Integrand size = 23, antiderivative size = 152 \[ \int \frac {\coth ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=-\frac {b^4}{4 a^3 (a+b)^2 d \left (b+a \cosh ^2(c+d x)\right )^2}+\frac {b^3 (2 a+b)}{a^3 (a+b)^3 d \left (b+a \cosh ^2(c+d x)\right )}-\frac {\text {csch}^2(c+d x)}{2 (a+b)^3 d}+\frac {b^2 \left (6 a^2+4 a b+b^2\right ) \log \left (b+a \cosh ^2(c+d x)\right )}{2 a^3 (a+b)^4 d}+\frac {(a+4 b) \log (\sinh (c+d x))}{(a+b)^4 d} \]
-1/4*b^4/a^3/(a+b)^2/d/(b+a*cosh(d*x+c)^2)^2+b^3*(2*a+b)/a^3/(a+b)^3/d/(b+ a*cosh(d*x+c)^2)-1/2*csch(d*x+c)^2/(a+b)^3/d+1/2*b^2*(6*a^2+4*a*b+b^2)*ln( b+a*cosh(d*x+c)^2)/a^3/(a+b)^4/d+(a+4*b)*ln(sinh(d*x+c))/(a+b)^4/d
Time = 1.30 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.13 \[ \int \frac {\coth ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=-\frac {(a+2 b+a \cosh (2 (c+d x)))^3 \text {sech}^6(c+d x) \left (2 (a+b) \text {csch}^2(c+d x)-4 (a+4 b) \log (\sinh (c+d x))-\frac {2 b^2 \left (6 a^2+4 a b+b^2\right ) \log \left (a+b+a \sinh ^2(c+d x)\right )}{a^3}+\frac {b^4 (a+b)^2}{a^3 \left (a+b+a \sinh ^2(c+d x)\right )^2}-\frac {4 b^3 (a+b) (2 a+b)}{a^3 \left (a+b+a \sinh ^2(c+d x)\right )}\right )}{32 (a+b)^4 d \left (a+b \text {sech}^2(c+d x)\right )^3} \]
-1/32*((a + 2*b + a*Cosh[2*(c + d*x)])^3*Sech[c + d*x]^6*(2*(a + b)*Csch[c + d*x]^2 - 4*(a + 4*b)*Log[Sinh[c + d*x]] - (2*b^2*(6*a^2 + 4*a*b + b^2)* Log[a + b + a*Sinh[c + d*x]^2])/a^3 + (b^4*(a + b)^2)/(a^3*(a + b + a*Sinh [c + d*x]^2)^2) - (4*b^3*(a + b)*(2*a + b))/(a^3*(a + b + a*Sinh[c + d*x]^ 2))))/((a + b)^4*d*(a + b*Sech[c + d*x]^2)^3)
Time = 0.41 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 26, 4626, 354, 99, 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.
| \(\displaystyle \int \frac {\coth ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {i}{\tan (i c+i d x)^3 \left (a+b \sec (i c+i d x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle -i \int \frac {1}{\left (b \sec (i c+i d x)^2+a\right )^3 \tan (i c+i d x)^3}dx\) |
| \(\Big \downarrow \) 4626 |
| \(\displaystyle \frac {\int \frac {\cosh ^9(c+d x)}{\left (1-\cosh ^2(c+d x)\right )^2 \left (a \cosh ^2(c+d x)+b\right )^3}d\cosh (c+d x)}{d}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {\int \frac {\cosh ^8(c+d x)}{\left (1-\cosh ^2(c+d x)\right )^2 \left (a \cosh ^2(c+d x)+b\right )^3}d\cosh ^2(c+d x)}{2 d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\int \left (\frac {b^4}{a^2 (a+b)^2 \left (a \cosh ^2(c+d x)+b\right )^3}-\frac {2 (2 a+b) b^3}{a^2 (a+b)^3 \left (a \cosh ^2(c+d x)+b\right )^2}+\frac {\left (6 a^2+4 b a+b^2\right ) b^2}{a^2 (a+b)^4 \left (a \cosh ^2(c+d x)+b\right )}+\frac {a+4 b}{(a+b)^4 \left (\cosh ^2(c+d x)-1\right )}+\frac {1}{(a+b)^3 \left (\cosh ^2(c+d x)-1\right )^2}\right )d\cosh ^2(c+d x)}{2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {b^4}{2 a^3 (a+b)^2 \left (a \cosh ^2(c+d x)+b\right )^2}+\frac {2 b^3 (2 a+b)}{a^3 (a+b)^3 \left (a \cosh ^2(c+d x)+b\right )}+\frac {b^2 \left (6 a^2+4 a b+b^2\right ) \log \left (a \cosh ^2(c+d x)+b\right )}{a^3 (a+b)^4}+\frac {1}{(a+b)^3 \left (1-\cosh ^2(c+d x)\right )}+\frac {(a+4 b) \log \left (1-\cosh ^2(c+d x)\right )}{(a+b)^4}}{2 d}\) |
(1/((a + b)^3*(1 - Cosh[c + d*x]^2)) - b^4/(2*a^3*(a + b)^2*(b + a*Cosh[c + d*x]^2)^2) + (2*b^3*(2*a + b))/(a^3*(a + b)^3*(b + a*Cosh[c + d*x]^2)) + ((a + 4*b)*Log[1 - Cosh[c + d*x]^2])/(a + b)^4 + (b^2*(6*a^2 + 4*a*b + b^ 2)*Log[b + a*Cosh[c + d*x]^2])/(a^3*(a + b)^4))/(2*d)
3.2.67.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_ )]^(m_.), x_Symbol] :> Module[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-(f *ff^(m + n*p - 1))^(-1) Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*(ff* x)^n)^p/x^(m + n*p)), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n} , x] && IntegerQ[(m - 1)/2] && IntegerQ[n] && IntegerQ[p]
Leaf count of result is larger than twice the leaf count of optimal. \(353\) vs. \(2(146)=292\).
Time = 191.06 (sec) , antiderivative size = 354, normalized size of antiderivative = 2.33
| method | result | size |
| derivativedivides | \(\frac {-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {1}{8 \left (a +b \right )^{3} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (4 a +16 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 \left (a +b \right )^{4}}-\frac {\ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}+\frac {b^{2} \left (\frac {\left (-8 a^{3} b -10 a^{2} b^{2}-2 a \,b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-4 \left (4 a^{2}-2 a b -b^{2}\right ) a b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2 \left (4 a^{2}+5 a b +b^{2}\right ) a b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (6 a^{2}+4 a b +b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}{2}\right )}{\left (a +b \right )^{4} a^{3}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}}{d}\) | \(354\) |
| default | \(\frac {-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {1}{8 \left (a +b \right )^{3} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (4 a +16 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 \left (a +b \right )^{4}}-\frac {\ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}+\frac {b^{2} \left (\frac {\left (-8 a^{3} b -10 a^{2} b^{2}-2 a \,b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-4 \left (4 a^{2}-2 a b -b^{2}\right ) a b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2 \left (4 a^{2}+5 a b +b^{2}\right ) a b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (6 a^{2}+4 a b +b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}{2}\right )}{\left (a +b \right )^{4} a^{3}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}}{d}\) | \(354\) |
| risch | \(\text {Expression too large to display}\) | \(1033\) |
1/d*(-1/8*tanh(1/2*d*x+1/2*c)^2/(a^3+3*a^2*b+3*a*b^2+b^3)-1/8/(a+b)^3/tanh (1/2*d*x+1/2*c)^2+1/4/(a+b)^4*(4*a+16*b)*ln(tanh(1/2*d*x+1/2*c))-1/a^3*ln( 1+tanh(1/2*d*x+1/2*c))+b^2/(a+b)^4/a^3*(((-8*a^3*b-10*a^2*b^2-2*a*b^3)*tan h(1/2*d*x+1/2*c)^6-4*(4*a^2-2*a*b-b^2)*a*b*tanh(1/2*d*x+1/2*c)^4-2*(4*a^2+ 5*a*b+b^2)*a*b*tanh(1/2*d*x+1/2*c)^2)/(tanh(1/2*d*x+1/2*c)^4*a+tanh(1/2*d* x+1/2*c)^4*b+2*tanh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)^2*b+a+b)^2+1/ 2*(6*a^2+4*a*b+b^2)*ln(tanh(1/2*d*x+1/2*c)^4*a+tanh(1/2*d*x+1/2*c)^4*b+2*t anh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)^2*b+a+b))-1/a^3*ln(tanh(1/2*d *x+1/2*c)-1))
Leaf count of result is larger than twice the leaf count of optimal. 10255 vs. \(2 (146) = 292\).
Time = 0.80 (sec) , antiderivative size = 10255, normalized size of antiderivative = 67.47 \[ \int \frac {\coth ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
\[ \int \frac {\coth ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int \frac {\coth ^{3}{\left (c + d x \right )}}{\left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{3}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 692 vs. \(2 (146) = 292\).
Time = 0.27 (sec) , antiderivative size = 692, normalized size of antiderivative = 4.55 \[ \int \frac {\coth ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\frac {{\left (6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \log \left (2 \, {\left (a + 2 \, b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a e^{\left (-4 \, d x - 4 \, c\right )} + a\right )}{2 \, {\left (a^{7} + 4 \, a^{6} b + 6 \, a^{5} b^{2} + 4 \, a^{4} b^{3} + a^{3} b^{4}\right )} d} + \frac {{\left (a + 4 \, b\right )} \log \left (e^{\left (-d x - c\right )} + 1\right )}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} d} + \frac {{\left (a + 4 \, b\right )} \log \left (e^{\left (-d x - c\right )} - 1\right )}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} d} - \frac {2 \, {\left ({\left (a^{5} - 4 \, a^{2} b^{3} - 2 \, a b^{4}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 2 \, {\left (2 \, a^{5} + 4 \, a^{4} b - 7 \, a b^{4} - 3 \, b^{5}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + 2 \, {\left (3 \, a^{5} + 8 \, a^{4} b + 8 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + 16 \, a b^{4} + 6 \, b^{5}\right )} e^{\left (-6 \, d x - 6 \, c\right )} + 2 \, {\left (2 \, a^{5} + 4 \, a^{4} b - 7 \, a b^{4} - 3 \, b^{5}\right )} e^{\left (-8 \, d x - 8 \, c\right )} + {\left (a^{5} - 4 \, a^{2} b^{3} - 2 \, a b^{4}\right )} e^{\left (-10 \, d x - 10 \, c\right )}\right )}}{{\left (a^{8} + 3 \, a^{7} b + 3 \, a^{6} b^{2} + a^{5} b^{3} + 2 \, {\left (a^{8} + 7 \, a^{7} b + 15 \, a^{6} b^{2} + 13 \, a^{5} b^{3} + 4 \, a^{4} b^{4}\right )} e^{\left (-2 \, d x - 2 \, c\right )} - {\left (a^{8} + 3 \, a^{7} b - 13 \, a^{6} b^{2} - 47 \, a^{5} b^{3} - 48 \, a^{4} b^{4} - 16 \, a^{3} b^{5}\right )} e^{\left (-4 \, d x - 4 \, c\right )} - 4 \, {\left (a^{8} + 7 \, a^{7} b + 23 \, a^{6} b^{2} + 37 \, a^{5} b^{3} + 28 \, a^{4} b^{4} + 8 \, a^{3} b^{5}\right )} e^{\left (-6 \, d x - 6 \, c\right )} - {\left (a^{8} + 3 \, a^{7} b - 13 \, a^{6} b^{2} - 47 \, a^{5} b^{3} - 48 \, a^{4} b^{4} - 16 \, a^{3} b^{5}\right )} e^{\left (-8 \, d x - 8 \, c\right )} + 2 \, {\left (a^{8} + 7 \, a^{7} b + 15 \, a^{6} b^{2} + 13 \, a^{5} b^{3} + 4 \, a^{4} b^{4}\right )} e^{\left (-10 \, d x - 10 \, c\right )} + {\left (a^{8} + 3 \, a^{7} b + 3 \, a^{6} b^{2} + a^{5} b^{3}\right )} e^{\left (-12 \, d x - 12 \, c\right )}\right )} d} + \frac {d x + c}{a^{3} d} \]
1/2*(6*a^2*b^2 + 4*a*b^3 + b^4)*log(2*(a + 2*b)*e^(-2*d*x - 2*c) + a*e^(-4 *d*x - 4*c) + a)/((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*d) + ( a + 4*b)*log(e^(-d*x - c) + 1)/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4 )*d) + (a + 4*b)*log(e^(-d*x - c) - 1)/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b ^3 + b^4)*d) - 2*((a^5 - 4*a^2*b^3 - 2*a*b^4)*e^(-2*d*x - 2*c) + 2*(2*a^5 + 4*a^4*b - 7*a*b^4 - 3*b^5)*e^(-4*d*x - 4*c) + 2*(3*a^5 + 8*a^4*b + 8*a^3 *b^2 + 4*a^2*b^3 + 16*a*b^4 + 6*b^5)*e^(-6*d*x - 6*c) + 2*(2*a^5 + 4*a^4*b - 7*a*b^4 - 3*b^5)*e^(-8*d*x - 8*c) + (a^5 - 4*a^2*b^3 - 2*a*b^4)*e^(-10* d*x - 10*c))/((a^8 + 3*a^7*b + 3*a^6*b^2 + a^5*b^3 + 2*(a^8 + 7*a^7*b + 15 *a^6*b^2 + 13*a^5*b^3 + 4*a^4*b^4)*e^(-2*d*x - 2*c) - (a^8 + 3*a^7*b - 13* a^6*b^2 - 47*a^5*b^3 - 48*a^4*b^4 - 16*a^3*b^5)*e^(-4*d*x - 4*c) - 4*(a^8 + 7*a^7*b + 23*a^6*b^2 + 37*a^5*b^3 + 28*a^4*b^4 + 8*a^3*b^5)*e^(-6*d*x - 6*c) - (a^8 + 3*a^7*b - 13*a^6*b^2 - 47*a^5*b^3 - 48*a^4*b^4 - 16*a^3*b^5) *e^(-8*d*x - 8*c) + 2*(a^8 + 7*a^7*b + 15*a^6*b^2 + 13*a^5*b^3 + 4*a^4*b^4 )*e^(-10*d*x - 10*c) + (a^8 + 3*a^7*b + 3*a^6*b^2 + a^5*b^3)*e^(-12*d*x - 12*c))*d) + (d*x + c)/(a^3*d)
\[ \int \frac {\coth ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int { \frac {\coth \left (d x + c\right )^{3}}{{\left (b \operatorname {sech}\left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {\coth ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^6\,{\mathrm {coth}\left (c+d\,x\right )}^3}{{\left (a\,{\mathrm {cosh}\left (c+d\,x\right )}^2+b\right )}^3} \,d x \]