3.49 \(\int \cosh ^4(c+d x) (a+b \text {sech}^2(c+d x)) \, dx\)

Optimal. Leaf size=61 \[ \frac {(3 a+4 b) \sinh (c+d x) \cosh (c+d x)}{8 d}+\frac {1}{8} x (3 a+4 b)+\frac {a \sinh (c+d x) \cosh ^3(c+d x)}{4 d} \]

[Out]

1/8*(3*a+4*b)*x+1/8*(3*a+4*b)*cosh(d*x+c)*sinh(d*x+c)/d+1/4*a*cosh(d*x+c)^3*sinh(d*x+c)/d

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Rubi [A]  time = 0.05, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4045, 2635, 8} \[ \frac {(3 a+4 b) \sinh (c+d x) \cosh (c+d x)}{8 d}+\frac {1}{8} x (3 a+4 b)+\frac {a \sinh (c+d x) \cosh ^3(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[c + d*x]^4*(a + b*Sech[c + d*x]^2),x]

[Out]

((3*a + 4*b)*x)/8 + ((3*a + 4*b)*Cosh[c + d*x]*Sinh[c + d*x])/(8*d) + (a*Cosh[c + d*x]^3*Sinh[c + d*x])/(4*d)

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 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
 + f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rubi steps

\begin {align*} \int \cosh ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right ) \, dx &=\frac {a \cosh ^3(c+d x) \sinh (c+d x)}{4 d}+\frac {1}{4} (3 a+4 b) \int \cosh ^2(c+d x) \, dx\\ &=\frac {(3 a+4 b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {a \cosh ^3(c+d x) \sinh (c+d x)}{4 d}+\frac {1}{8} (3 a+4 b) \int 1 \, dx\\ &=\frac {1}{8} (3 a+4 b) x+\frac {(3 a+4 b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {a \cosh ^3(c+d x) \sinh (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 45, normalized size = 0.74 \[ \frac {4 (3 a+4 b) (c+d x)+8 (a+b) \sinh (2 (c+d x))+a \sinh (4 (c+d x))}{32 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[c + d*x]^4*(a + b*Sech[c + d*x]^2),x]

[Out]

(4*(3*a + 4*b)*(c + d*x) + 8*(a + b)*Sinh[2*(c + d*x)] + a*Sinh[4*(c + d*x)])/(32*d)

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fricas [A]  time = 0.40, size = 61, normalized size = 1.00 \[ \frac {a \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (3 \, a + 4 \, b\right )} d x + {\left (a \cosh \left (d x + c\right )^{3} + 4 \, {\left (a + b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^4*(a+b*sech(d*x+c)^2),x, algorithm="fricas")

[Out]

1/8*(a*cosh(d*x + c)*sinh(d*x + c)^3 + (3*a + 4*b)*d*x + (a*cosh(d*x + c)^3 + 4*(a + b)*cosh(d*x + c))*sinh(d*
x + c))/d

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giac [B]  time = 0.14, size = 116, normalized size = 1.90 \[ \frac {8 \, {\left (d x + c\right )} {\left (3 \, a + 4 \, b\right )} + a e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a e^{\left (2 \, d x + 2 \, c\right )} + 8 \, b e^{\left (2 \, d x + 2 \, c\right )} - {\left (18 \, a e^{\left (4 \, d x + 4 \, c\right )} + 24 \, b e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a e^{\left (2 \, d x + 2 \, c\right )} + 8 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^4*(a+b*sech(d*x+c)^2),x, algorithm="giac")

[Out]

1/64*(8*(d*x + c)*(3*a + 4*b) + a*e^(4*d*x + 4*c) + 8*a*e^(2*d*x + 2*c) + 8*b*e^(2*d*x + 2*c) - (18*a*e^(4*d*x
 + 4*c) + 24*b*e^(4*d*x + 4*c) + 8*a*e^(2*d*x + 2*c) + 8*b*e^(2*d*x + 2*c) + a)*e^(-4*d*x - 4*c))/d

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maple [A]  time = 0.46, size = 66, normalized size = 1.08 \[ \frac {a \left (\left (\frac {\left (\cosh ^{3}\left (d x +c \right )\right )}{4}+\frac {3 \cosh \left (d x +c \right )}{8}\right ) \sinh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+b \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(d*x+c)^4*(a+b*sech(d*x+c)^2),x)

[Out]

1/d*(a*((1/4*cosh(d*x+c)^3+3/8*cosh(d*x+c))*sinh(d*x+c)+3/8*d*x+3/8*c)+b*(1/2*cosh(d*x+c)*sinh(d*x+c)+1/2*d*x+
1/2*c))

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maxima [A]  time = 0.31, size = 97, normalized size = 1.59 \[ \frac {1}{64} \, a {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} + \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + \frac {1}{8} \, b {\left (4 \, x + \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^4*(a+b*sech(d*x+c)^2),x, algorithm="maxima")

[Out]

1/64*a*(24*x + e^(4*d*x + 4*c)/d + 8*e^(2*d*x + 2*c)/d - 8*e^(-2*d*x - 2*c)/d - e^(-4*d*x - 4*c)/d) + 1/8*b*(4
*x + e^(2*d*x + 2*c)/d - e^(-2*d*x - 2*c)/d)

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mupad [B]  time = 0.13, size = 50, normalized size = 0.82 \[ \frac {\frac {a\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )}{4}+\frac {a\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )}{32}+\frac {b\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )}{4}}{d}+\frac {3\,a\,x}{8}+\frac {b\,x}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(c + d*x)^4*(a + b/cosh(c + d*x)^2),x)

[Out]

((a*sinh(2*c + 2*d*x))/4 + (a*sinh(4*c + 4*d*x))/32 + (b*sinh(2*c + 2*d*x))/4)/d + (3*a*x)/8 + (b*x)/2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right ) \cosh ^{4}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)**4*(a+b*sech(d*x+c)**2),x)

[Out]

Integral((a + b*sech(c + d*x)**2)*cosh(c + d*x)**4, x)

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