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

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

[Out]

2/7*x/b/sech(b*x+a)^(7/2)-4/49*sinh(b*x+a)/b^2/sech(b*x+a)^(5/2)-20/147*sinh(b*x+a)/b^2/sech(b*x+a)^(1/2)+20/1
47*I*(cosh(1/2*a+1/2*b*x)^2)^(1/2)/cosh(1/2*a+1/2*b*x)*EllipticF(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 = 0.07, antiderivative size = 107, normalized size of antiderivative = 1.00, 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 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]

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 {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*}

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Mathematica [A]  time = 0.29, size = 93, normalized size = 0.87 \[ \frac {\sqrt {\text {sech}(a+b x)} \left (-52 \sinh (2 (a+b x))-6 \sinh (4 (a+b x))+84 b x \cosh (2 (a+b x))+21 b x \cosh (4 (a+b x))+80 i \sqrt {\cosh (a+b x)} F\left (\left .\frac {1}{2} i (a+b x)\right |2\right )+63 b x\right )}{588 b^2} \]

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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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)^(5/2),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 \[ \int \frac {x \sinh \left (b x + a\right )}{\operatorname {sech}\left (b x + a\right )^{\frac {5}{2}}}\,{d x} \]

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)

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maple [F]  time = 0.17, size = 0, normalized size = 0.00 \[ \int \frac {x \sinh \left (b x +a \right )}{\mathrm {sech}\left (b x +a \right )^{\frac {5}{2}}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \sinh \left (b x + a\right )}{\operatorname {sech}\left (b x + a\right )^{\frac {5}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,\mathrm {sinh}\left (a+b\,x\right )}{{\left (\frac {1}{\mathrm {cosh}\left (a+b\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*sinh(a + b*x))/(1/cosh(a + b*x))^(5/2),x)

[Out]

int((x*sinh(a + b*x))/(1/cosh(a + b*x))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \sinh {\left (a + b x \right )}}{\operatorname {sech}^{\frac {5}{2}}{\left (a + b x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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