Integrand size = 21, antiderivative size = 61 \[ \int \frac {\sinh ^3(c+d x)}{a+b \cosh (c+d x)} \, dx=-\frac {a \cosh (c+d x)}{b^2 d}+\frac {\cosh ^2(c+d x)}{2 b d}+\frac {\left (a^2-b^2\right ) \log (a+b \cosh (c+d x))}{b^3 d} \] Output:
-a*cosh(d*x+c)/b^2/d+1/2*cosh(d*x+c)^2/b/d+(a^2-b^2)*ln(a+b*cosh(d*x+c))/b ^3/d
Time = 0.25 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.90 \[ \int \frac {\sinh ^3(c+d x)}{a+b \cosh (c+d x)} \, dx=\frac {-4 a b \cosh (c+d x)+b^2 \cosh (2 (c+d x))+4 \left (a^2-b^2\right ) \log (a+b \cosh (c+d x))}{4 b^3 d} \] Input:
Integrate[Sinh[c + d*x]^3/(a + b*Cosh[c + d*x]),x]
Output:
(-4*a*b*Cosh[c + d*x] + b^2*Cosh[2*(c + d*x)] + 4*(a^2 - b^2)*Log[a + b*Co sh[c + d*x]])/(4*b^3*d)
Time = 0.32 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 26, 3147, 476, 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.
\(\displaystyle \int \frac {\sinh ^3(c+d x)}{a+b \cosh (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i \cos \left (i c+i d x-\frac {\pi }{2}\right )^3}{a-b \sin \left (i c+i d x-\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {\cos \left (\frac {1}{2} (2 i c-\pi )+i d x\right )^3}{a-b \sin \left (\frac {1}{2} (2 i c-\pi )+i d x\right )}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle -\frac {\int \frac {b^2-b^2 \cosh ^2(c+d x)}{a+b \cosh (c+d x)}d(b \cosh (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 476 |
\(\displaystyle -\frac {\int \left (a-b \cosh (c+d x)+\frac {b^2-a^2}{a+b \cosh (c+d x)}\right )d(b \cosh (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\left (a^2-b^2\right ) \log (a+b \cosh (c+d x))+a b \cosh (c+d x)-\frac {1}{2} b^2 \cosh ^2(c+d x)}{b^3 d}\) |
Input:
Int[Sinh[c + d*x]^3/(a + b*Cosh[c + d*x]),x]
Output:
-((a*b*Cosh[c + d*x] - (b^2*Cosh[c + d*x]^2)/2 - (a^2 - b^2)*Log[a + b*Cos h[c + d*x]])/(b^3*d))
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 7.57 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.90
method | result | size |
derivativedivides | \(\frac {-\frac {-\frac {b \cosh \left (d x +c \right )^{2}}{2}+a \cosh \left (d x +c \right )}{b^{2}}+\frac {\left (a^{2}-b^{2}\right ) \ln \left (a +b \cosh \left (d x +c \right )\right )}{b^{3}}}{d}\) | \(55\) |
default | \(\frac {-\frac {-\frac {b \cosh \left (d x +c \right )^{2}}{2}+a \cosh \left (d x +c \right )}{b^{2}}+\frac {\left (a^{2}-b^{2}\right ) \ln \left (a +b \cosh \left (d x +c \right )\right )}{b^{3}}}{d}\) | \(55\) |
risch | \(-\frac {x \,a^{2}}{b^{3}}+\frac {x}{b}+\frac {{\mathrm e}^{2 d x +2 c}}{8 d b}-\frac {a \,{\mathrm e}^{d x +c}}{2 d \,b^{2}}-\frac {a \,{\mathrm e}^{-d x -c}}{2 d \,b^{2}}+\frac {{\mathrm e}^{-2 d x -2 c}}{8 d b}-\frac {2 a^{2} c}{d \,b^{3}}+\frac {2 c}{b d}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{b}+1\right ) a^{2}}{d \,b^{3}}-\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{b}+1\right )}{b d}\) | \(170\) |
Input:
int(sinh(d*x+c)^3/(a+b*cosh(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(-1/b^2*(-1/2*b*cosh(d*x+c)^2+a*cosh(d*x+c))+(a^2-b^2)/b^3*ln(a+b*cosh (d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 340 vs. \(2 (59) = 118\).
Time = 0.09 (sec) , antiderivative size = 340, normalized size of antiderivative = 5.57 \[ \int \frac {\sinh ^3(c+d x)}{a+b \cosh (c+d x)} \, dx=\frac {b^{2} \cosh \left (d x + c\right )^{4} + b^{2} \sinh \left (d x + c\right )^{4} - 8 \, {\left (a^{2} - b^{2}\right )} d x \cosh \left (d x + c\right )^{2} - 4 \, a b \cosh \left (d x + c\right )^{3} + 4 \, {\left (b^{2} \cosh \left (d x + c\right ) - a b\right )} \sinh \left (d x + c\right )^{3} - 4 \, a b \cosh \left (d x + c\right ) + 2 \, {\left (3 \, b^{2} \cosh \left (d x + c\right )^{2} - 4 \, {\left (a^{2} - b^{2}\right )} d x - 6 \, a b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + b^{2} + 8 \, {\left ({\left (a^{2} - b^{2}\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (a^{2} - b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \sinh \left (d x + c\right )^{2}\right )} \log \left (\frac {2 \, {\left (b \cosh \left (d x + c\right ) + a\right )}}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + 4 \, {\left (b^{2} \cosh \left (d x + c\right )^{3} - 4 \, {\left (a^{2} - b^{2}\right )} d x \cosh \left (d x + c\right ) - 3 \, a b \cosh \left (d x + c\right )^{2} - a b\right )} \sinh \left (d x + c\right )}{8 \, {\left (b^{3} d \cosh \left (d x + c\right )^{2} + 2 \, b^{3} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{3} d \sinh \left (d x + c\right )^{2}\right )}} \] Input:
integrate(sinh(d*x+c)^3/(a+b*cosh(d*x+c)),x, algorithm="fricas")
Output:
1/8*(b^2*cosh(d*x + c)^4 + b^2*sinh(d*x + c)^4 - 8*(a^2 - b^2)*d*x*cosh(d* x + c)^2 - 4*a*b*cosh(d*x + c)^3 + 4*(b^2*cosh(d*x + c) - a*b)*sinh(d*x + c)^3 - 4*a*b*cosh(d*x + c) + 2*(3*b^2*cosh(d*x + c)^2 - 4*(a^2 - b^2)*d*x - 6*a*b*cosh(d*x + c))*sinh(d*x + c)^2 + b^2 + 8*((a^2 - b^2)*cosh(d*x + c )^2 + 2*(a^2 - b^2)*cosh(d*x + c)*sinh(d*x + c) + (a^2 - b^2)*sinh(d*x + c )^2)*log(2*(b*cosh(d*x + c) + a)/(cosh(d*x + c) - sinh(d*x + c))) + 4*(b^2 *cosh(d*x + c)^3 - 4*(a^2 - b^2)*d*x*cosh(d*x + c) - 3*a*b*cosh(d*x + c)^2 - a*b)*sinh(d*x + c))/(b^3*d*cosh(d*x + c)^2 + 2*b^3*d*cosh(d*x + c)*sinh (d*x + c) + b^3*d*sinh(d*x + c)^2)
Timed out. \[ \int \frac {\sinh ^3(c+d x)}{a+b \cosh (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(sinh(d*x+c)**3/(a+b*cosh(d*x+c)),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 130 vs. \(2 (59) = 118\).
Time = 0.05 (sec) , antiderivative size = 130, normalized size of antiderivative = 2.13 \[ \int \frac {\sinh ^3(c+d x)}{a+b \cosh (c+d x)} \, dx=-\frac {{\left (4 \, a e^{\left (-d x - c\right )} - b\right )} e^{\left (2 \, d x + 2 \, c\right )}}{8 \, b^{2} d} + \frac {{\left (a^{2} - b^{2}\right )} {\left (d x + c\right )}}{b^{3} d} - \frac {4 \, a e^{\left (-d x - c\right )} - b e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, b^{2} d} + \frac {{\left (a^{2} - b^{2}\right )} \log \left (2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} + b\right )}{b^{3} d} \] Input:
integrate(sinh(d*x+c)^3/(a+b*cosh(d*x+c)),x, algorithm="maxima")
Output:
-1/8*(4*a*e^(-d*x - c) - b)*e^(2*d*x + 2*c)/(b^2*d) + (a^2 - b^2)*(d*x + c )/(b^3*d) - 1/8*(4*a*e^(-d*x - c) - b*e^(-2*d*x - 2*c))/(b^2*d) + (a^2 - b ^2)*log(2*a*e^(-d*x - c) + b*e^(-2*d*x - 2*c) + b)/(b^3*d)
Time = 0.12 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.44 \[ \int \frac {\sinh ^3(c+d x)}{a+b \cosh (c+d x)} \, dx=\frac {\frac {b {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{2} - 4 \, a {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}}{b^{2}} + \frac {8 \, {\left (a^{2} - b^{2}\right )} \log \left ({\left | b {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{b^{3}}}{8 \, d} \] Input:
integrate(sinh(d*x+c)^3/(a+b*cosh(d*x+c)),x, algorithm="giac")
Output:
1/8*((b*(e^(d*x + c) + e^(-d*x - c))^2 - 4*a*(e^(d*x + c) + e^(-d*x - c))) /b^2 + 8*(a^2 - b^2)*log(abs(b*(e^(d*x + c) + e^(-d*x - c)) + 2*a))/b^3)/d
Time = 2.21 (sec) , antiderivative size = 122, normalized size of antiderivative = 2.00 \[ \int \frac {\sinh ^3(c+d x)}{a+b \cosh (c+d x)} \, dx=\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,b\,d}+\frac {{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,b\,d}-\frac {x\,\left (a^2-b^2\right )}{b^3}+\frac {\ln \left (b+2\,a\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\right )\,\left (a^2-b^2\right )}{b^3\,d}-\frac {a\,{\mathrm {e}}^{-c-d\,x}}{2\,b^2\,d}-\frac {a\,{\mathrm {e}}^{c+d\,x}}{2\,b^2\,d} \] Input:
int(sinh(c + d*x)^3/(a + b*cosh(c + d*x)),x)
Output:
exp(- 2*c - 2*d*x)/(8*b*d) + exp(2*c + 2*d*x)/(8*b*d) - (x*(a^2 - b^2))/b^ 3 + (log(b + 2*a*exp(d*x)*exp(c) + b*exp(2*c)*exp(2*d*x))*(a^2 - b^2))/(b^ 3*d) - (a*exp(- c - d*x))/(2*b^2*d) - (a*exp(c + d*x))/(2*b^2*d)
Time = 0.23 (sec) , antiderivative size = 177, normalized size of antiderivative = 2.90 \[ \int \frac {\sinh ^3(c+d x)}{a+b \cosh (c+d x)} \, dx=\frac {e^{4 d x +4 c} b^{2}-4 e^{3 d x +3 c} a b +8 e^{2 d x +2 c} \mathrm {log}\left (e^{2 d x +2 c} b +2 e^{d x +c} a +b \right ) a^{2}-8 e^{2 d x +2 c} \mathrm {log}\left (e^{2 d x +2 c} b +2 e^{d x +c} a +b \right ) b^{2}-8 e^{2 d x +2 c} a^{2} d x +8 e^{2 d x +2 c} b^{2} d x -4 e^{d x +c} a b +b^{2}}{8 e^{2 d x +2 c} b^{3} d} \] Input:
int(sinh(d*x+c)^3/(a+b*cosh(d*x+c)),x)
Output:
(e**(4*c + 4*d*x)*b**2 - 4*e**(3*c + 3*d*x)*a*b + 8*e**(2*c + 2*d*x)*log(e **(2*c + 2*d*x)*b + 2*e**(c + d*x)*a + b)*a**2 - 8*e**(2*c + 2*d*x)*log(e* *(2*c + 2*d*x)*b + 2*e**(c + d*x)*a + b)*b**2 - 8*e**(2*c + 2*d*x)*a**2*d* x + 8*e**(2*c + 2*d*x)*b**2*d*x - 4*e**(c + d*x)*a*b + b**2)/(8*e**(2*c + 2*d*x)*b**3*d)