### 3.61 $$\int \text{csch}(c+d x) (a+b \tanh ^3(c+d x))^2 \, dx$$

Optimal. Leaf size=98 $-\frac{a^2 \tanh ^{-1}(\cosh (c+d x))}{d}+\frac{a b \tan ^{-1}(\sinh (c+d x))}{d}-\frac{a b \tanh (c+d x) \text{sech}(c+d x)}{d}-\frac{b^2 \text{sech}^5(c+d x)}{5 d}+\frac{2 b^2 \text{sech}^3(c+d x)}{3 d}-\frac{b^2 \text{sech}(c+d x)}{d}$

[Out]

(a*b*ArcTan[Sinh[c + d*x]])/d - (a^2*ArcTanh[Cosh[c + d*x]])/d - (b^2*Sech[c + d*x])/d + (2*b^2*Sech[c + d*x]^
3)/(3*d) - (b^2*Sech[c + d*x]^5)/(5*d) - (a*b*Sech[c + d*x]*Tanh[c + d*x])/d

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Rubi [A]  time = 0.136202, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.238, Rules used = {3666, 3770, 2611, 2606, 194} $-\frac{a^2 \tanh ^{-1}(\cosh (c+d x))}{d}+\frac{a b \tan ^{-1}(\sinh (c+d x))}{d}-\frac{a b \tanh (c+d x) \text{sech}(c+d x)}{d}-\frac{b^2 \text{sech}^5(c+d x)}{5 d}+\frac{2 b^2 \text{sech}^3(c+d x)}{3 d}-\frac{b^2 \text{sech}(c+d x)}{d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(a*b*ArcTan[Sinh[c + d*x]])/d - (a^2*ArcTanh[Cosh[c + d*x]])/d - (b^2*Sech[c + d*x])/d + (2*b^2*Sech[c + d*x]^
3)/(3*d) - (b^2*Sech[c + d*x]^5)/(5*d) - (a*b*Sech[c + d*x]*Tanh[c + d*x])/d

Rule 3666

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> Int[ExpandTrig[(d*sin[e + f*x])^m*(a + b*(c*tan[e + f*x])^n)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n},
x] && IGtQ[p, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 0.156975, size = 106, normalized size = 1.08 $\frac{a^2 \log \left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{2 a b \tan ^{-1}\left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )}{d}-\frac{a b \tanh (c+d x) \text{sech}(c+d x)}{d}-\frac{b^2 \text{sech}^5(c+d x)}{5 d}+\frac{2 b^2 \text{sech}^3(c+d x)}{3 d}-\frac{b^2 \text{sech}(c+d x)}{d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.072, size = 180, normalized size = 1.8 \begin{align*} -2\,{\frac{{a}^{2}{\it Artanh} \left ({{\rm e}^{dx+c}} \right ) }{d}}-2\,{\frac{ab\sinh \left ( dx+c \right ) }{d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}+{\frac{ab{\rm sech} \left (dx+c\right )\tanh \left ( dx+c \right ) }{d}}+2\,{\frac{ab\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}}-{\frac{{b}^{2} \left ( \sinh \left ( dx+c \right ) \right ) ^{4}}{d \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}-{\frac{4\,{b}^{2} \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{5\,d \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}+{\frac{8\,{b}^{2} \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{15\,d \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}+{\frac{8\,{b}^{2} \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{15\,d\cosh \left ( dx+c \right ) }}-{\frac{8\,{b}^{2}\cosh \left ( dx+c \right ) }{15\,d}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d*a^2*arctanh(exp(d*x+c))-2/d*a*b*sinh(d*x+c)/cosh(d*x+c)^2+1/d*a*b*sech(d*x+c)*tanh(d*x+c)+2/d*a*b*arctan(
exp(d*x+c))-1/d*b^2*sinh(d*x+c)^4/cosh(d*x+c)^5-4/5/d*b^2*sinh(d*x+c)^2/cosh(d*x+c)^5+8/15/d*b^2*sinh(d*x+c)^2
/cosh(d*x+c)^3+8/15/d*b^2*sinh(d*x+c)^2/cosh(d*x+c)-8/15*b^2*cosh(d*x+c)/d

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Maxima [B]  time = 1.53543, size = 603, normalized size = 6.15 \begin{align*} -2 \, a b{\left (\frac{\arctan \left (e^{\left (-d x - c\right )}\right )}{d} + \frac{e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d{\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} - \frac{2}{15} \, b^{2}{\left (\frac{15 \, e^{\left (-d x - c\right )}}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac{20 \, e^{\left (-3 \, d x - 3 \, c\right )}}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac{58 \, e^{\left (-5 \, d x - 5 \, c\right )}}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac{20 \, e^{\left (-7 \, d x - 7 \, c\right )}}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac{15 \, e^{\left (-9 \, d x - 9 \, c\right )}}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + \frac{a^{2} \log \left (\tanh \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}{d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*b*(arctan(e^(-d*x - c))/d + (e^(-d*x - c) - e^(-3*d*x - 3*c))/(d*(2*e^(-2*d*x - 2*c) + e^(-4*d*x - 4*c) +
1))) - 2/15*b^2*(15*e^(-d*x - c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8
*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 20*e^(-3*d*x - 3*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10
*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 58*e^(-5*d*x - 5*c)/(d*(5*e^(-2*d*x - 2*c)
+ 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 20*e^(-7*d*x -
7*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10
*c) + 1)) + 15*e^(-9*d*x - 9*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d
*x - 8*c) + e^(-10*d*x - 10*c) + 1))) + a^2*log(tanh(1/2*d*x + 1/2*c))/d

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Fricas [B]  time = 2.67318, size = 6643, normalized size = 67.79 \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)*(a+b*tanh(d*x+c)^3)^2,x, algorithm="fricas")

[Out]

-1/15*(30*(a*b + b^2)*cosh(d*x + c)^9 + 270*(a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^8 + 30*(a*b + b^2)*sinh(d*
x + c)^9 + 20*(3*a*b + 2*b^2)*cosh(d*x + c)^7 + 20*(54*(a*b + b^2)*cosh(d*x + c)^2 + 3*a*b + 2*b^2)*sinh(d*x +
c)^7 + 116*b^2*cosh(d*x + c)^5 + 140*(18*(a*b + b^2)*cosh(d*x + c)^3 + (3*a*b + 2*b^2)*cosh(d*x + c))*sinh(d*
x + c)^6 + 4*(945*(a*b + b^2)*cosh(d*x + c)^4 + 105*(3*a*b + 2*b^2)*cosh(d*x + c)^2 + 29*b^2)*sinh(d*x + c)^5
+ 20*(189*(a*b + b^2)*cosh(d*x + c)^5 + 35*(3*a*b + 2*b^2)*cosh(d*x + c)^3 + 29*b^2*cosh(d*x + c))*sinh(d*x +
c)^4 - 20*(3*a*b - 2*b^2)*cosh(d*x + c)^3 + 20*(126*(a*b + b^2)*cosh(d*x + c)^6 + 35*(3*a*b + 2*b^2)*cosh(d*x
+ c)^4 + 58*b^2*cosh(d*x + c)^2 - 3*a*b + 2*b^2)*sinh(d*x + c)^3 + 20*(54*(a*b + b^2)*cosh(d*x + c)^7 + 21*(3*
a*b + 2*b^2)*cosh(d*x + c)^5 + 58*b^2*cosh(d*x + c)^3 - 3*(3*a*b - 2*b^2)*cosh(d*x + c))*sinh(d*x + c)^2 - 30*
(a*b*cosh(d*x + c)^10 + 10*a*b*cosh(d*x + c)*sinh(d*x + c)^9 + a*b*sinh(d*x + c)^10 + 5*a*b*cosh(d*x + c)^8 +
5*(9*a*b*cosh(d*x + c)^2 + a*b)*sinh(d*x + c)^8 + 10*a*b*cosh(d*x + c)^6 + 40*(3*a*b*cosh(d*x + c)^3 + a*b*cos
h(d*x + c))*sinh(d*x + c)^7 + 10*(21*a*b*cosh(d*x + c)^4 + 14*a*b*cosh(d*x + c)^2 + a*b)*sinh(d*x + c)^6 + 10*
a*b*cosh(d*x + c)^4 + 4*(63*a*b*cosh(d*x + c)^5 + 70*a*b*cosh(d*x + c)^3 + 15*a*b*cosh(d*x + c))*sinh(d*x + c)
^5 + 10*(21*a*b*cosh(d*x + c)^6 + 35*a*b*cosh(d*x + c)^4 + 15*a*b*cosh(d*x + c)^2 + a*b)*sinh(d*x + c)^4 + 5*a
*b*cosh(d*x + c)^2 + 40*(3*a*b*cosh(d*x + c)^7 + 7*a*b*cosh(d*x + c)^5 + 5*a*b*cosh(d*x + c)^3 + a*b*cosh(d*x
+ c))*sinh(d*x + c)^3 + 5*(9*a*b*cosh(d*x + c)^8 + 28*a*b*cosh(d*x + c)^6 + 30*a*b*cosh(d*x + c)^4 + 12*a*b*co
sh(d*x + c)^2 + a*b)*sinh(d*x + c)^2 + a*b + 10*(a*b*cosh(d*x + c)^9 + 4*a*b*cosh(d*x + c)^7 + 6*a*b*cosh(d*x
+ c)^5 + 4*a*b*cosh(d*x + c)^3 + a*b*cosh(d*x + c))*sinh(d*x + c))*arctan(cosh(d*x + c) + sinh(d*x + c)) - 30*
(a*b - b^2)*cosh(d*x + c) + 15*(a^2*cosh(d*x + c)^10 + 10*a^2*cosh(d*x + c)*sinh(d*x + c)^9 + a^2*sinh(d*x + c
)^10 + 5*a^2*cosh(d*x + c)^8 + 5*(9*a^2*cosh(d*x + c)^2 + a^2)*sinh(d*x + c)^8 + 10*a^2*cosh(d*x + c)^6 + 40*(
3*a^2*cosh(d*x + c)^3 + a^2*cosh(d*x + c))*sinh(d*x + c)^7 + 10*(21*a^2*cosh(d*x + c)^4 + 14*a^2*cosh(d*x + c)
^2 + a^2)*sinh(d*x + c)^6 + 10*a^2*cosh(d*x + c)^4 + 4*(63*a^2*cosh(d*x + c)^5 + 70*a^2*cosh(d*x + c)^3 + 15*a
^2*cosh(d*x + c))*sinh(d*x + c)^5 + 10*(21*a^2*cosh(d*x + c)^6 + 35*a^2*cosh(d*x + c)^4 + 15*a^2*cosh(d*x + c)
^2 + a^2)*sinh(d*x + c)^4 + 5*a^2*cosh(d*x + c)^2 + 40*(3*a^2*cosh(d*x + c)^7 + 7*a^2*cosh(d*x + c)^5 + 5*a^2*
cosh(d*x + c)^3 + a^2*cosh(d*x + c))*sinh(d*x + c)^3 + 5*(9*a^2*cosh(d*x + c)^8 + 28*a^2*cosh(d*x + c)^6 + 30*
a^2*cosh(d*x + c)^4 + 12*a^2*cosh(d*x + c)^2 + a^2)*sinh(d*x + c)^2 + a^2 + 10*(a^2*cosh(d*x + c)^9 + 4*a^2*co
sh(d*x + c)^7 + 6*a^2*cosh(d*x + c)^5 + 4*a^2*cosh(d*x + c)^3 + a^2*cosh(d*x + c))*sinh(d*x + c))*log(cosh(d*x
+ c) + sinh(d*x + c) + 1) - 15*(a^2*cosh(d*x + c)^10 + 10*a^2*cosh(d*x + c)*sinh(d*x + c)^9 + a^2*sinh(d*x +
c)^10 + 5*a^2*cosh(d*x + c)^8 + 5*(9*a^2*cosh(d*x + c)^2 + a^2)*sinh(d*x + c)^8 + 10*a^2*cosh(d*x + c)^6 + 40*
(3*a^2*cosh(d*x + c)^3 + a^2*cosh(d*x + c))*sinh(d*x + c)^7 + 10*(21*a^2*cosh(d*x + c)^4 + 14*a^2*cosh(d*x + c
)^2 + a^2)*sinh(d*x + c)^6 + 10*a^2*cosh(d*x + c)^4 + 4*(63*a^2*cosh(d*x + c)^5 + 70*a^2*cosh(d*x + c)^3 + 15*
a^2*cosh(d*x + c))*sinh(d*x + c)^5 + 10*(21*a^2*cosh(d*x + c)^6 + 35*a^2*cosh(d*x + c)^4 + 15*a^2*cosh(d*x + c
)^2 + a^2)*sinh(d*x + c)^4 + 5*a^2*cosh(d*x + c)^2 + 40*(3*a^2*cosh(d*x + c)^7 + 7*a^2*cosh(d*x + c)^5 + 5*a^2
*cosh(d*x + c)^3 + a^2*cosh(d*x + c))*sinh(d*x + c)^3 + 5*(9*a^2*cosh(d*x + c)^8 + 28*a^2*cosh(d*x + c)^6 + 30
*a^2*cosh(d*x + c)^4 + 12*a^2*cosh(d*x + c)^2 + a^2)*sinh(d*x + c)^2 + a^2 + 10*(a^2*cosh(d*x + c)^9 + 4*a^2*c
osh(d*x + c)^7 + 6*a^2*cosh(d*x + c)^5 + 4*a^2*cosh(d*x + c)^3 + a^2*cosh(d*x + c))*sinh(d*x + c))*log(cosh(d*
x + c) + sinh(d*x + c) - 1) + 10*(27*(a*b + b^2)*cosh(d*x + c)^8 + 14*(3*a*b + 2*b^2)*cosh(d*x + c)^6 + 58*b^2
*cosh(d*x + c)^4 - 6*(3*a*b - 2*b^2)*cosh(d*x + c)^2 - 3*a*b + 3*b^2)*sinh(d*x + c))/(d*cosh(d*x + c)^10 + 10*
d*cosh(d*x + c)*sinh(d*x + c)^9 + d*sinh(d*x + c)^10 + 5*d*cosh(d*x + c)^8 + 5*(9*d*cosh(d*x + c)^2 + d)*sinh(
d*x + c)^8 + 40*(3*d*cosh(d*x + c)^3 + d*cosh(d*x + c))*sinh(d*x + c)^7 + 10*d*cosh(d*x + c)^6 + 10*(21*d*cosh
(d*x + c)^4 + 14*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^6 + 4*(63*d*cosh(d*x + c)^5 + 70*d*cosh(d*x + c)^3 + 15*
d*cosh(d*x + c))*sinh(d*x + c)^5 + 10*d*cosh(d*x + c)^4 + 10*(21*d*cosh(d*x + c)^6 + 35*d*cosh(d*x + c)^4 + 15
*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^4 + 40*(3*d*cosh(d*x + c)^7 + 7*d*cosh(d*x + c)^5 + 5*d*cosh(d*x + c)^3
+ d*cosh(d*x + c))*sinh(d*x + c)^3 + 5*d*cosh(d*x + c)^2 + 5*(9*d*cosh(d*x + c)^8 + 28*d*cosh(d*x + c)^6 + 30*
d*cosh(d*x + c)^4 + 12*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^2 + 10*(d*cosh(d*x + c)^9 + 4*d*cosh(d*x + c)^7 +
6*d*cosh(d*x + c)^5 + 4*d*cosh(d*x + c)^3 + d*cosh(d*x + c))*sinh(d*x + c) + d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tanh ^{3}{\left (c + d x \right )}\right )^{2} \operatorname{csch}{\left (c + d x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*tanh(c + d*x)**3)**2*csch(c + d*x), x)

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Giac [A]  time = 1.51089, size = 240, normalized size = 2.45 \begin{align*} \frac{30 \, a b \arctan \left (e^{\left (d x + c\right )}\right ) - 15 \, a^{2} \log \left (e^{\left (d x + c\right )} + 1\right ) + 15 \, a^{2} \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right ) - \frac{2 \,{\left (15 \, a b e^{\left (9 \, d x + 9 \, c\right )} + 15 \, b^{2} e^{\left (9 \, d x + 9 \, c\right )} + 30 \, a b e^{\left (7 \, d x + 7 \, c\right )} + 20 \, b^{2} e^{\left (7 \, d x + 7 \, c\right )} + 58 \, b^{2} e^{\left (5 \, d x + 5 \, c\right )} - 30 \, a b e^{\left (3 \, d x + 3 \, c\right )} + 20 \, b^{2} e^{\left (3 \, d x + 3 \, c\right )} - 15 \, a b e^{\left (d x + c\right )} + 15 \, b^{2} e^{\left (d x + c\right )}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{15 \, d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(30*a*b*arctan(e^(d*x + c)) - 15*a^2*log(e^(d*x + c) + 1) + 15*a^2*log(abs(e^(d*x + c) - 1)) - 2*(15*a*b*
e^(9*d*x + 9*c) + 15*b^2*e^(9*d*x + 9*c) + 30*a*b*e^(7*d*x + 7*c) + 20*b^2*e^(7*d*x + 7*c) + 58*b^2*e^(5*d*x +
5*c) - 30*a*b*e^(3*d*x + 3*c) + 20*b^2*e^(3*d*x + 3*c) - 15*a*b*e^(d*x + c) + 15*b^2*e^(d*x + c))/(e^(2*d*x +
2*c) + 1)^5)/d