### 3.542 $$\int \frac{x \sinh (a+b x)}{\text{sech}^{\frac{3}{2}}(a+b x)} \, dx$$

Optimal. Leaf size=84 $-\frac{4 \sinh (a+b x)}{25 b^2 \text{sech}^{\frac{3}{2}}(a+b x)}+\frac{12 i \sqrt{\cosh (a+b x)} \sqrt{\text{sech}(a+b x)} E\left (\left .\frac{1}{2} i (a+b x)\right |2\right )}{25 b^2}+\frac{2 x}{5 b \text{sech}^{\frac{5}{2}}(a+b x)}$

[Out]

(2*x)/(5*b*Sech[a + b*x]^(5/2)) + (((12*I)/25)*Sqrt[Cosh[a + b*x]]*EllipticE[(I/2)*(a + b*x), 2]*Sqrt[Sech[a +
b*x]])/b^2 - (4*Sinh[a + b*x])/(25*b^2*Sech[a + b*x]^(3/2))

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Rubi [A]  time = 0.0534484, 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, 3769, 3771, 2639} $-\frac{4 \sinh (a+b x)}{25 b^2 \text{sech}^{\frac{3}{2}}(a+b x)}+\frac{12 i \sqrt{\cosh (a+b x)} \sqrt{\text{sech}(a+b x)} E\left (\left .\frac{1}{2} i (a+b x)\right |2\right )}{25 b^2}+\frac{2 x}{5 b \text{sech}^{\frac{5}{2}}(a+b x)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(x*Sinh[a + b*x])/Sech[a + b*x]^(3/2),x]

[Out]

(2*x)/(5*b*Sech[a + b*x]^(5/2)) + (((12*I)/25)*Sqrt[Cosh[a + b*x]]*EllipticE[(I/2)*(a + b*x), 2]*Sqrt[Sech[a +
b*x]])/b^2 - (4*Sinh[a + b*x])/(25*b^2*Sech[a + b*x]^(3/2))

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 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 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(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{3}{2}}(a+b x)} \, dx &=\frac{2 x}{5 b \text{sech}^{\frac{5}{2}}(a+b x)}-\frac{2 \int \frac{1}{\text{sech}^{\frac{5}{2}}(a+b x)} \, dx}{5 b}\\ &=\frac{2 x}{5 b \text{sech}^{\frac{5}{2}}(a+b x)}-\frac{4 \sinh (a+b x)}{25 b^2 \text{sech}^{\frac{3}{2}}(a+b x)}-\frac{6 \int \frac{1}{\sqrt{\text{sech}(a+b x)}} \, dx}{25 b}\\ &=\frac{2 x}{5 b \text{sech}^{\frac{5}{2}}(a+b x)}-\frac{4 \sinh (a+b x)}{25 b^2 \text{sech}^{\frac{3}{2}}(a+b x)}-\frac{\left (6 \sqrt{\cosh (a+b x)} \sqrt{\text{sech}(a+b x)}\right ) \int \sqrt{\cosh (a+b x)} \, dx}{25 b}\\ &=\frac{2 x}{5 b \text{sech}^{\frac{5}{2}}(a+b x)}+\frac{12 i \sqrt{\cosh (a+b x)} E\left (\left .\frac{1}{2} i (a+b x)\right |2\right ) \sqrt{\text{sech}(a+b x)}}{25 b^2}-\frac{4 \sinh (a+b x)}{25 b^2 \text{sech}^{\frac{3}{2}}(a+b x)}\\ \end{align*}

Mathematica [C]  time = 2.13397, size = 125, normalized size = 1.49 $\frac{e^{-3 (a+b x)} \left (48 e^{2 (a+b x)} \sqrt{e^{2 (a+b x)}+1} \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-e^{2 (a+b x)}\right )+\left (e^{2 (a+b x)}+1\right ) \left (2 (5 b x-12) e^{2 (a+b x)}+(5 b x-2) e^{4 (a+b x)}+5 b x+2\right )\right ) \sqrt{\text{sech}(a+b x)}}{100 b^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(x*Sinh[a + b*x])/Sech[a + b*x]^(3/2),x]

[Out]

(((1 + E^(2*(a + b*x)))*(2 + 5*b*x + 2*E^(2*(a + b*x))*(-12 + 5*b*x) + E^(4*(a + b*x))*(-2 + 5*b*x)) + 48*E^(2
*(a + b*x))*Sqrt[1 + E^(2*(a + b*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^(2*(a + b*x))])*Sqrt[Sech[a + b*x]]
)/(100*b^2*E^(3*(a + b*x)))

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Maple [F]  time = 0.033, size = 0, normalized size = 0. \begin{align*} \int{x\sinh \left ( bx+a \right ) \left ({\rm sech} \left (bx+a\right ) \right ) ^{-{\frac{3}{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)^(3/2),x)

[Out]

int(x*sinh(b*x+a)/sech(b*x+a)^(3/2),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{3}{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)^(3/2),x, algorithm="maxima")

[Out]

integrate(x*sinh(b*x + a)/sech(b*x + a)^(3/2), 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)^(3/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \sinh{\left (a + b x \right )}}{\operatorname{sech}^{\frac{3}{2}}{\left (a + b x \right )}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sinh(b*x+a)/sech(b*x+a)**(3/2),x)

[Out]

Integral(x*sinh(a + b*x)/sech(a + b*x)**(3/2), x)

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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{3}{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)^(3/2),x, algorithm="giac")

[Out]

integrate(x*sinh(b*x + a)/sech(b*x + a)^(3/2), x)