3.159 \(\int (c+d x) (a+b \sinh (e+f x)) \, dx\)

Optimal. Leaf size=45 \[ \frac{a (c+d x)^2}{2 d}+\frac{b (c+d x) \cosh (e+f x)}{f}-\frac{b d \sinh (e+f x)}{f^2} \]

[Out]

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

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Rubi [A]  time = 0.047893, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {3317, 3296, 2637} \[ \frac{a (c+d x)^2}{2 d}+\frac{b (c+d x) \cosh (e+f x)}{f}-\frac{b d \sinh (e+f x)}{f^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*(a + b*Sinh[e + f*x]),x]

[Out]

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

Rule 3317

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] ||
IGtQ[m, 0] || NeQ[a^2 - b^2, 0])

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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.111022, size = 43, normalized size = 0.96 \[ \frac{1}{2} a x (2 c+d x)+\frac{b (c+d x) \cosh (e+f x)}{f}-\frac{b d \sinh (e+f x)}{f^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)*(a + b*Sinh[e + f*x]),x]

[Out]

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

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Maple [B]  time = 0.007, size = 91, normalized size = 2. \begin{align*}{\frac{1}{f} \left ({\frac{da \left ( fx+e \right ) ^{2}}{2\,f}}+{\frac{bd \left ( \left ( fx+e \right ) \cosh \left ( fx+e \right ) -\sinh \left ( fx+e \right ) \right ) }{f}}-{\frac{dea \left ( fx+e \right ) }{f}}-{\frac{deb\cosh \left ( fx+e \right ) }{f}}+ca \left ( fx+e \right ) +cb\cosh \left ( fx+e \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)*(a+b*sinh(f*x+e)),x)

[Out]

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

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Maxima [A]  time = 1.30236, size = 88, normalized size = 1.96 \begin{align*} \frac{1}{2} \, a d x^{2} + a c x + \frac{1}{2} \, b d{\left (\frac{{\left (f x e^{e} - e^{e}\right )} e^{\left (f x\right )}}{f^{2}} + \frac{{\left (f x + 1\right )} e^{\left (-f x - e\right )}}{f^{2}}\right )} + \frac{b c \cosh \left (f x + e\right )}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.46015, size = 128, normalized size = 2.84 \begin{align*} \frac{a d f^{2} x^{2} + 2 \, a c f^{2} x - 2 \, b d \sinh \left (f x + e\right ) + 2 \,{\left (b d f x + b c f\right )} \cosh \left (f x + e\right )}{2 \, f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.36472, size = 68, normalized size = 1.51 \begin{align*} \begin{cases} a c x + \frac{a d x^{2}}{2} + \frac{b c \cosh{\left (e + f x \right )}}{f} + \frac{b d x \cosh{\left (e + f x \right )}}{f} - \frac{b d \sinh{\left (e + f x \right )}}{f^{2}} & \text{for}\: f \neq 0 \\\left (a + b \sinh{\left (e \right )}\right ) \left (c x + \frac{d x^{2}}{2}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.28085, size = 89, normalized size = 1.98 \begin{align*} \frac{1}{2} \, a d x^{2} + a c x + \frac{{\left (b d f x + b c f - b d\right )} e^{\left (f x + e\right )}}{2 \, f^{2}} + \frac{{\left (b d f x + b c f + b d\right )} e^{\left (-f x - e\right )}}{2 \, f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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