### 3.41 $$\int \text{csch}^4(a+b x) \text{sech}^3(a+b x) \, dx$$

Optimal. Leaf size=66 $-\frac{5 \text{csch}^3(a+b x)}{6 b}+\frac{5 \text{csch}(a+b x)}{2 b}+\frac{5 \tan ^{-1}(\sinh (a+b x))}{2 b}+\frac{\text{csch}^3(a+b x) \text{sech}^2(a+b x)}{2 b}$

[Out]

(5*ArcTan[Sinh[a + b*x]])/(2*b) + (5*Csch[a + b*x])/(2*b) - (5*Csch[a + b*x]^3)/(6*b) + (Csch[a + b*x]^3*Sech[
a + b*x]^2)/(2*b)

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Rubi [A]  time = 0.0432956, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.235, Rules used = {2621, 288, 302, 207} $-\frac{5 \text{csch}^3(a+b x)}{6 b}+\frac{5 \text{csch}(a+b x)}{2 b}+\frac{5 \tan ^{-1}(\sinh (a+b x))}{2 b}+\frac{\text{csch}^3(a+b x) \text{sech}^2(a+b x)}{2 b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Csch[a + b*x]^4*Sech[a + b*x]^3,x]

[Out]

(5*ArcTan[Sinh[a + b*x]])/(2*b) + (5*Csch[a + b*x])/(2*b) - (5*Csch[a + b*x]^3)/(6*b) + (Csch[a + b*x]^3*Sech[
a + b*x]^2)/(2*b)

Rule 2621

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

Rule 288

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

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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}^4(a+b x) \text{sech}^3(a+b x) \, dx &=\frac{i \operatorname{Subst}\left (\int \frac{x^6}{\left (-1+x^2\right )^2} \, dx,x,-i \text{csch}(a+b x)\right )}{b}\\ &=\frac{\text{csch}^3(a+b x) \text{sech}^2(a+b x)}{2 b}+\frac{(5 i) \operatorname{Subst}\left (\int \frac{x^4}{-1+x^2} \, dx,x,-i \text{csch}(a+b x)\right )}{2 b}\\ &=\frac{\text{csch}^3(a+b x) \text{sech}^2(a+b x)}{2 b}+\frac{(5 i) \operatorname{Subst}\left (\int \left (1+x^2+\frac{1}{-1+x^2}\right ) \, dx,x,-i \text{csch}(a+b x)\right )}{2 b}\\ &=\frac{5 \text{csch}(a+b x)}{2 b}-\frac{5 \text{csch}^3(a+b x)}{6 b}+\frac{\text{csch}^3(a+b x) \text{sech}^2(a+b x)}{2 b}+\frac{(5 i) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,-i \text{csch}(a+b x)\right )}{2 b}\\ &=\frac{5 \tan ^{-1}(\sinh (a+b x))}{2 b}+\frac{5 \text{csch}(a+b x)}{2 b}-\frac{5 \text{csch}^3(a+b x)}{6 b}+\frac{\text{csch}^3(a+b x) \text{sech}^2(a+b x)}{2 b}\\ \end{align*}

Mathematica [C]  time = 0.0158004, size = 33, normalized size = 0.5 $-\frac{\text{csch}^3(a+b x) \, _2F_1\left (-\frac{3}{2},2;-\frac{1}{2};-\sinh ^2(a+b x)\right )}{3 b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Csch[a + b*x]^4*Sech[a + b*x]^3,x]

[Out]

-(Csch[a + b*x]^3*Hypergeometric2F1[-3/2, 2, -1/2, -Sinh[a + b*x]^2])/(3*b)

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Maple [A]  time = 0.025, size = 73, normalized size = 1.1 \begin{align*} -{\frac{1}{3\,b \left ( \sinh \left ( bx+a \right ) \right ) ^{3} \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}}+{\frac{5}{3\,b\sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}}+{\frac{5\,{\rm sech} \left (bx+a\right )\tanh \left ( bx+a \right ) }{2\,b}}+5\,{\frac{\arctan \left ({{\rm e}^{bx+a}} \right ) }{b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(csch(b*x+a)^4*sech(b*x+a)^3,x)

[Out]

-1/3/b/sinh(b*x+a)^3/cosh(b*x+a)^2+5/3/b/sinh(b*x+a)/cosh(b*x+a)^2+5/2*sech(b*x+a)*tanh(b*x+a)/b+5*arctan(exp(
b*x+a))/b

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Maxima [B]  time = 1.78328, size = 178, normalized size = 2.7 \begin{align*} -\frac{5 \, \arctan \left (e^{\left (-b x - a\right )}\right )}{b} - \frac{15 \, e^{\left (-b x - a\right )} - 20 \, e^{\left (-3 \, b x - 3 \, a\right )} - 22 \, e^{\left (-5 \, b x - 5 \, a\right )} - 20 \, e^{\left (-7 \, b x - 7 \, a\right )} + 15 \, e^{\left (-9 \, b x - 9 \, a\right )}}{3 \, b{\left (e^{\left (-2 \, b x - 2 \, a\right )} + 2 \, e^{\left (-4 \, b x - 4 \, a\right )} - 2 \, e^{\left (-6 \, b x - 6 \, a\right )} - e^{\left (-8 \, b x - 8 \, a\right )} + e^{\left (-10 \, b x - 10 \, a\right )} - 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)^4*sech(b*x+a)^3,x, algorithm="maxima")

[Out]

-5*arctan(e^(-b*x - a))/b - 1/3*(15*e^(-b*x - a) - 20*e^(-3*b*x - 3*a) - 22*e^(-5*b*x - 5*a) - 20*e^(-7*b*x -
7*a) + 15*e^(-9*b*x - 9*a))/(b*(e^(-2*b*x - 2*a) + 2*e^(-4*b*x - 4*a) - 2*e^(-6*b*x - 6*a) - e^(-8*b*x - 8*a)
+ e^(-10*b*x - 10*a) - 1))

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Fricas [B]  time = 2.54743, size = 3283, normalized size = 49.74 \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)^4*sech(b*x+a)^3,x, algorithm="fricas")

[Out]

1/3*(15*cosh(b*x + a)^9 + 135*cosh(b*x + a)*sinh(b*x + a)^8 + 15*sinh(b*x + a)^9 + 20*(27*cosh(b*x + a)^2 - 1)
*sinh(b*x + a)^7 - 20*cosh(b*x + a)^7 + 140*(9*cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a)^6 + 2*(945*cosh(
b*x + a)^4 - 210*cosh(b*x + a)^2 - 11)*sinh(b*x + a)^5 - 22*cosh(b*x + a)^5 + 10*(189*cosh(b*x + a)^5 - 70*cos
h(b*x + a)^3 - 11*cosh(b*x + a))*sinh(b*x + a)^4 + 20*(63*cosh(b*x + a)^6 - 35*cosh(b*x + a)^4 - 11*cosh(b*x +
a)^2 - 1)*sinh(b*x + a)^3 - 20*cosh(b*x + a)^3 + 20*(27*cosh(b*x + a)^7 - 21*cosh(b*x + a)^5 - 11*cosh(b*x +
a)^3 - 3*cosh(b*x + a))*sinh(b*x + a)^2 + 15*(cosh(b*x + a)^10 + 10*cosh(b*x + a)*sinh(b*x + a)^9 + sinh(b*x +
a)^10 + (45*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^8 - cosh(b*x + a)^8 + 8*(15*cosh(b*x + a)^3 - cosh(b*x + a))*s
inh(b*x + a)^7 + 2*(105*cosh(b*x + a)^4 - 14*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^6 - 2*cosh(b*x + a)^6 + 4*(63*
cosh(b*x + a)^5 - 14*cosh(b*x + a)^3 - 3*cosh(b*x + a))*sinh(b*x + a)^5 + 2*(105*cosh(b*x + a)^6 - 35*cosh(b*x
+ a)^4 - 15*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^4 + 2*cosh(b*x + a)^4 + 8*(15*cosh(b*x + a)^7 - 7*cosh(b*x + a
)^5 - 5*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a)^3 + (45*cosh(b*x + a)^8 - 28*cosh(b*x + a)^6 - 30*cosh(
b*x + a)^4 + 12*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + cosh(b*x + a)^2 + 2*(5*cosh(b*x + a)^9 - 4*cosh(b*x + a
)^7 - 6*cosh(b*x + a)^5 + 4*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a) - 1)*arctan(cosh(b*x + a) + sinh(b*
x + a)) + 5*(27*cosh(b*x + a)^8 - 28*cosh(b*x + a)^6 - 22*cosh(b*x + a)^4 - 12*cosh(b*x + a)^2 + 3)*sinh(b*x +
a) + 15*cosh(b*x + a))/(b*cosh(b*x + a)^10 + 10*b*cosh(b*x + a)*sinh(b*x + a)^9 + b*sinh(b*x + a)^10 - b*cosh
(b*x + a)^8 + (45*b*cosh(b*x + a)^2 - b)*sinh(b*x + a)^8 + 8*(15*b*cosh(b*x + a)^3 - b*cosh(b*x + a))*sinh(b*x
+ a)^7 - 2*b*cosh(b*x + a)^6 + 2*(105*b*cosh(b*x + a)^4 - 14*b*cosh(b*x + a)^2 - b)*sinh(b*x + a)^6 + 4*(63*b
*cosh(b*x + a)^5 - 14*b*cosh(b*x + a)^3 - 3*b*cosh(b*x + a))*sinh(b*x + a)^5 + 2*b*cosh(b*x + a)^4 + 2*(105*b*
cosh(b*x + a)^6 - 35*b*cosh(b*x + a)^4 - 15*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^4 + 8*(15*b*cosh(b*x + a)^7 -
7*b*cosh(b*x + a)^5 - 5*b*cosh(b*x + a)^3 + b*cosh(b*x + a))*sinh(b*x + a)^3 + b*cosh(b*x + a)^2 + (45*b*cosh
(b*x + a)^8 - 28*b*cosh(b*x + a)^6 - 30*b*cosh(b*x + a)^4 + 12*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^2 + 2*(5*b
*cosh(b*x + a)^9 - 4*b*cosh(b*x + a)^7 - 6*b*cosh(b*x + a)^5 + 4*b*cosh(b*x + a)^3 + b*cosh(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}^{4}{\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)**4*sech(b*x+a)**3,x)

[Out]

Integral(csch(a + b*x)**4*sech(a + b*x)**3, x)

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Giac [B]  time = 1.16499, size = 173, normalized size = 2.62 \begin{align*} \frac{5 \,{\left (\pi + 2 \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-b x - a\right )}\right )\right )}}{4 \, b} + \frac{e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}}{{\left ({\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{2} + 4\right )} b} + \frac{4 \,{\left (3 \,{\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{2} - 2\right )}}{3 \, b{\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)^4*sech(b*x+a)^3,x, algorithm="giac")

[Out]

5/4*(pi + 2*arctan(1/2*(e^(2*b*x + 2*a) - 1)*e^(-b*x - a)))/b + (e^(b*x + a) - e^(-b*x - a))/(((e^(b*x + a) -
e^(-b*x - a))^2 + 4)*b) + 4/3*(3*(e^(b*x + a) - e^(-b*x - a))^2 - 2)/(b*(e^(b*x + a) - e^(-b*x - a))^3)