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

Optimal. Leaf size=30 \[ a x+\frac{b \sinh (c+d x) \cosh (c+d x)}{2 d}-\frac{b x}{2} \]

[Out]

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

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Rubi [A]  time = 0.0160623, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2635, 8} \[ a x+\frac{b \sinh (c+d x) \cosh (c+d x)}{2 d}-\frac{b x}{2} \]

Antiderivative was successfully verified.

[In]

Int[a + b*Sinh[c + d*x]^2,x]

[Out]

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

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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.0316776, size = 36, normalized size = 1.2 \[ a x+\frac{b (-c-d x)}{2 d}+\frac{b \sinh (2 (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[a + b*Sinh[c + d*x]^2,x]

[Out]

a*x + (b*(-c - d*x))/(2*d) + (b*Sinh[2*(c + d*x)])/(4*d)

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Maple [A]  time = 0.007, size = 32, normalized size = 1.1 \begin{align*} ax+{\frac{b}{d} \left ({\frac{\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }{2}}-{\frac{dx}{2}}-{\frac{c}{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.04378, size = 51, normalized size = 1.7 \begin{align*} -\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 )} + a x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.84118, size = 74, normalized size = 2.47 \begin{align*} \frac{{\left (2 \, a - b\right )} d x + b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.408983, size = 51, normalized size = 1.7 \begin{align*} a x + b \left (\begin{cases} \frac{x \sinh ^{2}{\left (c + d x \right )}}{2} - \frac{x \cosh ^{2}{\left (c + d x \right )}}{2} + \frac{\sinh{\left (c + d x \right )} \cosh{\left (c + d x \right )}}{2 d} & \text{for}\: d \neq 0 \\x \sinh ^{2}{\left (c \right )} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.30807, size = 72, normalized size = 2.4 \begin{align*} a x - \frac{{\left (4 \, d x -{\left (2 \, e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 4 \, c - e^{\left (2 \, d x + 2 \, c\right )}\right )} b}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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