Integrand size = 23, antiderivative size = 52 \[ \int \coth ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {a^2 \coth ^2(c+d x)}{2 d}+\frac {(a+b)^2 \log (\cosh (c+d x))}{d}+\frac {a (a+2 b) \log (\tanh (c+d x))}{d} \] Output:
-1/2*a^2*coth(d*x+c)^2/d+(a+b)^2*ln(cosh(d*x+c))/d+a*(a+2*b)*ln(tanh(d*x+c ))/d
Time = 0.08 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.10 \[ \int \coth ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {b^2 \log (\cosh (c+d x))}{d}-\frac {a^2 \left (\text {csch}^2(c+d x)-2 \log (\sinh (c+d x))\right )}{2 d}+\frac {2 a b \log (\sinh (c+d x))}{d} \] Input:
Integrate[Coth[c + d*x]^3*(a + b*Tanh[c + d*x]^2)^2,x]
Output:
(b^2*Log[Cosh[c + d*x]])/d - (a^2*(Csch[c + d*x]^2 - 2*Log[Sinh[c + d*x]]) )/(2*d) + (2*a*b*Log[Sinh[c + d*x]])/d
Time = 0.51 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.06, 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.
\(\displaystyle \int \coth ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {i \left (a-b \tan (i c+i d x)^2\right )^2}{\tan (i c+i d x)^3}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {\left (a-b \tan (i c+i d x)^2\right )^2}{\tan (i c+i d x)^3}dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle -\frac {i \int \frac {i \coth ^3(c+d x) \left (b \tanh ^2(c+d x)+a\right )^2}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\int \frac {\coth ^3(c+d x) \left (b \tanh ^2(c+d x)+a\right )^2}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {\int \frac {\coth ^2(c+d x) \left (b \tanh ^2(c+d x)+a\right )^2}{1-\tanh ^2(c+d x)}d\tanh ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (-\frac {(a+b)^2}{\tanh ^2(c+d x)-1}+a^2 \coth ^2(c+d x)+a (a+2 b) \coth (c+d x)\right )d\tanh ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 (-\coth (c+d x))+a (a+2 b) \log \left (\tanh ^2(c+d x)\right )-(a+b)^2 \log \left (1-\tanh ^2(c+d x)\right )}{2 d}\) |
Input:
Int[Coth[c + d*x]^3*(a + b*Tanh[c + d*x]^2)^2,x]
Output:
(-(a^2*Coth[c + d*x]) + a*(a + 2*b)*Log[Tanh[c + d*x]^2] - (a + b)^2*Log[1 - Tanh[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[((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 = 0.15 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.19
method | result | size |
parallelrisch | \(\frac {-2 \left (a +b \right )^{2} \ln \left (1-\tanh \left (d x +c \right )\right )+2 a \left (a +2 b \right ) \ln \left (\tanh \left (d x +c \right )\right )-\coth \left (d x +c \right )^{2} a^{2}-2 d x \left (a +b \right )^{2}}{2 d}\) | \(62\) |
derivativedivides | \(-\frac {\frac {a^{2}}{2 \tanh \left (d x +c \right )^{2}}-a \left (a +2 b \right ) \ln \left (\tanh \left (d x +c \right )\right )+\left (\frac {1}{2} a^{2}+a b +\frac {1}{2} b^{2}\right ) \ln \left (1+\tanh \left (d x +c \right )\right )+\left (\frac {1}{2} a^{2}+a b +\frac {1}{2} b^{2}\right ) \ln \left (-1+\tanh \left (d x +c \right )\right )}{d}\) | \(83\) |
default | \(-\frac {\frac {a^{2}}{2 \tanh \left (d x +c \right )^{2}}-a \left (a +2 b \right ) \ln \left (\tanh \left (d x +c \right )\right )+\left (\frac {1}{2} a^{2}+a b +\frac {1}{2} b^{2}\right ) \ln \left (1+\tanh \left (d x +c \right )\right )+\left (\frac {1}{2} a^{2}+a b +\frac {1}{2} b^{2}\right ) \ln \left (-1+\tanh \left (d x +c \right )\right )}{d}\) | \(83\) |
risch | \(-a^{2} x -2 a b x -b^{2} x -\frac {2 a^{2} c}{d}-\frac {4 a b c}{d}-\frac {2 b^{2} c}{d}-\frac {2 a^{2} {\mathrm e}^{2 d x +2 c}}{d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}+\frac {a^{2} \ln \left ({\mathrm e}^{2 d x +2 c}-1\right )}{d}+\frac {2 a \ln \left ({\mathrm e}^{2 d x +2 c}-1\right ) b}{d}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+1\right ) b^{2}}{d}\) | \(132\) |
Input:
int(coth(d*x+c)^3*(a+b*tanh(d*x+c)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/2*(-2*(a+b)^2*ln(1-tanh(d*x+c))+2*a*(a+2*b)*ln(tanh(d*x+c))-coth(d*x+c)^ 2*a^2-2*d*x*(a+b)^2)/d
Leaf count of result is larger than twice the leaf count of optimal. 677 vs. \(2 (50) = 100\).
Time = 0.11 (sec) , antiderivative size = 677, normalized size of antiderivative = 13.02 \[ \int \coth ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx =\text {Too large to display} \] Input:
integrate(coth(d*x+c)^3*(a+b*tanh(d*x+c)^2)^2,x, algorithm="fricas")
Output:
-((a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b + b^2)*d*x*cosh (d*x + c)*sinh(d*x + c)^3 + (a^2 + 2*a*b + b^2)*d*x*sinh(d*x + c)^4 + (a^2 + 2*a*b + b^2)*d*x - 2*((a^2 + 2*a*b + b^2)*d*x - a^2)*cosh(d*x + c)^2 + 2*(3*(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)*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))*sinh(d*x + c))*log(2*cosh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) - ((a^2 + 2*a*b)*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 + 2*a*b)*sinh(d*x + c)^4 - 2*(a^2 + 2*a*b)*cosh(d*x + c)^2 + 2*(3*(a^2 + 2*a*b)*cosh(d*x + c)^2 - a^2 - 2*a*b)*sinh(d*x + c)^2 + a^2 + 2*a*b + 4*((a^2 + 2*a*b)*cosh(d*x + c)^3 - (a^2 + 2*a*b)*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 + 2*a*b + b^2)*d*x*cosh(d*x + c)^3 - ((a^2 + 2*a*b + b^2)*d*x - a^2)*c osh(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)
\[ \int \coth ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{2} \coth ^{3}{\left (c + d x \right )}\, dx \] Input:
integrate(coth(d*x+c)**3*(a+b*tanh(d*x+c)**2)**2,x)
Output:
Integral((a + b*tanh(c + d*x)**2)**2*coth(c + d*x)**3, x)
Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (50) = 100\).
Time = 0.05 (sec) , antiderivative size = 134, normalized size of antiderivative = 2.58 \[ \int \coth ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=a^{2} {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} + \frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )}}\right )} + \frac {b^{2} \log \left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}{d} + \frac {2 \, a b \log \left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}{d} \] Input:
integrate(coth(d*x+c)^3*(a+b*tanh(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) + e^(-d*x - c))/d + 2*a*b*log(e^(d*x + c) - e^(-d*x - c))/d
Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (50) = 100\).
Time = 0.20 (sec) , antiderivative size = 141, normalized size of antiderivative = 2.71 \[ \int \coth ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {b^{2} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} + 2\right ) + {\left (a^{2} + 2 \, a b\right )} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} - 2\right ) - \frac {a^{2} {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 2 \, a b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 2 \, a^{2} - 4 \, a b}{e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} - 2}}{2 \, d} \] Input:
integrate(coth(d*x+c)^3*(a+b*tanh(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 + 2*a*b)*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)) + 2*a*b*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) + 2*a^2 - 4*a*b)/(e^ (2*d*x + 2*c) + e^(-2*d*x - 2*c) - 2))/d
Time = 0.25 (sec) , antiderivative size = 211, normalized size of antiderivative = 4.06 \[ \int \coth ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {\ln \left ({\mathrm {e}}^{4\,c+4\,d\,x}-1\right )\,\left (d\,\left (a^2+2\,b\,a\right )+b^2\,d\right )}{2\,d^2}-\frac {2\,a^2}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {2\,a^2}{d\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-x\,{\left (a+b\right )}^2-\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\left (a^2\,\sqrt {-d^2}-b^2\,\sqrt {-d^2}+2\,a\,b\,\sqrt {-d^2}\right )}{d\,\sqrt {a^4+4\,a^3\,b+2\,a^2\,b^2-4\,a\,b^3+b^4}}\right )\,\sqrt {a^4+4\,a^3\,b+2\,a^2\,b^2-4\,a\,b^3+b^4}}{\sqrt {-d^2}} \] Input:
int(coth(c + d*x)^3*(a + b*tanh(c + d*x)^2)^2,x)
Output:
(log(exp(4*c + 4*d*x) - 1)*(d*(2*a*b + a^2) + b^2*d))/(2*d^2) - (2*a^2)/(d *(exp(2*c + 2*d*x) - 1)) - (2*a^2)/(d*(exp(4*c + 4*d*x) - 2*exp(2*c + 2*d* x) + 1)) - x*(a + b)^2 - (atan((exp(2*c)*exp(2*d*x)*(a^2*(-d^2)^(1/2) - b^ 2*(-d^2)^(1/2) + 2*a*b*(-d^2)^(1/2)))/(d*(4*a^3*b - 4*a*b^3 + a^4 + b^4 + 2*a^2*b^2)^(1/2)))*(4*a^3*b - 4*a*b^3 + a^4 + b^4 + 2*a^2*b^2)^(1/2))/(-d^ 2)^(1/2)
Time = 0.26 (sec) , antiderivative size = 493, normalized size of antiderivative = 9.48 \[ \int \coth ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {2 e^{2 d x +2 c} a^{2} d x +2 \,\mathrm {log}\left (e^{d x +c}-1\right ) a b +2 \,\mathrm {log}\left (e^{d x +c}+1\right ) a b -e^{4 d x +4 c} a^{2} d x -e^{4 d x +4 c} b^{2} d x +e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}-1\right ) a^{2}+e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}+1\right ) a^{2}-2 e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}-1\right ) a^{2}-2 e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}+1\right ) a^{2}-a^{2}+4 e^{2 d x +2 c} a b d x -2 e^{4 d x +4 c} a b d x +\mathrm {log}\left (e^{2 d x +2 c}+1\right ) b^{2}+2 e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}-1\right ) a b +2 e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}+1\right ) a b -4 e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}-1\right ) a b -4 e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}+1\right ) a b -2 a b d x -a^{2} d x -b^{2} d x +e^{4 d x +4 c} \mathrm {log}\left (e^{2 d x +2 c}+1\right ) b^{2}-2 e^{2 d x +2 c} \mathrm {log}\left (e^{2 d x +2 c}+1\right ) b^{2}+2 e^{2 d x +2 c} b^{2} d x +\mathrm {log}\left (e^{d x +c}-1\right ) a^{2}+\mathrm {log}\left (e^{d x +c}+1\right ) a^{2}-e^{4 d x +4 c} a^{2}}{d \left (e^{4 d x +4 c}-2 e^{2 d x +2 c}+1\right )} \] Input:
int(coth(d*x+c)^3*(a+b*tanh(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 + 2*e**(4*c + 4*d*x)*log(e**(c + d*x) - 1)*a*b + e**( 4*c + 4*d*x)*log(e**(c + d*x) + 1)*a**2 + 2*e**(4*c + 4*d*x)*log(e**(c + d *x) + 1)*a*b - 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*d*x - e**(4*c + 4*d*x)*b**2*d*x - 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 - 4*e**(2*c + 2*d*x)*log(e**(c + d*x) - 1)*a*b - 2*e**(2*c + 2*d*x)*log(e* *(c + d*x) + 1)*a**2 - 4*e**(2*c + 2*d*x)*log(e**(c + d*x) + 1)*a*b + 2*e* *(2*c + 2*d*x)*a**2*d*x + 4*e**(2*c + 2*d*x)*a*b*d*x + 2*e**(2*c + 2*d*x)* b**2*d*x + log(e**(2*c + 2*d*x) + 1)*b**2 + log(e**(c + d*x) - 1)*a**2 + 2 *log(e**(c + d*x) - 1)*a*b + log(e**(c + d*x) + 1)*a**2 + 2*log(e**(c + d* x) + 1)*a*b - a**2*d*x - a**2 - 2*a*b*d*x - b**2*d*x)/(d*(e**(4*c + 4*d*x) - 2*e**(2*c + 2*d*x) + 1))