Integrand size = 15, antiderivative size = 92 \[ \int (a-a \cosh (c+d x))^{5/2} \, dx=-\frac {64 a^3 \sinh (c+d x)}{15 d \sqrt {a-a \cosh (c+d x)}}-\frac {16 a^2 \sqrt {a-a \cosh (c+d x)} \sinh (c+d x)}{15 d}-\frac {2 a (a-a \cosh (c+d x))^{3/2} \sinh (c+d x)}{5 d} \] Output:
-64/15*a^3*sinh(d*x+c)/d/(a-a*cosh(d*x+c))^(1/2)-16/15*a^2*(a-a*cosh(d*x+c ))^(1/2)*sinh(d*x+c)/d-2/5*a*(a-a*cosh(d*x+c))^(3/2)*sinh(d*x+c)/d
Time = 0.19 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.78 \[ \int (a-a \cosh (c+d x))^{5/2} \, dx=\frac {a^2 \sqrt {a-a \cosh (c+d x)} \left (150 \cosh \left (\frac {1}{2} (c+d x)\right )-25 \cosh \left (\frac {3}{2} (c+d x)\right )+3 \cosh \left (\frac {5}{2} (c+d x)\right )\right ) \text {csch}\left (\frac {1}{2} (c+d x)\right )}{30 d} \] Input:
Integrate[(a - a*Cosh[c + d*x])^(5/2),x]
Output:
(a^2*Sqrt[a - a*Cosh[c + d*x]]*(150*Cosh[(c + d*x)/2] - 25*Cosh[(3*(c + d* x))/2] + 3*Cosh[(5*(c + d*x))/2])*Csch[(c + d*x)/2])/(30*d)
Time = 0.39 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3126, 3042, 3126, 3042, 3125}
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.
\(\displaystyle \int (a-a \cosh (c+d x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a-a \sin \left (i c+i d x+\frac {\pi }{2}\right )\right )^{5/2}dx\) |
\(\Big \downarrow \) 3126 |
\(\displaystyle \frac {8}{5} a \int (a-a \cosh (c+d x))^{3/2}dx-\frac {2 a \sinh (c+d x) (a-a \cosh (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 a \sinh (c+d x) (a-a \cosh (c+d x))^{3/2}}{5 d}+\frac {8}{5} a \int \left (a-a \sin \left (i c+i d x+\frac {\pi }{2}\right )\right )^{3/2}dx\) |
\(\Big \downarrow \) 3126 |
\(\displaystyle \frac {8}{5} a \left (\frac {4}{3} a \int \sqrt {a-a \cosh (c+d x)}dx-\frac {2 a \sinh (c+d x) \sqrt {a-a \cosh (c+d x)}}{3 d}\right )-\frac {2 a \sinh (c+d x) (a-a \cosh (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 a \sinh (c+d x) (a-a \cosh (c+d x))^{3/2}}{5 d}+\frac {8}{5} a \left (-\frac {2 a \sinh (c+d x) \sqrt {a-a \cosh (c+d x)}}{3 d}+\frac {4}{3} a \int \sqrt {a-a \sin \left (i c+i d x+\frac {\pi }{2}\right )}dx\right )\) |
\(\Big \downarrow \) 3125 |
\(\displaystyle \frac {8}{5} a \left (-\frac {8 a^2 \sinh (c+d x)}{3 d \sqrt {a-a \cosh (c+d x)}}-\frac {2 a \sinh (c+d x) \sqrt {a-a \cosh (c+d x)}}{3 d}\right )-\frac {2 a \sinh (c+d x) (a-a \cosh (c+d x))^{3/2}}{5 d}\) |
Input:
Int[(a - a*Cosh[c + d*x])^(5/2),x]
Output:
(-2*a*(a - a*Cosh[c + d*x])^(3/2)*Sinh[c + d*x])/(5*d) + (8*a*((-8*a^2*Sin h[c + d*x])/(3*d*Sqrt[a - a*Cosh[c + d*x]]) - (2*a*Sqrt[a - a*Cosh[c + d*x ]]*Sinh[c + d*x])/(3*d)))/5
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos [c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]])), x] /; FreeQ[{a, b, c, d}, x] && Eq Q[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos [c + d*x]*((a + b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[a*((2*n - 1)/n) Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[ a^2 - b^2, 0] && IGtQ[n - 1/2, 0]
Time = 0.52 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.77
method | result | size |
default | \(-\frac {16 \sinh \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \cosh \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (3 \sinh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-4 \sinh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+8\right )}{15 \sqrt {-2 \sinh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, d}\) | \(71\) |
Input:
int((a-a*cosh(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
-16/15*sinh(1/2*d*x+1/2*c)*a^3*cosh(1/2*d*x+1/2*c)*(3*sinh(1/2*d*x+1/2*c)^ 4-4*sinh(1/2*d*x+1/2*c)^2+8)/(-2*sinh(1/2*d*x+1/2*c)^2*a)^(1/2)/d
Leaf count of result is larger than twice the leaf count of optimal. 328 vs. \(2 (80) = 160\).
Time = 0.12 (sec) , antiderivative size = 328, normalized size of antiderivative = 3.57 \[ \int (a-a \cosh (c+d x))^{5/2} \, dx=\frac {\sqrt {\frac {1}{2}} {\left (3 \, a^{2} \cosh \left (d x + c\right )^{5} + 3 \, a^{2} \sinh \left (d x + c\right )^{5} - 25 \, a^{2} \cosh \left (d x + c\right )^{4} + 150 \, a^{2} \cosh \left (d x + c\right )^{3} + 5 \, {\left (3 \, a^{2} \cosh \left (d x + c\right ) - 5 \, a^{2}\right )} \sinh \left (d x + c\right )^{4} + 150 \, a^{2} \cosh \left (d x + c\right )^{2} + 10 \, {\left (3 \, a^{2} \cosh \left (d x + c\right )^{2} - 10 \, a^{2} \cosh \left (d x + c\right ) + 15 \, a^{2}\right )} \sinh \left (d x + c\right )^{3} - 25 \, a^{2} \cosh \left (d x + c\right ) + 30 \, {\left (a^{2} \cosh \left (d x + c\right )^{3} - 5 \, a^{2} \cosh \left (d x + c\right )^{2} + 15 \, a^{2} \cosh \left (d x + c\right ) + 5 \, a^{2}\right )} \sinh \left (d x + c\right )^{2} + 3 \, a^{2} + 5 \, {\left (3 \, a^{2} \cosh \left (d x + c\right )^{4} - 20 \, a^{2} \cosh \left (d x + c\right )^{3} + 90 \, a^{2} \cosh \left (d x + c\right )^{2} + 60 \, a^{2} \cosh \left (d x + c\right ) - 5 \, a^{2}\right )} \sinh \left (d x + c\right )\right )} \sqrt {-\frac {a}{\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )}}}{30 \, {\left (d \cosh \left (d x + c\right )^{2} + 2 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + d \sinh \left (d x + c\right )^{2}\right )}} \] Input:
integrate((a-a*cosh(d*x+c))^(5/2),x, algorithm="fricas")
Output:
1/30*sqrt(1/2)*(3*a^2*cosh(d*x + c)^5 + 3*a^2*sinh(d*x + c)^5 - 25*a^2*cos h(d*x + c)^4 + 150*a^2*cosh(d*x + c)^3 + 5*(3*a^2*cosh(d*x + c) - 5*a^2)*s inh(d*x + c)^4 + 150*a^2*cosh(d*x + c)^2 + 10*(3*a^2*cosh(d*x + c)^2 - 10* a^2*cosh(d*x + c) + 15*a^2)*sinh(d*x + c)^3 - 25*a^2*cosh(d*x + c) + 30*(a ^2*cosh(d*x + c)^3 - 5*a^2*cosh(d*x + c)^2 + 15*a^2*cosh(d*x + c) + 5*a^2) *sinh(d*x + c)^2 + 3*a^2 + 5*(3*a^2*cosh(d*x + c)^4 - 20*a^2*cosh(d*x + c) ^3 + 90*a^2*cosh(d*x + c)^2 + 60*a^2*cosh(d*x + c) - 5*a^2)*sinh(d*x + c)) *sqrt(-a/(cosh(d*x + c) + sinh(d*x + c)))/(d*cosh(d*x + c)^2 + 2*d*cosh(d* x + c)*sinh(d*x + c) + d*sinh(d*x + c)^2)
Timed out. \[ \int (a-a \cosh (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:
integrate((a-a*cosh(d*x+c))**(5/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 190 vs. \(2 (80) = 160\).
Time = 0.14 (sec) , antiderivative size = 190, normalized size of antiderivative = 2.07 \[ \int (a-a \cosh (c+d x))^{5/2} \, dx=\frac {5 \, \sqrt {2} a^{\frac {5}{2}} e^{\left (-d x - c\right )}}{12 \, d \left (-e^{\left (-d x - c\right )}\right )^{\frac {5}{2}}} - \frac {5 \, \sqrt {2} a^{\frac {5}{2}} e^{\left (-2 \, d x - 2 \, c\right )}}{2 \, d \left (-e^{\left (-d x - c\right )}\right )^{\frac {5}{2}}} - \frac {5 \, \sqrt {2} a^{\frac {5}{2}} e^{\left (-3 \, d x - 3 \, c\right )}}{2 \, d \left (-e^{\left (-d x - c\right )}\right )^{\frac {5}{2}}} + \frac {5 \, \sqrt {2} a^{\frac {5}{2}} e^{\left (-4 \, d x - 4 \, c\right )}}{12 \, d \left (-e^{\left (-d x - c\right )}\right )^{\frac {5}{2}}} - \frac {\sqrt {2} a^{\frac {5}{2}} e^{\left (-5 \, d x - 5 \, c\right )}}{20 \, d \left (-e^{\left (-d x - c\right )}\right )^{\frac {5}{2}}} - \frac {\sqrt {2} a^{\frac {5}{2}}}{20 \, d \left (-e^{\left (-d x - c\right )}\right )^{\frac {5}{2}}} \] Input:
integrate((a-a*cosh(d*x+c))^(5/2),x, algorithm="maxima")
Output:
5/12*sqrt(2)*a^(5/2)*e^(-d*x - c)/(d*(-e^(-d*x - c))^(5/2)) - 5/2*sqrt(2)* a^(5/2)*e^(-2*d*x - 2*c)/(d*(-e^(-d*x - c))^(5/2)) - 5/2*sqrt(2)*a^(5/2)*e ^(-3*d*x - 3*c)/(d*(-e^(-d*x - c))^(5/2)) + 5/12*sqrt(2)*a^(5/2)*e^(-4*d*x - 4*c)/(d*(-e^(-d*x - c))^(5/2)) - 1/20*sqrt(2)*a^(5/2)*e^(-5*d*x - 5*c)/ (d*(-e^(-d*x - c))^(5/2)) - 1/20*sqrt(2)*a^(5/2)/(d*(-e^(-d*x - c))^(5/2))
Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (80) = 160\).
Time = 0.13 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.11 \[ \int (a-a \cosh (c+d x))^{5/2} \, dx=-\frac {\sqrt {2} {\left (3 \, \sqrt {-a e^{\left (d x + c\right )}} a^{2} e^{\left (2 \, d x + 2 \, c\right )} \mathrm {sgn}\left (-e^{\left (d x + c\right )} + 1\right ) - 25 \, \sqrt {-a e^{\left (d x + c\right )}} a^{2} e^{\left (d x + c\right )} \mathrm {sgn}\left (-e^{\left (d x + c\right )} + 1\right ) + 150 \, \sqrt {-a e^{\left (d x + c\right )}} a^{2} \mathrm {sgn}\left (-e^{\left (d x + c\right )} + 1\right ) - \frac {{\left (150 \, a^{5} e^{\left (2 \, d x + 2 \, c\right )} \mathrm {sgn}\left (-e^{\left (d x + c\right )} + 1\right ) - 25 \, a^{5} e^{\left (d x + c\right )} \mathrm {sgn}\left (-e^{\left (d x + c\right )} + 1\right ) + 3 \, a^{5} \mathrm {sgn}\left (-e^{\left (d x + c\right )} + 1\right )\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{\sqrt {-a e^{\left (d x + c\right )}} a^{2}}\right )}}{60 \, d} \] Input:
integrate((a-a*cosh(d*x+c))^(5/2),x, algorithm="giac")
Output:
-1/60*sqrt(2)*(3*sqrt(-a*e^(d*x + c))*a^2*e^(2*d*x + 2*c)*sgn(-e^(d*x + c) + 1) - 25*sqrt(-a*e^(d*x + c))*a^2*e^(d*x + c)*sgn(-e^(d*x + c) + 1) + 15 0*sqrt(-a*e^(d*x + c))*a^2*sgn(-e^(d*x + c) + 1) - (150*a^5*e^(2*d*x + 2*c )*sgn(-e^(d*x + c) + 1) - 25*a^5*e^(d*x + c)*sgn(-e^(d*x + c) + 1) + 3*a^5 *sgn(-e^(d*x + c) + 1))*e^(-2*d*x - 2*c)/(sqrt(-a*e^(d*x + c))*a^2))/d
Timed out. \[ \int (a-a \cosh (c+d x))^{5/2} \, dx=\int {\left (a-a\,\mathrm {cosh}\left (c+d\,x\right )\right )}^{5/2} \,d x \] Input:
int((a - a*cosh(c + d*x))^(5/2),x)
Output:
int((a - a*cosh(c + d*x))^(5/2), x)
\[ \int (a-a \cosh (c+d x))^{5/2} \, dx=\sqrt {a}\, a^{2} \left (\int \sqrt {-\cosh \left (d x +c \right )+1}d x -2 \left (\int \sqrt {-\cosh \left (d x +c \right )+1}\, \cosh \left (d x +c \right )d x \right )+\int \sqrt {-\cosh \left (d x +c \right )+1}\, \cosh \left (d x +c \right )^{2}d x \right ) \] Input:
int((a-a*cosh(d*x+c))^(5/2),x)
Output:
sqrt(a)*a**2*(int(sqrt( - cosh(c + d*x) + 1),x) - 2*int(sqrt( - cosh(c + d *x) + 1)*cosh(c + d*x),x) + int(sqrt( - cosh(c + d*x) + 1)*cosh(c + d*x)** 2,x))