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

Optimal. Leaf size=107 \[ \frac {3 \tan ^{-1}\left (\frac {\sqrt {a} \sinh (c+d x)}{\sqrt {2} \sqrt {a \cosh (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {3 \sinh (c+d x)}{16 a d (a \cosh (c+d x)+a)^{3/2}}+\frac {\sinh (c+d x)}{4 d (a \cosh (c+d x)+a)^{5/2}} \]

[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 = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2650, 2649, 206} \[ \frac {3 \tan ^{-1}\left (\frac {\sqrt {a} \sinh (c+d x)}{\sqrt {2} \sqrt {a \cosh (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {3 \sinh (c+d x)}{16 a d (a \cosh (c+d x)+a)^{3/2}}+\frac {\sinh (c+d x)}{4 d (a \cosh (c+d x)+a)^{5/2}} \]

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 = 0.28, size = 91, normalized size = 0.85 \[ \frac {\cosh ^5\left (\frac {1}{2} (c+d x)\right ) \left (32 \sinh ^5\left (\frac {1}{2} (c+d x)\right ) \text {csch}^4(c+d x)+3 \left (\tan ^{-1}\left (\sinh \left (\frac {1}{2} (c+d x)\right )\right )+\tanh \left (\frac {1}{2} (c+d x)\right ) \text {sech}\left (\frac {1}{2} (c+d x)\right )\right )\right )}{4 d (a (\cosh (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cosh[(c + d*x)/2]^5*(32*Csch[c + d*x]^4*Sinh[(c + d*x)/2]^5 + 3*(ArcTan[Sinh[(c + d*x)/2]] + Sech[(c + d*x)/2
]*Tanh[(c + d*x)/2])))/(4*d*(a*(1 + Cosh[c + d*x]))^(5/2))

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fricas [B]  time = 0.42, size = 522, normalized size = 4.88 \[ -\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} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {\frac {a}{\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )}}}{\sqrt {a}}\right ) - 2 \, \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 )}}}{16 \, {\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]

-1/16*(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)*arctan(sqrt(2)*sqrt(1/2)*sqrt
(a/(cosh(d*x + c) + sinh(d*x + c)))/sqrt(a)) - 2*sqrt(1/2)*(3*cosh(d*x + c)^4 + (12*cosh(d*x + c) + 11)*sinh(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*cosh(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 = 0.27, size = 97, normalized size = 0.91 \[ \frac {\sqrt {2} {\left (\frac {3 \, \arctan \left (e^{\left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}\right )}{a^{\frac {5}{2}}} + \frac {3 \, a^{\frac {7}{2}} e^{\left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )} + 11 \, a^{\frac {7}{2}} e^{\left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )} - 11 \, a^{\frac {7}{2}} e^{\left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )} - 3 \, a^{\frac {7}{2}} e^{\left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}}{{\left (a e^{\left (d x + c\right )} + a\right )}^{4} a^{2}}\right )}}{16 \, d} \]

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(e^(1/2*d*x + 1/2*c))/a^(5/2) + (3*a^(7/2)*e^(7/2*d*x + 7/2*c) + 11*a^(7/2)*e^(5/2*d*x +
5/2*c) - 11*a^(7/2)*e^(3/2*d*x + 3/2*c) - 3*a^(7/2)*e^(1/2*d*x + 1/2*c))/((a*e^(d*x + c) + a)^4*a^2))/d

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maple [B]  time = 0.29, size = 178, normalized size = 1.66 \[ -\frac {\sqrt {a \left (\sinh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (3 \ln \left (\frac {2 \sqrt {a \left (\sinh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {-a}-2 a}{\cosh \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \left (\cosh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 \sqrt {a \left (\sinh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\cosh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {-a}-2 \sqrt {a \left (\sinh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {-a}\right ) \sqrt {2}}{32 a^{3} \cosh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {-a}\, \sinh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cosh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

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

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maxima [B]  time = 0.51, size = 250, normalized size = 2.34 \[ \frac {1}{80} \, \sqrt {2} {\left (\frac {15 \, e^{\left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )} + 70 \, e^{\left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )} + 128 \, e^{\left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )} - 70 \, e^{\left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )} - 15 \, e^{\left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}}{{\left (a^{\frac {5}{2}} e^{\left (5 \, d x + 5 \, c\right )} + 5 \, a^{\frac {5}{2}} e^{\left (4 \, d x + 4 \, c\right )} + 10 \, a^{\frac {5}{2}} e^{\left (3 \, d x + 3 \, c\right )} + 10 \, a^{\frac {5}{2}} e^{\left (2 \, d x + 2 \, c\right )} + 5 \, a^{\frac {5}{2}} e^{\left (d x + c\right )} + a^{\frac {5}{2}}\right )} d} + \frac {15 \, \arctan \left (e^{\left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}\right )}{a^{\frac {5}{2}} d}\right )} - \frac {8 \, \sqrt {2} e^{\left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )}}{5 \, {\left (a^{\frac {5}{2}} d e^{\left (5 \, d x + 5 \, c\right )} + 5 \, a^{\frac {5}{2}} d e^{\left (4 \, d x + 4 \, c\right )} + 10 \, a^{\frac {5}{2}} d e^{\left (3 \, d x + 3 \, c\right )} + 10 \, a^{\frac {5}{2}} d e^{\left (2 \, d x + 2 \, c\right )} + 5 \, a^{\frac {5}{2}} d e^{\left (d x + c\right )} + a^{\frac {5}{2}} d\right )}} \]

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]

1/80*sqrt(2)*((15*e^(9/2*d*x + 9/2*c) + 70*e^(7/2*d*x + 7/2*c) + 128*e^(5/2*d*x + 5/2*c) - 70*e^(3/2*d*x + 3/2
*c) - 15*e^(1/2*d*x + 1/2*c))/((a^(5/2)*e^(5*d*x + 5*c) + 5*a^(5/2)*e^(4*d*x + 4*c) + 10*a^(5/2)*e^(3*d*x + 3*
c) + 10*a^(5/2)*e^(2*d*x + 2*c) + 5*a^(5/2)*e^(d*x + c) + a^(5/2))*d) + 15*arctan(e^(1/2*d*x + 1/2*c))/(a^(5/2
)*d)) - 8/5*sqrt(2)*e^(5/2*d*x + 5/2*c)/(a^(5/2)*d*e^(5*d*x + 5*c) + 5*a^(5/2)*d*e^(4*d*x + 4*c) + 10*a^(5/2)*
d*e^(3*d*x + 3*c) + 10*a^(5/2)*d*e^(2*d*x + 2*c) + 5*a^(5/2)*d*e^(d*x + c) + a^(5/2)*d)

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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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