\(\int e^{c (a+b x)} \sinh ^2(a c+b c x)^{5/2} \, dx\) [329]

Optimal result

Integrand size = 25, antiderivative size = 250 \[ \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{5/2} \, dx=\frac {e^{-4 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{128 b c}-\frac {5 e^{-2 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{64 b c}+\frac {5 e^{2 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{32 b c}-\frac {5 e^{4 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{128 b c}+\frac {e^{6 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{192 b c}-\frac {5}{16} x \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)} \]

[Out]

1/128*csch(b*c*x+a*c)*(sinh(b*c*x+a*c)^2)^(1/2)/b/c/exp(4*c*(b*x+a))-5/64*csch(b*c*x+a*c)*(sinh(b*c*x+a*c)^2)^
(1/2)/b/c/exp(2*c*(b*x+a))+5/32*exp(2*c*(b*x+a))*csch(b*c*x+a*c)*(sinh(b*c*x+a*c)^2)^(1/2)/b/c-5/128*exp(4*c*(
b*x+a))*csch(b*c*x+a*c)*(sinh(b*c*x+a*c)^2)^(1/2)/b/c+1/192*exp(6*c*(b*x+a))*csch(b*c*x+a*c)*(sinh(b*c*x+a*c)^
2)^(1/2)/b/c-5/16*x*csch(b*c*x+a*c)*(sinh(b*c*x+a*c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 250, 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} \[ \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{5/2} \, dx=\frac {e^{-4 c (a+b x)} \sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x)}{128 b c}-\frac {5 e^{-2 c (a+b x)} \sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x)}{64 b c}+\frac {5 e^{2 c (a+b x)} \sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x)}{32 b c}-\frac {5 e^{4 c (a+b x)} \sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x)}{128 b c}+\frac {e^{6 c (a+b x)} \sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x)}{192 b c}-\frac {5}{16} x \sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x) \]

[In]

Int[E^(c*(a + b*x))*(Sinh[a*c + b*c*x]^2)^(5/2),x]

[Out]

(Csch[a*c + b*c*x]*Sqrt[Sinh[a*c + b*c*x]^2])/(128*b*c*E^(4*c*(a + b*x))) - (5*Csch[a*c + b*c*x]*Sqrt[Sinh[a*c
+ b*c*x]^2])/(64*b*c*E^(2*c*(a + b*x))) + (5*E^(2*c*(a + b*x))*Csch[a*c + b*c*x]*Sqrt[Sinh[a*c + b*c*x]^2])/(
32*b*c) - (5*E^(4*c*(a + b*x))*Csch[a*c + b*c*x]*Sqrt[Sinh[a*c + b*c*x]^2])/(128*b*c) + (E^(6*c*(a + b*x))*Csc
h[a*c + b*c*x]*Sqrt[Sinh[a*c + b*c*x]^2])/(192*b*c) - (5*x*Csch[a*c + b*c*x]*Sqrt[Sinh[a*c + b*c*x]^2])/16

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 (\text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}\right ) \int e^{c (a+b x)} \sinh ^5(a c+b c x) \, dx \\ & = \frac {\left (\text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}\right ) \text {Subst}\left (\int \frac {\left (-1+x^2\right )^5}{32 x^5} \, dx,x,e^{c (a+b x)}\right )}{b c} \\ & = \frac {\left (\text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}\right ) \text {Subst}\left (\int \frac {\left (-1+x^2\right )^5}{x^5} \, dx,x,e^{c (a+b x)}\right )}{32 b c} \\ & = \frac {\left (\text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}\right ) \text {Subst}\left (\int \frac {(-1+x)^5}{x^3} \, dx,x,e^{2 c (a+b x)}\right )}{64 b c} \\ & = \frac {\left (\text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}\right ) \text {Subst}\left (\int \left (10-\frac {1}{x^3}+\frac {5}{x^2}-\frac {10}{x}-5 x+x^2\right ) \, dx,x,e^{2 c (a+b x)}\right )}{64 b c} \\ & = \frac {e^{-4 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{128 b c}-\frac {5 e^{-2 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{64 b c}+\frac {5 e^{2 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{32 b c}-\frac {5 e^{4 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{128 b c}+\frac {e^{6 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{192 b c}-\frac {5}{16} x \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.44 \[ \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{5/2} \, dx=\frac {\left (\frac {1}{128} e^{-4 c (a+b x)}-\frac {5}{64} e^{-2 c (a+b x)}+\frac {5}{32} e^{2 c (a+b x)}-\frac {5}{128} e^{4 c (a+b x)}+\frac {1}{192} e^{6 c (a+b x)}-\frac {5 b c x}{16}\right ) \text {csch}^5(c (a+b x)) \sinh ^2(c (a+b x))^{5/2}}{b c} \]

[In]

Integrate[E^(c*(a + b*x))*(Sinh[a*c + b*c*x]^2)^(5/2),x]

[Out]

((1/(128*E^(4*c*(a + b*x))) - 5/(64*E^(2*c*(a + b*x))) + (5*E^(2*c*(a + b*x)))/32 - (5*E^(4*c*(a + b*x)))/128
+ E^(6*c*(a + b*x))/192 - (5*b*c*x)/16)*Csch[c*(a + b*x)]^5*(Sinh[c*(a + b*x)]^2)^(5/2))/(b*c)

Maple [C] (warning: unable to verify)

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

Time = 6.26 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.35

method	result	size
default	\(\frac {\operatorname {csgn}\left (\sinh \left (c \left (b x +a \right )\right )\right ) \left (\frac {\sinh \left (b c x +a c \right )^{6}}{6}+\left (\frac {\sinh \left (b c x +a c \right )^{5}}{6}-\frac {5 \sinh \left (b c x +a c \right )^{3}}{24}+\frac {5 \sinh \left (b c x +a c \right )}{16}\right ) \cosh \left (b c x +a c \right )-\frac {5 b c x}{16}-\frac {5 a c}{16}\right )}{c b}\)	\(88\)
risch	\(-\frac {5 x \sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{c \left (b x +a \right )}}{16 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )}+\frac {\sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{7 c \left (b x +a \right )}}{192 c b \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )}-\frac {5 \sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{5 c \left (b x +a \right )}}{128 c b \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )}+\frac {5 \sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{3 c \left (b x +a \right )}}{32 c b \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )}-\frac {5 \sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{-c \left (b x +a \right )}}{64 c b \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )}+\frac {\sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{-3 c \left (b x +a \right )}}{128 c b \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )}\)	\(326\)

[In]

int(exp(c*(b*x+a))*(sinh(b*c*x+a*c)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

csgn(sinh(c*(b*x+a)))/c/b*(1/6*sinh(b*c*x+a*c)^6+(1/6*sinh(b*c*x+a*c)^5-5/24*sinh(b*c*x+a*c)^3+5/16*sinh(b*c*x
+a*c))*cosh(b*c*x+a*c)-5/16*b*c*x-5/16*a*c)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.87 \[ \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{5/2} \, dx=\frac {5 \, \cosh \left (b c x + a c\right )^{5} + 25 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{4} - \sinh \left (b c x + a c\right )^{5} - 5 \, {\left (2 \, \cosh \left (b c x + a c\right )^{2} - 3\right )} \sinh \left (b c x + a c\right )^{3} - 45 \, \cosh \left (b c x + a c\right )^{3} + 5 \, {\left (10 \, \cosh \left (b c x + a c\right )^{3} - 27 \, \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right )^{2} - 60 \, {\left (2 \, b c x - 1\right )} \cosh \left (b c x + a c\right ) - 5 \, {\left (\cosh \left (b c x + a c\right )^{4} - 24 \, b c x - 9 \, \cosh \left (b c x + a c\right )^{2} - 12\right )} \sinh \left (b c x + a c\right )}{384 \, {\left (b c \cosh \left (b c x + a c\right ) - b c \sinh \left (b c x + a c\right )\right )}} \]

[In]

integrate(exp(c*(b*x+a))*(sinh(b*c*x+a*c)^2)^(5/2),x, algorithm="fricas")

[Out]

1/384*(5*cosh(b*c*x + a*c)^5 + 25*cosh(b*c*x + a*c)*sinh(b*c*x + a*c)^4 - sinh(b*c*x + a*c)^5 - 5*(2*cosh(b*c*
x + a*c)^2 - 3)*sinh(b*c*x + a*c)^3 - 45*cosh(b*c*x + a*c)^3 + 5*(10*cosh(b*c*x + a*c)^3 - 27*cosh(b*c*x + a*c
))*sinh(b*c*x + a*c)^2 - 60*(2*b*c*x - 1)*cosh(b*c*x + a*c) - 5*(cosh(b*c*x + a*c)^4 - 24*b*c*x - 9*cosh(b*c*x
+ a*c)^2 - 12)*sinh(b*c*x + a*c))/(b*c*cosh(b*c*x + a*c) - b*c*sinh(b*c*x + a*c))

Sympy [F(-1)]

Timed out. \[ \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{5/2} \, dx=\text {Timed out} \]

[In]

integrate(exp(c*(b*x+a))*(sinh(b*c*x+a*c)**2)**(5/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.36 \[ \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{5/2} \, dx=\frac {{\left (2 \, e^{\left (10 \, b c x + 10 \, a c\right )} - 15 \, e^{\left (8 \, b c x + 8 \, a c\right )} + 60 \, e^{\left (6 \, b c x + 6 \, a c\right )} - 30 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 3\right )} e^{\left (-4 \, b c x - 4 \, a c\right )}}{384 \, b c} - \frac {5 \, {\left (b c x + a c\right )}}{16 \, b c} \]

[In]

integrate(exp(c*(b*x+a))*(sinh(b*c*x+a*c)^2)^(5/2),x, algorithm="maxima")

[Out]

1/384*(2*e^(10*b*c*x + 10*a*c) - 15*e^(8*b*c*x + 8*a*c) + 60*e^(6*b*c*x + 6*a*c) - 30*e^(2*b*c*x + 2*a*c) + 3)
*e^(-4*b*c*x - 4*a*c)/(b*c) - 5/16*(b*c*x + a*c)/(b*c)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.08 \[ \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{5/2} \, dx=-\frac {120 \, b c x \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) - 3 \, {\left (30 \, e^{\left (4 \, b c x + 4 \, a c\right )} \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) - 10 \, e^{\left (2 \, b c x + 2 \, a c\right )} \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) + \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right )\right )} e^{\left (-4 \, b c x - 4 \, a c\right )} - {\left (2 \, e^{\left (6 \, b c x + 18 \, a c\right )} \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) - 15 \, e^{\left (4 \, b c x + 16 \, a c\right )} \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) + 60 \, e^{\left (2 \, b c x + 14 \, a c\right )} \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right )\right )} e^{\left (-12 \, a c\right )}}{384 \, b c} \]

[In]

integrate(exp(c*(b*x+a))*(sinh(b*c*x+a*c)^2)^(5/2),x, algorithm="giac")

[Out]

-1/384*(120*b*c*x*sgn(e^(b*c*x + a*c) - e^(-b*c*x - a*c)) - 3*(30*e^(4*b*c*x + 4*a*c)*sgn(e^(b*c*x + a*c) - e^
(-b*c*x - a*c)) - 10*e^(2*b*c*x + 2*a*c)*sgn(e^(b*c*x + a*c) - e^(-b*c*x - a*c)) + sgn(e^(b*c*x + a*c) - e^(-b
*c*x - a*c)))*e^(-4*b*c*x - 4*a*c) - (2*e^(6*b*c*x + 18*a*c)*sgn(e^(b*c*x + a*c) - e^(-b*c*x - a*c)) - 15*e^(4
*b*c*x + 16*a*c)*sgn(e^(b*c*x + a*c) - e^(-b*c*x - a*c)) + 60*e^(2*b*c*x + 14*a*c)*sgn(e^(b*c*x + a*c) - e^(-b
*c*x - a*c)))*e^(-12*a*c))/(b*c)

Mupad [F(-1)]

Timed out. \[ \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{5/2} \, dx=\int {\mathrm {e}}^{c\,\left (a+b\,x\right )}\,{\left ({\mathrm {sinh}\left (a\,c+b\,c\,x\right )}^2\right )}^{5/2} \,d x \]

[In]

int(exp(c*(a + b*x))*(sinh(a*c + b*c*x)^2)^(5/2),x)

[Out]

int(exp(c*(a + b*x))*(sinh(a*c + b*c*x)^2)^(5/2), x)