3.4 \(\int (c+d x) \cosh (a+b x) \, dx\)

Optimal. Leaf size=28 \[ \frac{(c+d x) \sinh (a+b x)}{b}-\frac{d \cosh (a+b x)}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.0195894, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3296, 2638} \[ \frac{(c+d x) \sinh (a+b x)}{b}-\frac{d \cosh (a+b x)}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Cosh[a + b*x],x]

[Out]

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

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.0527064, size = 27, normalized size = 0.96 \[ \frac{b (c+d x) \sinh (a+b x)-d \cosh (a+b x)}{b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)*Cosh[a + b*x],x]

[Out]

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

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Maple [A]  time = 0.007, size = 53, normalized size = 1.9 \begin{align*}{\frac{1}{b} \left ({\frac{d \left ( \left ( bx+a \right ) \sinh \left ( bx+a \right ) -\cosh \left ( bx+a \right ) \right ) }{b}}-{\frac{da\sinh \left ( bx+a \right ) }{b}}+c\sinh \left ( bx+a \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.10975, size = 92, normalized size = 3.29 \begin{align*} \frac{c e^{\left (b x + a\right )}}{2 \, b} + \frac{{\left (b x e^{a} - e^{a}\right )} d e^{\left (b x\right )}}{2 \, b^{2}} - \frac{c e^{\left (-b x - a\right )}}{2 \, b} - \frac{{\left (b x + 1\right )} d e^{\left (-b x - a\right )}}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.7661, size = 73, normalized size = 2.61 \begin{align*} -\frac{d \cosh \left (b x + a\right ) -{\left (b d x + b c\right )} \sinh \left (b x + a\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.275544, size = 46, normalized size = 1.64 \begin{align*} \begin{cases} \frac{c \sinh{\left (a + b x \right )}}{b} + \frac{d x \sinh{\left (a + b x \right )}}{b} - \frac{d \cosh{\left (a + b x \right )}}{b^{2}} & \text{for}\: b \neq 0 \\\left (c x + \frac{d x^{2}}{2}\right ) \cosh{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.26977, size = 62, normalized size = 2.21 \begin{align*} \frac{{\left (b d x + b c - d\right )} e^{\left (b x + a\right )}}{2 \, b^{2}} - \frac{{\left (b d x + b c + d\right )} e^{\left (-b x - a\right )}}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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