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]

2/5*x/b/sech(b*x+a)^(5/2)-4/25*sinh(b*x+a)/b^2/sech(b*x+a)^(3/2)+12/25*I*(cosh(1/2*a+1/2*b*x)^2)^(1/2)/cosh(1/
2*a+1/2*b*x)*EllipticE(I*sinh(1/2*a+1/2*b*x),2^(1/2))*cosh(b*x+a)^(1/2)*sech(b*x+a)^(1/2)/b^2

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Rubi [A]  time = 0.05, antiderivative size = 84, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [C]  time = 2.43, 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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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \sinh \left (b x + a\right )}{\operatorname {sech}\left (b x + a\right )^{\frac {3}{2}}}\,{d x} \]

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)

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maple [F]  time = 0.18, size = 0, normalized size = 0.00 \[ \int \frac {x \sinh \left (b x +a \right )}{\mathrm {sech}\left (b x +a \right )^{\frac {3}{2}}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \sinh \left (b x + a\right )}{\operatorname {sech}\left (b x + a\right )^{\frac {3}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,\mathrm {sinh}\left (a+b\,x\right )}{{\left (\frac {1}{\mathrm {cosh}\left (a+b\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*sinh(a + b*x))/(1/cosh(a + b*x))^(3/2),x)

[Out]

int((x*sinh(a + b*x))/(1/cosh(a + b*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \sinh {\left (a + b x \right )}}{\operatorname {sech}^{\frac {3}{2}}{\left (a + b x \right )}}\, dx \]

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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