Integrand size = 27, antiderivative size = 95 \[ \int \frac {\cosh ^2(c+d x) \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\left (2 a^2+b^2\right ) x}{2 b^3}+\frac {2 a \sqrt {a^2+b^2} \text {arctanh}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{b^3 d}-\frac {\cosh (c+d x) (2 a-b \sinh (c+d x))}{2 b^2 d} \] Output:
1/2*(2*a^2+b^2)*x/b^3+2*a*(a^2+b^2)^(1/2)*arctanh((b-a*tanh(1/2*d*x+1/2*c) )/(a^2+b^2)^(1/2))/b^3/d-1/2*cosh(d*x+c)*(2*a-b*sinh(d*x+c))/b^2/d
Time = 0.45 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.15 \[ \int \frac {\cosh ^2(c+d x) \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {4 a^2 c+2 b^2 c+4 a^2 d x+2 b^2 d x+8 a \sqrt {-a^2-b^2} \arctan \left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )-4 a b \cosh (c+d x)+b^2 \sinh (2 (c+d x))}{4 b^3 d} \] Input:
Integrate[(Cosh[c + d*x]^2*Sinh[c + d*x])/(a + b*Sinh[c + d*x]),x]
Output:
(4*a^2*c + 2*b^2*c + 4*a^2*d*x + 2*b^2*d*x + 8*a*Sqrt[-a^2 - b^2]*ArcTan[( b - a*Tanh[(c + d*x)/2])/Sqrt[-a^2 - b^2]] - 4*a*b*Cosh[c + d*x] + b^2*Sin h[2*(c + d*x)])/(4*b^3*d)
Result contains complex when optimal does not.
Time = 0.56 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.13, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {3042, 26, 3344, 26, 3042, 3214, 3042, 3139, 1083, 217}
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.
| \(\displaystyle \int \frac {\sinh (c+d x) \cosh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {i \sin (i c+i d x) \cos (i c+i d x)^2}{a-i b \sin (i c+i d x)}dx\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle -i \int \frac {\cos (i c+i d x)^2 \sin (i c+i d x)}{a-i b \sin (i c+i d x)}dx\) |
| \(\Big \downarrow \) 3344 |
| \(\displaystyle -i \left (-\frac {\int \frac {i \left (a b-\left (2 a^2+b^2\right ) \sinh (c+d x)\right )}{a+b \sinh (c+d x)}dx}{2 b^2}-\frac {i \cosh (c+d x) (2 a-b \sinh (c+d x))}{2 b^2 d}\right )\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle -i \left (-\frac {i \int \frac {a b-\left (2 a^2+b^2\right ) \sinh (c+d x)}{a+b \sinh (c+d x)}dx}{2 b^2}-\frac {i \cosh (c+d x) (2 a-b \sinh (c+d x))}{2 b^2 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -i \left (-\frac {i \int \frac {a b+i \left (2 a^2+b^2\right ) \sin (i c+i d x)}{a-i b \sin (i c+i d x)}dx}{2 b^2}-\frac {i \cosh (c+d x) (2 a-b \sinh (c+d x))}{2 b^2 d}\right )\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle -i \left (-\frac {i \left (\frac {2 a \left (a^2+b^2\right ) \int \frac {1}{a+b \sinh (c+d x)}dx}{b}-\frac {x \left (2 a^2+b^2\right )}{b}\right )}{2 b^2}-\frac {i \cosh (c+d x) (2 a-b \sinh (c+d x))}{2 b^2 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -i \left (-\frac {i \left (-\frac {x \left (2 a^2+b^2\right )}{b}+\frac {2 a \left (a^2+b^2\right ) \int \frac {1}{a-i b \sin (i c+i d x)}dx}{b}\right )}{2 b^2}-\frac {i \cosh (c+d x) (2 a-b \sinh (c+d x))}{2 b^2 d}\right )\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle -i \left (-\frac {i \left (-\frac {x \left (2 a^2+b^2\right )}{b}-\frac {4 i a \left (a^2+b^2\right ) \int \frac {1}{-a \tanh ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tanh \left (\frac {1}{2} (c+d x)\right )+a}d\left (i \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{b d}\right )}{2 b^2}-\frac {i \cosh (c+d x) (2 a-b \sinh (c+d x))}{2 b^2 d}\right )\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle -i \left (-\frac {i \left (-\frac {x \left (2 a^2+b^2\right )}{b}+\frac {8 i a \left (a^2+b^2\right ) \int \frac {1}{\tanh ^2\left (\frac {1}{2} (c+d x)\right )-4 \left (a^2+b^2\right )}d\left (2 i a \tanh \left (\frac {1}{2} (c+d x)\right )-2 i b\right )}{b d}\right )}{2 b^2}-\frac {i \cosh (c+d x) (2 a-b \sinh (c+d x))}{2 b^2 d}\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle -i \left (-\frac {i \left (\frac {4 a \sqrt {a^2+b^2} \text {arctanh}\left (\frac {\tanh \left (\frac {1}{2} (c+d x)\right )}{2 \sqrt {a^2+b^2}}\right )}{b d}-\frac {x \left (2 a^2+b^2\right )}{b}\right )}{2 b^2}-\frac {i \cosh (c+d x) (2 a-b \sinh (c+d x))}{2 b^2 d}\right )\) |
Input:
Int[(Cosh[c + d*x]^2*Sinh[c + d*x])/(a + b*Sinh[c + d*x]),x]
Output:
(-I)*(((-1/2*I)*(-(((2*a^2 + b^2)*x)/b) + (4*a*Sqrt[a^2 + b^2]*ArcTanh[Tan h[(c + d*x)/2]/(2*Sqrt[a^2 + b^2])])/(b*d)))/b^2 - ((I/2)*Cosh[c + d*x]*(2 *a - b*Sinh[c + d*x]))/(b^2*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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ [a^2 - b^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g*(g* Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c*(m + p + 1) - a*d* p + b*d*(m + p)*Sin[e + f*x])/(b^2*f*(m + p)*(m + p + 1))), x] + Simp[g^2*( (p - 1)/(b^2*(m + p)*(m + p + 1))) Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Si n[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1) - d*(a^ 2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1 , 0] && IntegerQ[2*m]
Time = 3.06 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.76
| method | result | size |
| risch | \(\frac {x \,a^{2}}{b^{3}}+\frac {x}{2 b}+\frac {{\mathrm e}^{2 d x +2 c}}{8 b d}-\frac {a \,{\mathrm e}^{d x +c}}{2 b^{2} d}-\frac {a \,{\mathrm e}^{-d x -c}}{2 b^{2} d}-\frac {{\mathrm e}^{-2 d x -2 c}}{8 b d}+\frac {\sqrt {a^{2}+b^{2}}\, a \ln \left ({\mathrm e}^{d x +c}+\frac {a +\sqrt {a^{2}+b^{2}}}{b}\right )}{d \,b^{3}}-\frac {\sqrt {a^{2}+b^{2}}\, a \ln \left ({\mathrm e}^{d x +c}-\frac {-a +\sqrt {a^{2}+b^{2}}}{b}\right )}{d \,b^{3}}\) | \(167\) |
| derivativedivides | \(\frac {-\frac {1}{2 b \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {-b +2 a}{2 b^{2} \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {\left (2 a^{2}+b^{2}\right ) \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 b^{3}}+\frac {1}{2 b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-b -2 a}{2 b^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-2 a^{2}-b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{3}}-\frac {2 a \sqrt {a^{2}+b^{2}}\, \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{b^{3}}}{d}\) | \(189\) |
| default | \(\frac {-\frac {1}{2 b \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {-b +2 a}{2 b^{2} \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {\left (2 a^{2}+b^{2}\right ) \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 b^{3}}+\frac {1}{2 b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-b -2 a}{2 b^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-2 a^{2}-b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{3}}-\frac {2 a \sqrt {a^{2}+b^{2}}\, \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{b^{3}}}{d}\) | \(189\) |
Input:
int(cosh(d*x+c)^2*sinh(d*x+c)/(a+b*sinh(d*x+c)),x,method=_RETURNVERBOSE)
Output:
x/b^3*a^2+1/2*x/b+1/8/b/d*exp(2*d*x+2*c)-1/2*a/b^2/d*exp(d*x+c)-1/2*a/b^2/ d*exp(-d*x-c)-1/8/b/d*exp(-2*d*x-2*c)+(a^2+b^2)^(1/2)*a/d/b^3*ln(exp(d*x+c )+(a+(a^2+b^2)^(1/2))/b)-(a^2+b^2)^(1/2)*a/d/b^3*ln(exp(d*x+c)-(-a+(a^2+b^ 2)^(1/2))/b)
Leaf count of result is larger than twice the leaf count of optimal. 446 vs. \(2 (87) = 174\).
Time = 0.10 (sec) , antiderivative size = 446, normalized size of antiderivative = 4.69 \[ \int \frac {\cosh ^2(c+d x) \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {b^{2} \cosh \left (d x + c\right )^{4} + b^{2} \sinh \left (d x + c\right )^{4} + 4 \, {\left (2 \, a^{2} + b^{2}\right )} d x \cosh \left (d x + c\right )^{2} - 4 \, a b \cosh \left (d x + c\right )^{3} + 4 \, {\left (b^{2} \cosh \left (d x + c\right ) - a b\right )} \sinh \left (d x + c\right )^{3} - 4 \, a b \cosh \left (d x + c\right ) + 2 \, {\left (3 \, b^{2} \cosh \left (d x + c\right )^{2} + 2 \, {\left (2 \, a^{2} + b^{2}\right )} d x - 6 \, a b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 8 \, {\left (a \cosh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a \sinh \left (d x + c\right )^{2}\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {b^{2} \cosh \left (d x + c\right )^{2} + b^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{b \cosh \left (d x + c\right )^{2} + b \sinh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) + 2 \, {\left (b \cosh \left (d x + c\right ) + a\right )} \sinh \left (d x + c\right ) - b}\right ) - b^{2} + 4 \, {\left (b^{2} \cosh \left (d x + c\right )^{3} + 2 \, {\left (2 \, a^{2} + b^{2}\right )} d x \cosh \left (d x + c\right ) - 3 \, a b \cosh \left (d x + c\right )^{2} - a b\right )} \sinh \left (d x + c\right )}{8 \, {\left (b^{3} d \cosh \left (d x + c\right )^{2} + 2 \, b^{3} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{3} d \sinh \left (d x + c\right )^{2}\right )}} \] Input:
integrate(cosh(d*x+c)^2*sinh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="fricas ")
Output:
1/8*(b^2*cosh(d*x + c)^4 + b^2*sinh(d*x + c)^4 + 4*(2*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 + 2*(2*a^2 + b^2)* d*x - 6*a*b*cosh(d*x + c))*sinh(d*x + c)^2 + 8*(a*cosh(d*x + c)^2 + 2*a*co sh(d*x + c)*sinh(d*x + c) + a*sinh(d*x + c)^2)*sqrt(a^2 + b^2)*log((b^2*co sh(d*x + c)^2 + b^2*sinh(d*x + c)^2 + 2*a*b*cosh(d*x + c) + 2*a^2 + b^2 + 2*(b^2*cosh(d*x + c) + a*b)*sinh(d*x + c) + 2*sqrt(a^2 + b^2)*(b*cosh(d*x + c) + b*sinh(d*x + c) + a))/(b*cosh(d*x + c)^2 + b*sinh(d*x + c)^2 + 2*a* cosh(d*x + c) + 2*(b*cosh(d*x + c) + a)*sinh(d*x + c) - b)) - b^2 + 4*(b^2 *cosh(d*x + c)^3 + 2*(2*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)*si nh(d*x + c) + b^3*d*sinh(d*x + c)^2)
Timed out. \[ \int \frac {\cosh ^2(c+d x) \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(cosh(d*x+c)**2*sinh(d*x+c)/(a+b*sinh(d*x+c)),x)
Output:
Timed out
Time = 0.13 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.68 \[ \int \frac {\cosh ^2(c+d x) \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {{\left (4 \, a e^{\left (-d x - c\right )} - b\right )} e^{\left (2 \, d x + 2 \, c\right )}}{8 \, b^{2} d} - \frac {\sqrt {a^{2} + b^{2}} a \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{b^{3} d} + \frac {{\left (2 \, a^{2} + b^{2}\right )} {\left (d x + c\right )}}{2 \, b^{3} d} - \frac {4 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, b^{2} d} \] Input:
integrate(cosh(d*x+c)^2*sinh(d*x+c)/(a+b*sinh(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) - sqrt(a^2 + b^2)*a*lo g((b*e^(-d*x - c) - a - sqrt(a^2 + b^2))/(b*e^(-d*x - c) - a + sqrt(a^2 + b^2)))/(b^3*d) + 1/2*(2*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)
Time = 0.15 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.63 \[ \int \frac {\cosh ^2(c+d x) \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\frac {4 \, {\left (2 \, a^{2} + b^{2}\right )} {\left (d x + c\right )}}{b^{3}} + \frac {b e^{\left (2 \, d x + 2 \, c\right )} - 4 \, a e^{\left (d x + c\right )}}{b^{2}} - \frac {{\left (4 \, a b e^{\left (d x + c\right )} + b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{b^{3}} - \frac {8 \, {\left (a^{3} + a b^{2}\right )} \log \left (\frac {{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b^{3}}}{8 \, d} \] Input:
integrate(cosh(d*x+c)^2*sinh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="giac")
Output:
1/8*(4*(2*a^2 + b^2)*(d*x + c)/b^3 + (b*e^(2*d*x + 2*c) - 4*a*e^(d*x + c)) /b^2 - (4*a*b*e^(d*x + c) + b^2)*e^(-2*d*x - 2*c)/b^3 - 8*(a^3 + a*b^2)*lo g(abs(2*b*e^(d*x + c) + 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*e^(d*x + c) + 2*a + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*b^3))/d
Time = 1.32 (sec) , antiderivative size = 212, normalized size of antiderivative = 2.23 \[ \int \frac {\cosh ^2(c+d x) \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,b\,d}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,b\,d}+\frac {x\,\left (2\,a^2+b^2\right )}{2\,b^3}-\frac {a\,{\mathrm {e}}^{-c-d\,x}}{2\,b^2\,d}-\frac {a\,{\mathrm {e}}^{c+d\,x}}{2\,b^2\,d}-\frac {a\,\ln \left (\frac {2\,a\,{\mathrm {e}}^{c+d\,x}\,\left (a^2+b^2\right )}{b^4}-\frac {2\,a\,\sqrt {a^2+b^2}\,\left (b-a\,{\mathrm {e}}^{c+d\,x}\right )}{b^4}\right )\,\sqrt {a^2+b^2}}{b^3\,d}+\frac {a\,\ln \left (\frac {2\,a\,\sqrt {a^2+b^2}\,\left (b-a\,{\mathrm {e}}^{c+d\,x}\right )}{b^4}+\frac {2\,a\,{\mathrm {e}}^{c+d\,x}\,\left (a^2+b^2\right )}{b^4}\right )\,\sqrt {a^2+b^2}}{b^3\,d} \] Input:
int((cosh(c + d*x)^2*sinh(c + d*x))/(a + b*sinh(c + d*x)),x)
Output:
exp(2*c + 2*d*x)/(8*b*d) - exp(- 2*c - 2*d*x)/(8*b*d) + (x*(2*a^2 + b^2))/ (2*b^3) - (a*exp(- c - d*x))/(2*b^2*d) - (a*exp(c + d*x))/(2*b^2*d) - (a*l og((2*a*exp(c + d*x)*(a^2 + b^2))/b^4 - (2*a*(a^2 + b^2)^(1/2)*(b - a*exp( c + d*x)))/b^4)*(a^2 + b^2)^(1/2))/(b^3*d) + (a*log((2*a*(a^2 + b^2)^(1/2) *(b - a*exp(c + d*x)))/b^4 + (2*a*exp(c + d*x)*(a^2 + b^2))/b^4)*(a^2 + b^ 2)^(1/2))/(b^3*d)
Time = 0.24 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.55 \[ \int \frac {\cosh ^2(c+d x) \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {-16 e^{2 d x +2 c} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{d x +c} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a i +e^{4 d x +4 c} b^{2}-4 e^{3 d x +3 c} a b +8 e^{2 d x +2 c} a^{2} d x +4 e^{2 d x +2 c} b^{2} d x -4 e^{d x +c} a b -b^{2}}{8 e^{2 d x +2 c} b^{3} d} \] Input:
int(cosh(d*x+c)^2*sinh(d*x+c)/(a+b*sinh(d*x+c)),x)
Output:
( - 16*e**(2*c + 2*d*x)*sqrt(a**2 + b**2)*atan((e**(c + d*x)*b*i + a*i)/sq rt(a**2 + b**2))*a*i + e**(4*c + 4*d*x)*b**2 - 4*e**(3*c + 3*d*x)*a*b + 8* e**(2*c + 2*d*x)*a**2*d*x + 4*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)