Integrand size = 25, antiderivative size = 319 \[ \int \frac {e^{c (a+b x)}}{\tanh ^2(a c+b c x)^{5/2}} \, dx=\frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}-\frac {4 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt {\tanh ^2(a c+b c x)}}+\frac {26 e^{c (a+b x)} \tanh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3 \sqrt {\tanh ^2(a c+b c x)}}-\frac {55 e^{c (a+b x)} \tanh (a c+b c x)}{6 b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt {\tanh ^2(a c+b c x)}}+\frac {25 e^{c (a+b x)} \tanh (a c+b c x)}{4 b c \left (1-e^{2 c (a+b x)}\right ) \sqrt {\tanh ^2(a c+b c x)}}-\frac {15 \text {arctanh}\left (e^{c (a+b x)}\right ) \tanh (a c+b c x)}{4 b c \sqrt {\tanh ^2(a c+b c x)}} \]
exp(c*(b*x+a))*tanh(b*c*x+a*c)/b/c/(tanh(b*c*x+a*c)^2)^(1/2)-4*exp(c*(b*x+ a))*tanh(b*c*x+a*c)/b/c/(1-exp(2*c*(b*x+a)))^4/(tanh(b*c*x+a*c)^2)^(1/2)+2 6/3*exp(c*(b*x+a))*tanh(b*c*x+a*c)/b/c/(1-exp(2*c*(b*x+a)))^3/(tanh(b*c*x+ a*c)^2)^(1/2)-55/6*exp(c*(b*x+a))*tanh(b*c*x+a*c)/b/c/(1-exp(2*c*(b*x+a))) ^2/(tanh(b*c*x+a*c)^2)^(1/2)+25/4*exp(c*(b*x+a))*tanh(b*c*x+a*c)/b/c/(1-ex p(2*c*(b*x+a)))/(tanh(b*c*x+a*c)^2)^(1/2)-15/4*arctanh(exp(c*(b*x+a)))*tan h(b*c*x+a*c)/b/c/(tanh(b*c*x+a*c)^2)^(1/2)
Time = 11.73 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.51 \[ \int \frac {e^{c (a+b x)}}{\tanh ^2(a c+b c x)^{5/2}} \, dx=\frac {\left (66 e^{c (a+b x)}-314 e^{3 c (a+b x)}+374 e^{5 c (a+b x)}-246 e^{7 c (a+b x)}+24 e^{9 c (a+b x)}+45 \left (-1+e^{2 c (a+b x)}\right )^4 \log \left (1-e^{c (a+b x)}\right )-45 \left (-1+e^{2 c (a+b x)}\right )^4 \log \left (1+e^{c (a+b x)}\right )\right ) \tanh (c (a+b x))}{24 b c \left (-1+e^{2 c (a+b x)}\right )^4 \sqrt {\tanh ^2(c (a+b x))}} \]
((66*E^(c*(a + b*x)) - 314*E^(3*c*(a + b*x)) + 374*E^(5*c*(a + b*x)) - 246 *E^(7*c*(a + b*x)) + 24*E^(9*c*(a + b*x)) + 45*(-1 + E^(2*c*(a + b*x)))^4* Log[1 - E^(c*(a + b*x))] - 45*(-1 + E^(2*c*(a + b*x)))^4*Log[1 + E^(c*(a + b*x))])*Tanh[c*(a + b*x)])/(24*b*c*(-1 + E^(2*c*(a + b*x)))^4*Sqrt[Tanh[c *(a + b*x)]^2])
Time = 1.54 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.53, 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.
| \(\displaystyle \int \frac {e^{c (a+b x)}}{\tanh ^2(a c+b c x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 7271 |
| \(\displaystyle \frac {\tanh (a c+b c x) \int e^{c (a+b x)} \coth ^5(a c+b x c)dx}{\sqrt {\tanh ^2(a c+b c x)}}\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {\tanh (a c+b c x) \int -\frac {\left (1+e^{2 c (a+b x)}\right )^5}{\left (1-e^{2 c (a+b x)}\right )^5}de^{c (a+b x)}}{b c \sqrt {\tanh ^2(a c+b c x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\tanh (a c+b c x) \int \frac {\left (1+e^{2 c (a+b x)}\right )^5}{\left (1-e^{2 c (a+b x)}\right )^5}de^{c (a+b x)}}{b c \sqrt {\tanh ^2(a c+b c x)}}\) |
| \(\Big \downarrow \) 300 |
| \(\displaystyle -\frac {\tanh (a c+b c x) \int \left (\frac {2 \left (1+10 e^{4 c (a+b x)}+5 e^{8 c (a+b x)}\right )}{\left (1-e^{2 c (a+b x)}\right )^5}-1\right )de^{c (a+b x)}}{b c \sqrt {\tanh ^2(a c+b c x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\left (-\frac {15}{4} \text {arctanh}\left (e^{c (a+b x)}\right )+e^{c (a+b x)}+\frac {25 e^{c (a+b x)}}{4 \left (1-e^{2 c (a+b x)}\right )}-\frac {55 e^{c (a+b x)}}{6 \left (1-e^{2 c (a+b x)}\right )^2}+\frac {26 e^{c (a+b x)}}{3 \left (1-e^{2 c (a+b x)}\right )^3}-\frac {4 e^{c (a+b x)}}{\left (1-e^{2 c (a+b x)}\right )^4}\right ) \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}\) |
((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*ArcTanh[E^(c*(a + b*x))])/4)*Tanh[a*c + b*c*x])/(b*c*Sqrt[Tanh[a*c + b*c*x]^2])
3.3.39.3.1 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 = 0.46 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.61
| method | result | size |
| default | \(\frac {\operatorname {csgn}\left (\tanh \left (c \left (b x +a \right )\right )\right ) \left (\frac {\cosh \left (b c x +a c \right )^{5}}{\sinh \left (b c x +a c \right )^{4}}-\frac {5 \cosh \left (b c x +a c \right )^{3}}{\sinh \left (b c x +a c \right )^{4}}+\frac {5 \cosh \left (b c x +a c \right )}{\sinh \left (b c x +a c \right )^{4}}+5 \left (-\frac {\operatorname {csch}\left (b c x +a c \right )^{3}}{4}+\frac {3 \,\operatorname {csch}\left (b c x +a c \right )}{8}\right ) \coth \left (b c x +a c \right )-\frac {15 \,\operatorname {arctanh}\left ({\mathrm e}^{b c x +a c}\right )}{4}+\frac {\cosh \left (b c x +a c \right )^{4}}{\sinh \left (b c x +a c \right )^{3}}-\frac {4 \cosh \left (b c x +a c \right )^{2}}{\sinh \left (b c x +a c \right )^{3}}+\frac {8}{3 \sinh \left (b c x +a c \right )^{3}}\right )}{c b}\) | \(195\) |
| risch | \(\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) {\mathrm e}^{c \left (b x +a \right )}}{\sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) b c}-\frac {{\mathrm e}^{c \left (b x +a \right )} \left (75 \,{\mathrm e}^{6 c \left (b x +a \right )}-115 \,{\mathrm e}^{4 c \left (b x +a \right )}+109 \,{\mathrm e}^{2 c \left (b x +a \right )}-21\right )}{12 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{3} \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, c b}-\frac {15 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \ln \left ({\mathrm e}^{c \left (b x +a \right )}+1\right )}{8 \sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) c b}+\frac {15 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \ln \left ({\mathrm e}^{c \left (b x +a \right )}-1\right )}{8 \sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) c b}\) | \(320\) |
csgn(tanh(c*(b*x+a)))/c/b*(cosh(b*c*x+a*c)^5/sinh(b*c*x+a*c)^4-5/sinh(b*c* x+a*c)^4*cosh(b*c*x+a*c)^3+5/sinh(b*c*x+a*c)^4*cosh(b*c*x+a*c)+5*(-1/4*csc h(b*c*x+a*c)^3+3/8*csch(b*c*x+a*c))*coth(b*c*x+a*c)-15/4*arctanh(exp(b*c*x +a*c))+1/sinh(b*c*x+a*c)^3*cosh(b*c*x+a*c)^4-4/sinh(b*c*x+a*c)^3*cosh(b*c* x+a*c)^2+8/3/sinh(b*c*x+a*c)^3)
Leaf count of result is larger than twice the leaf count of optimal. 1617 vs. \(2 (281) = 562\).
Time = 0.27 (sec) , antiderivative size = 1617, normalized size of antiderivative = 5.07 \[ \int \frac {e^{c (a+b x)}}{\tanh ^2(a c+b c x)^{5/2}} \, dx=\text {Too large to display} \]
1/24*(24*cosh(b*c*x + a*c)^9 + 216*cosh(b*c*x + a*c)*sinh(b*c*x + a*c)^8 + 24*sinh(b*c*x + a*c)^9 + 6*(144*cosh(b*c*x + a*c)^2 - 41)*sinh(b*c*x + a* c)^7 - 246*cosh(b*c*x + a*c)^7 + 42*(48*cosh(b*c*x + a*c)^3 - 41*cosh(b*c* x + a*c))*sinh(b*c*x + a*c)^6 + 2*(1512*cosh(b*c*x + a*c)^4 - 2583*cosh(b* c*x + a*c)^2 + 187)*sinh(b*c*x + a*c)^5 + 374*cosh(b*c*x + a*c)^5 + 2*(151 2*cosh(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 + 2*(1008*cosh(b*c*x + a*c)^6 - 4305*cosh(b*c*x + a*c) ^4 + 1870*cosh(b*c*x + a*c)^2 - 157)*sinh(b*c*x + a*c)^3 - 314*cosh(b*c*x + a*c)^3 + 2*(432*cosh(b*c*x + a*c)^7 - 2583*cosh(b*c*x + a*c)^5 + 1870*co sh(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 - 15*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)*lo g(cosh(b*c*x + a*c) + sinh(b*c*x + a*c) + 1) + 45*(cosh(b*c*x + a*c)^8 ...
Timed out. \[ \int \frac {e^{c (a+b x)}}{\tanh ^2(a c+b c x)^{5/2}} \, dx=\text {Timed out} \]
Time = 0.31 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.52 \[ \int \frac {e^{c (a+b x)}}{\tanh ^2(a c+b c x)^{5/2}} \, dx=-\frac {15 \, \log \left (e^{\left (b c x + a c\right )} + 1\right )}{8 \, b c} + \frac {15 \, \log \left (e^{\left (b c x + a c\right )} - 1\right )}{8 \, b c} + \frac {12 \, e^{\left (9 \, b c x + 9 \, a c\right )} - 123 \, e^{\left (7 \, b c x + 7 \, a c\right )} + 187 \, e^{\left (5 \, b c x + 5 \, a c\right )} - 157 \, e^{\left (3 \, b c x + 3 \, a c\right )} + 33 \, e^{\left (b c x + a c\right )}}{12 \, b c {\left (e^{\left (8 \, b c x + 8 \, a c\right )} - 4 \, e^{\left (6 \, b c x + 6 \, a c\right )} + 6 \, e^{\left (4 \, b c x + 4 \, a c\right )} - 4 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} \]
-15/8*log(e^(b*c*x + a*c) + 1)/(b*c) + 15/8*log(e^(b*c*x + a*c) - 1)/(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) + 33*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 = 0.42 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.67 \[ \int \frac {e^{c (a+b x)}}{\tanh ^2(a c+b c x)^{5/2}} \, dx=\frac {24 \, e^{\left (b c x + a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - 45 \, \log \left (e^{\left (b c x + a c\right )} + 1\right ) \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) + 45 \, \log \left ({\left | e^{\left (b c x + a c\right )} - 1 \right |}\right ) \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - \frac {2 \, {\left (75 \, e^{\left (7 \, b c x + 7 \, a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - 115 \, e^{\left (5 \, b c x + 5 \, a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) + 109 \, e^{\left (3 \, b c x + 3 \, a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - 21 \, e^{\left (b c x + a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )\right )}}{{\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}^{4}}}{24 \, b c} \]
1/24*(24*e^(b*c*x + a*c)*sgn(e^(2*b*c*x + 2*a*c) - 1) - 45*log(e^(b*c*x + a*c) + 1)*sgn(e^(2*b*c*x + 2*a*c) - 1) + 45*log(abs(e^(b*c*x + a*c) - 1))* sgn(e^(2*b*c*x + 2*a*c) - 1) - 2*(75*e^(7*b*c*x + 7*a*c)*sgn(e^(2*b*c*x + 2*a*c) - 1) - 115*e^(5*b*c*x + 5*a*c)*sgn(e^(2*b*c*x + 2*a*c) - 1) + 109*e ^(3*b*c*x + 3*a*c)*sgn(e^(2*b*c*x + 2*a*c) - 1) - 21*e^(b*c*x + a*c)*sgn(e ^(2*b*c*x + 2*a*c) - 1))/(e^(2*b*c*x + 2*a*c) - 1)^4)/(b*c)
Timed out. \[ \int \frac {e^{c (a+b x)}}{\tanh ^2(a c+b c x)^{5/2}} \, dx=\int \frac {{\mathrm {e}}^{c\,\left (a+b\,x\right )}}{{\left ({\mathrm {tanh}\left (a\,c+b\,c\,x\right )}^2\right )}^{5/2}} \,d x \]