### 3.543 $$\int \frac{x \sinh (a+b x)}{\text{sech}^{\frac{5}{2}}(a+b x)} \, dx$$

Optimal. Leaf size=107 $\frac{20 i \sqrt{\cosh (a+b x)} \sqrt{\text{sech}(a+b x)} \text{EllipticF}\left (\frac{1}{2} i (a+b x),2\right )}{147 b^2}-\frac{4 \sinh (a+b x)}{49 b^2 \text{sech}^{\frac{5}{2}}(a+b x)}-\frac{20 \sinh (a+b x)}{147 b^2 \sqrt{\text{sech}(a+b x)}}+\frac{2 x}{7 b \text{sech}^{\frac{7}{2}}(a+b x)}$

[Out]

(2*x)/(7*b*Sech[a + b*x]^(7/2)) + (((20*I)/147)*Sqrt[Cosh[a + b*x]]*EllipticF[(I/2)*(a + b*x), 2]*Sqrt[Sech[a
+ b*x]])/b^2 - (4*Sinh[a + b*x])/(49*b^2*Sech[a + b*x]^(5/2)) - (20*Sinh[a + b*x])/(147*b^2*Sqrt[Sech[a + b*x]
])

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Rubi [A]  time = 0.069102, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.222, Rules used = {5444, 3769, 3771, 2641} $-\frac{4 \sinh (a+b x)}{49 b^2 \text{sech}^{\frac{5}{2}}(a+b x)}-\frac{20 \sinh (a+b x)}{147 b^2 \sqrt{\text{sech}(a+b x)}}+\frac{20 i \sqrt{\cosh (a+b x)} \sqrt{\text{sech}(a+b x)} F\left (\left .\frac{1}{2} i (a+b x)\right |2\right )}{147 b^2}+\frac{2 x}{7 b \text{sech}^{\frac{7}{2}}(a+b x)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(x*Sinh[a + b*x])/Sech[a + b*x]^(5/2),x]

[Out]

(2*x)/(7*b*Sech[a + b*x]^(7/2)) + (((20*I)/147)*Sqrt[Cosh[a + b*x]]*EllipticF[(I/2)*(a + b*x), 2]*Sqrt[Sech[a
+ b*x]])/b^2 - (4*Sinh[a + b*x])/(49*b^2*Sech[a + b*x]^(5/2)) - (20*Sinh[a + b*x])/(147*b^2*Sqrt[Sech[a + b*x]
])

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 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[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 \frac{x \sinh (a+b x)}{\text{sech}^{\frac{5}{2}}(a+b x)} \, dx &=\frac{2 x}{7 b \text{sech}^{\frac{7}{2}}(a+b x)}-\frac{2 \int \frac{1}{\text{sech}^{\frac{7}{2}}(a+b x)} \, dx}{7 b}\\ &=\frac{2 x}{7 b \text{sech}^{\frac{7}{2}}(a+b x)}-\frac{4 \sinh (a+b x)}{49 b^2 \text{sech}^{\frac{5}{2}}(a+b x)}-\frac{10 \int \frac{1}{\text{sech}^{\frac{3}{2}}(a+b x)} \, dx}{49 b}\\ &=\frac{2 x}{7 b \text{sech}^{\frac{7}{2}}(a+b x)}-\frac{4 \sinh (a+b x)}{49 b^2 \text{sech}^{\frac{5}{2}}(a+b x)}-\frac{20 \sinh (a+b x)}{147 b^2 \sqrt{\text{sech}(a+b x)}}-\frac{10 \int \sqrt{\text{sech}(a+b x)} \, dx}{147 b}\\ &=\frac{2 x}{7 b \text{sech}^{\frac{7}{2}}(a+b x)}-\frac{4 \sinh (a+b x)}{49 b^2 \text{sech}^{\frac{5}{2}}(a+b x)}-\frac{20 \sinh (a+b x)}{147 b^2 \sqrt{\text{sech}(a+b x)}}-\frac{\left (10 \sqrt{\cosh (a+b x)} \sqrt{\text{sech}(a+b x)}\right ) \int \frac{1}{\sqrt{\cosh (a+b x)}} \, dx}{147 b}\\ &=\frac{2 x}{7 b \text{sech}^{\frac{7}{2}}(a+b x)}+\frac{20 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)}}{147 b^2}-\frac{4 \sinh (a+b x)}{49 b^2 \text{sech}^{\frac{5}{2}}(a+b x)}-\frac{20 \sinh (a+b x)}{147 b^2 \sqrt{\text{sech}(a+b x)}}\\ \end{align*}

Mathematica [A]  time = 0.273311, size = 93, normalized size = 0.87 $\frac{\sqrt{\text{sech}(a+b x)} \left (80 i \sqrt{\cosh (a+b x)} \text{EllipticF}\left (\frac{1}{2} i (a+b x),2\right )-52 \sinh (2 (a+b x))-6 \sinh (4 (a+b x))+84 b x \cosh (2 (a+b x))+21 b x \cosh (4 (a+b x))+63 b x\right )}{588 b^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(x*Sinh[a + b*x])/Sech[a + b*x]^(5/2),x]

[Out]

(Sqrt[Sech[a + b*x]]*(63*b*x + 84*b*x*Cosh[2*(a + b*x)] + 21*b*x*Cosh[4*(a + b*x)] + (80*I)*Sqrt[Cosh[a + b*x]
]*EllipticF[(I/2)*(a + b*x), 2] - 52*Sinh[2*(a + b*x)] - 6*Sinh[4*(a + b*x)]))/(588*b^2)

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Maple [F]  time = 0.032, size = 0, normalized size = 0. \begin{align*} \int{x\sinh \left ( bx+a \right ) \left ({\rm sech} \left (bx+a\right ) \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x*sinh(b*x+a)/sech(b*x+a)^(5/2),x)

[Out]

int(x*sinh(b*x+a)/sech(b*x+a)^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \sinh \left (b x + a\right )}{\operatorname{sech}\left (b x + a\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sinh(b*x+a)/sech(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

integrate(x*sinh(b*x + a)/sech(b*x + a)^(5/2), x)

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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*sinh(b*x+a)/sech(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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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*sinh(b*x+a)/sech(b*x+a)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \sinh \left (b x + a\right )}{\operatorname{sech}\left (b x + a\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sinh(b*x+a)/sech(b*x+a)^(5/2),x, algorithm="giac")

[Out]

integrate(x*sinh(b*x + a)/sech(b*x + a)^(5/2), x)