Integrand size = 21, antiderivative size = 84 \[ \int \coth (c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=-\frac {b \left (3 a^2+3 a b+b^2\right ) \log (\cosh (c+d x))}{d}+\frac {(a+b)^3 \log (\sinh (c+d x))}{d}+\frac {b^2 (3 a+b) \text {sech}^2(c+d x)}{2 d}+\frac {b^3 \text {sech}^4(c+d x)}{4 d} \]
-b*(3*a^2+3*a*b+b^2)*ln(cosh(d*x+c))/d+(a+b)^3*ln(sinh(d*x+c))/d+1/2*b^2*( 3*a+b)*sech(d*x+c)^2/d+1/4*b^3*sech(d*x+c)^4/d
Time = 0.69 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.36 \[ \int \coth (c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=-\frac {2 \cosh ^6(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \left (4 b \left (3 a^2+3 a b+b^2\right ) \log (\cosh (c+d x))-4 (a+b)^3 \log (\sinh (c+d x))-2 b^2 (3 a+b) \text {sech}^2(c+d x)-b^3 \text {sech}^4(c+d x)\right )}{d (a+2 b+a \cosh (2 c+2 d x))^3} \]
(-2*Cosh[c + d*x]^6*(a + b*Sech[c + d*x]^2)^3*(4*b*(3*a^2 + 3*a*b + b^2)*L og[Cosh[c + d*x]] - 4*(a + b)^3*Log[Sinh[c + d*x]] - 2*b^2*(3*a + b)*Sech[ c + d*x]^2 - b^3*Sech[c + d*x]^4))/(d*(a + 2*b + a*Cosh[2*c + 2*d*x])^3)
Time = 0.31 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, 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.
\(\displaystyle \int \coth (c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i \left (a+b \sec (i c+i d x)^2\right )^3}{\tan (i c+i d x)}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {\left (b \sec (i c+i d x)^2+a\right )^3}{\tan (i c+i d x)}dx\) |
\(\Big \downarrow \) 4626 |
\(\displaystyle -\frac {\int \frac {\left (a \cosh ^2(c+d x)+b\right )^3 \text {sech}^5(c+d x)}{1-\cosh ^2(c+d x)}d\cosh (c+d x)}{d}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle -\frac {\int \frac {\left (a \cosh ^2(c+d x)+b\right )^3 \text {sech}^3(c+d x)}{1-\cosh ^2(c+d x)}d\cosh ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {\int \left (-\frac {(a+b)^3}{\cosh ^2(c+d x)-1}+b^3 \text {sech}^3(c+d x)+b^2 (3 a+b) \text {sech}^2(c+d x)+b \left (3 a^2+3 b a+b^2\right ) \text {sech}(c+d x)\right )d\cosh ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {b \left (3 a^2+3 a b+b^2\right ) \log \left (\cosh ^2(c+d x)\right )-b^2 (3 a+b) \text {sech}(c+d x)-(a+b)^3 \log \left (1-\cosh ^2(c+d x)\right )-\frac {1}{2} b^3 \text {sech}^2(c+d x)}{2 d}\) |
-1/2*(b*(3*a^2 + 3*a*b + b^2)*Log[Cosh[c + d*x]^2] - (a + b)^3*Log[1 - Cos h[c + d*x]^2] - b^2*(3*a + b)*Sech[c + d*x] - (b^3*Sech[c + d*x]^2)/2)/d
3.2.29.3.1 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]
Time = 46.66 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.02
method | result | size |
derivativedivides | \(\frac {a^{3} \ln \left (\sinh \left (d x +c \right )\right )+3 \ln \left (\tanh \left (d x +c \right )\right ) a^{2} b +3 a \,b^{2} \left (\frac {1}{2 \cosh \left (d x +c \right )^{2}}+\ln \left (\tanh \left (d x +c \right )\right )\right )+b^{3} \left (\frac {1}{4 \cosh \left (d x +c \right )^{4}}+\frac {1}{2 \cosh \left (d x +c \right )^{2}}+\ln \left (\tanh \left (d x +c \right )\right )\right )}{d}\) | \(86\) |
default | \(\frac {a^{3} \ln \left (\sinh \left (d x +c \right )\right )+3 \ln \left (\tanh \left (d x +c \right )\right ) a^{2} b +3 a \,b^{2} \left (\frac {1}{2 \cosh \left (d x +c \right )^{2}}+\ln \left (\tanh \left (d x +c \right )\right )\right )+b^{3} \left (\frac {1}{4 \cosh \left (d x +c \right )^{4}}+\frac {1}{2 \cosh \left (d x +c \right )^{2}}+\ln \left (\tanh \left (d x +c \right )\right )\right )}{d}\) | \(86\) |
risch | \(-a^{3} x -\frac {2 a^{3} c}{d}+\frac {2 b^{2} {\mathrm e}^{2 d x +2 c} \left (3 a \,{\mathrm e}^{4 d x +4 c}+b \,{\mathrm e}^{4 d x +4 c}+6 \,{\mathrm e}^{2 d x +2 c} a +4 b \,{\mathrm e}^{2 d x +2 c}+3 a +b \right )}{d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{4}}-\frac {3 b \ln \left ({\mathrm e}^{2 d x +2 c}+1\right ) a^{2}}{d}-\frac {3 b^{2} \ln \left ({\mathrm e}^{2 d x +2 c}+1\right ) a}{d}-\frac {b^{3} \ln \left ({\mathrm e}^{2 d x +2 c}+1\right )}{d}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}-1\right ) a^{3}}{d}+\frac {3 \ln \left ({\mathrm e}^{2 d x +2 c}-1\right ) a^{2} b}{d}+\frac {3 \ln \left ({\mathrm e}^{2 d x +2 c}-1\right ) a \,b^{2}}{d}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}-1\right ) b^{3}}{d}\) | \(241\) |
1/d*(a^3*ln(sinh(d*x+c))+3*ln(tanh(d*x+c))*a^2*b+3*a*b^2*(1/2/cosh(d*x+c)^ 2+ln(tanh(d*x+c)))+b^3*(1/4/cosh(d*x+c)^4+1/2/cosh(d*x+c)^2+ln(tanh(d*x+c) )))
Leaf count of result is larger than twice the leaf count of optimal. 2376 vs. \(2 (80) = 160\).
Time = 0.28 (sec) , antiderivative size = 2376, normalized size of antiderivative = 28.29 \[ \int \coth (c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\text {Too large to display} \]
-(a^3*d*x*cosh(d*x + c)^8 + 8*a^3*d*x*cosh(d*x + c)*sinh(d*x + c)^7 + a^3* d*x*sinh(d*x + c)^8 + 2*(2*a^3*d*x - 3*a*b^2 - b^3)*cosh(d*x + c)^6 + 2*(1 4*a^3*d*x*cosh(d*x + c)^2 + 2*a^3*d*x - 3*a*b^2 - b^3)*sinh(d*x + c)^6 + 4 *(14*a^3*d*x*cosh(d*x + c)^3 + 3*(2*a^3*d*x - 3*a*b^2 - b^3)*cosh(d*x + c) )*sinh(d*x + c)^5 + a^3*d*x + 2*(3*a^3*d*x - 6*a*b^2 - 4*b^3)*cosh(d*x + c )^4 + 2*(35*a^3*d*x*cosh(d*x + c)^4 + 3*a^3*d*x - 6*a*b^2 - 4*b^3 + 15*(2* a^3*d*x - 3*a*b^2 - b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 8*(7*a^3*d*x*c osh(d*x + c)^5 + 5*(2*a^3*d*x - 3*a*b^2 - b^3)*cosh(d*x + c)^3 + (3*a^3*d* x - 6*a*b^2 - 4*b^3)*cosh(d*x + c))*sinh(d*x + c)^3 + 2*(2*a^3*d*x - 3*a*b ^2 - b^3)*cosh(d*x + c)^2 + 2*(14*a^3*d*x*cosh(d*x + c)^6 + 2*a^3*d*x + 15 *(2*a^3*d*x - 3*a*b^2 - b^3)*cosh(d*x + c)^4 - 3*a*b^2 - b^3 + 6*(3*a^3*d* x - 6*a*b^2 - 4*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + ((3*a^2*b + 3*a*b^ 2 + b^3)*cosh(d*x + c)^8 + 8*(3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)*sinh( d*x + c)^7 + (3*a^2*b + 3*a*b^2 + b^3)*sinh(d*x + c)^8 + 4*(3*a^2*b + 3*a* b^2 + b^3)*cosh(d*x + c)^6 + 4*(3*a^2*b + 3*a*b^2 + b^3 + 7*(3*a^2*b + 3*a *b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^6 + 8*(7*(3*a^2*b + 3*a*b^2 + b ^3)*cosh(d*x + c)^3 + 3*(3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c))*sinh(d*x + c)^5 + 6*(3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^4 + 2*(35*(3*a^2*b + 3* a*b^2 + b^3)*cosh(d*x + c)^4 + 9*a^2*b + 9*a*b^2 + 3*b^3 + 30*(3*a^2*b + 3 *a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 8*(7*(3*a^2*b + 3*a*b^...
\[ \int \coth (c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\int \left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{3} \coth {\left (c + d x \right )}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (80) = 160\).
Time = 0.28 (sec) , antiderivative size = 300, normalized size of antiderivative = 3.57 \[ \int \coth (c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=b^{3} {\left (\frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} - \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} + \frac {2 \, {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 4 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} + 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )} + 1\right )}}\right )} + 3 \, a b^{2} {\left (\frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} - \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} + \frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} + 3 \, a^{2} b {\left (\frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} - \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d}\right )} + \frac {a^{3} \log \left (\sinh \left (d x + c\right )\right )}{d} \]
b^3*(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) + 4*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c))/ (d*(4*e^(-2*d*x - 2*c) + 6*e^(-4*d*x - 4*c) + 4*e^(-6*d*x - 6*c) + e^(-8*d *x - 8*c) + 1))) + 3*a*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))) + 3*a^2*b*(log(e^(-d*x - c) + 1)/d + log(e^( -d*x - c) - 1)/d - log(e^(-2*d*x - 2*c) + 1)/d) + a^3*log(sinh(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (80) = 160\).
Time = 0.32 (sec) , antiderivative size = 283, normalized size of antiderivative = 3.37 \[ \int \coth (c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=-\frac {2 \, {\left (3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} + 2\right ) - 2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} - 2\right ) - \frac {9 \, a^{2} b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )}^{2} + 9 \, a b^{2} {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )}^{2} + 3 \, b^{3} {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )}^{2} + 36 \, a^{2} b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 60 \, a b^{2} {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 20 \, b^{3} {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 36 \, a^{2} b + 84 \, a b^{2} + 44 \, b^{3}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} + 2\right )}^{2}}}{4 \, d} \]
-1/4*(2*(3*a^2*b + 3*a*b^2 + b^3)*log(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c) + 2) - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*log(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c) - 2) - (9*a^2*b*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c))^2 + 9*a*b^2*(e^( 2*d*x + 2*c) + e^(-2*d*x - 2*c))^2 + 3*b^3*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c))^2 + 36*a^2*b*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) + 60*a*b^2*(e^(2*d *x + 2*c) + e^(-2*d*x - 2*c)) + 20*b^3*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c) ) + 36*a^2*b + 84*a*b^2 + 44*b^3)/(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c) + 2) ^2)/d
Time = 2.33 (sec) , antiderivative size = 360, normalized size of antiderivative = 4.29 \[ \int \coth (c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\frac {2\,\left (b^3+3\,a\,b^2\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-a^3\,x-\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\left (a^3\,\sqrt {-d^2}+2\,b^3\,\sqrt {-d^2}+6\,a\,b^2\,\sqrt {-d^2}+6\,a^2\,b\,\sqrt {-d^2}\right )}{d\,\sqrt {a^6+12\,a^5\,b+48\,a^4\,b^2+76\,a^3\,b^3+60\,a^2\,b^4+24\,a\,b^5+4\,b^6}}\right )\,\sqrt {a^6+12\,a^5\,b+48\,a^4\,b^2+76\,a^3\,b^3+60\,a^2\,b^4+24\,a\,b^5+4\,b^6}}{\sqrt {-d^2}}+\frac {a^3\,\ln \left ({\mathrm {e}}^{4\,c+4\,d\,x}-1\right )}{2\,d}-\frac {8\,b^3}{d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )}-\frac {2\,\left (3\,a\,b^2-b^3\right )}{d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )}+\frac {4\,b^3}{d\,\left (4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1\right )} \]
(2*(3*a*b^2 + b^3))/(d*(exp(2*c + 2*d*x) + 1)) - a^3*x - (atan((exp(2*c)*e xp(2*d*x)*(a^3*(-d^2)^(1/2) + 2*b^3*(-d^2)^(1/2) + 6*a*b^2*(-d^2)^(1/2) + 6*a^2*b*(-d^2)^(1/2)))/(d*(24*a*b^5 + 12*a^5*b + a^6 + 4*b^6 + 60*a^2*b^4 + 76*a^3*b^3 + 48*a^4*b^2)^(1/2)))*(24*a*b^5 + 12*a^5*b + a^6 + 4*b^6 + 60 *a^2*b^4 + 76*a^3*b^3 + 48*a^4*b^2)^(1/2))/(-d^2)^(1/2) + (a^3*log(exp(4*c + 4*d*x) - 1))/(2*d) - (8*b^3)/(d*(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x ) + exp(6*c + 6*d*x) + 1)) - (2*(3*a*b^2 - b^3))/(d*(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1)) + (4*b^3)/(d*(4*exp(2*c + 2*d*x) + 6*exp(4*c + 4*d* x) + 4*exp(6*c + 6*d*x) + exp(8*c + 8*d*x) + 1))