Integrand size = 21, antiderivative size = 61 \[ \int \cosh ^2(c+d x) \left (a+b \sinh ^2(c+d x)\right ) \, dx=\frac {1}{8} (4 a-b) x+\frac {(4 a-b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {b \cosh ^3(c+d x) \sinh (c+d x)}{4 d} \] Output:
1/8*(4*a-b)*x+1/8*(4*a-b)*cosh(d*x+c)*sinh(d*x+c)/d+1/4*b*cosh(d*x+c)^3*si nh(d*x+c)/d
Time = 0.07 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.70 \[ \int \cosh ^2(c+d x) \left (a+b \sinh ^2(c+d x)\right ) \, dx=\frac {16 a c+16 a d x-4 b d x+8 a \sinh (2 (c+d x))+b \sinh (4 (c+d x))}{32 d} \] Input:
Integrate[Cosh[c + d*x]^2*(a + b*Sinh[c + d*x]^2),x]
Output:
(16*a*c + 16*a*d*x - 4*b*d*x + 8*a*Sinh[2*(c + d*x)] + b*Sinh[4*(c + d*x)] )/(32*d)
Time = 0.26 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.26, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3670, 298, 215, 219}
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.
\(\displaystyle \int \cosh ^2(c+d x) \left (a+b \sinh ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (i c+i d x)^2 \left (a-b \sin (i c+i d x)^2\right )dx\) |
\(\Big \downarrow \) 3670 |
\(\displaystyle \frac {\int \frac {a-(a-b) \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right )^3}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle \frac {\frac {1}{4} (4 a-b) \int \frac {1}{\left (1-\tanh ^2(c+d x)\right )^2}d\tanh (c+d x)+\frac {b \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {1}{4} (4 a-b) \left (\frac {1}{2} \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)+\frac {\tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {b \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {1}{4} (4 a-b) \left (\frac {1}{2} \text {arctanh}(\tanh (c+d x))+\frac {\tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {b \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
Input:
Int[Cosh[c + d*x]^2*(a + b*Sinh[c + d*x]^2),x]
Output:
((b*Tanh[c + d*x])/(4*(1 - Tanh[c + d*x]^2)^2) + ((4*a - b)*(ArcTanh[Tanh[ c + d*x]]/2 + Tanh[c + d*x]/(2*(1 - Tanh[c + d*x]^2))))/4)/d
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Su bst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]
Time = 10.33 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.15
method | result | size |
derivativedivides | \(\frac {a \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b \left (\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )^{3}}{4}-\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{8}-\frac {d x}{8}-\frac {c}{8}\right )}{d}\) | \(70\) |
default | \(\frac {a \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b \left (\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )^{3}}{4}-\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{8}-\frac {d x}{8}-\frac {c}{8}\right )}{d}\) | \(70\) |
risch | \(\frac {a x}{2}-\frac {b x}{8}+\frac {{\mathrm e}^{4 d x +4 c} b}{64 d}+\frac {{\mathrm e}^{2 d x +2 c} a}{8 d}-\frac {{\mathrm e}^{-2 d x -2 c} a}{8 d}-\frac {{\mathrm e}^{-4 d x -4 c} b}{64 d}\) | \(70\) |
orering | \(x \cosh \left (d x +c \right )^{2} \left (a +b \sinh \left (d x +c \right )^{2}\right )+\frac {\frac {5 \cosh \left (d x +c \right ) \left (a +b \sinh \left (d x +c \right )^{2}\right ) d \sinh \left (d x +c \right )}{8}+\frac {5 \cosh \left (d x +c \right )^{3} b \sinh \left (d x +c \right ) d}{8}}{d^{2}}-\frac {5 x \left (2 d^{2} \sinh \left (d x +c \right )^{2} \left (a +b \sinh \left (d x +c \right )^{2}\right )+10 \cosh \left (d x +c \right )^{2} \sinh \left (d x +c \right )^{2} b \,d^{2}+2 \cosh \left (d x +c \right )^{2} \left (a +b \sinh \left (d x +c \right )^{2}\right ) d^{2}+2 \cosh \left (d x +c \right )^{4} b \,d^{2}\right )}{16 d^{2}}-\frac {8 d^{3} \sinh \left (d x +c \right ) \left (a +b \sinh \left (d x +c \right )^{2}\right ) \cosh \left (d x +c \right )+24 \cosh \left (d x +c \right ) \sinh \left (d x +c \right )^{3} b \,d^{3}+32 \cosh \left (d x +c \right )^{3} \sinh \left (d x +c \right ) b \,d^{3}}{64 d^{4}}+\frac {x \left (8 d^{4} \cosh \left (d x +c \right )^{2} \left (a +b \sinh \left (d x +c \right )^{2}\right )+184 d^{4} \sinh \left (d x +c \right )^{2} b \cosh \left (d x +c \right )^{2}+8 d^{4} \sinh \left (d x +c \right )^{2} \left (a +b \sinh \left (d x +c \right )^{2}\right )+24 d^{4} \sinh \left (d x +c \right )^{4} b +32 \cosh \left (d x +c \right )^{4} d^{4} b \right )}{64 d^{4}}\) | \(350\) |
Input:
int(cosh(d*x+c)^2*(a+b*sinh(d*x+c)^2),x,method=_RETURNVERBOSE)
Output:
1/d*(a*(1/2*cosh(d*x+c)*sinh(d*x+c)+1/2*d*x+1/2*c)+b*(1/4*sinh(d*x+c)*cosh (d*x+c)^3-1/8*cosh(d*x+c)*sinh(d*x+c)-1/8*d*x-1/8*c))
Time = 0.09 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.97 \[ \int \cosh ^2(c+d x) \left (a+b \sinh ^2(c+d x)\right ) \, dx=\frac {b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (4 \, a - b\right )} d x + {\left (b \cosh \left (d x + c\right )^{3} + 4 \, a \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{8 \, d} \] Input:
integrate(cosh(d*x+c)^2*(a+b*sinh(d*x+c)^2),x, algorithm="fricas")
Output:
1/8*(b*cosh(d*x + c)*sinh(d*x + c)^3 + (4*a - b)*d*x + (b*cosh(d*x + c)^3 + 4*a*cosh(d*x + c))*sinh(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 150 vs. \(2 (49) = 98\).
Time = 0.17 (sec) , antiderivative size = 150, normalized size of antiderivative = 2.46 \[ \int \cosh ^2(c+d x) \left (a+b \sinh ^2(c+d x)\right ) \, dx=\begin {cases} - \frac {a x \sinh ^{2}{\left (c + d x \right )}}{2} + \frac {a x \cosh ^{2}{\left (c + d x \right )}}{2} + \frac {a \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{2 d} - \frac {b x \sinh ^{4}{\left (c + d x \right )}}{8} + \frac {b x \sinh ^{2}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{4} - \frac {b x \cosh ^{4}{\left (c + d x \right )}}{8} + \frac {b \sinh ^{3}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{8 d} + \frac {b \sinh {\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (a + b \sinh ^{2}{\left (c \right )}\right ) \cosh ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cosh(d*x+c)**2*(a+b*sinh(d*x+c)**2),x)
Output:
Piecewise((-a*x*sinh(c + d*x)**2/2 + a*x*cosh(c + d*x)**2/2 + a*sinh(c + d *x)*cosh(c + d*x)/(2*d) - b*x*sinh(c + d*x)**4/8 + b*x*sinh(c + d*x)**2*co sh(c + d*x)**2/4 - b*x*cosh(c + d*x)**4/8 + b*sinh(c + d*x)**3*cosh(c + d* x)/(8*d) + b*sinh(c + d*x)*cosh(c + d*x)**3/(8*d), Ne(d, 0)), (x*(a + b*si nh(c)**2)*cosh(c)**2, True))
Time = 0.04 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.25 \[ \int \cosh ^2(c+d x) \left (a+b \sinh ^2(c+d x)\right ) \, dx=\frac {1}{8} \, a {\left (4 \, x + \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac {1}{64} \, b {\left (\frac {8 \, {\left (d x + c\right )}}{d} - \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} + \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} \] Input:
integrate(cosh(d*x+c)^2*(a+b*sinh(d*x+c)^2),x, algorithm="maxima")
Output:
1/8*a*(4*x + e^(2*d*x + 2*c)/d - e^(-2*d*x - 2*c)/d) - 1/64*b*(8*(d*x + c) /d - e^(4*d*x + 4*c)/d + e^(-4*d*x - 4*c)/d)
Time = 0.13 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.16 \[ \int \cosh ^2(c+d x) \left (a+b \sinh ^2(c+d x)\right ) \, dx=\frac {1}{8} \, {\left (4 \, a - b\right )} x + \frac {b e^{\left (4 \, d x + 4 \, c\right )}}{64 \, d} + \frac {a e^{\left (2 \, d x + 2 \, c\right )}}{8 \, d} - \frac {a e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, d} - \frac {b e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, d} \] Input:
integrate(cosh(d*x+c)^2*(a+b*sinh(d*x+c)^2),x, algorithm="giac")
Output:
1/8*(4*a - b)*x + 1/64*b*e^(4*d*x + 4*c)/d + 1/8*a*e^(2*d*x + 2*c)/d - 1/8 *a*e^(-2*d*x - 2*c)/d - 1/64*b*e^(-4*d*x - 4*c)/d
Time = 0.11 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.62 \[ \int \cosh ^2(c+d x) \left (a+b \sinh ^2(c+d x)\right ) \, dx=\frac {a\,x}{2}-\frac {b\,x}{8}+\frac {\frac {a\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )}{4}+\frac {b\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )}{32}}{d} \] Input:
int(cosh(c + d*x)^2*(a + b*sinh(c + d*x)^2),x)
Output:
(a*x)/2 - (b*x)/8 + ((a*sinh(2*c + 2*d*x))/4 + (b*sinh(4*c + 4*d*x))/32)/d
Time = 0.17 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.46 \[ \int \cosh ^2(c+d x) \left (a+b \sinh ^2(c+d x)\right ) \, dx=\frac {e^{8 d x +8 c} b +8 e^{6 d x +6 c} a +32 e^{4 d x +4 c} a d x -8 e^{4 d x +4 c} b d x -8 e^{2 d x +2 c} a -b}{64 e^{4 d x +4 c} d} \] Input:
int(cosh(d*x+c)^2*(a+b*sinh(d*x+c)^2),x)
Output:
(e**(8*c + 8*d*x)*b + 8*e**(6*c + 6*d*x)*a + 32*e**(4*c + 4*d*x)*a*d*x - 8 *e**(4*c + 4*d*x)*b*d*x - 8*e**(2*c + 2*d*x)*a - b)/(64*e**(4*c + 4*d*x)*d )