3.22 \(\int \frac{1}{(b \text{sech}(c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=104 \[ -\frac{10 i \sqrt{\cosh (c+d x)} \text{EllipticF}\left (\frac{1}{2} i (c+d x),2\right ) \sqrt{b \text{sech}(c+d x)}}{21 b^4 d}+\frac{10 \sinh (c+d x)}{21 b^3 d \sqrt{b \text{sech}(c+d x)}}+\frac{2 \sinh (c+d x)}{7 b d (b \text{sech}(c+d x))^{5/2}} \]

[Out]

(((-10*I)/21)*Sqrt[Cosh[c + d*x]]*EllipticF[(I/2)*(c + d*x), 2]*Sqrt[b*Sech[c + d*x]])/(b^4*d) + (2*Sinh[c + d
*x])/(7*b*d*(b*Sech[c + d*x])^(5/2)) + (10*Sinh[c + d*x])/(21*b^3*d*Sqrt[b*Sech[c + d*x]])

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Rubi [A]  time = 0.0581648, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3769, 3771, 2641} \[ \frac{10 \sinh (c+d x)}{21 b^3 d \sqrt{b \text{sech}(c+d x)}}-\frac{10 i \sqrt{\cosh (c+d x)} F\left (\left .\frac{1}{2} i (c+d x)\right |2\right ) \sqrt{b \text{sech}(c+d x)}}{21 b^4 d}+\frac{2 \sinh (c+d x)}{7 b d (b \text{sech}(c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sech[c + d*x])^(-7/2),x]

[Out]

(((-10*I)/21)*Sqrt[Cosh[c + d*x]]*EllipticF[(I/2)*(c + d*x), 2]*Sqrt[b*Sech[c + d*x]])/(b^4*d) + (2*Sinh[c + d
*x])/(7*b*d*(b*Sech[c + d*x])^(5/2)) + (10*Sinh[c + d*x])/(21*b^3*d*Sqrt[b*Sech[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 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{1}{(b \text{sech}(c+d x))^{7/2}} \, dx &=\frac{2 \sinh (c+d x)}{7 b d (b \text{sech}(c+d x))^{5/2}}+\frac{5 \int \frac{1}{(b \text{sech}(c+d x))^{3/2}} \, dx}{7 b^2}\\ &=\frac{2 \sinh (c+d x)}{7 b d (b \text{sech}(c+d x))^{5/2}}+\frac{10 \sinh (c+d x)}{21 b^3 d \sqrt{b \text{sech}(c+d x)}}+\frac{5 \int \sqrt{b \text{sech}(c+d x)} \, dx}{21 b^4}\\ &=\frac{2 \sinh (c+d x)}{7 b d (b \text{sech}(c+d x))^{5/2}}+\frac{10 \sinh (c+d x)}{21 b^3 d \sqrt{b \text{sech}(c+d x)}}+\frac{\left (5 \sqrt{\cosh (c+d x)} \sqrt{b \text{sech}(c+d x)}\right ) \int \frac{1}{\sqrt{\cosh (c+d x)}} \, dx}{21 b^4}\\ &=-\frac{10 i \sqrt{\cosh (c+d x)} F\left (\left .\frac{1}{2} i (c+d x)\right |2\right ) \sqrt{b \text{sech}(c+d x)}}{21 b^4 d}+\frac{2 \sinh (c+d x)}{7 b d (b \text{sech}(c+d x))^{5/2}}+\frac{10 \sinh (c+d x)}{21 b^3 d \sqrt{b \text{sech}(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.126529, size = 70, normalized size = 0.67 \[ \frac{\sqrt{b \text{sech}(c+d x)} \left (-40 i \sqrt{\cosh (c+d x)} \text{EllipticF}\left (\frac{1}{2} i (c+d x),2\right )+26 \sinh (2 (c+d x))+3 \sinh (4 (c+d x))\right )}{84 b^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Sech[c + d*x])^(-7/2),x]

[Out]

(Sqrt[b*Sech[c + d*x]]*((-40*I)*Sqrt[Cosh[c + d*x]]*EllipticF[(I/2)*(c + d*x), 2] + 26*Sinh[2*(c + d*x)] + 3*S
inh[4*(c + d*x)]))/(84*b^4*d)

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Maple [F]  time = 0.082, size = 0, normalized size = 0. \begin{align*} \int \left ( b{\rm sech} \left (dx+c\right ) \right ) ^{-{\frac{7}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*sech(d*x+c))^(7/2),x)

[Out]

int(1/(b*sech(d*x+c))^(7/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \operatorname{sech}\left (d x + c\right )\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sech(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate((b*sech(d*x + c))^(-7/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \operatorname{sech}\left (d x + c\right )}}{b^{4} \operatorname{sech}\left (d x + c\right )^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sech(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sech(d*x + c))/(b^4*sech(d*x + c)^4), x)

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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(1/(b*sech(d*x+c))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \operatorname{sech}\left (d x + c\right )\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sech(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((b*sech(d*x + c))^(-7/2), x)