Integrand size = 25, antiderivative size = 162 \[ \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{3/2} \, dx=\frac {e^{-2 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{16 b c}-\frac {3 e^{2 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{16 b c}+\frac {e^{4 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{32 b c}+\frac {3}{8} x \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)} \]
1/16*csch(b*c*x+a*c)*(sinh(b*c*x+a*c)^2)^(1/2)/b/c/exp(2*c*(b*x+a))-3/16*e xp(2*c*(b*x+a))*csch(b*c*x+a*c)*(sinh(b*c*x+a*c)^2)^(1/2)/b/c+1/32*exp(4*c *(b*x+a))*csch(b*c*x+a*c)*(sinh(b*c*x+a*c)^2)^(1/2)/b/c+3/8*x*csch(b*c*x+a *c)*(sinh(b*c*x+a*c)^2)^(1/2)
Time = 0.04 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.47 \[ \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{3/2} \, dx=\frac {\left (e^{-2 c (a+b x)}-3 e^{2 c (a+b x)}+\frac {1}{2} e^{4 c (a+b x)}+6 b c x\right ) \text {csch}^3(c (a+b x)) \sinh ^2(c (a+b x))^{3/2}}{16 b c} \]
((E^(-2*c*(a + b*x)) - 3*E^(2*c*(a + b*x)) + E^(4*c*(a + b*x))/2 + 6*b*c*x )*Csch[c*(a + b*x)]^3*(Sinh[c*(a + b*x)]^2)^(3/2))/(16*b*c)
Time = 0.38 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.46, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {7271, 2720, 27, 243, 49, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{3/2} \, dx\) |
\(\Big \downarrow \) 7271 |
\(\displaystyle \sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x) \int e^{c (a+b x)} \sinh ^3(a c+b x c)dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x) \int -\frac {1}{8} e^{-3 c (a+b x)} \left (1-e^{2 c (a+b x)}\right )^3de^{c (a+b x)}}{b c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x) \int e^{-3 c (a+b x)} \left (1-e^{2 c (a+b x)}\right )^3de^{c (a+b x)}}{8 b c}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle -\frac {\sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x) \int e^{-2 c (a+b x)} \left (1-e^{2 c (a+b x)}\right )^3de^{2 c (a+b x)}}{16 b c}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {\sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x) \int \left (3+e^{-2 c (a+b x)}-3 e^{-c (a+b x)}-e^{2 c (a+b x)}\right )de^{2 c (a+b x)}}{16 b c}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\left (-e^{-c (a+b x)}+\frac {5}{2} e^{2 c (a+b x)}-3 \log \left (e^{2 c (a+b x)}\right )\right ) \sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x)}{16 b c}\) |
-1/16*(Csch[a*c + b*c*x]*(-E^(-(c*(a + b*x))) + (5*E^(2*c*(a + b*x)))/2 - 3*Log[E^(2*c*(a + b*x))])*Sqrt[Sinh[a*c + b*c*x]^2])/(b*c)
3.4.30.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Simp[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]) && !(Eq Q[v, x] && EqQ[m, 1])
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.57 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.46
method | result | size |
default | \(\frac {\operatorname {csgn}\left (\sinh \left (c \left (b x +a \right )\right )\right ) \left (\frac {\sinh \left (b c x +a c \right )^{4}}{4}+\left (\frac {\sinh \left (b c x +a c \right )^{3}}{4}-\frac {3 \sinh \left (b c x +a c \right )}{8}\right ) \cosh \left (b c x +a c \right )+\frac {3 b c x}{8}+\frac {3 a c}{8}\right )}{c b}\) | \(75\) |
risch | \(\frac {3 x \sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{c \left (b x +a \right )}}{8 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )}+\frac {\sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{5 c \left (b x +a \right )}}{32 c b \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )}-\frac {3 \sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{3 c \left (b x +a \right )}}{16 c b \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )}+\frac {\sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{-c \left (b x +a \right )}}{16 c b \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )}\) | \(216\) |
csgn(sinh(c*(b*x+a)))/c/b*(1/4*sinh(b*c*x+a*c)^4+(1/4*sinh(b*c*x+a*c)^3-3/ 8*sinh(b*c*x+a*c))*cosh(b*c*x+a*c)+3/8*b*c*x+3/8*a*c)
Time = 0.29 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.78 \[ \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{3/2} \, dx=\frac {3 \, \cosh \left (b c x + a c\right )^{3} + 9 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{2} - \sinh \left (b c x + a c\right )^{3} + 6 \, {\left (2 \, b c x - 1\right )} \cosh \left (b c x + a c\right ) - 3 \, {\left (4 \, b c x + \cosh \left (b c x + a c\right )^{2} + 2\right )} \sinh \left (b c x + a c\right )}{32 \, {\left (b c \cosh \left (b c x + a c\right ) - b c \sinh \left (b c x + a c\right )\right )}} \]
1/32*(3*cosh(b*c*x + a*c)^3 + 9*cosh(b*c*x + a*c)*sinh(b*c*x + a*c)^2 - si nh(b*c*x + a*c)^3 + 6*(2*b*c*x - 1)*cosh(b*c*x + a*c) - 3*(4*b*c*x + cosh( b*c*x + a*c)^2 + 2)*sinh(b*c*x + a*c))/(b*c*cosh(b*c*x + a*c) - b*c*sinh(b *c*x + a*c))
Timed out. \[ \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{3/2} \, dx=\text {Timed out} \]
Time = 0.31 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.38 \[ \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{3/2} \, dx=\frac {{\left (e^{\left (6 \, b c x + 6 \, a c\right )} - 6 \, e^{\left (4 \, b c x + 4 \, a c\right )} + 2\right )} e^{\left (-2 \, b c x - 2 \, a c\right )}}{32 \, b c} + \frac {3 \, {\left (b c x + a c\right )}}{8 \, b c} \]
1/32*(e^(6*b*c*x + 6*a*c) - 6*e^(4*b*c*x + 4*a*c) + 2)*e^(-2*b*c*x - 2*a*c )/(b*c) + 3/8*(b*c*x + a*c)/(b*c)
Time = 0.30 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.20 \[ \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{3/2} \, dx=\frac {12 \, b c x \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) - 2 \, {\left (3 \, e^{\left (2 \, b c x + 2 \, a c\right )} \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) - \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right )\right )} e^{\left (-2 \, b c x - 2 \, a c\right )} + {\left (e^{\left (4 \, b c x + 8 \, a c\right )} \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) - 6 \, e^{\left (2 \, b c x + 6 \, a c\right )} \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right )\right )} e^{\left (-4 \, a c\right )}}{32 \, b c} \]
1/32*(12*b*c*x*sgn(e^(b*c*x + a*c) - e^(-b*c*x - a*c)) - 2*(3*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^(-2*b*c*x - 2*a*c) + (e^(4*b*c*x + 8*a*c)*sgn(e^(b*c*x + a*c) - e^(-b*c*x - a*c)) - 6*e^(2*b*c*x + 6*a*c)*sgn(e^(b*c*x + a*c) - e ^(-b*c*x - a*c)))*e^(-4*a*c))/(b*c)
Timed out. \[ \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{3/2} \, dx=\int {\mathrm {e}}^{c\,\left (a+b\,x\right )}\,{\left ({\mathrm {sinh}\left (a\,c+b\,c\,x\right )}^2\right )}^{3/2} \,d x \]