Integrand size = 31, antiderivative size = 109 \[ \int e^{\frac {5}{3} (a+b x)} (g \cosh (d+b x)+f \sinh (d+b x))^2 \, dx=-\frac {3 e^{\frac {5 (a-d)}{3}+\frac {1}{3} (-d-b x)} (f-g)^2}{4 b}+\frac {3 e^{\frac {5 (a-d)}{3}+\frac {11}{3} (d+b x)} (f+g)^2}{44 b}-\frac {3 e^{\frac {5 (a-d)}{3}+\frac {5}{3} (d+b x)} \left (f^2-g^2\right )}{10 b} \] Output:
-3/4*exp(5/3*a-2*d-1/3*b*x)*(f-g)^2/b+3/44*exp(5/3*a+2*d+11/3*b*x)*(f+g)^2 /b-3/10*exp(5/3*b*x+5/3*a)*(f^2-g^2)/b
Time = 0.66 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.63 \[ \int e^{\frac {5}{3} (a+b x)} (g \cosh (d+b x)+f \sinh (d+b x))^2 \, dx=\frac {3 e^{\frac {5 a}{3}-2 d-\frac {b x}{3}} \left (-55 (f-g)^2-22 e^{2 (d+b x)} (f-g) (f+g)+5 e^{4 (d+b x)} (f+g)^2\right )}{220 b} \] Input:
Integrate[E^((5*(a + b*x))/3)*(g*Cosh[d + b*x] + f*Sinh[d + b*x])^2,x]
Output:
(3*E^((5*a)/3 - 2*d - (b*x)/3)*(-55*(f - g)^2 - 22*E^(2*(d + b*x))*(f - g) *(f + g) + 5*E^(4*(d + b*x))*(f + g)^2))/(220*b)
Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.64, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, 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.
| \(\displaystyle \int e^{\frac {5}{3} (a+b x)} (f \sinh (b x+d)+g \cosh (b x+d))^2 \, dx\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {3 \int \frac {1}{4} e^{\frac {5 a}{3}-\frac {2 b x}{3}} \left (f-g-e^{2 b x} (f+g)\right )^2de^{\frac {b x}{3}}}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3 e^{5 a/3} \int e^{-\frac {2 b x}{3}} \left (f-g-e^{2 b x} (f+g)\right )^2de^{\frac {b x}{3}}}{4 b}\) |
| \(\Big \downarrow \) 802 |
| \(\displaystyle \frac {3 e^{5 a/3} \int \left (e^{-\frac {2 b x}{3}} (f-g)^2+e^{\frac {10 b x}{3}} (f+g)^2-2 e^{\frac {4 b x}{3}} \left (f^2-g^2\right )\right )de^{\frac {b x}{3}}}{4 b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3 e^{5 a/3} \left (-\frac {2}{5} e^{\frac {5 b x}{3}} \left (f^2-g^2\right )-e^{-\frac {b x}{3}} (f-g)^2+\frac {1}{11} e^{\frac {11 b x}{3}} (f+g)^2\right )}{4 b}\) |
Input:
Int[E^((5*(a + b*x))/3)*(g*Cosh[d + b*x] + f*Sinh[d + b*x])^2,x]
Output:
(3*E^((5*a)/3)*(-((f - g)^2/E^((b*x)/3)) + (E^((11*b*x)/3)*(f + g)^2)/11 - (2*E^((5*b*x)/3)*(f^2 - g^2))/5))/(4*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]]]
Leaf count of result is larger than twice the leaf count of optimal. \(153\) vs. \(2(70)=140\).
Time = 28.95 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.41
| method | result | size |
| risch | \(\frac {3 \,{\mathrm e}^{\frac {5 a}{3}+2 d +\frac {11 b x}{3}} f^{2}}{44 b}+\frac {3 \,{\mathrm e}^{\frac {5 a}{3}+2 d +\frac {11 b x}{3}} f g}{22 b}+\frac {3 \,{\mathrm e}^{\frac {5 a}{3}+2 d +\frac {11 b x}{3}} g^{2}}{44 b}-\frac {3 \,{\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}} f^{2}}{10 b}+\frac {3 \,{\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}} g^{2}}{10 b}-\frac {3 \,{\mathrm e}^{\frac {5 a}{3}-2 d -\frac {b x}{3}} f^{2}}{4 b}+\frac {3 \,{\mathrm e}^{\frac {5 a}{3}-2 d -\frac {b x}{3}} f g}{2 b}-\frac {3 \,{\mathrm e}^{\frac {5 a}{3}-2 d -\frac {b x}{3}} g^{2}}{4 b}\) | \(154\) |
| default | \(\frac {3 \left (\frac {g^{2}}{2}-\frac {f^{2}}{2}\right ) \sinh \left (\frac {5 b x}{3}+\frac {5 a}{3}\right )}{5 b}+\frac {3 \left (\frac {g^{2}}{2}-\frac {f^{2}}{2}\right ) \cosh \left (\frac {5 b x}{3}+\frac {5 a}{3}\right )}{5 b}-\frac {3 \left (\frac {1}{4} g^{2}+\frac {1}{4} f^{2}-\frac {1}{2} f g \right ) \sinh \left (\frac {5 a}{3}-2 d -\frac {b x}{3}\right )}{b}+\frac {3 \left (-\frac {1}{4} f^{2}+\frac {1}{2} f g -\frac {1}{4} g^{2}\right ) \cosh \left (\frac {5 a}{3}-2 d -\frac {b x}{3}\right )}{b}+\frac {3 \left (\frac {1}{4} g^{2}+\frac {1}{4} f^{2}+\frac {1}{2} f g \right ) \sinh \left (\frac {5 a}{3}+2 d +\frac {11 b x}{3}\right )}{11 b}+\frac {3 \left (\frac {1}{4} g^{2}+\frac {1}{4} f^{2}+\frac {1}{2} f g \right ) \cosh \left (\frac {5 a}{3}+2 d +\frac {11 b x}{3}\right )}{11 b}\) | \(180\) |
| parts | \(f^{2} \left (-\frac {3 \sinh \left (\frac {5 b x}{3}+\frac {5 a}{3}\right )}{10 b}-\frac {3 \sinh \left (\frac {5 a}{3}-2 d -\frac {b x}{3}\right )}{4 b}+\frac {3 \sinh \left (\frac {5 a}{3}+2 d +\frac {11 b x}{3}\right )}{44 b}-\frac {3 \cosh \left (\frac {5 b x}{3}+\frac {5 a}{3}\right )}{10 b}-\frac {3 \cosh \left (\frac {5 a}{3}-2 d -\frac {b x}{3}\right )}{4 b}+\frac {3 \cosh \left (\frac {5 a}{3}+2 d +\frac {11 b x}{3}\right )}{44 b}\right )+g^{2} \left (\frac {3 \sinh \left (\frac {5 b x}{3}+\frac {5 a}{3}\right )}{10 b}-\frac {3 \sinh \left (\frac {5 a}{3}-2 d -\frac {b x}{3}\right )}{4 b}+\frac {3 \sinh \left (\frac {5 a}{3}+2 d +\frac {11 b x}{3}\right )}{44 b}+\frac {3 \cosh \left (\frac {5 b x}{3}+\frac {5 a}{3}\right )}{10 b}-\frac {3 \cosh \left (\frac {5 a}{3}-2 d -\frac {b x}{3}\right )}{4 b}+\frac {3 \cosh \left (\frac {5 a}{3}+2 d +\frac {11 b x}{3}\right )}{44 b}\right )+2 f g \left (\frac {3 \sinh \left (\frac {5 a}{3}-2 d -\frac {b x}{3}\right )}{4 b}+\frac {3 \sinh \left (\frac {5 a}{3}+2 d +\frac {11 b x}{3}\right )}{44 b}+\frac {3 \cosh \left (\frac {5 a}{3}-2 d -\frac {b x}{3}\right )}{4 b}+\frac {3 \cosh \left (\frac {5 a}{3}+2 d +\frac {11 b x}{3}\right )}{44 b}\right )\) | \(277\) |
| orering | \(-\frac {117 \,{\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}} \left (g \cosh \left (b x +d \right )+f \sinh \left (b x +d \right )\right )^{2}}{55 b}+\frac {\frac {45 b \,{\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}} \left (g \cosh \left (b x +d \right )+f \sinh \left (b x +d \right )\right )^{2}}{11}+\frac {54 \,{\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}} \left (g \cosh \left (b x +d \right )+f \sinh \left (b x +d \right )\right ) \left (g b \sinh \left (b x +d \right )+f b \cosh \left (b x +d \right )\right )}{11}}{b^{2}}-\frac {27 \left (\frac {25 b^{2} {\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}} \left (g \cosh \left (b x +d \right )+f \sinh \left (b x +d \right )\right )^{2}}{9}+\frac {20 b \,{\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}} \left (g \cosh \left (b x +d \right )+f \sinh \left (b x +d \right )\right ) \left (g b \sinh \left (b x +d \right )+f b \cosh \left (b x +d \right )\right )}{3}+2 \,{\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}} \left (g b \sinh \left (b x +d \right )+f b \cosh \left (b x +d \right )\right )^{2}+2 \,{\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}} \left (g \cosh \left (b x +d \right )+f \sinh \left (b x +d \right )\right ) \left (g \,b^{2} \cosh \left (b x +d \right )+f \,b^{2} \sinh \left (b x +d \right )\right )\right )}{55 b^{3}}\) | \(289\) |
Input:
int(exp(5/3*b*x+5/3*a)*(g*cosh(b*x+d)+f*sinh(b*x+d))^2,x,method=_RETURNVER BOSE)
Output:
3/44/b*exp(5/3*a+2*d+11/3*b*x)*f^2+3/22/b*exp(5/3*a+2*d+11/3*b*x)*f*g+3/44 /b*exp(5/3*a+2*d+11/3*b*x)*g^2-3/10*exp(5/3*b*x+5/3*a)/b*f^2+3/10*exp(5/3* b*x+5/3*a)/b*g^2-3/4/b*exp(5/3*a-2*d-1/3*b*x)*f^2+3/2/b*exp(5/3*a-2*d-1/3* b*x)*f*g-3/4/b*exp(5/3*a-2*d-1/3*b*x)*g^2
Leaf count of result is larger than twice the leaf count of optimal. 715 vs. \(2 (70) = 140\).
Time = 0.09 (sec) , antiderivative size = 715, normalized size of antiderivative = 6.56 \[ \int e^{\frac {5}{3} (a+b x)} (g \cosh (d+b x)+f \sinh (d+b x))^2 \, dx=\text {Too large to display} \] Input:
integrate(exp(5/3*b*x+5/3*a)*(g*cosh(b*x+d)+f*sinh(b*x+d))^2,x, algorithm= "fricas")
Output:
-3/110*(5*(5*f^2 - 12*f*g + 5*g^2)*cosh(1/3*b*x + 1/3*d)^6*cosh(-5/3*a + 5 /3*d) + 5*((5*f^2 - 12*f*g + 5*g^2)*cosh(-5/3*a + 5/3*d) - (5*f^2 - 12*f*g + 5*g^2)*sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d)^6 - 60*((3*f^2 - 5*f *g + 3*g^2)*cosh(1/3*b*x + 1/3*d)*cosh(-5/3*a + 5/3*d) - (3*f^2 - 5*f*g + 3*g^2)*cosh(1/3*b*x + 1/3*d)*sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d)^5 + 75*((5*f^2 - 12*f*g + 5*g^2)*cosh(1/3*b*x + 1/3*d)^2*cosh(-5/3*a + 5/3* d) - (5*f^2 - 12*f*g + 5*g^2)*cosh(1/3*b*x + 1/3*d)^2*sinh(-5/3*a + 5/3*d) )*sinh(1/3*b*x + 1/3*d)^4 - 200*((3*f^2 - 5*f*g + 3*g^2)*cosh(1/3*b*x + 1/ 3*d)^3*cosh(-5/3*a + 5/3*d) - (3*f^2 - 5*f*g + 3*g^2)*cosh(1/3*b*x + 1/3*d )^3*sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d)^3 + 75*((5*f^2 - 12*f*g + 5*g^2)*cosh(1/3*b*x + 1/3*d)^4*cosh(-5/3*a + 5/3*d) - (5*f^2 - 12*f*g + 5* g^2)*cosh(1/3*b*x + 1/3*d)^4*sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d)^2 + 11*(f^2 - g^2)*cosh(-5/3*a + 5/3*d) - 60*((3*f^2 - 5*f*g + 3*g^2)*cosh( 1/3*b*x + 1/3*d)^5*cosh(-5/3*a + 5/3*d) - (3*f^2 - 5*f*g + 3*g^2)*cosh(1/3 *b*x + 1/3*d)^5*sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d) - (5*(5*f^2 - 12*f*g + 5*g^2)*cosh(1/3*b*x + 1/3*d)^6 + 11*f^2 - 11*g^2)*sinh(-5/3*a + 5 /3*d))/(b*cosh(1/3*b*x + 1/3*d)^5 - 5*b*cosh(1/3*b*x + 1/3*d)^4*sinh(1/3*b *x + 1/3*d) + 10*b*cosh(1/3*b*x + 1/3*d)^3*sinh(1/3*b*x + 1/3*d)^2 - 10*b* cosh(1/3*b*x + 1/3*d)^2*sinh(1/3*b*x + 1/3*d)^3 + 5*b*cosh(1/3*b*x + 1/3*d )*sinh(1/3*b*x + 1/3*d)^4 - b*sinh(1/3*b*x + 1/3*d)^5)
Leaf count of result is larger than twice the leaf count of optimal. 325 vs. \(2 (76) = 152\).
Time = 0.51 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.98 \[ \int e^{\frac {5}{3} (a+b x)} (g \cosh (d+b x)+f \sinh (d+b x))^2 \, dx=\begin {cases} - \frac {21 f^{2} e^{\frac {5 a}{3}} e^{\frac {5 b x}{3}} \sinh ^{2}{\left (b x + d \right )}}{55 b} + \frac {18 f^{2} e^{\frac {5 a}{3}} e^{\frac {5 b x}{3}} \sinh {\left (b x + d \right )} \cosh {\left (b x + d \right )}}{11 b} - \frac {54 f^{2} e^{\frac {5 a}{3}} e^{\frac {5 b x}{3}} \cosh ^{2}{\left (b x + d \right )}}{55 b} + \frac {18 f g e^{\frac {5 a}{3}} e^{\frac {5 b x}{3}} \sinh ^{2}{\left (b x + d \right )}}{11 b} - \frac {30 f g e^{\frac {5 a}{3}} e^{\frac {5 b x}{3}} \sinh {\left (b x + d \right )} \cosh {\left (b x + d \right )}}{11 b} + \frac {18 f g e^{\frac {5 a}{3}} e^{\frac {5 b x}{3}} \cosh ^{2}{\left (b x + d \right )}}{11 b} - \frac {54 g^{2} e^{\frac {5 a}{3}} e^{\frac {5 b x}{3}} \sinh ^{2}{\left (b x + d \right )}}{55 b} + \frac {18 g^{2} e^{\frac {5 a}{3}} e^{\frac {5 b x}{3}} \sinh {\left (b x + d \right )} \cosh {\left (b x + d \right )}}{11 b} - \frac {21 g^{2} e^{\frac {5 a}{3}} e^{\frac {5 b x}{3}} \cosh ^{2}{\left (b x + d \right )}}{55 b} & \text {for}\: b \neq 0 \\x \left (f \sinh {\left (d \right )} + g \cosh {\left (d \right )}\right )^{2} e^{\frac {5 a}{3}} & \text {otherwise} \end {cases} \] Input:
integrate(exp(5/3*b*x+5/3*a)*(g*cosh(b*x+d)+f*sinh(b*x+d))**2,x)
Output:
Piecewise((-21*f**2*exp(5*a/3)*exp(5*b*x/3)*sinh(b*x + d)**2/(55*b) + 18*f **2*exp(5*a/3)*exp(5*b*x/3)*sinh(b*x + d)*cosh(b*x + d)/(11*b) - 54*f**2*e xp(5*a/3)*exp(5*b*x/3)*cosh(b*x + d)**2/(55*b) + 18*f*g*exp(5*a/3)*exp(5*b *x/3)*sinh(b*x + d)**2/(11*b) - 30*f*g*exp(5*a/3)*exp(5*b*x/3)*sinh(b*x + d)*cosh(b*x + d)/(11*b) + 18*f*g*exp(5*a/3)*exp(5*b*x/3)*cosh(b*x + d)**2/ (11*b) - 54*g**2*exp(5*a/3)*exp(5*b*x/3)*sinh(b*x + d)**2/(55*b) + 18*g**2 *exp(5*a/3)*exp(5*b*x/3)*sinh(b*x + d)*cosh(b*x + d)/(11*b) - 21*g**2*exp( 5*a/3)*exp(5*b*x/3)*cosh(b*x + d)**2/(55*b), Ne(b, 0)), (x*(f*sinh(d) + g* cosh(d))**2*exp(5*a/3), True))
Leaf count of result is larger than twice the leaf count of optimal. 143 vs. \(2 (70) = 140\).
Time = 0.05 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.31 \[ \int e^{\frac {5}{3} (a+b x)} (g \cosh (d+b x)+f \sinh (d+b x))^2 \, dx=\frac {3}{220} \, g^{2} {\left (\frac {{\left (22 \, e^{\left (-2 \, b x - 2 \, d\right )} + 5\right )} e^{\left (\frac {11}{3} \, b x + \frac {5}{3} \, a + 2 \, d\right )}}{b} - \frac {55 \, e^{\left (-\frac {1}{3} \, b x + \frac {5}{3} \, a - 2 \, d\right )}}{b}\right )} - \frac {3}{220} \, f^{2} {\left (\frac {{\left (22 \, e^{\left (-2 \, b x - 2 \, d\right )} - 5\right )} e^{\left (\frac {11}{3} \, b x + \frac {5}{3} \, a + 2 \, d\right )}}{b} + \frac {55 \, e^{\left (-\frac {1}{3} \, b x + \frac {5}{3} \, a - 2 \, d\right )}}{b}\right )} + \frac {3}{22} \, f g {\left (\frac {e^{\left (\frac {11}{3} \, b x + \frac {5}{3} \, a + 2 \, d\right )}}{b} + \frac {11 \, e^{\left (-\frac {1}{3} \, b x + \frac {5}{3} \, a - 2 \, d\right )}}{b}\right )} \] Input:
integrate(exp(5/3*b*x+5/3*a)*(g*cosh(b*x+d)+f*sinh(b*x+d))^2,x, algorithm= "maxima")
Output:
3/220*g^2*((22*e^(-2*b*x - 2*d) + 5)*e^(11/3*b*x + 5/3*a + 2*d)/b - 55*e^( -1/3*b*x + 5/3*a - 2*d)/b) - 3/220*f^2*((22*e^(-2*b*x - 2*d) - 5)*e^(11/3* b*x + 5/3*a + 2*d)/b + 55*e^(-1/3*b*x + 5/3*a - 2*d)/b) + 3/22*f*g*(e^(11/ 3*b*x + 5/3*a + 2*d)/b + 11*e^(-1/3*b*x + 5/3*a - 2*d)/b)
Time = 0.11 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.16 \[ \int e^{\frac {5}{3} (a+b x)} (g \cosh (d+b x)+f \sinh (d+b x))^2 \, dx=\frac {3 \, {\left (5 \, f^{2} e^{\left (\frac {11}{3} \, b x + \frac {5}{3} \, a + 4 \, d\right )} + 10 \, f g e^{\left (\frac {11}{3} \, b x + \frac {5}{3} \, a + 4 \, d\right )} + 5 \, g^{2} e^{\left (\frac {11}{3} \, b x + \frac {5}{3} \, a + 4 \, d\right )} - 22 \, f^{2} e^{\left (\frac {5}{3} \, b x + \frac {5}{3} \, a + 2 \, d\right )} + 22 \, g^{2} e^{\left (\frac {5}{3} \, b x + \frac {5}{3} \, a + 2 \, d\right )} - 55 \, {\left (f^{2} e^{\left (\frac {5}{3} \, a\right )} - 2 \, f g e^{\left (\frac {5}{3} \, a\right )} + g^{2} e^{\left (\frac {5}{3} \, a\right )}\right )} e^{\left (-\frac {1}{3} \, b x\right )}\right )} e^{\left (-2 \, d\right )}}{220 \, b} \] Input:
integrate(exp(5/3*b*x+5/3*a)*(g*cosh(b*x+d)+f*sinh(b*x+d))^2,x, algorithm= "giac")
Output:
3/220*(5*f^2*e^(11/3*b*x + 5/3*a + 4*d) + 10*f*g*e^(11/3*b*x + 5/3*a + 4*d ) + 5*g^2*e^(11/3*b*x + 5/3*a + 4*d) - 22*f^2*e^(5/3*b*x + 5/3*a + 2*d) + 22*g^2*e^(5/3*b*x + 5/3*a + 2*d) - 55*(f^2*e^(5/3*a) - 2*f*g*e^(5/3*a) + g ^2*e^(5/3*a))*e^(-1/3*b*x))*e^(-2*d)/b
Time = 0.26 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.93 \[ \int e^{\frac {5}{3} (a+b x)} (g \cosh (d+b x)+f \sinh (d+b x))^2 \, dx=\frac {3\,{\mathrm {e}}^{\frac {5\,a}{3}-2\,d-\frac {b\,x}{3}}\,\left (110\,f\,g-55\,f^2-55\,g^2-22\,f^2\,{\mathrm {e}}^{2\,d+2\,b\,x}+5\,f^2\,{\mathrm {e}}^{4\,d+4\,b\,x}+22\,g^2\,{\mathrm {e}}^{2\,d+2\,b\,x}+5\,g^2\,{\mathrm {e}}^{4\,d+4\,b\,x}+10\,f\,g\,{\mathrm {e}}^{4\,d+4\,b\,x}\right )}{220\,b} \] Input:
int(exp((5*a)/3 + (5*b*x)/3)*(g*cosh(d + b*x) + f*sinh(d + b*x))^2,x)
Output:
(3*exp((5*a)/3 - 2*d - (b*x)/3)*(110*f*g - 55*f^2 - 55*g^2 - 22*f^2*exp(2* d + 2*b*x) + 5*f^2*exp(4*d + 4*b*x) + 22*g^2*exp(2*d + 2*b*x) + 5*g^2*exp( 4*d + 4*b*x) + 10*f*g*exp(4*d + 4*b*x)))/(220*b)
Time = 0.23 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.30 \[ \int e^{\frac {5}{3} (a+b x)} (g \cosh (d+b x)+f \sinh (d+b x))^2 \, dx=\frac {3 e^{\frac {5 b x}{3}+\frac {5 a}{3}} \left (-18 \cosh \left (b x +d \right )^{2} f^{2}+30 \cosh \left (b x +d \right )^{2} f g -7 \cosh \left (b x +d \right )^{2} g^{2}+30 \cosh \left (b x +d \right ) \sinh \left (b x +d \right ) f^{2}-50 \cosh \left (b x +d \right ) \sinh \left (b x +d \right ) f g +30 \cosh \left (b x +d \right ) \sinh \left (b x +d \right ) g^{2}-7 \sinh \left (b x +d \right )^{2} f^{2}+30 \sinh \left (b x +d \right )^{2} f g -18 \sinh \left (b x +d \right )^{2} g^{2}\right )}{55 b} \] Input:
int(exp(5/3*b*x+5/3*a)*(g*cosh(b*x+d)+f*sinh(b*x+d))^2,x)
Output:
(3*e**((5*a + 5*b*x)/3)*( - 18*cosh(b*x + d)**2*f**2 + 30*cosh(b*x + d)**2 *f*g - 7*cosh(b*x + d)**2*g**2 + 30*cosh(b*x + d)*sinh(b*x + d)*f**2 - 50* cosh(b*x + d)*sinh(b*x + d)*f*g + 30*cosh(b*x + d)*sinh(b*x + d)*g**2 - 7* sinh(b*x + d)**2*f**2 + 30*sinh(b*x + d)**2*f*g - 18*sinh(b*x + d)**2*g**2 ))/(55*b)