### 3.184 $$\int \frac{\tanh (c+d x)}{(a+b \tanh ^2(c+d x))^2} \, dx$$

Optimal. Leaf size=68 $-\frac{1}{2 d (a+b) \left (a+b \tanh ^2(c+d x)\right )}+\frac{\log \left (a+b \tanh ^2(c+d x)\right )}{2 d (a+b)^2}+\frac{\log (\cosh (c+d x))}{d (a+b)^2}$

[Out]

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

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Rubi [A]  time = 0.0857955, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.143, Rules used = {3670, 444, 44} $-\frac{1}{2 d (a+b) \left (a+b \tanh ^2(c+d x)\right )}+\frac{\log \left (a+b \tanh ^2(c+d x)\right )}{2 d (a+b)^2}+\frac{\log (\cosh (c+d x))}{d (a+b)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 3670

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]))

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\tanh (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x}{\left (1-x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{(1-x) (a+b x)^2} \, dx,x,\tanh ^2(c+d x)\right )}{2 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{(a+b)^2 (-1+x)}+\frac{b}{(a+b) (a+b x)^2}+\frac{b}{(a+b)^2 (a+b x)}\right ) \, dx,x,\tanh ^2(c+d x)\right )}{2 d}\\ &=\frac{\log (\cosh (c+d x))}{(a+b)^2 d}+\frac{\log \left (a+b \tanh ^2(c+d x)\right )}{2 (a+b)^2 d}-\frac{1}{2 (a+b) d \left (a+b \tanh ^2(c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.371088, size = 55, normalized size = 0.81 $-\frac{\frac{a+b}{a+b \tanh ^2(c+d x)}-\log \left (a+b \tanh ^2(c+d x)\right )-2 \log (\cosh (c+d x))}{2 d (a+b)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.027, size = 113, normalized size = 1.7 \begin{align*} -{\frac{\ln \left ( \tanh \left ( dx+c \right ) +1 \right ) }{2\,d \left ( a+b \right ) ^{2}}}-{\frac{a}{2\,d \left ( a+b \right ) ^{2} \left ( a+b \left ( \tanh \left ( dx+c \right ) \right ) ^{2} \right ) }}-{\frac{b}{2\,d \left ( a+b \right ) ^{2} \left ( a+b \left ( \tanh \left ( dx+c \right ) \right ) ^{2} \right ) }}+{\frac{\ln \left ( a+b \left ( \tanh \left ( dx+c \right ) \right ) ^{2} \right ) }{2\,d \left ( a+b \right ) ^{2}}}-{\frac{\ln \left ( \tanh \left ( dx+c \right ) -1 \right ) }{2\,d \left ( a+b \right ) ^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.09235, size = 230, normalized size = 3.38 \begin{align*} -\frac{2 \, b e^{\left (-2 \, d x - 2 \, c\right )}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3} + 2 \,{\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} +{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} + \frac{d x + c}{{\left (a^{2} + 2 \, a b + b^{2}\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^{2} + 2 \, a b + b^{2}\right )} d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.95637, size = 1611, normalized size = 23.69 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(a + b)*d*x*cosh(d*x + c)^4 + 8*(a + b)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 2*(a + b)*d*x*sinh(d*x + c
)^4 + 2*(a + b)*d*x + 4*((a - b)*d*x + b)*cosh(d*x + c)^2 + 4*(3*(a + b)*d*x*cosh(d*x + c)^2 + (a - b)*d*x + b
)*sinh(d*x + c)^2 - ((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)*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)) + 8*((a + b)*d*x*cosh(d*x + c)^
3 + ((a - b)*d*x + b)*cosh(d*x + c))*sinh(d*x + c))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c)^4 + 4*(a^
3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*sinh(d*x + c)
^4 + 2*(a^3 + a^2*b - a*b^2 - b^3)*d*cosh(d*x + c)^2 + 2*(3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c)^2
+ (a^3 + a^2*b - a*b^2 - b^3)*d)*sinh(d*x + c)^2 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d + 4*((a^3 + 3*a^2*b + 3*a
*b^2 + b^3)*d*cosh(d*x + c)^3 + (a^3 + a^2*b - a*b^2 - b^3)*d*cosh(d*x + c))*sinh(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.22318, size = 208, normalized size = 3.06 \begin{align*} \frac{\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 )}{2 \,{\left (a^{2} d + 2 \, a b d + b^{2} d\right )}} - \frac{e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} + 2}{2 \,{\left (a d + b d\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 )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*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^2
*d + 2*a*b*d + b^2*d) - 1/2*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c) + 2)/((a*d + b*d)*(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))