Integrand size = 25, antiderivative size = 141 \[ \int e^{c (a+b x)} \text {sech}^2(a c+b c x)^{5/2} \, dx=-\frac {4 \cosh (a c+b c x) \sqrt {\text {sech}^2(a c+b c x)}}{b c \left (1+e^{2 c (a+b x)}\right )^4}+\frac {32 \cosh (a c+b c x) \sqrt {\text {sech}^2(a c+b c x)}}{3 b c \left (1+e^{2 c (a+b x)}\right )^3}-\frac {8 \cosh (a c+b c x) \sqrt {\text {sech}^2(a c+b c x)}}{b c \left (1+e^{2 c (a+b x)}\right )^2} \]
-4*cosh(b*c*x+a*c)*(sech(b*c*x+a*c)^2)^(1/2)/b/c/(1+exp(2*c*(b*x+a)))^4+32 /3*cosh(b*c*x+a*c)*(sech(b*c*x+a*c)^2)^(1/2)/b/c/(1+exp(2*c*(b*x+a)))^3-8* cosh(b*c*x+a*c)*(sech(b*c*x+a*c)^2)^(1/2)/b/c/(1+exp(2*c*(b*x+a)))^2
Time = 0.09 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.51 \[ \int e^{c (a+b x)} \text {sech}^2(a c+b c x)^{5/2} \, dx=-\frac {4 \left (1+4 e^{2 c (a+b x)}+6 e^{4 c (a+b x)}\right ) \cosh (c (a+b x)) \sqrt {\text {sech}^2(c (a+b x))}}{3 b c \left (1+e^{2 c (a+b x)}\right )^4} \]
(-4*(1 + 4*E^(2*c*(a + b*x)) + 6*E^(4*c*(a + b*x)))*Cosh[c*(a + b*x)]*Sqrt [Sech[c*(a + b*x)]^2])/(3*b*c*(1 + E^(2*c*(a + b*x)))^4)
Time = 0.36 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.62, 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, 53, 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.
\(\displaystyle \int e^{c (a+b x)} \text {sech}^2(a c+b c x)^{5/2} \, dx\) |
\(\Big \downarrow \) 7271 |
\(\displaystyle \cosh (a c+b c x) \sqrt {\text {sech}^2(a c+b c x)} \int e^{c (a+b x)} \text {sech}^5(a c+b x c)dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\cosh (a c+b c x) \sqrt {\text {sech}^2(a c+b c x)} \int \frac {32 e^{5 c (a+b x)}}{\left (1+e^{2 c (a+b x)}\right )^5}de^{c (a+b x)}}{b c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {32 \cosh (a c+b c x) \sqrt {\text {sech}^2(a c+b c x)} \int \frac {e^{5 c (a+b x)}}{\left (1+e^{2 c (a+b x)}\right )^5}de^{c (a+b x)}}{b c}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {16 \cosh (a c+b c x) \sqrt {\text {sech}^2(a c+b c x)} \int \frac {e^{2 c (a+b x)}}{\left (1+e^{2 c (a+b x)}\right )^5}de^{2 c (a+b x)}}{b c}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {16 \cosh (a c+b c x) \sqrt {\text {sech}^2(a c+b c x)} \int \left (\frac {1}{\left (1+e^{2 c (a+b x)}\right )^3}-\frac {2}{\left (1+e^{2 c (a+b x)}\right )^4}+\frac {1}{\left (1+e^{2 c (a+b x)}\right )^5}\right )de^{2 c (a+b x)}}{b c}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {16 \left (-\frac {1}{2 \left (e^{2 c (a+b x)}+1\right )^2}+\frac {2}{3 \left (e^{2 c (a+b x)}+1\right )^3}-\frac {1}{4 \left (e^{2 c (a+b x)}+1\right )^4}\right ) \cosh (a c+b c x) \sqrt {\text {sech}^2(a c+b c x)}}{b c}\) |
(16*(-1/4*1/(1 + E^(2*c*(a + b*x)))^4 + 2/(3*(1 + E^(2*c*(a + b*x)))^3) - 1/(2*(1 + E^(2*c*(a + b*x)))^2))*Cosh[a*c + b*c*x]*Sqrt[Sech[a*c + b*c*x]^ 2])/(b*c)
3.2.52.3.1 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, n}, x] && 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])
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 = 140.89 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.46
method | result | size |
default | \(-\frac {\operatorname {csgn}\left (\operatorname {sech}\left (c \left (b x +a \right )\right )\right ) \left (\frac {\tanh \left (c \left (b x +a \right )\right )^{4}}{4}+\frac {\tanh \left (c \left (b x +a \right )\right )^{3}}{3}-\frac {\tanh \left (c \left (b x +a \right )\right )^{2}}{2}-\tanh \left (c \left (b x +a \right )\right )\right )}{c b}\) | \(65\) |
risch | \(-\frac {4 \left (6 \,{\mathrm e}^{4 c \left (b x +a \right )}+4 \,{\mathrm e}^{2 c \left (b x +a \right )}+1\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, {\mathrm e}^{-c \left (b x +a \right )}}{3 b c \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{3}}\) | \(80\) |
-csgn(sech(c*(b*x+a)))/c/b*(1/4*tanh(c*(b*x+a))^4+1/3*tanh(c*(b*x+a))^3-1/ 2*tanh(c*(b*x+a))^2-tanh(c*(b*x+a)))
Leaf count of result is larger than twice the leaf count of optimal. 315 vs. \(2 (130) = 260\).
Time = 0.27 (sec) , antiderivative size = 315, normalized size of antiderivative = 2.23 \[ \int e^{c (a+b x)} \text {sech}^2(a c+b c x)^{5/2} \, dx=-\frac {4 \, {\left (7 \, \cosh \left (b c x + a c\right )^{2} + 10 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right ) + 7 \, \sinh \left (b c x + a c\right )^{2} + 4\right )}}{3 \, {\left (b c \cosh \left (b c x + a c\right )^{6} + 6 \, b c \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{5} + b c \sinh \left (b c x + a c\right )^{6} + 4 \, b c \cosh \left (b c x + a c\right )^{4} + {\left (15 \, b c \cosh \left (b c x + a c\right )^{2} + 4 \, b c\right )} \sinh \left (b c x + a c\right )^{4} + 7 \, b c \cosh \left (b c x + a c\right )^{2} + 4 \, {\left (5 \, b c \cosh \left (b c x + a c\right )^{3} + 4 \, b c \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right )^{3} + {\left (15 \, b c \cosh \left (b c x + a c\right )^{4} + 24 \, b c \cosh \left (b c x + a c\right )^{2} + 7 \, b c\right )} \sinh \left (b c x + a c\right )^{2} + 4 \, b c + 2 \, {\left (3 \, b c \cosh \left (b c x + a c\right )^{5} + 8 \, b c \cosh \left (b c x + a c\right )^{3} + 5 \, b c \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right )\right )}} \]
-4/3*(7*cosh(b*c*x + a*c)^2 + 10*cosh(b*c*x + a*c)*sinh(b*c*x + a*c) + 7*s inh(b*c*x + a*c)^2 + 4)/(b*c*cosh(b*c*x + a*c)^6 + 6*b*c*cosh(b*c*x + a*c) *sinh(b*c*x + a*c)^5 + b*c*sinh(b*c*x + a*c)^6 + 4*b*c*cosh(b*c*x + a*c)^4 + (15*b*c*cosh(b*c*x + a*c)^2 + 4*b*c)*sinh(b*c*x + a*c)^4 + 7*b*c*cosh(b *c*x + a*c)^2 + 4*(5*b*c*cosh(b*c*x + a*c)^3 + 4*b*c*cosh(b*c*x + a*c))*si nh(b*c*x + a*c)^3 + (15*b*c*cosh(b*c*x + a*c)^4 + 24*b*c*cosh(b*c*x + a*c) ^2 + 7*b*c)*sinh(b*c*x + a*c)^2 + 4*b*c + 2*(3*b*c*cosh(b*c*x + a*c)^5 + 8 *b*c*cosh(b*c*x + a*c)^3 + 5*b*c*cosh(b*c*x + a*c))*sinh(b*c*x + a*c))
Timed out. \[ \int e^{c (a+b x)} \text {sech}^2(a c+b c x)^{5/2} \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.48 \[ \int e^{c (a+b x)} \text {sech}^2(a c+b c x)^{5/2} \, dx=-\frac {8 \, e^{\left (4 \, b c x + 4 \, a c\right )}}{b c {\left (e^{\left (8 \, b c x + 8 \, a c\right )} + 4 \, e^{\left (6 \, b c x + 6 \, a c\right )} + 6 \, e^{\left (4 \, b c x + 4 \, a c\right )} + 4 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} - \frac {16 \, e^{\left (2 \, b c x + 2 \, a c\right )}}{3 \, b c {\left (e^{\left (8 \, b c x + 8 \, a c\right )} + 4 \, e^{\left (6 \, b c x + 6 \, a c\right )} + 6 \, e^{\left (4 \, b c x + 4 \, a c\right )} + 4 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} - \frac {4}{3 \, b c {\left (e^{\left (8 \, b c x + 8 \, a c\right )} + 4 \, e^{\left (6 \, b c x + 6 \, a c\right )} + 6 \, e^{\left (4 \, b c x + 4 \, a c\right )} + 4 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} \]
-8*e^(4*b*c*x + 4*a*c)/(b*c*(e^(8*b*c*x + 8*a*c) + 4*e^(6*b*c*x + 6*a*c) + 6*e^(4*b*c*x + 4*a*c) + 4*e^(2*b*c*x + 2*a*c) + 1)) - 16/3*e^(2*b*c*x + 2 *a*c)/(b*c*(e^(8*b*c*x + 8*a*c) + 4*e^(6*b*c*x + 6*a*c) + 6*e^(4*b*c*x + 4 *a*c) + 4*e^(2*b*c*x + 2*a*c) + 1)) - 4/3/(b*c*(e^(8*b*c*x + 8*a*c) + 4*e^ (6*b*c*x + 6*a*c) + 6*e^(4*b*c*x + 4*a*c) + 4*e^(2*b*c*x + 2*a*c) + 1))
Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.36 \[ \int e^{c (a+b x)} \text {sech}^2(a c+b c x)^{5/2} \, dx=-\frac {4 \, {\left (6 \, e^{\left (4 \, b c x + 4 \, a c\right )} + 4 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}}{3 \, b c {\left (e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}^{4}} \]
Time = 2.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.65 \[ \int e^{c (a+b x)} \text {sech}^2(a c+b c x)^{5/2} \, dx=-\frac {2\,{\mathrm {e}}^{-a\,c-b\,c\,x}\,\sqrt {\frac {1}{{\left (\frac {{\mathrm {e}}^{a\,c+b\,c\,x}}{2}+\frac {{\mathrm {e}}^{-a\,c-b\,c\,x}}{2}\right )}^2}}\,\left (4\,{\mathrm {e}}^{2\,a\,c+2\,b\,c\,x}+6\,{\mathrm {e}}^{4\,a\,c+4\,b\,c\,x}+1\right )}{3\,b\,c\,{\left ({\mathrm {e}}^{2\,a\,c+2\,b\,c\,x}+1\right )}^3} \]