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

Optimal. Leaf size=102 \[ \frac {6 i b^4 E\left (\left .\frac {1}{2} i (c+d x)\right |2\right )}{5 d \sqrt {\cosh (c+d x)} \sqrt {b \text {sech}(c+d x)}}+\frac {6 b^3 \sinh (c+d x) \sqrt {b \text {sech}(c+d x)}}{5 d}+\frac {2 b \sinh (c+d x) (b \text {sech}(c+d x))^{5/2}}{5 d} \]

[Out]

2/5*b*(b*sech(d*x+c))^(5/2)*sinh(d*x+c)/d+6/5*I*b^4*(cosh(1/2*d*x+1/2*c)^2)^(1/2)/cosh(1/2*d*x+1/2*c)*Elliptic
E(I*sinh(1/2*d*x+1/2*c),2^(1/2))/d/cosh(d*x+c)^(1/2)/(b*sech(d*x+c))^(1/2)+6/5*b^3*sinh(d*x+c)*(b*sech(d*x+c))
^(1/2)/d

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Rubi [A]  time = 0.06, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3768, 3771, 2639} \[ \frac {6 b^3 \sinh (c+d x) \sqrt {b \text {sech}(c+d x)}}{5 d}+\frac {6 i b^4 E\left (\left .\frac {1}{2} i (c+d x)\right |2\right )}{5 d \sqrt {\cosh (c+d x)} \sqrt {b \text {sech}(c+d x)}}+\frac {2 b \sinh (c+d x) (b \text {sech}(c+d x))^{5/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sech[c + d*x])^(7/2),x]

[Out]

(((6*I)/5)*b^4*EllipticE[(I/2)*(c + d*x), 2])/(d*Sqrt[Cosh[c + d*x]]*Sqrt[b*Sech[c + d*x]]) + (6*b^3*Sqrt[b*Se
ch[c + d*x]]*Sinh[c + d*x])/(5*d) + (2*b*(b*Sech[c + d*x])^(5/2)*Sinh[c + d*x])/(5*d)

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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[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]

Rubi steps

\begin {align*} \int (b \text {sech}(c+d x))^{7/2} \, dx &=\frac {2 b (b \text {sech}(c+d x))^{5/2} \sinh (c+d x)}{5 d}+\frac {1}{5} \left (3 b^2\right ) \int (b \text {sech}(c+d x))^{3/2} \, dx\\ &=\frac {6 b^3 \sqrt {b \text {sech}(c+d x)} \sinh (c+d x)}{5 d}+\frac {2 b (b \text {sech}(c+d x))^{5/2} \sinh (c+d x)}{5 d}-\frac {1}{5} \left (3 b^4\right ) \int \frac {1}{\sqrt {b \text {sech}(c+d x)}} \, dx\\ &=\frac {6 b^3 \sqrt {b \text {sech}(c+d x)} \sinh (c+d x)}{5 d}+\frac {2 b (b \text {sech}(c+d x))^{5/2} \sinh (c+d x)}{5 d}-\frac {\left (3 b^4\right ) \int \sqrt {\cosh (c+d x)} \, dx}{5 \sqrt {\cosh (c+d x)} \sqrt {b \text {sech}(c+d x)}}\\ &=\frac {6 i b^4 E\left (\left .\frac {1}{2} i (c+d x)\right |2\right )}{5 d \sqrt {\cosh (c+d x)} \sqrt {b \text {sech}(c+d x)}}+\frac {6 b^3 \sqrt {b \text {sech}(c+d x)} \sinh (c+d x)}{5 d}+\frac {2 b (b \text {sech}(c+d x))^{5/2} \sinh (c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 68, normalized size = 0.67 \[ \frac {b^2 (b \text {sech}(c+d x))^{3/2} \left (3 \sinh (2 (c+d x))+2 \tanh (c+d x)+6 i \cosh ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} i (c+d x)\right |2\right )\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Sech[c + d*x])^(7/2),x]

[Out]

(b^2*(b*Sech[c + d*x])^(3/2)*((6*I)*Cosh[c + d*x]^(3/2)*EllipticE[(I/2)*(c + d*x), 2] + 3*Sinh[2*(c + d*x)] +
2*Tanh[c + d*x]))/(5*d)

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fricas [F]  time = 1.11, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \operatorname {sech}\left (d x + c\right )} b^{3} \operatorname {sech}\left (d x + c\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*sech(d*x + c))*b^3*sech(d*x + c)^3, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \operatorname {sech}\left (d x + c\right )\right )^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sech(d*x + c))^(7/2), x)

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maple [F]  time = 0.32, size = 0, normalized size = 0.00 \[ \int \left (b \,\mathrm {sech}\left (d x +c \right )\right )^{\frac {7}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \operatorname {sech}\left (d x + c\right )\right )^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sech(d*x + c))^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (\frac {b}{\mathrm {cosh}\left (c+d\,x\right )}\right )}^{7/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b/cosh(c + d*x))^(7/2),x)

[Out]

int((b/cosh(c + d*x))^(7/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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