3.53 \(\int \frac {1}{(a-a \cosh (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=110 \[ -\frac {3 \tan ^{-1}\left (\frac {\sqrt {a} \sinh (c+d x)}{\sqrt {2} \sqrt {a-a \cosh (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {3 \sinh (c+d x)}{16 a d (a-a \cosh (c+d x))^{3/2}}-\frac {\sinh (c+d x)}{4 d (a-a \cosh (c+d x))^{5/2}} \]

[Out]

-1/4*sinh(d*x+c)/d/(a-a*cosh(d*x+c))^(5/2)-3/16*sinh(d*x+c)/a/d/(a-a*cosh(d*x+c))^(3/2)-3/32*arctan(1/2*sinh(d
*x+c)*a^(1/2)*2^(1/2)/(a-a*cosh(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2650, 2649, 206} \[ -\frac {3 \tan ^{-1}\left (\frac {\sqrt {a} \sinh (c+d x)}{\sqrt {2} \sqrt {a-a \cosh (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {3 \sinh (c+d x)}{16 a d (a-a \cosh (c+d x))^{3/2}}-\frac {\sinh (c+d x)}{4 d (a-a \cosh (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a - a*Cosh[c + d*x])^(-5/2),x]

[Out]

(-3*ArcTan[(Sqrt[a]*Sinh[c + d*x])/(Sqrt[2]*Sqrt[a - a*Cosh[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - Sinh[c + d*x
]/(4*d*(a - a*Cosh[c + d*x])^(5/2)) - (3*Sinh[c + d*x])/(16*a*d*(a - a*Cosh[c + d*x])^(3/2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rubi steps

\begin {align*} \int \frac {1}{(a-a \cosh (c+d x))^{5/2}} \, dx &=-\frac {\sinh (c+d x)}{4 d (a-a \cosh (c+d x))^{5/2}}+\frac {3 \int \frac {1}{(a-a \cosh (c+d x))^{3/2}} \, dx}{8 a}\\ &=-\frac {\sinh (c+d x)}{4 d (a-a \cosh (c+d x))^{5/2}}-\frac {3 \sinh (c+d x)}{16 a d (a-a \cosh (c+d x))^{3/2}}+\frac {3 \int \frac {1}{\sqrt {a-a \cosh (c+d x)}} \, dx}{32 a^2}\\ &=-\frac {\sinh (c+d x)}{4 d (a-a \cosh (c+d x))^{5/2}}-\frac {3 \sinh (c+d x)}{16 a d (a-a \cosh (c+d x))^{3/2}}+\frac {(3 i) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {i a \sinh (c+d x)}{\sqrt {a-a \cosh (c+d x)}}\right )}{16 a^2 d}\\ &=-\frac {3 \tan ^{-1}\left (\frac {\sqrt {a} \sinh (c+d x)}{\sqrt {2} \sqrt {a-a \cosh (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {\sinh (c+d x)}{4 d (a-a \cosh (c+d x))^{5/2}}-\frac {3 \sinh (c+d x)}{16 a d (a-a \cosh (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 115, normalized size = 1.05 \[ \frac {\sinh ^5\left (\frac {1}{2} (c+d x)\right ) \left (-\text {csch}^4\left (\frac {1}{4} (c+d x)\right )+6 \text {csch}^2\left (\frac {1}{4} (c+d x)\right )+\text {sech}^4\left (\frac {1}{4} (c+d x)\right )+6 \text {sech}^2\left (\frac {1}{4} (c+d x)\right )+24 \log \left (\tanh \left (\frac {1}{4} (c+d x)\right )\right )\right )}{32 a^2 d (\cosh (c+d x)-1)^2 \sqrt {a-a \cosh (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*Csch[(c + d*x)/4]^2 - Csch[(c + d*x)/4]^4 + 24*Log[Tanh[(c + d*x)/4]] + 6*Sech[(c + d*x)/4]^2 + Sech[(c +
d*x)/4]^4)*Sinh[(c + d*x)/2]^5)/(32*a^2*d*(-1 + Cosh[c + d*x])^2*Sqrt[a - a*Cosh[c + d*x]])

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fricas [B]  time = 1.25, size = 580, normalized size = 5.27 \[ -\frac {3 \, \sqrt {2} {\left (\cosh \left (d x + c\right )^{4} + 4 \, {\left (\cosh \left (d x + c\right ) - 1\right )} \sinh \left (d x + c\right )^{3} + \sinh \left (d x + c\right )^{4} - 4 \, \cosh \left (d x + c\right )^{3} + 6 \, {\left (\cosh \left (d x + c\right )^{2} - 2 \, \cosh \left (d x + c\right ) + 1\right )} \sinh \left (d x + c\right )^{2} + 6 \, \cosh \left (d x + c\right )^{2} + 4 \, {\left (\cosh \left (d x + c\right )^{3} - 3 \, \cosh \left (d x + c\right )^{2} + 3 \, \cosh \left (d x + c\right ) - 1\right )} \sinh \left (d x + c\right ) - 4 \, \cosh \left (d x + c\right ) + 1\right )} \sqrt {-a} \log \left (-\frac {2 \, \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-a} \sqrt {-\frac {a}{\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )}} {\left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right )} + a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) + a}{\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) - 1}\right ) + 4 \, \sqrt {\frac {1}{2}} {\left (3 \, \cosh \left (d x + c\right )^{4} + {\left (12 \, \cosh \left (d x + c\right ) - 11\right )} \sinh \left (d x + c\right )^{3} + 3 \, \sinh \left (d x + c\right )^{4} - 11 \, \cosh \left (d x + c\right )^{3} + {\left (18 \, \cosh \left (d x + c\right )^{2} - 33 \, \cosh \left (d x + c\right ) - 11\right )} \sinh \left (d x + c\right )^{2} - 11 \, \cosh \left (d x + c\right )^{2} + {\left (12 \, \cosh \left (d x + c\right )^{3} - 33 \, \cosh \left (d x + c\right )^{2} - 22 \, \cosh \left (d x + c\right ) + 3\right )} \sinh \left (d x + c\right ) + 3 \, \cosh \left (d x + c\right )\right )} \sqrt {-\frac {a}{\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )}}}{32 \, {\left (a^{3} d \cosh \left (d x + c\right )^{4} + a^{3} d \sinh \left (d x + c\right )^{4} - 4 \, a^{3} d \cosh \left (d x + c\right )^{3} + 6 \, a^{3} d \cosh \left (d x + c\right )^{2} - 4 \, a^{3} d \cosh \left (d x + c\right ) + a^{3} d + 4 \, {\left (a^{3} d \cosh \left (d x + c\right ) - a^{3} d\right )} \sinh \left (d x + c\right )^{3} + 6 \, {\left (a^{3} d \cosh \left (d x + c\right )^{2} - 2 \, a^{3} d \cosh \left (d x + c\right ) + a^{3} d\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left (a^{3} d \cosh \left (d x + c\right )^{3} - 3 \, a^{3} d \cosh \left (d x + c\right )^{2} + 3 \, a^{3} d \cosh \left (d x + c\right ) - a^{3} d\right )} \sinh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*cosh(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/32*(3*sqrt(2)*(cosh(d*x + c)^4 + 4*(cosh(d*x + c) - 1)*sinh(d*x + c)^3 + sinh(d*x + c)^4 - 4*cosh(d*x + c)^
3 + 6*(cosh(d*x + c)^2 - 2*cosh(d*x + c) + 1)*sinh(d*x + c)^2 + 6*cosh(d*x + c)^2 + 4*(cosh(d*x + c)^3 - 3*cos
h(d*x + c)^2 + 3*cosh(d*x + c) - 1)*sinh(d*x + c) - 4*cosh(d*x + c) + 1)*sqrt(-a)*log(-(2*sqrt(2)*sqrt(1/2)*sq
rt(-a)*sqrt(-a/(cosh(d*x + c) + sinh(d*x + c)))*(cosh(d*x + c) + sinh(d*x + c)) + a*cosh(d*x + c) + a*sinh(d*x
+ c) + a)/(cosh(d*x + c) + sinh(d*x + c) - 1)) + 4*sqrt(1/2)*(3*cosh(d*x + c)^4 + (12*cosh(d*x + c) - 11)*sin
h(d*x + c)^3 + 3*sinh(d*x + c)^4 - 11*cosh(d*x + c)^3 + (18*cosh(d*x + c)^2 - 33*cosh(d*x + c) - 11)*sinh(d*x
+ c)^2 - 11*cosh(d*x + c)^2 + (12*cosh(d*x + c)^3 - 33*cosh(d*x + c)^2 - 22*cosh(d*x + c) + 3)*sinh(d*x + c) +
3*cosh(d*x + c))*sqrt(-a/(cosh(d*x + c) + sinh(d*x + c))))/(a^3*d*cosh(d*x + c)^4 + a^3*d*sinh(d*x + c)^4 - 4
*a^3*d*cosh(d*x + c)^3 + 6*a^3*d*cosh(d*x + c)^2 - 4*a^3*d*cosh(d*x + c) + a^3*d + 4*(a^3*d*cosh(d*x + c) - a^
3*d)*sinh(d*x + c)^3 + 6*(a^3*d*cosh(d*x + c)^2 - 2*a^3*d*cosh(d*x + c) + a^3*d)*sinh(d*x + c)^2 + 4*(a^3*d*co
sh(d*x + c)^3 - 3*a^3*d*cosh(d*x + c)^2 + 3*a^3*d*cosh(d*x + c) - a^3*d)*sinh(d*x + c))

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giac [A]  time = 0.30, size = 164, normalized size = 1.49 \[ -\frac {\sqrt {2} {\left (\frac {3 \, \arctan \left (\frac {\sqrt {-a e^{\left (d x + c\right )}}}{\sqrt {a}}\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (-e^{\left (d x + c\right )} + 1\right )} - \frac {3 \, \sqrt {-a e^{\left (d x + c\right )}} a^{3} e^{\left (3 \, d x + 3 \, c\right )} - 11 \, \sqrt {-a e^{\left (d x + c\right )}} a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 11 \, \sqrt {-a e^{\left (d x + c\right )}} a^{3} e^{\left (d x + c\right )} + 3 \, \sqrt {-a e^{\left (d x + c\right )}} a^{3}}{{\left (a e^{\left (d x + c\right )} - a\right )}^{4} a^{2} \mathrm {sgn}\left (-e^{\left (d x + c\right )} + 1\right )}\right )}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*cosh(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-1/16*sqrt(2)*(3*arctan(sqrt(-a*e^(d*x + c))/sqrt(a))/(a^(5/2)*sgn(-e^(d*x + c) + 1)) - (3*sqrt(-a*e^(d*x + c)
)*a^3*e^(3*d*x + 3*c) - 11*sqrt(-a*e^(d*x + c))*a^3*e^(2*d*x + 2*c) - 11*sqrt(-a*e^(d*x + c))*a^3*e^(d*x + c)
+ 3*sqrt(-a*e^(d*x + c))*a^3)/((a*e^(d*x + c) - a)^4*a^2*sgn(-e^(d*x + c) + 1)))/d

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maple [A]  time = 0.34, size = 137, normalized size = 1.25 \[ \frac {6 \left (\sinh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cosh \left (\frac {d x}{2}+\frac {c}{2}\right )-4 \cosh \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (3 \ln \left (-1+\cosh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 \ln \left (\cosh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )\right ) \left (\sinh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 a^{2} \left (\cosh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \left (-1+\cosh \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sinh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 a \left (\sinh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a-a*cosh(d*x+c))^(5/2),x)

[Out]

1/32/a^2*(6*sinh(1/2*d*x+1/2*c)^2*cosh(1/2*d*x+1/2*c)-4*cosh(1/2*d*x+1/2*c)+(3*ln(-1+cosh(1/2*d*x+1/2*c))-3*ln
(cosh(1/2*d*x+1/2*c)+1))*sinh(1/2*d*x+1/2*c)^4)/(cosh(1/2*d*x+1/2*c)+1)/(-1+cosh(1/2*d*x+1/2*c))/sinh(1/2*d*x+
1/2*c)/(-2*a*sinh(1/2*d*x+1/2*c)^2)^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-a \cosh \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*cosh(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a - a*cosh(c + d*x))^(5/2),x)

[Out]

int(1/(a - a*cosh(c + d*x))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (- a \cosh {\left (c + d x \right )} + a\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*cosh(d*x+c))**(5/2),x)

[Out]

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

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