Integrand size = 25, antiderivative size = 193 \[ \int e^{c (a+b x)} \tanh ^2(a c+b c x)^{3/2} \, dx=\frac {e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c}-\frac {2 e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c \left (1+e^{2 c (a+b x)}\right )^2}+\frac {3 e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c \left (1+e^{2 c (a+b x)}\right )}-\frac {3 \arctan \left (e^{c (a+b x)}\right ) \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c} \] Output:
exp(c*(b*x+a))*coth(b*c*x+a*c)*(tanh(b*c*x+a*c)^2)^(1/2)/b/c-2*exp(c*(b*x+ a))*coth(b*c*x+a*c)*(tanh(b*c*x+a*c)^2)^(1/2)/b/c/(1+exp(2*c*(b*x+a)))^2+3 *exp(c*(b*x+a))*coth(b*c*x+a*c)*(tanh(b*c*x+a*c)^2)^(1/2)/b/c/(1+exp(2*c*( b*x+a)))-3*arctan(exp(c*(b*x+a)))*coth(b*c*x+a*c)*(tanh(b*c*x+a*c)^2)^(1/2 )/b/c
Time = 0.12 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.54 \[ \int e^{c (a+b x)} \tanh ^2(a c+b c x)^{3/2} \, dx=\frac {\left (e^{c (a+b x)} \left (2+5 e^{2 c (a+b x)}+e^{4 c (a+b x)}\right )-3 \left (1+e^{2 c (a+b x)}\right )^2 \arctan \left (e^{c (a+b x)}\right )\right ) \coth (c (a+b x)) \sqrt {\tanh ^2(c (a+b x))}}{b c \left (1+e^{2 c (a+b x)}\right )^2} \] Input:
Integrate[E^(c*(a + b*x))*(Tanh[a*c + b*c*x]^2)^(3/2),x]
Output:
((E^(c*(a + b*x))*(2 + 5*E^(2*c*(a + b*x)) + E^(4*c*(a + b*x))) - 3*(1 + E ^(2*c*(a + b*x)))^2*ArcTan[E^(c*(a + b*x))])*Coth[c*(a + b*x)]*Sqrt[Tanh[c *(a + b*x)]^2])/(b*c*(1 + E^(2*c*(a + b*x)))^2)
Time = 0.49 (sec) , antiderivative size = 103, 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.
\(\displaystyle \int e^{c (a+b x)} \tanh ^2(a c+b c x)^{3/2} \, dx\) |
\(\Big \downarrow \) 7271 |
\(\displaystyle \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x) \int e^{c (a+b x)} \tanh ^3(a c+b x c)dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x) \int -\frac {\left (1-e^{2 c (a+b x)}\right )^3}{\left (1+e^{2 c (a+b x)}\right )^3}de^{c (a+b x)}}{b c}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x) \int \frac {\left (1-e^{2 c (a+b x)}\right )^3}{\left (1+e^{2 c (a+b x)}\right )^3}de^{c (a+b x)}}{b c}\) |
\(\Big \downarrow \) 300 |
\(\displaystyle -\frac {\sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x) \int \left (\frac {2 \left (1+3 e^{4 c (a+b x)}\right )}{\left (1+e^{2 c (a+b x)}\right )^3}-1\right )de^{c (a+b x)}}{b c}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (-3 \arctan \left (e^{c (a+b x)}\right )+e^{c (a+b x)}+\frac {3 e^{c (a+b x)}}{e^{2 c (a+b x)}+1}-\frac {2 e^{c (a+b x)}}{\left (e^{2 c (a+b x)}+1\right )^2}\right ) \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x)}{b c}\) |
Input:
Int[E^(c*(a + b*x))*(Tanh[a*c + b*c*x]^2)^(3/2),x]
Output:
((E^(c*(a + b*x)) - (2*E^(c*(a + b*x)))/(1 + E^(2*c*(a + b*x)))^2 + (3*E^( c*(a + b*x)))/(1 + E^(2*c*(a + b*x))) - 3*ArcTan[E^(c*(a + b*x))])*Coth[a* c + b*c*x]*Sqrt[Tanh[a*c + b*c*x]^2])/(b*c)
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 = 1.61 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.68
method | result | size |
default | \(\frac {\operatorname {csgn}\left (\tanh \left (c \left (b x +a \right )\right )\right ) \left (\frac {\sinh \left (b c x +a c \right )^{3}}{\cosh \left (b c x +a c \right )^{2}}+\frac {3 \sinh \left (b c x +a c \right )}{\cosh \left (b c x +a c \right )^{2}}-\frac {3 \,\operatorname {sech}\left (b c x +a c \right ) \tanh \left (b c x +a c \right )}{2}-3 \arctan \left ({\mathrm e}^{b c x +a c}\right )+\frac {\sinh \left (b c x +a c \right )^{2}}{\cosh \left (b c x +a c \right )}+\frac {2}{\cosh \left (b c x +a c \right )}\right )}{c b}\) | \(131\) |
risch | \(\frac {\sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \left (3 i \ln \left ({\mathrm e}^{c \left (b x +a \right )}-i\right ) {\mathrm e}^{4 c \left (b x +a \right )}-3 i \ln \left ({\mathrm e}^{c \left (b x +a \right )}+i\right ) {\mathrm e}^{4 c \left (b x +a \right )}+2 \,{\mathrm e}^{5 c \left (b x +a \right )}+6 i \ln \left ({\mathrm e}^{c \left (b x +a \right )}-i\right ) {\mathrm e}^{2 c \left (b x +a \right )}-6 i \ln \left ({\mathrm e}^{c \left (b x +a \right )}+i\right ) {\mathrm e}^{2 c \left (b x +a \right )}+10 \,{\mathrm e}^{3 c \left (b x +a \right )}+3 i \ln \left ({\mathrm e}^{c \left (b x +a \right )}-i\right )-3 i \ln \left ({\mathrm e}^{c \left (b x +a \right )}+i\right )+4 \,{\mathrm e}^{c \left (b x +a \right )}\right )}{2 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) c b}\) | \(223\) |
Input:
int(exp(c*(b*x+a))*(tanh(b*c*x+a*c)^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
csgn(tanh(c*(b*x+a)))/c/b*(sinh(b*c*x+a*c)^3/cosh(b*c*x+a*c)^2+3*sinh(b*c* x+a*c)/cosh(b*c*x+a*c)^2-3/2*sech(b*c*x+a*c)*tanh(b*c*x+a*c)-3*arctan(exp( b*c*x+a*c))+sinh(b*c*x+a*c)^2/cosh(b*c*x+a*c)+2/cosh(b*c*x+a*c))
Leaf count of result is larger than twice the leaf count of optimal. 458 vs. \(2 (179) = 358\).
Time = 0.11 (sec) , antiderivative size = 458, normalized size of antiderivative = 2.37 \[ \int e^{c (a+b x)} \tanh ^2(a c+b c x)^{3/2} \, dx=\frac {\cosh \left (b c x + a c\right )^{5} + 5 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{4} + \sinh \left (b c x + a c\right )^{5} + 5 \, {\left (2 \, \cosh \left (b c x + a c\right )^{2} + 1\right )} \sinh \left (b c x + a c\right )^{3} + 5 \, \cosh \left (b c x + a c\right )^{3} + 5 \, {\left (2 \, \cosh \left (b c x + a c\right )^{3} + 3 \, \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right )^{2} - 3 \, {\left (\cosh \left (b c x + a c\right )^{4} + 4 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{3} + \sinh \left (b c x + a c\right )^{4} + 2 \, {\left (3 \, \cosh \left (b c x + a c\right )^{2} + 1\right )} \sinh \left (b c x + a c\right )^{2} + 2 \, \cosh \left (b c x + a c\right )^{2} + 4 \, {\left (\cosh \left (b c x + a c\right )^{3} + \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right ) + 1\right )} \arctan \left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right ) + {\left (5 \, \cosh \left (b c x + a c\right )^{4} + 15 \, \cosh \left (b c x + a c\right )^{2} + 2\right )} \sinh \left (b c x + a c\right ) + 2 \, \cosh \left (b c x + a c\right )}{b c \cosh \left (b c x + a c\right )^{4} + 4 \, b c \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{3} + b c \sinh \left (b c x + a c\right )^{4} + 2 \, b c \cosh \left (b c x + a c\right )^{2} + 2 \, {\left (3 \, b c \cosh \left (b c x + a c\right )^{2} + b c\right )} \sinh \left (b c x + a c\right )^{2} + b c + 4 \, {\left (b c \cosh \left (b c x + a c\right )^{3} + b c \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right )} \] Input:
integrate(exp(c*(b*x+a))*(tanh(b*c*x+a*c)^2)^(3/2),x, algorithm="fricas")
Output:
(cosh(b*c*x + a*c)^5 + 5*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 + 1)*sinh(b*c*x + a*c)^3 + 5*cosh(b* c*x + a*c)^3 + 5*(2*cosh(b*c*x + a*c)^3 + 3*cosh(b*c*x + a*c))*sinh(b*c*x + a*c)^2 - 3*(cosh(b*c*x + a*c)^4 + 4*cosh(b*c*x + a*c)*sinh(b*c*x + a*c)^ 3 + sinh(b*c*x + a*c)^4 + 2*(3*cosh(b*c*x + a*c)^2 + 1)*sinh(b*c*x + a*c)^ 2 + 2*cosh(b*c*x + a*c)^2 + 4*(cosh(b*c*x + a*c)^3 + cosh(b*c*x + a*c))*si nh(b*c*x + a*c) + 1)*arctan(cosh(b*c*x + a*c) + sinh(b*c*x + a*c)) + (5*co sh(b*c*x + a*c)^4 + 15*cosh(b*c*x + a*c)^2 + 2)*sinh(b*c*x + a*c) + 2*cosh (b*c*x + a*c))/(b*c*cosh(b*c*x + a*c)^4 + 4*b*c*cosh(b*c*x + a*c)*sinh(b*c *x + a*c)^3 + b*c*sinh(b*c*x + a*c)^4 + 2*b*c*cosh(b*c*x + a*c)^2 + 2*(3*b *c*cosh(b*c*x + a*c)^2 + b*c)*sinh(b*c*x + a*c)^2 + b*c + 4*(b*c*cosh(b*c* x + a*c)^3 + b*c*cosh(b*c*x + a*c))*sinh(b*c*x + a*c))
Timed out. \[ \int e^{c (a+b x)} \tanh ^2(a c+b c x)^{3/2} \, dx=\text {Timed out} \] Input:
integrate(exp(c*(b*x+a))*(tanh(b*c*x+a*c)**2)**(3/2),x)
Output:
Timed out
Time = 0.14 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.47 \[ \int e^{c (a+b x)} \tanh ^2(a c+b c x)^{3/2} \, dx=-\frac {3 \, \arctan \left (e^{\left (b c x + a c\right )}\right )}{b c} + \frac {e^{\left (5 \, b c x + 5 \, a c\right )} + 5 \, e^{\left (3 \, b c x + 3 \, a c\right )} + 2 \, e^{\left (b c x + a c\right )}}{b c {\left (e^{\left (4 \, b c x + 4 \, a c\right )} + 2 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} \] Input:
integrate(exp(c*(b*x+a))*(tanh(b*c*x+a*c)^2)^(3/2),x, algorithm="maxima")
Output:
-3*arctan(e^(b*c*x + a*c))/(b*c) + (e^(5*b*c*x + 5*a*c) + 5*e^(3*b*c*x + 3 *a*c) + 2*e^(b*c*x + a*c))/(b*c*(e^(4*b*c*x + 4*a*c) + 2*e^(2*b*c*x + 2*a* c) + 1))
Time = 0.13 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.71 \[ \int e^{c (a+b x)} \tanh ^2(a c+b c x)^{3/2} \, dx=-\frac {3 \, \arctan \left (e^{\left (b c x + a c\right )}\right ) \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}{b c} + \frac {e^{\left (b c x + a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}{b c} + \frac {3 \, e^{\left (3 \, b c x + 3 \, a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) + e^{\left (b c x + a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}{b c {\left (e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}^{2}} \] Input:
integrate(exp(c*(b*x+a))*(tanh(b*c*x+a*c)^2)^(3/2),x, algorithm="giac")
Output:
-3*arctan(e^(b*c*x + a*c))*sgn(e^(2*b*c*x + 2*a*c) - 1)/(b*c) + e^(b*c*x + a*c)*sgn(e^(2*b*c*x + 2*a*c) - 1)/(b*c) + (3*e^(3*b*c*x + 3*a*c)*sgn(e^(2 *b*c*x + 2*a*c) - 1) + e^(b*c*x + a*c)*sgn(e^(2*b*c*x + 2*a*c) - 1))/(b*c* (e^(2*b*c*x + 2*a*c) + 1)^2)
Timed out. \[ \int e^{c (a+b x)} \tanh ^2(a c+b c x)^{3/2} \, dx=\int {\mathrm {e}}^{c\,\left (a+b\,x\right )}\,{\left ({\mathrm {tanh}\left (a\,c+b\,c\,x\right )}^2\right )}^{3/2} \,d x \] Input:
int(exp(c*(a + b*x))*(tanh(a*c + b*c*x)^2)^(3/2),x)
Output:
int(exp(c*(a + b*x))*(tanh(a*c + b*c*x)^2)^(3/2), x)
Time = 0.26 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.72 \[ \int e^{c (a+b x)} \tanh ^2(a c+b c x)^{3/2} \, dx=\frac {-3 e^{4 b c x +4 a c} \mathit {atan} \left (e^{b c x +a c}\right )-6 e^{2 b c x +2 a c} \mathit {atan} \left (e^{b c x +a c}\right )-3 \mathit {atan} \left (e^{b c x +a c}\right )+e^{5 b c x +5 a c}+5 e^{3 b c x +3 a c}+2 e^{b c x +a c}}{b c \left (e^{4 b c x +4 a c}+2 e^{2 b c x +2 a c}+1\right )} \] Input:
int(exp(c*(b*x+a))*(tanh(b*c*x+a*c)^2)^(3/2),x)
Output:
( - 3*e**(4*a*c + 4*b*c*x)*atan(e**(a*c + b*c*x)) - 6*e**(2*a*c + 2*b*c*x) *atan(e**(a*c + b*c*x)) - 3*atan(e**(a*c + b*c*x)) + e**(5*a*c + 5*b*c*x) + 5*e**(3*a*c + 3*b*c*x) + 2*e**(a*c + b*c*x))/(b*c*(e**(4*a*c + 4*b*c*x) + 2*e**(2*a*c + 2*b*c*x) + 1))