3.89 \(\int \text {sech}^{3+n}(a+b x) \sinh ^3(a+b x) \, dx\)

Optimal. Leaf size=36 \[ \frac {\text {sech}^{n+2}(a+b x)}{b (n+2)}-\frac {\text {sech}^n(a+b x)}{b n} \]

[Out]

-sech(b*x+a)^n/b/n+sech(b*x+a)^(2+n)/b/(2+n)

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Rubi [A]  time = 0.05, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2622, 14} \[ \frac {\text {sech}^{n+2}(a+b x)}{b (n+2)}-\frac {\text {sech}^n(a+b x)}{b n} \]

Antiderivative was successfully verified.

[In]

Int[Sech[a + b*x]^(3 + n)*Sinh[a + b*x]^3,x]

[Out]

-(Sech[a + b*x]^n/(b*n)) + Sech[a + b*x]^(2 + n)/(b*(2 + n))

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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

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Mathematica [A]  time = 0.13, size = 32, normalized size = 0.89 \[ \frac {\text {sech}^n(a+b x) \left (\frac {\text {sech}^2(a+b x)}{n+2}-\frac {1}{n}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[a + b*x]^(3 + n)*Sinh[a + b*x]^3,x]

[Out]

(Sech[a + b*x]^n*(-n^(-1) + Sech[a + b*x]^2/(2 + n)))/b

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fricas [B]  time = 0.45, size = 219, normalized size = 6.08 \[ -\frac {{\left ({\left (n + 2\right )} \cosh \left (b x + a\right )^{2} + {\left (n + 2\right )} \sinh \left (b x + a\right )^{2} - n + 2\right )} \cosh \left (n \log \left (\frac {2 \, {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}}{\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1}\right )\right ) + {\left ({\left (n + 2\right )} \cosh \left (b x + a\right )^{2} + {\left (n + 2\right )} \sinh \left (b x + a\right )^{2} - n + 2\right )} \sinh \left (n \log \left (\frac {2 \, {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}}{\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1}\right )\right )}{b n^{2} + {\left (b n^{2} + 2 \, b n\right )} \cosh \left (b x + a\right )^{2} + {\left (b n^{2} + 2 \, b n\right )} \sinh \left (b x + a\right )^{2} + 2 \, b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(((n + 2)*cosh(b*x + a)^2 + (n + 2)*sinh(b*x + a)^2 - n + 2)*cosh(n*log(2*(cosh(b*x + a) + sinh(b*x + a))/(co
sh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 + 1))) + ((n + 2)*cosh(b*x + a)^2 + (n + 2)*si
nh(b*x + a)^2 - n + 2)*sinh(n*log(2*(cosh(b*x + a) + sinh(b*x + a))/(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*
x + a) + sinh(b*x + a)^2 + 1))))/(b*n^2 + (b*n^2 + 2*b*n)*cosh(b*x + a)^2 + (b*n^2 + 2*b*n)*sinh(b*x + a)^2 +
2*b*n)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {sech}\left (b x + a\right )^{n} \tanh \left (b x + a\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sech(b*x + a)^n*tanh(b*x + a)^3, x)

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maple [C]  time = 0.45, size = 275, normalized size = 7.64 \[ -\frac {\left (n \,{\mathrm e}^{4 b x +4 a}+2 \,{\mathrm e}^{4 b x +4 a}-2 \,{\mathrm e}^{2 b x +2 a} n +4 \,{\mathrm e}^{2 b x +2 a}+n +2\right ) {\mathrm e}^{\frac {n \left (-i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{b x +a}}{1+{\mathrm e}^{2 b x +2 a}}\right )^{3}+i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{b x +a}}{1+{\mathrm e}^{2 b x +2 a}}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )+i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{b x +a}}{1+{\mathrm e}^{2 b x +2 a}}\right )^{2} \mathrm {csgn}\left (\frac {i}{1+{\mathrm e}^{2 b x +2 a}}\right )-i \pi \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{b x +a}}{1+{\mathrm e}^{2 b x +2 a}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (\frac {i}{1+{\mathrm e}^{2 b x +2 a}}\right )-2 \ln \left (1+{\mathrm e}^{2 b x +2 a}\right )+2 \ln \relax (2)+2 \ln \left ({\mathrm e}^{b x +a}\right )\right )}{2}}}{b n \left (n +2\right ) \left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(n*exp(4*b*x+4*a)+2*exp(4*b*x+4*a)-2*exp(2*b*x+2*a)*n+4*exp(2*b*x+2*a)+n+2)/b/n/(n+2)/(1+exp(2*b*x+2*a))^2*ex
p(1/2*n*(-I*Pi*csgn(I*exp(b*x+a)/(1+exp(2*b*x+2*a)))^3+I*Pi*csgn(I*exp(b*x+a)/(1+exp(2*b*x+2*a)))^2*csgn(I*exp
(b*x+a))+I*Pi*csgn(I*exp(b*x+a)/(1+exp(2*b*x+2*a)))^2*csgn(I/(1+exp(2*b*x+2*a)))-I*Pi*csgn(I*exp(b*x+a)/(1+exp
(2*b*x+2*a)))*csgn(I*exp(b*x+a))*csgn(I/(1+exp(2*b*x+2*a)))-2*ln(1+exp(2*b*x+2*a))+2*ln(2)+2*ln(exp(b*x+a))))

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maxima [B]  time = 0.54, size = 345, normalized size = 9.58 \[ -\frac {2^{n} n e^{\left (-{\left (b x + a\right )} n - n \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )\right )}}{{\left (n^{2} + 2 \, {\left (n^{2} + 2 \, n\right )} e^{\left (-2 \, b x - 2 \, a\right )} + {\left (n^{2} + 2 \, n\right )} e^{\left (-4 \, b x - 4 \, a\right )} + 2 \, n\right )} b} + \frac {{\left (2^{n + 1} n - 2^{n + 2}\right )} e^{\left (-{\left (b x + a\right )} n - 2 \, b x - n \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right ) - 2 \, a\right )}}{{\left (n^{2} + 2 \, {\left (n^{2} + 2 \, n\right )} e^{\left (-2 \, b x - 2 \, a\right )} + {\left (n^{2} + 2 \, n\right )} e^{\left (-4 \, b x - 4 \, a\right )} + 2 \, n\right )} b} - \frac {{\left (2^{n} n + 2^{n + 1}\right )} e^{\left (-{\left (b x + a\right )} n - 4 \, b x - n \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right ) - 4 \, a\right )}}{{\left (n^{2} + 2 \, {\left (n^{2} + 2 \, n\right )} e^{\left (-2 \, b x - 2 \, a\right )} + {\left (n^{2} + 2 \, n\right )} e^{\left (-4 \, b x - 4 \, a\right )} + 2 \, n\right )} b} - \frac {2^{n + 1} e^{\left (-{\left (b x + a\right )} n - n \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )\right )}}{{\left (n^{2} + 2 \, {\left (n^{2} + 2 \, n\right )} e^{\left (-2 \, b x - 2 \, a\right )} + {\left (n^{2} + 2 \, n\right )} e^{\left (-4 \, b x - 4 \, a\right )} + 2 \, n\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.57, size = 101, normalized size = 2.81 \[ -\frac {{\left (\frac {1}{\frac {{\mathrm {e}}^{a+b\,x}}{2}+\frac {{\mathrm {e}}^{-a-b\,x}}{2}}\right )}^n\,\left (\frac {1}{b\,n}+\frac {{\mathrm {e}}^{4\,a+4\,b\,x}}{b\,n}-\frac {{\mathrm {e}}^{2\,a+2\,b\,x}\,\left (2\,n-4\right )}{b\,n\,\left (n+2\right )}\right )}{2\,{\mathrm {e}}^{2\,a+2\,b\,x}+{\mathrm {e}}^{4\,a+4\,b\,x}+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \begin {cases} x \tanh ^{3}{\relax (a )} \operatorname {sech}^{n}{\relax (a )} & \text {for}\: b = 0 \\\int \frac {\tanh ^{3}{\left (a + b x \right )}}{\operatorname {sech}^{2}{\left (a + b x \right )}}\, dx & \text {for}\: n = -2 \\x - \frac {\log {\left (\tanh {\left (a + b x \right )} + 1 \right )}}{b} - \frac {\tanh ^{2}{\left (a + b x \right )}}{2 b} & \text {for}\: n = 0 \\- \frac {n \tanh ^{2}{\left (a + b x \right )} \operatorname {sech}^{n}{\left (a + b x \right )}}{b n^{2} + 2 b n} - \frac {2 \operatorname {sech}^{n}{\left (a + b x \right )}}{b n^{2} + 2 b n} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)**n*tanh(b*x+a)**3,x)

[Out]

Piecewise((x*tanh(a)**3*sech(a)**n, Eq(b, 0)), (Integral(tanh(a + b*x)**3/sech(a + b*x)**2, x), Eq(n, -2)), (x
- log(tanh(a + b*x) + 1)/b - tanh(a + b*x)**2/(2*b), Eq(n, 0)), (-n*tanh(a + b*x)**2*sech(a + b*x)**n/(b*n**2
+ 2*b*n) - 2*sech(a + b*x)**n/(b*n**2 + 2*b*n), True))

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