3.537 $$\int x \text{sech}^{\frac{7}{2}}(a+b x) \sinh (a+b x) \, dx$$

Optimal. Leaf size=84 $-\frac{4 i \sqrt{\cosh (a+b x)} \sqrt{\text{sech}(a+b x)} \text{EllipticF}\left (\frac{1}{2} i (a+b x),2\right )}{15 b^2}+\frac{4 \sinh (a+b x) \text{sech}^{\frac{3}{2}}(a+b x)}{15 b^2}-\frac{2 x \text{sech}^{\frac{5}{2}}(a+b x)}{5 b}$

[Out]

(((-4*I)/15)*Sqrt[Cosh[a + b*x]]*EllipticF[(I/2)*(a + b*x), 2]*Sqrt[Sech[a + b*x]])/b^2 - (2*x*Sech[a + b*x]^(
5/2))/(5*b) + (4*Sech[a + b*x]^(3/2)*Sinh[a + b*x])/(15*b^2)

________________________________________________________________________________________

Rubi [A]  time = 0.0522645, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.222, Rules used = {5444, 3768, 3771, 2641} $\frac{4 \sinh (a+b x) \text{sech}^{\frac{3}{2}}(a+b x)}{15 b^2}-\frac{4 i \sqrt{\cosh (a+b x)} \sqrt{\text{sech}(a+b x)} F\left (\left .\frac{1}{2} i (a+b x)\right |2\right )}{15 b^2}-\frac{2 x \text{sech}^{\frac{5}{2}}(a+b x)}{5 b}$

Antiderivative was successfully veriﬁed.

[In]

Int[x*Sech[a + b*x]^(7/2)*Sinh[a + b*x],x]

[Out]

(((-4*I)/15)*Sqrt[Cosh[a + b*x]]*EllipticF[(I/2)*(a + b*x), 2]*Sqrt[Sech[a + b*x]])/b^2 - (2*x*Sech[a + b*x]^(
5/2))/(5*b) + (4*Sech[a + b*x]^(3/2)*Sinh[a + b*x])/(15*b^2)

Rule 5444

Int[(x_)^(m_.)*Sech[(a_.) + (b_.)*(x_)^(n_.)]^(p_)*Sinh[(a_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> -Simp[(x^(m -
n + 1)*Sech[a + b*x^n]^(p - 1))/(b*n*(p - 1)), x] + Dist[(m - n + 1)/(b*n*(p - 1)), Int[x^(m - n)*Sech[a + b*x
^n]^(p - 1), x], x] /; FreeQ[{a, b, p}, x] && IntegerQ[n] && GeQ[m - n, 0] && NeQ[p, 1]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int x \text{sech}^{\frac{7}{2}}(a+b x) \sinh (a+b x) \, dx &=-\frac{2 x \text{sech}^{\frac{5}{2}}(a+b x)}{5 b}+\frac{2 \int \text{sech}^{\frac{5}{2}}(a+b x) \, dx}{5 b}\\ &=-\frac{2 x \text{sech}^{\frac{5}{2}}(a+b x)}{5 b}+\frac{4 \text{sech}^{\frac{3}{2}}(a+b x) \sinh (a+b x)}{15 b^2}+\frac{2 \int \sqrt{\text{sech}(a+b x)} \, dx}{15 b}\\ &=-\frac{2 x \text{sech}^{\frac{5}{2}}(a+b x)}{5 b}+\frac{4 \text{sech}^{\frac{3}{2}}(a+b x) \sinh (a+b x)}{15 b^2}+\frac{\left (2 \sqrt{\cosh (a+b x)} \sqrt{\text{sech}(a+b x)}\right ) \int \frac{1}{\sqrt{\cosh (a+b x)}} \, dx}{15 b}\\ &=-\frac{4 i \sqrt{\cosh (a+b x)} F\left (\left .\frac{1}{2} i (a+b x)\right |2\right ) \sqrt{\text{sech}(a+b x)}}{15 b^2}-\frac{2 x \text{sech}^{\frac{5}{2}}(a+b x)}{5 b}+\frac{4 \text{sech}^{\frac{3}{2}}(a+b x) \sinh (a+b x)}{15 b^2}\\ \end{align*}

Mathematica [A]  time = 0.240138, size = 65, normalized size = 0.77 $-\frac{2 \sqrt{\text{sech}(a+b x)} \left (2 i \sqrt{\cosh (a+b x)} \text{EllipticF}\left (\frac{1}{2} i (a+b x),2\right )-2 \tanh (a+b x)+3 b x \text{sech}^2(a+b x)\right )}{15 b^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x*Sech[a + b*x]^(7/2)*Sinh[a + b*x],x]

[Out]

(-2*Sqrt[Sech[a + b*x]]*((2*I)*Sqrt[Cosh[a + b*x]]*EllipticF[(I/2)*(a + b*x), 2] + 3*b*x*Sech[a + b*x]^2 - 2*T
anh[a + b*x]))/(15*b^2)

________________________________________________________________________________________

Maple [F]  time = 0.033, size = 0, normalized size = 0. \begin{align*} \int x \left ({\rm sech} \left (bx+a\right ) \right ) ^{{\frac{7}{2}}}\sinh \left ( bx+a \right ) \, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x*sech(b*x+a)^(7/2)*sinh(b*x+a),x)

[Out]

int(x*sech(b*x+a)^(7/2)*sinh(b*x+a),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{sech}\left (b x + a\right )^{\frac{7}{2}} \sinh \left (b x + a\right )\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sech(b*x+a)^(7/2)*sinh(b*x+a),x, algorithm="maxima")

[Out]

integrate(x*sech(b*x + a)^(7/2)*sinh(b*x + a), x)

________________________________________________________________________________________

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sech(b*x+a)^(7/2)*sinh(b*x+a),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sech(b*x+a)**(7/2)*sinh(b*x+a),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{sech}\left (b x + a\right )^{\frac{7}{2}} \sinh \left (b x + a\right )\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sech(b*x+a)^(7/2)*sinh(b*x+a),x, algorithm="giac")

[Out]

integrate(x*sech(b*x + a)^(7/2)*sinh(b*x + a), x)