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

Optimal. Leaf size=43 \[ \frac {\tanh ^2(a+b x)}{2 b}-\frac {\coth ^2(a+b x)}{2 b}-\frac {2 \log (\tanh (a+b x))}{b} \]

[Out]

-1/2*coth(b*x+a)^2/b-2*ln(tanh(b*x+a))/b+1/2*tanh(b*x+a)^2/b

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Rubi [A]  time = 0.04, antiderivative size = 43, normalized size of antiderivative = 1.00, 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} \[ \frac {\tanh ^2(a+b x)}{2 b}-\frac {\coth ^2(a+b x)}{2 b}-\frac {2 \log (\tanh (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 47, normalized size = 1.09 \[ 8 \left (-\frac {\text {csch}^2(a+b x)}{16 b}-\frac {\text {sech}^2(a+b x)}{16 b}-\frac {\log (\tanh (a+b x))}{4 b}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

8*(-1/16*Csch[a + b*x]^2/b - Log[Tanh[a + b*x]]/(4*b) - Sech[a + b*x]^2/(16*b))

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fricas [B]  time = 0.42, size = 774, normalized size = 18.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(2*cosh(b*x + a)^6 + 40*cosh(b*x + a)^3*sinh(b*x + a)^3 + 30*cosh(b*x + a)^2*sinh(b*x + a)^4 + 12*cosh(b*x
+ a)*sinh(b*x + a)^5 + 2*sinh(b*x + a)^6 + 2*(15*cosh(b*x + a)^4 + 1)*sinh(b*x + a)^2 + 2*cosh(b*x + a)^2 - (c
osh(b*x + a)^8 + 56*cosh(b*x + a)^3*sinh(b*x + a)^5 + 28*cosh(b*x + a)^2*sinh(b*x + a)^6 + 8*cosh(b*x + a)*sin
h(b*x + a)^7 + sinh(b*x + a)^8 + 2*(35*cosh(b*x + a)^4 - 1)*sinh(b*x + a)^4 - 2*cosh(b*x + a)^4 + 8*(7*cosh(b*
x + a)^5 - cosh(b*x + a))*sinh(b*x + a)^3 + 4*(7*cosh(b*x + a)^6 - 3*cosh(b*x + a)^2)*sinh(b*x + a)^2 + 8*(cos
h(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))) + (cos
h(b*x + a)^8 + 56*cosh(b*x + a)^3*sinh(b*x + a)^5 + 28*cosh(b*x + a)^2*sinh(b*x + a)^6 + 8*cosh(b*x + a)*sinh(
b*x + a)^7 + sinh(b*x + a)^8 + 2*(35*cosh(b*x + a)^4 - 1)*sinh(b*x + a)^4 - 2*cosh(b*x + a)^4 + 8*(7*cosh(b*x
+ a)^5 - cosh(b*x + a))*sinh(b*x + a)^3 + 4*(7*cosh(b*x + a)^6 - 3*cosh(b*x + a)^2)*sinh(b*x + a)^2 + 8*(cosh(
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*(3*c
osh(b*x + a)^5 + cosh(b*x + a))*sinh(b*x + a))/(b*cosh(b*x + a)^8 + 56*b*cosh(b*x + a)^3*sinh(b*x + a)^5 + 28*
b*cosh(b*x + a)^2*sinh(b*x + a)^6 + 8*b*cosh(b*x + a)*sinh(b*x + a)^7 + b*sinh(b*x + a)^8 - 2*b*cosh(b*x + a)^
4 + 2*(35*b*cosh(b*x + a)^4 - b)*sinh(b*x + a)^4 + 8*(7*b*cosh(b*x + a)^5 - b*cosh(b*x + a))*sinh(b*x + a)^3 +
4*(7*b*cosh(b*x + a)^6 - 3*b*cosh(b*x + a)^2)*sinh(b*x + a)^2 + 8*(b*cosh(b*x + a)^7 - b*cosh(b*x + a)^3)*sin
h(b*x + a) + b)

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giac [B]  time = 0.14, size = 96, normalized size = 2.23 \[ -\frac {\frac {4 \, {\left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )}\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )}\right )}^{2} - 4} - \log \left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )} + 2\right ) + \log \left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )} - 2\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.22, size = 48, normalized size = 1.12 \[ -\frac {1}{2 b \sinh \left (b x +a \right )^{2} \cosh \left (b x +a \right )^{2}}-\frac {1}{b \cosh \left (b x +a \right )^{2}}-\frac {2 \ln \left (\tanh \left (b x +a \right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.41, size = 102, normalized size = 2.37 \[ -\frac {2 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{b} - \frac {2 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{b} + \frac {2 \, \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )}{b} + \frac {4 \, {\left (e^{\left (-2 \, b x - 2 \, a\right )} + e^{\left (-6 \, b x - 6 \, a\right )}\right )}}{b {\left (2 \, e^{\left (-4 \, b x - 4 \, a\right )} - e^{\left (-8 \, b x - 8 \, a\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.44, size = 96, normalized size = 2.23 \[ \frac {4\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}\,\sqrt {-b^2}}{b}\right )}{\sqrt {-b^2}}-\frac {4\,{\mathrm {e}}^{2\,a+2\,b\,x}}{b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-1\right )}-\frac {8\,{\mathrm {e}}^{2\,a+2\,b\,x}}{b\,\left ({\mathrm {e}}^{8\,a+8\,b\,x}-2\,{\mathrm {e}}^{4\,a+4\,b\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {csch}^{3}{\left (a + b x \right )} \operatorname {sech}^{3}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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