Integrand size = 23, antiderivative size = 55 \[ \int \coth ^3(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=-\frac {(a+b)^2 \text {csch}^2(c+d x)}{2 d}+\frac {b^2 \log (\cosh (c+d x))}{d}+\frac {\left (a^2-b^2\right ) \log (\sinh (c+d x))}{d} \] Output:
-1/2*(a+b)^2*csch(d*x+c)^2/d+b^2*ln(cosh(d*x+c))/d+(a^2-b^2)*ln(sinh(d*x+c ))/d
Time = 0.15 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.49 \[ \int \coth ^3(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=-\frac {2 \left (b+a \cosh ^2(c+d x)\right )^2 \left ((a+b)^2 \text {csch}^2(c+d x)-2 \left (b^2 \log (\cosh (c+d x))+\left (a^2-b^2\right ) \log (\sinh (c+d x))\right )\right )}{d (a+2 b+a \cosh (2 (c+d x)))^2} \] Input:
Integrate[Coth[c + d*x]^3*(a + b*Sech[c + d*x]^2)^2,x]
Output:
(-2*(b + a*Cosh[c + d*x]^2)^2*((a + b)^2*Csch[c + d*x]^2 - 2*(b^2*Log[Cosh [c + d*x]] + (a^2 - b^2)*Log[Sinh[c + d*x]])))/(d*(a + 2*b + a*Cosh[2*(c + d*x)])^2)
Time = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.16, 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.
\(\displaystyle \int \coth ^3(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {i \left (a+b \sec (i c+i d x)^2\right )^2}{\tan (i c+i d x)^3}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {\left (b \sec (i c+i d x)^2+a\right )^2}{\tan (i c+i d x)^3}dx\) |
\(\Big \downarrow \) 4626 |
\(\displaystyle \frac {\int \frac {\left (a \cosh ^2(c+d x)+b\right )^2 \text {sech}(c+d x)}{\left (1-\cosh ^2(c+d x)\right )^2}d\cosh (c+d x)}{d}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {\int \frac {\left (a \cosh ^2(c+d x)+b\right )^2 \text {sech}(c+d x)}{\left (1-\cosh ^2(c+d x)\right )^2}d\cosh ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\text {sech}(c+d x) b^2+\frac {a^2-b^2}{\cosh ^2(c+d x)-1}+\frac {(a+b)^2}{\left (\cosh ^2(c+d x)-1\right )^2}\right )d\cosh ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (a^2-b^2\right ) \log \left (1-\cosh ^2(c+d x)\right )+\frac {(a+b)^2}{1-\cosh ^2(c+d x)}+b^2 \log \left (\cosh ^2(c+d x)\right )}{2 d}\) |
Input:
Int[Coth[c + d*x]^3*(a + b*Sech[c + d*x]^2)^2,x]
Output:
((a + b)^2/(1 - Cosh[c + d*x]^2) + b^2*Log[Cosh[c + d*x]^2] + (a^2 - b^2)* Log[1 - Cosh[c + d*x]^2])/(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_))^(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]
Time = 10.51 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.16
method | result | size |
derivativedivides | \(\frac {a^{2} \left (\ln \left (\sinh \left (d x +c \right )\right )-\frac {\coth \left (d x +c \right )^{2}}{2}\right )-\frac {a b}{\sinh \left (d x +c \right )^{2}}+b^{2} \left (-\frac {1}{2 \sinh \left (d x +c \right )^{2}}-\ln \left (\tanh \left (d x +c \right )\right )\right )}{d}\) | \(64\) |
default | \(\frac {a^{2} \left (\ln \left (\sinh \left (d x +c \right )\right )-\frac {\coth \left (d x +c \right )^{2}}{2}\right )-\frac {a b}{\sinh \left (d x +c \right )^{2}}+b^{2} \left (-\frac {1}{2 \sinh \left (d x +c \right )^{2}}-\ln \left (\tanh \left (d x +c \right )\right )\right )}{d}\) | \(64\) |
risch | \(-a^{2} x -\frac {2 a^{2} c}{d}-\frac {2 \,{\mathrm e}^{2 d x +2 c} \left (a^{2}+2 a b +b^{2}\right )}{d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}-1\right ) a^{2}}{d}-\frac {\ln \left ({\mathrm e}^{2 d x +2 c}-1\right ) b^{2}}{d}+\frac {b^{2} \ln \left ({\mathrm e}^{2 d x +2 c}+1\right )}{d}\) | \(113\) |
Input:
int(coth(d*x+c)^3*(a+b*sech(d*x+c)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/d*(a^2*(ln(sinh(d*x+c))-1/2*coth(d*x+c)^2)-a*b/sinh(d*x+c)^2+b^2*(-1/2/s inh(d*x+c)^2-ln(tanh(d*x+c))))
Leaf count of result is larger than twice the leaf count of optimal. 637 vs. \(2 (53) = 106\).
Time = 0.27 (sec) , antiderivative size = 637, normalized size of antiderivative = 11.58 \[ \int \coth ^3(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx =\text {Too large to display} \] Input:
integrate(coth(d*x+c)^3*(a+b*sech(d*x+c)^2)^2,x, algorithm="fricas")
Output:
-(a^2*d*x*cosh(d*x + c)^4 + 4*a^2*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + a^2* d*x*sinh(d*x + c)^4 + a^2*d*x - 2*(a^2*d*x - a^2 - 2*a*b - b^2)*cosh(d*x + c)^2 + 2*(3*a^2*d*x*cosh(d*x + c)^2 - a^2*d*x + a^2 + 2*a*b + b^2)*sinh(d *x + c)^2 - (b^2*cosh(d*x + c)^4 + 4*b^2*cosh(d*x + c)*sinh(d*x + c)^3 + b ^2*sinh(d*x + c)^4 - 2*b^2*cosh(d*x + c)^2 + 2*(3*b^2*cosh(d*x + c)^2 - b^ 2)*sinh(d*x + c)^2 + b^2 + 4*(b^2*cosh(d*x + c)^3 - b^2*cosh(d*x + c))*sin h(d*x + c))*log(2*cosh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) - ((a^2 - b^2)*cosh(d*x + c)^4 + 4*(a^2 - b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 - b^2)*sinh(d*x + c)^4 - 2*(a^2 - b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 - b^2) *cosh(d*x + c)^2 - a^2 + b^2)*sinh(d*x + c)^2 + a^2 - b^2 + 4*((a^2 - b^2) *cosh(d*x + c)^3 - (a^2 - b^2)*cosh(d*x + c))*sinh(d*x + c))*log(2*sinh(d* x + c)/(cosh(d*x + c) - sinh(d*x + c))) + 4*(a^2*d*x*cosh(d*x + c)^3 - (a^ 2*d*x - a^2 - 2*a*b - b^2)*cosh(d*x + c))*sinh(d*x + c))/(d*cosh(d*x + c)^ 4 + 4*d*cosh(d*x + c)*sinh(d*x + c)^3 + d*sinh(d*x + c)^4 - 2*d*cosh(d*x + c)^2 + 2*(3*d*cosh(d*x + c)^2 - d)*sinh(d*x + c)^2 + 4*(d*cosh(d*x + c)^3 - d*cosh(d*x + c))*sinh(d*x + c) + d)
Timed out. \[ \int \coth ^3(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=\text {Timed out} \] Input:
integrate(coth(d*x+c)**3*(a+b*sech(d*x+c)**2)**2,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 206 vs. \(2 (53) = 106\).
Time = 0.13 (sec) , antiderivative size = 206, normalized size of antiderivative = 3.75 \[ \int \coth ^3(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=a^{2} {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} + \frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )}}\right )} - b^{2} {\left (\frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} - \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} - \frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )}}\right )} - \frac {4 \, a b}{d {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2}} \] Input:
integrate(coth(d*x+c)^3*(a+b*sech(d*x+c)^2)^2,x, algorithm="maxima")
Output:
a^2*(x + c/d + log(e^(-d*x - c) + 1)/d + log(e^(-d*x - c) - 1)/d + 2*e^(-2 *d*x - 2*c)/(d*(2*e^(-2*d*x - 2*c) - e^(-4*d*x - 4*c) - 1))) - b^2*(log(e^ (-d*x - c) + 1)/d + log(e^(-d*x - c) - 1)/d - log(e^(-2*d*x - 2*c) + 1)/d - 2*e^(-2*d*x - 2*c)/(d*(2*e^(-2*d*x - 2*c) - e^(-4*d*x - 4*c) - 1))) - 4* a*b/(d*(e^(d*x + c) - e^(-d*x - c))^2)
Leaf count of result is larger than twice the leaf count of optimal. 148 vs. \(2 (53) = 106\).
Time = 0.18 (sec) , antiderivative size = 148, normalized size of antiderivative = 2.69 \[ \int \coth ^3(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=\frac {b^{2} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} + 2\right ) + {\left (a^{2} - b^{2}\right )} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} - 2\right ) - \frac {a^{2} {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} - b^{2} {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 2 \, a^{2} + 8 \, a b + 6 \, b^{2}}{e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} - 2}}{2 \, d} \] Input:
integrate(coth(d*x+c)^3*(a+b*sech(d*x+c)^2)^2,x, algorithm="giac")
Output:
1/2*(b^2*log(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c) + 2) + (a^2 - b^2)*log(e^( 2*d*x + 2*c) + e^(-2*d*x - 2*c) - 2) - (a^2*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) - b^2*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) + 2*a^2 + 8*a*b + 6*b^2) /(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c) - 2))/d
Time = 0.20 (sec) , antiderivative size = 240, normalized size of antiderivative = 4.36 \[ \int \coth ^3(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=\frac {\ln \left ({\mathrm {e}}^{4\,c+4\,d\,x}-1\right )\,\left (d\,\left (a^2-b^2\right )+b^2\,d\right )}{2\,d^2}-a^2\,x-\frac {2\,\left (a^2+2\,a\,b+b^2\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\left (a^4\,\sqrt {-d^2}+4\,b^4\,\sqrt {-d^2}-4\,a^2\,b^2\,\sqrt {-d^2}\right )}{a^2\,d\,\sqrt {a^4-4\,a^2\,b^2+4\,b^4}-2\,b^2\,d\,\sqrt {a^4-4\,a^2\,b^2+4\,b^4}}\right )\,\sqrt {a^4-4\,a^2\,b^2+4\,b^4}}{\sqrt {-d^2}}-\frac {2\,\left (a^2+2\,a\,b+b^2\right )}{d\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \] Input:
int(coth(c + d*x)^3*(a + b/cosh(c + d*x)^2)^2,x)
Output:
(log(exp(4*c + 4*d*x) - 1)*(d*(a^2 - b^2) + b^2*d))/(2*d^2) - a^2*x - (2*( 2*a*b + a^2 + b^2))/(d*(exp(2*c + 2*d*x) - 1)) - (atan((exp(2*c)*exp(2*d*x )*(a^4*(-d^2)^(1/2) + 4*b^4*(-d^2)^(1/2) - 4*a^2*b^2*(-d^2)^(1/2)))/(a^2*d *(a^4 + 4*b^4 - 4*a^2*b^2)^(1/2) - 2*b^2*d*(a^4 + 4*b^4 - 4*a^2*b^2)^(1/2) ))*(a^4 + 4*b^4 - 4*a^2*b^2)^(1/2))/(-d^2)^(1/2) - (2*(2*a*b + a^2 + b^2)) /(d*(exp(4*c + 4*d*x) - 2*exp(2*c + 2*d*x) + 1))
Time = 0.20 (sec) , antiderivative size = 458, normalized size of antiderivative = 8.33 \[ \int \coth ^3(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=\frac {e^{4 d x +4 c} \mathrm {log}\left (e^{2 d x +2 c}+1\right ) b^{2}+e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}-1\right ) a^{2}-e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}-1\right ) b^{2}+e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}+1\right ) a^{2}-e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}+1\right ) b^{2}-e^{4 d x +4 c} a^{2} d x -e^{4 d x +4 c} a^{2}-2 e^{4 d x +4 c} a b -e^{4 d x +4 c} b^{2}-2 e^{2 d x +2 c} \mathrm {log}\left (e^{2 d x +2 c}+1\right ) b^{2}-2 e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}-1\right ) a^{2}+2 e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}-1\right ) b^{2}-2 e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}+1\right ) a^{2}+2 e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}+1\right ) b^{2}+2 e^{2 d x +2 c} a^{2} d x +\mathrm {log}\left (e^{2 d x +2 c}+1\right ) b^{2}+\mathrm {log}\left (e^{d x +c}-1\right ) a^{2}-\mathrm {log}\left (e^{d x +c}-1\right ) b^{2}+\mathrm {log}\left (e^{d x +c}+1\right ) a^{2}-\mathrm {log}\left (e^{d x +c}+1\right ) b^{2}-a^{2} d x -a^{2}-2 a b -b^{2}}{d \left (e^{4 d x +4 c}-2 e^{2 d x +2 c}+1\right )} \] Input:
int(coth(d*x+c)^3*(a+b*sech(d*x+c)^2)^2,x)
Output:
(e**(4*c + 4*d*x)*log(e**(2*c + 2*d*x) + 1)*b**2 + e**(4*c + 4*d*x)*log(e* *(c + d*x) - 1)*a**2 - e**(4*c + 4*d*x)*log(e**(c + d*x) - 1)*b**2 + e**(4 *c + 4*d*x)*log(e**(c + d*x) + 1)*a**2 - e**(4*c + 4*d*x)*log(e**(c + d*x) + 1)*b**2 - e**(4*c + 4*d*x)*a**2*d*x - e**(4*c + 4*d*x)*a**2 - 2*e**(4*c + 4*d*x)*a*b - e**(4*c + 4*d*x)*b**2 - 2*e**(2*c + 2*d*x)*log(e**(2*c + 2 *d*x) + 1)*b**2 - 2*e**(2*c + 2*d*x)*log(e**(c + d*x) - 1)*a**2 + 2*e**(2* c + 2*d*x)*log(e**(c + d*x) - 1)*b**2 - 2*e**(2*c + 2*d*x)*log(e**(c + d*x ) + 1)*a**2 + 2*e**(2*c + 2*d*x)*log(e**(c + d*x) + 1)*b**2 + 2*e**(2*c + 2*d*x)*a**2*d*x + log(e**(2*c + 2*d*x) + 1)*b**2 + log(e**(c + d*x) - 1)*a **2 - log(e**(c + d*x) - 1)*b**2 + log(e**(c + d*x) + 1)*a**2 - log(e**(c + d*x) + 1)*b**2 - a**2*d*x - a**2 - 2*a*b - b**2)/(d*(e**(4*c + 4*d*x) - 2*e**(2*c + 2*d*x) + 1))