Integrand size = 20, antiderivative size = 113 \[ \int e^{\frac {5}{3} (a+b x)} \sinh ^3(d+b x) \, dx=\frac {3 e^{\frac {5 (a-d)}{3}-\frac {4}{3} (d+b x)}}{32 b}+\frac {9 e^{\frac {5 (a-d)}{3}+\frac {2}{3} (d+b x)}}{16 b}-\frac {9 e^{\frac {5 (a-d)}{3}+\frac {8}{3} (d+b x)}}{64 b}+\frac {3 e^{\frac {5 (a-d)}{3}+\frac {14}{3} (d+b x)}}{112 b} \] Output:
3/32*exp(5/3*a-3*d-4/3*b*x)/b+9/16*exp(5/3*a-d+2/3*b*x)/b-9/64*exp(5/3*a+d +8/3*b*x)/b+3/112*exp(5/3*a+3*d+14/3*b*x)/b
Time = 0.17 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.83 \[ \int e^{\frac {5}{3} (a+b x)} \sinh ^3(d+b x) \, dx=\frac {3 e^{\frac {5 a}{3}-\frac {4 b x}{3}} \left (-21 e^{2 b x} \left (-4+e^{2 b x}\right ) \cosh (d)+2 \left (7+2 e^{6 b x}\right ) \cosh (3 d)-84 e^{2 b x} \sinh (d)-21 e^{4 b x} \sinh (d)-14 \sinh (3 d)+4 e^{6 b x} \sinh (3 d)\right )}{448 b} \] Input:
Integrate[E^((5*(a + b*x))/3)*Sinh[d + b*x]^3,x]
Output:
(3*E^((5*a)/3 - (4*b*x)/3)*(-21*E^(2*b*x)*(-4 + E^(2*b*x))*Cosh[d] + 2*(7 + 2*E^(6*b*x))*Cosh[3*d] - 84*E^(2*b*x)*Sinh[d] - 21*E^(4*b*x)*Sinh[d] - 1 4*Sinh[3*d] + 4*E^(6*b*x)*Sinh[3*d]))/(448*b)
Time = 0.22 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.56, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2720, 27, 802, 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.
\(\displaystyle \int e^{\frac {5}{3} (a+b x)} \sinh ^3(b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {3 \int -\frac {1}{8} e^{\frac {5 a}{3}-\frac {5 b x}{3}} \left (1-e^{2 b x}\right )^3de^{\frac {b x}{3}}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {3 e^{5 a/3} \int e^{-\frac {5 b x}{3}} \left (1-e^{2 b x}\right )^3de^{\frac {b x}{3}}}{8 b}\) |
\(\Big \downarrow \) 802 |
\(\displaystyle -\frac {3 e^{5 a/3} \int \left (e^{-\frac {5 b x}{3}}-3 e^{\frac {b x}{3}}+3 e^{\frac {7 b x}{3}}-e^{\frac {13 b x}{3}}\right )de^{\frac {b x}{3}}}{8 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3 e^{5 a/3} \left (-\frac {1}{4} e^{-\frac {4 b x}{3}}-\frac {3}{2} e^{\frac {2 b x}{3}}+\frac {3}{8} e^{\frac {8 b x}{3}}-\frac {1}{14} e^{\frac {14 b x}{3}}\right )}{8 b}\) |
Input:
Int[E^((5*(a + b*x))/3)*Sinh[d + b*x]^3,x]
Output:
(-3*E^((5*a)/3)*(-1/4*1/E^((4*b*x)/3) - (3*E^((2*b*x)/3))/2 + (3*E^((8*b*x )/3))/8 - E^((14*b*x)/3)/14))/(8*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_)^(n_))^(p_.), x_Symbol] :> Int[Exp andIntegrand[(c*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n}, 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 = 0.78 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.48
method | result | size |
parallelrisch | \(-\frac {3 \,{\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}} \left (-18 \cosh \left (3 b x +3 d \right )-63 \cosh \left (b x +d \right )+10 \sinh \left (3 b x +3 d \right )+105 \sinh \left (b x +d \right )\right )}{448 b}\) | \(54\) |
risch | \(\frac {3 \,{\mathrm e}^{\frac {5 a}{3}-3 d -\frac {4 b x}{3}}}{32 b}+\frac {9 \,{\mathrm e}^{\frac {5 a}{3}-d +\frac {2 b x}{3}}}{16 b}-\frac {9 \,{\mathrm e}^{\frac {5 a}{3}+d +\frac {8 b x}{3}}}{64 b}+\frac {3 \,{\mathrm e}^{\frac {5 a}{3}+3 d +\frac {14 b x}{3}}}{112 b}\) | \(68\) |
default | \(\frac {3 \sinh \left (\frac {5 a}{3}-3 d -\frac {4 b x}{3}\right )}{32 b}+\frac {9 \sinh \left (\frac {5 a}{3}-d +\frac {2 b x}{3}\right )}{16 b}-\frac {9 \sinh \left (\frac {5 a}{3}+d +\frac {8 b x}{3}\right )}{64 b}+\frac {3 \sinh \left (\frac {5 a}{3}+3 d +\frac {14 b x}{3}\right )}{112 b}+\frac {3 \cosh \left (\frac {5 a}{3}-3 d -\frac {4 b x}{3}\right )}{32 b}+\frac {9 \cosh \left (\frac {5 a}{3}-d +\frac {2 b x}{3}\right )}{16 b}-\frac {9 \cosh \left (\frac {5 a}{3}+d +\frac {8 b x}{3}\right )}{64 b}+\frac {3 \cosh \left (\frac {5 a}{3}+3 d +\frac {14 b x}{3}\right )}{112 b}\) | \(134\) |
orering | \(\frac {75 \,{\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}} \sinh \left (b x +d \right )^{3}}{56 b}+\frac {\frac {225 b \,{\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}} \sinh \left (b x +d \right )^{3}}{224}+\frac {405 \,{\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}} \sinh \left (b x +d \right )^{2} b \cosh \left (b x +d \right )}{224}}{b^{2}}-\frac {135 \left (\frac {52 b^{2} {\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}} \sinh \left (b x +d \right )^{3}}{9}+10 b^{2} {\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}} \sinh \left (b x +d \right )^{2} \cosh \left (b x +d \right )+6 \,{\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}} \sinh \left (b x +d \right ) b^{2} \cosh \left (b x +d \right )^{2}\right )}{224 b^{3}}+\frac {\frac {795 b^{3} {\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}} \sinh \left (b x +d \right )^{3}}{448}+\frac {1863 b^{3} {\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}} \sinh \left (b x +d \right )^{2} \cosh \left (b x +d \right )}{448}+\frac {1215 b^{3} {\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}} \sinh \left (b x +d \right ) \cosh \left (b x +d \right )^{2}}{448}+\frac {243 \,{\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}} b^{3} \cosh \left (b x +d \right )^{3}}{448}}{b^{4}}\) | \(266\) |
Input:
int(exp(5/3*b*x+5/3*a)*sinh(b*x+d)^3,x,method=_RETURNVERBOSE)
Output:
-3/448*exp(5/3*b*x+5/3*a)/b*(-18*cosh(3*b*x+3*d)-63*cosh(b*x+d)+10*sinh(3* b*x+3*d)+105*sinh(b*x+d))
Leaf count of result is larger than twice the leaf count of optimal. 759 vs. \(2 (67) = 134\).
Time = 0.10 (sec) , antiderivative size = 759, normalized size of antiderivative = 6.72 \[ \int e^{\frac {5}{3} (a+b x)} \sinh ^3(d+b x) \, dx=\text {Too large to display} \] Input:
integrate(exp(5/3*b*x+5/3*a)*sinh(b*x+d)^3,x, algorithm="fricas")
Output:
3/448*(18*cosh(1/3*b*x + 1/3*d)^9*cosh(-5/3*a + 5/3*d) - 10*(cosh(-5/3*a + 5/3*d) - sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d)^9 + 162*(cosh(1/3*b* x + 1/3*d)*cosh(-5/3*a + 5/3*d) - cosh(1/3*b*x + 1/3*d)*sinh(-5/3*a + 5/3* d))*sinh(1/3*b*x + 1/3*d)^8 - 360*(cosh(1/3*b*x + 1/3*d)^2*cosh(-5/3*a + 5 /3*d) - cosh(1/3*b*x + 1/3*d)^2*sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d )^7 + 1512*(cosh(1/3*b*x + 1/3*d)^3*cosh(-5/3*a + 5/3*d) - cosh(1/3*b*x + 1/3*d)^3*sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d)^6 - 1260*(cosh(1/3*b* x + 1/3*d)^4*cosh(-5/3*a + 5/3*d) - cosh(1/3*b*x + 1/3*d)^4*sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d)^5 + 2268*(cosh(1/3*b*x + 1/3*d)^5*cosh(-5/3* a + 5/3*d) - cosh(1/3*b*x + 1/3*d)^5*sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d)^4 + 63*cosh(1/3*b*x + 1/3*d)^3*cosh(-5/3*a + 5/3*d) - 105*(8*cosh(1 /3*b*x + 1/3*d)^6*cosh(-5/3*a + 5/3*d) - (8*cosh(1/3*b*x + 1/3*d)^6 + 1)*s inh(-5/3*a + 5/3*d) + cosh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d)^3 + 27*( 24*cosh(1/3*b*x + 1/3*d)^7*cosh(-5/3*a + 5/3*d) + 7*cosh(1/3*b*x + 1/3*d)* cosh(-5/3*a + 5/3*d) - (24*cosh(1/3*b*x + 1/3*d)^7 + 7*cosh(1/3*b*x + 1/3* d))*sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d)^2 - 45*(2*cosh(1/3*b*x + 1 /3*d)^8*cosh(-5/3*a + 5/3*d) + 7*cosh(1/3*b*x + 1/3*d)^2*cosh(-5/3*a + 5/3 *d) - (2*cosh(1/3*b*x + 1/3*d)^8 + 7*cosh(1/3*b*x + 1/3*d)^2)*sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d) - 9*(2*cosh(1/3*b*x + 1/3*d)^9 + 7*cosh(1/ 3*b*x + 1/3*d)^3)*sinh(-5/3*a + 5/3*d))/(b*cosh(1/3*b*x + 1/3*d)^5 - 5*...
Time = 0.92 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.25 \[ \int e^{\frac {5}{3} (a+b x)} \sinh ^3(d+b x) \, dx=\begin {cases} \frac {285 e^{\frac {5 a}{3}} e^{\frac {5 b x}{3}} \sinh ^{3}{\left (b x + d \right )}}{448 b} - \frac {27 e^{\frac {5 a}{3}} e^{\frac {5 b x}{3}} \sinh ^{2}{\left (b x + d \right )} \cosh {\left (b x + d \right )}}{448 b} - \frac {405 e^{\frac {5 a}{3}} e^{\frac {5 b x}{3}} \sinh {\left (b x + d \right )} \cosh ^{2}{\left (b x + d \right )}}{448 b} + \frac {243 e^{\frac {5 a}{3}} e^{\frac {5 b x}{3}} \cosh ^{3}{\left (b x + d \right )}}{448 b} & \text {for}\: b \neq 0 \\x e^{\frac {5 a}{3}} \sinh ^{3}{\left (d \right )} & \text {otherwise} \end {cases} \] Input:
integrate(exp(5/3*b*x+5/3*a)*sinh(b*x+d)**3,x)
Output:
Piecewise((285*exp(5*a/3)*exp(5*b*x/3)*sinh(b*x + d)**3/(448*b) - 27*exp(5 *a/3)*exp(5*b*x/3)*sinh(b*x + d)**2*cosh(b*x + d)/(448*b) - 405*exp(5*a/3) *exp(5*b*x/3)*sinh(b*x + d)*cosh(b*x + d)**2/(448*b) + 243*exp(5*a/3)*exp( 5*b*x/3)*cosh(b*x + d)**3/(448*b), Ne(b, 0)), (x*exp(5*a/3)*sinh(d)**3, Tr ue))
Time = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.52 \[ \int e^{\frac {5}{3} (a+b x)} \sinh ^3(d+b x) \, dx=-\frac {3 \, {\left (21 \, e^{\left (-2 \, b x - 2 \, d\right )} - 84 \, e^{\left (-4 \, b x - 4 \, d\right )} - 4\right )} e^{\left (\frac {14}{3} \, b x + \frac {5}{3} \, a + 3 \, d\right )}}{448 \, b} + \frac {3 \, e^{\left (-\frac {4}{3} \, b x + \frac {5}{3} \, a - 3 \, d\right )}}{32 \, b} \] Input:
integrate(exp(5/3*b*x+5/3*a)*sinh(b*x+d)^3,x, algorithm="maxima")
Output:
-3/448*(21*e^(-2*b*x - 2*d) - 84*e^(-4*b*x - 4*d) - 4)*e^(14/3*b*x + 5/3*a + 3*d)/b + 3/32*e^(-4/3*b*x + 5/3*a - 3*d)/b
Time = 0.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.56 \[ \int e^{\frac {5}{3} (a+b x)} \sinh ^3(d+b x) \, dx=\frac {3 \, {\left (4 \, e^{\left (\frac {14}{3} \, b x + \frac {5}{3} \, a + 6 \, d\right )} - 21 \, e^{\left (\frac {8}{3} \, b x + \frac {5}{3} \, a + 4 \, d\right )} + 84 \, e^{\left (\frac {2}{3} \, b x + \frac {5}{3} \, a + 2 \, d\right )} + 14 \, e^{\left (-\frac {4}{3} \, b x + \frac {5}{3} \, a\right )}\right )} e^{\left (-3 \, d\right )}}{448 \, b} \] Input:
integrate(exp(5/3*b*x+5/3*a)*sinh(b*x+d)^3,x, algorithm="giac")
Output:
3/448*(4*e^(14/3*b*x + 5/3*a + 6*d) - 21*e^(8/3*b*x + 5/3*a + 4*d) + 84*e^ (2/3*b*x + 5/3*a + 2*d) + 14*e^(-4/3*b*x + 5/3*a))*e^(-3*d)/b
Time = 0.33 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.81 \[ \int e^{\frac {5}{3} (a+b x)} \sinh ^3(d+b x) \, dx=\frac {3\,{\mathrm {e}}^{\frac {5\,a}{3}}\,{\mathrm {e}}^{-3\,d}\,{\mathrm {e}}^{-3\,b\,x}\,{\mathrm {e}}^{\frac {5\,b\,x}{3}}}{32\,b}-\frac {9\,{\mathrm {e}}^{\frac {5\,a}{3}}\,{\mathrm {e}}^{-3\,b\,x}\,{\mathrm {e}}^{\frac {17\,b\,x}{3}}\,{\mathrm {e}}^d}{64\,b}+\frac {9\,{\mathrm {e}}^{\frac {5\,a}{3}}\,{\mathrm {e}}^{-d}\,{\mathrm {e}}^{-3\,b\,x}\,{\mathrm {e}}^{\frac {11\,b\,x}{3}}}{16\,b}+\frac {3\,{\mathrm {e}}^{\frac {5\,a}{3}}\,{\mathrm {e}}^{3\,d}\,{\mathrm {e}}^{-3\,b\,x}\,{\mathrm {e}}^{\frac {23\,b\,x}{3}}}{112\,b} \] Input:
int(exp((5*a)/3 + (5*b*x)/3)*sinh(d + b*x)^3,x)
Output:
(3*exp((5*a)/3)*exp(-3*d)*exp(-3*b*x)*exp((5*b*x)/3))/(32*b) - (9*exp((5*a )/3)*exp(-3*b*x)*exp((17*b*x)/3)*exp(d))/(64*b) + (9*exp((5*a)/3)*exp(-d)* exp(-3*b*x)*exp((11*b*x)/3))/(16*b) + (3*exp((5*a)/3)*exp(3*d)*exp(-3*b*x) *exp((23*b*x)/3))/(112*b)
Time = 0.24 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.58 \[ \int e^{\frac {5}{3} (a+b x)} \sinh ^3(d+b x) \, dx=\frac {3 e^{\frac {5 b x}{3}+\frac {5 a}{3}} \left (4 e^{6 b x +6 d}-21 e^{4 b x +4 d}+84 e^{2 b x +2 d}+14\right )}{448 e^{3 b x +3 d} b} \] Input:
int(exp(5/3*b*x+5/3*a)*sinh(b*x+d)^3,x)
Output:
(3*e**((5*a + 5*b*x)/3)*(4*e**(6*b*x + 6*d) - 21*e**(4*b*x + 4*d) + 84*e** (2*b*x + 2*d) + 14))/(448*e**(3*b*x + 3*d)*b)