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

Optimal. Leaf size=84 \[ -\frac{b \left (3 a^2+3 a b+b^2\right ) \text{sech}(c+d x)}{d}-\frac{a^3 \tanh ^{-1}(\cosh (c+d x))}{d}+\frac{b^2 (3 a+2 b) \text{sech}^3(c+d x)}{3 d}-\frac{b^3 \text{sech}^5(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.0839635, antiderivative size = 84, 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 = {3664, 390, 207} \[ -\frac{b \left (3 a^2+3 a b+b^2\right ) \text{sech}(c+d x)}{d}-\frac{a^3 \tanh ^{-1}(\cosh (c+d x))}{d}+\frac{b^2 (3 a+2 b) \text{sech}^3(c+d x)}{3 d}-\frac{b^3 \text{sech}^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3664

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a - b + b*ff^2*x^2)^p)/x^(
m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 207

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

Rubi steps

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

Mathematica [A]  time = 0.302332, size = 79, normalized size = 0.94 \[ \frac{-15 b \left (3 a^2+3 a b+b^2\right ) \text{sech}(c+d x)+15 a^3 \log \left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )+5 b^2 (3 a+2 b) \text{sech}^3(c+d x)-3 b^3 \text{sech}^5(c+d x)}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*a^3*Log[Tanh[(c + d*x)/2]] - 15*b*(3*a^2 + 3*a*b + b^2)*Sech[c + d*x] + 5*b^2*(3*a + 2*b)*Sech[c + d*x]^3
- 3*b^3*Sech[c + d*x]^5)/(15*d)

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Maple [B]  time = 0.052, size = 186, normalized size = 2.2 \begin{align*}{\frac{1}{d} \left ( -2\,{a}^{3}{\it Artanh} \left ({{\rm e}^{dx+c}} \right ) +3\,{a}^{2}b \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{\cosh \left ( dx+c \right ) }}-\cosh \left ( dx+c \right ) \right ) +3\,a{b}^{2} \left ( -1/3\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{ \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}+2/3\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{\cosh \left ( dx+c \right ) }}-2/3\,\cosh \left ( dx+c \right ) \right ) +{b}^{3} \left ( -{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{4}}{ \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}-{\frac{4\, \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{5\, \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}+{\frac{8\, \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{15\, \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}+{\frac{8\, \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{15\,\cosh \left ( dx+c \right ) }}-{\frac{8\,\cosh \left ( dx+c \right ) }{15}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.05016, size = 756, normalized size = 9. \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)^2)^3,x, algorithm="maxima")

[Out]

-2/15*b^3*(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))) - 2*a*b^2*(3*e^(-d*x - c)/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6*d
*x - 6*c) + 1)) + 2*e^(-3*d*x - 3*c)/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) + 1)) + 3*
e^(-5*d*x - 5*c)/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) + 1))) + a^3*log(tanh(1/2*d*x
+ 1/2*c))/d - 6*a^2*b/(d*(e^(d*x + c) + e^(-d*x - c)))

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Fricas [B]  time = 2.25076, size = 5895, normalized size = 70.18 \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)^2)^3,x, algorithm="fricas")

[Out]

-1/15*(30*(3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^9 + 270*(3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)*sinh(d*x + c
)^8 + 30*(3*a^2*b + 3*a*b^2 + b^3)*sinh(d*x + c)^9 + 40*(9*a^2*b + 6*a*b^2 + b^3)*cosh(d*x + c)^7 + 40*(9*a^2*
b + 6*a*b^2 + b^3 + 27*(3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^7 + 280*(9*(3*a^2*b + 3*a*b^2
+ b^3)*cosh(d*x + c)^3 + (9*a^2*b + 6*a*b^2 + b^3)*cosh(d*x + c))*sinh(d*x + c)^6 + 4*(135*a^2*b + 75*a*b^2 +
29*b^3)*cosh(d*x + c)^5 + 4*(945*(3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^4 + 135*a^2*b + 75*a*b^2 + 29*b^3 + 2
10*(9*a^2*b + 6*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^5 + 20*(189*(3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c
)^5 + 70*(9*a^2*b + 6*a*b^2 + b^3)*cosh(d*x + c)^3 + (135*a^2*b + 75*a*b^2 + 29*b^3)*cosh(d*x + c))*sinh(d*x +
c)^4 + 40*(9*a^2*b + 6*a*b^2 + b^3)*cosh(d*x + c)^3 + 40*(63*(3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^6 + 35*(
9*a^2*b + 6*a*b^2 + b^3)*cosh(d*x + c)^4 + 9*a^2*b + 6*a*b^2 + b^3 + (135*a^2*b + 75*a*b^2 + 29*b^3)*cosh(d*x
+ c)^2)*sinh(d*x + c)^3 + 40*(27*(3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^7 + 21*(9*a^2*b + 6*a*b^2 + b^3)*cosh
(d*x + c)^5 + (135*a^2*b + 75*a*b^2 + 29*b^3)*cosh(d*x + c)^3 + 3*(9*a^2*b + 6*a*b^2 + b^3)*cosh(d*x + c))*sin
h(d*x + c)^2 + 30*(3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c) + 15*(a^3*cosh(d*x + c)^10 + 10*a^3*cosh(d*x + c)*si
nh(d*x + c)^9 + a^3*sinh(d*x + c)^10 + 5*a^3*cosh(d*x + c)^8 + 10*a^3*cosh(d*x + c)^6 + 5*(9*a^3*cosh(d*x + c)
^2 + a^3)*sinh(d*x + c)^8 + 40*(3*a^3*cosh(d*x + c)^3 + a^3*cosh(d*x + c))*sinh(d*x + c)^7 + 10*a^3*cosh(d*x +
c)^4 + 10*(21*a^3*cosh(d*x + c)^4 + 14*a^3*cosh(d*x + c)^2 + a^3)*sinh(d*x + c)^6 + 4*(63*a^3*cosh(d*x + c)^5
+ 70*a^3*cosh(d*x + c)^3 + 15*a^3*cosh(d*x + c))*sinh(d*x + c)^5 + 5*a^3*cosh(d*x + c)^2 + 10*(21*a^3*cosh(d*
x + c)^6 + 35*a^3*cosh(d*x + c)^4 + 15*a^3*cosh(d*x + c)^2 + a^3)*sinh(d*x + c)^4 + 40*(3*a^3*cosh(d*x + c)^7
+ 7*a^3*cosh(d*x + c)^5 + 5*a^3*cosh(d*x + c)^3 + a^3*cosh(d*x + c))*sinh(d*x + c)^3 + a^3 + 5*(9*a^3*cosh(d*x
+ c)^8 + 28*a^3*cosh(d*x + c)^6 + 30*a^3*cosh(d*x + c)^4 + 12*a^3*cosh(d*x + c)^2 + a^3)*sinh(d*x + c)^2 + 10
*(a^3*cosh(d*x + c)^9 + 4*a^3*cosh(d*x + c)^7 + 6*a^3*cosh(d*x + c)^5 + 4*a^3*cosh(d*x + c)^3 + a^3*cosh(d*x +
c))*sinh(d*x + c))*log(cosh(d*x + c) + sinh(d*x + c) + 1) - 15*(a^3*cosh(d*x + c)^10 + 10*a^3*cosh(d*x + c)*s
inh(d*x + c)^9 + a^3*sinh(d*x + c)^10 + 5*a^3*cosh(d*x + c)^8 + 10*a^3*cosh(d*x + c)^6 + 5*(9*a^3*cosh(d*x + c
)^2 + a^3)*sinh(d*x + c)^8 + 40*(3*a^3*cosh(d*x + c)^3 + a^3*cosh(d*x + c))*sinh(d*x + c)^7 + 10*a^3*cosh(d*x
+ c)^4 + 10*(21*a^3*cosh(d*x + c)^4 + 14*a^3*cosh(d*x + c)^2 + a^3)*sinh(d*x + c)^6 + 4*(63*a^3*cosh(d*x + c)^
5 + 70*a^3*cosh(d*x + c)^3 + 15*a^3*cosh(d*x + c))*sinh(d*x + c)^5 + 5*a^3*cosh(d*x + c)^2 + 10*(21*a^3*cosh(d
*x + c)^6 + 35*a^3*cosh(d*x + c)^4 + 15*a^3*cosh(d*x + c)^2 + a^3)*sinh(d*x + c)^4 + 40*(3*a^3*cosh(d*x + c)^7
+ 7*a^3*cosh(d*x + c)^5 + 5*a^3*cosh(d*x + c)^3 + a^3*cosh(d*x + c))*sinh(d*x + c)^3 + a^3 + 5*(9*a^3*cosh(d*
x + c)^8 + 28*a^3*cosh(d*x + c)^6 + 30*a^3*cosh(d*x + c)^4 + 12*a^3*cosh(d*x + c)^2 + a^3)*sinh(d*x + c)^2 + 1
0*(a^3*cosh(d*x + c)^9 + 4*a^3*cosh(d*x + c)^7 + 6*a^3*cosh(d*x + c)^5 + 4*a^3*cosh(d*x + c)^3 + a^3*cosh(d*x
+ c))*sinh(d*x + c))*log(cosh(d*x + c) + sinh(d*x + c) - 1) + 10*(27*(3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^8
+ 28*(9*a^2*b + 6*a*b^2 + b^3)*cosh(d*x + c)^6 + 2*(135*a^2*b + 75*a*b^2 + 29*b^3)*cosh(d*x + c)^4 + 9*a^2*b
+ 9*a*b^2 + 3*b^3 + 12*(9*a^2*b + 6*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c))/(d*cosh(d*x + c)^10 + 10*d*co
sh(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*co
sh(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*c
osh(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*co
sh(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 ^{2}{\left (c + d x \right )}\right )^{3} \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)**2)**3,x)

[Out]

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

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Giac [B]  time = 1.64338, size = 354, normalized size = 4.21 \begin{align*} -\frac{15 \, a^{3} \log \left (e^{\left (d x + c\right )} + 1\right ) - 15 \, a^{3} \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right ) + \frac{2 \,{\left (45 \, a^{2} b e^{\left (9 \, d x + 9 \, c\right )} + 45 \, a b^{2} e^{\left (9 \, d x + 9 \, c\right )} + 15 \, b^{3} e^{\left (9 \, d x + 9 \, c\right )} + 180 \, a^{2} b e^{\left (7 \, d x + 7 \, c\right )} + 120 \, a b^{2} e^{\left (7 \, d x + 7 \, c\right )} + 20 \, b^{3} e^{\left (7 \, d x + 7 \, c\right )} + 270 \, a^{2} b e^{\left (5 \, d x + 5 \, c\right )} + 150 \, a b^{2} e^{\left (5 \, d x + 5 \, c\right )} + 58 \, b^{3} e^{\left (5 \, d x + 5 \, c\right )} + 180 \, a^{2} b e^{\left (3 \, d x + 3 \, c\right )} + 120 \, a b^{2} e^{\left (3 \, d x + 3 \, c\right )} + 20 \, b^{3} e^{\left (3 \, d x + 3 \, c\right )} + 45 \, a^{2} b e^{\left (d x + c\right )} + 45 \, a b^{2} e^{\left (d x + c\right )} + 15 \, b^{3} 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)^2)^3,x, algorithm="giac")

[Out]

-1/15*(15*a^3*log(e^(d*x + c) + 1) - 15*a^3*log(abs(e^(d*x + c) - 1)) + 2*(45*a^2*b*e^(9*d*x + 9*c) + 45*a*b^2
*e^(9*d*x + 9*c) + 15*b^3*e^(9*d*x + 9*c) + 180*a^2*b*e^(7*d*x + 7*c) + 120*a*b^2*e^(7*d*x + 7*c) + 20*b^3*e^(
7*d*x + 7*c) + 270*a^2*b*e^(5*d*x + 5*c) + 150*a*b^2*e^(5*d*x + 5*c) + 58*b^3*e^(5*d*x + 5*c) + 180*a^2*b*e^(3
*d*x + 3*c) + 120*a*b^2*e^(3*d*x + 3*c) + 20*b^3*e^(3*d*x + 3*c) + 45*a^2*b*e^(d*x + c) + 45*a*b^2*e^(d*x + c)
+ 15*b^3*e^(d*x + c))/(e^(2*d*x + 2*c) + 1)^5)/d