Integrand size = 16, antiderivative size = 83 \[ \int e^{a+b x} \sinh ^4(a+b x) \, dx=-\frac {e^{-3 a-3 b x}}{48 b}+\frac {e^{-a-b x}}{4 b}+\frac {3 e^{a+b x}}{8 b}-\frac {e^{3 a+3 b x}}{12 b}+\frac {e^{5 a+5 b x}}{80 b} \] Output:
-1/48*exp(-3*b*x-3*a)/b+1/4*exp(-b*x-a)/b+3/8*exp(b*x+a)/b-1/12*exp(3*b*x+ 3*a)/b+1/80*exp(5*b*x+5*a)/b
Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.75 \[ \int e^{a+b x} \sinh ^4(a+b x) \, dx=\frac {e^{-3 (a+b x)} \left (-5+60 e^{2 (a+b x)}+90 e^{4 (a+b x)}-20 e^{6 (a+b x)}+3 e^{8 (a+b x)}\right )}{240 b} \] Input:
Integrate[E^(a + b*x)*Sinh[a + b*x]^4,x]
Output:
(-5 + 60*E^(2*(a + b*x)) + 90*E^(4*(a + b*x)) - 20*E^(6*(a + b*x)) + 3*E^( 8*(a + b*x)))/(240*b*E^(3*(a + b*x)))
Time = 0.22 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.86, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2720, 27, 244, 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.
\(\displaystyle \int e^{a+b x} \sinh ^4(a+b x) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int \frac {1}{16} e^{-4 a-4 b x} \left (1-e^{2 a+2 b x}\right )^4de^{a+b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int e^{-4 a-4 b x} \left (1-e^{2 a+2 b x}\right )^4de^{a+b x}}{16 b}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\int \left (6+e^{-4 a-4 b x}-4 e^{-2 a-2 b x}-4 e^{2 a+2 b x}+e^{4 a+4 b x}\right )de^{a+b x}}{16 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{3} e^{-3 a-3 b x}+4 e^{-a-b x}+6 e^{a+b x}-\frac {4}{3} e^{3 a+3 b x}+\frac {1}{5} e^{5 a+5 b x}}{16 b}\) |
Input:
Int[E^(a + b*x)*Sinh[a + b*x]^4,x]
Output:
(-1/3*E^(-3*a - 3*b*x) + 4*E^(-a - b*x) + 6*E^(a + b*x) - (4*E^(3*a + 3*b* x))/3 + E^(5*a + 5*b*x)/5)/(16*b)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
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]]]
Time = 1.36 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.54
method | result | size |
derivativedivides | \(\frac {\left (\frac {8}{15}+\frac {\sinh \left (b x +a \right )^{4}}{5}-\frac {4 \sinh \left (b x +a \right )^{2}}{15}\right ) \cosh \left (b x +a \right )+\frac {\sinh \left (b x +a \right )^{5}}{5}}{b}\) | \(45\) |
default | \(\frac {\left (\frac {8}{15}+\frac {\sinh \left (b x +a \right )^{4}}{5}-\frac {4 \sinh \left (b x +a \right )^{2}}{15}\right ) \cosh \left (b x +a \right )+\frac {\sinh \left (b x +a \right )^{5}}{5}}{b}\) | \(45\) |
risch | \(-\frac {{\mathrm e}^{-3 b x -3 a}}{48 b}+\frac {{\mathrm e}^{-b x -a}}{4 b}+\frac {3 \,{\mathrm e}^{b x +a}}{8 b}-\frac {{\mathrm e}^{3 b x +3 a}}{12 b}+\frac {{\mathrm e}^{5 b x +5 a}}{80 b}\) | \(69\) |
parallelrisch | \(\frac {16 \,{\mathrm e}^{b x +a} \left (6 \tanh \left (\frac {b x}{2}+\frac {a}{2}\right )^{3}-2 \tanh \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-2 \tanh \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}{15 b \left (-1+\tanh \left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{4} \left (1+\tanh \left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{4}}\) | \(77\) |
orering | \(\frac {{\mathrm e}^{b x +a} \sinh \left (b x +a \right )^{4}}{5 b}+\frac {\frac {10 b \,{\mathrm e}^{b x +a} \sinh \left (b x +a \right )^{4}}{9}+\frac {40 \,{\mathrm e}^{b x +a} \sinh \left (b x +a \right )^{3} b \cosh \left (b x +a \right )}{9}}{b^{2}}-\frac {2 \left (5 b^{2} {\mathrm e}^{b x +a} \sinh \left (b x +a \right )^{4}+8 b^{2} {\mathrm e}^{b x +a} \sinh \left (b x +a \right )^{3} \cosh \left (b x +a \right )+12 \,{\mathrm e}^{b x +a} \sinh \left (b x +a \right )^{2} b^{2} \cosh \left (b x +a \right )^{2}\right )}{9 b^{3}}-\frac {13 b^{3} {\mathrm e}^{b x +a} \sinh \left (b x +a \right )^{4}+52 b^{3} {\mathrm e}^{b x +a} \sinh \left (b x +a \right )^{3} \cosh \left (b x +a \right )+36 b^{3} {\mathrm e}^{b x +a} \sinh \left (b x +a \right )^{2} \cosh \left (b x +a \right )^{2}+24 \,{\mathrm e}^{b x +a} \sinh \left (b x +a \right ) b^{3} \cosh \left (b x +a \right )^{3}}{9 b^{4}}+\frac {65 b^{4} {\mathrm e}^{b x +a} \sinh \left (b x +a \right )^{4}+176 b^{4} {\mathrm e}^{b x +a} \sinh \left (b x +a \right )^{3} \cosh \left (b x +a \right )+264 b^{4} {\mathrm e}^{b x +a} \sinh \left (b x +a \right )^{2} \cosh \left (b x +a \right )^{2}+96 b^{4} {\mathrm e}^{b x +a} \sinh \left (b x +a \right ) \cosh \left (b x +a \right )^{3}+24 \,{\mathrm e}^{b x +a} b^{4} \cosh \left (b x +a \right )^{4}}{45 b^{5}}\) | \(366\) |
Input:
int(exp(b*x+a)*sinh(b*x+a)^4,x,method=_RETURNVERBOSE)
Output:
1/b*((8/15+1/5*sinh(b*x+a)^4-4/15*sinh(b*x+a)^2)*cosh(b*x+a)+1/5*sinh(b*x+ a)^5)
Time = 0.11 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.36 \[ \int e^{a+b x} \sinh ^4(a+b x) \, dx=-\frac {\cosh \left (b x + a\right )^{4} - 16 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} - 10\right )} \sinh \left (b x + a\right )^{2} - 20 \, \cosh \left (b x + a\right )^{2} - 16 \, {\left (\cosh \left (b x + a\right )^{3} - 5 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) - 45}{120 \, {\left (b \cosh \left (b x + a\right ) - b \sinh \left (b x + a\right )\right )}} \] Input:
integrate(exp(b*x+a)*sinh(b*x+a)^4,x, algorithm="fricas")
Output:
-1/120*(cosh(b*x + a)^4 - 16*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a) ^4 + 2*(3*cosh(b*x + a)^2 - 10)*sinh(b*x + a)^2 - 20*cosh(b*x + a)^2 - 16* (cosh(b*x + a)^3 - 5*cosh(b*x + a))*sinh(b*x + a) - 45)/(b*cosh(b*x + a) - b*sinh(b*x + a))
Leaf count of result is larger than twice the leaf count of optimal. 139 vs. \(2 (65) = 130\).
Time = 2.37 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.67 \[ \int e^{a+b x} \sinh ^4(a+b x) \, dx=\begin {cases} \frac {e^{a} e^{b x} \sinh ^{4}{\left (a + b x \right )}}{5 b} + \frac {4 e^{a} e^{b x} \sinh ^{3}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{5 b} - \frac {4 e^{a} e^{b x} \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{5 b} - \frac {8 e^{a} e^{b x} \sinh {\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{15 b} + \frac {8 e^{a} e^{b x} \cosh ^{4}{\left (a + b x \right )}}{15 b} & \text {for}\: b \neq 0 \\x e^{a} \sinh ^{4}{\left (a \right )} & \text {otherwise} \end {cases} \] Input:
integrate(exp(b*x+a)*sinh(b*x+a)**4,x)
Output:
Piecewise((exp(a)*exp(b*x)*sinh(a + b*x)**4/(5*b) + 4*exp(a)*exp(b*x)*sinh (a + b*x)**3*cosh(a + b*x)/(5*b) - 4*exp(a)*exp(b*x)*sinh(a + b*x)**2*cosh (a + b*x)**2/(5*b) - 8*exp(a)*exp(b*x)*sinh(a + b*x)*cosh(a + b*x)**3/(15* b) + 8*exp(a)*exp(b*x)*cosh(a + b*x)**4/(15*b), Ne(b, 0)), (x*exp(a)*sinh( a)**4, True))
Time = 0.04 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.82 \[ \int e^{a+b x} \sinh ^4(a+b x) \, dx=\frac {e^{\left (5 \, b x + 5 \, a\right )}}{80 \, b} - \frac {e^{\left (3 \, b x + 3 \, a\right )}}{12 \, b} + \frac {3 \, e^{\left (b x + a\right )}}{8 \, b} + \frac {e^{\left (-b x - a\right )}}{4 \, b} - \frac {e^{\left (-3 \, b x - 3 \, a\right )}}{48 \, b} \] Input:
integrate(exp(b*x+a)*sinh(b*x+a)^4,x, algorithm="maxima")
Output:
1/80*e^(5*b*x + 5*a)/b - 1/12*e^(3*b*x + 3*a)/b + 3/8*e^(b*x + a)/b + 1/4* e^(-b*x - a)/b - 1/48*e^(-3*b*x - 3*a)/b
Time = 0.12 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.72 \[ \int e^{a+b x} \sinh ^4(a+b x) \, dx=\frac {5 \, {\left (12 \, e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-3 \, b x - 3 \, a\right )} + 3 \, e^{\left (5 \, b x + 5 \, a\right )} - 20 \, e^{\left (3 \, b x + 3 \, a\right )} + 90 \, e^{\left (b x + a\right )}}{240 \, b} \] Input:
integrate(exp(b*x+a)*sinh(b*x+a)^4,x, algorithm="giac")
Output:
1/240*(5*(12*e^(2*b*x + 2*a) - 1)*e^(-3*b*x - 3*a) + 3*e^(5*b*x + 5*a) - 2 0*e^(3*b*x + 3*a) + 90*e^(b*x + a))/b
Time = 0.53 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.70 \[ \int e^{a+b x} \sinh ^4(a+b x) \, dx=\frac {90\,{\mathrm {e}}^{a+b\,x}+60\,{\mathrm {e}}^{-a-b\,x}-5\,{\mathrm {e}}^{-3\,a-3\,b\,x}-20\,{\mathrm {e}}^{3\,a+3\,b\,x}+3\,{\mathrm {e}}^{5\,a+5\,b\,x}}{240\,b} \] Input:
int(exp(a + b*x)*sinh(a + b*x)^4,x)
Output:
(90*exp(a + b*x) + 60*exp(- a - b*x) - 5*exp(- 3*a - 3*b*x) - 20*exp(3*a + 3*b*x) + 3*exp(5*a + 5*b*x))/(240*b)
Time = 0.15 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.81 \[ \int e^{a+b x} \sinh ^4(a+b x) \, dx=\frac {3 e^{8 b x +8 a}-20 e^{6 b x +6 a}+90 e^{4 b x +4 a}+60 e^{2 b x +2 a}-5}{240 e^{3 b x +3 a} b} \] Input:
int(exp(b*x+a)*sinh(b*x+a)^4,x)
Output:
(3*e**(8*a + 8*b*x) - 20*e**(6*a + 6*b*x) + 90*e**(4*a + 4*b*x) + 60*e**(2 *a + 2*b*x) - 5)/(240*e**(3*a + 3*b*x)*b)