Integrand size = 15, antiderivative size = 44 \[ \int \sinh (a+b x) \tanh ^3(a+b x) \, dx=-\frac {3 \arctan (\sinh (a+b x))}{2 b}+\frac {\sinh (a+b x)}{b}+\frac {\text {sech}(a+b x) \tanh (a+b x)}{2 b} \] Output:
-3/2*arctan(sinh(b*x+a))/b+sinh(b*x+a)/b+1/2*sech(b*x+a)*tanh(b*x+a)/b
Time = 0.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.18 \[ \int \sinh (a+b x) \tanh ^3(a+b x) \, dx=-\frac {3 \arctan (\sinh (a+b x))}{2 b}+\frac {3 \text {sech}(a+b x) \tanh (a+b x)}{2 b}+\frac {\sinh (a+b x) \tanh ^2(a+b x)}{b} \] Input:
Integrate[Sinh[a + b*x]*Tanh[a + b*x]^3,x]
Output:
(-3*ArcTan[Sinh[a + b*x]])/(2*b) + (3*Sech[a + b*x]*Tanh[a + b*x])/(2*b) + (Sinh[a + b*x]*Tanh[a + b*x]^2)/b
Time = 0.21 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3072, 252, 262, 216}
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.
\(\displaystyle \int \sinh (a+b x) \tanh ^3(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (i a+i b x) \tan (i a+i b x)^3dx\) |
\(\Big \downarrow \) 3072 |
\(\displaystyle \frac {\int \frac {\sinh ^4(a+b x)}{\left (\sinh ^2(a+b x)+1\right )^2}d\sinh (a+b x)}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\frac {3}{2} \int \frac {\sinh ^2(a+b x)}{\sinh ^2(a+b x)+1}d\sinh (a+b x)-\frac {\sinh ^3(a+b x)}{2 \left (\sinh ^2(a+b x)+1\right )}}{b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {\frac {3}{2} \left (\sinh (a+b x)-\int \frac {1}{\sinh ^2(a+b x)+1}d\sinh (a+b x)\right )-\frac {\sinh ^3(a+b x)}{2 \left (\sinh ^2(a+b x)+1\right )}}{b}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {3}{2} (\sinh (a+b x)-\arctan (\sinh (a+b x)))-\frac {\sinh ^3(a+b x)}{2 \left (\sinh ^2(a+b x)+1\right )}}{b}\) |
Input:
Int[Sinh[a + b*x]*Tanh[a + b*x]^3,x]
Output:
((3*(-ArcTan[Sinh[a + b*x]] + Sinh[a + b*x]))/2 - Sinh[a + b*x]^3/(2*(1 + Sinh[a + b*x]^2)))/b
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ (ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)], x ]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]
Time = 0.14 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.41
method | result | size |
derivativedivides | \(\frac {\frac {\sinh \left (b x +a \right )^{3}}{\cosh \left (b x +a \right )^{2}}+\frac {3 \sinh \left (b x +a \right )}{\cosh \left (b x +a \right )^{2}}-\frac {3 \,\operatorname {sech}\left (b x +a \right ) \tanh \left (b x +a \right )}{2}-3 \arctan \left ({\mathrm e}^{b x +a}\right )}{b}\) | \(62\) |
default | \(\frac {\frac {\sinh \left (b x +a \right )^{3}}{\cosh \left (b x +a \right )^{2}}+\frac {3 \sinh \left (b x +a \right )}{\cosh \left (b x +a \right )^{2}}-\frac {3 \,\operatorname {sech}\left (b x +a \right ) \tanh \left (b x +a \right )}{2}-3 \arctan \left ({\mathrm e}^{b x +a}\right )}{b}\) | \(62\) |
risch | \(\frac {{\mathrm e}^{b x +a}}{2 b}-\frac {{\mathrm e}^{-b x -a}}{2 b}+\frac {{\mathrm e}^{b x +a} \left ({\mathrm e}^{2 b x +2 a}-1\right )}{b \left ({\mathrm e}^{2 b x +2 a}+1\right )^{2}}+\frac {3 i \ln \left ({\mathrm e}^{b x +a}-i\right )}{2 b}-\frac {3 i \ln \left ({\mathrm e}^{b x +a}+i\right )}{2 b}\) | \(93\) |
Input:
int(sinh(b*x+a)*tanh(b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
1/b*(sinh(b*x+a)^3/cosh(b*x+a)^2+3*sinh(b*x+a)/cosh(b*x+a)^2-3/2*sech(b*x+ a)*tanh(b*x+a)-3*arctan(exp(b*x+a)))
Leaf count of result is larger than twice the leaf count of optimal. 463 vs. \(2 (40) = 80\).
Time = 0.13 (sec) , antiderivative size = 463, normalized size of antiderivative = 10.52 \[ \int \sinh (a+b x) \tanh ^3(a+b x) \, dx=\frac {\cosh \left (b x + a\right )^{6} + 6 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} + \sinh \left (b x + a\right )^{6} + 3 \, {\left (5 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{4} + 3 \, \cosh \left (b x + a\right )^{4} + 4 \, {\left (5 \, \cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} + 3 \, {\left (5 \, \cosh \left (b x + a\right )^{4} + 6 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 6 \, {\left (\cosh \left (b x + a\right )^{5} + 5 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + \sinh \left (b x + a\right )^{5} + 2 \, {\left (5 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{3} + 2 \, \cosh \left (b x + a\right )^{3} + 2 \, {\left (5 \, \cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} + {\left (5 \, \cosh \left (b x + a\right )^{4} + 6 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right ) + \cosh \left (b x + a\right )\right )} \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 3 \, \cosh \left (b x + a\right )^{2} + 6 \, {\left (\cosh \left (b x + a\right )^{5} + 2 \, \cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) - 1}{2 \, {\left (b \cosh \left (b x + a\right )^{5} + 5 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + b \sinh \left (b x + a\right )^{5} + 2 \, b \cosh \left (b x + a\right )^{3} + 2 \, {\left (5 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )^{3} + 2 \, {\left (5 \, b \cosh \left (b x + a\right )^{3} + 3 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} + b \cosh \left (b x + a\right ) + {\left (5 \, b \cosh \left (b x + a\right )^{4} + 6 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )\right )}} \] Input:
integrate(sinh(b*x+a)*tanh(b*x+a)^3,x, algorithm="fricas")
Output:
1/2*(cosh(b*x + a)^6 + 6*cosh(b*x + a)*sinh(b*x + a)^5 + sinh(b*x + a)^6 + 3*(5*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^4 + 3*cosh(b*x + a)^4 + 4*(5*cosh (b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^3 + 3*(5*cosh(b*x + a)^4 + 6* cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 6*(cosh(b*x + a)^5 + 5*cosh(b*x + a )*sinh(b*x + a)^4 + sinh(b*x + a)^5 + 2*(5*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^3 + 2*cosh(b*x + a)^3 + 2*(5*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b *x + a)^2 + (5*cosh(b*x + a)^4 + 6*cosh(b*x + a)^2 + 1)*sinh(b*x + a) + co sh(b*x + a))*arctan(cosh(b*x + a) + sinh(b*x + a)) - 3*cosh(b*x + a)^2 + 6 *(cosh(b*x + a)^5 + 2*cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) - 1)/ (b*cosh(b*x + a)^5 + 5*b*cosh(b*x + a)*sinh(b*x + a)^4 + b*sinh(b*x + a)^5 + 2*b*cosh(b*x + a)^3 + 2*(5*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^3 + 2*( 5*b*cosh(b*x + a)^3 + 3*b*cosh(b*x + a))*sinh(b*x + a)^2 + b*cosh(b*x + a) + (5*b*cosh(b*x + a)^4 + 6*b*cosh(b*x + a)^2 + b)*sinh(b*x + a))
\[ \int \sinh (a+b x) \tanh ^3(a+b x) \, dx=\int \sinh {\left (a + b x \right )} \tanh ^{3}{\left (a + b x \right )}\, dx \] Input:
integrate(sinh(b*x+a)*tanh(b*x+a)**3,x)
Output:
Integral(sinh(a + b*x)*tanh(a + b*x)**3, x)
Leaf count of result is larger than twice the leaf count of optimal. 91 vs. \(2 (40) = 80\).
Time = 0.12 (sec) , antiderivative size = 91, normalized size of antiderivative = 2.07 \[ \int \sinh (a+b x) \tanh ^3(a+b x) \, dx=\frac {3 \, \arctan \left (e^{\left (-b x - a\right )}\right )}{b} - \frac {e^{\left (-b x - a\right )}}{2 \, b} + \frac {4 \, e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (-4 \, b x - 4 \, a\right )} + 1}{2 \, b {\left (e^{\left (-b x - a\right )} + 2 \, e^{\left (-3 \, b x - 3 \, a\right )} + e^{\left (-5 \, b x - 5 \, a\right )}\right )}} \] Input:
integrate(sinh(b*x+a)*tanh(b*x+a)^3,x, algorithm="maxima")
Output:
3*arctan(e^(-b*x - a))/b - 1/2*e^(-b*x - a)/b + 1/2*(4*e^(-2*b*x - 2*a) - e^(-4*b*x - 4*a) + 1)/(b*(e^(-b*x - a) + 2*e^(-3*b*x - 3*a) + e^(-5*b*x - 5*a)))
Leaf count of result is larger than twice the leaf count of optimal. 97 vs. \(2 (40) = 80\).
Time = 0.12 (sec) , antiderivative size = 97, normalized size of antiderivative = 2.20 \[ \int \sinh (a+b x) \tanh ^3(a+b x) \, dx=-\frac {3 \, \pi - \frac {4 \, {\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}}{{\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{2} + 4} + 6 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-b x - a\right )}\right ) - 2 \, e^{\left (b x + a\right )} + 2 \, e^{\left (-b x - a\right )}}{4 \, b} \] Input:
integrate(sinh(b*x+a)*tanh(b*x+a)^3,x, algorithm="giac")
Output:
-1/4*(3*pi - 4*(e^(b*x + a) - e^(-b*x - a))/((e^(b*x + a) - e^(-b*x - a))^ 2 + 4) + 6*arctan(1/2*(e^(2*b*x + 2*a) - 1)*e^(-b*x - a)) - 2*e^(b*x + a) + 2*e^(-b*x - a))/b
Time = 0.98 (sec) , antiderivative size = 107, normalized size of antiderivative = 2.43 \[ \int \sinh (a+b x) \tanh ^3(a+b x) \, dx=\frac {{\mathrm {e}}^{a+b\,x}}{2\,b}-\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {b^2}}{b}\right )}{\sqrt {b^2}}-\frac {{\mathrm {e}}^{-a-b\,x}}{2\,b}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left (2\,{\mathrm {e}}^{2\,a+2\,b\,x}+{\mathrm {e}}^{4\,a+4\,b\,x}+1\right )}+\frac {{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \] Input:
int(sinh(a + b*x)*tanh(a + b*x)^3,x)
Output:
exp(a + b*x)/(2*b) - (3*atan((exp(b*x)*exp(a)*(b^2)^(1/2))/b))/(b^2)^(1/2) - exp(- a - b*x)/(2*b) - (2*exp(a + b*x))/(b*(2*exp(2*a + 2*b*x) + exp(4* a + 4*b*x) + 1)) + exp(a + b*x)/(b*(exp(2*a + 2*b*x) + 1))
Time = 0.24 (sec) , antiderivative size = 104, normalized size of antiderivative = 2.36 \[ \int \sinh (a+b x) \tanh ^3(a+b x) \, dx=\frac {-12 e^{b x +a} \mathit {atan} \left (e^{b x +a}\right )-2 e^{b x +a} \cosh \left (b x +a \right ) \tanh \left (b x +a \right )+3 e^{2 b x +2 a}-2 e^{b x +a} \sinh \left (b x +a \right ) \tanh \left (b x +a \right )^{2}+2 e^{b x +a} \sinh \left (b x +a \right )-3}{4 e^{b x +a} b} \] Input:
int(sinh(b*x+a)*tanh(b*x+a)^3,x)
Output:
( - 12*e**(a + b*x)*atan(e**(a + b*x)) - 2*e**(a + b*x)*cosh(a + b*x)*tanh (a + b*x) + 3*e**(2*a + 2*b*x) - 2*e**(a + b*x)*sinh(a + b*x)*tanh(a + b*x )**2 + 2*e**(a + b*x)*sinh(a + b*x) - 3)/(4*e**(a + b*x)*b)