3.38 \(\int \text {csch}^3(a+b x) \text {sech}^5(a+b x) \, dx\)
Optimal. Leaf size=58 \[ -\frac {\tanh ^4(a+b x)}{4 b}+\frac {3 \tanh ^2(a+b x)}{2 b}-\frac {\coth ^2(a+b x)}{2 b}-\frac {3 \log (\tanh (a+b x))}{b} \]
[Out]
-1/2*coth(b*x+a)^2/b-3*ln(tanh(b*x+a))/b+3/2*tanh(b*x+a)^2/b-1/4*tanh(b*x+a)^4/b
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Rubi [A] time = 0.05, antiderivative size = 58, 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} \[ -\frac {\tanh ^4(a+b x)}{4 b}+\frac {3 \tanh ^2(a+b x)}{2 b}-\frac {\coth ^2(a+b x)}{2 b}-\frac {3 \log (\tanh (a+b x))}{b} \]
Antiderivative was successfully verified.
[In]
Int[Csch[a + b*x]^3*Sech[a + b*x]^5,x]
[Out]
-Coth[a + b*x]^2/(2*b) - (3*Log[Tanh[a + b*x]])/b + (3*Tanh[a + b*x]^2)/(2*b) - Tanh[a + b*x]^4/(4*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}^5(a+b x) \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^3}{x^3} \, dx,x,i \tanh (a+b x)\right )}{b}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {(1+x)^3}{x^2} \, dx,x,-\tanh ^2(a+b x)\right )}{2 b}\\ &=-\frac {\operatorname {Subst}\left (\int \left (3+\frac {1}{x^2}+\frac {3}{x}+x\right ) \, dx,x,-\tanh ^2(a+b x)\right )}{2 b}\\ &=-\frac {\coth ^2(a+b x)}{2 b}-\frac {3 \log (\tanh (a+b x))}{b}+\frac {3 \tanh ^2(a+b x)}{2 b}-\frac {\tanh ^4(a+b x)}{4 b}\\ \end {align*}
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Mathematica [A] time = 0.45, size = 54, normalized size = 0.93 \[ -\frac {2 \text {csch}^2(a+b x)+\text {sech}^4(a+b x)+4 \text {sech}^2(a+b x)+12 \log (\sinh (a+b x))-12 \log (\cosh (a+b x))}{4 b} \]
Antiderivative was successfully verified.
[In]
Integrate[Csch[a + b*x]^3*Sech[a + b*x]^5,x]
[Out]
-1/4*(2*Csch[a + b*x]^2 - 12*Log[Cosh[a + b*x]] + 12*Log[Sinh[a + b*x]] + 4*Sech[a + b*x]^2 + Sech[a + b*x]^4)
/b
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fricas [B] time = 0.42, size = 2103, normalized size = 36.26 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csch(b*x+a)^3*sech(b*x+a)^5,x, algorithm="fricas")
[Out]
-(6*cosh(b*x + a)^10 + 60*cosh(b*x + a)*sinh(b*x + a)^9 + 6*sinh(b*x + a)^10 + 6*(45*cosh(b*x + a)^2 + 2)*sinh
(b*x + a)^8 + 12*cosh(b*x + a)^8 + 48*(15*cosh(b*x + a)^3 + 2*cosh(b*x + a))*sinh(b*x + a)^7 + 4*(315*cosh(b*x
+ a)^4 + 84*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^6 - 4*cosh(b*x + a)^6 + 24*(63*cosh(b*x + a)^5 + 28*cosh(b*x +
a)^3 - cosh(b*x + a))*sinh(b*x + a)^5 + 12*(105*cosh(b*x + a)^6 + 70*cosh(b*x + a)^4 - 5*cosh(b*x + a)^2 + 1)
*sinh(b*x + a)^4 + 12*cosh(b*x + a)^4 + 16*(45*cosh(b*x + a)^7 + 42*cosh(b*x + a)^5 - 5*cosh(b*x + a)^3 + 3*co
sh(b*x + a))*sinh(b*x + a)^3 + 6*(45*cosh(b*x + a)^8 + 56*cosh(b*x + a)^6 - 10*cosh(b*x + a)^4 + 12*cosh(b*x +
a)^2 + 1)*sinh(b*x + a)^2 + 6*cosh(b*x + a)^2 - 3*(cosh(b*x + a)^12 + 12*cosh(b*x + a)*sinh(b*x + a)^11 + sin
h(b*x + a)^12 + 2*(33*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^10 + 2*cosh(b*x + a)^10 + 20*(11*cosh(b*x + a)^3 + co
sh(b*x + a))*sinh(b*x + a)^9 + (495*cosh(b*x + a)^4 + 90*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^8 - cosh(b*x + a)^
8 + 8*(99*cosh(b*x + a)^5 + 30*cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a)^7 + 4*(231*cosh(b*x + a)^6 + 105
*cosh(b*x + a)^4 - 7*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^6 - 4*cosh(b*x + a)^6 + 8*(99*cosh(b*x + a)^7 + 63*cos
h(b*x + a)^5 - 7*cosh(b*x + a)^3 - 3*cosh(b*x + a))*sinh(b*x + a)^5 + (495*cosh(b*x + a)^8 + 420*cosh(b*x + a)
^6 - 70*cosh(b*x + a)^4 - 60*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^4 - cosh(b*x + a)^4 + 4*(55*cosh(b*x + a)^9 +
60*cosh(b*x + a)^7 - 14*cosh(b*x + a)^5 - 20*cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a)^3 + 2*(33*cosh(b*x
+ a)^10 + 45*cosh(b*x + a)^8 - 14*cosh(b*x + a)^6 - 30*cosh(b*x + a)^4 - 3*cosh(b*x + a)^2 + 1)*sinh(b*x + a)
^2 + 2*cosh(b*x + a)^2 + 4*(3*cosh(b*x + a)^11 + 5*cosh(b*x + a)^9 - 2*cosh(b*x + a)^7 - 6*cosh(b*x + a)^5 - c
osh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a) + 1)*log(2*cosh(b*x + a)/(cosh(b*x + a) - sinh(b*x + a))) + 3*(c
osh(b*x + a)^12 + 12*cosh(b*x + a)*sinh(b*x + a)^11 + sinh(b*x + a)^12 + 2*(33*cosh(b*x + a)^2 + 1)*sinh(b*x +
a)^10 + 2*cosh(b*x + a)^10 + 20*(11*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a)^9 + (495*cosh(b*x + a)^4 +
90*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^8 - cosh(b*x + a)^8 + 8*(99*cosh(b*x + a)^5 + 30*cosh(b*x + a)^3 - cosh
(b*x + a))*sinh(b*x + a)^7 + 4*(231*cosh(b*x + a)^6 + 105*cosh(b*x + a)^4 - 7*cosh(b*x + a)^2 - 1)*sinh(b*x +
a)^6 - 4*cosh(b*x + a)^6 + 8*(99*cosh(b*x + a)^7 + 63*cosh(b*x + a)^5 - 7*cosh(b*x + a)^3 - 3*cosh(b*x + a))*s
inh(b*x + a)^5 + (495*cosh(b*x + a)^8 + 420*cosh(b*x + a)^6 - 70*cosh(b*x + a)^4 - 60*cosh(b*x + a)^2 - 1)*sin
h(b*x + a)^4 - cosh(b*x + a)^4 + 4*(55*cosh(b*x + a)^9 + 60*cosh(b*x + a)^7 - 14*cosh(b*x + a)^5 - 20*cosh(b*x
+ a)^3 - cosh(b*x + a))*sinh(b*x + a)^3 + 2*(33*cosh(b*x + a)^10 + 45*cosh(b*x + a)^8 - 14*cosh(b*x + a)^6 -
30*cosh(b*x + a)^4 - 3*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 2*cosh(b*x + a)^2 + 4*(3*cosh(b*x + a)^11 + 5*co
sh(b*x + a)^9 - 2*cosh(b*x + a)^7 - 6*cosh(b*x + a)^5 - cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a) + 1)*lo
g(2*sinh(b*x + a)/(cosh(b*x + a) - sinh(b*x + a))) + 12*(5*cosh(b*x + a)^9 + 8*cosh(b*x + a)^7 - 2*cosh(b*x +
a)^5 + 4*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a))/(b*cosh(b*x + a)^12 + 12*b*cosh(b*x + a)*sinh(b*x + a
)^11 + b*sinh(b*x + a)^12 + 2*b*cosh(b*x + a)^10 + 2*(33*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^10 + 20*(11*b*co
sh(b*x + a)^3 + b*cosh(b*x + a))*sinh(b*x + a)^9 - b*cosh(b*x + a)^8 + (495*b*cosh(b*x + a)^4 + 90*b*cosh(b*x
+ a)^2 - b)*sinh(b*x + a)^8 + 8*(99*b*cosh(b*x + a)^5 + 30*b*cosh(b*x + a)^3 - b*cosh(b*x + a))*sinh(b*x + a)^
7 - 4*b*cosh(b*x + a)^6 + 4*(231*b*cosh(b*x + a)^6 + 105*b*cosh(b*x + a)^4 - 7*b*cosh(b*x + a)^2 - b)*sinh(b*x
+ a)^6 + 8*(99*b*cosh(b*x + a)^7 + 63*b*cosh(b*x + a)^5 - 7*b*cosh(b*x + a)^3 - 3*b*cosh(b*x + a))*sinh(b*x +
a)^5 - b*cosh(b*x + a)^4 + (495*b*cosh(b*x + a)^8 + 420*b*cosh(b*x + a)^6 - 70*b*cosh(b*x + a)^4 - 60*b*cosh(
b*x + a)^2 - b)*sinh(b*x + a)^4 + 4*(55*b*cosh(b*x + a)^9 + 60*b*cosh(b*x + a)^7 - 14*b*cosh(b*x + a)^5 - 20*b
*cosh(b*x + a)^3 - b*cosh(b*x + a))*sinh(b*x + a)^3 + 2*b*cosh(b*x + a)^2 + 2*(33*b*cosh(b*x + a)^10 + 45*b*co
sh(b*x + a)^8 - 14*b*cosh(b*x + a)^6 - 30*b*cosh(b*x + a)^4 - 3*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^2 + 4*(3*
b*cosh(b*x + a)^11 + 5*b*cosh(b*x + a)^9 - 2*b*cosh(b*x + a)^7 - 6*b*cosh(b*x + a)^5 - b*cosh(b*x + a)^3 + b*c
osh(b*x + a))*sinh(b*x + a) + b)
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giac [B] time = 0.17, size = 171, normalized size = 2.95 \[ \frac {\frac {2 \, {\left (3 \, e^{\left (2 \, b x + 2 \, a\right )} + 3 \, e^{\left (-2 \, b x - 2 \, a\right )} - 10\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )} - 2} - \frac {9 \, {\left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )}\right )}^{2} + 52 \, e^{\left (2 \, b x + 2 \, a\right )} + 52 \, e^{\left (-2 \, b x - 2 \, a\right )} + 84}{{\left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )} + 2\right )}^{2}} + 6 \, \log \left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )} + 2\right ) - 6 \, \log \left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )} - 2\right )}{4 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csch(b*x+a)^3*sech(b*x+a)^5,x, algorithm="giac")
[Out]
1/4*(2*(3*e^(2*b*x + 2*a) + 3*e^(-2*b*x - 2*a) - 10)/(e^(2*b*x + 2*a) + e^(-2*b*x - 2*a) - 2) - (9*(e^(2*b*x +
2*a) + e^(-2*b*x - 2*a))^2 + 52*e^(2*b*x + 2*a) + 52*e^(-2*b*x - 2*a) + 84)/(e^(2*b*x + 2*a) + e^(-2*b*x - 2*
a) + 2)^2 + 6*log(e^(2*b*x + 2*a) + e^(-2*b*x - 2*a) + 2) - 6*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 = 61, normalized size = 1.05 \[ -\frac {1}{2 b \sinh \left (b x +a \right )^{2} \cosh \left (b x +a \right )^{4}}-\frac {3}{4 b \cosh \left (b x +a \right )^{4}}-\frac {3}{2 b \cosh \left (b x +a \right )^{2}}-\frac {3 \ln \left (\tanh \left (b x +a \right )\right )}{b} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(csch(b*x+a)^3*sech(b*x+a)^5,x)
[Out]
-1/2/b/sinh(b*x+a)^2/cosh(b*x+a)^4-3/4/b/cosh(b*x+a)^4-3/2/b/cosh(b*x+a)^2-3*ln(tanh(b*x+a))/b
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maxima [B] time = 0.41, size = 181, normalized size = 3.12 \[ -\frac {3 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{b} - \frac {3 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{b} + \frac {3 \, \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )}{b} - \frac {2 \, {\left (3 \, e^{\left (-2 \, b x - 2 \, a\right )} + 6 \, e^{\left (-4 \, b x - 4 \, a\right )} - 2 \, e^{\left (-6 \, b x - 6 \, a\right )} + 6 \, e^{\left (-8 \, b x - 8 \, a\right )} + 3 \, e^{\left (-10 \, b x - 10 \, a\right )}\right )}}{b {\left (2 \, e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (-4 \, b x - 4 \, a\right )} - 4 \, e^{\left (-6 \, b x - 6 \, a\right )} - e^{\left (-8 \, b x - 8 \, a\right )} + 2 \, e^{\left (-10 \, b x - 10 \, a\right )} + e^{\left (-12 \, b x - 12 \, a\right )} + 1\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csch(b*x+a)^3*sech(b*x+a)^5,x, algorithm="maxima")
[Out]
-3*log(e^(-b*x - a) + 1)/b - 3*log(e^(-b*x - a) - 1)/b + 3*log(e^(-2*b*x - 2*a) + 1)/b - 2*(3*e^(-2*b*x - 2*a)
+ 6*e^(-4*b*x - 4*a) - 2*e^(-6*b*x - 6*a) + 6*e^(-8*b*x - 8*a) + 3*e^(-10*b*x - 10*a))/(b*(2*e^(-2*b*x - 2*a)
- e^(-4*b*x - 4*a) - 4*e^(-6*b*x - 6*a) - e^(-8*b*x - 8*a) + 2*e^(-10*b*x - 10*a) + e^(-12*b*x - 12*a) + 1))
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mupad [B] time = 0.08, size = 187, normalized size = 3.22 \[ \frac {6\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}\,\sqrt {-b^2}}{b}\right )}{\sqrt {-b^2}}-\frac {4}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}-\frac {2}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )}-\frac {2}{b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}+\frac {8}{b\,\left (3\,{\mathrm {e}}^{2\,a+2\,b\,x}+3\,{\mathrm {e}}^{4\,a+4\,b\,x}+{\mathrm {e}}^{6\,a+6\,b\,x}+1\right )}-\frac {4}{b\,\left (4\,{\mathrm {e}}^{2\,a+2\,b\,x}+6\,{\mathrm {e}}^{4\,a+4\,b\,x}+4\,{\mathrm {e}}^{6\,a+6\,b\,x}+{\mathrm {e}}^{8\,a+8\,b\,x}+1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(cosh(a + b*x)^5*sinh(a + b*x)^3),x)
[Out]
(6*atan((exp(2*a)*exp(2*b*x)*(-b^2)^(1/2))/b))/(-b^2)^(1/2) - 4/(b*(exp(2*a + 2*b*x) + 1)) - 2/(b*(exp(2*a + 2
*b*x) - 1)) - 2/(b*(exp(4*a + 4*b*x) - 2*exp(2*a + 2*b*x) + 1)) + 8/(b*(3*exp(2*a + 2*b*x) + 3*exp(4*a + 4*b*x
) + exp(6*a + 6*b*x) + 1)) - 4/(b*(4*exp(2*a + 2*b*x) + 6*exp(4*a + 4*b*x) + 4*exp(6*a + 6*b*x) + exp(8*a + 8*
b*x) + 1))
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {csch}^{3}{\left (a + b x \right )} \operatorname {sech}^{5}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csch(b*x+a)**3*sech(b*x+a)**5,x)
[Out]
Integral(csch(a + b*x)**3*sech(a + b*x)**5, x)
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