Integrand size = 25, antiderivative size = 250 \[ \int \frac {e^{c (a+b x)}}{\text {sech}^2(a c+b c x)^{5/2}} \, dx=-\frac {e^{-4 c (a+b x)} \text {sech}(a c+b c x)}{128 b c \sqrt {\text {sech}^2(a c+b c x)}}-\frac {5 e^{-2 c (a+b x)} \text {sech}(a c+b c x)}{64 b c \sqrt {\text {sech}^2(a c+b c x)}}+\frac {5 e^{2 c (a+b x)} \text {sech}(a c+b c x)}{32 b c \sqrt {\text {sech}^2(a c+b c x)}}+\frac {5 e^{4 c (a+b x)} \text {sech}(a c+b c x)}{128 b c \sqrt {\text {sech}^2(a c+b c x)}}+\frac {e^{6 c (a+b x)} \text {sech}(a c+b c x)}{192 b c \sqrt {\text {sech}^2(a c+b c x)}}+\frac {5 x \text {sech}(a c+b c x)}{16 \sqrt {\text {sech}^2(a c+b c x)}} \]
-1/128*sech(b*c*x+a*c)/b/c/exp(4*c*(b*x+a))/(sech(b*c*x+a*c)^2)^(1/2)-5/64 *sech(b*c*x+a*c)/b/c/exp(2*c*(b*x+a))/(sech(b*c*x+a*c)^2)^(1/2)+5/32*exp(2 *c*(b*x+a))*sech(b*c*x+a*c)/b/c/(sech(b*c*x+a*c)^2)^(1/2)+5/128*exp(4*c*(b *x+a))*sech(b*c*x+a*c)/b/c/(sech(b*c*x+a*c)^2)^(1/2)+1/192*exp(6*c*(b*x+a) )*sech(b*c*x+a*c)/b/c/(sech(b*c*x+a*c)^2)^(1/2)+5/16*x*sech(b*c*x+a*c)/(se ch(b*c*x+a*c)^2)^(1/2)
Time = 0.16 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.44 \[ \int \frac {e^{c (a+b x)}}{\text {sech}^2(a c+b c x)^{5/2}} \, dx=\frac {\left (-\frac {1}{128} e^{-4 c (a+b x)}-\frac {5}{64} e^{-2 c (a+b x)}+\frac {5}{32} e^{2 c (a+b x)}+\frac {5}{128} e^{4 c (a+b x)}+\frac {1}{192} e^{6 c (a+b x)}+\frac {5 b c x}{16}\right ) \text {sech}^5(c (a+b x))}{b c \text {sech}^2(c (a+b x))^{5/2}} \]
((-1/128*1/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)/1 6)*Sech[c*(a + b*x)]^5)/(b*c*(Sech[c*(a + b*x)]^2)^(5/2))
Time = 0.42 (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.
| \(\displaystyle \int \frac {e^{c (a+b x)}}{\text {sech}^2(a c+b c x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 7271 |
| \(\displaystyle \frac {\text {sech}(a c+b c x) \int e^{c (a+b x)} \cosh ^5(a c+b x c)dx}{\sqrt {\text {sech}^2(a c+b c x)}}\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {\text {sech}(a c+b c x) \int \frac {1}{32} e^{-5 c (a+b x)} \left (1+e^{2 c (a+b x)}\right )^5de^{c (a+b x)}}{b c \sqrt {\text {sech}^2(a c+b c x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\text {sech}(a c+b c x) \int e^{-5 c (a+b x)} \left (1+e^{2 c (a+b x)}\right )^5de^{c (a+b x)}}{32 b c \sqrt {\text {sech}^2(a c+b c x)}}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {\text {sech}(a c+b c x) \int e^{-3 c (a+b x)} \left (1+e^{2 c (a+b x)}\right )^5de^{2 c (a+b x)}}{64 b c \sqrt {\text {sech}^2(a c+b c x)}}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle \frac {\text {sech}(a c+b c x) \int \left (10+e^{-3 c (a+b x)}+5 e^{-2 c (a+b x)}+10 e^{-c (a+b x)}+6 e^{2 c (a+b x)}\right )de^{2 c (a+b x)}}{64 b c \sqrt {\text {sech}^2(a c+b c x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\left (-\frac {1}{2} e^{-2 c (a+b x)}-5 e^{-c (a+b x)}+\frac {25}{2} e^{2 c (a+b x)}+\frac {1}{3} e^{3 c (a+b x)}+10 \log \left (e^{2 c (a+b x)}\right )\right ) \text {sech}(a c+b c x)}{64 b c \sqrt {\text {sech}^2(a c+b c x)}}\) |
((-1/2*1/E^(2*c*(a + b*x)) - 5/E^(c*(a + b*x)) + (25*E^(2*c*(a + b*x)))/2 + E^(3*c*(a + b*x))/3 + 10*Log[E^(2*c*(a + b*x))])*Sech[a*c + b*c*x])/(64* b*c*Sqrt[Sech[a*c + b*c*x]^2])
3.2.57.3.1 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 = 0.67 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.35
| method | result | size |
| default | \(\frac {\operatorname {csgn}\left (\operatorname {sech}\left (c \left (b x +a \right )\right )\right ) \left (\frac {\cosh \left (b c x +a c \right )^{6}}{6}+\left (\frac {\cosh \left (b c x +a c \right )^{5}}{6}+\frac {5 \cosh \left (b c x +a c \right )^{3}}{24}+\frac {5 \cosh \left (b c x +a c \right )}{16}\right ) \sinh \left (b c x +a c \right )+\frac {5 b c x}{16}+\frac {5 a c}{16}\right )}{c b}\) | \(88\) |
| risch | \(\frac {5 x \,{\mathrm e}^{c \left (b x +a \right )}}{16 \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}}+\frac {{\mathrm e}^{7 c \left (b x +a \right )}}{192 b c \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}}+\frac {5 \,{\mathrm e}^{5 c \left (b x +a \right )}}{128 b c \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}}+\frac {5 \,{\mathrm e}^{3 c \left (b x +a \right )}}{32 b c \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}}-\frac {5 \,{\mathrm e}^{-c \left (b x +a \right )}}{64 b c \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}}-\frac {{\mathrm e}^{-3 c \left (b x +a \right )}}{128 b c \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}}\) | \(326\) |
csgn(sech(c*(b*x+a)))/c/b*(1/6*cosh(b*c*x+a*c)^6+(1/6*cosh(b*c*x+a*c)^5+5/ 24*cosh(b*c*x+a*c)^3+5/16*cosh(b*c*x+a*c))*sinh(b*c*x+a*c)+5/16*b*c*x+5/16 *a*c)
Time = 0.26 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.87 \[ \int \frac {e^{c (a+b x)}}{\text {sech}^2(a c+b c x)^{5/2}} \, dx=-\frac {\cosh \left (b c x + a c\right )^{5} + 5 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{4} - 5 \, \sinh \left (b c x + a c\right )^{5} - 5 \, {\left (10 \, \cosh \left (b c x + a c\right )^{2} + 9\right )} \sinh \left (b c x + a c\right )^{3} + 15 \, \cosh \left (b c x + a c\right )^{3} + 5 \, {\left (2 \, \cosh \left (b c x + a c\right )^{3} + 9 \, \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right )^{2} - 60 \, {\left (2 \, b c x + 1\right )} \cosh \left (b c x + a c\right ) - 5 \, {\left (5 \, \cosh \left (b c x + a c\right )^{4} - 24 \, b c x + 27 \, \cosh \left (b c x + a c\right )^{2} + 12\right )} \sinh \left (b c x + a c\right )}{384 \, {\left (b c \cosh \left (b c x + a c\right ) - b c \sinh \left (b c x + a c\right )\right )}} \]
-1/384*(cosh(b*c*x + a*c)^5 + 5*cosh(b*c*x + a*c)*sinh(b*c*x + a*c)^4 - 5* sinh(b*c*x + a*c)^5 - 5*(10*cosh(b*c*x + a*c)^2 + 9)*sinh(b*c*x + a*c)^3 + 15*cosh(b*c*x + a*c)^3 + 5*(2*cosh(b*c*x + a*c)^3 + 9*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*(5*cosh(b*c*x + a*c)^4 - 24*b*c*x + 27*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))
Leaf count of result is larger than twice the leaf count of optimal. 530 vs. \(2 (238) = 476\).
Time = 86.60 (sec) , antiderivative size = 530, normalized size of antiderivative = 2.12 \[ \int \frac {e^{c (a+b x)}}{\text {sech}^2(a c+b c x)^{5/2}} \, dx=\begin {cases} x & \text {for}\: b = 0 \wedge c = 0 \\\frac {x e^{a c}}{\left (\operatorname {sech}^{2}{\left (a c \right )}\right )^{\frac {5}{2}}} & \text {for}\: b = 0 \\x & \text {for}\: c = 0 \\- \frac {5 x e^{a c} e^{b c x} \tanh ^{5}{\left (a c + b c x \right )}}{16 \left (\operatorname {sech}^{2}{\left (a c + b c x \right )}\right )^{\frac {5}{2}}} + \frac {5 x e^{a c} e^{b c x} \tanh ^{4}{\left (a c + b c x \right )}}{16 \left (\operatorname {sech}^{2}{\left (a c + b c x \right )}\right )^{\frac {5}{2}}} + \frac {5 x e^{a c} e^{b c x} \tanh ^{3}{\left (a c + b c x \right )}}{8 \left (\operatorname {sech}^{2}{\left (a c + b c x \right )}\right )^{\frac {5}{2}}} - \frac {5 x e^{a c} e^{b c x} \tanh ^{2}{\left (a c + b c x \right )}}{8 \left (\operatorname {sech}^{2}{\left (a c + b c x \right )}\right )^{\frac {5}{2}}} - \frac {5 x e^{a c} e^{b c x} \tanh {\left (a c + b c x \right )}}{16 \left (\operatorname {sech}^{2}{\left (a c + b c x \right )}\right )^{\frac {5}{2}}} + \frac {5 x e^{a c} e^{b c x}}{16 \left (\operatorname {sech}^{2}{\left (a c + b c x \right )}\right )^{\frac {5}{2}}} + \frac {8 e^{a c} e^{b c x} \tanh ^{5}{\left (a c + b c x \right )}}{15 b c \left (\operatorname {sech}^{2}{\left (a c + b c x \right )}\right )^{\frac {5}{2}}} - \frac {53 e^{a c} e^{b c x} \tanh ^{4}{\left (a c + b c x \right )}}{240 b c \left (\operatorname {sech}^{2}{\left (a c + b c x \right )}\right )^{\frac {5}{2}}} - \frac {331 e^{a c} e^{b c x} \tanh ^{3}{\left (a c + b c x \right )}}{240 b c \left (\operatorname {sech}^{2}{\left (a c + b c x \right )}\right )^{\frac {5}{2}}} + \frac {131 e^{a c} e^{b c x} \tanh ^{2}{\left (a c + b c x \right )}}{240 b c \left (\operatorname {sech}^{2}{\left (a c + b c x \right )}\right )^{\frac {5}{2}}} + \frac {253 e^{a c} e^{b c x} \tanh {\left (a c + b c x \right )}}{240 b c \left (\operatorname {sech}^{2}{\left (a c + b c x \right )}\right )^{\frac {5}{2}}} - \frac {11 e^{a c} e^{b c x}}{30 b c \left (\operatorname {sech}^{2}{\left (a c + b c x \right )}\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((x, Eq(b, 0) & Eq(c, 0)), (x*exp(a*c)/(sech(a*c)**2)**(5/2), Eq( b, 0)), (x, Eq(c, 0)), (-5*x*exp(a*c)*exp(b*c*x)*tanh(a*c + b*c*x)**5/(16* (sech(a*c + b*c*x)**2)**(5/2)) + 5*x*exp(a*c)*exp(b*c*x)*tanh(a*c + b*c*x) **4/(16*(sech(a*c + b*c*x)**2)**(5/2)) + 5*x*exp(a*c)*exp(b*c*x)*tanh(a*c + b*c*x)**3/(8*(sech(a*c + b*c*x)**2)**(5/2)) - 5*x*exp(a*c)*exp(b*c*x)*ta nh(a*c + b*c*x)**2/(8*(sech(a*c + b*c*x)**2)**(5/2)) - 5*x*exp(a*c)*exp(b* c*x)*tanh(a*c + b*c*x)/(16*(sech(a*c + b*c*x)**2)**(5/2)) + 5*x*exp(a*c)*e xp(b*c*x)/(16*(sech(a*c + b*c*x)**2)**(5/2)) + 8*exp(a*c)*exp(b*c*x)*tanh( a*c + b*c*x)**5/(15*b*c*(sech(a*c + b*c*x)**2)**(5/2)) - 53*exp(a*c)*exp(b *c*x)*tanh(a*c + b*c*x)**4/(240*b*c*(sech(a*c + b*c*x)**2)**(5/2)) - 331*e xp(a*c)*exp(b*c*x)*tanh(a*c + b*c*x)**3/(240*b*c*(sech(a*c + b*c*x)**2)**( 5/2)) + 131*exp(a*c)*exp(b*c*x)*tanh(a*c + b*c*x)**2/(240*b*c*(sech(a*c + b*c*x)**2)**(5/2)) + 253*exp(a*c)*exp(b*c*x)*tanh(a*c + b*c*x)/(240*b*c*(s ech(a*c + b*c*x)**2)**(5/2)) - 11*exp(a*c)*exp(b*c*x)/(30*b*c*(sech(a*c + b*c*x)**2)**(5/2)), True))
Time = 0.19 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.45 \[ \int \frac {e^{c (a+b x)}}{\text {sech}^2(a c+b c x)^{5/2}} \, dx=\frac {5 \, {\left (b c x + a c\right )}}{16 \, b c} + \frac {e^{\left (6 \, b c x + 6 \, a c\right )}}{192 \, b c} + \frac {5 \, e^{\left (4 \, b c x + 4 \, a c\right )}}{128 \, b c} + \frac {5 \, e^{\left (2 \, b c x + 2 \, a c\right )}}{32 \, b c} - \frac {5 \, e^{\left (-2 \, b c x - 2 \, a c\right )}}{64 \, b c} - \frac {e^{\left (-4 \, b c x - 4 \, a c\right )}}{128 \, b c} \]
5/16*(b*c*x + a*c)/(b*c) + 1/192*e^(6*b*c*x + 6*a*c)/(b*c) + 5/128*e^(4*b* c*x + 4*a*c)/(b*c) + 5/32*e^(2*b*c*x + 2*a*c)/(b*c) - 5/64*e^(-2*b*c*x - 2 *a*c)/(b*c) - 1/128*e^(-4*b*c*x - 4*a*c)/(b*c)
Time = 0.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.44 \[ \int \frac {e^{c (a+b x)}}{\text {sech}^2(a c+b c x)^{5/2}} \, dx=\frac {{\left (120 \, b c x e^{\left (-a c\right )} - 3 \, {\left (30 \, e^{\left (4 \, b c x + 4 \, a c\right )} + 10 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )} e^{\left (-4 \, b c x - 5 \, a c\right )} + {\left (2 \, e^{\left (6 \, b c x + 20 \, a c\right )} + 15 \, e^{\left (4 \, b c x + 18 \, a c\right )} + 60 \, e^{\left (2 \, b c x + 16 \, a c\right )}\right )} e^{\left (-15 \, a c\right )}\right )} e^{\left (a c\right )}}{384 \, b c} \]
1/384*(120*b*c*x*e^(-a*c) - 3*(30*e^(4*b*c*x + 4*a*c) + 10*e^(2*b*c*x + 2* a*c) + 1)*e^(-4*b*c*x - 5*a*c) + (2*e^(6*b*c*x + 20*a*c) + 15*e^(4*b*c*x + 18*a*c) + 60*e^(2*b*c*x + 16*a*c))*e^(-15*a*c))*e^(a*c)/(b*c)
Timed out. \[ \int \frac {e^{c (a+b x)}}{\text {sech}^2(a c+b c x)^{5/2}} \, dx=\int \frac {{\mathrm {e}}^{c\,\left (a+b\,x\right )}}{{\left (\frac {1}{{\mathrm {cosh}\left (a\,c+b\,c\,x\right )}^2}\right )}^{5/2}} \,d x \]