\(\int (c+d x) \cosh ^2(a+b x) \, dx\) [11]

Optimal result

Integrand size = 14, antiderivative size = 55 \[ \int (c+d x) \cosh ^2(a+b x) \, dx=\frac {c x}{2}+\frac {d x^2}{4}-\frac {d \cosh ^2(a+b x)}{4 b^2}+\frac {(c+d x) \cosh (a+b x) \sinh (a+b x)}{2 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {3391} \[ \int (c+d x) \cosh ^2(a+b x) \, dx=-\frac {d \cosh ^2(a+b x)}{4 b^2}+\frac {(c+d x) \sinh (a+b x) \cosh (a+b x)}{2 b}+\frac {c x}{2}+\frac {d x^2}{4} \]

[In]

Int[(c + d*x)*Cosh[a + b*x]^2,x]

[Out]

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

Rule 3391

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int (c+d x) \cosh ^2(a+b x) \, dx=\frac {-d \cosh (2 (a+b x))+2 b (2 a c+b x (2 c+d x)+(c+d x) \sinh (2 (a+b x)))}{8 b^2} \]

[In]

Integrate[(c + d*x)*Cosh[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)

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.95

method	result	size
parallelrisch	\(\frac {2 b \sinh \left (2 b x +2 a \right ) \left (d x +c \right )-d \cosh \left (2 b x +2 a \right )+\left (2 d \,x^{2}+4 c x \right ) b^{2}+d}{8 b^{2}}\)	\(52\)
risch	\(\frac {d \,x^{2}}{4}+\frac {c x}{2}+\frac {\left (2 d x b +2 c b -d \right ) {\mathrm e}^{2 b x +2 a}}{16 b^{2}}-\frac {\left (2 d x b +2 c b +d \right ) {\mathrm e}^{-2 b x -2 a}}{16 b^{2}}\)	\(64\)
derivativedivides	\(\frac {\frac {d \left (\frac {\left (b x +a \right ) \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{2}+\frac {\left (b x +a \right )^{2}}{4}-\frac {\cosh \left (b x +a \right )^{2}}{4}\right )}{b}-\frac {d a \left (\frac {\cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{b}+c \left (\frac {\cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{b}\)	\(103\)
default	\(\frac {\frac {d \left (\frac {\left (b x +a \right ) \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{2}+\frac {\left (b x +a \right )^{2}}{4}-\frac {\cosh \left (b x +a \right )^{2}}{4}\right )}{b}-\frac {d a \left (\frac {\cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{b}+c \left (\frac {\cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{b}\)	\(103\)

[In]

int((d*x+c)*cosh(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.20 \[ \int (c+d x) \cosh ^2(a+b x) \, dx=\frac {2 \, b^{2} d x^{2} + 4 \, b^{2} c x - d \cosh \left (b x + a\right )^{2} + 4 \, {\left (b d x + b c\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - d \sinh \left (b x + a\right )^{2}}{8 \, b^{2}} \]

[In]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 126 vs. \(2 (49) = 98\).

Time = 0.21 (sec) , antiderivative size = 126, normalized size of antiderivative = 2.29 \[ \int (c+d x) \cosh ^2(a+b x) \, dx=\begin {cases} - \frac {c x \sinh ^{2}{\left (a + b x \right )}}{2} + \frac {c x \cosh ^{2}{\left (a + b x \right )}}{2} - \frac {d x^{2} \sinh ^{2}{\left (a + b x \right )}}{4} + \frac {d x^{2} \cosh ^{2}{\left (a + b x \right )}}{4} + \frac {c \sinh {\left (a + b x \right )} \cosh {\left (a + b x \right )}}{2 b} + \frac {d x \sinh {\left (a + b x \right )} \cosh {\left (a + b x \right )}}{2 b} - \frac {d \cosh ^{2}{\left (a + b x \right )}}{4 b^{2}} & \text {for}\: b \neq 0 \\\left (c x + \frac {d x^{2}}{2}\right ) \cosh ^{2}{\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.60 \[ \int (c+d x) \cosh ^2(a+b x) \, dx=\frac {1}{16} \, {\left (4 \, x^{2} + \frac {{\left (2 \, b x e^{\left (2 \, a\right )} - e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{b^{2}} - \frac {{\left (2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{b^{2}}\right )} d + \frac {1}{8} \, c {\left (4 \, x + \frac {e^{\left (2 \, b x + 2 \, a\right )}}{b} - \frac {e^{\left (-2 \, b x - 2 \, a\right )}}{b}\right )} \]

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.15 \[ \int (c+d x) \cosh ^2(a+b x) \, dx=\frac {1}{4} \, d x^{2} + \frac {1}{2} \, c x + \frac {{\left (2 \, b d x + 2 \, b c - d\right )} e^{\left (2 \, b x + 2 \, a\right )}}{16 \, b^{2}} - \frac {{\left (2 \, b d x + 2 \, b c + d\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{16 \, b^{2}} \]

[In]

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

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.05 \[ \int (c+d x) \cosh ^2(a+b x) \, dx=\frac {b^2\,d\,x^2-\frac {d\,\mathrm {cosh}\left (2\,a+2\,b\,x\right )}{2}+b\,c\,\mathrm {sinh}\left (2\,a+2\,b\,x\right )+2\,b^2\,c\,x+b\,d\,x\,\mathrm {sinh}\left (2\,a+2\,b\,x\right )}{4\,b^2} \]

[In]

int(cosh(a + b*x)^2*(c + d*x),x)

[Out]

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