\(\int \frac {\tanh ^4(c+d x)}{(a+b \tanh ^2(c+d x))^2} \, dx\) [181]

Optimal result

Integrand size = 23, antiderivative size = 89 \[ \int \frac {\tanh ^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {x}{(a+b)^2}-\frac {\sqrt {a} (a+3 b) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{2 b^{3/2} (a+b)^2 d}+\frac {a \tanh (c+d x)}{2 b (a+b) d \left (a+b \tanh ^2(c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3751, 481, 536, 212, 211} \[ \int \frac {\tanh ^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=-\frac {\sqrt {a} (a+3 b) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{2 b^{3/2} d (a+b)^2}+\frac {a \tanh (c+d x)}{2 b d (a+b) \left (a+b \tanh ^2(c+d x)\right )}+\frac {x}{(a+b)^2} \]

[In]

Int[Tanh[c + d*x]^4/(a + b*Tanh[c + d*x]^2)^2,x]

[Out]

x/(a + b)^2 - (Sqrt[a]*(a + 3*b)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(2*b^(3/2)*(a + b)^2*d) + (a*Tanh[c
+ d*x])/(2*b*(a + b)*d*(a + b*Tanh[c + d*x]^2))

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 481

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-a)*e^(
2*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*n*(b*c - a*d)*(p + 1))), x] + Dist[e^
(2*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1)
+ (a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0]
&& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 536

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

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^4}{\left (1-x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d} \\ & = \frac {a \tanh (c+d x)}{2 b (a+b) d \left (a+b \tanh ^2(c+d x)\right )}-\frac {\text {Subst}\left (\int \frac {a+(-a-2 b) x^2}{\left (1-x^2\right ) \left (a+b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{2 b (a+b) d} \\ & = \frac {a \tanh (c+d x)}{2 b (a+b) d \left (a+b \tanh ^2(c+d x)\right )}+\frac {\text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{(a+b)^2 d}-\frac {(a (a+3 b)) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tanh (c+d x)\right )}{2 b (a+b)^2 d} \\ & = \frac {x}{(a+b)^2}-\frac {\sqrt {a} (a+3 b) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{2 b^{3/2} (a+b)^2 d}+\frac {a \tanh (c+d x)}{2 b (a+b) d \left (a+b \tanh ^2(c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.01 \[ \int \frac {\tanh ^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {2 (c+d x)-\frac {\sqrt {a} (a+3 b) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{b^{3/2}}+\frac {a (a+b) \sinh (2 (c+d x))}{b (a-b+(a+b) \cosh (2 (c+d x)))}}{2 (a+b)^2 d} \]

[In]

Integrate[Tanh[c + d*x]^4/(a + b*Tanh[c + d*x]^2)^2,x]

[Out]

(2*(c + d*x) - (Sqrt[a]*(a + 3*b)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/b^(3/2) + (a*(a + b)*Sinh[2*(c + d*
x)])/(b*(a - b + (a + b)*Cosh[2*(c + d*x)])))/(2*(a + b)^2*d)

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.17

method	result	size
derivativedivides	\(\frac {-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{2}}-\frac {a \left (-\frac {\left (a +b \right ) \tanh \left (d x +c \right )}{2 b \left (a +b \tanh \left (d x +c \right )^{2}\right )}+\frac {\left (a +3 b \right ) \arctan \left (\frac {b \tanh \left (d x +c \right )}{\sqrt {a b}}\right )}{2 b \sqrt {a b}}\right )}{\left (a +b \right )^{2}}+\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2 \left (a +b \right )^{2}}}{d}\)	\(104\)
default	\(\frac {-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{2}}-\frac {a \left (-\frac {\left (a +b \right ) \tanh \left (d x +c \right )}{2 b \left (a +b \tanh \left (d x +c \right )^{2}\right )}+\frac {\left (a +3 b \right ) \arctan \left (\frac {b \tanh \left (d x +c \right )}{\sqrt {a b}}\right )}{2 b \sqrt {a b}}\right )}{\left (a +b \right )^{2}}+\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2 \left (a +b \right )^{2}}}{d}\)	\(104\)
risch	\(\frac {x}{a^{2}+2 a b +b^{2}}-\frac {a \left ({\mathrm e}^{2 d x +2 c} a -b \,{\mathrm e}^{2 d x +2 c}+a +b \right )}{d \left (a +b \right )^{2} b \left (a \,{\mathrm e}^{4 d x +4 c}+b \,{\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{2 d x +2 c} a -2 b \,{\mathrm e}^{2 d x +2 c}+a +b \right )}+\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-a b}-a +b}{a +b}\right ) a}{4 b^{2} \left (a +b \right )^{2} d}+\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-a b}-a +b}{a +b}\right )}{4 b \left (a +b \right )^{2} d}-\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-a b}+a -b}{a +b}\right ) a}{4 b^{2} \left (a +b \right )^{2} d}-\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-a b}+a -b}{a +b}\right )}{4 b \left (a +b \right )^{2} d}\)	\(308\)

[In]

int(tanh(d*x+c)^4/(a+b*tanh(d*x+c)^2)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 814 vs. \(2 (77) = 154\).

Time = 0.32 (sec) , antiderivative size = 1950, normalized size of antiderivative = 21.91 \[ \int \frac {\tanh ^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \]

[In]

integrate(tanh(d*x+c)^4/(a+b*tanh(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

[1/4*(4*(a*b + b^2)*d*x*cosh(d*x + c)^4 + 16*(a*b + b^2)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 4*(a*b + b^2)*d*x
*sinh(d*x + c)^4 + 4*(a*b + b^2)*d*x + 4*(2*(a*b - b^2)*d*x - a^2 + a*b)*cosh(d*x + c)^2 + 4*(6*(a*b + b^2)*d*
x*cosh(d*x + c)^2 + 2*(a*b - b^2)*d*x - a^2 + a*b)*sinh(d*x + c)^2 + ((a^2 + 4*a*b + 3*b^2)*cosh(d*x + c)^4 +
4*(a^2 + 4*a*b + 3*b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 + 4*a*b + 3*b^2)*sinh(d*x + c)^4 + 2*(a^2 + 2*a*b
- 3*b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 + 4*a*b + 3*b^2)*cosh(d*x + c)^2 + a^2 + 2*a*b - 3*b^2)*sinh(d*x + c)^2
+ a^2 + 4*a*b + 3*b^2 + 4*((a^2 + 4*a*b + 3*b^2)*cosh(d*x + c)^3 + (a^2 + 2*a*b - 3*b^2)*cosh(d*x + c))*sinh(d
*x + c))*sqrt(-a/b)*log(((a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b + b^2)*cosh(d*x + c)*sinh(d*x +
c)^3 + (a^2 + 2*a*b + b^2)*sinh(d*x + c)^4 + 2*(a^2 - b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 + 2*a*b + b^2)*cosh(d*x
+ c)^2 + a^2 - b^2)*sinh(d*x + c)^2 + a^2 - 6*a*b + b^2 + 4*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^3 + (a^2 - b^2
)*cosh(d*x + c))*sinh(d*x + c) - 4*((a*b + b^2)*cosh(d*x + c)^2 + 2*(a*b + b^2)*cosh(d*x + c)*sinh(d*x + c) +
(a*b + b^2)*sinh(d*x + c)^2 + a*b - b^2)*sqrt(-a/b))/((a + b)*cosh(d*x + c)^4 + 4*(a + b)*cosh(d*x + c)*sinh(d
*x + c)^3 + (a + b)*sinh(d*x + c)^4 + 2*(a - b)*cosh(d*x + c)^2 + 2*(3*(a + b)*cosh(d*x + c)^2 + a - b)*sinh(d
*x + c)^2 + 4*((a + b)*cosh(d*x + c)^3 + (a - b)*cosh(d*x + c))*sinh(d*x + c) + a + b)) - 4*a^2 - 4*a*b + 8*(2
*(a*b + b^2)*d*x*cosh(d*x + c)^3 + (2*(a*b - b^2)*d*x - a^2 + a*b)*cosh(d*x + c))*sinh(d*x + c))/((a^3*b + 3*a
^2*b^2 + 3*a*b^3 + b^4)*d*cosh(d*x + c)^4 + 4*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*cosh(d*x + c)*sinh(d*x + c
)^3 + (a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*sinh(d*x + c)^4 + 2*(a^3*b + a^2*b^2 - a*b^3 - b^4)*d*cosh(d*x + c
)^2 + 2*(3*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*cosh(d*x + c)^2 + (a^3*b + a^2*b^2 - a*b^3 - b^4)*d)*sinh(d*x
+ c)^2 + (a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d + 4*((a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*cosh(d*x + c)^3 + (
a^3*b + a^2*b^2 - a*b^3 - b^4)*d*cosh(d*x + c))*sinh(d*x + c)), 1/2*(2*(a*b + b^2)*d*x*cosh(d*x + c)^4 + 8*(a*
b + b^2)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 2*(a*b + b^2)*d*x*sinh(d*x + c)^4 + 2*(a*b + b^2)*d*x + 2*(2*(a*b
- b^2)*d*x - a^2 + a*b)*cosh(d*x + c)^2 + 2*(6*(a*b + b^2)*d*x*cosh(d*x + c)^2 + 2*(a*b - b^2)*d*x - a^2 + a*
b)*sinh(d*x + c)^2 - ((a^2 + 4*a*b + 3*b^2)*cosh(d*x + c)^4 + 4*(a^2 + 4*a*b + 3*b^2)*cosh(d*x + c)*sinh(d*x +
c)^3 + (a^2 + 4*a*b + 3*b^2)*sinh(d*x + c)^4 + 2*(a^2 + 2*a*b - 3*b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 + 4*a*b +
3*b^2)*cosh(d*x + c)^2 + a^2 + 2*a*b - 3*b^2)*sinh(d*x + c)^2 + a^2 + 4*a*b + 3*b^2 + 4*((a^2 + 4*a*b + 3*b^2)
*cosh(d*x + c)^3 + (a^2 + 2*a*b - 3*b^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt(a/b)*arctan(1/2*((a + b)*cosh(d*x
+ c)^2 + 2*(a + b)*cosh(d*x + c)*sinh(d*x + c) + (a + b)*sinh(d*x + c)^2 + a - b)*sqrt(a/b)/a) - 2*a^2 - 2*a*b
+ 4*(2*(a*b + b^2)*d*x*cosh(d*x + c)^3 + (2*(a*b - b^2)*d*x - a^2 + a*b)*cosh(d*x + c))*sinh(d*x + c))/((a^3*
b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*cosh(d*x + c)^4 + 4*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*cosh(d*x + c)*sinh(
d*x + c)^3 + (a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*sinh(d*x + c)^4 + 2*(a^3*b + a^2*b^2 - a*b^3 - b^4)*d*cosh(
d*x + c)^2 + 2*(3*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*cosh(d*x + c)^2 + (a^3*b + a^2*b^2 - a*b^3 - b^4)*d)*s
inh(d*x + c)^2 + (a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d + 4*((a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*cosh(d*x + c
)^3 + (a^3*b + a^2*b^2 - a*b^3 - b^4)*d*cosh(d*x + c))*sinh(d*x + c))]

Sympy [F(-1)]

Timed out. \[ \int \frac {\tanh ^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\text {Timed out} \]

[In]

integrate(tanh(d*x+c)**4/(a+b*tanh(d*x+c)**2)**2,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1010 vs. \(2 (77) = 154\).

Time = 0.52 (sec) , antiderivative size = 1010, normalized size of antiderivative = 11.35 \[ \int \frac {\tanh ^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \]

[In]

integrate(tanh(d*x+c)^4/(a+b*tanh(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

-1/32*(a^3 + 9*a^2*b - 9*a*b^2 - b^3)*arctan(1/2*((a + b)*e^(2*d*x + 2*c) + a - b)/sqrt(a*b))/((a^3*b + 2*a^2*
b^2 + a*b^3)*sqrt(a*b)*d) + 1/32*(a^3 + 9*a^2*b - 9*a*b^2 - b^3)*arctan(1/2*((a + b)*e^(-2*d*x - 2*c) + a - b)
/sqrt(a*b))/((a^3*b + 2*a^2*b^2 + a*b^3)*sqrt(a*b)*d) - 1/16*(a^3 - 5*a^2*b - 5*a*b^2 + b^3 + (a^3 - 15*a^2*b
+ 15*a*b^2 - b^3)*e^(2*d*x + 2*c))/((a^4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4 + (a^4*b + 3*a^3*b^2 + 3*a^2*b^3 +
a*b^4)*e^(4*d*x + 4*c) + 2*(a^4*b + a^3*b^2 - a^2*b^3 - a*b^4)*e^(2*d*x + 2*c))*d) + 1/16*(a^3 - 5*a^2*b - 5*a
*b^2 + b^3 + (a^3 - 15*a^2*b + 15*a*b^2 - b^3)*e^(-2*d*x - 2*c))/((a^4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4 + 2*(
a^4*b + a^3*b^2 - a^2*b^3 - a*b^4)*e^(-2*d*x - 2*c) + (a^4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*e^(-4*d*x - 4*c)
)*d) - 1/4*(a^2 - b^2 + (a^2 - 6*a*b + b^2)*e^(2*d*x + 2*c))/((a^3*b + 2*a^2*b^2 + a*b^3 + (a^3*b + 2*a^2*b^2
+ a*b^3)*e^(4*d*x + 4*c) + 2*(a^3*b - a*b^3)*e^(2*d*x + 2*c))*d) + 1/4*(a^2 - b^2 + (a^2 - 6*a*b + b^2)*e^(-2*
d*x - 2*c))/((a^3*b + 2*a^2*b^2 + a*b^3 + 2*(a^3*b - a*b^3)*e^(-2*d*x - 2*c) + (a^3*b + 2*a^2*b^2 + a*b^3)*e^(
-4*d*x - 4*c))*d) + 3/8*((a - b)*e^(-2*d*x - 2*c) + a + b)/((a^2*b + a*b^2 + 2*(a^2*b - a*b^2)*e^(-2*d*x - 2*c
) + (a^2*b + a*b^2)*e^(-4*d*x - 4*c))*d) + 1/4*log((a + b)*e^(4*d*x + 4*c) + 2*(a - b)*e^(2*d*x + 2*c) + a + b
)/((a^2 + 2*a*b + b^2)*d) - 1/4*log(2*(a - b)*e^(-2*d*x - 2*c) + (a + b)*e^(-4*d*x - 4*c) + a + b)/((a^2 + 2*a
*b + b^2)*d) - 1/8*(a + b)*arctan(1/2*((a + b)*e^(2*d*x + 2*c) + a - b)/sqrt(a*b))/(sqrt(a*b)*a*b*d) + 1/8*(a
+ b)*arctan(1/2*((a + b)*e^(-2*d*x - 2*c) + a - b)/sqrt(a*b))/(sqrt(a*b)*a*b*d) + 3/16*(a - b)*arctan(1/2*((a
+ b)*e^(-2*d*x - 2*c) + a - b)/sqrt(a*b))/(sqrt(a*b)*a*b*d)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 195 vs. \(2 (77) = 154\).

Time = 0.38 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.19 \[ \int \frac {\tanh ^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=-\frac {\frac {{\left (a^{2} + 3 \, a b\right )} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \sqrt {a b}} - \frac {2 \, {\left (d x + c\right )}}{a^{2} + 2 \, a b + b^{2}} + \frac {2 \, {\left (a^{2} e^{\left (2 \, d x + 2 \, c\right )} - a b e^{\left (2 \, d x + 2 \, c\right )} + a^{2} + a b\right )}}{{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} {\left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}}}{2 \, d} \]

[In]

integrate(tanh(d*x+c)^4/(a+b*tanh(d*x+c)^2)^2,x, algorithm="giac")

[Out]

-1/2*((a^2 + 3*a*b)*arctan(1/2*(a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) + a - b)/sqrt(a*b))/((a^2*b + 2*a*b^2 +
b^3)*sqrt(a*b)) - 2*(d*x + c)/(a^2 + 2*a*b + b^2) + 2*(a^2*e^(2*d*x + 2*c) - a*b*e^(2*d*x + 2*c) + a^2 + a*b)/
((a^2*b + 2*a*b^2 + b^3)*(a*e^(4*d*x + 4*c) + b*e^(4*d*x + 4*c) + 2*a*e^(2*d*x + 2*c) - 2*b*e^(2*d*x + 2*c) +
a + b)))/d

Mupad [B] (verification not implemented)

Time = 2.44 (sec) , antiderivative size = 1655, normalized size of antiderivative = 18.60 \[ \int \frac {\tanh ^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \]

[In]

int(tanh(c + d*x)^4/(a + b*tanh(c + d*x)^2)^2,x)

[Out]

log(tanh(c + d*x) + 1)/(2*a^2*d + 2*b^2*d + 4*a*b*d) - log(tanh(c + d*x) - 1)/(2*d*(a + b)^2) - (atan((((a + 3
*b)*(-a*b^3)^(1/2)*((tanh(c + d*x)*(6*a^3*b + a^4 + 4*b^4 + 9*a^2*b^2))/(2*(b^3*d^2 + 2*a*b^2*d^2 + a^2*b*d^2)
) + ((a + 3*b)*(-a*b^3)^(1/2)*((2*a*b^6*d^2 + 8*a^2*b^5*d^2 + 12*a^3*b^4*d^2 + 8*a^4*b^3*d^2 + 2*a^5*b^2*d^2)/
(b^4*d^3 + 3*a*b^3*d^3 + a^3*b*d^3 + 3*a^2*b^2*d^3) - (tanh(c + d*x)*(a + 3*b)*(-a*b^3)^(1/2)*(16*b^8*d^2 + 48
*a*b^7*d^2 + 32*a^2*b^6*d^2 - 32*a^3*b^5*d^2 - 48*a^4*b^4*d^2 - 16*a^5*b^3*d^2))/(8*(b^5*d + a^2*b^3*d + 2*a*b
^4*d)*(b^3*d^2 + 2*a*b^2*d^2 + a^2*b*d^2))))/(4*(b^5*d + a^2*b^3*d + 2*a*b^4*d)))*1i)/(4*(b^5*d + a^2*b^3*d +
2*a*b^4*d)) + ((a + 3*b)*(-a*b^3)^(1/2)*((tanh(c + d*x)*(6*a^3*b + a^4 + 4*b^4 + 9*a^2*b^2))/(2*(b^3*d^2 + 2*a
*b^2*d^2 + a^2*b*d^2)) - ((a + 3*b)*(-a*b^3)^(1/2)*((2*a*b^6*d^2 + 8*a^2*b^5*d^2 + 12*a^3*b^4*d^2 + 8*a^4*b^3*
d^2 + 2*a^5*b^2*d^2)/(b^4*d^3 + 3*a*b^3*d^3 + a^3*b*d^3 + 3*a^2*b^2*d^3) + (tanh(c + d*x)*(a + 3*b)*(-a*b^3)^(
1/2)*(16*b^8*d^2 + 48*a*b^7*d^2 + 32*a^2*b^6*d^2 - 32*a^3*b^5*d^2 - 48*a^4*b^4*d^2 - 16*a^5*b^3*d^2))/(8*(b^5*
d + a^2*b^3*d + 2*a*b^4*d)*(b^3*d^2 + 2*a*b^2*d^2 + a^2*b*d^2))))/(4*(b^5*d + a^2*b^3*d + 2*a*b^4*d)))*1i)/(4*
(b^5*d + a^2*b^3*d + 2*a*b^4*d)))/((3*a*b^2 + (5*a^2*b)/2 + a^3/2)/(b^4*d^3 + 3*a*b^3*d^3 + a^3*b*d^3 + 3*a^2*
b^2*d^3) - ((a + 3*b)*(-a*b^3)^(1/2)*((tanh(c + d*x)*(6*a^3*b + a^4 + 4*b^4 + 9*a^2*b^2))/(2*(b^3*d^2 + 2*a*b^
2*d^2 + a^2*b*d^2)) + ((a + 3*b)*(-a*b^3)^(1/2)*((2*a*b^6*d^2 + 8*a^2*b^5*d^2 + 12*a^3*b^4*d^2 + 8*a^4*b^3*d^2
+ 2*a^5*b^2*d^2)/(b^4*d^3 + 3*a*b^3*d^3 + a^3*b*d^3 + 3*a^2*b^2*d^3) - (tanh(c + d*x)*(a + 3*b)*(-a*b^3)^(1/2
)*(16*b^8*d^2 + 48*a*b^7*d^2 + 32*a^2*b^6*d^2 - 32*a^3*b^5*d^2 - 48*a^4*b^4*d^2 - 16*a^5*b^3*d^2))/(8*(b^5*d +
a^2*b^3*d + 2*a*b^4*d)*(b^3*d^2 + 2*a*b^2*d^2 + a^2*b*d^2))))/(4*(b^5*d + a^2*b^3*d + 2*a*b^4*d))))/(4*(b^5*d
+ a^2*b^3*d + 2*a*b^4*d)) + ((a + 3*b)*(-a*b^3)^(1/2)*((tanh(c + d*x)*(6*a^3*b + a^4 + 4*b^4 + 9*a^2*b^2))/(2
*(b^3*d^2 + 2*a*b^2*d^2 + a^2*b*d^2)) - ((a + 3*b)*(-a*b^3)^(1/2)*((2*a*b^6*d^2 + 8*a^2*b^5*d^2 + 12*a^3*b^4*d
^2 + 8*a^4*b^3*d^2 + 2*a^5*b^2*d^2)/(b^4*d^3 + 3*a*b^3*d^3 + a^3*b*d^3 + 3*a^2*b^2*d^3) + (tanh(c + d*x)*(a +
3*b)*(-a*b^3)^(1/2)*(16*b^8*d^2 + 48*a*b^7*d^2 + 32*a^2*b^6*d^2 - 32*a^3*b^5*d^2 - 48*a^4*b^4*d^2 - 16*a^5*b^3
*d^2))/(8*(b^5*d + a^2*b^3*d + 2*a*b^4*d)*(b^3*d^2 + 2*a*b^2*d^2 + a^2*b*d^2))))/(4*(b^5*d + a^2*b^3*d + 2*a*b
^4*d))))/(4*(b^5*d + a^2*b^3*d + 2*a*b^4*d))))*(a + 3*b)*(-a*b^3)^(1/2)*1i)/(2*(b^5*d + a^2*b^3*d + 2*a*b^4*d)
) + (a*tanh(c + d*x))/(2*b*(a + b)*(a*d + b*d*tanh(c + d*x)^2))