Integrand size = 25, antiderivative size = 311 \[ \int \frac {e^{c (a+b x)}}{\coth ^2(a c+b c x)^{5/2}} \, dx=\frac {e^{c (a+b x)} \coth (a c+b c x)}{b c \sqrt {\coth ^2(a c+b c x)}}-\frac {4 e^{c (a+b x)} \coth (a c+b c x)}{b c \left (1+e^{2 c (a+b x)}\right )^4 \sqrt {\coth ^2(a c+b c x)}}+\frac {26 e^{c (a+b x)} \coth (a c+b c x)}{3 b c \left (1+e^{2 c (a+b x)}\right )^3 \sqrt {\coth ^2(a c+b c x)}}-\frac {55 e^{c (a+b x)} \coth (a c+b c x)}{6 b c \left (1+e^{2 c (a+b x)}\right )^2 \sqrt {\coth ^2(a c+b c x)}}+\frac {25 e^{c (a+b x)} \coth (a c+b c x)}{4 b c \left (1+e^{2 c (a+b x)}\right ) \sqrt {\coth ^2(a c+b c x)}}-\frac {15 \arctan \left (e^{c (a+b x)}\right ) \coth (a c+b c x)}{4 b c \sqrt {\coth ^2(a c+b c x)}} \]
exp(c*(b*x+a))*coth(b*c*x+a*c)/b/c/(coth(b*c*x+a*c)^2)^(1/2)-4*exp(c*(b*x+ a))*coth(b*c*x+a*c)/b/c/(1+exp(2*c*(b*x+a)))^4/(coth(b*c*x+a*c)^2)^(1/2)+2 6/3*exp(c*(b*x+a))*coth(b*c*x+a*c)/b/c/(1+exp(2*c*(b*x+a)))^3/(coth(b*c*x+ a*c)^2)^(1/2)-55/6*exp(c*(b*x+a))*coth(b*c*x+a*c)/b/c/(1+exp(2*c*(b*x+a))) ^2/(coth(b*c*x+a*c)^2)^(1/2)+25/4*exp(c*(b*x+a))*coth(b*c*x+a*c)/b/c/(1+ex p(2*c*(b*x+a)))/(coth(b*c*x+a*c)^2)^(1/2)-15/4*arctan(exp(c*(b*x+a)))*coth (b*c*x+a*c)/b/c/(coth(b*c*x+a*c)^2)^(1/2)
Time = 0.52 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.43 \[ \int \frac {e^{c (a+b x)}}{\coth ^2(a c+b c x)^{5/2}} \, dx=\frac {\left (e^{c (a+b x)} \left (33+157 e^{2 c (a+b x)}+187 e^{4 c (a+b x)}+123 e^{6 c (a+b x)}+12 e^{8 c (a+b x)}\right )-45 \left (1+e^{2 c (a+b x)}\right )^4 \arctan \left (e^{c (a+b x)}\right )\right ) \coth (c (a+b x))}{12 b c \left (1+e^{2 c (a+b x)}\right )^4 \sqrt {\coth ^2(c (a+b x))}} \]
((E^(c*(a + b*x))*(33 + 157*E^(2*c*(a + b*x)) + 187*E^(4*c*(a + b*x)) + 12 3*E^(6*c*(a + b*x)) + 12*E^(8*c*(a + b*x))) - 45*(1 + E^(2*c*(a + b*x)))^4 *ArcTan[E^(c*(a + b*x))])*Coth[c*(a + b*x)])/(12*b*c*(1 + E^(2*c*(a + b*x) ))^4*Sqrt[Coth[c*(a + b*x)]^2])
Time = 1.53 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.52, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {7271, 2720, 25, 300, 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.
| \(\displaystyle \int \frac {e^{c (a+b x)}}{\coth ^2(a c+b c x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 7271 |
| \(\displaystyle \frac {\coth (a c+b c x) \int e^{c (a+b x)} \tanh ^5(a c+b x c)dx}{\sqrt {\coth ^2(a c+b c x)}}\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {\coth (a c+b c x) \int -\frac {\left (1-e^{2 c (a+b x)}\right )^5}{\left (1+e^{2 c (a+b x)}\right )^5}de^{c (a+b x)}}{b c \sqrt {\coth ^2(a c+b c x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\coth (a c+b c x) \int \frac {\left (1-e^{2 c (a+b x)}\right )^5}{\left (1+e^{2 c (a+b x)}\right )^5}de^{c (a+b x)}}{b c \sqrt {\coth ^2(a c+b c x)}}\) |
| \(\Big \downarrow \) 300 |
| \(\displaystyle -\frac {\coth (a c+b c x) \int \left (\frac {2 \left (1+10 e^{4 c (a+b x)}+5 e^{8 c (a+b x)}\right )}{\left (1+e^{2 c (a+b x)}\right )^5}-1\right )de^{c (a+b x)}}{b c \sqrt {\coth ^2(a c+b c x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\left (-\frac {15}{4} \arctan \left (e^{c (a+b x)}\right )+e^{c (a+b x)}+\frac {25 e^{c (a+b x)}}{4 \left (e^{2 c (a+b x)}+1\right )}-\frac {55 e^{c (a+b x)}}{6 \left (e^{2 c (a+b x)}+1\right )^2}+\frac {26 e^{c (a+b x)}}{3 \left (e^{2 c (a+b x)}+1\right )^3}-\frac {4 e^{c (a+b x)}}{\left (e^{2 c (a+b x)}+1\right )^4}\right ) \coth (a c+b c x)}{b c \sqrt {\coth ^2(a c+b c x)}}\) |
((E^(c*(a + b*x)) - (4*E^(c*(a + b*x)))/(1 + E^(2*c*(a + b*x)))^4 + (26*E^ (c*(a + b*x)))/(3*(1 + E^(2*c*(a + b*x)))^3) - (55*E^(c*(a + b*x)))/(6*(1 + E^(2*c*(a + b*x)))^2) + (25*E^(c*(a + b*x)))/(4*(1 + E^(2*c*(a + b*x)))) - (15*ArcTan[E^(c*(a + b*x))])/4)*Coth[a*c + b*c*x])/(b*c*Sqrt[Coth[a*c + b*c*x]^2])
3.3.16.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int [PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c , d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
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 = 0.59 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.63
| method | result | size |
| default | \(\frac {\operatorname {csgn}\left (\coth \left (c \left (b x +a \right )\right )\right ) \left (\frac {\sinh \left (b c x +a c \right )^{4}}{\cosh \left (b c x +a c \right )^{3}}+\frac {4 \sinh \left (b c x +a c \right )^{2}}{\cosh \left (b c x +a c \right )^{3}}+\frac {8}{3 \cosh \left (b c x +a c \right )^{3}}+\frac {\sinh \left (b c x +a c \right )^{5}}{\cosh \left (b c x +a c \right )^{4}}+\frac {5 \sinh \left (b c x +a c \right )^{3}}{\cosh \left (b c x +a c \right )^{4}}+\frac {5 \sinh \left (b c x +a c \right )}{\cosh \left (b c x +a c \right )^{4}}-5 \left (\frac {\operatorname {sech}\left (b c x +a c \right )^{3}}{4}+\frac {3 \,\operatorname {sech}\left (b c x +a c \right )}{8}\right ) \tanh \left (b c x +a c \right )-\frac {15 \arctan \left ({\mathrm e}^{b c x +a c}\right )}{4}\right )}{c b}\) | \(195\) |
| risch | \(\frac {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) {\mathrm e}^{c \left (b x +a \right )}}{\sqrt {\frac {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}}\, \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) b c}+\frac {{\mathrm e}^{c \left (b x +a \right )} \left (75 \,{\mathrm e}^{6 c \left (b x +a \right )}+115 \,{\mathrm e}^{4 c \left (b x +a \right )}+109 \,{\mathrm e}^{2 c \left (b x +a \right )}+21\right )}{12 \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{3} \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \sqrt {\frac {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}}\, c b}+\frac {15 i \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \ln \left ({\mathrm e}^{c \left (b x +a \right )}-i\right )}{8 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \sqrt {\frac {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}}\, c b}-\frac {15 i \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \ln \left ({\mathrm e}^{c \left (b x +a \right )}+i\right )}{8 \sqrt {\frac {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}}\, \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) c b}\) | \(324\) |
csgn(coth(c*(b*x+a)))/c/b*(sinh(b*c*x+a*c)^4/cosh(b*c*x+a*c)^3+4*sinh(b*c* x+a*c)^2/cosh(b*c*x+a*c)^3+8/3/cosh(b*c*x+a*c)^3+sinh(b*c*x+a*c)^5/cosh(b* c*x+a*c)^4+5*sinh(b*c*x+a*c)^3/cosh(b*c*x+a*c)^4+5*sinh(b*c*x+a*c)/cosh(b* c*x+a*c)^4-5*(1/4*sech(b*c*x+a*c)^3+3/8*sech(b*c*x+a*c))*tanh(b*c*x+a*c)-1 5/4*arctan(exp(b*c*x+a*c)))
Leaf count of result is larger than twice the leaf count of optimal. 1226 vs. \(2 (281) = 562\).
Time = 0.28 (sec) , antiderivative size = 1226, normalized size of antiderivative = 3.94 \[ \int \frac {e^{c (a+b x)}}{\coth ^2(a c+b c x)^{5/2}} \, dx=\text {Too large to display} \]
1/12*(12*cosh(b*c*x + a*c)^9 + 108*cosh(b*c*x + a*c)*sinh(b*c*x + a*c)^8 + 12*sinh(b*c*x + a*c)^9 + 3*(144*cosh(b*c*x + a*c)^2 + 41)*sinh(b*c*x + a* c)^7 + 123*cosh(b*c*x + a*c)^7 + 21*(48*cosh(b*c*x + a*c)^3 + 41*cosh(b*c* x + a*c))*sinh(b*c*x + a*c)^6 + (1512*cosh(b*c*x + a*c)^4 + 2583*cosh(b*c* x + a*c)^2 + 187)*sinh(b*c*x + a*c)^5 + 187*cosh(b*c*x + a*c)^5 + (1512*co sh(b*c*x + a*c)^5 + 4305*cosh(b*c*x + a*c)^3 + 935*cosh(b*c*x + a*c))*sinh (b*c*x + a*c)^4 + (1008*cosh(b*c*x + a*c)^6 + 4305*cosh(b*c*x + a*c)^4 + 1 870*cosh(b*c*x + a*c)^2 + 157)*sinh(b*c*x + a*c)^3 + 157*cosh(b*c*x + a*c) ^3 + (432*cosh(b*c*x + a*c)^7 + 2583*cosh(b*c*x + a*c)^5 + 1870*cosh(b*c*x + a*c)^3 + 471*cosh(b*c*x + a*c))*sinh(b*c*x + a*c)^2 - 45*(cosh(b*c*x + a*c)^8 + 8*cosh(b*c*x + a*c)*sinh(b*c*x + a*c)^7 + sinh(b*c*x + a*c)^8 + 4 *(7*cosh(b*c*x + a*c)^2 + 1)*sinh(b*c*x + a*c)^6 + 4*cosh(b*c*x + a*c)^6 + 8*(7*cosh(b*c*x + a*c)^3 + 3*cosh(b*c*x + a*c))*sinh(b*c*x + a*c)^5 + 2*( 35*cosh(b*c*x + a*c)^4 + 30*cosh(b*c*x + a*c)^2 + 3)*sinh(b*c*x + a*c)^4 + 6*cosh(b*c*x + a*c)^4 + 8*(7*cosh(b*c*x + a*c)^5 + 10*cosh(b*c*x + a*c)^3 + 3*cosh(b*c*x + a*c))*sinh(b*c*x + a*c)^3 + 4*(7*cosh(b*c*x + a*c)^6 + 1 5*cosh(b*c*x + a*c)^4 + 9*cosh(b*c*x + a*c)^2 + 1)*sinh(b*c*x + a*c)^2 + 4 *cosh(b*c*x + a*c)^2 + 8*(cosh(b*c*x + a*c)^7 + 3*cosh(b*c*x + a*c)^5 + 3* cosh(b*c*x + a*c)^3 + cosh(b*c*x + a*c))*sinh(b*c*x + a*c) + 1)*arctan(cos h(b*c*x + a*c) + sinh(b*c*x + a*c)) + (108*cosh(b*c*x + a*c)^8 + 861*co...
Timed out. \[ \int \frac {e^{c (a+b x)}}{\coth ^2(a c+b c x)^{5/2}} \, dx=\text {Timed out} \]
Time = 0.31 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.47 \[ \int \frac {e^{c (a+b x)}}{\coth ^2(a c+b c x)^{5/2}} \, dx=-\frac {15 \, \arctan \left (e^{\left (b c x + a c\right )}\right )}{4 \, b c} + \frac {12 \, e^{\left (9 \, b c x + 9 \, a c\right )} + 123 \, e^{\left (7 \, b c x + 7 \, a c\right )} + 187 \, e^{\left (5 \, b c x + 5 \, a c\right )} + 157 \, e^{\left (3 \, b c x + 3 \, a c\right )} + 33 \, e^{\left (b c x + a c\right )}}{12 \, b c {\left (e^{\left (8 \, b c x + 8 \, a c\right )} + 4 \, e^{\left (6 \, b c x + 6 \, a c\right )} + 6 \, e^{\left (4 \, b c x + 4 \, a c\right )} + 4 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} \]
-15/4*arctan(e^(b*c*x + a*c))/(b*c) + 1/12*(12*e^(9*b*c*x + 9*a*c) + 123*e ^(7*b*c*x + 7*a*c) + 187*e^(5*b*c*x + 5*a*c) + 157*e^(3*b*c*x + 3*a*c) + 3 3*e^(b*c*x + 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))
Time = 0.30 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.59 \[ \int \frac {e^{c (a+b x)}}{\coth ^2(a c+b c x)^{5/2}} \, dx=-\frac {{\left (45 \, \arctan \left (e^{\left (b c x + a c\right )}\right ) e^{\left (-a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - 12 \, e^{\left (b c x\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - \frac {75 \, e^{\left (7 \, b c x + 6 \, a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) + 115 \, e^{\left (5 \, b c x + 4 \, a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) + 109 \, e^{\left (3 \, b c x + 2 \, a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) + 21 \, e^{\left (b c x\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}{{\left (e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}^{4}}\right )} e^{\left (a c\right )}}{12 \, b c} \]
-1/12*(45*arctan(e^(b*c*x + a*c))*e^(-a*c)*sgn(e^(2*b*c*x + 2*a*c) - 1) - 12*e^(b*c*x)*sgn(e^(2*b*c*x + 2*a*c) - 1) - (75*e^(7*b*c*x + 6*a*c)*sgn(e^ (2*b*c*x + 2*a*c) - 1) + 115*e^(5*b*c*x + 4*a*c)*sgn(e^(2*b*c*x + 2*a*c) - 1) + 109*e^(3*b*c*x + 2*a*c)*sgn(e^(2*b*c*x + 2*a*c) - 1) + 21*e^(b*c*x)* sgn(e^(2*b*c*x + 2*a*c) - 1))/(e^(2*b*c*x + 2*a*c) + 1)^4)*e^(a*c)/(b*c)
Timed out. \[ \int \frac {e^{c (a+b x)}}{\coth ^2(a c+b c x)^{5/2}} \, dx=\int \frac {{\mathrm {e}}^{c\,\left (a+b\,x\right )}}{{\left ({\mathrm {coth}\left (a\,c+b\,c\,x\right )}^2\right )}^{5/2}} \,d x \]