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

Optimal. Leaf size=66 \[ -\frac {5 \text {sech}^3(a+b x)}{6 b}-\frac {5 \text {sech}(a+b x)}{2 b}+\frac {5 \tanh ^{-1}(\cosh (a+b x))}{2 b}-\frac {\text {csch}^2(a+b x) \text {sech}^3(a+b x)}{2 b} \]

[Out]

5/2*arctanh(cosh(b*x+a))/b-5/2*sech(b*x+a)/b-5/6*sech(b*x+a)^3/b-1/2*csch(b*x+a)^2*sech(b*x+a)^3/b

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Rubi [A]  time = 0.05, antiderivative size = 66, 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, 302, 207} \[ -\frac {5 \text {sech}^3(a+b x)}{6 b}-\frac {5 \text {sech}(a+b x)}{2 b}+\frac {5 \tanh ^{-1}(\cosh (a+b x))}{2 b}-\frac {\text {csch}^2(a+b x) \text {sech}^3(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*ArcTanh[Cosh[a + b*x]])/(2*b) - (5*Sech[a + b*x])/(2*b) - (5*Sech[a + b*x]^3)/(6*b) - (Csch[a + b*x]^2*Sech
[a + b*x]^3)/(2*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 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 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}^3(a+b x) \text {sech}^4(a+b x) \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {x^6}{\left (-1+x^2\right )^2} \, dx,x,\text {sech}(a+b x)\right )}{b}\\ &=-\frac {\text {csch}^2(a+b x) \text {sech}^3(a+b x)}{2 b}-\frac {5 \operatorname {Subst}\left (\int \frac {x^4}{-1+x^2} \, dx,x,\text {sech}(a+b x)\right )}{2 b}\\ &=-\frac {\text {csch}^2(a+b x) \text {sech}^3(a+b x)}{2 b}-\frac {5 \operatorname {Subst}\left (\int \left (1+x^2+\frac {1}{-1+x^2}\right ) \, dx,x,\text {sech}(a+b x)\right )}{2 b}\\ &=-\frac {5 \text {sech}(a+b x)}{2 b}-\frac {5 \text {sech}^3(a+b x)}{6 b}-\frac {\text {csch}^2(a+b x) \text {sech}^3(a+b x)}{2 b}-\frac {5 \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\text {sech}(a+b x)\right )}{2 b}\\ &=\frac {5 \tanh ^{-1}(\cosh (a+b x))}{2 b}-\frac {5 \text {sech}(a+b x)}{2 b}-\frac {5 \text {sech}^3(a+b x)}{6 b}-\frac {\text {csch}^2(a+b x) \text {sech}^3(a+b x)}{2 b}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 83, normalized size = 1.26 \[ -\frac {\text {csch}^2\left (\frac {1}{2} (a+b x)\right )}{8 b}-\frac {\text {sech}^3(a+b x)}{3 b}-\frac {\text {sech}^2\left (\frac {1}{2} (a+b x)\right )}{8 b}-\frac {2 \text {sech}(a+b x)}{b}-\frac {5 \log \left (\tanh \left (\frac {1}{2} (a+b x)\right )\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*Csch[(a + b*x)/2]^2/b - (5*Log[Tanh[(a + b*x)/2]])/(2*b) - Sech[(a + b*x)/2]^2/(8*b) - (2*Sech[a + b*x])/
b - Sech[a + b*x]^3/(3*b)

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fricas [B]  time = 0.43, size = 1573, normalized size = 23.83 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(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 + 1
)*sinh(b*x + a)^7 + 40*cosh(b*x + a)^7 + 280*(9*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a)^6 + 4*(945*cosh
(b*x + a)^4 + 210*cosh(b*x + a)^2 - 11)*sinh(b*x + a)^5 - 44*cosh(b*x + a)^5 + 20*(189*cosh(b*x + a)^5 + 70*co
sh(b*x + a)^3 - 11*cosh(b*x + a))*sinh(b*x + a)^4 + 40*(63*cosh(b*x + a)^6 + 35*cosh(b*x + a)^4 - 11*cosh(b*x
+ a)^2 + 1)*sinh(b*x + a)^3 + 40*cosh(b*x + a)^3 + 40*(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))*
sinh(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)*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 + (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))*sinh(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*co
sh(b*x + a)^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 + 28*cosh(b*x + a)^6 - 22*cosh(b*x + a)^4 + 12*cosh(b*x + a)^2 + 3)*sinh(b*x + a) + 30*cosh(b*x + a))/(b*c
osh(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*c
osh(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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giac [B]  time = 0.14, size = 128, normalized size = 1.94 \[ -\frac {\frac {12 \, {\left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}\right )}}{{\left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}\right )}^{2} - 4} + \frac {16 \, {\left (3 \, {\left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}\right )}^{2} + 2\right )}}{{\left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}\right )}^{3}} - 15 \, \log \left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )} + 2\right ) + 15 \, \log \left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )} - 2\right )}{12 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(12*(e^(b*x + a) + e^(-b*x - a))/((e^(b*x + a) + e^(-b*x - a))^2 - 4) + 16*(3*(e^(b*x + a) + e^(-b*x - a
))^2 + 2)/(e^(b*x + a) + e^(-b*x - a))^3 - 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 = 0.22, size = 53, normalized size = 0.80 \[ \frac {-\frac {1}{2 \sinh \left (b x +a \right )^{2} \cosh \left (b x +a \right )^{3}}-\frac {5}{6 \cosh \left (b x +a \right )^{3}}-\frac {5}{2 \cosh \left (b x +a \right )}+5 \arctanh \left ({\mathrm e}^{b x +a}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(-1/2/sinh(b*x+a)^2/cosh(b*x+a)^3-5/6/cosh(b*x+a)^3-5/2/cosh(b*x+a)+5*arctanh(exp(b*x+a)))

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maxima [B]  time = 0.33, size = 149, normalized size = 2.26 \[ \frac {5 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{2 \, b} - \frac {5 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{2 \, b} - \frac {15 \, e^{\left (-b x - a\right )} + 20 \, e^{\left (-3 \, b x - 3 \, a\right )} - 22 \, e^{\left (-5 \, b x - 5 \, a\right )} + 20 \, e^{\left (-7 \, b x - 7 \, a\right )} + 15 \, e^{\left (-9 \, b x - 9 \, a\right )}}{3 \, b {\left (e^{\left (-2 \, b x - 2 \, a\right )} - 2 \, e^{\left (-4 \, b x - 4 \, a\right )} - 2 \, e^{\left (-6 \, b x - 6 \, a\right )} + e^{\left (-8 \, b x - 8 \, a\right )} + e^{\left (-10 \, b x - 10 \, a\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/2*log(e^(-b*x - a) + 1)/b - 5/2*log(e^(-b*x - a) - 1)/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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mupad [B]  time = 1.44, size = 192, normalized size = 2.91 \[ \frac {5\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {-b^2}}{b}\right )}{\sqrt {-b^2}}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}-\frac {8\,{\mathrm {e}}^{a+b\,x}}{3\,b\,\left (2\,{\mathrm {e}}^{2\,a+2\,b\,x}+{\mathrm {e}}^{4\,a+4\,b\,x}+1\right )}+\frac {8\,{\mathrm {e}}^{a+b\,x}}{3\,b\,\left (3\,{\mathrm {e}}^{2\,a+2\,b\,x}+3\,{\mathrm {e}}^{4\,a+4\,b\,x}+{\mathrm {e}}^{6\,a+6\,b\,x}+1\right )}-\frac {{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )}-\frac {4\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(a + b*x)^4*sinh(a + b*x)^3),x)

[Out]

(5*atan((exp(b*x)*exp(a)*(-b^2)^(1/2))/b))/(-b^2)^(1/2) - (2*exp(a + b*x))/(b*(exp(4*a + 4*b*x) - 2*exp(2*a +
2*b*x) + 1)) - (8*exp(a + b*x))/(3*b*(2*exp(2*a + 2*b*x) + exp(4*a + 4*b*x) + 1)) + (8*exp(a + b*x))/(3*b*(3*e
xp(2*a + 2*b*x) + 3*exp(4*a + 4*b*x) + exp(6*a + 6*b*x) + 1)) - exp(a + b*x)/(b*(exp(2*a + 2*b*x) - 1)) - (4*e
xp(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 \[ \int \operatorname {csch}^{3}{\left (a + b x \right )} \operatorname {sech}^{4}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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