3.138 \(\int \frac {1}{(a \cosh ^4(x))^{3/2}} \, dx\)

Optimal. Leaf size=67 \[ \frac {\sinh (x) \cosh (x)}{a \sqrt {a \cosh ^4(x)}}+\frac {\sinh ^2(x) \tanh ^3(x)}{5 a \sqrt {a \cosh ^4(x)}}-\frac {2 \sinh ^2(x) \tanh (x)}{3 a \sqrt {a \cosh ^4(x)}} \]

[Out]

cosh(x)*sinh(x)/a/(a*cosh(x)^4)^(1/2)-2/3*sinh(x)^2*tanh(x)/a/(a*cosh(x)^4)^(1/2)+1/5*sinh(x)^2*tanh(x)^3/a/(a
*cosh(x)^4)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3207, 3767} \[ \frac {\sinh (x) \cosh (x)}{a \sqrt {a \cosh ^4(x)}}+\frac {\sinh ^2(x) \tanh ^3(x)}{5 a \sqrt {a \cosh ^4(x)}}-\frac {2 \sinh ^2(x) \tanh (x)}{3 a \sqrt {a \cosh ^4(x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cosh[x]^4)^(-3/2),x]

[Out]

(Cosh[x]*Sinh[x])/(a*Sqrt[a*Cosh[x]^4]) - (2*Sinh[x]^2*Tanh[x])/(3*a*Sqrt[a*Cosh[x]^4]) + (Sinh[x]^2*Tanh[x]^3
)/(5*a*Sqrt[a*Cosh[x]^4])

Rule 3207

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin {align*} \int \frac {1}{\left (a \cosh ^4(x)\right )^{3/2}} \, dx &=\frac {\cosh ^2(x) \int \text {sech}^6(x) \, dx}{a \sqrt {a \cosh ^4(x)}}\\ &=\frac {\left (i \cosh ^2(x)\right ) \operatorname {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-i \tanh (x)\right )}{a \sqrt {a \cosh ^4(x)}}\\ &=\frac {\cosh (x) \sinh (x)}{a \sqrt {a \cosh ^4(x)}}-\frac {2 \sinh ^2(x) \tanh (x)}{3 a \sqrt {a \cosh ^4(x)}}+\frac {\sinh ^2(x) \tanh ^3(x)}{5 a \sqrt {a \cosh ^4(x)}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 30, normalized size = 0.45 \[ \frac {\sinh (x) \cosh (x) (6 \cosh (2 x)+\cosh (4 x)+8)}{15 \left (a \cosh ^4(x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Cosh[x]^4)^(-3/2),x]

[Out]

(Cosh[x]*(8 + 6*Cosh[2*x] + Cosh[4*x])*Sinh[x])/(15*(a*Cosh[x]^4)^(3/2))

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fricas [B]  time = 1.33, size = 1137, normalized size = 16.97 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(x)^4)^(3/2),x, algorithm="fricas")

[Out]

-16/15*(40*cosh(x)*e^(2*x)*sinh(x)^3 + 10*e^(2*x)*sinh(x)^4 + 5*(12*cosh(x)^2 + 1)*e^(2*x)*sinh(x)^2 + 10*(4*c
osh(x)^3 + cosh(x))*e^(2*x)*sinh(x) + (10*cosh(x)^4 + 5*cosh(x)^2 + 1)*e^(2*x))*sqrt(a*e^(8*x) + 4*a*e^(6*x) +
6*a*e^(4*x) + 4*a*e^(2*x) + a)*e^(-2*x)/(a^2*cosh(x)^10 + (a^2*e^(4*x) + 2*a^2*e^(2*x) + a^2)*sinh(x)^10 + 5*
a^2*cosh(x)^8 + 10*(a^2*cosh(x)*e^(4*x) + 2*a^2*cosh(x)*e^(2*x) + a^2*cosh(x))*sinh(x)^9 + 5*(9*a^2*cosh(x)^2
+ a^2 + (9*a^2*cosh(x)^2 + a^2)*e^(4*x) + 2*(9*a^2*cosh(x)^2 + a^2)*e^(2*x))*sinh(x)^8 + 10*a^2*cosh(x)^6 + 40
*(3*a^2*cosh(x)^3 + a^2*cosh(x) + (3*a^2*cosh(x)^3 + a^2*cosh(x))*e^(4*x) + 2*(3*a^2*cosh(x)^3 + a^2*cosh(x))*
e^(2*x))*sinh(x)^7 + 10*(21*a^2*cosh(x)^4 + 14*a^2*cosh(x)^2 + a^2 + (21*a^2*cosh(x)^4 + 14*a^2*cosh(x)^2 + a^
2)*e^(4*x) + 2*(21*a^2*cosh(x)^4 + 14*a^2*cosh(x)^2 + a^2)*e^(2*x))*sinh(x)^6 + 10*a^2*cosh(x)^4 + 4*(63*a^2*c
osh(x)^5 + 70*a^2*cosh(x)^3 + 15*a^2*cosh(x) + (63*a^2*cosh(x)^5 + 70*a^2*cosh(x)^3 + 15*a^2*cosh(x))*e^(4*x)
+ 2*(63*a^2*cosh(x)^5 + 70*a^2*cosh(x)^3 + 15*a^2*cosh(x))*e^(2*x))*sinh(x)^5 + 10*(21*a^2*cosh(x)^6 + 35*a^2*
cosh(x)^4 + 15*a^2*cosh(x)^2 + a^2 + (21*a^2*cosh(x)^6 + 35*a^2*cosh(x)^4 + 15*a^2*cosh(x)^2 + a^2)*e^(4*x) +
2*(21*a^2*cosh(x)^6 + 35*a^2*cosh(x)^4 + 15*a^2*cosh(x)^2 + a^2)*e^(2*x))*sinh(x)^4 + 5*a^2*cosh(x)^2 + 40*(3*
a^2*cosh(x)^7 + 7*a^2*cosh(x)^5 + 5*a^2*cosh(x)^3 + a^2*cosh(x) + (3*a^2*cosh(x)^7 + 7*a^2*cosh(x)^5 + 5*a^2*c
osh(x)^3 + a^2*cosh(x))*e^(4*x) + 2*(3*a^2*cosh(x)^7 + 7*a^2*cosh(x)^5 + 5*a^2*cosh(x)^3 + a^2*cosh(x))*e^(2*x
))*sinh(x)^3 + 5*(9*a^2*cosh(x)^8 + 28*a^2*cosh(x)^6 + 30*a^2*cosh(x)^4 + 12*a^2*cosh(x)^2 + a^2 + (9*a^2*cosh
(x)^8 + 28*a^2*cosh(x)^6 + 30*a^2*cosh(x)^4 + 12*a^2*cosh(x)^2 + a^2)*e^(4*x) + 2*(9*a^2*cosh(x)^8 + 28*a^2*co
sh(x)^6 + 30*a^2*cosh(x)^4 + 12*a^2*cosh(x)^2 + a^2)*e^(2*x))*sinh(x)^2 + a^2 + (a^2*cosh(x)^10 + 5*a^2*cosh(x
)^8 + 10*a^2*cosh(x)^6 + 10*a^2*cosh(x)^4 + 5*a^2*cosh(x)^2 + a^2)*e^(4*x) + 2*(a^2*cosh(x)^10 + 5*a^2*cosh(x)
^8 + 10*a^2*cosh(x)^6 + 10*a^2*cosh(x)^4 + 5*a^2*cosh(x)^2 + a^2)*e^(2*x) + 10*(a^2*cosh(x)^9 + 4*a^2*cosh(x)^
7 + 6*a^2*cosh(x)^5 + 4*a^2*cosh(x)^3 + a^2*cosh(x) + (a^2*cosh(x)^9 + 4*a^2*cosh(x)^7 + 6*a^2*cosh(x)^5 + 4*a
^2*cosh(x)^3 + a^2*cosh(x))*e^(4*x) + 2*(a^2*cosh(x)^9 + 4*a^2*cosh(x)^7 + 6*a^2*cosh(x)^5 + 4*a^2*cosh(x)^3 +
a^2*cosh(x))*e^(2*x))*sinh(x))

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giac [A]  time = 0.16, size = 27, normalized size = 0.40 \[ -\frac {16 \, {\left (10 \, e^{\left (4 \, x\right )} + 5 \, e^{\left (2 \, x\right )} + 1\right )}}{15 \, a^{\frac {3}{2}} {\left (e^{\left (2 \, x\right )} + 1\right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(x)^4)^(3/2),x, algorithm="giac")

[Out]

-16/15*(10*e^(4*x) + 5*e^(2*x) + 1)/(a^(3/2)*(e^(2*x) + 1)^5)

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maple [A]  time = 0.39, size = 80, normalized size = 1.19 \[ \frac {\sqrt {8}\, \sqrt {2}\, \left (2 \left (\cosh ^{2}\left (2 x \right )\right )+6 \cosh \left (2 x \right )+7\right ) \sqrt {a \left (\sinh ^{2}\left (2 x \right )\right )}\, \sqrt {a \left (-1+\cosh \left (2 x \right )\right ) \left (\cosh \left (2 x \right )+1\right )}}{15 a^{2} \left (\cosh \left (2 x \right )+1\right )^{2} \sinh \left (2 x \right ) \sqrt {\left (\cosh \left (2 x \right )+1\right )^{2} a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cosh(x)^4)^(3/2),x)

[Out]

1/15*8^(1/2)*2^(1/2)/a^2*(2*cosh(2*x)^2+6*cosh(2*x)+7)*(a*sinh(2*x)^2)^(1/2)*(a*(-1+cosh(2*x))*(cosh(2*x)+1))^
(1/2)/(cosh(2*x)+1)^2/sinh(2*x)/((cosh(2*x)+1)^2*a)^(1/2)

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maxima [B]  time = 0.43, size = 165, normalized size = 2.46 \[ \frac {16 \, e^{\left (-2 \, x\right )}}{3 \, {\left (5 \, a^{\frac {3}{2}} e^{\left (-2 \, x\right )} + 10 \, a^{\frac {3}{2}} e^{\left (-4 \, x\right )} + 10 \, a^{\frac {3}{2}} e^{\left (-6 \, x\right )} + 5 \, a^{\frac {3}{2}} e^{\left (-8 \, x\right )} + a^{\frac {3}{2}} e^{\left (-10 \, x\right )} + a^{\frac {3}{2}}\right )}} + \frac {32 \, e^{\left (-4 \, x\right )}}{3 \, {\left (5 \, a^{\frac {3}{2}} e^{\left (-2 \, x\right )} + 10 \, a^{\frac {3}{2}} e^{\left (-4 \, x\right )} + 10 \, a^{\frac {3}{2}} e^{\left (-6 \, x\right )} + 5 \, a^{\frac {3}{2}} e^{\left (-8 \, x\right )} + a^{\frac {3}{2}} e^{\left (-10 \, x\right )} + a^{\frac {3}{2}}\right )}} + \frac {16}{15 \, {\left (5 \, a^{\frac {3}{2}} e^{\left (-2 \, x\right )} + 10 \, a^{\frac {3}{2}} e^{\left (-4 \, x\right )} + 10 \, a^{\frac {3}{2}} e^{\left (-6 \, x\right )} + 5 \, a^{\frac {3}{2}} e^{\left (-8 \, x\right )} + a^{\frac {3}{2}} e^{\left (-10 \, x\right )} + a^{\frac {3}{2}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(x)^4)^(3/2),x, algorithm="maxima")

[Out]

16/3*e^(-2*x)/(5*a^(3/2)*e^(-2*x) + 10*a^(3/2)*e^(-4*x) + 10*a^(3/2)*e^(-6*x) + 5*a^(3/2)*e^(-8*x) + a^(3/2)*e
^(-10*x) + a^(3/2)) + 32/3*e^(-4*x)/(5*a^(3/2)*e^(-2*x) + 10*a^(3/2)*e^(-4*x) + 10*a^(3/2)*e^(-6*x) + 5*a^(3/2
)*e^(-8*x) + a^(3/2)*e^(-10*x) + a^(3/2)) + 16/15/(5*a^(3/2)*e^(-2*x) + 10*a^(3/2)*e^(-4*x) + 10*a^(3/2)*e^(-6
*x) + 5*a^(3/2)*e^(-8*x) + a^(3/2)*e^(-10*x) + a^(3/2))

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mupad [B]  time = 0.97, size = 48, normalized size = 0.72 \[ -\frac {64\,{\mathrm {e}}^{2\,x}\,\sqrt {a\,{\left (\frac {{\mathrm {e}}^{-x}}{2}+\frac {{\mathrm {e}}^x}{2}\right )}^4}\,\left (5\,{\mathrm {e}}^{2\,x}+10\,{\mathrm {e}}^{4\,x}+1\right )}{15\,a^2\,{\left ({\mathrm {e}}^{2\,x}+1\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cosh(x)^4)^(3/2),x)

[Out]

-(64*exp(2*x)*(a*(exp(-x)/2 + exp(x)/2)^4)^(1/2)*(5*exp(2*x) + 10*exp(4*x) + 1))/(15*a^2*(exp(2*x) + 1)^7)

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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(1/(a*cosh(x)**4)**(3/2),x)

[Out]

Timed out

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