\(\int (c+d x) (a+b \cosh (e+f x)) \, dx\) [158]

Optimal result

Integrand size = 16, antiderivative size = 45 \[ \int (c+d x) (a+b \cosh (e+f x)) \, dx=\frac {a (c+d x)^2}{2 d}-\frac {b d \cosh (e+f x)}{f^2}+\frac {b (c+d x) \sinh (e+f x)}{f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3398, 3377, 2718} \[ \int (c+d x) (a+b \cosh (e+f x)) \, dx=\frac {a (c+d x)^2}{2 d}+\frac {b (c+d x) \sinh (e+f x)}{f}-\frac {b d \cosh (e+f x)}{f^2} \]

[In]

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

[Out]

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

Rule 2718

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

Rule 3377

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 3398

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])

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.02 \[ \int (c+d x) (a+b \cosh (e+f x)) \, dx=\frac {-2 b d \cosh (e+f x)+f (a f x (2 c+d x)+2 b (c+d x) \sinh (e+f x))}{2 f^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.02

method	result	size
parallelrisch	\(\frac {\left (d x +c \right ) b f \sinh \left (f x +e \right )-\cosh \left (f x +e \right ) b d +x \left (\frac {d x}{2}+c \right ) a \,f^{2}-b d}{f^{2}}\)	\(46\)
risch	\(\frac {a d \,x^{2}}{2}+a c x +\frac {b \left (d x f +c f -d \right ) {\mathrm e}^{f x +e}}{2 f^{2}}-\frac {b \left (d x f +c f +d \right ) {\mathrm e}^{-f x -e}}{2 f^{2}}\)	\(60\)
parts	\(a \left (\frac {1}{2} d \,x^{2}+c x \right )+\frac {b \left (\frac {d \left (\left (f x +e \right ) \sinh \left (f x +e \right )-\cosh \left (f x +e \right )\right )}{f}-\frac {d e \sinh \left (f x +e \right )}{f}+c \sinh \left (f x +e \right )\right )}{f}\)	\(67\)
derivativedivides	\(\frac {\frac {d a \left (f x +e \right )^{2}}{2 f}+\frac {d b \left (\left (f x +e \right ) \sinh \left (f x +e \right )-\cosh \left (f x +e \right )\right )}{f}-\frac {d e a \left (f x +e \right )}{f}-\frac {d e b \sinh \left (f x +e \right )}{f}+c a \left (f x +e \right )+c b \sinh \left (f x +e \right )}{f}\)	\(91\)
default	\(\frac {\frac {d a \left (f x +e \right )^{2}}{2 f}+\frac {d b \left (\left (f x +e \right ) \sinh \left (f x +e \right )-\cosh \left (f x +e \right )\right )}{f}-\frac {d e a \left (f x +e \right )}{f}-\frac {d e b \sinh \left (f x +e \right )}{f}+c a \left (f x +e \right )+c b \sinh \left (f x +e \right )}{f}\)	\(91\)

[In]

int((d*x+c)*(a+b*cosh(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.13 \[ \int (c+d x) (a+b \cosh (e+f x)) \, dx=\frac {a d f^{2} x^{2} + 2 \, a c f^{2} x - 2 \, b d \cosh \left (f x + e\right ) + 2 \, {\left (b d f x + b c f\right )} \sinh \left (f x + e\right )}{2 \, f^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.51 \[ \int (c+d x) (a+b \cosh (e+f x)) \, dx=\begin {cases} a c x + \frac {a d x^{2}}{2} + \frac {b c \sinh {\left (e + f x \right )}}{f} + \frac {b d x \sinh {\left (e + f x \right )}}{f} - \frac {b d \cosh {\left (e + f x \right )}}{f^{2}} & \text {for}\: f \neq 0 \\\left (a + b \cosh {\left (e \right )}\right ) \left (c x + \frac {d x^{2}}{2}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.47 \[ \int (c+d x) (a+b \cosh (e+f x)) \, dx=\frac {1}{2} \, a d x^{2} + a c x + \frac {1}{2} \, b d {\left (\frac {{\left (f x e^{e} - e^{e}\right )} e^{\left (f x\right )}}{f^{2}} - \frac {{\left (f x + 1\right )} e^{\left (-f x - e\right )}}{f^{2}}\right )} + \frac {b c \sinh \left (f x + e\right )}{f} \]

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.42 \[ \int (c+d x) (a+b \cosh (e+f x)) \, dx=\frac {1}{2} \, a d x^{2} + a c x + \frac {{\left (b d f x + b c f - b d\right )} e^{\left (f x + e\right )}}{2 \, f^{2}} - \frac {{\left (b d f x + b c f + b d\right )} e^{\left (-f x - e\right )}}{2 \, f^{2}} \]

[In]

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

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.09 \[ \int (c+d x) (a+b \cosh (e+f x)) \, dx=\frac {f\,\left (b\,c\,\mathrm {sinh}\left (e+f\,x\right )+b\,d\,x\,\mathrm {sinh}\left (e+f\,x\right )\right )-b\,d\,\mathrm {cosh}\left (e+f\,x\right )}{f^2}+a\,c\,x+\frac {a\,d\,x^2}{2} \]

[In]

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

[Out]

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