3.855 \(\int (a+b \cosh (c+d x) \sinh (c+d x))^3 \, dx\)

Optimal. Leaf size=109 \[ \frac {b \left (16 a^2-b^2\right ) \cosh (2 c+2 d x)}{24 d}+\frac {1}{8} a x \left (8 a^2-3 b^2\right )+\frac {5 a b^2 \sinh (2 c+2 d x) \cosh (2 c+2 d x)}{48 d}+\frac {b \cosh (2 c+2 d x) (2 a+b \sinh (2 c+2 d x))^2}{48 d} \]

[Out]

1/8*a*(8*a^2-3*b^2)*x+1/24*b*(16*a^2-b^2)*cosh(2*d*x+2*c)/d+5/48*a*b^2*cosh(2*d*x+2*c)*sinh(2*d*x+2*c)/d+1/48*
b*cosh(2*d*x+2*c)*(2*a+b*sinh(2*d*x+2*c))^2/d

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Rubi [A]  time = 0.10, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2666, 2656, 2734} \[ \frac {b \left (16 a^2-b^2\right ) \cosh (2 c+2 d x)}{24 d}+\frac {1}{8} a x \left (8 a^2-3 b^2\right )+\frac {5 a b^2 \sinh (2 c+2 d x) \cosh (2 c+2 d x)}{48 d}+\frac {b \cosh (2 c+2 d x) (2 a+b \sinh (2 c+2 d x))^2}{48 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cosh[c + d*x]*Sinh[c + d*x])^3,x]

[Out]

(a*(8*a^2 - 3*b^2)*x)/8 + (b*(16*a^2 - b^2)*Cosh[2*c + 2*d*x])/(24*d) + (5*a*b^2*Cosh[2*c + 2*d*x]*Sinh[2*c +
2*d*x])/(48*d) + (b*Cosh[2*c + 2*d*x]*(2*a + b*Sinh[2*c + 2*d*x])^2)/(48*d)

Rule 2656

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 2666

Int[((a_) + cos[(c_.) + (d_.)*(x_)]*(b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[(a + (b*Sin[2*c + 2*
d*x])/2)^n, x] /; FreeQ[{a, b, c, d, n}, x]

Rule 2734

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) \sinh (c+d x))^3 \, dx &=\int \left (a+\frac {1}{2} b \sinh (2 c+2 d x)\right )^3 \, dx\\ &=\frac {b \cosh (2 c+2 d x) (2 a+b \sinh (2 c+2 d x))^2}{48 d}+\frac {1}{3} \int \left (a+\frac {1}{2} b \sinh (2 c+2 d x)\right ) \left (\frac {1}{2} \left (6 a^2-b^2\right )+\frac {5}{2} a b \sinh (2 c+2 d x)\right ) \, dx\\ &=\frac {1}{8} a \left (8 a^2-3 b^2\right ) x+\frac {b \left (16 a^2-b^2\right ) \cosh (2 c+2 d x)}{24 d}+\frac {5 a b^2 \cosh (2 c+2 d x) \sinh (2 c+2 d x)}{48 d}+\frac {b \cosh (2 c+2 d x) (2 a+b \sinh (2 c+2 d x))^2}{48 d}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 77, normalized size = 0.71 \[ \frac {9 \left (16 a^2 b-b^3\right ) \cosh (2 (c+d x))+6 a \left (4 \left (8 a^2-3 b^2\right ) (c+d x)+3 b^2 \sinh (4 (c+d x))\right )+b^3 \cosh (6 (c+d x))}{192 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cosh[c + d*x]*Sinh[c + d*x])^3,x]

[Out]

(9*(16*a^2*b - b^3)*Cosh[2*(c + d*x)] + b^3*Cosh[6*(c + d*x)] + 6*a*(4*(8*a^2 - 3*b^2)*(c + d*x) + 3*b^2*Sinh[
4*(c + d*x)]))/(192*d)

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fricas [A]  time = 0.68, size = 164, normalized size = 1.50 \[ \frac {b^{3} \cosh \left (d x + c\right )^{6} + 15 \, b^{3} \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{4} + b^{3} \sinh \left (d x + c\right )^{6} + 72 \, a b^{2} \cosh \left (d x + c\right )^{3} \sinh \left (d x + c\right ) + 72 \, a b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + 24 \, {\left (8 \, a^{3} - 3 \, a b^{2}\right )} d x + 9 \, {\left (16 \, a^{2} b - b^{3}\right )} \cosh \left (d x + c\right )^{2} + 3 \, {\left (5 \, b^{3} \cosh \left (d x + c\right )^{4} + 48 \, a^{2} b - 3 \, b^{3}\right )} \sinh \left (d x + c\right )^{2}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(b^3*cosh(d*x + c)^6 + 15*b^3*cosh(d*x + c)^2*sinh(d*x + c)^4 + b^3*sinh(d*x + c)^6 + 72*a*b^2*cosh(d*x
+ c)^3*sinh(d*x + c) + 72*a*b^2*cosh(d*x + c)*sinh(d*x + c)^3 + 24*(8*a^3 - 3*a*b^2)*d*x + 9*(16*a^2*b - b^3)*
cosh(d*x + c)^2 + 3*(5*b^3*cosh(d*x + c)^4 + 48*a^2*b - 3*b^3)*sinh(d*x + c)^2)/d

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giac [A]  time = 0.12, size = 138, normalized size = 1.27 \[ \frac {b^{3} e^{\left (6 \, d x + 6 \, c\right )}}{384 \, d} + \frac {3 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )}}{64 \, d} - \frac {3 \, a b^{2} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, d} + \frac {b^{3} e^{\left (-6 \, d x - 6 \, c\right )}}{384 \, d} + \frac {1}{8} \, {\left (8 \, a^{3} - 3 \, a b^{2}\right )} x + \frac {3 \, {\left (16 \, a^{2} b - b^{3}\right )} e^{\left (2 \, d x + 2 \, c\right )}}{128 \, d} + \frac {3 \, {\left (16 \, a^{2} b - b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{128 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*b^3*e^(6*d*x + 6*c)/d + 3/64*a*b^2*e^(4*d*x + 4*c)/d - 3/64*a*b^2*e^(-4*d*x - 4*c)/d + 1/384*b^3*e^(-6*d
*x - 6*c)/d + 1/8*(8*a^3 - 3*a*b^2)*x + 3/128*(16*a^2*b - b^3)*e^(2*d*x + 2*c)/d + 3/128*(16*a^2*b - b^3)*e^(-
2*d*x - 2*c)/d

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maple [A]  time = 0.44, size = 106, normalized size = 0.97 \[ \frac {b^{3} \left (\frac {\left (\sinh ^{2}\left (d x +c \right )\right ) \left (\cosh ^{4}\left (d x +c \right )\right )}{6}-\frac {\left (\cosh ^{4}\left (d x +c \right )\right )}{12}\right )+3 a \,b^{2} \left (\frac {\sinh \left (d x +c \right ) \left (\cosh ^{3}\left (d x +c \right )\right )}{4}-\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{8}-\frac {d x}{8}-\frac {c}{8}\right )+\frac {3 a^{2} b \left (\cosh ^{2}\left (d x +c \right )\right )}{2}+a^{3} \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cosh(d*x+c)*sinh(d*x+c))^3,x)

[Out]

1/d*(b^3*(1/6*sinh(d*x+c)^2*cosh(d*x+c)^4-1/12*cosh(d*x+c)^4)+3*a*b^2*(1/4*sinh(d*x+c)*cosh(d*x+c)^3-1/8*cosh(
d*x+c)*sinh(d*x+c)-1/8*d*x-1/8*c)+3/2*a^2*b*cosh(d*x+c)^2+a^3*(d*x+c))

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maxima [A]  time = 0.33, size = 126, normalized size = 1.16 \[ a^{3} x - \frac {1}{384} \, b^{3} {\left (\frac {{\left (9 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )} e^{\left (6 \, d x + 6 \, c\right )}}{d} + \frac {9 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-6 \, d x - 6 \, c\right )}}{d}\right )} - \frac {3}{64} \, a b^{2} {\left (\frac {8 \, {\left (d x + c\right )}}{d} - \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} + \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + \frac {3 \, a^{2} b \cosh \left (d x + c\right )^{2}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*x - 1/384*b^3*((9*e^(-4*d*x - 4*c) - 1)*e^(6*d*x + 6*c)/d + (9*e^(-2*d*x - 2*c) - e^(-6*d*x - 6*c))/d) - 3
/64*a*b^2*(8*(d*x + c)/d - e^(4*d*x + 4*c)/d + e^(-4*d*x - 4*c)/d) + 3/2*a^2*b*cosh(d*x + c)^2/d

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mupad [B]  time = 1.87, size = 79, normalized size = 0.72 \[ \frac {\frac {b^3\,\mathrm {cosh}\left (6\,c+6\,d\,x\right )}{8}-\frac {9\,b^3\,\mathrm {cosh}\left (2\,c+2\,d\,x\right )}{8}+18\,a^2\,b\,\mathrm {cosh}\left (2\,c+2\,d\,x\right )+\frac {9\,a\,b^2\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )}{4}+24\,a^3\,d\,x-9\,a\,b^2\,d\,x}{24\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*cosh(c + d*x)*sinh(c + d*x))^3,x)

[Out]

((b^3*cosh(6*c + 6*d*x))/8 - (9*b^3*cosh(2*c + 2*d*x))/8 + 18*a^2*b*cosh(2*c + 2*d*x) + (9*a*b^2*sinh(4*c + 4*
d*x))/4 + 24*a^3*d*x - 9*a*b^2*d*x)/(24*d)

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sympy [A]  time = 3.25, size = 190, normalized size = 1.74 \[ \begin {cases} a^{3} x + \frac {3 a^{2} b \sinh ^{2}{\left (c + d x \right )}}{2 d} - \frac {3 a b^{2} x \sinh ^{4}{\left (c + d x \right )}}{8} + \frac {3 a b^{2} x \sinh ^{2}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{4} - \frac {3 a b^{2} x \cosh ^{4}{\left (c + d x \right )}}{8} + \frac {3 a b^{2} \sinh ^{3}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{8 d} + \frac {3 a b^{2} \sinh {\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{8 d} + \frac {b^{3} \sinh ^{2}{\left (c + d x \right )} \cosh ^{4}{\left (c + d x \right )}}{4 d} - \frac {b^{3} \cosh ^{6}{\left (c + d x \right )}}{12 d} & \text {for}\: d \neq 0 \\x \left (a + b \sinh {\relax (c )} \cosh {\relax (c )}\right )^{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**3*x + 3*a**2*b*sinh(c + d*x)**2/(2*d) - 3*a*b**2*x*sinh(c + d*x)**4/8 + 3*a*b**2*x*sinh(c + d*x)
**2*cosh(c + d*x)**2/4 - 3*a*b**2*x*cosh(c + d*x)**4/8 + 3*a*b**2*sinh(c + d*x)**3*cosh(c + d*x)/(8*d) + 3*a*b
**2*sinh(c + d*x)*cosh(c + d*x)**3/(8*d) + b**3*sinh(c + d*x)**2*cosh(c + d*x)**4/(4*d) - b**3*cosh(c + d*x)**
6/(12*d), Ne(d, 0)), (x*(a + b*sinh(c)*cosh(c))**3, True))

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