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

Optimal. Leaf size=55 \[ -\frac{d \sinh ^2(a+b x)}{4 b^2}+\frac{(c+d x) \sinh (a+b x) \cosh (a+b x)}{2 b}-\frac{c x}{2}-\frac{d x^2}{4} \]

[Out]

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

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Rubi [A]  time = 0.0262664, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {3310} \[ -\frac{d \sinh ^2(a+b x)}{4 b^2}+\frac{(c+d x) \sinh (a+b x) \cosh (a+b x)}{2 b}-\frac{c x}{2}-\frac{d x^2}{4} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Sinh[a + b*x]^2,x]

[Out]

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

Rule 3310

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

Rubi steps

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

Mathematica [A]  time = 0.161799, size = 52, normalized size = 0.95 \[ \frac{2 b ((c+d x) \sinh (2 (a+b x))-2 a c-b x (2 c+d x))-d \cosh (2 (a+b x))}{8 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.006, size = 103, normalized size = 1.9 \begin{align*}{\frac{1}{b} \left ({\frac{d}{b} \left ({\frac{ \left ( bx+a \right ) \cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{2}}-{\frac{ \left ( bx+a \right ) ^{2}}{4}}-{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{4}} \right ) }-{\frac{da}{b} \left ({\frac{\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{2}}-{\frac{bx}{2}}-{\frac{a}{2}} \right ) }+c \left ({\frac{\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{2}}-{\frac{bx}{2}}-{\frac{a}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.21832, size = 119, normalized size = 2.16 \begin{align*} -\frac{1}{16} \,{\left (4 \, x^{2} - \frac{{\left (2 \, b x e^{\left (2 \, a\right )} - e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{b^{2}} + \frac{{\left (2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{b^{2}}\right )} d - \frac{1}{8} \, c{\left (4 \, x - \frac{e^{\left (2 \, b x + 2 \, a\right )}}{b} + \frac{e^{\left (-2 \, b x - 2 \, a\right )}}{b}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.62637, size = 165, normalized size = 3. \begin{align*} -\frac{2 \, b^{2} d x^{2} + 4 \, b^{2} c x + d \cosh \left (b x + a\right )^{2} - 4 \,{\left (b d x + b c\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + d \sinh \left (b x + a\right )^{2}}{8 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.800126, size = 126, normalized size = 2.29 \begin{align*} \begin{cases} \frac{c x \sinh ^{2}{\left (a + b x \right )}}{2} - \frac{c x \cosh ^{2}{\left (a + b x \right )}}{2} + \frac{d x^{2} \sinh ^{2}{\left (a + b x \right )}}{4} - \frac{d x^{2} \cosh ^{2}{\left (a + b x \right )}}{4} + \frac{c \sinh{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{2 b} + \frac{d x \sinh{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{2 b} - \frac{d \sinh ^{2}{\left (a + b x \right )}}{4 b^{2}} & \text{for}\: b \neq 0 \\\left (c x + \frac{d x^{2}}{2}\right ) \sinh ^{2}{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.21919, size = 85, normalized size = 1.55 \begin{align*} -\frac{1}{4} \, d x^{2} - \frac{1}{2} \, c x + \frac{{\left (2 \, b d x + 2 \, b c - d\right )} e^{\left (2 \, b x + 2 \, a\right )}}{16 \, b^{2}} - \frac{{\left (2 \, b d x + 2 \, b c + d\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{16 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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