Integrand size = 23, antiderivative size = 72 \[ \int \coth ^5(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {a (a+2 b) \coth ^2(c+d x)}{2 d}-\frac {a^2 \coth ^4(c+d x)}{4 d}+\frac {(a+b)^2 \log (\cosh (c+d x))}{d}+\frac {(a+b)^2 \log (\tanh (c+d x))}{d} \]
-1/2*a*(a+2*b)*coth(d*x+c)^2/d-1/4*a^2*coth(d*x+c)^4/d+(a+b)^2*ln(cosh(d*x +c))/d+(a+b)^2*ln(tanh(d*x+c))/d
Time = 0.47 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.81 \[ \int \coth ^5(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {2 a (a+2 b) \coth ^2(c+d x)+a^2 \coth ^4(c+d x)-4 (a+b)^2 (\log (\cosh (c+d x))+\log (\tanh (c+d x)))}{4 d} \]
-1/4*(2*a*(a + 2*b)*Coth[c + d*x]^2 + a^2*Coth[c + d*x]^4 - 4*(a + b)^2*(L og[Cosh[c + d*x]] + Log[Tanh[c + d*x]]))/d
Time = 0.31 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 26, 4153, 26, 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.
\(\displaystyle \int \coth ^5(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i \left (a-b \tan (i c+i d x)^2\right )^2}{\tan (i c+i d x)^5}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {\left (a-b \tan (i c+i d x)^2\right )^2}{\tan (i c+i d x)^5}dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle \frac {i \int -\frac {i \coth ^5(c+d x) \left (b \tanh ^2(c+d x)+a\right )^2}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\int \frac {\coth ^5(c+d x) \left (b \tanh ^2(c+d x)+a\right )^2}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {\int \frac {\coth ^3(c+d x) \left (b \tanh ^2(c+d x)+a\right )^2}{1-\tanh ^2(c+d x)}d\tanh ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (a^2 \coth ^3(c+d x)+a (a+2 b) \coth ^2(c+d x)+(a+b)^2 \coth (c+d x)-\frac {(a+b)^2}{\tanh ^2(c+d x)-1}\right )d\tanh ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{2} a^2 \coth ^2(c+d x)-a (a+2 b) \coth (c+d x)+(a+b)^2 \log \left (\tanh ^2(c+d x)\right )-(a+b)^2 \log \left (1-\tanh ^2(c+d x)\right )}{2 d}\) |
(-(a*(a + 2*b)*Coth[c + d*x]) - (a^2*Coth[c + d*x]^2)/2 + (a + b)^2*Log[Ta nh[c + d*x]^2] - (a + b)^2*Log[1 - Tanh[c + d*x]^2])/(2*d)
3.2.53.3.1 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[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[c*(ff/f) Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio nalQ[n]))
Time = 0.22 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.07
method | result | size |
parallelrisch | \(\frac {-4 \left (a +b \right )^{2} \ln \left (1-\tanh \left (d x +c \right )\right )+4 \left (a +b \right )^{2} \ln \left (\tanh \left (d x +c \right )\right )-\coth \left (d x +c \right )^{4} a^{2}-2 a \coth \left (d x +c \right )^{2} \left (a +2 b \right )-4 d x \left (a +b \right )^{2}}{4 d}\) | \(77\) |
derivativedivides | \(-\frac {\left (\frac {1}{2} a^{2}+a b +\frac {1}{2} b^{2}\right ) \ln \left (\tanh \left (d x +c \right )+1\right )+\left (-a^{2}-2 a b -b^{2}\right ) \ln \left (\tanh \left (d x +c \right )\right )+\frac {a^{2}}{4 \tanh \left (d x +c \right )^{4}}+\frac {a \left (a +2 b \right )}{2 \tanh \left (d x +c \right )^{2}}+\left (\frac {1}{2} a^{2}+a b +\frac {1}{2} b^{2}\right ) \ln \left (\tanh \left (d x +c \right )-1\right )}{d}\) | \(107\) |
default | \(-\frac {\left (\frac {1}{2} a^{2}+a b +\frac {1}{2} b^{2}\right ) \ln \left (\tanh \left (d x +c \right )+1\right )+\left (-a^{2}-2 a b -b^{2}\right ) \ln \left (\tanh \left (d x +c \right )\right )+\frac {a^{2}}{4 \tanh \left (d x +c \right )^{4}}+\frac {a \left (a +2 b \right )}{2 \tanh \left (d x +c \right )^{2}}+\left (\frac {1}{2} a^{2}+a b +\frac {1}{2} b^{2}\right ) \ln \left (\tanh \left (d x +c \right )-1\right )}{d}\) | \(107\) |
risch | \(-a^{2} x -2 a b x -b^{2} x -\frac {2 a^{2} c}{d}-\frac {4 a b c}{d}-\frac {2 c \,b^{2}}{d}-\frac {4 a \,{\mathrm e}^{2 d x +2 c} \left (a \,{\mathrm e}^{4 d x +4 c}+b \,{\mathrm e}^{4 d x +4 c}-{\mathrm e}^{2 d x +2 c} a -2 b \,{\mathrm e}^{2 d x +2 c}+a +b \right )}{d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{4}}+\frac {a^{2} \ln \left ({\mathrm e}^{2 d x +2 c}-1\right )}{d}+\frac {2 a \ln \left ({\mathrm e}^{2 d x +2 c}-1\right ) b}{d}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}-1\right ) b^{2}}{d}\) | \(179\) |
1/4*(-4*(a+b)^2*ln(1-tanh(d*x+c))+4*(a+b)^2*ln(tanh(d*x+c))-coth(d*x+c)^4* a^2-2*a*coth(d*x+c)^2*(a+2*b)-4*d*x*(a+b)^2)/d
Leaf count of result is larger than twice the leaf count of optimal. 1649 vs. \(2 (68) = 136\).
Time = 0.27 (sec) , antiderivative size = 1649, normalized size of antiderivative = 22.90 \[ \int \coth ^5(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\text {Too large to display} \]
-((a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^8 + 8*(a^2 + 2*a*b + b^2)*d*x*cosh (d*x + c)*sinh(d*x + c)^7 + (a^2 + 2*a*b + b^2)*d*x*sinh(d*x + c)^8 - 4*(( a^2 + 2*a*b + b^2)*d*x - a^2 - a*b)*cosh(d*x + c)^6 + 4*(7*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^2 - (a^2 + 2*a*b + b^2)*d*x + a^2 + a*b)*sinh(d*x + c)^6 + 8*(7*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^3 - 3*((a^2 + 2*a*b + b ^2)*d*x - a^2 - a*b)*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(3*(a^2 + 2*a*b + b^2)*d*x - 2*a^2 - 4*a*b)*cosh(d*x + c)^4 + 2*(35*(a^2 + 2*a*b + b^2)*d*x* cosh(d*x + c)^4 + 3*(a^2 + 2*a*b + b^2)*d*x - 30*((a^2 + 2*a*b + b^2)*d*x - a^2 - a*b)*cosh(d*x + c)^2 - 2*a^2 - 4*a*b)*sinh(d*x + c)^4 + 8*(7*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^5 - 10*((a^2 + 2*a*b + b^2)*d*x - a^2 - a *b)*cosh(d*x + c)^3 + (3*(a^2 + 2*a*b + b^2)*d*x - 2*a^2 - 4*a*b)*cosh(d*x + c))*sinh(d*x + c)^3 + (a^2 + 2*a*b + b^2)*d*x - 4*((a^2 + 2*a*b + b^2)* d*x - a^2 - a*b)*cosh(d*x + c)^2 + 4*(7*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^6 - 15*((a^2 + 2*a*b + b^2)*d*x - a^2 - a*b)*cosh(d*x + c)^4 - (a^2 + 2*a*b + b^2)*d*x + 3*(3*(a^2 + 2*a*b + b^2)*d*x - 2*a^2 - 4*a*b)*cosh(d*x + c)^2 + a^2 + a*b)*sinh(d*x + c)^2 - ((a^2 + 2*a*b + b^2)*cosh(d*x + c)^8 + 8*(a^2 + 2*a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^7 + (a^2 + 2*a*b + b^ 2)*sinh(d*x + c)^8 - 4*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^6 + 4*(7*(a^2 + 2 *a*b + b^2)*cosh(d*x + c)^2 - a^2 - 2*a*b - b^2)*sinh(d*x + c)^6 + 8*(7*(a ^2 + 2*a*b + b^2)*cosh(d*x + c)^3 - 3*(a^2 + 2*a*b + b^2)*cosh(d*x + c)...
\[ \int \coth ^5(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{2} \coth ^{5}{\left (c + d x \right )}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 236 vs. \(2 (68) = 136\).
Time = 0.19 (sec) , antiderivative size = 236, normalized size of antiderivative = 3.28 \[ \int \coth ^5(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=a^{2} {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} + \frac {4 \, {\left (e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} - 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}}\right )} + 2 \, a b {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} + \frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )}}\right )} + \frac {b^{2} \log \left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}{d} \]
a^2*(x + c/d + log(e^(-d*x - c) + 1)/d + log(e^(-d*x - c) - 1)/d + 4*(e^(- 2*d*x - 2*c) - 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))) + 2*a *b*(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) - e^(-d*x - c))/d
Time = 0.42 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.64 \[ \int \coth ^5(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (d x + c\right )} - {\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right ) + \frac {4 \, {\left ({\left (a^{2} + a b\right )} e^{\left (6 \, d x + 6 \, c\right )} - {\left (a^{2} + 2 \, a b\right )} e^{\left (4 \, d x + 4 \, c\right )} + {\left (a^{2} + a b\right )} e^{\left (2 \, d x + 2 \, c\right )}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{4}}}{d} \]
-((a^2 + 2*a*b + b^2)*(d*x + c) - (a^2 + 2*a*b + b^2)*log(abs(e^(2*d*x + 2 *c) - 1)) + 4*((a^2 + a*b)*e^(6*d*x + 6*c) - (a^2 + 2*a*b)*e^(4*d*x + 4*c) + (a^2 + a*b)*e^(2*d*x + 2*c))/(e^(2*d*x + 2*c) - 1)^4)/d
Time = 1.92 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.74 \[ \int \coth ^5(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {\ln \left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}-1\right )\,\left (a^2+2\,a\,b+b^2\right )}{d}-x\,{\left (a+b\right )}^2-\frac {4\,\left (2\,a^2+b\,a\right )}{d\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {8\,a^2}{d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1\right )}-\frac {4\,a^2}{d\,\left (6\,{\mathrm {e}}^{4\,c+4\,d\,x}-4\,{\mathrm {e}}^{2\,c+2\,d\,x}-4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1\right )}-\frac {4\,\left (a^2+b\,a\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )} \]
(log(exp(2*c)*exp(2*d*x) - 1)*(2*a*b + a^2 + b^2))/d - x*(a + b)^2 - (4*(a *b + 2*a^2))/(d*(exp(4*c + 4*d*x) - 2*exp(2*c + 2*d*x) + 1)) - (8*a^2)/(d* (3*exp(2*c + 2*d*x) - 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) - 1)) - (4*a^2 )/(d*(6*exp(4*c + 4*d*x) - 4*exp(2*c + 2*d*x) - 4*exp(6*c + 6*d*x) + exp(8 *c + 8*d*x) + 1)) - (4*(a*b + a^2))/(d*(exp(2*c + 2*d*x) - 1))