3.1.64 \(\int (a+b \cosh (c+d x))^3 \, dx\) [64]

Optimal. Leaf size=90 \[ \frac {1}{2} a \left (2 a^2+3 b^2\right ) x+\frac {2 b \left (4 a^2+b^2\right ) \sinh (c+d x)}{3 d}+\frac {5 a b^2 \cosh (c+d x) \sinh (c+d x)}{6 d}+\frac {b (a+b \cosh (c+d x))^2 \sinh (c+d x)}{3 d} \]

[Out]

1/2*a*(2*a^2+3*b^2)*x+2/3*b*(4*a^2+b^2)*sinh(d*x+c)/d+5/6*a*b^2*cosh(d*x+c)*sinh(d*x+c)/d+1/3*b*(a+b*cosh(d*x+
c))^2*sinh(d*x+c)/d

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Rubi [A]
time = 0.05, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2735, 2813} \begin {gather*} \frac {2 b \left (4 a^2+b^2\right ) \sinh (c+d x)}{3 d}+\frac {1}{2} a x \left (2 a^2+3 b^2\right )+\frac {5 a b^2 \sinh (c+d x) \cosh (c+d x)}{6 d}+\frac {b \sinh (c+d x) (a+b \cosh (c+d x))^2}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Cosh[c + d*x])^3,x]

[Out]

(a*(2*a^2 + 3*b^2)*x)/2 + (2*b*(4*a^2 + b^2)*Sinh[c + d*x])/(3*d) + (5*a*b^2*Cosh[c + d*x]*Sinh[c + d*x])/(6*d
) + (b*(a + b*Cosh[c + d*x])^2*Sinh[c + d*x])/(3*d)

Rule 2735

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((a + b*Sin[c + d*x])^(n
- 1)/(d*n)), x] + Dist[1/n, Int[(a + b*Sin[c + d*x])^(n - 2)*Simp[a^2*n + b^2*(n - 1) + a*b*(2*n - 1)*Sin[c +
d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]

Rule 2813

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(2*a*c +
b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Cos[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rubi steps

\begin {align*} \int (a+b \cosh (c+d x))^3 \, dx &=\frac {b (a+b \cosh (c+d x))^2 \sinh (c+d x)}{3 d}+\frac {1}{3} \int (a+b \cosh (c+d x)) \left (3 a^2+2 b^2+5 a b \cosh (c+d x)\right ) \, dx\\ &=\frac {1}{2} a \left (2 a^2+3 b^2\right ) x+\frac {2 b \left (4 a^2+b^2\right ) \sinh (c+d x)}{3 d}+\frac {5 a b^2 \cosh (c+d x) \sinh (c+d x)}{6 d}+\frac {b (a+b \cosh (c+d x))^2 \sinh (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 80, normalized size = 0.89 \begin {gather*} \frac {12 a^3 c+18 a b^2 c+12 a^3 d x+18 a b^2 d x+9 b \left (4 a^2+b^2\right ) \sinh (c+d x)+9 a b^2 \sinh (2 (c+d x))+b^3 \sinh (3 (c+d x))}{12 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cosh[c + d*x])^3,x]

[Out]

(12*a^3*c + 18*a*b^2*c + 12*a^3*d*x + 18*a*b^2*d*x + 9*b*(4*a^2 + b^2)*Sinh[c + d*x] + 9*a*b^2*Sinh[2*(c + d*x
)] + b^3*Sinh[3*(c + d*x)])/(12*d)

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Maple [A]
time = 1.11, size = 71, normalized size = 0.79

method	result	size
default	\(a^{3} x +\frac {\left (\frac {3}{4} b^{3}+3 a^{2} b \right ) \sinh \left (d x +c \right )}{d}+\frac {3 a \,b^{2} x}{2}+\frac {b^{3} \sinh \left (3 d x +3 c \right )}{12 d}+\frac {3 a \,b^{2} \sinh \left (2 d x +2 c \right )}{4 d}\)	\(71\)
risch	\(a^{3} x +\frac {3 a \,b^{2} x}{2}+\frac {b^{3} {\mathrm e}^{3 d x +3 c}}{24 d}+\frac {3 \,{\mathrm e}^{2 d x +2 c} a \,b^{2}}{8 d}+\frac {3 b \,{\mathrm e}^{d x +c} a^{2}}{2 d}+\frac {3 b^{3} {\mathrm e}^{d x +c}}{8 d}-\frac {3 b \,{\mathrm e}^{-d x -c} a^{2}}{2 d}-\frac {3 b^{3} {\mathrm e}^{-d x -c}}{8 d}-\frac {3 \,{\mathrm e}^{-2 d x -2 c} a \,b^{2}}{8 d}-\frac {b^{3} {\mathrm e}^{-3 d x -3 c}}{24 d}\)	\(148\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cosh(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

a^3*x+(3/4*b^3+3*a^2*b)/d*sinh(d*x+c)+3/2*a*b^2*x+1/12*b^3/d*sinh(3*d*x+3*c)+3/4*a*b^2/d*sinh(2*d*x+2*c)

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Maxima [A]
time = 0.27, size = 116, normalized size = 1.29 \begin {gather*} \frac {3}{8} \, a b^{2} {\left (4 \, x + \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} + a^{3} x + \frac {1}{24} \, b^{3} {\left (\frac {e^{\left (3 \, d x + 3 \, c\right )}}{d} + \frac {9 \, e^{\left (d x + c\right )}}{d} - \frac {9 \, e^{\left (-d x - c\right )}}{d} - \frac {e^{\left (-3 \, d x - 3 \, c\right )}}{d}\right )} + \frac {3 \, a^{2} b \sinh \left (d x + c\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cosh(d*x+c))^3,x, algorithm="maxima")

[Out]

3/8*a*b^2*(4*x + e^(2*d*x + 2*c)/d - e^(-2*d*x - 2*c)/d) + a^3*x + 1/24*b^3*(e^(3*d*x + 3*c)/d + 9*e^(d*x + c)
/d - 9*e^(-d*x - c)/d - e^(-3*d*x - 3*c)/d) + 3*a^2*b*sinh(d*x + c)/d

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Fricas [A]
time = 0.39, size = 78, normalized size = 0.87 \begin {gather*} \frac {b^{3} \sinh \left (d x + c\right )^{3} + 6 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} d x + 3 \, {\left (b^{3} \cosh \left (d x + c\right )^{2} + 6 \, a b^{2} \cosh \left (d x + c\right ) + 12 \, a^{2} b + 3 \, b^{3}\right )} \sinh \left (d x + c\right )}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cosh(d*x+c))^3,x, algorithm="fricas")

[Out]

1/12*(b^3*sinh(d*x + c)^3 + 6*(2*a^3 + 3*a*b^2)*d*x + 3*(b^3*cosh(d*x + c)^2 + 6*a*b^2*cosh(d*x + c) + 12*a^2*
b + 3*b^3)*sinh(d*x + c))/d

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Sympy [A]
time = 0.13, size = 128, normalized size = 1.42 \begin {gather*} \begin {cases} a^{3} x + \frac {3 a^{2} b \sinh {\left (c + d x \right )}}{d} - \frac {3 a b^{2} x \sinh ^{2}{\left (c + d x \right )}}{2} + \frac {3 a b^{2} x \cosh ^{2}{\left (c + d x \right )}}{2} + \frac {3 a b^{2} \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{2 d} - \frac {2 b^{3} \sinh ^{3}{\left (c + d x \right )}}{3 d} + \frac {b^{3} \sinh {\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \cosh {\left (c \right )}\right )^{3} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cosh(d*x+c))**3,x)

[Out]

Piecewise((a**3*x + 3*a**2*b*sinh(c + d*x)/d - 3*a*b**2*x*sinh(c + d*x)**2/2 + 3*a*b**2*x*cosh(c + d*x)**2/2 +
3*a*b**2*sinh(c + d*x)*cosh(c + d*x)/(2*d) - 2*b**3*sinh(c + d*x)**3/(3*d) + b**3*sinh(c + d*x)*cosh(c + d*x)
**2/d, Ne(d, 0)), (x*(a + b*cosh(c))**3, True))

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Giac [A]
time = 0.42, size = 131, normalized size = 1.46 \begin {gather*} \frac {b^{3} e^{\left (3 \, d x + 3 \, c\right )}}{24 \, d} + \frac {3 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )}}{8 \, d} - \frac {3 \, a b^{2} e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, d} - \frac {b^{3} e^{\left (-3 \, d x - 3 \, c\right )}}{24 \, d} + \frac {1}{2} \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} x + \frac {3 \, {\left (4 \, a^{2} b + b^{3}\right )} e^{\left (d x + c\right )}}{8 \, d} - \frac {3 \, {\left (4 \, a^{2} b + b^{3}\right )} e^{\left (-d x - c\right )}}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cosh(d*x+c))^3,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.95, size = 73, normalized size = 0.81 \begin {gather*} \frac {\frac {9\,b^3\,\mathrm {sinh}\left (c+d\,x\right )}{2}+\frac {b^3\,\mathrm {sinh}\left (3\,c+3\,d\,x\right )}{2}+\frac {9\,a\,b^2\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )}{2}+18\,a^2\,b\,\mathrm {sinh}\left (c+d\,x\right )+6\,a^3\,d\,x+9\,a\,b^2\,d\,x}{6\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*cosh(c + d*x))^3,x)

[Out]

((9*b^3*sinh(c + d*x))/2 + (b^3*sinh(3*c + 3*d*x))/2 + (9*a*b^2*sinh(2*c + 2*d*x))/2 + 18*a^2*b*sinh(c + d*x)
+ 6*a^3*d*x + 9*a*b^2*d*x)/(6*d)

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