3.48 \(\int (a-a \cosh (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=92 \[ -\frac {64 a^3 \sinh (c+d x)}{15 d \sqrt {a-a \cosh (c+d x)}}-\frac {16 a^2 \sinh (c+d x) \sqrt {a-a \cosh (c+d x)}}{15 d}-\frac {2 a \sinh (c+d x) (a-a \cosh (c+d x))^{3/2}}{5 d} \]

[Out]

-2/5*a*(a-a*cosh(d*x+c))^(3/2)*sinh(d*x+c)/d-64/15*a^3*sinh(d*x+c)/d/(a-a*cosh(d*x+c))^(1/2)-16/15*a^2*sinh(d*
x+c)*(a-a*cosh(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.05, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2647, 2646} \[ -\frac {64 a^3 \sinh (c+d x)}{15 d \sqrt {a-a \cosh (c+d x)}}-\frac {16 a^2 \sinh (c+d x) \sqrt {a-a \cosh (c+d x)}}{15 d}-\frac {2 a \sinh (c+d x) (a-a \cosh (c+d x))^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a - a*Cosh[c + d*x])^(5/2),x]

[Out]

(-64*a^3*Sinh[c + d*x])/(15*d*Sqrt[a - a*Cosh[c + d*x]]) - (16*a^2*Sqrt[a - a*Cosh[c + d*x]]*Sinh[c + d*x])/(1
5*d) - (2*a*(a - a*Cosh[c + d*x])^(3/2)*Sinh[c + d*x])/(5*d)

Rule 2646

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] && EqQ[a^2 - b^2, 0]

Rule 2647

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] + Dist[(a*(2*n - 1))/n, Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && Eq
Q[a^2 - b^2, 0] && IGtQ[n - 1/2, 0]

Rubi steps

\begin {align*} \int (a-a \cosh (c+d x))^{5/2} \, dx &=-\frac {2 a (a-a \cosh (c+d x))^{3/2} \sinh (c+d x)}{5 d}+\frac {1}{5} (8 a) \int (a-a \cosh (c+d x))^{3/2} \, dx\\ &=-\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}+\frac {1}{15} \left (32 a^2\right ) \int \sqrt {a-a \cosh (c+d x)} \, 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}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 72, normalized size = 0.78 \[ \frac {a^2 \left (150 \cosh \left (\frac {1}{2} (c+d x)\right )-25 \cosh \left (\frac {3}{2} (c+d x)\right )+3 \cosh \left (\frac {5}{2} (c+d x)\right )\right ) \text {csch}\left (\frac {1}{2} (c+d x)\right ) \sqrt {a-a \cosh (c+d x)}}{30 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - a*Cosh[c + d*x])^(5/2),x]

[Out]

(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])*Cs
ch[(c + d*x)/2])/(30*d)

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fricas [B]  time = 0.51, size = 328, normalized size = 3.57 \[ \frac {\sqrt {\frac {1}{2}} {\left (3 \, a^{2} \cosh \left (d x + c\right )^{5} + 3 \, a^{2} \sinh \left (d x + c\right )^{5} - 25 \, a^{2} \cosh \left (d x + c\right )^{4} + 150 \, a^{2} \cosh \left (d x + c\right )^{3} + 5 \, {\left (3 \, a^{2} \cosh \left (d x + c\right ) - 5 \, a^{2}\right )} \sinh \left (d x + c\right )^{4} + 150 \, a^{2} \cosh \left (d x + c\right )^{2} + 10 \, {\left (3 \, a^{2} \cosh \left (d x + c\right )^{2} - 10 \, a^{2} \cosh \left (d x + c\right ) + 15 \, a^{2}\right )} \sinh \left (d x + c\right )^{3} - 25 \, a^{2} \cosh \left (d x + c\right ) + 30 \, {\left (a^{2} \cosh \left (d x + c\right )^{3} - 5 \, a^{2} \cosh \left (d x + c\right )^{2} + 15 \, a^{2} \cosh \left (d x + c\right ) + 5 \, a^{2}\right )} \sinh \left (d x + c\right )^{2} + 3 \, a^{2} + 5 \, {\left (3 \, a^{2} \cosh \left (d x + c\right )^{4} - 20 \, a^{2} \cosh \left (d x + c\right )^{3} + 90 \, a^{2} \cosh \left (d x + c\right )^{2} + 60 \, a^{2} \cosh \left (d x + c\right ) - 5 \, a^{2}\right )} \sinh \left (d x + c\right )\right )} \sqrt {-\frac {a}{\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )}}}{30 \, {\left (d \cosh \left (d x + c\right )^{2} + 2 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + d \sinh \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*cosh(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/30*sqrt(1/2)*(3*a^2*cosh(d*x + c)^5 + 3*a^2*sinh(d*x + c)^5 - 25*a^2*cosh(d*x + c)^4 + 150*a^2*cosh(d*x + c)
^3 + 5*(3*a^2*cosh(d*x + c) - 5*a^2)*sinh(d*x + c)^4 + 150*a^2*cosh(d*x + c)^2 + 10*(3*a^2*cosh(d*x + c)^2 - 1
0*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)

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giac [B]  time = 0.16, size = 189, normalized size = 2.05 \[ -\frac {\sqrt {2} {\left (3 \, \sqrt {-a} a^{2} e^{\left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )} \mathrm {sgn}\left (-e^{\left (d x + c\right )} + 1\right ) - 25 \, \sqrt {-a} a^{2} e^{\left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )} \mathrm {sgn}\left (-e^{\left (d x + c\right )} + 1\right ) + 150 \, \sqrt {-a} a^{2} e^{\left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} \mathrm {sgn}\left (-e^{\left (d x + c\right )} + 1\right ) + 150 \, \sqrt {-a} a^{2} e^{\left (-\frac {1}{2} \, d x - \frac {1}{2} \, c\right )} \mathrm {sgn}\left (-e^{\left (d x + c\right )} + 1\right ) - 25 \, \sqrt {-a} a^{2} e^{\left (-\frac {3}{2} \, d x - \frac {3}{2} \, c\right )} \mathrm {sgn}\left (-e^{\left (d x + c\right )} + 1\right ) + 3 \, \sqrt {-a} a^{2} e^{\left (-\frac {5}{2} \, d x - \frac {5}{2} \, c\right )} \mathrm {sgn}\left (-e^{\left (d x + c\right )} + 1\right )\right )}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*cosh(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-1/60*sqrt(2)*(3*sqrt(-a)*a^2*e^(5/2*d*x + 5/2*c)*sgn(-e^(d*x + c) + 1) - 25*sqrt(-a)*a^2*e^(3/2*d*x + 3/2*c)*
sgn(-e^(d*x + c) + 1) + 150*sqrt(-a)*a^2*e^(1/2*d*x + 1/2*c)*sgn(-e^(d*x + c) + 1) + 150*sqrt(-a)*a^2*e^(-1/2*
d*x - 1/2*c)*sgn(-e^(d*x + c) + 1) - 25*sqrt(-a)*a^2*e^(-3/2*d*x - 3/2*c)*sgn(-e^(d*x + c) + 1) + 3*sqrt(-a)*a
^2*e^(-5/2*d*x - 5/2*c)*sgn(-e^(d*x + c) + 1))/d

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maple [A]  time = 0.26, size = 71, normalized size = 0.77 \[ -\frac {16 \sinh \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \cosh \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (3 \left (\sinh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\sinh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8\right )}{15 \sqrt {-2 a \left (\sinh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a-a*cosh(d*x+c))^(5/2),x)

[Out]

-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*a*s
inh(1/2*d*x+1/2*c)^2)^(1/2)/d

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maxima [B]  time = 0.42, size = 190, normalized size = 2.07 \[ \frac {5 \, \sqrt {2} a^{\frac {5}{2}} e^{\left (-d x - c\right )}}{12 \, d \left (-e^{\left (-d x - c\right )}\right )^{\frac {5}{2}}} - \frac {5 \, \sqrt {2} a^{\frac {5}{2}} e^{\left (-2 \, d x - 2 \, c\right )}}{2 \, d \left (-e^{\left (-d x - c\right )}\right )^{\frac {5}{2}}} - \frac {5 \, \sqrt {2} a^{\frac {5}{2}} e^{\left (-3 \, d x - 3 \, c\right )}}{2 \, d \left (-e^{\left (-d x - c\right )}\right )^{\frac {5}{2}}} + \frac {5 \, \sqrt {2} a^{\frac {5}{2}} e^{\left (-4 \, d x - 4 \, c\right )}}{12 \, d \left (-e^{\left (-d x - c\right )}\right )^{\frac {5}{2}}} - \frac {\sqrt {2} a^{\frac {5}{2}} e^{\left (-5 \, d x - 5 \, c\right )}}{20 \, d \left (-e^{\left (-d x - c\right )}\right )^{\frac {5}{2}}} - \frac {\sqrt {2} a^{\frac {5}{2}}}{20 \, d \left (-e^{\left (-d x - c\right )}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*cosh(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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*s
qrt(2)*a^(5/2)/(d*(-e^(-d*x - c))^(5/2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a-a\,\mathrm {cosh}\left (c+d\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - a*cosh(c + d*x))^(5/2),x)

[Out]

int((a - a*cosh(c + d*x))^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*cosh(d*x+c))**(5/2),x)

[Out]

Timed out

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