3.107 \(\int (c+d x) (a+a \cosh (e+f x))^2 \, dx\)

Optimal. Leaf size=118 \[ \frac{2 a^2 (c+d x) \sinh (e+f x)}{f}+\frac{a^2 (c+d x) \sinh (e+f x) \cosh (e+f x)}{2 f}+\frac{a^2 (c+d x)^2}{2 d}+\frac{1}{2} a^2 c x-\frac{a^2 d \cosh ^2(e+f x)}{4 f^2}-\frac{2 a^2 d \cosh (e+f x)}{f^2}+\frac{1}{4} a^2 d x^2 \]

[Out]

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

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Rubi [A]  time = 0.099404, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3317, 3296, 2638, 3310} \[ \frac{2 a^2 (c+d x) \sinh (e+f x)}{f}+\frac{a^2 (c+d x) \sinh (e+f x) \cosh (e+f x)}{2 f}+\frac{a^2 (c+d x)^2}{2 d}+\frac{1}{2} a^2 c x-\frac{a^2 d \cosh ^2(e+f x)}{4 f^2}-\frac{2 a^2 d \cosh (e+f x)}{f^2}+\frac{1}{4} a^2 d x^2 \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*(a + a*Cosh[e + f*x])^2,x]

[Out]

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

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 2638

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

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

Mathematica [A]  time = 0.478236, size = 81, normalized size = 0.69 \[ \frac{a^2 (-6 (e+f x) (d (e-f x)-2 c f)+16 f (c+d x) \sinh (e+f x)+2 f (c+d x) \sinh (2 (e+f x))-16 d \cosh (e+f x)-d \cosh (2 (e+f x)))}{8 f^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)*(a + a*Cosh[e + f*x])^2,x]

[Out]

(a^2*(-6*(e + f*x)*(-2*c*f + d*(e - f*x)) - 16*d*Cosh[e + f*x] - d*Cosh[2*(e + f*x)] + 16*f*(c + d*x)*Sinh[e +
f*x] + 2*f*(c + d*x)*Sinh[2*(e + f*x)]))/(8*f^2)

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Maple [A]  time = 0.013, size = 211, normalized size = 1.8 \begin{align*}{\frac{1}{f} \left ({\frac{d{a}^{2} \left ( fx+e \right ) ^{2}}{2\,f}}+2\,{\frac{d{a}^{2} \left ( \left ( fx+e \right ) \sinh \left ( fx+e \right ) -\cosh \left ( fx+e \right ) \right ) }{f}}+{\frac{d{a}^{2}}{f} \left ({\frac{ \left ( fx+e \right ) \cosh \left ( fx+e \right ) \sinh \left ( fx+e \right ) }{2}}+{\frac{ \left ( fx+e \right ) ^{2}}{4}}-{\frac{ \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}{4}} \right ) }-{\frac{de{a}^{2} \left ( fx+e \right ) }{f}}-2\,{\frac{de{a}^{2}\sinh \left ( fx+e \right ) }{f}}-{\frac{de{a}^{2}}{f} \left ({\frac{\sinh \left ( fx+e \right ) \cosh \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) }+c{a}^{2} \left ( fx+e \right ) +2\,c{a}^{2}\sinh \left ( fx+e \right ) +c{a}^{2} \left ({\frac{\sinh \left ( fx+e \right ) \cosh \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)*(a+a*cosh(f*x+e))^2,x)

[Out]

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

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Maxima [A]  time = 1.07958, size = 225, normalized size = 1.91 \begin{align*} \frac{1}{2} \, a^{2} d x^{2} + \frac{1}{16} \,{\left (4 \, x^{2} + \frac{{\left (2 \, f x e^{\left (2 \, e\right )} - e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{f^{2}} - \frac{{\left (2 \, f x + 1\right )} e^{\left (-2 \, f x - 2 \, e\right )}}{f^{2}}\right )} a^{2} d + \frac{1}{8} \, a^{2} c{\left (4 \, x + \frac{e^{\left (2 \, f x + 2 \, e\right )}}{f} - \frac{e^{\left (-2 \, f x - 2 \, e\right )}}{f}\right )} + a^{2} c x + a^{2} 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{2 \, a^{2} c \sinh \left (f x + e\right )}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*(a+a*cosh(f*x+e))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.0235, size = 269, normalized size = 2.28 \begin{align*} \frac{6 \, a^{2} d f^{2} x^{2} + 12 \, a^{2} c f^{2} x - a^{2} d \cosh \left (f x + e\right )^{2} - a^{2} d \sinh \left (f x + e\right )^{2} - 16 \, a^{2} d \cosh \left (f x + e\right ) + 4 \,{\left (4 \, a^{2} d f x + 4 \, a^{2} c f +{\left (a^{2} d f x + a^{2} c f\right )} \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right )}{8 \, f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*(a+a*cosh(f*x+e))^2,x, algorithm="fricas")

[Out]

1/8*(6*a^2*d*f^2*x^2 + 12*a^2*c*f^2*x - a^2*d*cosh(f*x + e)^2 - a^2*d*sinh(f*x + e)^2 - 16*a^2*d*cosh(f*x + e)
+ 4*(4*a^2*d*f*x + 4*a^2*c*f + (a^2*d*f*x + a^2*c*f)*cosh(f*x + e))*sinh(f*x + e))/f^2

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Sympy [A]  time = 1.39805, size = 219, normalized size = 1.86 \begin{align*} \begin{cases} - \frac{a^{2} c x \sinh ^{2}{\left (e + f x \right )}}{2} + \frac{a^{2} c x \cosh ^{2}{\left (e + f x \right )}}{2} + a^{2} c x + \frac{a^{2} c \sinh{\left (e + f x \right )} \cosh{\left (e + f x \right )}}{2 f} + \frac{2 a^{2} c \sinh{\left (e + f x \right )}}{f} - \frac{a^{2} d x^{2} \sinh ^{2}{\left (e + f x \right )}}{4} + \frac{a^{2} d x^{2} \cosh ^{2}{\left (e + f x \right )}}{4} + \frac{a^{2} d x^{2}}{2} + \frac{a^{2} d x \sinh{\left (e + f x \right )} \cosh{\left (e + f x \right )}}{2 f} + \frac{2 a^{2} d x \sinh{\left (e + f x \right )}}{f} - \frac{a^{2} d \sinh ^{2}{\left (e + f x \right )}}{4 f^{2}} - \frac{2 a^{2} d \cosh{\left (e + f x \right )}}{f^{2}} & \text{for}\: f \neq 0 \\\left (a \cosh{\left (e \right )} + a\right )^{2} \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+a*cosh(f*x+e))**2,x)

[Out]

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

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Giac [A]  time = 1.32986, size = 209, normalized size = 1.77 \begin{align*} \frac{3}{4} \, a^{2} d x^{2} + \frac{3}{2} \, a^{2} c x + \frac{{\left (2 \, a^{2} d f x + 2 \, a^{2} c f - a^{2} d\right )} e^{\left (2 \, f x + 2 \, e\right )}}{16 \, f^{2}} + \frac{{\left (a^{2} d f x + a^{2} c f - a^{2} d\right )} e^{\left (f x + e\right )}}{f^{2}} - \frac{{\left (a^{2} d f x + a^{2} c f + a^{2} d\right )} e^{\left (-f x - e\right )}}{f^{2}} - \frac{{\left (2 \, a^{2} d f x + 2 \, a^{2} c f + a^{2} d\right )} e^{\left (-2 \, f x - 2 \, e\right )}}{16 \, f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*(a+a*cosh(f*x+e))^2,x, algorithm="giac")

[Out]

3/4*a^2*d*x^2 + 3/2*a^2*c*x + 1/16*(2*a^2*d*f*x + 2*a^2*c*f - a^2*d)*e^(2*f*x + 2*e)/f^2 + (a^2*d*f*x + a^2*c*
f - a^2*d)*e^(f*x + e)/f^2 - (a^2*d*f*x + a^2*c*f + a^2*d)*e^(-f*x - e)/f^2 - 1/16*(2*a^2*d*f*x + 2*a^2*c*f +
a^2*d)*e^(-2*f*x - 2*e)/f^2