3.121 \(\int (a \cosh ^2(x))^{7/2} \, dx\)

Optimal. Leaf size=72 \[ \frac {16}{35} a^3 \tanh (x) \sqrt {a \cosh ^2(x)}+\frac {8}{35} a^2 \tanh (x) \left (a \cosh ^2(x)\right )^{3/2}+\frac {1}{7} \tanh (x) \left (a \cosh ^2(x)\right )^{7/2}+\frac {6}{35} a \tanh (x) \left (a \cosh ^2(x)\right )^{5/2} \]

[Out]

8/35*a^2*(a*cosh(x)^2)^(3/2)*tanh(x)+6/35*a*(a*cosh(x)^2)^(5/2)*tanh(x)+1/7*(a*cosh(x)^2)^(7/2)*tanh(x)+16/35*
a^3*(a*cosh(x)^2)^(1/2)*tanh(x)

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Rubi [A]  time = 0.05, antiderivative size = 72, 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 = {3203, 3207, 2637} \[ \frac {8}{35} a^2 \tanh (x) \left (a \cosh ^2(x)\right )^{3/2}+\frac {16}{35} a^3 \tanh (x) \sqrt {a \cosh ^2(x)}+\frac {1}{7} \tanh (x) \left (a \cosh ^2(x)\right )^{7/2}+\frac {6}{35} a \tanh (x) \left (a \cosh ^2(x)\right )^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cosh[x]^2)^(7/2),x]

[Out]

(16*a^3*Sqrt[a*Cosh[x]^2]*Tanh[x])/35 + (8*a^2*(a*Cosh[x]^2)^(3/2)*Tanh[x])/35 + (6*a*(a*Cosh[x]^2)^(5/2)*Tanh
[x])/35 + ((a*Cosh[x]^2)^(7/2)*Tanh[x])/7

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3203

Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> -Simp[(Cot[e + f*x]*(b*Sin[e + f*x]^2)^p)/(2*f*p), x]
+ Dist[(b*(2*p - 1))/(2*p), Int[(b*Sin[e + f*x]^2)^(p - 1), x], x] /; FreeQ[{b, e, f}, x] &&  !IntegerQ[p] &&
GtQ[p, 1]

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 ^2(x)\right )^{7/2} \, dx &=\frac {1}{7} \left (a \cosh ^2(x)\right )^{7/2} \tanh (x)+\frac {1}{7} (6 a) \int \left (a \cosh ^2(x)\right )^{5/2} \, dx\\ &=\frac {6}{35} a \left (a \cosh ^2(x)\right )^{5/2} \tanh (x)+\frac {1}{7} \left (a \cosh ^2(x)\right )^{7/2} \tanh (x)+\frac {1}{35} \left (24 a^2\right ) \int \left (a \cosh ^2(x)\right )^{3/2} \, dx\\ &=\frac {8}{35} a^2 \left (a \cosh ^2(x)\right )^{3/2} \tanh (x)+\frac {6}{35} a \left (a \cosh ^2(x)\right )^{5/2} \tanh (x)+\frac {1}{7} \left (a \cosh ^2(x)\right )^{7/2} \tanh (x)+\frac {1}{35} \left (16 a^3\right ) \int \sqrt {a \cosh ^2(x)} \, dx\\ &=\frac {8}{35} a^2 \left (a \cosh ^2(x)\right )^{3/2} \tanh (x)+\frac {6}{35} a \left (a \cosh ^2(x)\right )^{5/2} \tanh (x)+\frac {1}{7} \left (a \cosh ^2(x)\right )^{7/2} \tanh (x)+\frac {1}{35} \left (16 a^3 \sqrt {a \cosh ^2(x)} \text {sech}(x)\right ) \int \cosh (x) \, dx\\ &=\frac {16}{35} a^3 \sqrt {a \cosh ^2(x)} \tanh (x)+\frac {8}{35} a^2 \left (a \cosh ^2(x)\right )^{3/2} \tanh (x)+\frac {6}{35} a \left (a \cosh ^2(x)\right )^{5/2} \tanh (x)+\frac {1}{7} \left (a \cosh ^2(x)\right )^{7/2} \tanh (x)\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 42, normalized size = 0.58 \[ \frac {a^3 (1225 \sinh (x)+245 \sinh (3 x)+49 \sinh (5 x)+5 \sinh (7 x)) \text {sech}(x) \sqrt {a \cosh ^2(x)}}{2240} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Cosh[x]^2)^(7/2),x]

[Out]

(a^3*Sqrt[a*Cosh[x]^2]*Sech[x]*(1225*Sinh[x] + 245*Sinh[3*x] + 49*Sinh[5*x] + 5*Sinh[7*x]))/2240

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fricas [B]  time = 2.76, size = 817, normalized size = 11.35 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cosh(x)^2)^(7/2),x, algorithm="fricas")

[Out]

1/4480*(70*a^3*cosh(x)*e^x*sinh(x)^13 + 5*a^3*e^x*sinh(x)^14 + 7*(65*a^3*cosh(x)^2 + 7*a^3)*e^x*sinh(x)^12 + 2
8*(65*a^3*cosh(x)^3 + 21*a^3*cosh(x))*e^x*sinh(x)^11 + 7*(715*a^3*cosh(x)^4 + 462*a^3*cosh(x)^2 + 35*a^3)*e^x*
sinh(x)^10 + 70*(143*a^3*cosh(x)^5 + 154*a^3*cosh(x)^3 + 35*a^3*cosh(x))*e^x*sinh(x)^9 + 35*(429*a^3*cosh(x)^6
+ 693*a^3*cosh(x)^4 + 315*a^3*cosh(x)^2 + 35*a^3)*e^x*sinh(x)^8 + 8*(2145*a^3*cosh(x)^7 + 4851*a^3*cosh(x)^5
+ 3675*a^3*cosh(x)^3 + 1225*a^3*cosh(x))*e^x*sinh(x)^7 + 7*(2145*a^3*cosh(x)^8 + 6468*a^3*cosh(x)^6 + 7350*a^3
*cosh(x)^4 + 4900*a^3*cosh(x)^2 - 175*a^3)*e^x*sinh(x)^6 + 14*(715*a^3*cosh(x)^9 + 2772*a^3*cosh(x)^7 + 4410*a
^3*cosh(x)^5 + 4900*a^3*cosh(x)^3 - 525*a^3*cosh(x))*e^x*sinh(x)^5 + 35*(143*a^3*cosh(x)^10 + 693*a^3*cosh(x)^
8 + 1470*a^3*cosh(x)^6 + 2450*a^3*cosh(x)^4 - 525*a^3*cosh(x)^2 - 7*a^3)*e^x*sinh(x)^4 + 140*(13*a^3*cosh(x)^1
1 + 77*a^3*cosh(x)^9 + 210*a^3*cosh(x)^7 + 490*a^3*cosh(x)^5 - 175*a^3*cosh(x)^3 - 7*a^3*cosh(x))*e^x*sinh(x)^
3 + 7*(65*a^3*cosh(x)^12 + 462*a^3*cosh(x)^10 + 1575*a^3*cosh(x)^8 + 4900*a^3*cosh(x)^6 - 2625*a^3*cosh(x)^4 -
210*a^3*cosh(x)^2 - 7*a^3)*e^x*sinh(x)^2 + 14*(5*a^3*cosh(x)^13 + 42*a^3*cosh(x)^11 + 175*a^3*cosh(x)^9 + 700
*a^3*cosh(x)^7 - 525*a^3*cosh(x)^5 - 70*a^3*cosh(x)^3 - 7*a^3*cosh(x))*e^x*sinh(x) + (5*a^3*cosh(x)^14 + 49*a^
3*cosh(x)^12 + 245*a^3*cosh(x)^10 + 1225*a^3*cosh(x)^8 - 1225*a^3*cosh(x)^6 - 245*a^3*cosh(x)^4 - 49*a^3*cosh(
x)^2 - 5*a^3)*e^x)*sqrt(a*e^(4*x) + 2*a*e^(2*x) + a)*e^(-x)/(cosh(x)^7*e^(2*x) + (e^(2*x) + 1)*sinh(x)^7 + cos
h(x)^7 + 7*(cosh(x)*e^(2*x) + cosh(x))*sinh(x)^6 + 21*(cosh(x)^2*e^(2*x) + cosh(x)^2)*sinh(x)^5 + 35*(cosh(x)^
3*e^(2*x) + cosh(x)^3)*sinh(x)^4 + 35*(cosh(x)^4*e^(2*x) + cosh(x)^4)*sinh(x)^3 + 21*(cosh(x)^5*e^(2*x) + cosh
(x)^5)*sinh(x)^2 + 7*(cosh(x)^6*e^(2*x) + cosh(x)^6)*sinh(x))

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giac [A]  time = 0.12, size = 79, normalized size = 1.10 \[ \frac {1}{4480} \, {\left (5 \, a^{3} e^{\left (7 \, x\right )} + 49 \, a^{3} e^{\left (5 \, x\right )} + 245 \, a^{3} e^{\left (3 \, x\right )} + 1225 \, a^{3} e^{x} - {\left (1225 \, a^{3} e^{\left (6 \, x\right )} + 245 \, a^{3} e^{\left (4 \, x\right )} + 49 \, a^{3} e^{\left (2 \, x\right )} + 5 \, a^{3}\right )} e^{\left (-7 \, x\right )}\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cosh(x)^2)^(7/2),x, algorithm="giac")

[Out]

1/4480*(5*a^3*e^(7*x) + 49*a^3*e^(5*x) + 245*a^3*e^(3*x) + 1225*a^3*e^x - (1225*a^3*e^(6*x) + 245*a^3*e^(4*x)
+ 49*a^3*e^(2*x) + 5*a^3)*e^(-7*x))*sqrt(a)

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maple [A]  time = 0.16, size = 38, normalized size = 0.53 \[ \frac {a^{4} \cosh \relax (x ) \sinh \relax (x ) \left (5 \left (\cosh ^{6}\relax (x )\right )+6 \left (\cosh ^{4}\relax (x )\right )+8 \left (\cosh ^{2}\relax (x )\right )+16\right )}{35 \sqrt {a \left (\cosh ^{2}\relax (x )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cosh(x)^2)^(7/2),x)

[Out]

1/35*a^4*cosh(x)*sinh(x)*(5*cosh(x)^6+6*cosh(x)^4+8*cosh(x)^2+16)/(a*cosh(x)^2)^(1/2)

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maxima [A]  time = 0.43, size = 71, normalized size = 0.99 \[ \frac {1}{896} \, a^{\frac {7}{2}} e^{\left (7 \, x\right )} + \frac {7}{640} \, a^{\frac {7}{2}} e^{\left (5 \, x\right )} + \frac {7}{128} \, a^{\frac {7}{2}} e^{\left (3 \, x\right )} - \frac {35}{128} \, a^{\frac {7}{2}} e^{\left (-x\right )} - \frac {7}{128} \, a^{\frac {7}{2}} e^{\left (-3 \, x\right )} - \frac {7}{640} \, a^{\frac {7}{2}} e^{\left (-5 \, x\right )} - \frac {1}{896} \, a^{\frac {7}{2}} e^{\left (-7 \, x\right )} + \frac {35}{128} \, a^{\frac {7}{2}} e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cosh(x)^2)^(7/2),x, algorithm="maxima")

[Out]

1/896*a^(7/2)*e^(7*x) + 7/640*a^(7/2)*e^(5*x) + 7/128*a^(7/2)*e^(3*x) - 35/128*a^(7/2)*e^(-x) - 7/128*a^(7/2)*
e^(-3*x) - 7/640*a^(7/2)*e^(-5*x) - 1/896*a^(7/2)*e^(-7*x) + 35/128*a^(7/2)*e^x

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a\,{\mathrm {cosh}\relax (x)}^2\right )}^{7/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cosh(x)^2)^(7/2),x)

[Out]

int((a*cosh(x)^2)^(7/2), 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)**2)**(7/2),x)

[Out]

Timed out

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