### 3.180 $$\int \frac{\tanh ^5(c+d x)}{(a+b \tanh ^2(c+d x))^2} \, dx$$

Optimal. Leaf size=83 $-\frac{a^2}{2 b^2 d (a+b) \left (a+b \tanh ^2(c+d x)\right )}-\frac{a (a+2 b) \log \left (a+b \tanh ^2(c+d x)\right )}{2 b^2 d (a+b)^2}+\frac{\log (\cosh (c+d x))}{d (a+b)^2}$

[Out]

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

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Rubi [A]  time = 0.150087, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.13, Rules used = {3670, 446, 88} $-\frac{a^2}{2 b^2 d (a+b) \left (a+b \tanh ^2(c+d x)\right )}-\frac{a (a+2 b) \log \left (a+b \tanh ^2(c+d x)\right )}{2 b^2 d (a+b)^2}+\frac{\log (\cosh (c+d x))}{d (a+b)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

Log[Cosh[c + d*x]]/((a + b)^2*d) - (a*(a + 2*b)*Log[a + b*Tanh[c + d*x]^2])/(2*b^2*(a + b)^2*d) - a^2/(2*b^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 446

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 88

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]))

Rubi steps

\begin{align*} \int \frac{\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^5}{\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{x^2}{(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{a^2}{b (a+b) (a+b x)^2}-\frac{a (a+2 b)}{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{a (a+2 b) \log \left (a+b \tanh ^2(c+d x)\right )}{2 b^2 (a+b)^2 d}-\frac{a^2}{2 b^2 (a+b) d \left (a+b \tanh ^2(c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.473557, size = 69, normalized size = 0.83 $-\frac{\frac{a^2 (a+b)}{b^2 \left (a+b \tanh ^2(c+d x)\right )}+\frac{a (a+2 b) \log \left (a+b \tanh ^2(c+d x)\right )}{b^2}-2 \log (\cosh (c+d x))}{2 d (a+b)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.027, size = 156, normalized size = 1.9 \begin{align*} -{\frac{\ln \left ( \tanh \left ( dx+c \right ) +1 \right ) }{2\,d \left ( a+b \right ) ^{2}}}-{\frac{{a}^{3}}{2\,d \left ( a+b \right ) ^{2}{b}^{2} \left ( a+b \left ( \tanh \left ( dx+c \right ) \right ) ^{2} \right ) }}-{\frac{{a}^{2}}{2\,d \left ( a+b \right ) ^{2}b \left ( a+b \left ( \tanh \left ( dx+c \right ) \right ) ^{2} \right ) }}-{\frac{{a}^{2}\ln \left ( a+b \left ( \tanh \left ( dx+c \right ) \right ) ^{2} \right ) }{2\,d \left ( a+b \right ) ^{2}{b}^{2}}}-{\frac{a\ln \left ( a+b \left ( \tanh \left ( dx+c \right ) \right ) ^{2} \right ) }{d \left ( a+b \right ) ^{2}b}}-{\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)^5/(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^3/b^2/(a+b*tanh(d*x+c)^2)-1/2/d/(a+b)^2*a^2/b/(a+b*tanh(d*x+c
)^2)-1/2/d/(a+b)^2*a^2/b^2*ln(a+b*tanh(d*x+c)^2)-1/d/(a+b)^2*a/b*ln(a+b*tanh(d*x+c)^2)-1/2/d/(a+b)^2*ln(tanh(d
*x+c)-1)

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Maxima [B]  time = 1.63801, size = 293, normalized size = 3.53 \begin{align*} -\frac{2 \, a^{2} e^{\left (-2 \, d x - 2 \, c\right )}}{{\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4} + 2 \,{\left (a^{3} b + a^{2} b^{2} - a b^{3} - b^{4}\right )} e^{\left (-2 \, d x - 2 \, c\right )} +{\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} - \frac{{\left (a^{2} + 2 \, a b\right )} \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} b^{2} + 2 \, a b^{3} + b^{4}\right )} d} + \frac{d x + c}{{\left (a^{2} + 2 \, a b + b^{2}\right )} d} + \frac{\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{b^{2} d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.61627, size = 2678, normalized size = 32.27 \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)^5/(a+b*tanh(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

-1/2*(2*(a*b^2 + b^3)*d*x*cosh(d*x + c)^4 + 8*(a*b^2 + b^3)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 2*(a*b^2 + b^3
)*d*x*sinh(d*x + c)^4 + 2*(a*b^2 + b^3)*d*x + 4*(a^2*b + (a*b^2 - b^3)*d*x)*cosh(d*x + c)^2 + 4*(3*(a*b^2 + b^
3)*d*x*cosh(d*x + c)^2 + a^2*b + (a*b^2 - b^3)*d*x)*sinh(d*x + c)^2 + ((a^3 + 3*a^2*b + 2*a*b^2)*cosh(d*x + c)
^4 + 4*(a^3 + 3*a^2*b + 2*a*b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^3 + 3*a^2*b + 2*a*b^2)*sinh(d*x + c)^4 + a
^3 + 3*a^2*b + 2*a*b^2 + 2*(a^3 + a^2*b - 2*a*b^2)*cosh(d*x + c)^2 + 2*(a^3 + a^2*b - 2*a*b^2 + 3*(a^3 + 3*a^2
*b + 2*a*b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*((a^3 + 3*a^2*b + 2*a*b^2)*cosh(d*x + c)^3 + (a^3 + a^2*b -
2*a*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)/(cos
h(d*x + c)^2 - 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2)) - 2*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x
+ c)^4 + 4*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*si
nh(d*x + c)^4 + a^3 + 3*a^2*b + 3*a*b^2 + b^3 + 2*(a^3 + a^2*b - a*b^2 - b^3)*cosh(d*x + c)^2 + 2*(a^3 + a^2*b
- a*b^2 - b^3 + 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*((a^3 + 3*a^2*b + 3*a*
b^2 + b^3)*cosh(d*x + c)^3 + (a^3 + a^2*b - a*b^2 - b^3)*cosh(d*x + c))*sinh(d*x + c))*log(2*cosh(d*x + c)/(co
sh(d*x + c) - sinh(d*x + c))) + 8*((a*b^2 + b^3)*d*x*cosh(d*x + c)^3 + (a^2*b + (a*b^2 - b^3)*d*x)*cosh(d*x +
c))*sinh(d*x + c))/((a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*d*cosh(d*x + c)^4 + 4*(a^3*b^2 + 3*a^2*b^3 + 3*a*b^4
+ b^5)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*d*sinh(d*x + c)^4 + 2*(a^3*b^2
+ a^2*b^3 - a*b^4 - b^5)*d*cosh(d*x + c)^2 + 2*(3*(a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*d*cosh(d*x + c)^2 + (
a^3*b^2 + a^2*b^3 - a*b^4 - b^5)*d)*sinh(d*x + c)^2 + (a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*d + 4*((a^3*b^2 +
3*a^2*b^3 + 3*a*b^4 + b^5)*d*cosh(d*x + c)^3 + (a^3*b^2 + a^2*b^3 - a*b^4 - b^5)*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)**5/(a+b*tanh(d*x+c)**2)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.26749, size = 275, normalized size = 3.31 \begin{align*} -\frac{{\left (a^{2} + 2 \, a b\right )} \log \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 \,{\left (a^{2} b^{2} d + 2 \, a b^{3} d + b^{4} d\right )}} - \frac{d x + c}{a^{2} d + 2 \, a b d + b^{2} d} - \frac{2 \, a^{2} e^{\left (2 \, d x + 2 \, c\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 )}{\left (a + b\right )}^{2} b d} + \frac{\log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}{b^{2} d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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