\(\int e^{c (a+b x)} \cosh ^2(a c+b c x)^{3/2} \, dx\) [293]

Optimal result

Integrand size = 25, antiderivative size = 162 \[ \int e^{c (a+b x)} \cosh ^2(a c+b c x)^{3/2} \, dx=-\frac {e^{-2 c (a+b x)} \sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x)}{16 b c}+\frac {3 e^{2 c (a+b x)} \sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x)}{16 b c}+\frac {e^{4 c (a+b x)} \sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x)}{32 b c}+\frac {3}{8} x \sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x) \]

[Out]

-1/16*sech(b*c*x+a*c)*(cosh(b*c*x+a*c)^2)^(1/2)/b/c/exp(2*c*(b*x+a))+3/16*exp(2*c*(b*x+a))*sech(b*c*x+a*c)*(co
sh(b*c*x+a*c)^2)^(1/2)/b/c+1/32*exp(4*c*(b*x+a))*sech(b*c*x+a*c)*(cosh(b*c*x+a*c)^2)^(1/2)/b/c+3/8*x*sech(b*c*
x+a*c)*(cosh(b*c*x+a*c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6852, 2320, 12, 272, 45} \[ \int e^{c (a+b x)} \cosh ^2(a c+b c x)^{3/2} \, dx=-\frac {e^{-2 c (a+b x)} \sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x)}{16 b c}+\frac {3 e^{2 c (a+b x)} \sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x)}{16 b c}+\frac {e^{4 c (a+b x)} \sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x)}{32 b c}+\frac {3}{8} x \sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x) \]

[In]

Int[E^(c*(a + b*x))*(Cosh[a*c + b*c*x]^2)^(3/2),x]

[Out]

-1/16*(Sqrt[Cosh[a*c + b*c*x]^2]*Sech[a*c + b*c*x])/(b*c*E^(2*c*(a + b*x))) + (3*E^(2*c*(a + b*x))*Sqrt[Cosh[a
*c + b*c*x]^2]*Sech[a*c + b*c*x])/(16*b*c) + (E^(4*c*(a + b*x))*Sqrt[Cosh[a*c + b*c*x]^2]*Sech[a*c + b*c*x])/(
32*b*c) + (3*x*Sqrt[Cosh[a*c + b*c*x]^2]*Sech[a*c + b*c*x])/8

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6852

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a*v^m)^FracPart[p]/v^(m*FracPart[p])), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x)\right ) \int e^{c (a+b x)} \cosh ^3(a c+b c x) \, dx \\ & = \frac {\left (\sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x)\right ) \text {Subst}\left (\int \frac {\left (1+x^2\right )^3}{8 x^3} \, dx,x,e^{c (a+b x)}\right )}{b c} \\ & = \frac {\left (\sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x)\right ) \text {Subst}\left (\int \frac {\left (1+x^2\right )^3}{x^3} \, dx,x,e^{c (a+b x)}\right )}{8 b c} \\ & = \frac {\left (\sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x)\right ) \text {Subst}\left (\int \frac {(1+x)^3}{x^2} \, dx,x,e^{2 c (a+b x)}\right )}{16 b c} \\ & = \frac {\left (\sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x)\right ) \text {Subst}\left (\int \left (3+\frac {1}{x^2}+\frac {3}{x}+x\right ) \, dx,x,e^{2 c (a+b x)}\right )}{16 b c} \\ & = -\frac {e^{-2 c (a+b x)} \sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x)}{16 b c}+\frac {3 e^{2 c (a+b x)} \sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x)}{16 b c}+\frac {e^{4 c (a+b x)} \sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x)}{32 b c}+\frac {3}{8} x \sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.50 \[ \int e^{c (a+b x)} \cosh ^2(a c+b c x)^{3/2} \, dx=\frac {\left (-\frac {1}{16} e^{-2 c (a+b x)}+\frac {3}{16} e^{2 c (a+b x)}+\frac {1}{32} e^{4 c (a+b x)}+\frac {3 b c x}{8}\right ) \cosh ^2(c (a+b x))^{3/2} \text {sech}^3(c (a+b x))}{b c} \]

[In]

Integrate[E^(c*(a + b*x))*(Cosh[a*c + b*c*x]^2)^(3/2),x]

[Out]

((-1/16*1/E^(2*c*(a + b*x)) + (3*E^(2*c*(a + b*x)))/16 + E^(4*c*(a + b*x))/32 + (3*b*c*x)/8)*(Cosh[c*(a + b*x)
]^2)^(3/2)*Sech[c*(a + b*x)]^3)/(b*c)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.26 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.46

method	result	size
default	\(\frac {\operatorname {csgn}\left (\cosh \left (c \left (b x +a \right )\right )\right ) \left (\frac {\cosh \left (b c x +a c \right )^{4}}{4}+\left (\frac {\cosh \left (b c x +a c \right )^{3}}{4}+\frac {3 \cosh \left (b c x +a c \right )}{8}\right ) \sinh \left (b c x +a c \right )+\frac {3 b c x}{8}+\frac {3 a c}{8}\right )}{c b}\)	\(75\)
risch	\(\frac {3 x \sqrt {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{c \left (b x +a \right )}}{8 \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )}+\frac {\sqrt {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{5 c \left (b x +a \right )}}{32 b c \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )}+\frac {3 \sqrt {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{3 c \left (b x +a \right )}}{16 b c \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )}-\frac {\sqrt {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{-c \left (b x +a \right )}}{16 b c \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )}\)	\(216\)

[In]

int(exp(c*(b*x+a))*(cosh(b*c*x+a*c)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

csgn(cosh(c*(b*x+a)))/c/b*(1/4*cosh(b*c*x+a*c)^4+(1/4*cosh(b*c*x+a*c)^3+3/8*cosh(b*c*x+a*c))*sinh(b*c*x+a*c)+3
/8*b*c*x+3/8*a*c)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.78 \[ \int e^{c (a+b x)} \cosh ^2(a c+b c x)^{3/2} \, dx=-\frac {\cosh \left (b c x + a c\right )^{3} + 3 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{2} - 3 \, \sinh \left (b c x + a c\right )^{3} - 6 \, {\left (2 \, b c x + 1\right )} \cosh \left (b c x + a c\right ) + 3 \, {\left (4 \, b c x - 3 \, \cosh \left (b c x + a c\right )^{2} - 2\right )} \sinh \left (b c x + a c\right )}{32 \, {\left (b c \cosh \left (b c x + a c\right ) - b c \sinh \left (b c x + a c\right )\right )}} \]

[In]

integrate(exp(c*(b*x+a))*(cosh(b*c*x+a*c)^2)^(3/2),x, algorithm="fricas")

[Out]

-1/32*(cosh(b*c*x + a*c)^3 + 3*cosh(b*c*x + a*c)*sinh(b*c*x + a*c)^2 - 3*sinh(b*c*x + a*c)^3 - 6*(2*b*c*x + 1)
*cosh(b*c*x + a*c) + 3*(4*b*c*x - 3*cosh(b*c*x + a*c)^2 - 2)*sinh(b*c*x + a*c))/(b*c*cosh(b*c*x + a*c) - b*c*s
inh(b*c*x + a*c))

Sympy [F(-1)]

Timed out. \[ \int e^{c (a+b x)} \cosh ^2(a c+b c x)^{3/2} \, dx=\text {Timed out} \]

[In]

integrate(exp(c*(b*x+a))*(cosh(b*c*x+a*c)**2)**(3/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.46 \[ \int e^{c (a+b x)} \cosh ^2(a c+b c x)^{3/2} \, dx=\frac {3 \, {\left (b c x + a c\right )}}{8 \, b c} + \frac {e^{\left (4 \, b c x + 4 \, a c\right )}}{32 \, b c} + \frac {3 \, e^{\left (2 \, b c x + 2 \, a c\right )}}{16 \, b c} - \frac {e^{\left (-2 \, b c x - 2 \, a c\right )}}{16 \, b c} \]

[In]

integrate(exp(c*(b*x+a))*(cosh(b*c*x+a*c)^2)^(3/2),x, algorithm="maxima")

[Out]

3/8*(b*c*x + a*c)/(b*c) + 1/32*e^(4*b*c*x + 4*a*c)/(b*c) + 3/16*e^(2*b*c*x + 2*a*c)/(b*c) - 1/16*e^(-2*b*c*x -
2*a*c)/(b*c)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.45 \[ \int e^{c (a+b x)} \cosh ^2(a c+b c x)^{3/2} \, dx=\frac {12 \, b c x - 2 \, {\left (3 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )} e^{\left (-2 \, b c x - 2 \, a c\right )} + {\left (e^{\left (4 \, b c x + 8 \, a c\right )} + 6 \, e^{\left (2 \, b c x + 6 \, a c\right )}\right )} e^{\left (-4 \, a c\right )}}{32 \, b c} \]

[In]

integrate(exp(c*(b*x+a))*(cosh(b*c*x+a*c)^2)^(3/2),x, algorithm="giac")

[Out]

1/32*(12*b*c*x - 2*(3*e^(2*b*c*x + 2*a*c) + 1)*e^(-2*b*c*x - 2*a*c) + (e^(4*b*c*x + 8*a*c) + 6*e^(2*b*c*x + 6*
a*c))*e^(-4*a*c))/(b*c)

Mupad [F(-1)]

Timed out. \[ \int e^{c (a+b x)} \cosh ^2(a c+b c x)^{3/2} \, dx=\int {\mathrm {e}}^{c\,\left (a+b\,x\right )}\,{\left ({\mathrm {cosh}\left (a\,c+b\,c\,x\right )}^2\right )}^{3/2} \,d x \]

[In]

int(exp(c*(a + b*x))*(cosh(a*c + b*c*x)^2)^(3/2),x)

[Out]

int(exp(c*(a + b*x))*(cosh(a*c + b*c*x)^2)^(3/2), x)