinh (x))^{5/2} \, dx=\int {\left (a\,\mathrm {cosh}\left (x\right )+b\,\mathrm {sinh}\left (x\right )\right )}^{5/2} \,d x \]