Integrand size = 21, antiderivative size = 113 \[ \int \frac {\sinh (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=-\frac {15 \sqrt {b} \arctan \left (\frac {\sqrt {a} \cosh (c+d x)}{\sqrt {b}}\right )}{8 a^{7/2} d}+\frac {\cosh (c+d x)}{a^3 d}-\frac {b^2 \cosh (c+d x)}{4 a^3 d \left (b+a \cosh ^2(c+d x)\right )^2}+\frac {9 b \cosh (c+d x)}{8 a^3 d \left (b+a \cosh ^2(c+d x)\right )} \] Output:
-15/8*b^(1/2)*arctan(a^(1/2)*cosh(d*x+c)/b^(1/2))/a^(7/2)/d+cosh(d*x+c)/a^ 3/d-1/4*b^2*cosh(d*x+c)/a^3/d/(b+a*cosh(d*x+c)^2)^2+9/8*b*cosh(d*x+c)/a^3/ d/(b+a*cosh(d*x+c)^2)
Result contains complex when optimal does not.
Time = 6.38 (sec) , antiderivative size = 453, normalized size of antiderivative = 4.01 \[ \int \frac {\sinh (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\frac {(a+2 b+a \cosh (2 (c+d x)))^3 \text {sech}^6(c+d x) \left (-\frac {15 \left (\left (a^3+64 b^3\right ) \arctan \left (\frac {\left (\sqrt {a}-i \sqrt {a+b} \sqrt {(\cosh (c)-\sinh (c))^2}\right ) \sinh (c) \tanh \left (\frac {d x}{2}\right )+\cosh (c) \left (\sqrt {a}-i \sqrt {a+b} \sqrt {(\cosh (c)-\sinh (c))^2} \tanh \left (\frac {d x}{2}\right )\right )}{\sqrt {b}}\right )+\left (a^3+64 b^3\right ) \arctan \left (\frac {\left (\sqrt {a}+i \sqrt {a+b} \sqrt {(\cosh (c)-\sinh (c))^2}\right ) \sinh (c) \tanh \left (\frac {d x}{2}\right )+\cosh (c) \left (\sqrt {a}+i \sqrt {a+b} \sqrt {(\cosh (c)-\sinh (c))^2} \tanh \left (\frac {d x}{2}\right )\right )}{\sqrt {b}}\right )-a^3 \left (\arctan \left (\frac {\sqrt {a}-i \sqrt {a+b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b}}\right )+\arctan \left (\frac {\sqrt {a}+i \sqrt {a+b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b}}\right )\right )\right )}{a^{7/2} b^{5/2}}+\frac {512 \cosh (c) \cosh (d x)}{a^3}+\frac {8 \cosh (c+d x) \left (16 b^3 (9 a+14 b)+3 \left (a^4+48 a b^3\right ) \cosh (2 (c+d x))\right )}{a^3 b^2 (a+2 b+a \cosh (2 (c+d x)))^2}+\frac {512 \sinh (c) \sinh (d x)}{a^3}-\frac {6 a \text {csch}(c+d x) \sinh (4 (c+d x))}{b^2 (a+2 b+a \cosh (2 (c+d x)))^2}\right )}{4096 d \left (a+b \text {sech}^2(c+d x)\right )^3} \] Input:
Integrate[Sinh[c + d*x]/(a + b*Sech[c + d*x]^2)^3,x]
Output:
((a + 2*b + a*Cosh[2*(c + d*x)])^3*Sech[c + d*x]^6*((-15*((a^3 + 64*b^3)*A rcTan[((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]*Ta nh[(d*x)/2]))/Sqrt[b]] + (a^3 + 64*b^3)*ArcTan[((Sqrt[a] + I*Sqrt[a + b]*S qrt[(Cosh[c] - Sinh[c])^2])*Sinh[c]*Tanh[(d*x)/2] + Cosh[c]*(Sqrt[a] + I*S qrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2]*Tanh[(d*x)/2]))/Sqrt[b]] - a^3*(Arc Tan[(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]])))/(a^(7/2)*b^(5/2)) + (512*C osh[c]*Cosh[d*x])/a^3 + (8*Cosh[c + d*x]*(16*b^3*(9*a + 14*b) + 3*(a^4 + 4 8*a*b^3)*Cosh[2*(c + d*x)]))/(a^3*b^2*(a + 2*b + a*Cosh[2*(c + d*x)])^2) + (512*Sinh[c]*Sinh[d*x])/a^3 - (6*a*Csch[c + d*x]*Sinh[4*(c + d*x)])/(b^2* (a + 2*b + a*Cosh[2*(c + d*x)])^2)))/(4096*d*(a + b*Sech[c + d*x]^2)^3)
Time = 0.27 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.05, 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 |
\(\displaystyle \frac {\frac {5 \left (\frac {3 \left (\frac {\cosh (c+d x)}{a}-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {a} \cosh (c+d x)}{\sqrt {b}}\right )}{a^{3/2}}\right )}{2 a}-\frac {\cosh ^3(c+d x)}{2 a \left (a \cosh ^2(c+d x)+b\right )}\right )}{4 a}-\frac {\cosh ^5(c+d x)}{4 a \left (a \cosh ^2(c+d x)+b\right )^2}}{d}\) |
Input:
Int[Sinh[c + d*x]/(a + b*Sech[c + d*x]^2)^3,x]
Output:
(-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
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 = 194.52 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.74
method | result | size |
derivativedivides | \(-\frac {-\frac {1}{a^{3} \operatorname {sech}\left (d x +c \right )}-\frac {b \left (\frac {\frac {7 \operatorname {sech}\left (d x +c \right )^{3} b}{8}+\frac {9 a \,\operatorname {sech}\left (d x +c \right )}{8}}{\left (a +b \operatorname {sech}\left (d x +c \right )^{2}\right )^{2}}+\frac {15 \arctan \left (\frac {b \,\operatorname {sech}\left (d x +c \right )}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{3}}}{d}\) | \(84\) |
default | \(-\frac {-\frac {1}{a^{3} \operatorname {sech}\left (d x +c \right )}-\frac {b \left (\frac {\frac {7 \operatorname {sech}\left (d x +c \right )^{3} b}{8}+\frac {9 a \,\operatorname {sech}\left (d x +c \right )}{8}}{\left (a +b \operatorname {sech}\left (d x +c \right )^{2}\right )^{2}}+\frac {15 \arctan \left (\frac {b \,\operatorname {sech}\left (d x +c \right )}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{3}}}{d}\) | \(84\) |
risch | \(\frac {{\mathrm e}^{d x +c}}{2 a^{3} d}+\frac {{\mathrm e}^{-d x -c}}{2 a^{3} d}+\frac {\left (9 \,{\mathrm e}^{6 d x +6 c} a +27 \,{\mathrm e}^{4 d x +4 c} a +28 \,{\mathrm e}^{4 d x +4 c} b +27 a \,{\mathrm e}^{2 d x +2 c}+28 b \,{\mathrm e}^{2 d x +2 c}+9 a \right ) b \,{\mathrm e}^{d x +c}}{4 a^{3} \left ({\mathrm e}^{4 d x +4 c} a +2 a \,{\mathrm e}^{2 d x +2 c}+4 b \,{\mathrm e}^{2 d x +2 c}+a \right )^{2} d}+\frac {15 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-a b}\, {\mathrm e}^{d x +c}}{a}+1\right )}{16 a^{4} d}-\frac {15 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-a b}\, {\mathrm e}^{d x +c}}{a}+1\right )}{16 a^{4} d}\) | \(237\) |
Input:
int(sinh(d*x+c)/(a+b*sech(d*x+c)^2)^3,x,method=_RETURNVERBOSE)
Output:
-1/d*(-1/a^3/sech(d*x+c)-1/a^3*b*((7/8*sech(d*x+c)^3*b+9/8*a*sech(d*x+c))/ (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. \(2 (99) = 198\).
Time = 0.24 (sec) , antiderivative size = 4829, normalized size of antiderivative = 42.73 \[ \int \frac {\sinh (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\text {Too large to display} \] Input:
integrate(sinh(d*x+c)/(a+b*sech(d*x+c)^2)^3,x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {\sinh (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\text {Timed out} \] Input:
integrate(sinh(d*x+c)/(a+b*sech(d*x+c)**2)**3,x)
Output:
Timed out
\[ \int \frac {\sinh (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int { \frac {\sinh \left (d x + c\right )}{{\left (b \operatorname {sech}\left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \] Input:
integrate(sinh(d*x+c)/(a+b*sech(d*x+c)^2)^3,x, algorithm="maxima")
Output:
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)
Exception generated. \[ \int \frac {\sinh (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(sinh(d*x+c)/(a+b*sech(d*x+c)^2)^3,x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Limit: Max order reached or unable to make series expansion Error: Bad Argument Value
Time = 2.45 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.91 \[ \int \frac {\sinh (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\frac {\frac {7\,b^2\,\mathrm {cosh}\left (c+d\,x\right )}{8}+\frac {9\,a\,b\,{\mathrm {cosh}\left (c+d\,x\right )}^3}{8}}{d\,a^5\,{\mathrm {cosh}\left (c+d\,x\right )}^4+2\,d\,a^4\,b\,{\mathrm {cosh}\left (c+d\,x\right )}^2+d\,a^3\,b^2}+\frac {\mathrm {cosh}\left (c+d\,x\right )}{a^3\,d}-\frac {15\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\mathrm {cosh}\left (c+d\,x\right )}{\sqrt {b}}\right )}{8\,a^{7/2}\,d} \] Input:
int(sinh(c + d*x)/(a + b/cosh(c + d*x)^2)^3,x)
Output:
((7*b^2*cosh(c + d*x))/8 + (9*a*b*cosh(c + d*x)^3)/8)/(a^5*d*cosh(c + d*x) ^4 + a^3*b^2*d + 2*a^4*b*d*cosh(c + d*x)^2) + cosh(c + d*x)/(a^3*d) - (15* b^(1/2)*atan((a^(1/2)*cosh(c + d*x))/b^(1/2)))/(8*a^(7/2)*d)
Time = 0.32 (sec) , antiderivative size = 3809, normalized size of antiderivative = 33.71 \[ \int \frac {\sinh (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx =\text {Too large to display} \] Input:
int(sinh(d*x+c)/(a+b*sech(d*x+c)^2)^3,x)
Output:
( - 30*e**(9*c + 9*d*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((e**(c + d*x)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*a**2 - 120*e**(7*c + 7*d*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt( 2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((e**(c + d*x)*a)/(sqrt(a)*sqrt(2*sqr t(b)*sqrt(a + b) + a + 2*b)))*a**2 - 240*e**(7*c + 7*d*x)*sqrt(b)*sqrt(a)* sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((e**(c + d*x)*a)/(s qrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*a*b - 180*e**(5*c + 5*d*x)* sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((e* *(c + d*x)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*a**2 - 480* e**(5*c + 5*d*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((e**(c + d*x)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2 *b)))*a*b - 480*e**(5*c + 5*d*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b )*sqrt(a + b) + a + 2*b)*atan((e**(c + d*x)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqr t(a + b) + a + 2*b)))*b**2 - 120*e**(3*c + 3*d*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((e**(c + d*x)*a)/(sqrt(a)*s qrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*a**2 - 240*e**(3*c + 3*d*x)*sqrt(b) *sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((e**(c + d *x)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*a*b - 30*e**(c + d *x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan ((e**(c + d*x)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*a**2...