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

Optimal. Leaf size=250 \[ \frac{e^{-4 c (a+b x)} \sqrt{\sinh ^2(a c+b c x)} \text{csch}(a c+b c x)}{128 b c}-\frac{5 e^{-2 c (a+b x)} \sqrt{\sinh ^2(a c+b c x)} \text{csch}(a c+b c x)}{64 b c}+\frac{5 e^{2 c (a+b x)} \sqrt{\sinh ^2(a c+b c x)} \text{csch}(a c+b c x)}{32 b c}-\frac{5 e^{4 c (a+b x)} \sqrt{\sinh ^2(a c+b c x)} \text{csch}(a c+b c x)}{128 b c}+\frac{e^{6 c (a+b x)} \sqrt{\sinh ^2(a c+b c x)} \text{csch}(a c+b c x)}{192 b c}-\frac{5}{16} x \sqrt{\sinh ^2(a c+b c x)} \text{csch}(a c+b c x) \]

[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

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Rubi [A]  time = 0.247979, antiderivative size = 250, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6720, 2282, 12, 266, 43} \[ \frac{e^{-4 c (a+b x)} \sqrt{\sinh ^2(a c+b c x)} \text{csch}(a c+b c x)}{128 b c}-\frac{5 e^{-2 c (a+b x)} \sqrt{\sinh ^2(a c+b c x)} \text{csch}(a c+b c x)}{64 b c}+\frac{5 e^{2 c (a+b x)} \sqrt{\sinh ^2(a c+b c x)} \text{csch}(a c+b c x)}{32 b c}-\frac{5 e^{4 c (a+b x)} \sqrt{\sinh ^2(a c+b c x)} \text{csch}(a c+b c x)}{128 b c}+\frac{e^{6 c (a+b x)} \sqrt{\sinh ^2(a c+b c x)} \text{csch}(a c+b c x)}{192 b c}-\frac{5}{16} x \sqrt{\sinh ^2(a c+b c x)} \text{csch}(a c+b c x) \]

Antiderivative was successfully verified.

[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 6720

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])

Rule 2282

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 12

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

Rule 266

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 43

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])

Rubi steps

\begin{align*} \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{5/2} \, dx &=\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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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]  time = 0.109684, size = 106, normalized size = 0.42 \[ \frac{\left (\frac{1}{2} e^{-4 c (a+b x)}-5 e^{-2 c (a+b x)}+10 e^{2 c (a+b x)}-\frac{5}{2} e^{4 c (a+b x)}+\frac{1}{3} e^{6 c (a+b x)}-20 b c x\right ) \sinh ^2(c (a+b x))^{5/2} \text{csch}^5(c (a+b x))}{64 b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1/(2*E^(4*c*(a + b*x))) - 5/E^(2*c*(a + b*x)) + 10*E^(2*c*(a + b*x)) - (5*E^(4*c*(a + b*x)))/2 + E^(6*c*(a +
b*x))/3 - 20*b*c*x)*Csch[c*(a + b*x)]^5*(Sinh[c*(a + b*x)]^2)^(5/2))/(64*b*c)

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Maple [A]  time = 0.163, size = 184, normalized size = 0.7 \begin{align*}{\frac{1}{48\,\sinh \left ( c \left ( bx+a \right ) \right ) cb} \left ( 8\,\sqrt{ \left ( \sinh \left ( c \left ( bx+a \right ) \right ) \right ) ^{2}}\cosh \left ( c \left ( bx+a \right ) \right ) \left ( \sinh \left ( c \left ( bx+a \right ) \right ) \right ) ^{5}+8\,\sqrt{ \left ( \sinh \left ( c \left ( bx+a \right ) \right ) \right ) ^{2}} \left ( \sinh \left ( c \left ( bx+a \right ) \right ) \right ) ^{6}-10\,\sqrt{ \left ( \sinh \left ( c \left ( bx+a \right ) \right ) \right ) ^{2}} \left ( \sinh \left ( c \left ( bx+a \right ) \right ) \right ) ^{3}\cosh \left ( c \left ( bx+a \right ) \right ) +15\,\cosh \left ( c \left ( bx+a \right ) \right ) \sqrt{ \left ( \sinh \left ( c \left ( bx+a \right ) \right ) \right ) ^{2}}\sinh \left ( c \left ( bx+a \right ) \right ) -15\,\ln \left ( \cosh \left ( c \left ( bx+a \right ) \right ) +\sqrt{ \left ( \sinh \left ( c \left ( bx+a \right ) \right ) \right ) ^{2}} \right ) \sinh \left ( c \left ( bx+a \right ) \right ) +8\,\sqrt{ \left ( \sinh \left ( c \left ( bx+a \right ) \right ) \right ) ^{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(8*(sinh(c*(b*x+a))^2)^(1/2)*cosh(c*(b*x+a))*sinh(c*(b*x+a))^5+8*(sinh(c*(b*x+a))^2)^(1/2)*sinh(c*(b*x+a)
)^6-10*(sinh(c*(b*x+a))^2)^(1/2)*sinh(c*(b*x+a))^3*cosh(c*(b*x+a))+15*cosh(c*(b*x+a))*(sinh(c*(b*x+a))^2)^(1/2
)*sinh(c*(b*x+a))-15*ln(cosh(c*(b*x+a))+(sinh(c*(b*x+a))^2)^(1/2))*sinh(c*(b*x+a))+8*(sinh(c*(b*x+a))^2)^(1/2)
)/sinh(c*(b*x+a))/c/b

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Maxima [A]  time = 1.59572, size = 122, normalized size = 0.49 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.85523, size = 559, normalized size = 2.24 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.23678, size = 363, normalized size = 1.45 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)