### 3.46 $$\int \text{csch}^5(a+b x) \text{sech}^3(a+b x) \, dx$$

Optimal. Leaf size=58 $-\frac{\tanh ^2(a+b x)}{2 b}-\frac{\coth ^4(a+b x)}{4 b}+\frac{3 \coth ^2(a+b x)}{2 b}+\frac{3 \log (\tanh (a+b x))}{b}$

[Out]

(3*Coth[a + b*x]^2)/(2*b) - Coth[a + b*x]^4/(4*b) + (3*Log[Tanh[a + b*x]])/b - Tanh[a + b*x]^2/(2*b)

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Rubi [A]  time = 0.0451177, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.176, Rules used = {2620, 266, 43} $-\frac{\tanh ^2(a+b x)}{2 b}-\frac{\coth ^4(a+b x)}{4 b}+\frac{3 \coth ^2(a+b x)}{2 b}+\frac{3 \log (\tanh (a+b x))}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Csch[a + b*x]^5*Sech[a + b*x]^3,x]

[Out]

(3*Coth[a + b*x]^2)/(2*b) - Coth[a + b*x]^4/(4*b) + (3*Log[Tanh[a + b*x]])/b - Tanh[a + b*x]^2/(2*b)

Rule 2620

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.194266, size = 56, normalized size = 0.97 $\frac{-\text{csch}^4(a+b x)+4 \text{csch}^2(a+b x)+2 \text{sech}^2(a+b x)+12 \log (\sinh (a+b x))-12 \log (\cosh (a+b x))}{4 b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Csch[a + b*x]^5*Sech[a + b*x]^3,x]

[Out]

(4*Csch[a + b*x]^2 - Csch[a + b*x]^4 - 12*Log[Cosh[a + b*x]] + 12*Log[Sinh[a + b*x]] + 2*Sech[a + b*x]^2)/(4*b
)

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Maple [A]  time = 0.022, size = 69, normalized size = 1.2 \begin{align*} -{\frac{1}{4\,b \left ( \sinh \left ( bx+a \right ) \right ) ^{4} \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}}+{\frac{3}{4\,b \left ( \sinh \left ( bx+a \right ) \right ) ^{2} \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}}+{\frac{3}{2\,b \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}}+3\,{\frac{\ln \left ( \tanh \left ( bx+a \right ) \right ) }{b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(csch(b*x+a)^5*sech(b*x+a)^3,x)

[Out]

-1/4/b/sinh(b*x+a)^4/cosh(b*x+a)^2+3/4/b/sinh(b*x+a)^2/cosh(b*x+a)^2+3/2/b/cosh(b*x+a)^2+3*ln(tanh(b*x+a))/b

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Maxima [B]  time = 1.56098, size = 242, normalized size = 4.17 \begin{align*} \frac{3 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{b} + \frac{3 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{b} - \frac{3 \, \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )}{b} - \frac{2 \,{\left (3 \, e^{\left (-2 \, b x - 2 \, a\right )} - 6 \, e^{\left (-4 \, b x - 4 \, a\right )} - 2 \, e^{\left (-6 \, b x - 6 \, a\right )} - 6 \, e^{\left (-8 \, b x - 8 \, a\right )} + 3 \, e^{\left (-10 \, b x - 10 \, a\right )}\right )}}{b{\left (2 \, e^{\left (-2 \, b x - 2 \, a\right )} + e^{\left (-4 \, b x - 4 \, a\right )} - 4 \, e^{\left (-6 \, b x - 6 \, a\right )} + e^{\left (-8 \, b x - 8 \, a\right )} + 2 \, e^{\left (-10 \, b x - 10 \, a\right )} - e^{\left (-12 \, b x - 12 \, a\right )} - 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)^5*sech(b*x+a)^3,x, algorithm="maxima")

[Out]

3*log(e^(-b*x - a) + 1)/b + 3*log(e^(-b*x - a) - 1)/b - 3*log(e^(-2*b*x - 2*a) + 1)/b - 2*(3*e^(-2*b*x - 2*a)
- 6*e^(-4*b*x - 4*a) - 2*e^(-6*b*x - 6*a) - 6*e^(-8*b*x - 8*a) + 3*e^(-10*b*x - 10*a))/(b*(2*e^(-2*b*x - 2*a)
+ e^(-4*b*x - 4*a) - 4*e^(-6*b*x - 6*a) + e^(-8*b*x - 8*a) + 2*e^(-10*b*x - 10*a) - e^(-12*b*x - 12*a) - 1))

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Fricas [B]  time = 2.58681, size = 5836, normalized size = 100.62 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)^5*sech(b*x+a)^3,x, algorithm="fricas")

[Out]

(6*cosh(b*x + a)^10 + 60*cosh(b*x + a)*sinh(b*x + a)^9 + 6*sinh(b*x + a)^10 + 6*(45*cosh(b*x + a)^2 - 2)*sinh(
b*x + a)^8 - 12*cosh(b*x + a)^8 + 48*(15*cosh(b*x + a)^3 - 2*cosh(b*x + a))*sinh(b*x + a)^7 + 4*(315*cosh(b*x
+ a)^4 - 84*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^6 - 4*cosh(b*x + a)^6 + 24*(63*cosh(b*x + a)^5 - 28*cosh(b*x +
a)^3 - cosh(b*x + a))*sinh(b*x + a)^5 + 12*(105*cosh(b*x + a)^6 - 70*cosh(b*x + a)^4 - 5*cosh(b*x + a)^2 - 1)*
sinh(b*x + a)^4 - 12*cosh(b*x + a)^4 + 16*(45*cosh(b*x + a)^7 - 42*cosh(b*x + a)^5 - 5*cosh(b*x + a)^3 - 3*cos
h(b*x + a))*sinh(b*x + a)^3 + 6*(45*cosh(b*x + a)^8 - 56*cosh(b*x + a)^6 - 10*cosh(b*x + a)^4 - 12*cosh(b*x +
a)^2 + 1)*sinh(b*x + a)^2 + 6*cosh(b*x + a)^2 - 3*(cosh(b*x + a)^12 + 12*cosh(b*x + a)*sinh(b*x + a)^11 + sinh
(b*x + a)^12 + 2*(33*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^10 - 2*cosh(b*x + a)^10 + 20*(11*cosh(b*x + a)^3 - cos
h(b*x + a))*sinh(b*x + a)^9 + (495*cosh(b*x + a)^4 - 90*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^8 - cosh(b*x + a)^8
+ 8*(99*cosh(b*x + a)^5 - 30*cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a)^7 + 4*(231*cosh(b*x + a)^6 - 105*
cosh(b*x + a)^4 - 7*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^6 + 4*cosh(b*x + a)^6 + 8*(99*cosh(b*x + a)^7 - 63*cosh
(b*x + a)^5 - 7*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^5 + (495*cosh(b*x + a)^8 - 420*cosh(b*x + a)^
6 - 70*cosh(b*x + a)^4 + 60*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^4 - cosh(b*x + a)^4 + 4*(55*cosh(b*x + a)^9 - 6
0*cosh(b*x + a)^7 - 14*cosh(b*x + a)^5 + 20*cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a)^3 + 2*(33*cosh(b*x
+ a)^10 - 45*cosh(b*x + a)^8 - 14*cosh(b*x + a)^6 + 30*cosh(b*x + a)^4 - 3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^
2 - 2*cosh(b*x + a)^2 + 4*(3*cosh(b*x + a)^11 - 5*cosh(b*x + a)^9 - 2*cosh(b*x + a)^7 + 6*cosh(b*x + a)^5 - co
sh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*log(2*cosh(b*x + a)/(cosh(b*x + a) - sinh(b*x + a))) + 3*(co
sh(b*x + a)^12 + 12*cosh(b*x + a)*sinh(b*x + a)^11 + sinh(b*x + a)^12 + 2*(33*cosh(b*x + a)^2 - 1)*sinh(b*x +
a)^10 - 2*cosh(b*x + a)^10 + 20*(11*cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a)^9 + (495*cosh(b*x + a)^4 -
90*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^8 - cosh(b*x + a)^8 + 8*(99*cosh(b*x + a)^5 - 30*cosh(b*x + a)^3 - cosh(
b*x + a))*sinh(b*x + a)^7 + 4*(231*cosh(b*x + a)^6 - 105*cosh(b*x + a)^4 - 7*cosh(b*x + a)^2 + 1)*sinh(b*x + a
)^6 + 4*cosh(b*x + a)^6 + 8*(99*cosh(b*x + a)^7 - 63*cosh(b*x + a)^5 - 7*cosh(b*x + a)^3 + 3*cosh(b*x + a))*si
nh(b*x + a)^5 + (495*cosh(b*x + a)^8 - 420*cosh(b*x + a)^6 - 70*cosh(b*x + a)^4 + 60*cosh(b*x + a)^2 - 1)*sinh
(b*x + a)^4 - cosh(b*x + a)^4 + 4*(55*cosh(b*x + a)^9 - 60*cosh(b*x + a)^7 - 14*cosh(b*x + a)^5 + 20*cosh(b*x
+ a)^3 - cosh(b*x + a))*sinh(b*x + a)^3 + 2*(33*cosh(b*x + a)^10 - 45*cosh(b*x + a)^8 - 14*cosh(b*x + a)^6 + 3
0*cosh(b*x + a)^4 - 3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 2*cosh(b*x + a)^2 + 4*(3*cosh(b*x + a)^11 - 5*cos
h(b*x + a)^9 - 2*cosh(b*x + a)^7 + 6*cosh(b*x + a)^5 - cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*log
(2*sinh(b*x + a)/(cosh(b*x + a) - sinh(b*x + a))) + 12*(5*cosh(b*x + a)^9 - 8*cosh(b*x + a)^7 - 2*cosh(b*x + a
)^5 - 4*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a))/(b*cosh(b*x + a)^12 + 12*b*cosh(b*x + a)*sinh(b*x + a)
^11 + b*sinh(b*x + a)^12 - 2*b*cosh(b*x + a)^10 + 2*(33*b*cosh(b*x + a)^2 - b)*sinh(b*x + a)^10 + 20*(11*b*cos
h(b*x + a)^3 - b*cosh(b*x + a))*sinh(b*x + a)^9 - b*cosh(b*x + a)^8 + (495*b*cosh(b*x + a)^4 - 90*b*cosh(b*x +
a)^2 - b)*sinh(b*x + a)^8 + 8*(99*b*cosh(b*x + a)^5 - 30*b*cosh(b*x + a)^3 - b*cosh(b*x + a))*sinh(b*x + a)^7
+ 4*b*cosh(b*x + a)^6 + 4*(231*b*cosh(b*x + a)^6 - 105*b*cosh(b*x + a)^4 - 7*b*cosh(b*x + a)^2 + b)*sinh(b*x
+ a)^6 + 8*(99*b*cosh(b*x + a)^7 - 63*b*cosh(b*x + a)^5 - 7*b*cosh(b*x + a)^3 + 3*b*cosh(b*x + a))*sinh(b*x +
a)^5 - b*cosh(b*x + a)^4 + (495*b*cosh(b*x + a)^8 - 420*b*cosh(b*x + a)^6 - 70*b*cosh(b*x + a)^4 + 60*b*cosh(b
*x + a)^2 - b)*sinh(b*x + a)^4 + 4*(55*b*cosh(b*x + a)^9 - 60*b*cosh(b*x + a)^7 - 14*b*cosh(b*x + a)^5 + 20*b*
cosh(b*x + a)^3 - b*cosh(b*x + a))*sinh(b*x + a)^3 - 2*b*cosh(b*x + a)^2 + 2*(33*b*cosh(b*x + a)^10 - 45*b*cos
h(b*x + a)^8 - 14*b*cosh(b*x + a)^6 + 30*b*cosh(b*x + a)^4 - 3*b*cosh(b*x + a)^2 - b)*sinh(b*x + a)^2 + 4*(3*b
*cosh(b*x + a)^11 - 5*b*cosh(b*x + a)^9 - 2*b*cosh(b*x + a)^7 + 6*b*cosh(b*x + a)^5 - b*cosh(b*x + a)^3 - b*co
sh(b*x + a))*sinh(b*x + a) + b)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{csch}^{5}{\left (a + b x \right )} \operatorname{sech}^{3}{\left (a + b x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)**5*sech(b*x+a)**3,x)

[Out]

Integral(csch(a + b*x)**5*sech(a + b*x)**3, x)

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Giac [B]  time = 1.19195, size = 240, normalized size = 4.14 \begin{align*} -\frac{3 \, \log \left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )} + 2\right )}{2 \, b} + \frac{3 \, \log \left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )} - 2\right )}{2 \, b} + \frac{3 \, e^{\left (2 \, b x + 2 \, a\right )} + 3 \, e^{\left (-2 \, b x - 2 \, a\right )} + 10}{2 \, b{\left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )} + 2\right )}} - \frac{9 \,{\left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )}\right )}^{2} - 52 \, e^{\left (2 \, b x + 2 \, a\right )} - 52 \, e^{\left (-2 \, b x - 2 \, a\right )} + 84}{4 \, b{\left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )} - 2\right )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)^5*sech(b*x+a)^3,x, algorithm="giac")

[Out]

-3/2*log(e^(2*b*x + 2*a) + e^(-2*b*x - 2*a) + 2)/b + 3/2*log(e^(2*b*x + 2*a) + e^(-2*b*x - 2*a) - 2)/b + 1/2*(
3*e^(2*b*x + 2*a) + 3*e^(-2*b*x - 2*a) + 10)/(b*(e^(2*b*x + 2*a) + e^(-2*b*x - 2*a) + 2)) - 1/4*(9*(e^(2*b*x +
2*a) + e^(-2*b*x - 2*a))^2 - 52*e^(2*b*x + 2*a) - 52*e^(-2*b*x - 2*a) + 84)/(b*(e^(2*b*x + 2*a) + e^(-2*b*x -
2*a) - 2)^2)