3.1.22 \(\int \frac {1}{(b \sinh (c+d x))^{7/2}} \, dx\) [22]

Optimal. Leaf size=118 \[ -\frac {2 \cosh (c+d x)}{5 b d (b \sinh (c+d x))^{5/2}}+\frac {6 \cosh (c+d x)}{5 b^3 d \sqrt {b \sinh (c+d x)}}+\frac {6 i E\left (\left .\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right )\right |2\right ) \sqrt {b \sinh (c+d x)}}{5 b^4 d \sqrt {i \sinh (c+d x)}} \]

[Out]

-2/5*cosh(d*x+c)/b/d/(b*sinh(d*x+c))^(5/2)+6/5*cosh(d*x+c)/b^3/d/(b*sinh(d*x+c))^(1/2)-6/5*I*(sin(1/2*I*c+1/4*
Pi+1/2*I*d*x)^2)^(1/2)/sin(1/2*I*c+1/4*Pi+1/2*I*d*x)*EllipticE(cos(1/2*I*c+1/4*Pi+1/2*I*d*x),2^(1/2))*(b*sinh(
d*x+c))^(1/2)/b^4/d/(I*sinh(d*x+c))^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 118, 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 = {2716, 2721, 2719} \begin {gather*} \frac {6 i E\left (\left .\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \sinh (c+d x)}}{5 b^4 d \sqrt {i \sinh (c+d x)}}+\frac {6 \cosh (c+d x)}{5 b^3 d \sqrt {b \sinh (c+d x)}}-\frac {2 \cosh (c+d x)}{5 b d (b \sinh (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Sinh[c + d*x])^(-7/2),x]

[Out]

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

Rule 2716

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

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2721

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

Rubi steps

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

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Mathematica [A]
time = 0.12, size = 79, normalized size = 0.67 \begin {gather*} -\frac {2 \left (-3 \cosh (c+d x)+\coth (c+d x) \text {csch}(c+d x)+3 E\left (\left .\frac {1}{4} (-2 i c+\pi -2 i d x)\right |2\right ) \sqrt {i \sinh (c+d x)}\right )}{5 b^3 d \sqrt {b \sinh (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Sinh[c + d*x])^(-7/2),x]

[Out]

(-2*(-3*Cosh[c + d*x] + Coth[c + d*x]*Csch[c + d*x] + 3*EllipticE[((-2*I)*c + Pi - (2*I)*d*x)/4, 2]*Sqrt[I*Sin
h[c + d*x]]))/(5*b^3*d*Sqrt[b*Sinh[c + d*x]])

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Maple [A]
time = 0.73, size = 205, normalized size = 1.74

method	result	size
default	\(-\frac {6 \sqrt {-i \left (\sinh \left (d x +c \right )+i\right )}\, \sqrt {2}\, \sqrt {-i \left (i-\sinh \left (d x +c \right )\right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \left (\sinh ^{2}\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-i \left (\sinh \left (d x +c \right )+i\right )}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-i \left (\sinh \left (d x +c \right )+i\right )}\, \sqrt {2}\, \sqrt {-i \left (i-\sinh \left (d x +c \right )\right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \left (\sinh ^{2}\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-i \left (\sinh \left (d x +c \right )+i\right )}, \frac {\sqrt {2}}{2}\right )-6 \left (\sinh ^{4}\left (d x +c \right )\right )-4 \left (\sinh ^{2}\left (d x +c \right )\right )+2}{5 b^{3} \sinh \left (d x +c \right )^{2} \cosh \left (d x +c \right ) \sqrt {b \sinh \left (d x +c \right )}\, d}\)	\(205\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*sinh(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

-1/5/b^3/sinh(d*x+c)^2*(6*(-I*(sinh(d*x+c)+I))^(1/2)*2^(1/2)*(-I*(I-sinh(d*x+c)))^(1/2)*(I*sinh(d*x+c))^(1/2)*
sinh(d*x+c)^2*EllipticE((-I*(sinh(d*x+c)+I))^(1/2),1/2*2^(1/2))-3*(-I*(sinh(d*x+c)+I))^(1/2)*2^(1/2)*(-I*(I-si
nh(d*x+c)))^(1/2)*(I*sinh(d*x+c))^(1/2)*sinh(d*x+c)^2*EllipticF((-I*(sinh(d*x+c)+I))^(1/2),1/2*2^(1/2))-6*sinh
(d*x+c)^4-4*sinh(d*x+c)^2+2)/cosh(d*x+c)/(b*sinh(d*x+c))^(1/2)/d

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.08, size = 675, normalized size = 5.72 \begin {gather*} \frac {2 \, {\left (3 \, {\left (\sqrt {2} \cosh \left (d x + c\right )^{6} + 6 \, \sqrt {2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + \sqrt {2} \sinh \left (d x + c\right )^{6} + 3 \, {\left (5 \, \sqrt {2} \cosh \left (d x + c\right )^{2} - \sqrt {2}\right )} \sinh \left (d x + c\right )^{4} - 3 \, \sqrt {2} \cosh \left (d x + c\right )^{4} + 4 \, {\left (5 \, \sqrt {2} \cosh \left (d x + c\right )^{3} - 3 \, \sqrt {2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 3 \, {\left (5 \, \sqrt {2} \cosh \left (d x + c\right )^{4} - 6 \, \sqrt {2} \cosh \left (d x + c\right )^{2} + \sqrt {2}\right )} \sinh \left (d x + c\right )^{2} + 3 \, \sqrt {2} \cosh \left (d x + c\right )^{2} + 6 \, {\left (\sqrt {2} \cosh \left (d x + c\right )^{5} - 2 \, \sqrt {2} \cosh \left (d x + c\right )^{3} + \sqrt {2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - \sqrt {2}\right )} \sqrt {b} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right )\right ) + 2 \, {\left (3 \, \cosh \left (d x + c\right )^{6} + 18 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + 3 \, \sinh \left (d x + c\right )^{6} + {\left (45 \, \cosh \left (d x + c\right )^{2} - 8\right )} \sinh \left (d x + c\right )^{4} - 8 \, \cosh \left (d x + c\right )^{4} + 4 \, {\left (15 \, \cosh \left (d x + c\right )^{3} - 8 \, \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + {\left (45 \, \cosh \left (d x + c\right )^{4} - 48 \, \cosh \left (d x + c\right )^{2} + 1\right )} \sinh \left (d x + c\right )^{2} + \cosh \left (d x + c\right )^{2} + 2 \, {\left (9 \, \cosh \left (d x + c\right )^{5} - 16 \, \cosh \left (d x + c\right )^{3} + \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )} \sqrt {b \sinh \left (d x + c\right )}\right )}}{5 \, {\left (b^{4} d \cosh \left (d x + c\right )^{6} + 6 \, b^{4} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + b^{4} d \sinh \left (d x + c\right )^{6} - 3 \, b^{4} d \cosh \left (d x + c\right )^{4} + 3 \, b^{4} d \cosh \left (d x + c\right )^{2} - b^{4} d + 3 \, {\left (5 \, b^{4} d \cosh \left (d x + c\right )^{2} - b^{4} d\right )} \sinh \left (d x + c\right )^{4} + 4 \, {\left (5 \, b^{4} d \cosh \left (d x + c\right )^{3} - 3 \, b^{4} d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 3 \, {\left (5 \, b^{4} d \cosh \left (d x + c\right )^{4} - 6 \, b^{4} d \cosh \left (d x + c\right )^{2} + b^{4} d\right )} \sinh \left (d x + c\right )^{2} + 6 \, {\left (b^{4} d \cosh \left (d x + c\right )^{5} - 2 \, b^{4} d \cosh \left (d x + c\right )^{3} + b^{4} d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(3*(sqrt(2)*cosh(d*x + c)^6 + 6*sqrt(2)*cosh(d*x + c)*sinh(d*x + c)^5 + sqrt(2)*sinh(d*x + c)^6 + 3*(5*sqr
t(2)*cosh(d*x + c)^2 - sqrt(2))*sinh(d*x + c)^4 - 3*sqrt(2)*cosh(d*x + c)^4 + 4*(5*sqrt(2)*cosh(d*x + c)^3 - 3
*sqrt(2)*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*sqrt(2)*cosh(d*x + c)^4 - 6*sqrt(2)*cosh(d*x + c)^2 + sqrt(2))*
sinh(d*x + c)^2 + 3*sqrt(2)*cosh(d*x + c)^2 + 6*(sqrt(2)*cosh(d*x + c)^5 - 2*sqrt(2)*cosh(d*x + c)^3 + sqrt(2)
*cosh(d*x + c))*sinh(d*x + c) - sqrt(2))*sqrt(b)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cosh(d*x + c)
+ sinh(d*x + c))) + 2*(3*cosh(d*x + c)^6 + 18*cosh(d*x + c)*sinh(d*x + c)^5 + 3*sinh(d*x + c)^6 + (45*cosh(d*
x + c)^2 - 8)*sinh(d*x + c)^4 - 8*cosh(d*x + c)^4 + 4*(15*cosh(d*x + c)^3 - 8*cosh(d*x + c))*sinh(d*x + c)^3 +
(45*cosh(d*x + c)^4 - 48*cosh(d*x + c)^2 + 1)*sinh(d*x + c)^2 + cosh(d*x + c)^2 + 2*(9*cosh(d*x + c)^5 - 16*c
osh(d*x + c)^3 + cosh(d*x + c))*sinh(d*x + c))*sqrt(b*sinh(d*x + c)))/(b^4*d*cosh(d*x + c)^6 + 6*b^4*d*cosh(d*
x + c)*sinh(d*x + c)^5 + b^4*d*sinh(d*x + c)^6 - 3*b^4*d*cosh(d*x + c)^4 + 3*b^4*d*cosh(d*x + c)^2 - b^4*d + 3
*(5*b^4*d*cosh(d*x + c)^2 - b^4*d)*sinh(d*x + c)^4 + 4*(5*b^4*d*cosh(d*x + c)^3 - 3*b^4*d*cosh(d*x + c))*sinh(
d*x + c)^3 + 3*(5*b^4*d*cosh(d*x + c)^4 - 6*b^4*d*cosh(d*x + c)^2 + b^4*d)*sinh(d*x + c)^2 + 6*(b^4*d*cosh(d*x
+ c)^5 - 2*b^4*d*cosh(d*x + c)^3 + b^4*d*cosh(d*x + c))*sinh(d*x + c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (b \sinh {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*sinh(c + d*x))**(-7/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (b\,\mathrm {sinh}\left (c+d\,x\right )\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*sinh(c + d*x))^(7/2),x)

[Out]

int(1/(b*sinh(c + d*x))^(7/2), x)

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