### 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]

((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)

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Rubi [A]  time = 0.0354427, antiderivative size = 63, normalized size of antiderivative = 1., 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 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 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]

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 = 0.113222, 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)

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Maple [A]  time = 0.039, size = 68, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({b}^{2} \left ({\frac{\sinh \left ( dx+c \right ) \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}{4}}-{\frac{\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }{8}}-{\frac{dx}{8}}-{\frac{c}{8}} \right ) +ab \left ( \cosh \left ( dx+c \right ) \right ) ^{2}+{a}^{2} \left ( dx+c \right ) \right ) } \end{align*}

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

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Maxima [A]  time = 1.16773, size = 85, normalized size = 1.35 \begin{align*} 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} \end{align*}

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

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Fricas [A]  time = 2.17857, size = 198, normalized size = 3.14 \begin{align*} \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} \end{align*}

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

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Sympy [A]  time = 1.46069, size = 129, normalized size = 2.05 \begin{align*} \begin{cases} a^{2} x + \frac{a b \cosh ^{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{\left (c \right )} \cosh{\left (c \right )}\right )^{2} & \text{otherwise} \end{cases} \end{align*}

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*cosh(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))

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Giac [A]  time = 1.18512, size = 143, normalized size = 2.27 \begin{align*} \frac{b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 16 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 8 \,{\left (8 \, a^{2} - b^{2}\right )}{\left (d x + c\right )} -{\left (48 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} - 6 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 16 \, a b e^{\left (2 \, d x + 2 \, c\right )} + b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, d} \end{align*}

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]

1/64*(b^2*e^(4*d*x + 4*c) + 16*a*b*e^(2*d*x + 2*c) + 8*(8*a^2 - b^2)*(d*x + c) - (48*a^2*e^(4*d*x + 4*c) - 6*b
^2*e^(4*d*x + 4*c) - 16*a*b*e^(2*d*x + 2*c) + b^2)*e^(-4*d*x - 4*c))/d