3.856 \(\int (a+b \cosh (c+d x) \sinh (c+d x))^2 \, dx\)

Optimal. Leaf size=63 \[ \frac {1}{8} x \left (8 a^2-b^2\right )+\frac {a b \cosh (2 c+2 d x)}{2 d}+\frac {b^2 \sinh (2 c+2 d x) \cosh (2 c+2 d x)}{16 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2666, 2644} \[ \frac {1}{8} x \left (8 a^2-b^2\right )+\frac {a b \cosh (2 c+2 d x)}{2 d}+\frac {b^2 \sinh (2 c+2 d x) \cosh (2 c+2 d x)}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2644

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

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.12, size = 50, normalized size = 0.79 \[ \frac {4 \left (8 a^2-b^2\right ) (c+d x)+16 a b \cosh (2 (c+d x))+b^2 \sinh (4 (c+d x))}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(8*a^2 - b^2)*(c + d*x) + 16*a*b*Cosh[2*(c + d*x)] + b^2*Sinh[4*(c + d*x)])/(32*d)

________________________________________________________________________________________

fricas [A]  time = 0.49, size = 80, normalized size = 1.27 \[ \frac {b^{2} \cosh \left (d x + c\right )^{3} \sinh \left (d x + c\right ) + b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + 4 \, a b \cosh \left (d x + c\right )^{2} + 4 \, a b \sinh \left (d x + c\right )^{2} + {\left (8 \, a^{2} - b^{2}\right )} d x}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.12, size = 81, normalized size = 1.29 \[ \frac {1}{8} \, {\left (8 \, a^{2} - b^{2}\right )} x + \frac {b^{2} e^{\left (4 \, d x + 4 \, c\right )}}{64 \, d} + \frac {a b e^{\left (2 \, d x + 2 \, c\right )}}{4 \, d} + \frac {a b e^{\left (-2 \, d x - 2 \, c\right )}}{4 \, d} - \frac {b^{2} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.43, size = 68, normalized size = 1.08 \[ \frac {b^{2} \left (\frac {\sinh \left (d x +c \right ) \left (\cosh ^{3}\left (d x +c \right )\right )}{4}-\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{8}-\frac {d x}{8}-\frac {c}{8}\right )+a \left (\cosh ^{2}\left (d x +c \right )\right ) b +a^{2} \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(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)+a*cosh(d*x+c)^2*b+a^2*(d*x+
c))

________________________________________________________________________________________

maxima [A]  time = 0.39, size = 63, normalized size = 1.00 \[ a^{2} x - \frac {1}{64} \, b^{2} {\left (\frac {8 \, {\left (d x + c\right )}}{d} - \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} + \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + \frac {a b \cosh \left (d x + c\right )^{2}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.14, size = 44, normalized size = 0.70 \[ \frac {\frac {\mathrm {sinh}\left (4\,c+4\,d\,x\right )\,b^2}{32}+\frac {a\,\mathrm {cosh}\left (2\,c+2\,d\,x\right )\,b}{2}}{d}+a^2\,x-\frac {b^2\,x}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b^2*sinh(4*c + 4*d*x))/32 + (a*b*cosh(2*c + 2*d*x))/2)/d + a^2*x - (b^2*x)/8

________________________________________________________________________________________

sympy [A]  time = 0.93, size = 129, normalized size = 2.05 \[ \begin {cases} a^{2} x + \frac {a b \sinh ^{2}{\left (c + d x \right )}}{d} - \frac {b^{2} x \sinh ^{4}{\left (c + d x \right )}}{8} + \frac {b^{2} x \sinh ^{2}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{4} - \frac {b^{2} x \cosh ^{4}{\left (c + d x \right )}}{8} + \frac {b^{2} \sinh ^{3}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{8 d} + \frac {b^{2} \sinh {\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (a + b \sinh {\relax (c )} \cosh {\relax (c )}\right )^{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________