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

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Rubi [A]  time = 0.0485654, antiderivative size = 36, normalized size of antiderivative = 1., 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 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])

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]]

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*}

Mathematica [A]  time = 0.122265, 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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Maple [C]  time = 0.134, size = 275, normalized size = 7.6 \begin{align*} -{\frac{n{{\rm e}^{4\,bx+4\,a}}+2\,{{\rm e}^{4\,bx+4\,a}}-2\,n{{\rm e}^{2\,bx+2\,a}}+4\,{{\rm e}^{2\,bx+2\,a}}+n+2}{ \left ( n+2 \right ) bn \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) ^{2}}{{\rm e}^{-{\frac{n}{2} \left ( i\pi \, \left ({\it csgn} \left ({\frac{i{{\rm e}^{bx+a}}}{1+{{\rm e}^{2\,bx+2\,a}}}} \right ) \right ) ^{3}-i\pi \, \left ({\it csgn} \left ({\frac{i{{\rm e}^{bx+a}}}{1+{{\rm e}^{2\,bx+2\,a}}}} \right ) \right ) ^{2}{\it csgn} \left ( i{{\rm e}^{bx+a}} \right ) -i\pi \, \left ({\it csgn} \left ({\frac{i{{\rm e}^{bx+a}}}{1+{{\rm e}^{2\,bx+2\,a}}}} \right ) \right ) ^{2}{\it csgn} \left ({\frac{i}{1+{{\rm e}^{2\,bx+2\,a}}}} \right ) +i\pi \,{\it csgn} \left ({\frac{i{{\rm e}^{bx+a}}}{1+{{\rm e}^{2\,bx+2\,a}}}} \right ){\it csgn} \left ( i{{\rm e}^{bx+a}} \right ){\it csgn} \left ({\frac{i}{1+{{\rm e}^{2\,bx+2\,a}}}} \right ) -2\,\ln \left ({{\rm e}^{bx+a}} \right ) +2\,\ln \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) -2\,\ln \left ( 2 \right ) \right ) }}}} \end{align*}

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*n*exp(2*b*x+2*a)+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(exp(b*x+a))+2*ln(1+exp(2*b*x+2*a))-2*ln(2)))

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Maxima [B]  time = 2.01708, size = 466, normalized size = 12.94 \begin{align*} -\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} \end{align*}

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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Fricas [B]  time = 1.85963, size = 605, normalized size = 16.81 \begin{align*} -\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} \end{align*}

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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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{sech}\left (b x + a\right )^{n} \tanh \left (b x + a\right )^{3}\,{d x} \end{align*}

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)