### 3.32 $$\int \frac{\text{csch}^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx$$

Optimal. Leaf size=70 $\frac{(a+b) \coth (c+d x)}{a^2 d}+\frac{\sqrt{b} (a+b) \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{a^{5/2} d}-\frac{\coth ^3(c+d x)}{3 a d}$

[Out]

(Sqrt[b]*(a + b)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(a^(5/2)*d) + ((a + b)*Coth[c + d*x])/(a^2*d) - Coth
[c + d*x]^3/(3*a*d)

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Rubi [A]  time = 0.0875878, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.174, Rules used = {3663, 453, 325, 205} $\frac{(a+b) \coth (c+d x)}{a^2 d}+\frac{\sqrt{b} (a+b) \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{a^{5/2} d}-\frac{\coth ^3(c+d x)}{3 a d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[b]*(a + b)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(a^(5/2)*d) + ((a + b)*Coth[c + d*x])/(a^2*d) - Coth
[c + d*x]^3/(3*a*d)

Rule 3663

Int[sin[(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^(m + 1))/f, Subst[Int[(x^m*(a + b*(ff*x)^n)^p)/(c^2 + ff^2*x^2
)^(m/2 + 1), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2]

Rule 453

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\text{csch}^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1-x^2}{x^4 \left (a+b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=-\frac{\coth ^3(c+d x)}{3 a d}-\frac{(a+b) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{a d}\\ &=\frac{(a+b) \coth (c+d x)}{a^2 d}-\frac{\coth ^3(c+d x)}{3 a d}+\frac{(b (a+b)) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tanh (c+d x)\right )}{a^2 d}\\ &=\frac{\sqrt{b} (a+b) \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{a^{5/2} d}+\frac{(a+b) \coth (c+d x)}{a^2 d}-\frac{\coth ^3(c+d x)}{3 a d}\\ \end{align*}

Mathematica [A]  time = 0.289309, size = 71, normalized size = 1.01 $\frac{3 \sqrt{b} (a+b) \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )+\sqrt{a} \coth (c+d x) \left (-a \text{csch}^2(c+d x)+2 a+3 b\right )}{3 a^{5/2} d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(3*Sqrt[b]*(a + b)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]] + Sqrt[a]*Coth[c + d*x]*(2*a + 3*b - a*Csch[c + d*x
]^2))/(3*a^(5/2)*d)

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Maple [B]  time = 0.089, size = 750, normalized size = 10.7 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24/d/a*tanh(1/2*d*x+1/2*c)^3+3/8/d/a*tanh(1/2*d*x+1/2*c)+1/2/d/a^2*tanh(1/2*d*x+1/2*c)*b-1/d*b/(b*(a+b))^(1
/2)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2))+1/d
*b/a/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2))-2/
d/(b*(a+b))^(1/2)/a/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*
b)*a)^(1/2))*b^2-1/d*b/(b*(a+b))^(1/2)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(b
*(a+b))^(1/2)+a+2*b)*a)^(1/2))-1/d*b/a/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(b
*(a+b))^(1/2)+a+2*b)*a)^(1/2))-2/d/(b*(a+b))^(1/2)/a/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x
+1/2*c)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2))*b^2+1/d/a^2*b^2/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2)*arctanh(a*tan
h(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2))-1/d/a^2*b^3/(b*(a+b))^(1/2)/((2*(b*(a+b))^(1/2)-a-2*b)*a
)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2))-1/d/a^2*b^2/((2*(b*(a+b))^(1/2)+a+2
*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2))-1/d/a^2*b^3/(b*(a+b))^(1/2)/((2
*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2))-1/24/d/a/ta
nh(1/2*d*x+1/2*c)^3+3/8/d/a/tanh(1/2*d*x+1/2*c)+1/2/d/a^2/tanh(1/2*d*x+1/2*c)*b

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.38569, size = 4307, normalized size = 61.53 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(12*b*cosh(d*x + c)^4 + 48*b*cosh(d*x + c)*sinh(d*x + c)^3 + 12*b*sinh(d*x + c)^4 - 24*(a + b)*cosh(d*x +
c)^2 + 24*(3*b*cosh(d*x + c)^2 - a - b)*sinh(d*x + c)^2 + 3*((a + b)*cosh(d*x + c)^6 + 6*(a + b)*cosh(d*x + c
)*sinh(d*x + c)^5 + (a + b)*sinh(d*x + c)^6 - 3*(a + b)*cosh(d*x + c)^4 + 3*(5*(a + b)*cosh(d*x + c)^2 - a - b
)*sinh(d*x + c)^4 + 4*(5*(a + b)*cosh(d*x + c)^3 - 3*(a + b)*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(a + b)*cosh(d
*x + c)^2 + 3*(5*(a + b)*cosh(d*x + c)^4 - 6*(a + b)*cosh(d*x + c)^2 + a + b)*sinh(d*x + c)^2 + 6*((a + b)*cos
h(d*x + c)^5 - 2*(a + b)*cosh(d*x + c)^3 + (a + b)*cosh(d*x + c))*sinh(d*x + c) - a - b)*sqrt(-b/a)*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)*sin
h(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^2 + a*b)*cosh(d*x + c)^2 + 2*(a^2 + a*b)*cosh(d*x + c)*sinh(d*x + c) + (a^2 + a*b)*sinh(d*x + c)^2 +
a^2 - a*b)*sqrt(-b/a))/((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)) + 48*(b*cosh(d*x + c)^3 - (a + b)*cosh(d*x + c))*sin
h(d*x + c) + 8*a + 12*b)/(a^2*d*cosh(d*x + c)^6 + 6*a^2*d*cosh(d*x + c)*sinh(d*x + c)^5 + a^2*d*sinh(d*x + c)^
6 - 3*a^2*d*cosh(d*x + c)^4 + 3*a^2*d*cosh(d*x + c)^2 + 3*(5*a^2*d*cosh(d*x + c)^2 - a^2*d)*sinh(d*x + c)^4 +
4*(5*a^2*d*cosh(d*x + c)^3 - 3*a^2*d*cosh(d*x + c))*sinh(d*x + c)^3 - a^2*d + 3*(5*a^2*d*cosh(d*x + c)^4 - 6*a
^2*d*cosh(d*x + c)^2 + a^2*d)*sinh(d*x + c)^2 + 6*(a^2*d*cosh(d*x + c)^5 - 2*a^2*d*cosh(d*x + c)^3 + a^2*d*cos
h(d*x + c))*sinh(d*x + c)), 1/3*(6*b*cosh(d*x + c)^4 + 24*b*cosh(d*x + c)*sinh(d*x + c)^3 + 6*b*sinh(d*x + c)^
4 - 12*(a + b)*cosh(d*x + c)^2 + 12*(3*b*cosh(d*x + c)^2 - a - b)*sinh(d*x + c)^2 + 3*((a + b)*cosh(d*x + c)^6
+ 6*(a + b)*cosh(d*x + c)*sinh(d*x + c)^5 + (a + b)*sinh(d*x + c)^6 - 3*(a + b)*cosh(d*x + c)^4 + 3*(5*(a + b
)*cosh(d*x + c)^2 - a - b)*sinh(d*x + c)^4 + 4*(5*(a + b)*cosh(d*x + c)^3 - 3*(a + b)*cosh(d*x + c))*sinh(d*x
+ c)^3 + 3*(a + b)*cosh(d*x + c)^2 + 3*(5*(a + b)*cosh(d*x + c)^4 - 6*(a + b)*cosh(d*x + c)^2 + a + b)*sinh(d*
x + c)^2 + 6*((a + b)*cosh(d*x + c)^5 - 2*(a + b)*cosh(d*x + c)^3 + (a + b)*cosh(d*x + c))*sinh(d*x + c) - a -
b)*sqrt(b/a)*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(b/a)/b) + 24*(b*cosh(d*x + c)^3 - (a + b)*cosh(d*x + c))*sinh(d*x + c) + 4*a + 6*b)/(a^2*d
*cosh(d*x + c)^6 + 6*a^2*d*cosh(d*x + c)*sinh(d*x + c)^5 + a^2*d*sinh(d*x + c)^6 - 3*a^2*d*cosh(d*x + c)^4 + 3
*a^2*d*cosh(d*x + c)^2 + 3*(5*a^2*d*cosh(d*x + c)^2 - a^2*d)*sinh(d*x + c)^4 + 4*(5*a^2*d*cosh(d*x + c)^3 - 3*
a^2*d*cosh(d*x + c))*sinh(d*x + c)^3 - a^2*d + 3*(5*a^2*d*cosh(d*x + c)^4 - 6*a^2*d*cosh(d*x + c)^2 + a^2*d)*s
inh(d*x + c)^2 + 6*(a^2*d*cosh(d*x + c)^5 - 2*a^2*d*cosh(d*x + c)^3 + a^2*d*cosh(d*x + c))*sinh(d*x + c))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{csch}^{4}{\left (c + d x \right )}}{a + b \tanh ^{2}{\left (c + d x \right )}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.4188, size = 178, normalized size = 2.54 \begin{align*} \frac{\frac{3 \,{\left (a b e^{\left (2 \, c\right )} + b^{2} e^{\left (2 \, c\right )}\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 ) e^{\left (-2 \, c\right )}}{\sqrt{a b} a^{2}} + \frac{2 \,{\left (3 \, b e^{\left (4 \, d x + 4 \, c\right )} - 6 \, a e^{\left (2 \, d x + 2 \, c\right )} - 6 \, b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a + 3 \, b\right )}}{a^{2}{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}}}{3 \, d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*(a*b*e^(2*c) + b^2*e^(2*c))*arctan(1/2*(a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) + a - b)/sqrt(a*b))*e^(-2
*c)/(sqrt(a*b)*a^2) + 2*(3*b*e^(4*d*x + 4*c) - 6*a*e^(2*d*x + 2*c) - 6*b*e^(2*d*x + 2*c) + 2*a + 3*b)/(a^2*(e^
(2*d*x + 2*c) - 1)^3))/d