3.45 \(\int \text {csch}^5(a+b x) \text {sech}^2(a+b x) \, dx\)

Optimal. Leaf size=70 \[ \frac {15 \text {sech}(a+b x)}{8 b}-\frac {15 \tanh ^{-1}(\cosh (a+b x))}{8 b}-\frac {\text {csch}^4(a+b x) \text {sech}(a+b x)}{4 b}+\frac {5 \text {csch}^2(a+b x) \text {sech}(a+b x)}{8 b} \]

[Out]

-15/8*arctanh(cosh(b*x+a))/b+15/8*sech(b*x+a)/b+5/8*csch(b*x+a)^2*sech(b*x+a)/b-1/4*csch(b*x+a)^4*sech(b*x+a)/
b

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Rubi [A]  time = 0.05, antiderivative size = 70, normalized size of antiderivative = 1.00, 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 = {2622, 288, 321, 207} \[ \frac {15 \text {sech}(a+b x)}{8 b}-\frac {15 \tanh ^{-1}(\cosh (a+b x))}{8 b}-\frac {\text {csch}^4(a+b x) \text {sech}(a+b x)}{4 b}+\frac {5 \text {csch}^2(a+b x) \text {sech}(a+b x)}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[Csch[a + b*x]^5*Sech[a + b*x]^2,x]

[Out]

(-15*ArcTanh[Cosh[a + b*x]])/(8*b) + (15*Sech[a + b*x])/(8*b) + (5*Csch[a + b*x]^2*Sech[a + b*x])/(8*b) - (Csc
h[a + b*x]^4*Sech[a + b*x])/(4*b)

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])

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 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2622

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

Rubi steps

\begin {align*} \int \text {csch}^5(a+b x) \text {sech}^2(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^6}{\left (-1+x^2\right )^3} \, dx,x,\text {sech}(a+b x)\right )}{b}\\ &=-\frac {\text {csch}^4(a+b x) \text {sech}(a+b x)}{4 b}+\frac {5 \operatorname {Subst}\left (\int \frac {x^4}{\left (-1+x^2\right )^2} \, dx,x,\text {sech}(a+b x)\right )}{4 b}\\ &=\frac {5 \text {csch}^2(a+b x) \text {sech}(a+b x)}{8 b}-\frac {\text {csch}^4(a+b x) \text {sech}(a+b x)}{4 b}+\frac {15 \operatorname {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\text {sech}(a+b x)\right )}{8 b}\\ &=\frac {15 \text {sech}(a+b x)}{8 b}+\frac {5 \text {csch}^2(a+b x) \text {sech}(a+b x)}{8 b}-\frac {\text {csch}^4(a+b x) \text {sech}(a+b x)}{4 b}+\frac {15 \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\text {sech}(a+b x)\right )}{8 b}\\ &=-\frac {15 \tanh ^{-1}(\cosh (a+b x))}{8 b}+\frac {15 \text {sech}(a+b x)}{8 b}+\frac {5 \text {csch}^2(a+b x) \text {sech}(a+b x)}{8 b}-\frac {\text {csch}^4(a+b x) \text {sech}(a+b x)}{4 b}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 105, normalized size = 1.50 \[ -\frac {\text {csch}^4\left (\frac {1}{2} (a+b x)\right )}{64 b}+\frac {7 \text {csch}^2\left (\frac {1}{2} (a+b x)\right )}{32 b}+\frac {\text {sech}^4\left (\frac {1}{2} (a+b x)\right )}{64 b}+\frac {7 \text {sech}^2\left (\frac {1}{2} (a+b x)\right )}{32 b}+\frac {\text {sech}(a+b x)}{b}+\frac {15 \log \left (\tanh \left (\frac {1}{2} (a+b x)\right )\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[a + b*x]^5*Sech[a + b*x]^2,x]

[Out]

(7*Csch[(a + b*x)/2]^2)/(32*b) - Csch[(a + b*x)/2]^4/(64*b) + (15*Log[Tanh[(a + b*x)/2]])/(8*b) + (7*Sech[(a +
b*x)/2]^2)/(32*b) + Sech[(a + b*x)/2]^4/(64*b) + Sech[a + b*x]/b

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fricas [B]  time = 0.43, size = 1591, normalized size = 22.73 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)^5*sech(b*x+a)^2,x, algorithm="fricas")

[Out]

1/8*(30*cosh(b*x + a)^9 + 270*cosh(b*x + a)*sinh(b*x + a)^8 + 30*sinh(b*x + a)^9 + 40*(27*cosh(b*x + a)^2 - 2)
*sinh(b*x + a)^7 - 80*cosh(b*x + a)^7 + 280*(9*cosh(b*x + a)^3 - 2*cosh(b*x + a))*sinh(b*x + a)^6 + 12*(315*co
sh(b*x + a)^4 - 140*cosh(b*x + a)^2 + 3)*sinh(b*x + a)^5 + 36*cosh(b*x + a)^5 + 20*(189*cosh(b*x + a)^5 - 140*
cosh(b*x + a)^3 + 9*cosh(b*x + a))*sinh(b*x + a)^4 + 40*(63*cosh(b*x + a)^6 - 70*cosh(b*x + a)^4 + 9*cosh(b*x
+ a)^2 - 2)*sinh(b*x + a)^3 - 80*cosh(b*x + a)^3 + 120*(9*cosh(b*x + a)^7 - 14*cosh(b*x + a)^5 + 3*cosh(b*x +
a)^3 - 2*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 + 3*(15*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^8 - 3*cosh(b*x + a)^8 + 24*(5*cosh(b*x + a)^3 - cosh(b*x + a
))*sinh(b*x + a)^7 + 2*(105*cosh(b*x + a)^4 - 42*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^6 + 2*cosh(b*x + a)^6 + 12
*(21*cosh(b*x + a)^5 - 14*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a)^5 + 2*(105*cosh(b*x + a)^6 - 105*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 - 21*cosh(b*
x + a)^5 + 5*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a)^3 + 3*(15*cosh(b*x + a)^8 - 28*cosh(b*x + a)^6 + 1
0*cosh(b*x + a)^4 + 4*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 3*cosh(b*x + a)^2 + 2*(5*cosh(b*x + a)^9 - 12*cos
h(b*x + a)^7 + 6*cosh(b*x + a)^5 + 4*cosh(b*x + a)^3 - 3*cosh(b*x + a))*sinh(b*x + a) + 1)*log(cosh(b*x + a) +
sinh(b*x + a) + 1) + 15*(cosh(b*x + a)^10 + 10*cosh(b*x + a)*sinh(b*x + a)^9 + sinh(b*x + a)^10 + 3*(15*cosh(
b*x + a)^2 - 1)*sinh(b*x + a)^8 - 3*cosh(b*x + a)^8 + 24*(5*cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a)^7 +
2*(105*cosh(b*x + a)^4 - 42*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^6 + 2*cosh(b*x + a)^6 + 12*(21*cosh(b*x + a)^5
- 14*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a)^5 + 2*(105*cosh(b*x + a)^6 - 105*cosh(b*x + a)^4 + 15*cos
h(b*x + a)^2 + 1)*sinh(b*x + a)^4 + 2*cosh(b*x + a)^4 + 8*(15*cosh(b*x + a)^7 - 21*cosh(b*x + a)^5 + 5*cosh(b*
x + a)^3 + cosh(b*x + a))*sinh(b*x + a)^3 + 3*(15*cosh(b*x + a)^8 - 28*cosh(b*x + a)^6 + 10*cosh(b*x + a)^4 +
4*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 3*cosh(b*x + a)^2 + 2*(5*cosh(b*x + a)^9 - 12*cosh(b*x + a)^7 + 6*cos
h(b*x + a)^5 + 4*cosh(b*x + a)^3 - 3*cosh(b*x + a))*sinh(b*x + a) + 1)*log(cosh(b*x + a) + sinh(b*x + a) - 1)
+ 10*(27*cosh(b*x + a)^8 - 56*cosh(b*x + a)^6 + 18*cosh(b*x + a)^4 - 24*cosh(b*x + a)^2 + 3)*sinh(b*x + a) + 3
0*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 - 3*b*cosh(b*x
+ a)^8 + 3*(15*b*cosh(b*x + a)^2 - b)*sinh(b*x + a)^8 + 24*(5*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 - 42*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^6 + 12*(21*b*c
osh(b*x + a)^5 - 14*b*cosh(b*x + a)^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 - 105*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 - 21
*b*cosh(b*x + a)^5 + 5*b*cosh(b*x + a)^3 + b*cosh(b*x + a))*sinh(b*x + a)^3 - 3*b*cosh(b*x + a)^2 + 3*(15*b*co
sh(b*x + a)^8 - 28*b*cosh(b*x + a)^6 + 10*b*cosh(b*x + a)^4 + 4*b*cosh(b*x + a)^2 - b)*sinh(b*x + a)^2 + 2*(5*
b*cosh(b*x + a)^9 - 12*b*cosh(b*x + a)^7 + 6*b*cosh(b*x + a)^5 + 4*b*cosh(b*x + a)^3 - 3*b*cosh(b*x + a))*sinh
(b*x + a) + b)

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giac [B]  time = 0.16, size = 130, normalized size = 1.86 \[ \frac {\frac {4 \, {\left (7 \, {\left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}\right )}^{3} - 36 \, e^{\left (b x + a\right )} - 36 \, e^{\left (-b x - a\right )}\right )}}{{\left ({\left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}\right )}^{2} - 4\right )}^{2}} + \frac {32}{e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}} - 15 \, \log \left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )} + 2\right ) + 15 \, \log \left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )} - 2\right )}{16 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)^5*sech(b*x+a)^2,x, algorithm="giac")

[Out]

1/16*(4*(7*(e^(b*x + a) + e^(-b*x - a))^3 - 36*e^(b*x + a) - 36*e^(-b*x - a))/((e^(b*x + a) + e^(-b*x - a))^2
- 4)^2 + 32/(e^(b*x + a) + e^(-b*x - a)) - 15*log(e^(b*x + a) + e^(-b*x - a) + 2) + 15*log(e^(b*x + a) + e^(-b
*x - a) - 2))/b

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maple [A]  time = 0.15, size = 61, normalized size = 0.87 \[ \frac {-\frac {1}{4 \sinh \left (b x +a \right )^{4} \cosh \left (b x +a \right )}+\frac {5}{8 \sinh \left (b x +a \right )^{2} \cosh \left (b x +a \right )}+\frac {15}{8 \cosh \left (b x +a \right )}-\frac {15 \arctanh \left ({\mathrm e}^{b x +a}\right )}{4}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(b*x+a)^5*sech(b*x+a)^2,x)

[Out]

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

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maxima [B]  time = 0.60, size = 155, normalized size = 2.21 \[ -\frac {15 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{8 \, b} + \frac {15 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{8 \, b} - \frac {15 \, e^{\left (-b x - a\right )} - 40 \, e^{\left (-3 \, b x - 3 \, a\right )} + 18 \, e^{\left (-5 \, b x - 5 \, a\right )} - 40 \, e^{\left (-7 \, b x - 7 \, a\right )} + 15 \, e^{\left (-9 \, b x - 9 \, a\right )}}{4 \, b {\left (3 \, e^{\left (-2 \, b x - 2 \, a\right )} - 2 \, e^{\left (-4 \, b x - 4 \, a\right )} - 2 \, e^{\left (-6 \, b x - 6 \, a\right )} + 3 \, e^{\left (-8 \, b x - 8 \, a\right )} - e^{\left (-10 \, b x - 10 \, a\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)^5*sech(b*x+a)^2,x, algorithm="maxima")

[Out]

-15/8*log(e^(-b*x - a) + 1)/b + 15/8*log(e^(-b*x - a) - 1)/b - 1/4*(15*e^(-b*x - a) - 40*e^(-3*b*x - 3*a) + 18
*e^(-5*b*x - 5*a) - 40*e^(-7*b*x - 7*a) + 15*e^(-9*b*x - 9*a))/(b*(3*e^(-2*b*x - 2*a) - 2*e^(-4*b*x - 4*a) - 2
*e^(-6*b*x - 6*a) + 3*e^(-8*b*x - 8*a) - e^(-10*b*x - 10*a) - 1))

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mupad [B]  time = 1.45, size = 214, normalized size = 3.06 \[ \frac {3\,{\mathrm {e}}^{a+b\,x}}{2\,b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}-\frac {15\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {-b^2}}{b}\right )}{4\,\sqrt {-b^2}}-\frac {6\,{\mathrm {e}}^{a+b\,x}}{b\,\left (3\,{\mathrm {e}}^{2\,a+2\,b\,x}-3\,{\mathrm {e}}^{4\,a+4\,b\,x}+{\mathrm {e}}^{6\,a+6\,b\,x}-1\right )}-\frac {4\,{\mathrm {e}}^{a+b\,x}}{b\,\left (6\,{\mathrm {e}}^{4\,a+4\,b\,x}-4\,{\mathrm {e}}^{2\,a+2\,b\,x}-4\,{\mathrm {e}}^{6\,a+6\,b\,x}+{\mathrm {e}}^{8\,a+8\,b\,x}+1\right )}+\frac {7\,{\mathrm {e}}^{a+b\,x}}{4\,b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )}+\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(a + b*x)^2*sinh(a + b*x)^5),x)

[Out]

(3*exp(a + b*x))/(2*b*(exp(4*a + 4*b*x) - 2*exp(2*a + 2*b*x) + 1)) - (15*atan((exp(b*x)*exp(a)*(-b^2)^(1/2))/b
))/(4*(-b^2)^(1/2)) - (6*exp(a + b*x))/(b*(3*exp(2*a + 2*b*x) - 3*exp(4*a + 4*b*x) + exp(6*a + 6*b*x) - 1)) -
(4*exp(a + b*x))/(b*(6*exp(4*a + 4*b*x) - 4*exp(2*a + 2*b*x) - 4*exp(6*a + 6*b*x) + exp(8*a + 8*b*x) + 1)) + (
7*exp(a + b*x))/(4*b*(exp(2*a + 2*b*x) - 1)) + (2*exp(a + b*x))/(b*(exp(2*a + 2*b*x) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {csch}^{5}{\left (a + b x \right )} \operatorname {sech}^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)**5*sech(b*x+a)**2,x)

[Out]

Integral(csch(a + b*x)**5*sech(a + b*x)**2, x)

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