Integrand size = 21, antiderivative size = 116 \[ \int \frac {\sinh (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=-\frac {15 \sqrt {b} \arctan \left (\frac {\sqrt {a} \cosh (c+d x)}{\sqrt {b}}\right )}{8 a^{7/2} d}+\frac {15 \cosh (c+d x)}{8 a^3 d}-\frac {\cosh ^5(c+d x)}{4 a d \left (b+a \cosh ^2(c+d x)\right )^2}-\frac {5 \cosh ^3(c+d x)}{8 a^2 d \left (b+a \cosh ^2(c+d x)\right )} \]
15/8*cosh(d*x+c)/a^3/d-1/4*cosh(d*x+c)^5/a/d/(b+a*cosh(d*x+c)^2)^2-5/8*cos h(d*x+c)^3/a^2/d/(b+a*cosh(d*x+c)^2)-15/8*arctan(cosh(d*x+c)*a^(1/2)/b^(1/ 2))*b^(1/2)/a^(7/2)/d
Result contains complex when optimal does not.
Time = 11.49 (sec) , antiderivative size = 1272, normalized size of antiderivative = 10.97 \[ \int \frac {\sinh (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx =\text {Too large to display} \]
(5*(3*(a^2 - 4*a*b + 16*b^2)*ArcTan[((Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cosh[c ] - Sinh[c])^2])*Sinh[c]*Tanh[(d*x)/2] + Cosh[c]*(Sqrt[a] - I*Sqrt[a + b]* Sqrt[(Cosh[c] - Sinh[c])^2]*Tanh[(d*x)/2]))/Sqrt[b]] + 3*(a^2 - 4*a*b + 16 *b^2)*ArcTan[((Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2])*Sinh[c ]*Tanh[(d*x)/2] + Cosh[c]*(Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c] )^2]*Tanh[(d*x)/2]))/Sqrt[b]] + (8*Sqrt[a]*b^(3/2)*(a^2 + 12*a*b + 16*b^2) *Cosh[c + d*x])/(a + 2*b + a*Cosh[2*(c + d*x)])^2 + (2*Sqrt[a]*Sqrt[b]*(3* a^2 - 12*a*b - 80*b^2)*Cosh[c + d*x])/(a + 2*b + a*Cosh[2*(c + d*x)]))*(a + 2*b + a*Cosh[2*c + 2*d*x])^3*Sech[c + d*x]^6)/(4096*a^(5/2)*b^(5/2)*d*(a + b*Sech[c + d*x]^2)^3) + (5*((3*(ArcTan[(Sqrt[a] - I*Sqrt[a + b]*Tanh[(c + d*x)/2])/Sqrt[b]] + ArcTan[(Sqrt[a] + I*Sqrt[a + b]*Tanh[(c + d*x)/2])/ Sqrt[b]]))/Sqrt[a] + (2*Sqrt[b]*Cosh[c + d*x]*(3*a + 10*b + 3*a*Cosh[2*(c + d*x)]))/(a + 2*b + a*Cosh[2*(c + d*x)])^2)*(a + 2*b + a*Cosh[2*c + 2*d*x ])^3*Sech[c + d*x]^6)/(4096*b^(5/2)*d*(a + b*Sech[c + d*x]^2)^3) + (9*(-(( 3*a - 4*b)*(ArcTan[((Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2])* Sinh[c]*Tanh[(d*x)/2] + Cosh[c]*(Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cosh[c] - S inh[c])^2]*Tanh[(d*x)/2]))/Sqrt[b]] + ArcTan[((Sqrt[a] + I*Sqrt[a + b]*Sqr t[(Cosh[c] - Sinh[c])^2])*Sinh[c]*Tanh[(d*x)/2] + Cosh[c]*(Sqrt[a] + I*Sqr t[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2]*Tanh[(d*x)/2]))/Sqrt[b]])) - (2*Sqrt[ a]*Sqrt[b]*Cosh[c + d*x]*(3*a^2 + 6*a*b + 8*b^2 + a*(3*a - 4*b)*Cosh[2*...
Time = 0.27 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 26, 4621, 252, 252, 262, 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.
| \(\displaystyle \int \frac {\sinh (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {i \sin (i c+i d x)}{\left (a+b \sec (i c+i d x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle -i \int \frac {\sin (i c+i d x)}{\left (b \sec (i c+i d x)^2+a\right )^3}dx\) |
| \(\Big \downarrow \) 4621 |
| \(\displaystyle \frac {\int \frac {\cosh ^6(c+d x)}{\left (a \cosh ^2(c+d x)+b\right )^3}d\cosh (c+d x)}{d}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {\frac {5 \int \frac {\cosh ^4(c+d x)}{\left (a \cosh ^2(c+d x)+b\right )^2}d\cosh (c+d x)}{4 a}-\frac {\cosh ^5(c+d x)}{4 a \left (a \cosh ^2(c+d x)+b\right )^2}}{d}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {\frac {5 \left (\frac {3 \int \frac {\cosh ^2(c+d x)}{a \cosh ^2(c+d x)+b}d\cosh (c+d x)}{2 a}-\frac {\cosh ^3(c+d x)}{2 a \left (a \cosh ^2(c+d x)+b\right )}\right )}{4 a}-\frac {\cosh ^5(c+d x)}{4 a \left (a \cosh ^2(c+d x)+b\right )^2}}{d}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {\frac {5 \left (\frac {3 \left (\frac {\cosh (c+d x)}{a}-\frac {b \int \frac {1}{a \cosh ^2(c+d x)+b}d\cosh (c+d x)}{a}\right )}{2 a}-\frac {\cosh ^3(c+d x)}{2 a \left (a \cosh ^2(c+d x)+b\right )}\right )}{4 a}-\frac {\cosh ^5(c+d x)}{4 a \left (a \cosh ^2(c+d x)+b\right )^2}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {5 \left (\frac {3 \left (\frac {\cosh (c+d x)}{a}-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {a} \cosh (c+d x)}{\sqrt {b}}\right )}{a^{3/2}}\right )}{2 a}-\frac {\cosh ^3(c+d x)}{2 a \left (a \cosh ^2(c+d x)+b\right )}\right )}{4 a}-\frac {\cosh ^5(c+d x)}{4 a \left (a \cosh ^2(c+d x)+b\right )^2}}{d}\) |
(-1/4*Cosh[c + d*x]^5/(a*(b + a*Cosh[c + d*x]^2)^2) + (5*((3*(-((Sqrt[b]*A rcTan[(Sqrt[a]*Cosh[c + d*x])/Sqrt[b]])/a^(3/2)) + Cosh[c + d*x]/a))/(2*a) - Cosh[c + d*x]^3/(2*a*(b + a*Cosh[c + d*x]^2))))/(4*a))/d
3.1.44.3.1 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)^(-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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ )]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*(ff*x)^n)^p/(ff*x)^(n*p)), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/ 2] && IntegerQ[n] && IntegerQ[p]
Time = 164.16 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.72
| method | result | size |
| derivativedivides | \(-\frac {-\frac {1}{a^{3} \operatorname {sech}\left (d x +c \right )}-\frac {b \left (\frac {\frac {7 \operatorname {sech}\left (d x +c \right )^{3} b}{8}+\frac {9 \,\operatorname {sech}\left (d x +c \right ) a}{8}}{\left (a +b \operatorname {sech}\left (d x +c \right )^{2}\right )^{2}}+\frac {15 \arctan \left (\frac {b \,\operatorname {sech}\left (d x +c \right )}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{3}}}{d}\) | \(84\) |
| default | \(-\frac {-\frac {1}{a^{3} \operatorname {sech}\left (d x +c \right )}-\frac {b \left (\frac {\frac {7 \operatorname {sech}\left (d x +c \right )^{3} b}{8}+\frac {9 \,\operatorname {sech}\left (d x +c \right ) a}{8}}{\left (a +b \operatorname {sech}\left (d x +c \right )^{2}\right )^{2}}+\frac {15 \arctan \left (\frac {b \,\operatorname {sech}\left (d x +c \right )}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{3}}}{d}\) | \(84\) |
| risch | \(\frac {{\mathrm e}^{d x +c}}{2 a^{3} d}+\frac {{\mathrm e}^{-d x -c}}{2 a^{3} d}+\frac {\left (9 a \,{\mathrm e}^{6 d x +6 c}+27 a \,{\mathrm e}^{4 d x +4 c}+28 b \,{\mathrm e}^{4 d x +4 c}+27 \,{\mathrm e}^{2 d x +2 c} a +28 b \,{\mathrm e}^{2 d x +2 c}+9 a \right ) {\mathrm e}^{d x +c} b}{4 a^{3} \left (a \,{\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{2 d x +2 c} a +4 b \,{\mathrm e}^{2 d x +2 c}+a \right )^{2} d}+\frac {15 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-a b}\, {\mathrm e}^{d x +c}}{a}+1\right )}{16 a^{4} d}-\frac {15 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-a b}\, {\mathrm e}^{d x +c}}{a}+1\right )}{16 a^{4} d}\) | \(237\) |
-1/d*(-1/a^3/sech(d*x+c)-1/a^3*b*((7/8*sech(d*x+c)^3*b+9/8*sech(d*x+c)*a)/ (a+b*sech(d*x+c)^2)^2+15/8/(a*b)^(1/2)*arctan(b*sech(d*x+c)/(a*b)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 2599 vs. \(2 (100) = 200\).
Time = 0.32 (sec) , antiderivative size = 4829, normalized size of antiderivative = 41.63 \[ \int \frac {\sinh (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
[1/16*(8*a^2*cosh(d*x + c)^10 + 80*a^2*cosh(d*x + c)*sinh(d*x + c)^9 + 8*a ^2*sinh(d*x + c)^10 + 20*(2*a^2 + 5*a*b)*cosh(d*x + c)^8 + 20*(18*a^2*cosh (d*x + c)^2 + 2*a^2 + 5*a*b)*sinh(d*x + c)^8 + 160*(6*a^2*cosh(d*x + c)^3 + (2*a^2 + 5*a*b)*cosh(d*x + c))*sinh(d*x + c)^7 + 20*(4*a^2 + 15*a*b + 12 *b^2)*cosh(d*x + c)^6 + 20*(84*a^2*cosh(d*x + c)^4 + 28*(2*a^2 + 5*a*b)*co sh(d*x + c)^2 + 4*a^2 + 15*a*b + 12*b^2)*sinh(d*x + c)^6 + 8*(252*a^2*cosh (d*x + c)^5 + 140*(2*a^2 + 5*a*b)*cosh(d*x + c)^3 + 15*(4*a^2 + 15*a*b + 1 2*b^2)*cosh(d*x + c))*sinh(d*x + c)^5 + 20*(4*a^2 + 15*a*b + 12*b^2)*cosh( d*x + c)^4 + 20*(84*a^2*cosh(d*x + c)^6 + 70*(2*a^2 + 5*a*b)*cosh(d*x + c) ^4 + 15*(4*a^2 + 15*a*b + 12*b^2)*cosh(d*x + c)^2 + 4*a^2 + 15*a*b + 12*b^ 2)*sinh(d*x + c)^4 + 80*(12*a^2*cosh(d*x + c)^7 + 14*(2*a^2 + 5*a*b)*cosh( d*x + c)^5 + 5*(4*a^2 + 15*a*b + 12*b^2)*cosh(d*x + c)^3 + (4*a^2 + 15*a*b + 12*b^2)*cosh(d*x + c))*sinh(d*x + c)^3 + 20*(2*a^2 + 5*a*b)*cosh(d*x + c)^2 + 20*(18*a^2*cosh(d*x + c)^8 + 28*(2*a^2 + 5*a*b)*cosh(d*x + c)^6 + 1 5*(4*a^2 + 15*a*b + 12*b^2)*cosh(d*x + c)^4 + 6*(4*a^2 + 15*a*b + 12*b^2)* cosh(d*x + c)^2 + 2*a^2 + 5*a*b)*sinh(d*x + c)^2 + 15*(a^2*cosh(d*x + c)^9 + 9*a^2*cosh(d*x + c)*sinh(d*x + c)^8 + a^2*sinh(d*x + c)^9 + 4*(a^2 + 2* a*b)*cosh(d*x + c)^7 + 4*(9*a^2*cosh(d*x + c)^2 + a^2 + 2*a*b)*sinh(d*x + c)^7 + 28*(3*a^2*cosh(d*x + c)^3 + (a^2 + 2*a*b)*cosh(d*x + c))*sinh(d*x + c)^6 + 2*(3*a^2 + 8*a*b + 8*b^2)*cosh(d*x + c)^5 + 2*(63*a^2*cosh(d*x ...
Timed out. \[ \int \frac {\sinh (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\text {Timed out} \]
\[ \int \frac {\sinh (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int { \frac {\sinh \left (d x + c\right )}{{\left (b \operatorname {sech}\left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
1/4*(2*a^2*e^(10*d*x + 10*c) + 2*a^2 + 5*(2*a^2*e^(8*c) + 5*a*b*e^(8*c))*e ^(8*d*x) + 5*(4*a^2*e^(6*c) + 15*a*b*e^(6*c) + 12*b^2*e^(6*c))*e^(6*d*x) + 5*(4*a^2*e^(4*c) + 15*a*b*e^(4*c) + 12*b^2*e^(4*c))*e^(4*d*x) + 5*(2*a^2* e^(2*c) + 5*a*b*e^(2*c))*e^(2*d*x))/(a^5*d*e^(9*d*x + 9*c) + a^5*d*e^(d*x + c) + 4*(a^5*d*e^(7*c) + 2*a^4*b*d*e^(7*c))*e^(7*d*x) + 2*(3*a^5*d*e^(5*c ) + 8*a^4*b*d*e^(5*c) + 8*a^3*b^2*d*e^(5*c))*e^(5*d*x) + 4*(a^5*d*e^(3*c) + 2*a^4*b*d*e^(3*c))*e^(3*d*x)) - 1/2*integrate(15/2*(b*e^(3*d*x + 3*c) - b*e^(d*x + c))/(a^4*e^(4*d*x + 4*c) + a^4 + 2*(a^4*e^(2*c) + 2*a^3*b*e^(2* c))*e^(2*d*x)), x)
\[ \int \frac {\sinh (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int { \frac {\sinh \left (d x + c\right )}{{\left (b \operatorname {sech}\left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
Time = 2.22 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.89 \[ \int \frac {\sinh (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\frac {\frac {7\,b^2\,\mathrm {cosh}\left (c+d\,x\right )}{8}+\frac {9\,a\,b\,{\mathrm {cosh}\left (c+d\,x\right )}^3}{8}}{d\,a^5\,{\mathrm {cosh}\left (c+d\,x\right )}^4+2\,d\,a^4\,b\,{\mathrm {cosh}\left (c+d\,x\right )}^2+d\,a^3\,b^2}+\frac {\mathrm {cosh}\left (c+d\,x\right )}{a^3\,d}-\frac {15\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\mathrm {cosh}\left (c+d\,x\right )}{\sqrt {b}}\right )}{8\,a^{7/2}\,d} \]