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

Optimal. Leaf size=78 \[ \frac {5}{24} a \sinh (x) \cosh (x) \sqrt {a \cosh ^4(x)}+\frac {5}{16} a \tanh (x) \sqrt {a \cosh ^4(x)}+\frac {5}{16} a x \text {sech}^2(x) \sqrt {a \cosh ^4(x)}+\frac {1}{6} a \sinh (x) \cosh ^3(x) \sqrt {a \cosh ^4(x)} \]

[Out]

5/16*a*x*sech(x)^2*(a*cosh(x)^4)^(1/2)+5/24*a*cosh(x)*sinh(x)*(a*cosh(x)^4)^(1/2)+1/6*a*cosh(x)^3*sinh(x)*(a*c
osh(x)^4)^(1/2)+5/16*a*(a*cosh(x)^4)^(1/2)*tanh(x)

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Rubi [A]  time = 0.04, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3207, 2635, 8} \[ \frac {1}{6} a \sinh (x) \cosh ^3(x) \sqrt {a \cosh ^4(x)}+\frac {5}{24} a \sinh (x) \cosh (x) \sqrt {a \cosh ^4(x)}+\frac {5}{16} a \tanh (x) \sqrt {a \cosh ^4(x)}+\frac {5}{16} a x \text {sech}^2(x) \sqrt {a \cosh ^4(x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a*x*Sqrt[a*Cosh[x]^4]*Sech[x]^2)/16 + (5*a*Cosh[x]*Sqrt[a*Cosh[x]^4]*Sinh[x])/24 + (a*Cosh[x]^3*Sqrt[a*Cosh
[x]^4]*Sinh[x])/6 + (5*a*Sqrt[a*Cosh[x]^4]*Tanh[x])/16

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2635

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

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
]])

Rubi steps

\begin {align*} \int \left (a \cosh ^4(x)\right )^{3/2} \, dx &=\left (a \sqrt {a \cosh ^4(x)} \text {sech}^2(x)\right ) \int \cosh ^6(x) \, dx\\ &=\frac {1}{6} a \cosh ^3(x) \sqrt {a \cosh ^4(x)} \sinh (x)+\frac {1}{6} \left (5 a \sqrt {a \cosh ^4(x)} \text {sech}^2(x)\right ) \int \cosh ^4(x) \, dx\\ &=\frac {5}{24} a \cosh (x) \sqrt {a \cosh ^4(x)} \sinh (x)+\frac {1}{6} a \cosh ^3(x) \sqrt {a \cosh ^4(x)} \sinh (x)+\frac {1}{8} \left (5 a \sqrt {a \cosh ^4(x)} \text {sech}^2(x)\right ) \int \cosh ^2(x) \, dx\\ &=\frac {5}{24} a \cosh (x) \sqrt {a \cosh ^4(x)} \sinh (x)+\frac {1}{6} a \cosh ^3(x) \sqrt {a \cosh ^4(x)} \sinh (x)+\frac {5}{16} a \sqrt {a \cosh ^4(x)} \tanh (x)+\frac {1}{16} \left (5 a \sqrt {a \cosh ^4(x)} \text {sech}^2(x)\right ) \int 1 \, dx\\ &=\frac {5}{16} a x \sqrt {a \cosh ^4(x)} \text {sech}^2(x)+\frac {5}{24} a \cosh (x) \sqrt {a \cosh ^4(x)} \sinh (x)+\frac {1}{6} a \cosh ^3(x) \sqrt {a \cosh ^4(x)} \sinh (x)+\frac {5}{16} a \sqrt {a \cosh ^4(x)} \tanh (x)\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 38, normalized size = 0.49 \[ \frac {1}{192} (60 x+45 \sinh (2 x)+9 \sinh (4 x)+\sinh (6 x)) \text {sech}^6(x) \left (a \cosh ^4(x)\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*Cosh[x]^4)^(3/2)*Sech[x]^6*(60*x + 45*Sinh[2*x] + 9*Sinh[4*x] + Sinh[6*x]))/192

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fricas [B]  time = 0.83, size = 659, normalized size = 8.45 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(12*a*cosh(x)*e^(2*x)*sinh(x)^11 + a*e^(2*x)*sinh(x)^12 + 3*(22*a*cosh(x)^2 + 3*a)*e^(2*x)*sinh(x)^10 +
10*(22*a*cosh(x)^3 + 9*a*cosh(x))*e^(2*x)*sinh(x)^9 + 45*(11*a*cosh(x)^4 + 9*a*cosh(x)^2 + a)*e^(2*x)*sinh(x)^
8 + 72*(11*a*cosh(x)^5 + 15*a*cosh(x)^3 + 5*a*cosh(x))*e^(2*x)*sinh(x)^7 + 6*(154*a*cosh(x)^6 + 315*a*cosh(x)^
4 + 210*a*cosh(x)^2 + 20*a*x)*e^(2*x)*sinh(x)^6 + 36*(22*a*cosh(x)^7 + 63*a*cosh(x)^5 + 70*a*cosh(x)^3 + 20*a*
x*cosh(x))*e^(2*x)*sinh(x)^5 + 45*(11*a*cosh(x)^8 + 42*a*cosh(x)^6 + 70*a*cosh(x)^4 + 40*a*x*cosh(x)^2 - a)*e^
(2*x)*sinh(x)^4 + 20*(11*a*cosh(x)^9 + 54*a*cosh(x)^7 + 126*a*cosh(x)^5 + 120*a*x*cosh(x)^3 - 9*a*cosh(x))*e^(
2*x)*sinh(x)^3 + 3*(22*a*cosh(x)^10 + 135*a*cosh(x)^8 + 420*a*cosh(x)^6 + 600*a*x*cosh(x)^4 - 90*a*cosh(x)^2 -
3*a)*e^(2*x)*sinh(x)^2 + 6*(2*a*cosh(x)^11 + 15*a*cosh(x)^9 + 60*a*cosh(x)^7 + 120*a*x*cosh(x)^5 - 30*a*cosh(
x)^3 - 3*a*cosh(x))*e^(2*x)*sinh(x) + (a*cosh(x)^12 + 9*a*cosh(x)^10 + 45*a*cosh(x)^8 + 120*a*x*cosh(x)^6 - 45
*a*cosh(x)^4 - 9*a*cosh(x)^2 - a)*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)/(cosh(x)^6*e^(4*x) + 2*cosh(x)^6*e^(2*x) + (e^(4*x) + 2*e^(2*x) + 1)*sinh(x)^6 + cosh(x)^6 + 6*(cosh(x)*e^
(4*x) + 2*cosh(x)*e^(2*x) + cosh(x))*sinh(x)^5 + 15*(cosh(x)^2*e^(4*x) + 2*cosh(x)^2*e^(2*x) + cosh(x)^2)*sinh
(x)^4 + 20*(cosh(x)^3*e^(4*x) + 2*cosh(x)^3*e^(2*x) + cosh(x)^3)*sinh(x)^3 + 15*(cosh(x)^4*e^(4*x) + 2*cosh(x)
^4*e^(2*x) + cosh(x)^4)*sinh(x)^2 + 6*(cosh(x)^5*e^(4*x) + 2*cosh(x)^5*e^(2*x) + cosh(x)^5)*sinh(x))

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giac [A]  time = 0.14, size = 52, normalized size = 0.67 \[ -\frac {1}{384} \, {\left ({\left (110 \, e^{\left (6 \, x\right )} + 45 \, e^{\left (4 \, x\right )} + 9 \, e^{\left (2 \, x\right )} + 1\right )} e^{\left (-6 \, x\right )} - 120 \, x - e^{\left (6 \, x\right )} - 9 \, e^{\left (4 \, x\right )} - 45 \, e^{\left (2 \, x\right )}\right )} a^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/384*((110*e^(6*x) + 45*e^(4*x) + 9*e^(2*x) + 1)*e^(-6*x) - 120*x - e^(6*x) - 9*e^(4*x) - 45*e^(2*x))*a^(3/2
)

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maple [B]  time = 0.50, size = 131, normalized size = 1.68 \[ \frac {\sqrt {8}\, \left (\cosh \left (2 x \right )+1\right ) \sqrt {a \left (-1+\cosh \left (2 x \right )\right ) \left (\cosh \left (2 x \right )+1\right )}\, \sqrt {2}\, \sqrt {a}\, \left (2 \sqrt {a \left (\sinh ^{2}\left (2 x \right )\right )}\, \sqrt {a}\, \left (\sinh ^{2}\left (2 x \right )\right )+9 \cosh \left (2 x \right ) \sqrt {a \left (\sinh ^{2}\left (2 x \right )\right )}\, \sqrt {a}+24 \sqrt {a \left (\sinh ^{2}\left (2 x \right )\right )}\, \sqrt {a}+15 \ln \left (\sqrt {a}\, \cosh \left (2 x \right )+\sqrt {a \left (\sinh ^{2}\left (2 x \right )\right )}\right ) a \right )}{384 \sinh \left (2 x \right ) \sqrt {\left (\cosh \left (2 x \right )+1\right )^{2} a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*8^(1/2)*(cosh(2*x)+1)*(a*(-1+cosh(2*x))*(cosh(2*x)+1))^(1/2)*2^(1/2)*a^(1/2)*(2*(a*sinh(2*x)^2)^(1/2)*a^
(1/2)*sinh(2*x)^2+9*cosh(2*x)*(a*sinh(2*x)^2)^(1/2)*a^(1/2)+24*(a*sinh(2*x)^2)^(1/2)*a^(1/2)+15*ln(a^(1/2)*cos
h(2*x)+(a*sinh(2*x)^2)^(1/2))*a)/sinh(2*x)/((cosh(2*x)+1)^2*a)^(1/2)

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maxima [A]  time = 0.42, size = 62, normalized size = 0.79 \[ \frac {5}{16} \, a^{\frac {3}{2}} x + \frac {1}{384} \, {\left (9 \, a^{\frac {3}{2}} e^{\left (-2 \, x\right )} + 45 \, a^{\frac {3}{2}} e^{\left (-4 \, x\right )} - 45 \, a^{\frac {3}{2}} e^{\left (-8 \, x\right )} - 9 \, a^{\frac {3}{2}} e^{\left (-10 \, x\right )} - a^{\frac {3}{2}} e^{\left (-12 \, x\right )} + a^{\frac {3}{2}}\right )} e^{\left (6 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/16*a^(3/2)*x + 1/384*(9*a^(3/2)*e^(-2*x) + 45*a^(3/2)*e^(-4*x) - 45*a^(3/2)*e^(-8*x) - 9*a^(3/2)*e^(-10*x) -
a^(3/2)*e^(-12*x) + a^(3/2))*e^(6*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a\,{\mathrm {cosh}\relax (x)}^4\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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