Integrand size = 25, antiderivative size = 250 \[ \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{5/2} \, dx=\frac {e^{-4 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{128 b c}-\frac {5 e^{-2 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{64 b c}+\frac {5 e^{2 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{32 b c}-\frac {5 e^{4 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{128 b c}+\frac {e^{6 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{192 b c}-\frac {5}{16} x \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)} \]
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)*(sin h(b*c*x+a*c)^2)^(1/2)
Time = 0.08 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.44 \[ \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{5/2} \, dx=\frac {\left (\frac {1}{128} e^{-4 c (a+b x)}-\frac {5}{64} e^{-2 c (a+b x)}+\frac {5}{32} e^{2 c (a+b x)}-\frac {5}{128} e^{4 c (a+b x)}+\frac {1}{192} e^{6 c (a+b x)}-\frac {5 b c x}{16}\right ) \text {csch}^5(c (a+b x)) \sinh ^2(c (a+b x))^{5/2}}{b c} \]
((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)
Time = 0.43 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.41, 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.
\(\displaystyle \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{5/2} \, dx\) |
\(\Big \downarrow \) 7271 |
\(\displaystyle \sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x) \int e^{c (a+b x)} \sinh ^5(a c+b x c)dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x) \int -\frac {1}{32} e^{-5 c (a+b x)} \left (1-e^{2 c (a+b x)}\right )^5de^{c (a+b x)}}{b c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x) \int e^{-5 c (a+b x)} \left (1-e^{2 c (a+b x)}\right )^5de^{c (a+b x)}}{32 b c}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle -\frac {\sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x) \int e^{-3 c (a+b x)} \left (1-e^{2 c (a+b x)}\right )^5de^{2 c (a+b x)}}{64 b c}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {\sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x) \int \left (-10+e^{-3 c (a+b x)}-5 e^{-2 c (a+b x)}+10 e^{-c (a+b x)}+4 e^{2 c (a+b x)}\right )de^{2 c (a+b x)}}{64 b c}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\left (-\frac {1}{2} e^{-2 c (a+b x)}+5 e^{-c (a+b x)}-\frac {15}{2} e^{2 c (a+b x)}-\frac {1}{3} e^{3 c (a+b x)}+10 \log \left (e^{2 c (a+b x)}\right )\right ) \sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x)}{64 b c}\) |
-1/64*(Csch[a*c + b*c*x]*(-1/2*1/E^(2*c*(a + b*x)) + 5/E^(c*(a + b*x)) - ( 15*E^(2*c*(a + b*x)))/2 - E^(3*c*(a + b*x))/3 + 10*Log[E^(2*c*(a + b*x))]) *Sqrt[Sinh[a*c + b*c*x]^2])/(b*c)
3.4.29.3.1 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 = 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 | \(-\frac {5 x \sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{c \left (b x +a \right )}}{16 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )}+\frac {\sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{7 c \left (b x +a \right )}}{192 c b \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )}-\frac {5 \sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{5 c \left (b x +a \right )}}{128 c b \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )}+\frac {5 \sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{3 c \left (b x +a \right )}}{32 c b \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )}-\frac {5 \sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{-c \left (b x +a \right )}}{64 c b \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )}+\frac {\sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{-3 c \left (b x +a \right )}}{128 c b \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )}\) | \(326\) |
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)
Time = 0.29 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.87 \[ \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{5/2} \, dx=\frac {5 \, \cosh \left (b c x + a c\right )^{5} + 25 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{4} - \sinh \left (b c x + a c\right )^{5} - 5 \, {\left (2 \, \cosh \left (b c x + a c\right )^{2} - 3\right )} \sinh \left (b c x + a c\right )^{3} - 45 \, \cosh \left (b c x + a c\right )^{3} + 5 \, {\left (10 \, \cosh \left (b c x + a c\right )^{3} - 27 \, \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right )^{2} - 60 \, {\left (2 \, b c x - 1\right )} \cosh \left (b c x + a c\right ) - 5 \, {\left (\cosh \left (b c x + a c\right )^{4} - 24 \, b c x - 9 \, \cosh \left (b c x + a c\right )^{2} - 12\right )} \sinh \left (b c x + a c\right )}{384 \, {\left (b c \cosh \left (b c x + a c\right ) - b c \sinh \left (b c x + a c\right )\right )}} \]
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))
Timed out. \[ \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{5/2} \, dx=\text {Timed out} \]
Time = 0.31 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.36 \[ \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{5/2} \, dx=\frac {{\left (2 \, e^{\left (10 \, b c x + 10 \, a c\right )} - 15 \, e^{\left (8 \, b c x + 8 \, a c\right )} + 60 \, e^{\left (6 \, b c x + 6 \, a c\right )} - 30 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 3\right )} e^{\left (-4 \, b c x - 4 \, a c\right )}}{384 \, b c} - \frac {5 \, {\left (b c x + a c\right )}}{16 \, b c} \]
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)
Time = 0.29 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.08 \[ \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{5/2} \, dx=-\frac {120 \, b c x \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) - 3 \, {\left (30 \, e^{\left (4 \, b c x + 4 \, a c\right )} \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) - 10 \, e^{\left (2 \, b c x + 2 \, a c\right )} \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) + \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right )\right )} e^{\left (-4 \, b c x - 4 \, a c\right )} - {\left (2 \, e^{\left (6 \, b c x + 18 \, a c\right )} \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) - 15 \, e^{\left (4 \, b c x + 16 \, a c\right )} \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) + 60 \, e^{\left (2 \, b c x + 14 \, a c\right )} \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right )\right )} e^{\left (-12 \, a c\right )}}{384 \, b c} \]
-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)
Timed out. \[ \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{5/2} \, dx=\int {\mathrm {e}}^{c\,\left (a+b\,x\right )}\,{\left ({\mathrm {sinh}\left (a\,c+b\,c\,x\right )}^2\right )}^{5/2} \,d x \]