\(\int \frac {\tanh ^5(c+d x)}{(a+b \tanh ^2(c+d x))^3} \, dx\) [191]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F(-1)]
   Maxima [B] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 23, antiderivative size = 109 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {\log (\cosh (c+d x))}{(a+b)^3 d}+\frac {\log \left (a+b \tanh ^2(c+d x)\right )}{2 (a+b)^3 d}-\frac {a^2}{4 b^2 (a+b) d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac {a (a+2 b)}{2 b^2 (a+b)^2 d \left (a+b \tanh ^2(c+d x)\right )} \]

[Out]

ln(cosh(d*x+c))/(a+b)^3/d+1/2*ln(a+b*tanh(d*x+c)^2)/(a+b)^3/d-1/4*a^2/b^2/(a+b)/d/(a+b*tanh(d*x+c)^2)^2+1/2*a*
(a+2*b)/b^2/(a+b)^2/d/(a+b*tanh(d*x+c)^2)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3751, 457, 90} \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=-\frac {a^2}{4 b^2 d (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}+\frac {a (a+2 b)}{2 b^2 d (a+b)^2 \left (a+b \tanh ^2(c+d x)\right )}+\frac {\log \left (a+b \tanh ^2(c+d x)\right )}{2 d (a+b)^3}+\frac {\log (\cosh (c+d x))}{d (a+b)^3} \]

[In]

Int[Tanh[c + d*x]^5/(a + b*Tanh[c + d*x]^2)^3,x]

[Out]

Log[Cosh[c + d*x]]/((a + b)^3*d) + Log[a + b*Tanh[c + d*x]^2]/(2*(a + b)^3*d) - a^2/(4*b^2*(a + b)*d*(a + b*Ta
nh[c + d*x]^2)^2) + (a*(a + 2*b))/(2*b^2*(a + b)^2*d*(a + b*Tanh[c + d*x]^2))

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(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]))

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
 NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 3751

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]}, Dist[c*(ff/f), Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2
 + ff^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] && RationalQ[n]))

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^5}{\left (1-x^2\right ) \left (a+b x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \frac {x^2}{(1-x) (a+b x)^3} \, dx,x,\tanh ^2(c+d x)\right )}{2 d} \\ & = \frac {\text {Subst}\left (\int \left (-\frac {1}{(a+b)^3 (-1+x)}+\frac {a^2}{b (a+b) (a+b x)^3}-\frac {a (a+2 b)}{b (a+b)^2 (a+b x)^2}+\frac {b}{(a+b)^3 (a+b x)}\right ) \, dx,x,\tanh ^2(c+d x)\right )}{2 d} \\ & = \frac {\log (\cosh (c+d x))}{(a+b)^3 d}+\frac {\log \left (a+b \tanh ^2(c+d x)\right )}{2 (a+b)^3 d}-\frac {a^2}{4 b^2 (a+b) d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac {a (a+2 b)}{2 b^2 (a+b)^2 d \left (a+b \tanh ^2(c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.22 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.83 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=-\frac {-4 \log (\cosh (c+d x))-2 \log \left (a+b \tanh ^2(c+d x)\right )+\frac {a^2 (a+b)^2}{b^2 \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {2 a (a+b) (a+2 b)}{b^2 \left (a+b \tanh ^2(c+d x)\right )}}{4 (a+b)^3 d} \]

[In]

Integrate[Tanh[c + d*x]^5/(a + b*Tanh[c + d*x]^2)^3,x]

[Out]

-1/4*(-4*Log[Cosh[c + d*x]] - 2*Log[a + b*Tanh[c + d*x]^2] + (a^2*(a + b)^2)/(b^2*(a + b*Tanh[c + d*x]^2)^2) -
 (2*a*(a + b)*(a + 2*b))/(b^2*(a + b*Tanh[c + d*x]^2)))/((a + b)^3*d)

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.17

method result size
derivativedivides \(\frac {-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{3}}-\frac {-\frac {a \left (a^{2}+3 a b +2 b^{2}\right )}{b^{2} \left (a +b \tanh \left (d x +c \right )^{2}\right )}+\frac {a^{2} \left (a^{2}+2 a b +b^{2}\right )}{2 b^{2} \left (a +b \tanh \left (d x +c \right )^{2}\right )^{2}}-\ln \left (a +b \tanh \left (d x +c \right )^{2}\right )}{2 \left (a +b \right )^{3}}-\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2 \left (a +b \right )^{3}}}{d}\) \(127\)
default \(\frac {-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{3}}-\frac {-\frac {a \left (a^{2}+3 a b +2 b^{2}\right )}{b^{2} \left (a +b \tanh \left (d x +c \right )^{2}\right )}+\frac {a^{2} \left (a^{2}+2 a b +b^{2}\right )}{2 b^{2} \left (a +b \tanh \left (d x +c \right )^{2}\right )^{2}}-\ln \left (a +b \tanh \left (d x +c \right )^{2}\right )}{2 \left (a +b \right )^{3}}-\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2 \left (a +b \right )^{3}}}{d}\) \(127\)
risch \(-\frac {x}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}-\frac {2 c}{d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {4 \left (a \,{\mathrm e}^{4 d x +4 c}+b \,{\mathrm e}^{4 d x +4 c}+{\mathrm e}^{2 d x +2 c} a -2 b \,{\mathrm e}^{2 d x +2 c}+a +b \right ) a \,{\mathrm e}^{2 d x +2 c}}{\left (a \,{\mathrm e}^{4 d x +4 c}+b \,{\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{2 d x +2 c} a -2 b \,{\mathrm e}^{2 d x +2 c}+a +b \right )^{2} d \left (a +b \right )^{3}}+\frac {\ln \left ({\mathrm e}^{4 d x +4 c}+\frac {2 \left (a -b \right ) {\mathrm e}^{2 d x +2 c}}{a +b}+1\right )}{2 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}\) \(231\)
parallelrisch \(-\frac {4 \ln \left (1-\tanh \left (d x +c \right )\right ) a^{2} b^{2}+4 \ln \left (1-\tanh \left (d x +c \right )\right ) \tanh \left (d x +c \right )^{4} b^{4}-2 \ln \left (a +b \tanh \left (d x +c \right )^{2}\right ) \tanh \left (d x +c \right )^{4} b^{4}-4 a \,b^{3} \tanh \left (d x +c \right )^{2}-2 \tanh \left (d x +c \right )^{2} a^{3} b -2 \ln \left (a +b \tanh \left (d x +c \right )^{2}\right ) a^{2} b^{2}+8 \ln \left (1-\tanh \left (d x +c \right )\right ) \tanh \left (d x +c \right )^{2} a \,b^{3}+4 b^{4} \tanh \left (d x +c \right )^{4} x d -a^{4}+4 a^{2} b^{2} d x -3 a^{2} b^{2}-4 a^{3} b -6 a^{2} b^{2} \tanh \left (d x +c \right )^{2}+8 a \,b^{3} \tanh \left (d x +c \right )^{2} x d -4 \ln \left (a +b \tanh \left (d x +c \right )^{2}\right ) \tanh \left (d x +c \right )^{2} a \,b^{3}}{4 \left (a +b \tanh \left (d x +c \right )^{2}\right )^{2} \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) b^{2} d}\) \(291\)

[In]

int(tanh(d*x+c)^5/(a+b*tanh(d*x+c)^2)^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(-1/2/(a+b)^3*ln(tanh(d*x+c)-1)-1/2/(a+b)^3*(-a*(a^2+3*a*b+2*b^2)/b^2/(a+b*tanh(d*x+c)^2)+1/2*a^2*(a^2+2*a
*b+b^2)/b^2/(a+b*tanh(d*x+c)^2)^2-ln(a+b*tanh(d*x+c)^2))-1/2/(a+b)^3*ln(tanh(d*x+c)+1))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2584 vs. \(2 (103) = 206\).

Time = 0.30 (sec) , antiderivative size = 2584, normalized size of antiderivative = 23.71 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]

[In]

integrate(tanh(d*x+c)^5/(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

-1/2*(2*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^8 + 16*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)*sinh(d*x + c)^7 + 2
*(a^2 + 2*a*b + b^2)*d*x*sinh(d*x + c)^8 + 8*((a^2 - b^2)*d*x - a^2 - a*b)*cosh(d*x + c)^6 + 8*(7*(a^2 + 2*a*b
 + b^2)*d*x*cosh(d*x + c)^2 + (a^2 - b^2)*d*x - a^2 - a*b)*sinh(d*x + c)^6 + 16*(7*(a^2 + 2*a*b + b^2)*d*x*cos
h(d*x + c)^3 + 3*((a^2 - b^2)*d*x - a^2 - a*b)*cosh(d*x + c))*sinh(d*x + c)^5 + 4*((3*a^2 - 2*a*b + 3*b^2)*d*x
 - 2*a^2 + 4*a*b)*cosh(d*x + c)^4 + 4*(35*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^4 + (3*a^2 - 2*a*b + 3*b^2)*d*
x + 30*((a^2 - b^2)*d*x - a^2 - a*b)*cosh(d*x + c)^2 - 2*a^2 + 4*a*b)*sinh(d*x + c)^4 + 16*(7*(a^2 + 2*a*b + b
^2)*d*x*cosh(d*x + c)^5 + 10*((a^2 - b^2)*d*x - a^2 - a*b)*cosh(d*x + c)^3 + ((3*a^2 - 2*a*b + 3*b^2)*d*x - 2*
a^2 + 4*a*b)*cosh(d*x + c))*sinh(d*x + c)^3 + 2*(a^2 + 2*a*b + b^2)*d*x + 8*((a^2 - b^2)*d*x - a^2 - a*b)*cosh
(d*x + c)^2 + 8*(7*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^6 + 15*((a^2 - b^2)*d*x - a^2 - a*b)*cosh(d*x + c)^4
+ (a^2 - b^2)*d*x + 3*((3*a^2 - 2*a*b + 3*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 - b^2)*cosh(d*x + c)^6 + 4*(7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + a^2 - b^2
)*sinh(d*x + c)^6 + 8*(7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^3 + 3*(a^2 - b^2)*cosh(d*x + c))*sinh(d*x + c)^5 +
2*(3*a^2 - 2*a*b + 3*b^2)*cosh(d*x + c)^4 + 2*(35*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 30*(a^2 - b^2)*cosh(d*
x + c)^2 + 3*a^2 - 2*a*b + 3*b^2)*sinh(d*x + c)^4 + 8*(7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^5 + 10*(a^2 - b^2)*
cosh(d*x + c)^3 + (3*a^2 - 2*a*b + 3*b^2)*cosh(d*x + c))*sinh(d*x + c)^3 + 4*(a^2 - b^2)*cosh(d*x + c)^2 + 4*(
7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^6 + 15*(a^2 - b^2)*cosh(d*x + c)^4 + 3*(3*a^2 - 2*a*b + 3*b^2)*cosh(d*x +
c)^2 + a^2 - b^2)*sinh(d*x + c)^2 + a^2 + 2*a*b + b^2 + 8*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^7 + 3*(a^2 - b^2)
*cosh(d*x + c)^5 + (3*a^2 - 2*a*b + 3*b^2)*cosh(d*x + c)^3 + (a^2 - b^2)*cosh(d*x + c))*sinh(d*x + c))*log(2*(
(a + b)*cosh(d*x + c)^2 + (a + b)*sinh(d*x + c)^2 + a - b)/(cosh(d*x + c)^2 - 2*cosh(d*x + c)*sinh(d*x + c) +
sinh(d*x + c)^2)) + 16*((a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^7 + 3*((a^2 - b^2)*d*x - a^2 - a*b)*cosh(d*x + c
)^5 + ((3*a^2 - 2*a*b + 3*b^2)*d*x - 2*a^2 + 4*a*b)*cosh(d*x + c)^3 + ((a^2 - b^2)*d*x - a^2 - a*b)*cosh(d*x +
 c))*sinh(d*x + c))/((a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*d*cosh(d*x + c)^8 + 8*(a^5 + 5*
a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*d*cosh(d*x + c)*sinh(d*x + c)^7 + (a^5 + 5*a^4*b + 10*a^3*b^2
 + 10*a^2*b^3 + 5*a*b^4 + b^5)*d*sinh(d*x + c)^8 + 4*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*d
*cosh(d*x + c)^6 + 4*(7*(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*d*cosh(d*x + c)^2 + (a^5 + 3
*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*d)*sinh(d*x + c)^6 + 2*(3*a^5 + 7*a^4*b + 6*a^3*b^2 + 6*a^2*b^
3 + 7*a*b^4 + 3*b^5)*d*cosh(d*x + c)^4 + 8*(7*(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*d*cosh
(d*x + c)^3 + 3*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*d*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(
35*(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*d*cosh(d*x + c)^4 + 30*(a^5 + 3*a^4*b + 2*a^3*b^2
 - 2*a^2*b^3 - 3*a*b^4 - b^5)*d*cosh(d*x + c)^2 + (3*a^5 + 7*a^4*b + 6*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 + 3*b^5)*
d)*sinh(d*x + c)^4 + 4*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*d*cosh(d*x + c)^2 + 8*(7*(a^5 +
 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*d*cosh(d*x + c)^5 + 10*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*
b^3 - 3*a*b^4 - b^5)*d*cosh(d*x + c)^3 + (3*a^5 + 7*a^4*b + 6*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 + 3*b^5)*d*cosh(d*
x + c))*sinh(d*x + c)^3 + 4*(7*(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*d*cosh(d*x + c)^6 + 1
5*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*d*cosh(d*x + c)^4 + 3*(3*a^5 + 7*a^4*b + 6*a^3*b^2 +
 6*a^2*b^3 + 7*a*b^4 + 3*b^5)*d*cosh(d*x + c)^2 + (a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*d)*s
inh(d*x + c)^2 + (a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*d + 8*((a^5 + 5*a^4*b + 10*a^3*b^2
+ 10*a^2*b^3 + 5*a*b^4 + b^5)*d*cosh(d*x + c)^7 + 3*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*d*
cosh(d*x + c)^5 + (3*a^5 + 7*a^4*b + 6*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 + 3*b^5)*d*cosh(d*x + c)^3 + (a^5 + 3*a^4
*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*d*cosh(d*x + c))*sinh(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Timed out} \]

[In]

integrate(tanh(d*x+c)**5/(a+b*tanh(d*x+c)**2)**3,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 376 vs. \(2 (103) = 206\).

Time = 0.24 (sec) , antiderivative size = 376, normalized size of antiderivative = 3.45 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {d x + c}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d} + \frac {4 \, {\left ({\left (a^{2} + a b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{2} - 2 \, a b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + {\left (a^{2} + a b\right )} e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{{\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5} + 4 \, {\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} - 3 \, a b^{4} - b^{5}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 2 \, {\left (3 \, a^{5} + 7 \, a^{4} b + 6 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 7 \, a b^{4} + 3 \, b^{5}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, {\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} - 3 \, a b^{4} - b^{5}\right )} e^{\left (-6 \, d x - 6 \, c\right )} + {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} e^{\left (-8 \, d x - 8 \, c\right )}\right )} d} + \frac {\log \left (2 \, {\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + a + b\right )}{2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d} \]

[In]

integrate(tanh(d*x+c)^5/(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

(d*x + c)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d) + 4*((a^2 + a*b)*e^(-2*d*x - 2*c) + (a^2 - 2*a*b)*e^(-4*d*x - 4*
c) + (a^2 + a*b)*e^(-6*d*x - 6*c))/((a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5 + 4*(a^5 + 3*a^4*
b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*e^(-2*d*x - 2*c) + 2*(3*a^5 + 7*a^4*b + 6*a^3*b^2 + 6*a^2*b^3 + 7*a
*b^4 + 3*b^5)*e^(-4*d*x - 4*c) + 4*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*e^(-6*d*x - 6*c) +
(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*e^(-8*d*x - 8*c))*d) + 1/2*log(2*(a - b)*e^(-2*d*x -
 2*c) + (a + b)*e^(-4*d*x - 4*c) + a + b)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 245 vs. \(2 (103) = 206\).

Time = 0.47 (sec) , antiderivative size = 245, normalized size of antiderivative = 2.25 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {\frac {2 \, \log \left ({\left | a {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 2 \, a - 2 \, b \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {3 \, a {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )}^{2} + 3 \, b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )}^{2} - 4 \, a {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} - 12 \, b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} - 4 \, a + 12 \, b}{{\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (a {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 2 \, a - 2 \, b\right )}^{2}}}{4 \, d} \]

[In]

integrate(tanh(d*x+c)^5/(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")

[Out]

1/4*(2*log(abs(a*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) + b*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) + 2*a - 2*b))/(
a^3 + 3*a^2*b + 3*a*b^2 + b^3) - (3*a*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c))^2 + 3*b*(e^(2*d*x + 2*c) + e^(-2*d*
x - 2*c))^2 - 4*a*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) - 12*b*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) - 4*a + 12*
b)/((a^2 + 2*a*b + b^2)*(a*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) + b*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) + 2*a
 - 2*b)^2))/d

Mupad [B] (verification not implemented)

Time = 0.74 (sec) , antiderivative size = 416, normalized size of antiderivative = 3.82 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {a^4+a^3\,b\,\left (2\,{\mathrm {tanh}\left (c+d\,x\right )}^2+4\right )-a\,b^3\,\left (-4\,{\mathrm {tanh}\left (c+d\,x\right )}^2+{\mathrm {tanh}\left (c+d\,x\right )}^2\,\mathrm {atan}\left (\frac {a\,{\mathrm {tanh}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+b\,{\mathrm {tanh}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{2\,a-a\,{\mathrm {tanh}\left (c+d\,x\right )}^2+b\,{\mathrm {tanh}\left (c+d\,x\right )}^2}\right )\,8{}\mathrm {i}\right )+a^2\,b^2\,\left (6\,{\mathrm {tanh}\left (c+d\,x\right )}^2+3-\mathrm {atan}\left (\frac {a\,{\mathrm {tanh}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+b\,{\mathrm {tanh}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{2\,a-a\,{\mathrm {tanh}\left (c+d\,x\right )}^2+b\,{\mathrm {tanh}\left (c+d\,x\right )}^2}\right )\,4{}\mathrm {i}\right )-b^4\,{\mathrm {tanh}\left (c+d\,x\right )}^4\,\mathrm {atan}\left (\frac {a\,{\mathrm {tanh}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+b\,{\mathrm {tanh}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{2\,a-a\,{\mathrm {tanh}\left (c+d\,x\right )}^2+b\,{\mathrm {tanh}\left (c+d\,x\right )}^2}\right )\,4{}\mathrm {i}}{4\,d\,a^5\,b^2+8\,d\,a^4\,b^3\,{\mathrm {tanh}\left (c+d\,x\right )}^2+12\,d\,a^4\,b^3+4\,d\,a^3\,b^4\,{\mathrm {tanh}\left (c+d\,x\right )}^4+24\,d\,a^3\,b^4\,{\mathrm {tanh}\left (c+d\,x\right )}^2+12\,d\,a^3\,b^4+12\,d\,a^2\,b^5\,{\mathrm {tanh}\left (c+d\,x\right )}^4+24\,d\,a^2\,b^5\,{\mathrm {tanh}\left (c+d\,x\right )}^2+4\,d\,a^2\,b^5+12\,d\,a\,b^6\,{\mathrm {tanh}\left (c+d\,x\right )}^4+8\,d\,a\,b^6\,{\mathrm {tanh}\left (c+d\,x\right )}^2+4\,d\,b^7\,{\mathrm {tanh}\left (c+d\,x\right )}^4} \]

[In]

int(tanh(c + d*x)^5/(a + b*tanh(c + d*x)^2)^3,x)

[Out]

(a^4 + a^3*b*(2*tanh(c + d*x)^2 + 4) - a*b^3*(tanh(c + d*x)^2*atan((a*tanh(c + d*x)^2*1i + b*tanh(c + d*x)^2*1
i)/(2*a - a*tanh(c + d*x)^2 + b*tanh(c + d*x)^2))*8i - 4*tanh(c + d*x)^2) + a^2*b^2*(6*tanh(c + d*x)^2 - atan(
(a*tanh(c + d*x)^2*1i + b*tanh(c + d*x)^2*1i)/(2*a - a*tanh(c + d*x)^2 + b*tanh(c + d*x)^2))*4i + 3) - b^4*tan
h(c + d*x)^4*atan((a*tanh(c + d*x)^2*1i + b*tanh(c + d*x)^2*1i)/(2*a - a*tanh(c + d*x)^2 + b*tanh(c + d*x)^2))
*4i)/(4*a^2*b^5*d + 12*a^3*b^4*d + 12*a^4*b^3*d + 4*a^5*b^2*d + 4*b^7*d*tanh(c + d*x)^4 + 24*a^2*b^5*d*tanh(c
+ d*x)^2 + 24*a^3*b^4*d*tanh(c + d*x)^2 + 8*a^4*b^3*d*tanh(c + d*x)^2 + 12*a^2*b^5*d*tanh(c + d*x)^4 + 4*a^3*b
^4*d*tanh(c + d*x)^4 + 8*a*b^6*d*tanh(c + d*x)^2 + 12*a*b^6*d*tanh(c + d*x)^4)