### 3.48 $$\int \text{csch}^5(a+b x) \text{sech}^5(a+b x) \, dx$$

Optimal. Leaf size=69 $\frac{\tanh ^4(a+b x)}{4 b}-\frac{2 \tanh ^2(a+b x)}{b}-\frac{\coth ^4(a+b x)}{4 b}+\frac{2 \coth ^2(a+b x)}{b}+\frac{6 \log (\tanh (a+b x))}{b}$

[Out]

(2*Coth[a + b*x]^2)/b - Coth[a + b*x]^4/(4*b) + (6*Log[Tanh[a + b*x]])/b - (2*Tanh[a + b*x]^2)/b + Tanh[a + b*
x]^4/(4*b)

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Rubi [A]  time = 0.0539318, antiderivative size = 69, 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} $\frac{\tanh ^4(a+b x)}{4 b}-\frac{2 \tanh ^2(a+b x)}{b}-\frac{\coth ^4(a+b x)}{4 b}+\frac{2 \coth ^2(a+b x)}{b}+\frac{6 \log (\tanh (a+b x))}{b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*Coth[a + b*x]^2)/b - Coth[a + b*x]^4/(4*b) + (6*Log[Tanh[a + b*x]])/b - (2*Tanh[a + b*x]^2)/b + Tanh[a + b*
x]^4/(4*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}^5(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^4}{x^5} \, dx,x,i \tanh (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(1+x)^4}{x^3} \, dx,x,-\tanh ^2(a+b x)\right )}{2 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (4+\frac{1}{x^3}+\frac{4}{x^2}+\frac{6}{x}+x\right ) \, dx,x,-\tanh ^2(a+b x)\right )}{2 b}\\ &=\frac{2 \coth ^2(a+b x)}{b}-\frac{\coth ^4(a+b x)}{4 b}+\frac{6 \log (\tanh (a+b x))}{b}-\frac{2 \tanh ^2(a+b x)}{b}+\frac{\tanh ^4(a+b x)}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.0248018, size = 77, normalized size = 1.12 $32 \left (-\frac{\text{csch}^4(a+b x)}{128 b}+\frac{3 \text{csch}^2(a+b x)}{64 b}+\frac{\text{sech}^4(a+b x)}{128 b}+\frac{3 \text{sech}^2(a+b x)}{64 b}+\frac{3 \log (\tanh (a+b x))}{16 b}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

32*((3*Csch[a + b*x]^2)/(64*b) - Csch[a + b*x]^4/(128*b) + (3*Log[Tanh[a + b*x]])/(16*b) + (3*Sech[a + b*x]^2)
/(64*b) + Sech[a + b*x]^4/(128*b))

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Maple [A]  time = 0.024, size = 81, 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 ) ^{4}}}+{\frac{1}{b \left ( \sinh \left ( bx+a \right ) \right ) ^{2} \left ( \cosh \left ( bx+a \right ) \right ) ^{4}}}+{\frac{3}{2\,b \left ( \cosh \left ( bx+a \right ) \right ) ^{4}}}+3\,{\frac{1}{b \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}}+6\,{\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)^5,x)

[Out]

-1/4/b/sinh(b*x+a)^4/cosh(b*x+a)^4+1/b/sinh(b*x+a)^2/cosh(b*x+a)^4+3/2/b/cosh(b*x+a)^4+3/b/cosh(b*x+a)^2+6*ln(
tanh(b*x+a))/b

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Maxima [B]  time = 1.65642, size = 203, normalized size = 2.94 \begin{align*} \frac{6 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{b} + \frac{6 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{b} - \frac{6 \, \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )}{b} - \frac{4 \,{\left (3 \, e^{\left (-2 \, b x - 2 \, a\right )} - 11 \, e^{\left (-6 \, b x - 6 \, a\right )} - 11 \, e^{\left (-10 \, b x - 10 \, a\right )} + 3 \, e^{\left (-14 \, b x - 14 \, a\right )}\right )}}{b{\left (4 \, e^{\left (-4 \, b x - 4 \, a\right )} - 6 \, e^{\left (-8 \, b x - 8 \, a\right )} + 4 \, e^{\left (-12 \, b x - 12 \, a\right )} - e^{\left (-16 \, b x - 16 \, 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)^5,x, algorithm="maxima")

[Out]

6*log(e^(-b*x - a) + 1)/b + 6*log(e^(-b*x - a) - 1)/b - 6*log(e^(-2*b*x - 2*a) + 1)/b - 4*(3*e^(-2*b*x - 2*a)
- 11*e^(-6*b*x - 6*a) - 11*e^(-10*b*x - 10*a) + 3*e^(-14*b*x - 14*a))/(b*(4*e^(-4*b*x - 4*a) - 6*e^(-8*b*x - 8
*a) + 4*e^(-12*b*x - 12*a) - e^(-16*b*x - 16*a) - 1))

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Fricas [B]  time = 2.74439, size = 6263, normalized size = 90.77 \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)^5,x, algorithm="fricas")

[Out]

2*(6*cosh(b*x + a)^14 + 2184*cosh(b*x + a)^3*sinh(b*x + a)^11 + 546*cosh(b*x + a)^2*sinh(b*x + a)^12 + 84*cosh
(b*x + a)*sinh(b*x + a)^13 + 6*sinh(b*x + a)^14 + 22*(273*cosh(b*x + a)^4 - 1)*sinh(b*x + a)^10 - 22*cosh(b*x
+ a)^10 + 44*(273*cosh(b*x + a)^5 - 5*cosh(b*x + a))*sinh(b*x + a)^9 + 198*(91*cosh(b*x + a)^6 - 5*cosh(b*x +
a)^2)*sinh(b*x + a)^8 + 528*(39*cosh(b*x + a)^7 - 5*cosh(b*x + a)^3)*sinh(b*x + a)^7 + 22*(819*cosh(b*x + a)^8
- 210*cosh(b*x + a)^4 - 1)*sinh(b*x + a)^6 - 22*cosh(b*x + a)^6 + 132*(91*cosh(b*x + a)^9 - 42*cosh(b*x + a)^
5 - cosh(b*x + a))*sinh(b*x + a)^5 + 66*(91*cosh(b*x + a)^10 - 70*cosh(b*x + a)^6 - 5*cosh(b*x + a)^2)*sinh(b*
x + a)^4 + 8*(273*cosh(b*x + a)^11 - 330*cosh(b*x + a)^7 - 55*cosh(b*x + a)^3)*sinh(b*x + a)^3 + 6*(91*cosh(b*
x + a)^12 - 165*cosh(b*x + a)^8 - 55*cosh(b*x + a)^4 + 1)*sinh(b*x + a)^2 + 6*cosh(b*x + a)^2 - 3*(cosh(b*x +
a)^16 + 560*cosh(b*x + a)^3*sinh(b*x + a)^13 + 120*cosh(b*x + a)^2*sinh(b*x + a)^14 + 16*cosh(b*x + a)*sinh(b*
x + a)^15 + sinh(b*x + a)^16 + 4*(455*cosh(b*x + a)^4 - 1)*sinh(b*x + a)^12 - 4*cosh(b*x + a)^12 + 48*(91*cosh
(b*x + a)^5 - cosh(b*x + a))*sinh(b*x + a)^11 + 88*(91*cosh(b*x + a)^6 - 3*cosh(b*x + a)^2)*sinh(b*x + a)^10 +
880*(13*cosh(b*x + a)^7 - cosh(b*x + a)^3)*sinh(b*x + a)^9 + 6*(2145*cosh(b*x + a)^8 - 330*cosh(b*x + a)^4 +
1)*sinh(b*x + a)^8 + 6*cosh(b*x + a)^8 + 16*(715*cosh(b*x + a)^9 - 198*cosh(b*x + a)^5 + 3*cosh(b*x + a))*sinh
(b*x + a)^7 + 56*(143*cosh(b*x + a)^10 - 66*cosh(b*x + a)^6 + 3*cosh(b*x + a)^2)*sinh(b*x + a)^6 + 48*(91*cosh
(b*x + a)^11 - 66*cosh(b*x + a)^7 + 7*cosh(b*x + a)^3)*sinh(b*x + a)^5 + 4*(455*cosh(b*x + a)^12 - 495*cosh(b*
x + a)^8 + 105*cosh(b*x + a)^4 - 1)*sinh(b*x + a)^4 - 4*cosh(b*x + a)^4 + 16*(35*cosh(b*x + a)^13 - 55*cosh(b*
x + a)^9 + 21*cosh(b*x + a)^5 - cosh(b*x + a))*sinh(b*x + a)^3 + 24*(5*cosh(b*x + a)^14 - 11*cosh(b*x + a)^10
+ 7*cosh(b*x + a)^6 - cosh(b*x + a)^2)*sinh(b*x + a)^2 + 16*(cosh(b*x + a)^15 - 3*cosh(b*x + a)^11 + 3*cosh(b*
x + a)^7 - cosh(b*x + a)^3)*sinh(b*x + a) + 1)*log(2*cosh(b*x + a)/(cosh(b*x + a) - sinh(b*x + a))) + 3*(cosh(
b*x + a)^16 + 560*cosh(b*x + a)^3*sinh(b*x + a)^13 + 120*cosh(b*x + a)^2*sinh(b*x + a)^14 + 16*cosh(b*x + a)*s
inh(b*x + a)^15 + sinh(b*x + a)^16 + 4*(455*cosh(b*x + a)^4 - 1)*sinh(b*x + a)^12 - 4*cosh(b*x + a)^12 + 48*(9
1*cosh(b*x + a)^5 - cosh(b*x + a))*sinh(b*x + a)^11 + 88*(91*cosh(b*x + a)^6 - 3*cosh(b*x + a)^2)*sinh(b*x + a
)^10 + 880*(13*cosh(b*x + a)^7 - cosh(b*x + a)^3)*sinh(b*x + a)^9 + 6*(2145*cosh(b*x + a)^8 - 330*cosh(b*x + a
)^4 + 1)*sinh(b*x + a)^8 + 6*cosh(b*x + a)^8 + 16*(715*cosh(b*x + a)^9 - 198*cosh(b*x + a)^5 + 3*cosh(b*x + a)
)*sinh(b*x + a)^7 + 56*(143*cosh(b*x + a)^10 - 66*cosh(b*x + a)^6 + 3*cosh(b*x + a)^2)*sinh(b*x + a)^6 + 48*(9
1*cosh(b*x + a)^11 - 66*cosh(b*x + a)^7 + 7*cosh(b*x + a)^3)*sinh(b*x + a)^5 + 4*(455*cosh(b*x + a)^12 - 495*c
osh(b*x + a)^8 + 105*cosh(b*x + a)^4 - 1)*sinh(b*x + a)^4 - 4*cosh(b*x + a)^4 + 16*(35*cosh(b*x + a)^13 - 55*c
osh(b*x + a)^9 + 21*cosh(b*x + a)^5 - cosh(b*x + a))*sinh(b*x + a)^3 + 24*(5*cosh(b*x + a)^14 - 11*cosh(b*x +
a)^10 + 7*cosh(b*x + a)^6 - cosh(b*x + a)^2)*sinh(b*x + a)^2 + 16*(cosh(b*x + a)^15 - 3*cosh(b*x + a)^11 + 3*c
osh(b*x + a)^7 - cosh(b*x + a)^3)*sinh(b*x + a) + 1)*log(2*sinh(b*x + a)/(cosh(b*x + a) - sinh(b*x + a))) + 4*
(21*cosh(b*x + a)^13 - 55*cosh(b*x + a)^9 - 33*cosh(b*x + a)^5 + 3*cosh(b*x + a))*sinh(b*x + a))/(b*cosh(b*x +
a)^16 + 560*b*cosh(b*x + a)^3*sinh(b*x + a)^13 + 120*b*cosh(b*x + a)^2*sinh(b*x + a)^14 + 16*b*cosh(b*x + a)*
sinh(b*x + a)^15 + b*sinh(b*x + a)^16 - 4*b*cosh(b*x + a)^12 + 4*(455*b*cosh(b*x + a)^4 - b)*sinh(b*x + a)^12
+ 48*(91*b*cosh(b*x + a)^5 - b*cosh(b*x + a))*sinh(b*x + a)^11 + 88*(91*b*cosh(b*x + a)^6 - 3*b*cosh(b*x + a)^
2)*sinh(b*x + a)^10 + 880*(13*b*cosh(b*x + a)^7 - b*cosh(b*x + a)^3)*sinh(b*x + a)^9 + 6*b*cosh(b*x + a)^8 + 6
*(2145*b*cosh(b*x + a)^8 - 330*b*cosh(b*x + a)^4 + b)*sinh(b*x + a)^8 + 16*(715*b*cosh(b*x + a)^9 - 198*b*cosh
(b*x + a)^5 + 3*b*cosh(b*x + a))*sinh(b*x + a)^7 + 56*(143*b*cosh(b*x + a)^10 - 66*b*cosh(b*x + a)^6 + 3*b*cos
h(b*x + a)^2)*sinh(b*x + a)^6 + 48*(91*b*cosh(b*x + a)^11 - 66*b*cosh(b*x + a)^7 + 7*b*cosh(b*x + a)^3)*sinh(b
*x + a)^5 - 4*b*cosh(b*x + a)^4 + 4*(455*b*cosh(b*x + a)^12 - 495*b*cosh(b*x + a)^8 + 105*b*cosh(b*x + a)^4 -
b)*sinh(b*x + a)^4 + 16*(35*b*cosh(b*x + a)^13 - 55*b*cosh(b*x + a)^9 + 21*b*cosh(b*x + a)^5 - b*cosh(b*x + a)
)*sinh(b*x + a)^3 + 24*(5*b*cosh(b*x + a)^14 - 11*b*cosh(b*x + a)^10 + 7*b*cosh(b*x + a)^6 - b*cosh(b*x + a)^2
)*sinh(b*x + a)^2 + 16*(b*cosh(b*x + a)^15 - 3*b*cosh(b*x + a)^11 + 3*b*cosh(b*x + a)^7 - b*cosh(b*x + a)^3)*s
inh(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}^{5}{\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)**5,x)

[Out]

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

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Giac [A]  time = 1.22274, size = 174, normalized size = 2.52 \begin{align*} -\frac{3 \, \log \left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )} + 2\right )}{b} + \frac{3 \, \log \left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )} - 2\right )}{b} + \frac{4 \,{\left (3 \,{\left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )}\right )}^{3} - 20 \, e^{\left (2 \, b x + 2 \, a\right )} - 20 \, e^{\left (-2 \, b x - 2 \, a\right )}\right )}}{{\left ({\left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )}\right )}^{2} - 4\right )}^{2} b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*log(e^(2*b*x + 2*a) + e^(-2*b*x - 2*a) + 2)/b + 3*log(e^(2*b*x + 2*a) + e^(-2*b*x - 2*a) - 2)/b + 4*(3*(e^(
2*b*x + 2*a) + e^(-2*b*x - 2*a))^3 - 20*e^(2*b*x + 2*a) - 20*e^(-2*b*x - 2*a))/(((e^(2*b*x + 2*a) + e^(-2*b*x
- 2*a))^2 - 4)^2*b)