Integrand size = 21, antiderivative size = 118 \[ \int \frac {\sinh (c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\frac {3 \arctan \left (\frac {\sqrt {b} \cosh (c+d x)}{\sqrt {a-b}}\right )}{8 (a-b)^{5/2} \sqrt {b} d}+\frac {\cosh (c+d x)}{4 (a-b) d \left (a-b+b \cosh ^2(c+d x)\right )^2}+\frac {3 \cosh (c+d x)}{8 (a-b)^2 d \left (a-b+b \cosh ^2(c+d x)\right )} \]
1/4*cosh(d*x+c)/(a-b)/d/(a-b+b*cosh(d*x+c)^2)^2+3/8*cosh(d*x+c)/(a-b)^2/d/ (a-b+b*cosh(d*x+c)^2)+3/8*arctan(cosh(d*x+c)*b^(1/2)/(a-b)^(1/2))/(a-b)^(5 /2)/d/b^(1/2)
Result contains complex when optimal does not.
Time = 0.83 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.26 \[ \int \frac {\sinh (c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\frac {\frac {3 \left (\arctan \left (\frac {\sqrt {b}-i \sqrt {a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a-b}}\right )+\arctan \left (\frac {\sqrt {b}+i \sqrt {a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a-b}}\right )\right )}{(a-b)^{5/2} \sqrt {b}}+\frac {2 \cosh (c+d x) (10 a-7 b+3 b \cosh (2 (c+d x)))}{(a-b)^2 (2 a-b+b \cosh (2 (c+d x)))^2}}{8 d} \]
((3*(ArcTan[(Sqrt[b] - I*Sqrt[a]*Tanh[(c + d*x)/2])/Sqrt[a - b]] + ArcTan[ (Sqrt[b] + I*Sqrt[a]*Tanh[(c + d*x)/2])/Sqrt[a - b]]))/((a - b)^(5/2)*Sqrt [b]) + (2*Cosh[c + d*x]*(10*a - 7*b + 3*b*Cosh[2*(c + d*x)]))/((a - b)^2*( 2*a - b + b*Cosh[2*(c + d*x)])^2))/(8*d)
Time = 0.27 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 26, 3665, 215, 215, 218}
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.
\(\displaystyle \int \frac {\sinh (c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {i \sin (i c+i d x)}{\left (a-b \sin (i c+i d x)^2\right )^3}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {\sin (i c+i d x)}{\left (a-b \sin (i c+i d x)^2\right )^3}dx\) |
\(\Big \downarrow \) 3665 |
\(\displaystyle \frac {\int \frac {1}{\left (b \cosh ^2(c+d x)+a-b\right )^3}d\cosh (c+d x)}{d}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {3 \int \frac {1}{\left (b \cosh ^2(c+d x)+a-b\right )^2}d\cosh (c+d x)}{4 (a-b)}+\frac {\cosh (c+d x)}{4 (a-b) \left (a+b \cosh ^2(c+d x)-b\right )^2}}{d}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {3 \left (\frac {\int \frac {1}{b \cosh ^2(c+d x)+a-b}d\cosh (c+d x)}{2 (a-b)}+\frac {\cosh (c+d x)}{2 (a-b) \left (a+b \cosh ^2(c+d x)-b\right )}\right )}{4 (a-b)}+\frac {\cosh (c+d x)}{4 (a-b) \left (a+b \cosh ^2(c+d x)-b\right )^2}}{d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {3 \left (\frac {\arctan \left (\frac {\sqrt {b} \cosh (c+d x)}{\sqrt {a-b}}\right )}{2 \sqrt {b} (a-b)^{3/2}}+\frac {\cosh (c+d x)}{2 (a-b) \left (a+b \cosh ^2(c+d x)-b\right )}\right )}{4 (a-b)}+\frac {\cosh (c+d x)}{4 (a-b) \left (a+b \cosh ^2(c+d x)-b\right )^2}}{d}\) |
(Cosh[c + d*x]/(4*(a - b)*(a - b + b*Cosh[c + d*x]^2)^2) + (3*(ArcTan[(Sqr t[b]*Cosh[c + d*x])/Sqrt[a - b]]/(2*(a - b)^(3/2)*Sqrt[b]) + Cosh[c + d*x] /(2*(a - b)*(a - b + b*Cosh[c + d*x]^2))))/(4*(a - b)))/d
3.1.54.3.1 Defintions of rubi rules used
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)^(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[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Leaf count of result is larger than twice the leaf count of optimal. \(236\) vs. \(2(104)=208\).
Time = 0.47 (sec) , antiderivative size = 237, normalized size of antiderivative = 2.01
method | result | size |
risch | \(\frac {{\mathrm e}^{d x +c} \left (3 b \,{\mathrm e}^{6 d x +6 c}+20 \,{\mathrm e}^{4 d x +4 c} a -11 b \,{\mathrm e}^{4 d x +4 c}+20 a \,{\mathrm e}^{2 d x +2 c}-11 b \,{\mathrm e}^{2 d x +2 c}+3 b \right )}{4 \left (a -b \right )^{2} d \left (b \,{\mathrm e}^{4 d x +4 c}+4 a \,{\mathrm e}^{2 d x +2 c}-2 b \,{\mathrm e}^{2 d x +2 c}+b \right )^{2}}-\frac {3 \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \left (a -b \right ) {\mathrm e}^{d x +c}}{\sqrt {-a b +b^{2}}}+1\right )}{16 \sqrt {-a b +b^{2}}\, \left (a -b \right )^{2} d}+\frac {3 \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \left (a -b \right ) {\mathrm e}^{d x +c}}{\sqrt {-a b +b^{2}}}+1\right )}{16 \sqrt {-a b +b^{2}}\, \left (a -b \right )^{2} d}\) | \(237\) |
derivativedivides | \(\frac {\frac {-\frac {\left (5 a^{2}-16 a b +8 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{4 a \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (15 a^{3}-46 a^{2} b +56 a \,b^{2}-16 b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4 a^{2} \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (15 a^{2}-32 a b +8 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{4 a \left (a^{2}-2 a b +b^{2}\right )}+\frac {2 \left (5 a -2 b \right )}{8 a^{2}-16 a b +8 b^{2}}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a \right )^{2}}+\frac {3 \arctan \left (\frac {2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 a +4 b}{4 \sqrt {a b -b^{2}}}\right )}{8 \left (a^{2}-2 a b +b^{2}\right ) \sqrt {a b -b^{2}}}}{d}\) | \(277\) |
default | \(\frac {\frac {-\frac {\left (5 a^{2}-16 a b +8 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{4 a \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (15 a^{3}-46 a^{2} b +56 a \,b^{2}-16 b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4 a^{2} \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (15 a^{2}-32 a b +8 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{4 a \left (a^{2}-2 a b +b^{2}\right )}+\frac {2 \left (5 a -2 b \right )}{8 a^{2}-16 a b +8 b^{2}}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a \right )^{2}}+\frac {3 \arctan \left (\frac {2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 a +4 b}{4 \sqrt {a b -b^{2}}}\right )}{8 \left (a^{2}-2 a b +b^{2}\right ) \sqrt {a b -b^{2}}}}{d}\) | \(277\) |
1/4*exp(d*x+c)*(3*b*exp(6*d*x+6*c)+20*exp(4*d*x+4*c)*a-11*b*exp(4*d*x+4*c) +20*a*exp(2*d*x+2*c)-11*b*exp(2*d*x+2*c)+3*b)/(a-b)^2/d/(b*exp(4*d*x+4*c)+ 4*a*exp(2*d*x+2*c)-2*b*exp(2*d*x+2*c)+b)^2-3/16/(-a*b+b^2)^(1/2)/(a-b)^2/d *ln(exp(2*d*x+2*c)-2*(a-b)/(-a*b+b^2)^(1/2)*exp(d*x+c)+1)+3/16/(-a*b+b^2)^ (1/2)/(a-b)^2/d*ln(exp(2*d*x+2*c)+2*(a-b)/(-a*b+b^2)^(1/2)*exp(d*x+c)+1)
Leaf count of result is larger than twice the leaf count of optimal. 2726 vs. \(2 (104) = 208\).
Time = 0.34 (sec) , antiderivative size = 5152, normalized size of antiderivative = 43.66 \[ \int \frac {\sinh (c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {\sinh (c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\text {Timed out} \]
\[ \int \frac {\sinh (c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\int { \frac {\sinh \left (d x + c\right )}{{\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
1/4*((20*a*e^(5*c) - 11*b*e^(5*c))*e^(5*d*x) + (20*a*e^(3*c) - 11*b*e^(3*c ))*e^(3*d*x) + 3*b*e^(7*d*x + 7*c) + 3*b*e^(d*x + c))/(a^2*b^2*d - 2*a*b^3 *d + b^4*d + (a^2*b^2*d*e^(8*c) - 2*a*b^3*d*e^(8*c) + b^4*d*e^(8*c))*e^(8* d*x) + 4*(2*a^3*b*d*e^(6*c) - 5*a^2*b^2*d*e^(6*c) + 4*a*b^3*d*e^(6*c) - b^ 4*d*e^(6*c))*e^(6*d*x) + 2*(8*a^4*d*e^(4*c) - 24*a^3*b*d*e^(4*c) + 27*a^2* b^2*d*e^(4*c) - 14*a*b^3*d*e^(4*c) + 3*b^4*d*e^(4*c))*e^(4*d*x) + 4*(2*a^3 *b*d*e^(2*c) - 5*a^2*b^2*d*e^(2*c) + 4*a*b^3*d*e^(2*c) - b^4*d*e^(2*c))*e^ (2*d*x)) + 1/2*integrate(3/2*(e^(3*d*x + 3*c) - e^(d*x + c))/(a^2*b - 2*a* b^2 + b^3 + (a^2*b*e^(4*c) - 2*a*b^2*e^(4*c) + b^3*e^(4*c))*e^(4*d*x) + 2* (2*a^3*e^(2*c) - 5*a^2*b*e^(2*c) + 4*a*b^2*e^(2*c) - b^3*e^(2*c))*e^(2*d*x )), x)
\[ \int \frac {\sinh (c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\int { \frac {\sinh \left (d x + c\right )}{{\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {\sinh (c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\int \frac {\mathrm {sinh}\left (c+d\,x\right )}{{\left (b\,{\mathrm {sinh}\left (c+d\,x\right )}^2+a\right )}^3} \,d x \]